Solution Manual for Partial Differential Equations for --Stanley J_ Farlow Dover Books on Mathematic

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SOLUTION MANUAL FOR

PARTIAL DIFFEREN TIAL EQ1JATION S FOR

SCIENTIST S & ENGINEER S

SOLU TION MANU AL FOR

PARTIAL DIFF ERE NTIA L

EQlJ ATIO NS FOR

SCIE NTIS TS & ENG INEE RS STANLEY J. FARLOW Professor Emeritus of Mathematics University of Maine

DOVER PUBLICAT IONS GARDEN CITY, NEW YORK

Copyright Copyright © 2020 by Stanley J. Farlow All rights reserved.

Bibliographical Note Solution Manual for Partial Differential Equations for Scientists and Engineers is a new work, first published by Dover Publications in 2020.

International Standard Book Number

ISBN-I 3: 978-0-486-84252-3 ISBN-I 0: 0-486-84252-5 Manufactured in the United States of America 84252503 www.doverpublications.com

Contents

Section 1: Introduction Lesson I: Introduction

Section 2: Diffusion Type Problems Lesson 2: Diffusion-Type Problems (Parabolic Equations)

6

Lesson 3: Boundary Conditions for Diffusion-Type Problems

11

Lesson 4: Derivation of the Heat Equation

15

Lesson 5: Separation of Variables

19

Lesson 6: Transforming Nonhomogeneous BCs to Homogeneous Ones Lesson 7: Solving More Complicated Problems by Separation of Variables

24 31

Lesson 8: Transforming Harder Problems into Easier Ones

37

Lesson 9: Solving Non homogeneous PD Es (Eigenfunction Expansions)

45

Lesson I 0: Integral Transforms (Sine and Cosine Transforms)

53

Lesson 11: The Fourier Series and Transform

59

Lesson 12: The Fourier Transform and Its Applications to PD Es

68

Lesson 13: The Laplace Transform

73

Lesson 14: Duhamel's Principle

78

Lesson 15: The Convection Term u in Diffusion Problems

'

84

Section 3: Hyperbolic Type Problems Lesson 16: The One-Dimensional Wave Equation (Hyperbolic Equations)

89

Lesson 17: The d'Alembert Solution of the Wave Equation

92

VIII

CONTENTS

Lesson 18: More on the d' Alembert Equation

96

Lesson 19: Boundary Conditions Associated with the Wave Equation

103

Lesson 20: The Finite Vibrating String (Standing Waves)

I07

Lesson 21: The Vibrating Beam (Fourth Order PDE)

117

Lesson 22: Dimensionless Problems

125

Lesson 23: Classification of PD Es (Canonical Form of Hyperbolic Equations)

132

Lesson 24: The Wave Equation in Two and Three Dimensions (Free Space)

138

Lesson 25: The Finite Fourier Transforms (Sine and Cosine Transforms)

144

Lesson 26: Superposition (The Backbone of Linear Systems)

152

Lesson 27: First-Order Equations (Method of Characteristics)

156

Lesson 28: Nonlinear First-Order Equations (Conservation Equations)

163

Lesson 29: Systems of PD Es

169

Lesson 30: The Vibrating Drumhead (Wave Equation in Polar Coordinates)

175

Section 4: Elliptic Type Problems Lesson 31: The Laplacian (an Intuitive Description)

183

Lesson 32: General Nature of Boundary Value Problems

192

Lesson 33: Interior Dirichlet Problem for a Circle

198

Lesson 34: The Dirichlet Problem in an Annulus

210

Lesson 35: Laplace's Equation in Spherical Coordinates (Spherical Harmonics)

218

Lesson 36: A Nonhomogeneous Dirichlet Problem (Green's Function)

228

Section 5: Numerical and Approximate Methods Lesson 37: Numerical Solutions (Elliptic Problems)

233

Lesson 38: An Explicit Finite-Difference Method

238

Lesson 39: An Implicit Finite-Difference Method (Crank-Nicolson Method)

242

Lesson 40: Analytic versus Numerical Solutions

244

Lesson 4 I: Classification of PD Es (Parabolic and Elliptic Equations)

248

CONTENTS

ix

Lesson 42: Monte Carlo Methods

254

Lesson 43: Monte Carlo Solutions of Partial Differential Equations

260

Lesson 44: Calculus of Variations (Euler-Lagrange Equations

265

Lesson 45: Variation Methods for Solving PDEs (Method of Ritz)

270

Lesson 46: Perturbation Methods for Solving PD Es

276

Lesson 47: Conformal Mapping Solutions of PDEs

282

SOLUTION MANUAL FOR

PARTIAL DIFFER ENTIAL EQ!JAT IONS FOR

SCIENT ISTS & ENGINE ERS

XII

Lesson 1: Introduction to Partial Differential Equations 1. Classify the following equations according to the properties we've discussed in Figure 1.1.

a) u, =u,._"( +2ux+u b) u, = uxx + e-, c) uxx + 3u_-cy + u>Y =sinx

d) u,, =uux.= + e _,

Solution: The general form of a second-order linear PDE in two independent variables is Au_0 . +Bu-" +Cuw +Dux +Eu.v +Fu=G

where A,B,C,D,E,F, G are functions of x,y (which of course allows them to be constants as well). They are classified according to B 2 - 4AC < 0 ⇒ equation is elliptic { B 2 - 4AC = 0 ⇒ equation is parabolic B2 - 4AC > 0 ⇒ equation is hyperbolic a) u, = u_~, + 2ux + u Second order, linear, homogeneous, constant coefficients, two independent variables x,t. Since the only nonzero secondorder derivative is u.O" , we have that B =0, C =0 and so B2 - 4A C = 0 and so the equation is parabolic. Parabolic equations describe diffusion problems such as heat flow or diffusion of substances in a medium. This equation is called the one-dimensional heat (or diffusion) equation since it only has 1 space dimension, in this case x. b) u, =ux.r +e- 1 Second order, linear, nonhomogeneous, constant coefficients, two independent variables x,t. Since the only second-order derivative is u,.,, we have B = 0, C = 0 , hence B2 - 4AC =0 and so the

2

Solution Manual

equation is called parabolic. Parabolic equations describe diffusion problems such as heat flow . This equation is called the one-dimensional heat (or diffusion) equation (dimension I since it only has I space dimension, in this case x ). The difference of this equation and the one in a) is that this equation has an external heat source e-' but does not have terms in u and u x which describe damping and heat flux, respectively. More on this later. c) uxx +3u_'), +u.,,y =sinx Second order, linear, nonhomogeneous, constant coefficients, two independent variables x, y. Here A= C = I, B =3 and so we have the equation is a hyperbolic PDE. d) u" = uux= +e-' Fourth order, nonlinear. The words homogenous or nonhomogeneous are not generally used in connection with nonlinear equations, nor are the words constant or nonconstant coefficients. The term that makes the equation non! in ear is uu_nxx . Almost all PD Es have either I st or 2nd (partial) derivatives, but rarely higher than two. The Koreweg-DeVries equation u, + u_= - 6uux = 0 u has a 3 rd order (partial) derivative but this is uncommon.

2. How many solutions to the PDE u, = u" can you find? Try to find a solution of the form u x,t = eax+b, .

Solution: Of course, any linear homogeneous PDE always has the zero solution u( x, t) = 0 . We could guess at exponential solutions of the form u(x,t) = ea.,+h,, but let's just plug this expression in the PDE and see if we can determine the constants a,b that give us a solution. Plugging this • 1 •m t he PDE gives • beax+ht = a 2 eax+bt or b = a 2 . I n other word s, any exponentia function of the form u( x,t) = e a,+a' , = e a'e 0 ' 1 is a solution, where a is a real (or complex) number. For example, u =exe ' ,u = e2xe 4' , u =e-xe ' , ... are all solutions. In fact if you let a be the complex constant a= i, then b =i 2 =- 1 which means the complex function

1. INTRODUCTION TO PARTIAL DIFFERENTIAL EQ!)ATIONS

u(x,t)=e'xe-1

= e- cosx + isinx] • = e cos x + te - t sin x 1

[

- I

•

is also a solution. In fact you can show (try it) that if a PDE has real coefficients (like the one in the problem) and if you have a complex solution, say u = v( x,y) + iw( x,y), then both the real and complex parts v and w are also solutions. In this problem you can verify that both (real) functions v(x,y) = e- 1 cosx and w( x,y) = e- 1 sinx are also solutions of the (one-dimensional heat) equation u, = uxx.

3. Suppose that if u 1( x,y) and u2( x,y) satisfy the homogeneous linear equation Auxx + Buxy + Cuyy + Dux + Eu.v +Fu= 0 where A,B,C,D,E, and F can either be constants or functions of x,y, then the sum u1 + u2 also satisfies the PDE.

Solution: Since u 1 and u 2 are solutions of the PDE, we have

A(cfu,)+B(15_)u =O ox2 oxay XY +c(ciay2u,)u +D(au,)+E(au,)+Fu ax ay JY

A(a

2 u 2) ox2

I

+ B( 8 u2 )u + c(a u 2 )u + D(au2) + E(au2) + Fu = o axay .,y ay2 yy ax ay 2 2

Adding these equations, we find

2

3

4

Solution Manual

and using the fact that the sum of the derivatives is the derivative of the sum, we have

which says that u1 +u2 is a solution of the PDE. Note that if the PDE was nonhomogeneous, i.e., G "# 0, the sum u1 + u2 would not be a solution. You can also check and see that the sum of two solutions of a nonlinear PDE is not (always) a solution.

4. Probably the easiest of all PDEs to solve is

au(x,y)=O ax Can you find all the solutions of this equation?

Solution: Integrating this equation with respect to x yields u( x,y) = g(y) where g is any (differentiable) function of y . You can see that u( x,y) =siny, u(x,y) = e>, u( x,y) = 10 are all solutions.

5. Find the solutions of the cross-partial PDE

Solution: We first integrate with respect to x, getting uY = g(y), and then integrate with respect to y, getting

u(x,y)= f g(y)dy+ f(x)=G(y)+ f(x)

1. INTRODUCTION TO PARTIAL DIFFERENTIAL EQ\)ATIONS

In other words, the general solution consists of sums of two (differentiable) functions f,G where f depends only on x and G depends only on y. For example u = x+ y,u = x 2 +siny, .... are solutions of the PDE (check them), whereas u( x,y) = .zy is not.

Note that this second-order PDE has two arbitrary functions in the solution compared with the second-order ODE

that has two arbitrary constants c1 and c2 . In other words, there are many more solutions of PD Es than of ODEs.

5

Lesson 2: Diffusion-Type Problems (Parabolic Equations)

1. If the initial temperature (IC) of the rod is IC: u(x,O)=sin(ru:)

o::;;x::;;l

and the BCs are BCs:

u(0,t)=0 u(l,t)=0

what is the behavior of the rod temperature u(x,t) at later values of time? HINT: Use the physical interpretation of the heat equation u, =o}uxx.

Solution: The rate u, (degrees/unit time) at which the temperature u increases or decreases (degrees per unit time) is proportional to the concavity of the curve u(x,t) as a function of x at every instant of time t. Hence the plots of the , temperature for different values of time are shown below in Figure 2.1. We will see later that the function u( x,t) that satisfies the PDE u, = a.2uxx with given IC and BC is u(x,t) =e-a', sin(n.x ), which evaluated at time t =0, O.l, 0.2,and 0.3.

For simplicity we pick the conductivity parameter a= 1. Note that the temperature is gradually going to zero everywhere in the interval O::;; x::;; 1. Not only is the temperature going to zero, but the 'profile' of the curve does not change, i.e., the profile sin ( n.x) is the function of x and it gets damped by the function of time

e

-a2,

.

Figure 2.1: Temperature at times t = 0,0. l,0.2,0.3

6

2. DIFFUSION-TYPE PROBLEMS

2. Suppose the rod in Problem I has a constant (internal) heat source everywhere along the rod so that the basic equation describing the heat flow within the rod is u, = a.2u_" + 1, 0 < x < I Suppose we fix the boundary temperatures as u(0,t) = 0, u(l,t) = 1. What will be the steady-state temperature of the rod (if there is one)? In other words, will the temperature of the rod approach a constant temperature U ( x) independent of time?

Solution: If there is a steady state solution it will happen when the temperature, which we call U(x), depends only on x and not time t , hence if there is a steady state temperature we must have u, =0 , which means the steady state temperature U(x) should satisfy the ODE

which has solutions

where c1 and c2 are arbitrary constants. lfwe determine the constants to fit the boundary conditions U ( 0) = 0, U ( 1) = I , we get the steady state 2 + ( 1+ - 1 -2 )x U(x) = - -1-x 2

2a

2a

When a= I this steady state is shown in Figure 2.2.

7

8

Solution Manual

1.0

0.8

0.6

0.4

0.2

0.2

0.4

-'-0.8

0.6

1.0

X

Figure 2.2: Steady state temperature of a rod

Starting with initial temperature of u(x,O) =x, the temperature at different values of time are shown in Figure 2.3.

1.0

0.8

0.6

0.4

02

02

0.4

0.6

0.8

Figure 2.3: Solutions going from IC u( x,O) = x to steady state

1.0

X

2 . DIFFUSION-TYPE PROBLEMS

9

3. Suppose a metal rod loses heat across the lateral boundary according to the equation u, = a.2u.,,. - f3u, 0 < x < I , f3 > 0 and suppose we keep the ends of the rod at u ( 0, 1) = 0 and u ( l,t) =1. Find the steady-state temperature of the rod. Where is heat flowing in this problem?

Solution: lfwe set u, =0 the PDE reduces to the ODE

and ifwe let the parameters a= f3 = 1 for simplicity we have the solution

and plugging this into the BC u ( 0) = 0, u ( l) = 1 one arrives at what we suspect is the steady state

We suspect that this is the steady state from the nature of the physical system.

Heat flows into the right at the right-hand side (x = 1) and flows through the rod but escapes on the lateral sides of the rod. However, keep in mind that at each point along the rod the net flow of heat in and out of the point is zero (i.e., net flow in= net flow out).

10

Solution Manual

• "11:a<fy 1.0

0.8

116

0.4

112

Figure 2.4: Steady state of u, = a 2 uxx -l3u,

(13 > 0) with BC u( 0,t) = 0, u(l,t) = 1

Although it is clear that the above function u( x, oo) satisfies the PDE as well as the boundary conditions, we must be careful using this technique of setting u, = 0 and solving the ODE since the solution may be unstable; i.e., nearby solutions move away from u( x, oo ). Recalling from ODE the equation y' - y =I, it is clear that y( x) = -1 is a constant solution, which we get from setting y' = 0 , but y( x) = -1 is not the steady state of the equation since the general solution is y( x) = cex - I; i.e., which means that all other solutions (for c i: 0) move away from y( x) = - 1 and so the equation has no steady state.

4. Suppose a laterally insulated rod of length L =1 has an initial temperature of u ( x, 0) =sin (3ru:) and has its left and right ends kept at u ( 0, t) =o°C and u(I,t) == I0°C respectively. What wouJd be the IBVP that describes this problem?

Solution: The initial boundary value problem (IBVP) would be PDE: u,, = a 2uxx 0 < X < I, 0 < t < 00

u(0,t)=0, BC· { • u(l,t) = I0,

0 <t <OO 0<t <OO

IC: u(x,0) =sin(31tX), 0 :S: x:S:l (Note that the PDE is defined inside the region O< x < I and O< t < oo since the derivatives u, and u.u are not defined on the boundaries.)

Lesson 3: Boundary Conditions for Diffusion-Type Problems 1. Draw rough sketches of the solution of the following IBVP at different values of time. Will there be a steady-state temperature, and if so what is it?

PDE u, =a 2uxx • 0<x<200, 0<t< oo

ux(0,t)=0 BC {

h ux(200,1) =-k[ u(200,1 )- 20]

IC u ( x, 0) =0°C

0 < t< oo

0 ~ X ~ 200

Solution: The IBVP describes a rod that is insulated at the left end. At the right end an external heat source of20 degrees is applied, which tends to keep the temperature at 20 degrees. With this in mind our intuition tells us that the temperature of the rod will eventually be a constant of20 degrees everywhere in the rod, i.e., u(x, oo) = 20, 0 ~ x ~ 200 . An engineer might say that the right end of the rod has a sensor so when the temperature is less than 20 degrees, heat is applied increasing the temperature, and when the temperature is greater than 20 degrees, a cooling device will lower the temperature, the net effect being that the temperature is held around 20 dewees. nµ..""mtl.l'C

31--- ------ ------ -15

10

X

Figure 3.1: Solutions u( x,0) =0 ➔ u( x, oo) = 20 11

12

Solution Manual

Note in the graphs in Figure 3.1 u_,(0,t) =0 for all time t and that ux(200,t)>0 for all time tsince u(200,t)<20.

2. What is your interpretation of the following IBVP: PDE u, = a 2u·"'' 0<x<l, 0<t<oo BC {u(O,t) = 0 ux(l,t)=l

0<t<oo

IC u(x,0)=sin(ru: ), 0$x $1 Draw rough sketches of the solution. Does the solution have a steady state, and if so, what do you think it will be?

Solution: This IBVP might model a rod of length 1 which has an initial temperature in the shape of a sine curve. The temperature at the left end of the rod is fixed at zero, and heat flows into the right end of the rod (inward flux of I unit of heat per unit time - maybe cal/sec) at a constant rate. The first derivative ux stands for flux in heat problems, if ux ( x,t) > 0 heat is flowing to the left at point x and time t (from HOT to COLD), and if ux ( x,t) < 0 heat is flowing to the right (from HOT to COLD). Intuitively, we suspect there will be a steady state, and if this is so then we must have u, = 0. Hence, the steady state would satisfy the boundary value problem d 2U dU(I) -= 0 U(0)=0 - - = l dx 2 , ' dx

which has the solution U ( x) = x. You can draw the temperature curves starting at u(x,0)=sin(ru: ) and going to u(x,oo)=x inFigure3.2.

3. BOUNDARY COND ITIONS FOR DIFFUSION -TYPE PROBLEMS

13

~

10

0.8

06

0.4

02

02

0.4

0.6

08

10

Figure 3.2: Temperature going from u(x,O) =sin( 7tX) to u(x,oo) =x

3. What is your physical interpretation of the following IBVP? PDE u, = a.2u.u, 0 < X < 1, 0 < t < 00 BC {u,. (O,t)=O

O<t<oo

u,.(1,t)=O IC u(x,O)=sin(n.x), 0:5x:51 Make a rough sketch of the solution at various values oftime. Does your intuition think there will be a steady-state solution, if so, what is it?

Solution: This IBVP might be a model for an insulated rod of length I, which has initial temperature u( x, 0) = sin ( nx) and is insulated at both the left and right ends. This being the case, there is nowhere for the heat energy in the rod (remember the rod is insulated on the outside) to go except to spread out, reaching a constant temperature, which should be the average value of the initial temperature (conservation of energy). That is

U(x)=u(x,oo)=-lf 1sin(n.x)dx=-2 7t 10 Try drawing yourself various temperature curves for values of time between the initial temperature u(x, 0) = sin ( nx) and steady state temperature u( x,oo) = 2/n (i.e., play Mother Nature yourself). Keep in mind that the

14

Solution Manual

solution mu st always have zero derivatives u_, = O at the end points x = 0 and x= I .

08

06

04

02

02

04

08

10

Figure 3.3: Fill in temperatures u( x,O) = sin ( ru:) ➔ u( x,oo) = 2/1t

4. Suppose a metal rod, laterally insulated, has an initial temperature of 20°C , but immediately thereafter has one end fixed at 50°C . The right end of the rod is immersed in a liquid solution of temperature 30°C. What would be the IBVP that describes this problem?

Solution: PDE u,=a 2u_,,,, O<x< l, O <t<oo BC {

u(O,t) =50 k ux (l,t) = -h[ u(l,t)-30]

O< t <oo

IC u(x,0)=20, 0 ~x ~I Note at the right end point x = I if the temperature inside (at the edge x = I ) is greater than 30 degrees, then ux( l,t) < 0 which (according to Fourier's law) heat flow out of the region proportional to h/ k .

Lesson 4: Derivation of the Heat Equation 1. Substitute the units of each quantity u,u, , ... into the equation u, =o.1u:a -f3u

to see that every term has the units of ·CI sec.

Solution: The term u, has units of deg/sec, and u,, has units of deg/cm 2 , and u has units of deg, so a 2 must have units of cm-2 • [ a 2] =deg/sec ~--= units deg/ cm 2 sec

and f3 (time constant) has units of units[f3] =deg/ sec= _I_ sec deg

2. Substitute the units of each quantity u,u, , ... into the equation

where u has units of velocity to see that every term has the same units.

Solution: The term u, clearly has units of deg/sec, u,, has units of deg/cm 2 , 11 has units of deg, and a 2 must have units of • [ a 2] = deg/sec, units deg/ cm -

cm 2 sec

and v has the correct units of ve locity units[f3]= deg/ sec= cm deg/ cm sec 15

16

Solution Manual

3. Derive the heat equation I a [K(x)u, ]+ f(x,I) cp ax

111 =- -

for the situation where the thermal conductivity K(x) depends on x.

Solution: The fmm of this PDE may look a little unusual to the reader, but this is a common way to write the partial derivative with respect to x when the proportionality constant k =k( x) is a variable function of x. If k were a constant, then the equation would be u, =I_u,, + f(x ,t) cp

which may seem more common to the reader. To prove the one-dimensional heat equation when the thermal conductivity k ( x) varies, we begin with the conservation of energy relation (eq 4.2 in the text) cpA C "-' u, ( s.t )ds = A [ k(x + & )u, (x + &,t )-k(x )u, (x,1)] + Acpr

ilr1 (s,t )ds

which we now modify since the conductivity k(x) is no longer a constant.

This above equation is the mathematical way of saying that the net change of energy in [x,x+&] is equal to the heat gain/loss across the boundaries at x and x+& plus the heat generated in the interval [x,x+&] by the heat source f(x,t). We now rewrite the above equation as k (x)[ u., (x+&,1)-u, (x,1)] =cpC"' [ u, (s ,I )-/ (s ,t )]ds

and apply the mean value theorem

J: f(x)dx = f (q)(b - a), a < q <h to the integral of this equation, getting

4. DERIVATION OF THE HEAT EQlJATlON

17

k(x )[ u-' (x+ Llx,t)-u-' (x,t)] =cp[u, (,;,t)- f (,;,t)]& We now solve for u, (,;,t) getting 1 f( ,;,t) u, (,;,t) = _I {k(x+ Llx )ux (x + !ll,t)-k(x)ux(x,t)}+cp Ill cp Ifwe now let & ➔ O, we get the desired result l 8 K(x)ux] + f(x,t) u, =--[ cp ox

Note: We could (although it is seldom written this way) write this last equation as

u]

o 2 + f(x,t) ou 1 [dk --+k(x)u, =cp dx 2

ox

ox

4. Suppose u(x,t) measures the concentration of the substance in a moving stream moving (to the right) with velocity v, where now the concentration u changes both by convection and diffusion. Derive the equation

from the fact that at any instant of time, the total mass of the material is not created or destroyed in the region x, x + & 1- Hint: Write the conservation equation

r

Change of mass inside [ x, x + & ] =Change due to diffusion across boundaries + Change due to material being carried across the boundaries

Solu tion : We now must modify the conservation of energy equation, which says Change of mass inside [ x, x + ~] = Change due to diffusion across boundaries + Change due to material being carried across the boundaries

18

Solution Manual

to account for the fact that heat is lost (or gained) across the boundaries as a result of convection, not just diffusion as it was before. Hence, we have the new equation cpA

r ''

u, ( s,I )ds = Ak [ux (x + &,! )-u , (x,t) ]- v[ u (x + &,t )-u ( x,I)]

where the last term

v[u(x + ~x,t)-u(x,t)

J

measures the heat gain or loss in [x,x+&] as a result of convection. We now rewrite this equation as

k[u, (x+ &,t)-u., (x,t )] =-v[ u(x+ &,t)-u(x,t)] + cpf+"' u, (s,t)ds and applying the mean value theorem to the integral, we have k[ u., (x + &,t)-u, (x,t )] = vcp[ u(x+ &,t )-u (x,t )]+c,ou, (,;,t )& Finally, solving for u,(,;,t), we get

{u(x+&,t)-u(x,t)}-b± ✓b 2 -4ac u ( ,; t ) =k- {u.,(x+&,t)-u,(x,t)} -~-~-~~ -v ' ' cp & & 2a and letting r x ➔ 0 which forces ,; ➔ x , we get the desired result

where a 2 =k/cp.

Lesson 5: Separation of Variables I. Show that

satisfies the PDE u, = a 2uxx for arbitrary A, B.

Solution: Simply compute the partial derivatives u, and uxx and you will see they satisfy u, = a 2u:o .

2. Prove the orthogonality property

io1sin( mm: )sin( nm: )dx ={ 1/20 mi:-n m=n for the sine function. Hint: Use the identity sin(mm: )sin(mtX )= .!.[ cos(m-n )x-cos(m+n )x] 2

Solution: We won't insult your intelligence by doing this. It is a simple integral.

3. Find the Fourier sine expansion of ~( x) =1, 0 5 x 51. Draw the first three or four terms.

Solution: The Fourier (sine) series is 00

~(x)= I= I a,, sin(nm:) n=l

19

20

Solution Manual

To find the coefficients an, we multiply each side of the equation by sin (m1!X) and integrate each side with respect to x from Oto 1. Then, using the orthogonality following property of the sines and cosine functions

f 1sin(m1tX)sin(mu)dx={

Jo

f 1cos( m1!X )cos( mrx )dx = {

Jo

° ° *= 1 m-:t-n

I12 m = n

m n I2 m n

J>in(m1tx)cos(mu)dx=O all m,n=0,1,2, .... we get

t

sin( m1!X )dx = amt sin 2( m1!X )dx

which we can solve for a111 , getting

-t

a111 -fi

o

sin(m1tx )dx -{ 0 m = 0,2,4, •••

sin 2 (m1tx)dx

-

4 m = I,3,5, ... mrc

-

Hence, the sine expansion of f ( x) = 1, 0 ~ x ~ 1 is 1 = i[sin( 7lX) + .!.sin(37!X) + .!.sin(57tX) + · · ·], 0 :S: x :S: 1 re 3 5 Plotting the first four terms of this series we have

21

5. SEPARATION OF VARIABLES

Figure 5.1 Graphs of terms of the Fourier series

Note, ifwe were to find the Fourier cosine series 1=

~:>ncos( mr.x) =a + a cos( 7tX) + a cos(21tx) + ••• "'

0

2

1

n=O

of the function f( x) = I, 0 :=; x :=;I, it would consist of one term, namely

a0 = I 4. Using the results of Problem 3, what is the solution of the lBVP PDE u, = a.2uxx, 0 < X < 1, 0 < t < CO BC {u(O,t)=O u(l,t) = 0

O<t<co

IC u(x,0)=1, o:=;x:=;I

Solution: We saw in Problem 3 that the IC u( x,O) =I has a Fourier sine series of 1 = -±[sin( 7tX) + .!.sin(31tx) + .!.sin(51tx) + ••·], 0 :=; x n

Hence, the solution of the IBVP is

3

5

~I

22

Solution Manual

Figure 5.2: Temperature u(x,0) = 1➔ u(x, oo)=O

5. What is the solution to the IBVP in Problem 4 if the IC is changed to

u( x,O) =sin (21tX) + !sin( 47tX) + !..sin( 67tX) 3

5

Solution: When the JC is already a sine series, as it is in this problem, the solution is trivial; we simply slip the appropriate exponential factors in each term, getting • ( 1 -(4n}', sm • ( ( -(6n)', • ( u ( x, t)= e- (h)' , sm 27tX ) + -e 47tX ) +-e sm 67tX ) 3 5

Tn general, the idea is to write the IC as a sine series and then include the factors of time in each term .

23

5. SEPARATION OF VARIABLES

6. What is the solution to the IBVP in Problem 4 if the IC is changed to U ( X, 0) = X - X 2 , 0 :$ X :$ 1

Solution: The first step is to expand the IC u( x,O) = x-x 2 as a Fourier sine series ,, 2 x-x = :~:>nsin(mu) 11:.:: J

and then multiplying each side of the equation by sin( mm:) and integrating to get

t (x - x )sin( mm: )dx = amt sin 2(m7tX )dx 2

or a = m

t (x-x2)sin(m1r.X)dx ={O 8 n=0,2,4, ••• tsin 2 (mm:)dx

m 31t3

m=I, 3, ...

Hence, the sine expansion of the IC is x - x 2 = ~[sin ( m:) + _!_sin(3m:) + - 1-sin (Sm:)+--·] 125 27 n 3

and so the solution is

• (5 1 -(5n) 2 1 sm • ( 3 1 - (h) 21 sm • ( m: ) +-e m: ) +--· ] 7tX ) +-e u ( x,t ) = 38 [ e- n' , sm 125 27 1t

Lesson 6: Transforming Nonhomogeneous BCs Into Homogeneous Ones 1. Solve the following IBVP by transforming it into a new IBVP with homogeneous BCs and then solving the transformed problem. Does the solution agree with your intuition?

PDE u r = a 2u.u'

O< x < I, 0 < t < oo

BC { u(O,t)=l ux ( l,t) + hu ( 1, t) =1

O<t <oo

IC u ( x, 0) =sin ( ru:) + l, 0 ::; x ::; 1

Solution: Since the rod is fixed at 1 at the left end, and heat is added or subtracted at the right end depending on whether 1-hu(l,t) is positive (heat applied since ux (l,t) > 0) or negative (heat extracted) since l,t) < 0 , our intuition tells us that the rod should reach a steady state temperature of the form u ( x , oo) = Ax+ 1 . Hence, we tly

u,(

u(x,t) = Ax+ 1+U(x,t)

which gives u, =U, u, (x ,t)=A+U, (x,t)

u(x,O) = Ax+ I+ U(x,O)

Plugging these values into the given IBVP gives PDE U, =a 2U 0 _,

O< x < l, O<t <oo

BC { U (O ,t) = 0 A+Ux(l ,t)+h[ A+l+U(l,t)]=I

O< t <oo

IC U ( x, 0) =sin ( ru: )- Ax, 0 ::; x::; 1 The idea now is to detennine the constant A which makes the BCs zero. To do this we require A + hA + h = 1 or 24

6. TRANSFORMING NONHOMOGENEOU S BCS

1-h A==1+h

Hence, we seek a solution of the form u(x,t)== ( -1-h) x+l+U(x,t) l+h

where U ( x,t) satisfies PDE U, ==a 2 Uu , O<x< l, O<t<oo BC {

U ( O,t) == 0 Ux(1,t)+hU(l,t)==0

O<t< oo

IC U(x,0)==sin(ru:)- ( -1-h) x , 0:<::;x:<::;1 l+h

We can solve this problem by trying U( x,t) == X(x )T(t), getting the two ODEs and ODE: X" + "),., 2X == 0 X(O)==O BCs· { X' ( 1) + hX ( l) == 0 •

The ODE in x has the solutions

X( X) == cl COSA-X + C2 sin°),.,x and plugging this in the BCs gives

25

26

Solution Manual

X(O)=O ⇒ c 1 =0

X' ( 1) + hX ( 1) = Acos A + h sin A = 0 ⇒ tan A = - _!_ A h

For simplicity, we choose h = l and finding the roots of

we get an infinite number of positive roots

illustrated in Figure 6.1.

l'

10

)

)

./

,t

- lO

--~ -:0

tanA. =-..4

Figure 6.1 : Graphical illustration of the roots of tan A= - A

Hence we have the solutions

X( x) = sin(Anx) T,,(t) = e-,., and thus "' u(x,t) = Z:a,,e - ).,,1 sin (\,x) n= l

6. TRANSFORMIN G NONHOMOGEN EOUS BCS

27

Note that we had to select only positive roots A.Of tan/\.=-/\. otherwise the above infinite series for u would blow up. To find the coefficients we multiply each side of the series solution by sin(\nx) and integrate each side of the equation from Oto I. Doing this (and letting h =1) we get

from which we get

f

sin(\,,x)dx a =--'-"-o_ _ __ "' f~sin 2 (A"'x)dx whose values are a, = 1.18, a2 = 0.32,Cl:J = 0.26,an = • •• Hence the solution is <X)

u( x,t) = :~::>ne - is,, sin (A,,X) = 1.l 8e-202 1 sin( 2.02x) + 0.32e-4 91 sin ( 4.9x) + .. • 2. Transform the following IBVP to an IBVP with zero BCs and solve the transformed problem. What does the solution look like for different values to time t . Does the solution agree with your intuition?

PDE u, = u_-a, 0 < x < 1, 0 < t < oo BC { u(O,t)=O u(l,t) = l

O<t<oo

IC u(x,O)=x2, 05x$]

Solution: Tfwe try a solution of the form u(x,t)=x+U(x ,t), we find u, = U, (x,t), ux( x,t) = 1 + Ux ( x,t ), uxx =Uxx (x,t ). Plugging these values into the IBVP, we obtain the new IBVP in U( x,t):

28

Solution Manual

PDE U, =Uxx' O<x<I, O<t<oo BC { U(O,t) = 0 U(l,t)=O

O<t< oo

IC U(x,0)=x 2 -x, O::::x::::l We now expand the IC u ( x, 0) = x 2 - x as a Fourier sine series 00

x 2 -x= z>nsin(mu:) n=I

and then multiply each side of the equation by sin( mnx) and integrate to get

t (x -x)sin( mnx )dx = a t sin mnx )dx 2

111

2(

or =

a,,,

f) x -x )sin(mm )d1: = { O n = o, 2, 4, ...

2

fl

Josin 2( mnx )dx

-8 m31t3

m = 1,3, •••

Hence, the sine expansion of the IC is x 2 -x =-.![sin( 7tX) + J_sin(3nx) +-1-sin (5nx) + · · ·] n3 27 125 and so the solution is

. (n:x )+-e 1 -tJ•J', sm • ( 37tX ) +-e 1 -ts.J', sm • ( 57tX ) +··· ] u (x,t ) =- 3s [e- ,,, sm 1t 27 125

6. TRANSFORMING NONHOMOGENEOUS BCS

29

3. Transform the following IBVP to one with zero BCs. Is the PDE in the transfo1med problem homogeneous?

PDE u, =uxx , O<x<I, O<t< oo BC { u., (O,t)=O ux (1,t) + hu(1,t) =1

O<t< oo

IC u(x,0)=sin(n.x), Q::;x:=;J

Solution: Since the left end of the rod is insulated, and at the right end a heat source is applied (or subtracted) depending on if I - hu( 1,t) is positive (heat applied since u x ( l,t) > 0) or negative (heat extracted) since u x ( l,t) < 0, our intuition tells us that the rod should reach a steady state temperature of the form u( x,oo) =Ax+ 1 . Our intuition tells us that the temperature should approach some constant, say u( x, oo) =A, thus we seek a solution as the sum of the steady state plus a transient. In other words

u(x,t)=A+U(x,t) The first step is to find the steady state A which transforms the initial problem in u to a new problem in U with zero BC, thus making the problem easier to solve and finding the steady state in the process.

Computing the derivatives of u, we have

u, (x,t) =U, (x,t) u_,(x,t)=Ux(x,t) uu (x,t)=U xx (x,t) u(x,O) =A+ U(x,O) and plugging them into the original IBVP, we find

30

Solution Manual

PDE U, =Ua, BC { IC

0 < X < I, 0 < ( < 00

Ux(O,t) = 0 O<t<oo Ux(l,t)+h[ A+U(I,t)]=I

A +U(x,O) = sin( 7tX)

If we now pick A =1I h, we see that the BC at x = 1 becomes zero, and so the new IBVP becomes PDE U, =U"', O<x<l, O<t<oo BC { U,(O,t)=O • O<t<oo Ux(l,t)+hU(l,t)=O IC U(x,O)=sin(nx)-.! ., Osxsl h

Hence, we now solve this problem for U ( x,t) and then the solution of the original IBVP is 1 u(x,t) = - + U(x,t) h

Lesson 7: Solving More Complicated Problems by Separation of Variables

1. Solve the following IBVP.

O<t<oo

IC u(x,0)=x, 0:s;x:s;l Solution: We have seen before that the Sturm-Liouville problem for this IBVP is X"+U=0

X(0)=0

X(l) =0 which has eigenfunctions

• (2n+l) - 2 - 7tX, n=0,1,2, ...

Xn(x)=sm The first three ( n =0, 1, 2) are shown in Figure 7 .1.

'

1.0

X

I 1.0

2

• ( 51t x ) • ( 31t x ) ,sm • ( 1t x ) , sm Figure 7.1: Eigenfunctions sm

2

31

2

32

Solution Manual

Hence, the solution of the IBVP is "'

u(x,t)= Ia,,e

n+

-( -2n+1) 'n·, , (2 I 2 sin - -

J

7tX

2

n=O

where the Fourier sine coefficients a,, satisfy the JC

J

~

• (2n+ X = L. a,, Sll1 - -I n=O

7tX

2

To find the coefficients a,, , we multiply both sides of this equation by • (2m +I SID -2

J

and integrate both sides with respect to x from Oto I, getting f 2(2m+l) - 2ru:dx = a"' Jo f xsm (2m+JJ ru:dx 2l

0

1 •

•

SID

-

or

J>sin(2m/ 1)ru:dx

a'" =

(-1)"' 8

-f-~s-in_2_(.,...2_m_2_+_1.,...)_m_dx_ = ( 2m + I)27t2 ' m = 0, I, 2, ...

(We know the functions • (2m + 1) , m=0,1,2, ...} { Sin 2since they are the eigenvectors of a Sturm-Liouville BVP and the eigenvectors of a Sturm-Liouville system are always orthogonal.) Hence, the solution is

33

7. SOLVING MORE COMPLICATED PROBLEMS

00

-

u(x,t)= ~a,,e 8

-(-,)' n ,1

-;?" [e 2

)' nl, (211+1

2

( 2 n+ ) ) sin - 2 - 7t.X

{5)', 57t - n·1 {3)' 37t - n·1, 7t )x- ... sin ( sin( )x+e 2 sinh·)x-e 2

2

2

l

2. What are the eigenvalues and eigenvectors of the following SturmLiouville problem?

ODE: X"+A.X=O, 0<x<1 _ {X(O)=O X'(O) = 0 BC. What are the functions p(x), q(x), and r(x)in the general Sturm-Liouville problem?

Solution: The general Sturm-Liouville (eigenvalue/vector) problem has the form

In the current problem, we have p(x) = 1, q( x) = O,r( x) = l. For the BCs we have a 1 = a 2 = 1, 13 1 = 132 = 0. ln the current problem, the general solution of the differential equation is

X (x) = c1 sin ./ix+ c2 cos ./ix and plugging this in the BC gives

x(o) = c2 = o X'(l) =c,.Ji cos.Ji - c2

./i sin .Ji= 0

34

Solution Manual

The second of the previous equations implies cos ✓ A = 0, which gives us the eigenvalues of the problem

r,:;- (2n+ -2-I) 7t, n = 0,1,2, ...

'\/1\, n =

(we call

,JC the eigenvalues, not their square An) and the eigenfunctions • (2n+ X (x)=sm - 2 - I) m, n=l,2, ...

11

The first three eigenfunctions are drawn in Figure 2.9 in Problem 1.

3. Solve the following problem with insulated boundaries. Does your solution agree with your interpretation of the problem? What is the steadystate solution and does it make sense?

O<t<oo IC u(x,0)=x, 0~x~1

Solution: The Sturm-Liouville problem is X"+AX=0 X'(0)= 0 X'(l) = 0 which has eigenfunctions

X n(x)=cos(mr.x), n=0,1,2, ...

The first three ( n = 0, l,2) are shown in Figure 7.2.

7. SOLVING MOR.E COMPLICATED PR.OBLEMS

35

e~ 1.0 f.....,.--=------------:,,.....--------,--

0.5

D0.5

Figure 7.2: Eigenfunctions l, cos( rr.x ),cos( 2rr.x ),cos( 3m ), ...

Hence, the solution of the IBVP is "'

'

2

u( X,f) = z>ne- (mt) 1 COS( mtX) = aO + ale- (mt) 1 COS( 7tX) + .. ' n; Q

where the Fourier cosine coefficients an satisfy the IC "'

u(x,O)=x= ~>ncos(nrr.x) n; Q

We find the Fourier coefficients an by multiplying both sides of this equation by cos ( m1tX) and integrating each side of the equation from Oto 1, getting xcos( mm )dx =amt cos 2( mrrx )dx

t

or a = m

m =1,3, ...

-4 (mn)2 m=0,2, ...

J,rt xcos ( mm )dx { 0

t

=

0

cos2(mm)dx

36

Solution Manual

m=O m=l,3, .. . m=2,4, .. .

Hence, the solution

u(x,t) =

f

a ,, e - (m, )' , cos(mtx)

n=O

Note as t ➔ oo the solution u( x,t) converges to the constant

which you know from intuition since both ends of the rod are insulated and so the initial heat u( x,O) = x must spread out along the rod at the average temperature, which is 1/2.

4. What are the eigenvalues and eigenvectors of the Sturm-Liouville problem?

ODE: X" + )} X =0, 0 < x < 1 BC· {

•

X'(0)=0 X'(O) =0

Solution: We have found the eigenvalues and eigenvectors in Problem 3 and found them to be

eigenvalues: A,= mt, n = 0,1, 2, .. . eigenvectors: X,, (x)=cos(mu), n=0,1,2, .. .

Lesson 8: Transforming Harder Problems to Easier Ones

l. Solve the following IBVP PDE U1 =Uxx -Ux, 0 < X <], 0 <{ <OO BC { u(O,t)=O u(I,t) = 0

O<t<oo

IC u(x,0)=ex12 , O::;;x::;;I by transforming it into an easier problem.

Solution: Since the IBVP describes the concentration u( x,t) of a substance in a moving stream (due to the term -ux) that is also diffusing (due to the term u ) in the medium, we might want to 'factor out' the amount of u that ·"' is due to the movement of the stream, and we do this by trying a solution of the form u( X,{) = e(anbt)u( X,{)

Computing the derivatives of u

u( X,{) = e(ax+br)u ( X,f) u, ( x,t) = be(ax+bt)u ( x,t) + e(ax+bt)u, ( x,t)

Ux ( X,f) = ae(ax+br)u ( X,f) + e(ax+bt)ux ( X,{) U:n ( X,f) = a 2e(ax+bt)U ( X,{) + 2ae(ax+bt)u x ( X,{) + e(ax+bt)u xx ( X,f)

we plug these values into u, = u_-a - ux getting

bU( x,t) + U, ( x,t) = a 2U ( x,t) + 2aUx ( x,t)

+ U _t, (x,t)-aU(x,t)- Ux (x,t) 37

38

Solution Manual

or U, =U xx + (2a - l}Ux + ( a 2 - a -b

)u

1 b =--.

1 H ence,the .

Wenowset2a-1=0,a 2 - a -b =O getting a=-, 2 4 transformation

u( x,t) = e(05x- 0.25t)u( x,t) transforms the original PDE to U, == U 0-, and the entire initial IBVP transforms to PDE u, = u<X, o< x < 1, o< t < oo BC { U(O,t)=O U(l,t)=O IC

O<t<oo

U(x,0)=1, 0;:S;x;:S;l

We solved this problem in Lesson 5 (Problem 4) and saw that the solution has the fonn

U(x,t)= fa,,e-(,m)' , sin(mtx) n= I

where the coefficients a,, are determined so U ( x,t) satisfy the lC <X)

U(x,0)= ~::>,,sin(mtx)=l n=I

We find the coefficients a by multiplying each side of this equation by sin(nmx) and integrating xfrom Oto l, and using the orthogonality ofthe sine functions, getting

t

sin(mnx )dx =amf~ sin 2 ( mnx )dx

_ f>in(m1tx)dx -{O

a,,, - f~ sin2 ( m1tx )dx -

Hence

m=0,2,4,

4/ mn m = 1,3, ...

8. TRANSFORMING HARDER PROBLEMS TO EASIER ONES

n:: ;: l

• ( l - (mr)', sm mu: ) L.. - e = -4 ~ 7C 11=1,3.... I n

= -±[sin( nx) + !sin (3nx) + !sin( 5nx) + ••·] n

5

3

Hence, the solution u(x,t) of the original IBVP is

u ( x,t) = e(05x- 025t)u ( x,t) 2

<X)

• ( nnx ) L..,a e- (nn) 1 sm = e(O.Sx-0.251)" 11

,r-::c J

• ( 4 (0.5x- 0.25t) ~ nnx ) L.., -1 e - (1m) 1 sm = -e 2

7C

11=1.3,... I

n

Note: In general, the PDE u, = a 2uxx -vu, that contains a convection term -vu, (we include a negative with the term since that means the substance described by u( x,t) moves in the direction of increasing x) can be transfonned to the PDE U, = a 2U.,. by the transformation I (•x v'r) u(x,t)=e" 1 T"'i" U(x,t)

The proof is identical to the steps carried out in this problem.

39

40

Solution Manual

2. Use steps a), b), and c) below to solve the IBVP PDE u, =uxx -u+x, 0 < X < 1, 0 < t < 00 BC { u(O,t)=O u(l,t)=l

O<t<oo

TC u(x,0)=0, Osxsl

a) transforming the nonhomogeneous BCs to homogeneous ones b) transforming into a new PDE without the term -u c) solving the resulting problem.

Solution: Since the temperature at the ends of the rod are fixed at O and 1, we suspect the temperature will reach a steady state of u ( x, oo) = x, Hence, we try a solution of the form u(x,t)=x+U(x,t) Plugging this value into the original IBVP, we arrive at (the x term in the PDE with cancel) PDE u, =U.u -U O<x < I, 0 <t <OO BC { U ( 0, t) = 0 U(1,t)=0

O<t<oo

IC U(x,0)=-x, Osxsl

Since the term -U in this PDE is a damping (or evaporation) term, we try to find a solution of this equation of the form U( x,t) = e-' w( x,t). Plugging this into the previous IBVP we find PDE w, =w.u

O<x<l, O<t<oo

BC { w(O,t)=O w(l,t) = 0

O<t<oo

IC w(x,0)=-x, Osxsl

8. TRANSFORMING HARDER PROBLEMS TO EASIER ONES

This IBVP has the solution

' ' sin(mtx )dx "' w(x,t) = ~:>ne-(mt) n=I

where we find the coefficients an so the JC "'

w(x,0)= ~:>nsin(mtx)dx=- x n=I

is satisfied. We do this by multiplying each side of the equation by sin (mm) and integrating each side with respect to x from O to 1, getting

-t a

xsin ( m1tx )dx = a f ~ sin 2( m1tx )dx

_ 111 - -

111

_ m+i 2 f>sin(m1tx)dx _ -(-1) - , m-1,2,3, ...

1 mrr. fosin 2 (mm)dx

Hence

n= I

) • { nrr.x L. -J {- l)n+I e- (n• )' , sm -_ -2 ~

Tr. n= I n

• (5 ) • ( 3 ) -e I - (Jn)' , sm rr.x +-- •] • sm rr.x + J -(Sn)', 5 3

• { rr.x ) - -e = -2 [ e - (n)', sm

rr.

So, FINALLY, the solution u(x,t) of the original lBVP is

u(x,t)=x+U(x,t) • (5 ) I -(Sn}', sm • (3 ) I -(h)' , sm . ( rr.x ) --e 2 _, [ e-(,)', sm rr.x +··· ] rr.x +-e =x+-e 5 3 rr.

41

42

Solution Manual

3. Solve PDE u, =uxx - u O< x < I, 0 < t < oo BC {

u(O,t)=O u(l,t)=O

IC u(x,O)=sin(ru:), 0$x$l directly by separation of variables without making any preliminary transformations. Does your solution agree with the solution you obtain if you make the transformation u( x,t) = e-'w( x,t) in advance?

Solution: Since the term -u in this PDE is a damping term, we could solve this problem by seeking a solution of the form u(x,t) =e-'U( x,t). This factors out the damping so we are only left with diffusion described by the variable U(x,t). Plugging this into the IBVP we find

PDE U, = U"'

0 < x <l, 0 < t < oo

BC { U(O,t)=O U(I,t) = 0

O<t< oo

JC U(x,O)=sin(ru:), 0 $ x$l

which has the solution

U ( x,t) = e-"'' sin( nx)

thus the original problem has the solution

=e' e

2 - 1( '

sin ( ru:)

=e-(n'-1)1 sm ( ru: )

8. TRANSFORMIN G HARDER PROBLEMS TO EASIER ONES

However, ifwe solve this problem without the preliminary transformation, we use separation of variables, seeking a solution of the form u( x,t) = X( x )T(t), and plugging this into the PDE we get

XI''=X'T-XT and dividing by XT, arriving at

T'

x•

-=--1=-A X T

2

or

The solution of the ODE in X ( x) is

X(x) =c1 cos( .J,-2 -1 t )+c2 sin( ~ t) and using the BCs give c1 = 0 and

X(I) = c2 sin(~)= 0 ⇒ ~=nrc ⇒ A 2 =(mr)2+1

⇒ A. 1 = ✓(nn) 2 + 1 Hence, we arrive at solutions

u (x,t) = X (x)J;,(t) 11

11

2 / +Ix =e-[(111,J' +1], sin ( -v(nn)

= e-' [e-(m,)' , sin(nnx)]

)

43

44

Solution Manual

and so we get a solution of the form co

u(x,t)= 2:a,,x n(x)T,,(t) n= I

=e-' i::a,,e ~(mt)' , sin(mtx) n= I

lf we find the coefficients an so that u ( x, 0) = sin ( 7tX) we find the only nonzero value is a 1 =1 , and so u(x,t)= e-' sin(1tX) which is the solution we got when we made the initial change of variables.

Note: ln general the PDE u, = a 2u,, -13u with decay term 13u can be transformed to the equation U, = a 2U,, with the transformation u( x,t) = e~P1U ( x,t) . The decay constant 13 is generally positive since then a positive value of u means that - 13u < 0, which makes u smaller.

Lesson 9: Solving Nonhomogeneous PDEs (Eigenfunction Expansions)

1. The solution of the IBVP PDE u, =u xx + sin (37tX ), 0 < X < l, 0 < t < 00 BC { u(O,t)=O u(l,t) = 0

O<t< oo

IC u(x,O)=sin(1tX) , 0:5:x:5:l is given in the book by

u( x,t) = e - (1ta) sin( 7tX) + -11

'

2

[1

-e- (J1ta)', ] sin(3nx)

(3mx) Does this solution agree with your intuition? What does the solution look like?

Solution: The solution starts off looking like u( x,O) = sin ( 7tX) and gradually morphs into the steady state

u( x,oo )=-1- 2 sin(31tX)

(3,ta)

2. Solve the following IBVP PDE u, =u_u +sin(1tX)+sin(21tX), 0 < X < 1, 0 < f < 00 BC {

u(O,t)=O u(1,t)=0

O<t< oo

IC u(x,O) =0, 0 :5: x :5:1 Solution: Step 1: We expand the nonhomogeneous terms of the PDE 45

46

Solution Manual

00

J( x ,t) = LJ,,(t )sin( mtx) =sin( TIX)+ sin (27tX) n= I

and so we identify

J;(t)= J; (t)=l J,,(t)=O, n=3,4, .. .' .

Step 2: We now seek a solution of the fonn 00

u(x,t)= LT,,(t)X,,(x) n= I

where (in this particular problem) the X n ( x) are the eigenfunctions of the Sturm-Liouville problem X" +"A, 2 X=0

x(o) = o X(l)=O and the T,, (t) are solutions of the initial-value problems

r; + ( nn )2 T,, = J,,(t) T,,(0)=0 Solving these problems, we find X,, (x)=sin(n1tX), n=l,2, ... [ - (nn)' ,] T,, (t ) = -J , n = 1,2.

2 1-e

(nn) T (t) =O, n=3,4, ...

11

Hence, the solution of the IBVP is

9. SOLVING NONHOMOGENEOU S PDES

<X)

u( x,t)= L7;,(t)X,, (x) n=I

3. Solve the following IBVP by the method of eigenfunction expansion.

PDE u, =un_ +sin(1tX) , O<x<l, O<t< oo O<t< oo

BC { u(O,t)=O u(l,t)=O IC u ( x, 0) = 1, 0 :S: x:::;; 1

Solution: Step I : We expand the nonhomogeneous terms of the PDE "' f ( x,t) = L/,,(t )sin ( mtx) =sin( 7rX) n= I

and so we identify

J;(t)=I J,, (t)=O, n=2,3, .. .' .

Step 2: We now seek a solution of the form 00

u(x,t)= LT,, (t)X (x) 11

n=I

where (in this particular problem) the X ,, ( x) are the eigenfunctions of the Sturm-Liouville problem

47

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Solution Manual

X" +t..2 X =0

x(o) = o X(l) = 0 and the T,,(t) are solutions of the initial-value problems

n= 1,2,3, ... where an are the Fourier coefficients in the Fourier series expansion of the IC u(x,0) = !. That is 00

I

u(x,O) = I= }::Onsin(xnx) ⇒ an== 2fosin( mtx )dx == ( n=I

4

)

2n - J 7t

For this problem, we have

1

1 4 7;(0)=2 sin(nx)dx=-

o

7t

n== 2,3, .. .

Solving these problems, we find X" (x)=sin(mtx), n=l,2, ... _4 e- rr ' 1 + 1 [] -e - rr' 1] .

T1 ( t ) 2 7t

7t

0, n = 2,4,6, ....

T,, (t)=

{

~ e - (nrr)' t n --3 , 5, 7 ,. n1t

9. SOLVING NONHOMOGENEOUS PDES

49

Hence, the solution of the IBVP is 00

u(x,t)= LTn(t)Xn(x) n= I

4. Solve the following IBVP by the method of eigenfunction expansion where 11. 1 is the first root of the equation tan A= -11, . What are the eigenfunctions X,. of the problem?

PDE u, =uxx + sin(11, 1x ), 0 < x < 1, 0 < t < oo

u(O,t)=O ux(1,t)+u(1,t)=0 IC u(x,0) =0, 0 :S: x ::,1

BC {

O<t<oo

Solution: Step I: We expand the nonhomogeneous terms of the PDE 00

f(x,t)= L J,,(t)X,,(x) n= I

where X,, ( x) are the eigenfunctions of

X" +t...2X=0 X(0) = 0 X'(l) + X(!) = 0 In this problem the nonhomogeneous term sin(11. 1x) is the first eigenfunction and so we have ft (t) = 1,J,, (t) = 0, n = 2,3, ... Hence, we seek a solution of the form

50

Solution Manual

"'

u(x,t)= L/n(t)Xn(x) n= I

where (in this particular problem) the X ,, ( x) are the eigenfunctions of the Sturm-Liouville problem

X" +).2 X=O

x(o) = o X' ( 1) + X ( 1) = 0 which are

X ,, (x) = sin1cnx where \, is the nth root of tan 1c = - A, and T,, (t) are solutions of the initial-value problems

~ n= 2,3, ...

T; + 1c ~Tn = J,, (t) ={

n=I

7;, (0) = a,, and a11 are the coefficients of the Fourier series expansion of the IC

u(x,0)=0. That is "'

u ( x, 0) = 0 = La,, sin ( XTIX) ⇒ a = 0 11

n=l

In this problem, we have

J;'+1c~:,; = 1 7;(0)=0 T' +1/ T =0 " nn ⇒ 7;, (t)=On=2,3, ....

T,,(0)=0 Hence, the solution of the IBVP is

9. SOLVING NONHOMOGENEOUS POES

5. Solve the IBVP PDE u, =ux.,, O<x<l, O<t<oo BC {

u(O,t) = 0 O<t<oo u(l,t)=cost

IC u ( x, 0) = x, 0 :::; x :::; l by a) transforming the problem to one with zero BCs b) solving the transfonned problem by expanding the solution in terms of eigenfunctions Solution: Trying a solution of the form u( x,t) = xcost + U (x,t) we see that U ( x,t) satisfies

U, =Ux.. +xsint U(O,t) = 0 U(l,t)=O U(x,O) = 0 We now expand the RHS xsint = J; (t )sin( m:) + J; (t )sin(2m:) +· •· from which we can find /,, ( t) to be

f

I,, (t) = 2 >sin(t )sin ( nnx )dx =(-1

r' (

n21Jsin(t ), n = 1,2, ...

We now can find the functions Tn (t) which are the solutions of the initialvalue problems

51

52

Solution Manual

r;+(mc)2T,, =(- 1t' (n2n}in(t) T,, (0)=0 which has solutions (-1)" 2n( e(- ,,n)' , +(nn) 2 sint-cost)

T,,(t)=

4

5

n n +n

Hence, the solution of the transformed IBVP is

., ,, (-1)"2n(e(-,m}', +(nn)2 sint-cost) U(x,t)= LT,, (t)sin(mr.x)= sin(mr.x) 4 5 n=I n=I 7t n + n

L

and so the solution of the original IBVP is

u(x,t) = xcost + U(x,t) "' (-1)"2n(e(-,m)', +(nn) 2 sint-cost)

=xcost+

L 11=1

4

7t n

5

+n

sin(nnx)

Lesson 10: Integral Transforms (Sine and Cosine Transforms) 1. Prove the identities

~,[f']=-ro::5c[f]

1.

2. '.3, [/"]=3_rof(O)-ro 2 '.3s[f] 7t

3. 3c[f']=3-wf(O)+w 2 '.3,[f] 7t

4. '.3c[f"]=-3_J'(O)-ro 2 '.3c[/] 7t

Solution: I. Using the definition of the Sine transform and integration by parts, we get

3, [f']=-2 i"' f'(t)sin(wt)dt 7t 0

= -2 [ f(t)sin(wt) 7t

JI"' -2ro - i"' f(t)cos(wt)dt 7t 0

O

= - -2w 7t

f"' f(t)cos(wt)dt 0

=-ro::\[J] In the above integration we have used the fact that we must assume f (t) ➔ 0 as t ➔ oo else the transform will not exist. The identities 2,3, and 4 are proven in much the same way, by integration by parts. We leave this simple exercise for the reader.

2. Solve the initial-value problem ODE : dU + ( aro )2 U = 2Aroa 2

dt

1t

IC: U(O)=O

53

54

Solution Manual

Solution: This problem is a simple review of what you learned in ordinary differential equations. The general solution ofthe ODE y' +ay=b where a,b are constants is

where c is an arbitrary constant. Plugging this general solution in the IC y( 0) = 0 gives c = -b I a, hence the solution of the initial-value problem ODE: y' +ay=b IC: IS

y(0) =0

b - at +-=b b ( l -e - at ) y ( t ) =--e

a

2

a

a

2Affia. 2

In our problem we have a= ( affi) and b = - - - and so the solution of 7t

the given problem is

10. INTEGRAL TRANSFORMS

55

3. Solve the following IBVP by means of the Sine or Cosine transform.

PDE u, =ciuxx, O<x<oo, O<t<oo BC

u(O,t)=O O<t<oo

IC

u(x,O)=H(l-x), O~x< oo

where H ( x) is the Heaviside (off-on) function defined by

H(x)={O x<O 1 x~O and hence the (on-off) function

x<l

H(l-x)={~ x~I

Figure 10.1: Heaviside (or step) function H(l-x)

Solution: Applying the (infinite) Fourier sine transform (with respect to the

variable x)

F,.[u]=-2 f"' u(x,t)sin(cox)dx 7t 0

we have

F,[u,] = 3. f"' u, (x,t)sin(cox) dx =~F, [u] = !:!.__U(t) dt ot 1t Jo F,[uxx ]= 3_ f"' uxx (x,t )sin(cox) dx=3.cou(O,t)-u/U =-ro 2U(t) 7t nJo

56

Solution Manual

Hence the PDE in x and t is now transformed to a first-order ODE in t (the variable x in the PDE is transformed to w , which now is considered to be a parameter in the ODE in t ). The IC IC

u{x, 0) = H(l-x) , 0 :::; x <oo

in the original IBVP is transformed to the constant ( constant insofar as the ordinary differential equation in t is concerned) f1 ( 1-cosOJ U (0) = 2Jf"' OJ ) 0 H(x)sin(wx) dx = 2J 0 sin(wx)dx=2

Hence, we now have the initial-value problem in variable t ODE: dU + OJ 2U = 0 dt

IC:

U(0) = 2c-:sOJ )

which can easily be solved and found to be U (t) = 2(1 - :sOJ )e-m',

Finally, taking the inverse Sine transform, gives f"'( --OJl - cos OJ ) e-"'', sin(OJx)dOJ u(x,t) =210

where er.f and e,fc are the error and complementary error functions, respectively.

The Mathematica statement for finding the above inverse Fourier sine transform is InverseFourierSinTransfo r m[2*((1 - Cos[w))/w)*Exp[ - w*w*t) , w, x)

10. INTEGRAL TRANSFORMS

57

The solution may not be in familiar terms to some readers, but it can be evaluated accurately for any values of x and t. Below we have plotting the solution as a function of x for different values of time t. We see that the heat is diffusing out across the rod for larger and larger x. Heat is being lost from the rod at the left end point x = 0. The steady-state temperature of the rod is u( x,oo) = 0 for all x. Note that when t =10 in Figure 10.1 c) the temperature is nearly 0.

,

u{x,0.1}

Figure IO.I a): Solution u(x,t) as a function of xwhen t=O.l

u(x,l) ,~

I

\

\ ()10

..,

\

Figure I 0.1b): Solution u( x,t) as a function of xwhen t = I

58

Solution Manual

"'"

'

u(x,10)

~-

o.cu

... ...,.,.

///

0.010

'""

/

/ X

Figure I 0.1 c): Solution u( x,t) as a function of x when t

=10

Lesson 11: The Fourier Series and Transform I. What is the Fourier series expansion of the square wave

f(x)=

{

-1 -l<x<0 +l 0<x<l

f (x + 2) =f (x) (periodic condition) '~

f(x)

1 ,___ X 0

1

-

Solution: The function is odd, i.e., f (x) =-f (-x) and so we know the Fourier series only contains sine terms, hence it has the form SJ ( X) =

"° sin( 1>n

n1tx ), n

=1,2, ...

n=I

(Note that the sine series starts with n = l since sin( 0) =0.) To find the Fourier coefficients an, n =I, 2, ... , we multiply each side of the equation by sin ( nmx) and integrate each side of the equation from -1 to 1, and because of the orthogonality relation

f sin( 1

mtX )sin( mm

-1

we get 59

{O m-:t-n

)dx =

1 m=n

60

Solution Manual

L/ (x )sin( mnx )dx = I a Lsin( nnx )sin( mnx )dx I

00

I

11

n= I

=a,,, Hence, for the given function the Fourier coefficients are a,,,= (f(x)sin(mnx)dx

=-f sin(mnx)dx+ Jof sin(mnx)dx 0 -I

1

J: J: sin(mnx)dx = 2 J: sin( mnx )dx = sin(mnx)dx+

= - ~) cos( mnx )I~) ~

={ mn

m = 1,3,5, ...

0, m=2,4, ..

and so the Fourier sine series of f is

4 1 S1 (x)=-I-sin[(2 n-l)nx] 7t ,,=1 2n -1 00

= ~[sin( nx) + .!.sin(3m) + .!.sin( 5m) + •••] 1t 3 5

The first few terms of this series are shown in Figure 11. l.

11. THE FOURIER SERIES AND TRANSFORM

1.0

0.5

1.0

Figure I I.I a): First term of the Fourier series

Figure 11.1 b): First 2 terms of the Fourier series

Figure I 1.1 c): First 3 terms of the Fourier series

61

62

Solution Manual

Figure 11.1 d): First 4 terms of the Fourier series

2. The sawtooth function

f(x)=x

-L<x~L

f (x + L) =f (x) (periodic condition)

f(x)

Ii i

i

i

i - Li i

X

i

[/ has a Fourier sine series given by

Show that if we differentiate this series term by term, we arrive at an infinite series that does not represent the derivative of the sawtooth function.

11. THE FOURIER SER.I ES AND TRANSFORM

63

Solution: The Fourier sine series Sf ( x) represents a sawtooth curve, whose derivative is 1 except at the integers where the sawtooth function is not continuous, hence not derivative. If we differentiate the sine series Sf ( x) term by term, we do not get a series that represents the derivative of the sawtooth function since

doesn't represent anything since the series does not converge for any value of x.

3. Graph the frequency spectrum of the following periodic functions. a) /(x)=sinx b) f(x) =sinx +cos2x c) / ( x) = sinx + cosx + .!.sin3x 2

Solution: Ifthe Fourier series F[/] ofa function f(x) is F[/] = f [ a cos( mu:)+ b sin( mu:)] 11

11

n=O

= a0 + I[ ancos(mu:) + b11 sin(mu:)] n=I

then its frequency spectrum is the function C11 , n = 0, 1,2, ... (or sequence if you like) defined on the non-negative integers by ~

en = vun -t-Dn ,

n = 0,J,2, ...

(Note: b0 is always O in a Fourier series.) Computing these values for each function, we have the following graphs. a)

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So/Wion Manual

a)

_

_.__

___......._._ _ _ _ _,.. D

Figure 11.2 a): Frequency spectrum off ( x) = sinx b)

_

_.__..___ _ _ _ _ _ _ _., D

0123456

Figure 11.2 b): Frequency spectrum of f ( x) = sinx +cos2x c)

Ji

1/2

0123456

-

D

Figure 11 .2 c): Frequency spectrum of f ( x) =sin x + cos x +

.!. sin 3x 2

11. THE FOURIER SERIES AND TRANSFORM

4. What is the Fourier transform F(I;) and frequency spectrum

C(I;) = IF(l;)i of the function 1 -l<x<l { f ( x) = 0 elsewhere

Solution: Computing the Fourier transform we find

Since the transform is a real function , its spectrum is simply its absolute value, or

5. Show that the absolute value of the complex function

1 F(I;)= I+il;

is

65

66

Solution Manual

Solution: The idea is to write the transform in the form F(s) = a(s) + ib(s) and then compute IF(s)I = ✓a 2 (s) + b 2 (s). Jn this problem, we do this by using the process of rationalizing the denominator, or

Hence

6. Verify the following orthogonality properties of the sin and cosine functions on the interval f-L, L

l,

- - sm (nrcx) dx= {O m::t:-n I,_sin. (mrcx).

L L L m=n - 1-

- - cos (nrcx) dx= {O m ::t:- n I'- cos (mrcx) L L L m =n -L

L cos (nrcx) L dx=O allm,n=l,2,3, ...

I-Lsm (mrcx) L

•

Solution: Straightforward integration. Hint: Use the identities

11. THE FOURJER SERIES AND TRANSFORM

67

sin0sin<j> = _!_[ cos(0-<j> )-cos(0 + <j> )] 2 cos0cos<j> = _!_[ cos(0-<j>) + cos(0 + <j>)] 2 sin0cos<j> = _!_[ sin(0 + <j>) + sin (0-<j> )] 2 We will prove the orthogonality condition

r

l

sin ( m~ )cos ( n~) dx

=_!_ f L[sin(m+n)n.x +sin(m-n)n.x]dx 2 -l

L

L

l

-L ) cos(m+n)1t.Xll 1t.Xl l =-2I [ (m+nn +-l [ ( -L ) cos(m - n)L -l 2 m-nn L -L ( - L ) [ cos ( m + n) 7t - cos ( m + n) 7t] 2 m+n 1t

+ ( -L) [(cos(m-n)n-cos(m-n)n)] 2 m-n 1t =0 allm,n=0,1,2, ...

l

Lesson 12: The Fourier Transform and Its Applications to PD Es

I. Find the Fourier transform of

f ( X) = {e~x X < 0

x2::0

Solution: From the definition of the Fourier transform, we have

=};_Jo"' I

e-xe- •;x dx

f"'

- (1 +,; )xd

= .Ji;, Jo e

x

_ -I- --1 - e - (l +1~ )x I"' .Ji;, 1+ ii; 0 1

1

= .Ji;, 1+ ii; To find the Fourier spectrum, we first write F

__ 1 _1 1-il; _ _ l .!..=.5_ _ _ 1 (I;) - ✓'hr 1+ ii; I - ii; - .Ji;, 1+ I; 2 - .Ji;,

Hence the Fourier spectrum is

68

[(-1 1+ I; 2

J+i(----=LJ] 1 + I; 2

12. THE FOURJER TRANSFORM AND ITS APPLICATIONS TO POES

l IF(l;)I= ✓'he I

(

2

l ) ( -/; ) + l+/;2 ~

69

2

I +/; 2

= ✓'he (1+1;2)2

1

1

= ✓'htN

2. Verify that the Fourier transform and inverse Fourier transform are linear transformations.

Solution: The calculus identity

F[cJ + c2g]

=;;.

J: [cJ(x)+c 2g(x)Je-,g dx

= c1{;;. J: [/(x)Je-,f.r dx }+c 2{;;. L: [g(x)Je-,g dx} = C13[/] + C23[g]

shows that the Fourier transform is a linear transfonn. Also, if F(i;) and G(I;) are the Fourier transforms of functions f(x) andg(x), respectively, then the inverse Fourier transform is also linear due to the following linear property of the integral:

r 1 [c 1F +cp]

=;;. J: [

c1F(i;)+cp(i;)]e't.' di;

=c1{;;. J: [F(i;)]e'~ d/;}+c2{;;. J: [G(i;)]e'g d/;}

=C1'.:r 1[ F] + C23- J [ G]

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Solution Manual

3. Solve the following initial-value problem by the Fourier transform.

PDE: u, =a 2 uxx IC:

-oo<x<oo, O< t <oo

u(x,O)=e-·'

2

-oo < x<oo

Solution: Taking the Fourier transform of the partial derivatives

~[ I f"" u, ( x,t )e - ,1;,x dx ...1 u, ] -- r::;v2n -

= !!___[

~ f"" u(x ,t)e-'~-' dr:]

dt v2rt -00

and plugging these values in the PDE and TC, we get ODE:

dU(t)= - (a~) 2 U(t) dt

IC:

U ( 0) =

:i[e-x' ] = Jz e-<~ >'

O<t<OO

12

Remembering your ODEs, we solve this initial-value problem and get

which has an inverse Fourier transform

12. THE FOURIER TRANSFORM AND ITS APPLICATIONS TO PDES

f"' -I- e- ('., 12J' e - (~)' , e~,,, dt":i 2 ~ -"' ✓

= -I-

I

= --

f "'

=

,~, d J:

- (~12 J' - (a{)' ,

2J;. _., e

e

e

~

~ e-c.:~,,) (not so straightforward integration) 1+4a 2/

Note: A good student might be able to evaluate the above integral by collecting terms involving <; in the exponent, completing the square, and then integrating. If not, you can always go to www.wolframalpha.com and enter (J /(2*sqrt(pi))* int exp(-(l /4)*z"2 - t*(a*z)"2 +l * z*x) dz z=-infinity .. infinity

4. Verify the following properties of the Fourier transform :i[ u,] = ii;:![ u] :i[u,... ]=-1; 2:l[u]

Solution: From the definition of the Fourier transform and applying the integration by parts formula f

h a

we have

udv =uvj h

f h

a

tJ

vdu

71

72

Solution Manual

F[ux] =

Jk

J: ux(x)e-ig dx

=u(x)e-if.'[.,, +iE,f: u(x)e-'g dx =ilf[u] We can now apply the above formula to find

F [u"' ]= if,F [ux] = if, [if,] F [u] = - f, 2 F [u]

5. Verify that the convolution of two functions f and g can be written either as

or

l J"' J(E,)g(x-f,)df, (f*g)(x)= ~ v21t -

Solution: Going from one convolution integral to the other is simply a change of variables. lfwe let z = x- f, in the first convolution integral, we get

I f"' f(z)g(x-z)dz = v2n ~ _,,,

1 J_ "' ('f,)g( x -'f,)d'f, =~ 00

/

Lesson 13: The Laplace Transform l. Verify the following formula for the transform of the partial derivative

[ u, ( x,t )] = sU(x,s )-u(x,O) Solution: Using the definition of the Laplace transform and integration by parts (which will take the derivative off u, and put it on e-s' ) , we have

L[u,]= [ 'u,(x,t) e-s'dt

= u( x,t )e-"[ ; + sJo'"u( x,t )e-" dt = -u(x,O) +sf; u(x,t )e-" dt 1

= s L[ u(x,t)]-u(x ,O) 2. Solve the following initial-value problem using the Laplace transform

- 00 < X < oo, 0 < t < CX)

PDE: u, = a.2uxx IC:

u(x,O)=sinx -oo<x< oo

Solution: Since

L[u,] =s U ( s) - u (x, 0) di

L[uxx ]= dx 2 U(s) where we denote U ( s) = U (x,s) = L[ u( x,t)], the PDE has independent variables x,t, and transforms to an ODE with independent variable x. The ODE also depends on the transfonned variable s, but s is held constant in the ODE, so we treat it like a parameter. Hence, we have 73

74

Solution Manual

ODE: sU(x )-sin x = a. 2

d 2U(x) 2 dx

Any student who has studied ODEs should be able to solve this 2 nd order, linear, nonhomogeneous ODE. If you can't solve this equation yourself, just go to www.wolframalpha.com and enter a"2 * U" (x) - s * U(x) = sin(x) and wolframalpha will respond with

The inverse Laplace transform of this expression can be found in a table of inverse transforms or by use of a computer program like Mathematica or Maple, and is

You can verify that this function satisfies both the PDE and ICs. You can also see that the solution is a 'decaying' sine curve. Does this solution make sense to you? It should; the energy gets spread out more and more on the real line.

3. Solve the following IBVP using the Laplace transform.

PDE: u, =uxx

O<x <oo, O<t<oo

BC: u(O,t)=sint

O<t<oo

IC:

O<x<oo

u(x,0)=0

What is the physical interpretation of the problem?

Solution: Taking the Laplace transform of this problem, we find

75

13. THE LAPLACE TRANSFORM

dl ODE: sU(x,s)=-2 U(x,s) dx IC :

U(O)=/+l

Note that this initial-value problem doesn't make sense inasmuch as the ordinary differential equation is 2 nd order and there is only one initial condition. We add a second condition from the physical fact we require the solution U(x) to remain bounded for large x. lfwe solve the above ordinary differential equation with the single initial condition, we will get one arbitrary constant c, in the solution. Solving this problem with wolframalpha.com we enter U"(x)=s*U(x), U(O)= l/(s"'2+ 1 ) and wolframalpha gives us e-F , ( c1s 2e 2.r:.., - c1s 2 + c1e 2 .f;_, - c, +

1)

U ( X ) = - ~ - - - -2 - - - - - ~ s +l

but the only bounded solution occurs when we pick c, = O, and so we have the solution

U( x,s ) =-2-I e-,/; s +I

and finding the inverse Laplace transform, we get u(x,t)= c' [u(x,s)] =

c' [-,-1-e-x.r:.] s~+I

= L-1[s2l+i] * c ' [e-x-"]

76

Solution Manual

The above convolution integral can be evaluated numerically for any ( x,t) to essentially any accuracy you please. The above inverse transform of e-x.rs was found with the help ofwolframal pha by entering the command inverse laplace transform exp(-x*sqrt(s)) which yields ;i.

xc-4' 2

..r;;

for X > 0

t 3/ 2

You might try to imagine what the solution looks like. The temperature of the semi-infinite rod is initially zero, but we force the left end of the rod to have the periodic temperature u( O,t) = sint. What does the temperature look like as you move down the rod?

4. Solve the following two-point boundary value problem for U ( x) (treat s as a parameter; i.e., a constant).

ODE:

0<x<l

BCs:

Solution: As you learned in ODE, the general solution of the ODE is

where c1,c2 are arbitrary constants. To find c1,c2 we plug the general solution in the BCs, getting

13. THE LAPLACE TRANSFORM

which yields the constants

Plugging these values into the general solution U(x) and simplifying, we get

77

Lesson 14: Duhamel's Principle Jean-Marie Duhamel (1797-1872) was a French mathematician and physicist who obtained a method of finding the solution of one PDE in terms of another.

1. Write the solution of the IBVP PDE: u, := u.x.,

0 < X < I, 0 < t < 00

BC: {u(O,t)=O u(l,t) = f(t)

u(x,0)=0

IC:

O<t<oo O<x<I

in terms of the solution of the IBVP PDE: w, := w.x.,

0 < X < I, 0 < t < 00

BC: {w(O,t) = O w(l,t) = o(t)

w(x,0)=0

IC:

O< t <oo O<x<l

Solution: Taking the Laplace transform of PDE: u, =U.x.r BC: { IC:

0 <x < I, O<t<oo

u(O,t) = 0 u(l,t)=f(t)

u(x,O) = 0

O<t<oo O<x<l

and PDE: w, = w.x., BC: { IC:

0<x<1, O<t<oo

w(O,t) = 0 w(l,t) = o(t)

w(x,0)=0

O<t<oo O<x<l

we find 78

14. DUHAMEL'S PRJNCIPLE

d2U(x) dx 2

sU(s)=O

u(o) =o U(l) =F(s)

_dz_w~(x~)-sW(s)=O dx 2

W(O)=O W(l)=l

which give solutions

✓s X] u (x,s) =F( s)[si'.111 ✓s X] w (x,s) = [si~ smh ✓s smh ✓s

Hence we have (watch carefully)

=w(x,t)* f(t) 2. Find the solution of the boundary value problem

d 2U(x)

-sU(s)=O dx 2 U(O)=O BCs { U(l)=F(s)

ODE:

Solution: The ODE has the general solution

79

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Solution Manual

To find the constants cPc2 we plug the general solution in the BCs, getting c 1 +c2 = 0

cle..{s + C2e- .Jsx = F( S)

and solving for ci,c2 gives

F(s)

CI = - .Js -=s~~ -✓ -= s s'

e

-e

F(s)

C2 = - - .Js-=s~~ - .Js -= s' e -e

Hence, the solution is U(x)-ce.fsx +c 2e- .fs., I =

F(s)

e✓s - e- .fs

[e.Jsx -e-.fs.' ]

✓sx

- .Jsx l

= F ( x) [ e .fs - e- .fs e s -e s

=

F(s)[si~ ✓s xl smh ✓s

3. What is the solution u(x,t) of the IBVP PDE: u1 = ux.,

0 < x < l, 0 < t < oo

BC: {u(O,t)=O u(l,t)=f(t) IC:

u(x,0)=0

O<t<oo Osxsl

in terms of the solution w(x,t) of PDE: w,=wx.r BC: {w(O,t) = 0

w(l,t) = g(t)

IC:

w(x,0)=0

O<x<l, O<t<oo 0 < t <oo Osxsl

81

14. DUHAMEL'S PRINCIPLE

Solution: Taking the Laplace transform of each problem gives

d 2U(x) • dx2

- s

u (s) = 0

U(O)=O U(I)=F(s)

d 2W(x) -~-sW(s)=O dx 2 W(O) = 0 W(l)=G(s)

which give solutions

U(x,s) = F(s)[si~h.Js x] W(x,s) = G(s )[si~1t] smh.Js

Hence, we can write

U(x,s) = F(s )[si~1t] s1nh s = F(s) G(s )[sinh J; x] sinh ✓s G(s) Hence, we have

smh s

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Solution Manual

4. Use the result in Problem 3 to find the solution of PDE: u, =u.u BC: {u(O,t)=O

O<x<I, O<t<oo O<t<oo

u(I,t)=sint

IC:

u(x,0)=0

0$;x$;1

in terms ofthe solution w(x,t) of PDE: w 1 =w,u

O<x<l, O<t<oo

BC: {w(O,t)=O

O<t<oo

w(1,t)=l IC:

w(x,0)=0

0$;x$;l

Solution: Here

f ( t) = sin t ⇒ F ( s) = I / ( s 2 + 1) g(t)=l ⇒ G(s)=Ils

and so

F(s)

s

G( s)

s2 + 1

1

s ] =smt .

--=-- ⇒ F - [ - -

s2 + l

Hence

u(x,t)=F- 1[ ~~:~}w(x,t) =sint*w(x,t)

=J: w(x,t-~)sin~d~ This is not part of the problem, but we could now observe that the steady state for w( x,t) will be w( x, oo ) = x and so letting

w( x,t) = x + U ( x,t)

14. DUHAMEL'S PRINCIPLE

and arrive at the IBVP for the transient U ( x,t):

« PDE: u, = u_

o< x < 1, o<, < oo

BC: {U(O,t)=O U(I,t) = 0

O<t<OO

U(x,0)=-x

O:-s;x:-s;I

IC:

which we can solve by separation of variables. Hence we have U(x,t) and thus w(x,t) and thus u(x,t). Although w(x,t) is an infinite series, we can use the first four or five terms to get an approximation of u( x,t).

83

Lesson 15: The Convection Term ux in Diffusion Problems l. Solve the initial-value problem PDE: u, = uxx - ux

- 00 < X < oo,

IC:

-oo<x<oo

u(x,O)=sinx

0 < t < 00

Solution: We saw in Lesson 8 that the transformation

transforms the PDE 2

u, =a uxx -vux

in dependent variable u(x,t) to the simpler PDE

in dependent variable U(x,t). But there is another way to eliminate the convection tenn -ux ( we let v =1 as in the problem) and that is by transforming the independent variables x,t to two new ones ~, t via the transformation "C =t

~=x-t

(We haven't really changed the time variable t, but it is more convenient to give it a new name, which we call , ). To find the new PDE with new independent variables ~' t we make use of the diagram in Figure 15.1 to find the following partial derivatives.

u, = ui;~, + ut,, = -ui; + ut ux = Ui; ~x + ut"C X = Ui; u.« =(ux) X =(Ui; ) X =U1;1; ~x + Ui;, "C X =U1;1; 84

15. THE CONVECTION TERM ux IN DIFFUSION PROBLEMS

'l =t

X

t

Figure 15 .1: Functional relationship of variables

Plugging these derivatives in the original PDE, we find

or the new problem u(~,O) =sin~ which has the solution

Plugging back in for the original variables gives the solution of the original problem to be

85

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Solution Manual

2. Solve the following diffusion convection equation PDE: u, = uxx - 2ux

IC:

u(x,O)=e ... sinx

- 00

< X < oo, 0 < t < 00

-oo<x<oo

by making a transformation like we did in Problem 1 in Lesson 8.

Solution: In Problem 1, Lesson 8 we showed the transfonnation that maps

IS

Hence, in this problem we let u( x,t) = e(x-,)u( x,t)

hence u, = -e(x-,)u ( x,t) + e(x-,)u, ( x,t)

u, =e(x-,)u(x,t)+e(x-,) ux(x,t) u_'C;( = e<x-,)u ( x,t) + 2e<x-,)ux ( x ,t) + i,•-,)u_u ( x,t)

Plugging these values in the equation

after some cancellation gives U, =Uxx • The initial condition

then becomes u(x,O) =exU(x,O) = e·' sinx

or U ( x, 0) = sin x. Hence, the transformed problem is

15. THE CONVECTION TERM ux IN DIFFUSION PROBLEMS

-oo < x<oo, 0 <t <oo

PDE: U, = Ux, IC:

87

-oo<x<oo

U(x,O)=sinx

which has the solution

U(x,t) =e-' sinx and thus the original problem has the solution

u( x,t) = /,-,Ju (x,t)

=e(x-t) e-' sinx = e(x- 2t) sin X 3. Solve PDE: u, = -2ux IC:

u(x,O)=e_,..,

- 00 < X < oo,

0 < t < 00

-oo<x<oo

Solution: In the next lesson, we will give a fonnal description of solving pure convection equations like this problem (no u X< term only a term in ux) .

For the time being, we can interpret this problem as describing the amount u(x,t) ofa substance moving to the right al~ng the x-axis with velocity 2, we can imagine the initial wave u ( x, 0) =e - x , we might be tempted to try the solution

u( x,t) =e(x- Ji)' You can verify by direct substitution that this function satisfies both the PDE and the IC.

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Solution Manual

4. What is the solution of the convection problem PDE: u, =uxx - Vux - oo < x < oo, 0 < t < oo IC: u(x,O)=e-x2 -oo<x<oo and check your _solution. Hint: Note that the transformation to moving coordinates allows us to essentially neglect the convection term -Vu,.

Solution: Let r = t, ( = x - Vt , using the following diagram to help with our partial derivatives. u

r =t

q = x - YI

X

Figure 15.2: Diagram to explain partial derivatives of compositions

We find

u, =u: ,;, +u,r, =u:

a

u,, = ax (u<) =u« ,;, +u¢,'x =u« u, =u<(-V)+u,r, =u,-Vu< Plugging these values in the original initial-value problem yields PDE: u, = u~, IC:

u(l;,O)=e-' 2

-oo < 1; < oo, 0 < t < oo -oo<l;<oo

We then solve this problem for u(,;,r) and then substitute back r = t, ( = x- Vt to get u as a function of x and t .

Lesson 16: The One-Dimensional Wave Equation (Hyperbolic Equations)

1. Derive the transmission line equation

v_"'" =CLvn +(CR+GL)v, +GRv for v from the following system of first-order PDEs ix + Cv 1 + Gv =0 v x + Li, + Ri = 0

Solution: Ifwe differentiate the first equation with respect to t and the second equation with respect to x , we get ix, + Cv11 + Gv, =0 v.u + Lix, + Rix = 0

We now multiply the first equation by L and subtract the second equation, getting LCv" +LGv, -v.a -Rix = 0 and finally solving for v"" ( along with plugging in for ix = - Cv, - Gv in the first equation) gives the desired result vxx =LCv" +LGv, -Rix =LCvu + LGv, -R(- Cv, -Gv) =LCvu + (CR+ LG)v, + GRv

89

90

Solution Manual

2. From your knowledge of the various terms of the wave equation, what would you expect the solution of the following problem to look like at various values of time?

PDE: u,, =uxx - u,

0<x<1 , O<t< oo

BCs: { u(O,t) = 0 u(l,t) =0

0 <t < 00

ICs:

{

u(x,O) =_sin(n.x) u, (x,0)-0

0 <x<l

Solution: The wave initially looks like the sine curve sin( n.x) but vibrates to zero due to the friction (or damping) term u,.

3. What would the solution of the following problem look like at various values of time? What is the physical interpretation of the problem?

PDE: ull=uxx

0 < X < 1, 0 < t < 00

BCs: { u(O,t)=O u(l,t) = sin ( nt) ICs:

{

u(x,O) = 0 u,(x,O) =0

O<t< oo 0 :s; x :S:l

Solution: You might imagine this problem describing the motion of a vibrating string, which is initially at rest, but a person standing at the right end moving the string up and down like a sine wave. The motion is complicated but one can imagine waves moving down the string from right to left, sometimes bouncing off the left-hand end, which is fixed at zero, and then coming back only to collide with waves moving from the other direction. One can solve this problem analytically by either the Laplace transform (transforming either x or t actually) or by Duhamel's principle.

16. THE ONE- DIMENSIO NAL WAVE EQ1JATION

4. How many solutions of u" =uxx can you find by looking for solutions of the form u( X,t) =e ax+bt

Solution: Substituting U(f,

f) =e x+ bt

in the wave equation

one gets b 2 = a 2 orb = ±a . Hence, we get solutions of the form

u (x,t) =e a(x+t ) where a is any constant. The sum is also a constant and so we have more solutions I

Later we will see that the wave equation has (many) more solutions than this.

91

Lesson 17: The d' Alembert Solution of the Wave Equation 1. Verify that general d'Alembert solution

f

1 H CI g(,;)# u(x,t)=-1 [ .f(x-ct)+ .f(x+ct)] +2 2c x- ct satisfies the initial-value problem -oo < X < oo,

!Cs·

•

{

u(x,O)=f(x) u, (x,O) = g(x)

0 <I< oo

-oo<x<oo

Solution: The general fonnula for the derivative of an integral when the limits of integration are not constants, but functions, is called the Leibniz formula given by

Remembering your multi variable calculus and how you compute partial derivatives, we see

u, ( x,t) =.!_[-c f'(x-ct) + cf ( x +ct)]+_!_[ c g(x + ct )-cg(x-ct)]

2

1111 (

2c

x,t) = ½[c 2 f"(x-ct) + c2 f"( x +ct)]+ ½[c g'(x +ct)+ c2g' (x -ct)]

u, (x,t) =½[f'(x-ct) + f'(x +ct)]+

L L

uu (x,t) = ½[f"(x - ct) + f"(x +ct)]+

[g(x +ct)+ g(x -ct)] [g'(x +ct)+ g'(x-ct)]

c2 ux., ( x,t) = ½[ c2 f"(x-ct) + c2 f"(x +ct)]+½[ cg'(x +ct)+ cg'(x-ct)]

92

17. THE D'ALEMBERT SOLUTION OF THE WAVE EQl)ATION

93

and upon close inspection, we see that u,, =c 2ux., . The initial conditions can also be easily verified.

2. Substitute } Ix g(;)d; 1 )-<l>(x)=-f(x

2

2c xo

\II( X) =.!. J (X) + _!_Ix g(;)d;

2

2c ·' o

in the general formula u( x,t) = <l>(x-ct) + lJJ( x+ct) to get the d'Alembert solution.

Solution: By direct substitution, we get

u( x,t) = <l>(x-ct) + \Jf (x + ct)

1 fx+ct g(;)d; ] 1 1 fx- c, 1 g(;)d; ] + [ -f(x+ct)+= [ -f(x-ct)-2c 2 2c

2

Xo

Xo

x+c, g(;)d~ - f x-c' g(;)d;] = _!_[/ (X - ct)+ J (X +ct)]+ _l_[f 2c

2

Xo

Xo

3. Solve the initial-value problem PDE: u,, = uxx

IC:

{u(x,O)=e-x' u, (x,O) =0

- 00 < X < oo, 0 < t < 00 -oo<x<oo

What does the solution look like for different values of time?

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Solution Manual

Solution: The d'Alembert solution gives 1[ 1 f x+ct u(x,t)= 2 f(x-ct)+ f(x+ct) ] + le x-c, g(~)d~

1 [ e - (x-1)' +e- (x+t)' ] =2

4. Solve the initial-value problem PDE: U11 =Uxx IC:

{

u(x,O) = 0

- 00

.

u, ( x,O) = xe-x·

< X < oo, 0 < f < 00 -oo<x<oo

What does the solution look like for different values oftime?

Solution: The d'Alembert solution gives

_ I [ e- (x-1)' -e - (x+1)' ] -4

17. THE D'ALEMBERT SOLUTION OF THE WAVE E~ATION

5. Algebraically solve for <!>( x) and \JI( x) from the equations

~(x) + \Jf(x) = f(x) -c<j)(x)+c\jl(x)= (g(~)d~ to arrive at I fx 1 g(~)d~ )-<!>(x)=-f(x 2c 2 .To

1 fx 1 g(~)d~ )+\Jf(x)=-f(x 2c 2 Xo

Solution: Straightforward algebra.

95

Lesson 18: More on the d'Alembert Solution

1. Solve the semi-infinite string problem PDE: u11 =c 2uxx

0 <x<oo, O<t<oo

BC:

u(O,t)=O

O<t<oo

!Cs:

{

u(x,O) = xe-x' u,(x,0) =0

O<x<oo

Solution: For the semi-infinite string we use the modified d'Alembert formula ] f x+cr -] [ f(x-ct)+ f(x+ct) ] +g(~)d~ 2 2c x- cr ( ) { u x,t = ][ ] J x+cl f(x+ct)-f(ct-x) + 2 c c,-x g(~)d~

I

2

Ct S. X

ct >x

= { ~[{ x-ct )e-~~:): +{ x + ct)e:::''.'] cl< x

2[(x+ct)e ( l -(t-cx)e ( . ) ]

ct>x

2. The solution of the semi-infinite string problem in Problem l can also be written by a) extending the IC to the entire x-axis by defining u(x,0)=-xe-x' -oo<x<O

b) averaging the left and right moving waves c) taking the solution for x 2:'.: 0 Use this strategy to find the solution for the IC in Figure 18.1.

96

97

18. MORE ON THE D'ALEMBERT SOLUTION

.-n

,. z -~-------

1

2

Figure 18.1: Initial condition u ( x, 0)

Solution: The left moving wave from the initial wave

f(x,O)={xe-x' O<x<oo 0 -oo<x<O is

u_(x,t) = f(x +ct) =(x +ct)e-(x+ci)' -oo <x < oo and the right moving wave that results from the odd extension

x' f(x,O)= { -xe- -oo<x<O O<x<oo 0 is

U+ ( X,t) = f (x - Ct)= -( X-Ct )e-(x-ct)' , -00 < X < 00 Hence, we have

u(x,t)= {

½[ (x + ct )e-(x+ci)' + ( x- ct )e-(x-ci)'] ct~x ½[ (x + ct )e-(x+ci)' -(t - ex )e-(,-a)']

ct >x

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Solution Manual

The values of x where ct ~ x are those values of x where the wave has not yet bounced off the wall, while the values of x where ct> x are those values of x where the wave has bounced off the wall.

Figure 18.2 shows the motion of an initial triangular wave with BCs u(O,t) = 0 by averaging the left and right moving waves of the function and its odd extension.

~L

0

~

'

:::::;,....-----; !

,,~

><:I

Figure 18.2: Average of the left and right moving waves

3. Solve the semi-infinite string problem 2 PDE: u11 =cu""

O<x<oo, O<t<oo

BC:

ux(O,t) =0

O<t<oo

ICs:

{

u(x,O) = f(x) u,(x,0)= 0

O<x<oo

in a manner analogous of the way the semi-infinite string problem was solved in the lesson. What is the interpretation of the problem?

18. MORE ON THE D'ALEMBERT SOLUTION

Solution: We define the even extension of f(x), 0::S:x<oo to the real line by

](x)={f(x) x~O f(-x) x<0 We also extend g( x) = 0, 0 ::s; x < oo to the entire real line by g(x) = 0, - oo < x < oo. The solution ii of the extended problem - 00 < X < 00,

BC:

iix(0,t)=0

ICs:

{

ii(x,0) = f(x) ii, (x,O) = 0

0 < t < 00

0<t<oo 0<x<oo

is

ii(x,t) = ½[J(x-ct )+ ](x +ct)] We now define the function

u(x,t)=ii(x,t), x~0 and show that u ( x, t) satisfies the IBVP 0 < X < OO, 0 < t < 00 BC:

ux(0,t)=0

0<t<oo

ICs:

{

0) = f ( X) u, (x,O) = 0

0<x<oo

U ( X,

The function u satisfies the PDE since ii does. The IC are also satisfied since

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Solution Manual

u( x,0) =u( x,O)

=½[J(x)+ ](x)] 1 2

=-[f(x)+ f(x)] =f(x) Finally, to show the BC ux(O,t) = 0, we use the fact that if extension then its derivative is an odd extension, i.e.,

]' (-ct) = - ]' (ct) and by computing the derivative

we have ux(O,t) = ½[J'(-ct) +]'(+ct)] = _!_ [ - /' (ct) + /' (+ct)] = 0

2

and thus ux(O,t)=ux(O,t)=O.

J is an even

101

18. M O RE ON THE D 'ALEMBERT SOLUTION

4. Suppose the vibration of a string is described by u11 = u""' and has an initial displacement given by the diagram in Figure 18.3.

u

1 -4-

-3

-2

-1

0

1

2

3

4

5

Figure 18.3: Initial wave form

Assuming the initial velocity of u, ( x,O) = 0, describe the solution of this problem in the xt plane. Note that the IC is discontinuous.

Solution: Figure 18.4 shows the movement of the waves in the tx plane.

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Solution Manual

t

0

0

O=o

1

0

1

0

~=1/2

0

•

1

=1

Figure 18.4: Movement of waves in the be plane

To visualize the solution, imagine two initial half waves, one moving to the left and the other to the right and adding the results. Try to visualize what happens to the wave on the interval [0, 1].

Lesson 19: Boundary Conditions Associated with the Wave Equation 1. From your intuition of the various kinds ofBCs, draw a rough sketch of the solution to the following IBVP for various values of time.

PDE: Utt =c 2Uxx

0 <x <1, O<t< oo

BC:

ux(O,t) = O

O~t< oo

ICs:

{

U ( X, 0) = f ( X)

u, (x,O) =0

O~x~1

Solution: The boundary condition ux( O,t) =0 at x =0 resembles waves moving to the left and hitting a wall at x = 0. The water moves in and down but the slope of the water at the wall is always zero. The waves then bounce back and move to the right. The reader can find good animations of this action online by 'googling' phrases like 'one-dimensional d'Alembert motion'.

2. From your intuition of the various kinds of BCs, draw a rough sketch of the solution to the following IBVP for various values of time.

0 < X < 1, 0 < t < 00 BC:

ICs:

Solution: We have seen that the solution of the wave equation utt = u xx when defined in 'free space', i.e., -oo < x<oo , can be interpreted as the average of two moving waves, one to the right and the other to the left.

However, when there are boundaries like in this problem, the waves hit the walls at x = 0 and x =1 and bounce back in some manner; exactly how they 103

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Solution Manual

react to the wall depends on the boundary conditions ( u( O,t) == 0 is like a rope tied down at the end, whereas ux ( 0, t) == 0 is the way water behaves as it reaches the wall). But waves bouncing back and fmth become difficult, hitting each other as they run into each other, and so a better way to analyze wave problems on finite intervals, like O< x < l, is in tem1s of stand waves.

In this problem the initial wave is zero at the left ( x == 0) and 1 at the right (x =1). The shape of solutions of this problem are the eigenfunctions of the Sturm-Liouville problem

X" + t,} X = 0 0 < x < 1

x(o) == o X'(l) == 0 which are cos( nrc/2) and since our IC is the first eigenfunction, we suspect a solution to be a simple vibration like

You can verify that this function satisfies the IBVP. 3. What is the general nature of the BC

when his small, and when h is large?

Solution: In conjunction with the wave equation, when h == 0, we have that ux ( O,t) == 0. Have you ever seen waves lapping up against a wall? Provided the waves don't splash, the slope of the waves against the wall is zero, hence simulating the BC u x ( O,t) =0. When h is large, the solution at x =0 pays more attention to whether u(l,t) == 0.

19. BOUNDARY CONDITIONS

105

4. From your intuition of the various kinds ofBCs, draw a rough sketch of the solution to the following IBVP for various values of time.

PDE: u,, =uxx BC:

{

ICs:

{

u(O,t)=O ux(l,t)=O

u(x,0)= x u,(x,0)=0

0 < X < l, 0 < t < 00

O<t<oo Q:-::;x:-::;1

Solution: The problem describes a string whose initial shape is u(x,O)=x that initially have no motion ( u, ( x, 0) = 0 ) . The left end is fixed at O but the right end is free to move up and down but has slope Oat the right. We know that the PDE and BC give rise to the Stmm-Liouville problem

X" + )} X = 0 0 < x < 1 X(O) = 0 X'(l)=O that has eigenfunctions X,,(x)=sin(n ;). n=1,2, ...

These are the functions that give rise to simple harmonic motion of the type

. (mu:) cos (mrx) 2

u,,(x,t)=aJ;,(t)X,,(x)=a,,sm 2

Hence, to solve this problem, we simply expand the IC

where

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Solution Manual

(mu)

? f a,, =~J 0 xsin 2 1

dx

and then the solution is

(mrx) cos (mrx) 2

• u(x,t)= f, L.,a11 sm n= I

2

Lesson 20: The Finite Vibrating String (Standing Waves) 1. Find the solution to the following TBVP. Is the solution periodic in time; if so, what is the period?

0 < X < L, 0 < I < 00 BC:

!Cs:

{u(0, I) = 0 u(L,t) =0

0 < I < 00

fu(x,0) =sin( 7, )+o.5sin( 3~)

1u, (x,0)

=0

Solution: The PDE and BCs of this problem give eigenfunctions

X" (x) = sin( n~x). n = 1,2, ... and so the solution has the form

f, • (nTrx)[ L -L- ) +b,,cos (nTCat)] L a,,sm• (nTCat u(x,t)= ::i'sm Since the initial velocity u, (x, 0) = 0, we have that all the coefficients a,;= 0 and that the b,; are the Fourier coefficients of the IC u(x,0) or

!

I n= I b,, = 1/2 n = 3 0 all other n

Hence, the solution is

f, • (nTrx)[ L )] -L- ) +b,,cos (nTrat L a,,sm. (nTCat u(x,r)= f,:tsm

I . (3Trx) cos (3Trat L )

• (Trx) L + 2sm L L cos (Trat) =sm 107

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Solution Manual

2. Find the solution to the following IBVP. ls the solution periodic in time; if so, what is the period?

0 < X < L, 0 < t < 00 BC:

u(o,,)= 0 {

ICs:

{

0<t<oo

u(L,t)=0 u(x,0)= 0

u, (x,O) = sin(37tX / L)

0sxsL

Solution: The PDE and BCs of this problem give eigenfunctions • (nlrXJ X( ,, x ) -- sm L , n -l?

- . -, ... and so the solution has the form • (L nlfx u(x,t)= ~ '.:1sm

J[a,,sm. (L J+bncos (L J] nlfat

nlfat

where the a,; are the Fourier coefficients of the initial velocity u, (x,O) and the b~ are the Fourier coefficients of the IC u(x.0). Hence we have all the b;; =0 and n=3

Hence, the solution is

J . (31fxJ sm. (31fa/J L

L sm L u(x.t)= ( 31fa

3. Show that for ,1,:?:0 in Figure 20.2 in the book, the solutions X(x)T(t) are either unbounded or zero.

Solution: When ,,1, = 0 both functions X(x) and T(t) are linear functions in x and t, respectively. The function T(t) must increase with respect to t and so goes to infinity, which means the product X(x)T(t) does not remain

109

20. THE FINITE VIBRATING STRJNG

bounded. When ,bO one of the terms of both X(x) and T(t) is an exponential growing exponent in x and t, respectively, and so the product X(x)T(t) does not remain bounded.

4. What is the solution of the IBVP 0 < x < L, 0 < t < CX)

BC:

{ u(O,t) = 0 u(L,t)=O

ICs:

{

0 <t <CXJ

u(x,0)==sin(3m/ L) u, (x, 0)==(3na l L)sin{3m / L)

Osxs L

Solution: With the given PDE and BCs the solution has the general form

L

n,rat ) +b,, cos ( mrat )] • ( -Lf, • ( n,rx )[ a,,sm u(x,t)== ;:tsm

L

where the coefficients a,, and b,, are the Fourier coefficients of the ICs. In this problem, we have a == n

{I

b,, == {

0

~

n == 3 n,d n == 3

n,;,3

and so the solution is

L

L

L

. (3,rat) +cos ( 3,ra/ )] • ( 3,rx)[sm u(x,t)==sm

The main observation to be made here is that the string vibrates with simple harmonic motion with circular frequency 3,ra/ L oscillations per 21r units of time, i.e., everywhere along the string vibrates with the same frequency and the vibrating string maintains its original 'profile'.

110

Solution Manual

5. A guitar string of length xis pulled upward at the middle so that it reaches a height of h. Assuming the position of the string is initially

u(x,O)=

{

2hx

0~x<0.5

2h(l-x) 0.5~x~l

as drawn in Figure 20.1, what is the subsequent motion of the string after it is released? u

i i

jb

I i

X

1/2

0

Figure 20. l: Initial position of the guitar string

Solution: The solution has the form

( ) . (mrx)[ . (mrat) L +b.cos (mrat)] L

~ L u x,t = :;tsm

a.sm

where the coefficients a. =O and the coefficients b. are the Fourier coefficients of the IC U

XO - {

( ' )-

2hx

O~ X <0.5

2h(!-x) 0.5~x~I

20. THE FINITE VIBRATING ST!tlNG

111

6. Solve the damped vibrating string problem

BCs:

{

ICs:

{

u(O,t} = 0

O<l<OO

u(l,t} = O

u(x,0} =_f (x) u, (x,0)-0

0-5,x-5,L

Solution: Looking for standing wave solutions u(x,t) =X(x)T(t)

the PDE becomes XI'"= a 2 X'T-f]XF'

or

Since the left-hand side of this equation depends only on t and the righthand side only on x, then in order to be equal for all x and t, they both must be equal to a constant, so setting both the left- and right-hand sides of this equation to a constant we call -,l.2, we have T" + /JT' +(a,1.)2 T =0 X"+,1, 2 X=0

We now have that X(x) must satisfy the BVP ODE: X" +il 2 X =0

BCs:

{

X(0)=0 ()

XI =0

which has eigenvalues il" = mr, n =1,2, ... and eigenfunctions X,, (x) =sin(mrx)

112

Solution Manual

and so the equation T" + /3T' + ( a,1, )2 T = 0

becomes whose general solution is

where c,,c2 are arbitrary constants. Using the initial conditions we arrive at the solution of the lBVP u(x,t)= 'I)ne-P' sin(mrx)co{Jf3 2 - 4mral) 11-"-1

where the coefficients b,, are the Fourier coefficients of the IC u(x.O) = f(x)

The reader can verify that this function is the solution of the IBVP. Note how the damping term - /Ju on the string makes the solution damp to zero as well as make the vibrations of the string slower.

Solution with extra term -/Ju: Solve the damped vibrating string problem PDE: u" =a 2u« -13u

O<x<l, O<t<oo

BCs:

{u(O,l)=O u(l,t)=O

O<t<oo

ICs:

{u(x,O)=f(x)

05.x5.L

u, (x,0) =0

Solution: Looking for standing wave solutions u(x,t)=X(x)T(t)

20. THE FINITE VIBRATING STRING

the PDE becomes

or

Since the left hand side of this equation depends only on t and the right hand since only on x, then in order they are equal for all x and t, they both must be equal to a constant, so setting both the left and right hand sides of this equation to a constant we call -k2, we have T"+(ak) 2 T=O x ·+ (e -~ )x= o Calling

we now have that X ( x) must satisfy the boundary-value problem ODE: X" + 1 2 X = 0

BCs:

{ X(O) ()

=0

XI =0

which has eigenvalues A-11 = me, n = 1,2, ... and eigenfunctions X ,, (x)=sin(mrx) . We now find the constant k from the equation

and so the equation becomes whose general solutions is

113

114

Solution Manual

T,, ( t) = c, cos (rt) + c2 sin (rt) where c1, c2 are arbitrary constants and Yn =(mra )2 + /J, where we assume /J > O is sufficiently small so that all the r,, > O for all n = 1, 2..... The solution of the IBVP then has the form

where the coefficients b,, are the Fourier coefficients of the IC

u( x,O) = f(x) The reader can verify that this function is the solution of the IBVP. Note how the extra forcing term - /Ju on the string (kind of like a Hooke's law trying to restore the string to zero) makes the vibrations of the string faster; i.e., compare the faster vibration

versus the slower vibration

7. How would you solve the nonhomogeneous PDE with given boundary and initial conditions?

PDE: u,, =a 2u_« + Kx

O<x<l, O<t<oo

u(O,t) = 0 u(l,t) =0

O<t<OO

u (x,O) = f (x) u,(x,0)=0

OSxSL

BCs:

{

ICs:

{

Solution: We seek a solution in the form

115

20. THE FINITE VIBRATING STRING

"'

"'

11=1

n= I

u(x,t)= ~);,(t)X,, (x)= ~);, (t)sin(mrx)

where the X,, (x) =sin(mrx)

are the eigenfunctions of the boundary-value problem ODE: X"+J'X=O

X(0)=0

BCs: { X () 1 =0

To find the functions of time T,, (t) we find the eigenfunction expansion of the nonhomogeneous term 00

Kx= LJ,,(t)sin(mrx) n=I

getting Fourier coefficients .

(-1)"+1 2K

J'Kxsin(mrx)d-r

./,,(t)= 0 I

=~--

fosin 2 (mrx)dx

mr

The first few of these (constant) functions are 2K -2K 2K -2K J (J; (t), J; (t),J; (t) ,/4 (t) ... )=( , - , - , - , ...

1r

21r

31r

41r

We can now find the functions 7;, (t) by solving ODE: IC:

where the constants a,, in the initial conditions are the Fourier coefficients in the series 00

f (x) = La,, sin (n1rx) n=I

which are found from

116

Solution Manual

f f(x)sin(mrx)dx J Ji f(x)sin(mrx)dx

a,, = 0

r1

1

fosin 2 (mrx)dx

-

2 °

We then solve for T,, (t) and plug these functions in

"' u(x,t)= ~);,(t)X,, (x)= ~);, (t)sin(mrx) ~

n=I

n=I

Lesson 21: The Vibrating Beam (Fourth Order PDE) l. Solve the following cantilever-beam problem.

l

0 <x< 1, 0 <t <OO

PDE: u11 +u= = 0

u(0,t) =0 ux(0,t) = 0

BCs:

ICs:

0<t <oo

u.a (l ,t)=0 u= (l ,t) = 0

{

u(x,0) = f(x) u, (x,0) =g(x)

0~x~I

Hint: Although the eigenfunctions X " ( x) are not the usual functions (like sines and cosines) as they were in the case of the heat or wave equations, we can still separate variables, arriving at a Sturm-Liouville eigenvalue problem that gives rise to the solution as a sum of eigenfunctions times functions of time.

Solution: Looking for solutions of the form u( x,t) = X( x )T(t), we arrive at

XT" +X""T=O and dividing by XI' gives

T" X"" -+-=0 X T or

T"

X""

T

X

-=---

and setting both sides of the above equation to a constant, which we judiciously pick as - /4 2 , we arrive at the two ODEs

117

118

Solution Manual

We choose the notation for the separation constant -A 2 since it must be less than or equal to zero, else T(t) is not periodic. The two ODEs have general solutions T(t) = Asin(At) + Bcos( At) X( x) = Csin( ✓ Jc, x) + Dcos( ✓ Jc, x) + Esinh ( ✓ Jc, x) + Fcosh ( ✓ Jc, x) To determine A, we plug the general solution X( t) in the BCs getting

x(o) = D+ F = o X'(O) = C.Ji, +E✓ Jc, =0 X"(l) =-CA. sin( JX) - D,1,cos( JX)+ £,1,sinh( JX)+FAcosh( JX) = 0 X'"(l) =-C,1, 312 sin( JX)+ DA. 312 cos( JX)+ EA- 312 sinh( JX)+ FA- 312 cosh( JX)= 0 The first two of these equations give us D=-F C=-E and plugging these values into the last two equation gives two equations, which we write in matrix fotm

[

sin (Ji,)+ sinh ( JX)

cos ( JX) +cosh ( JX)

l[£] [OJ

✓ Jc, sin( Ji,)+ ✓ Jc, sinh( JX) -✓ Jc, cos( JX)+ ✓ Jc, cosh( JX) F - 0

From these two simultaneous equations, we must have that the determinant of the coefficient matrix is zero, else E = F = 0 which would imply all the

119

21. THE VIBRATING BEAM

four values C = D = E = F = 0, giving the unaccepted zero solution X(x) = 0. So, setting the detem1inant to zero; i.e., cos (Ji)+ cosh ( Ji) (Ji)+ sinh ( Ji) =0 Ji) ( cosh JI (Ji)+ cos -Ji Ji) ( sinh Ji Ji sin (Ji)+ sin

we arrive at a (horrible) equation for which we can solve (using a computer n = 1, 2, ... , which tum out to be a of course) for the eigenvalues sequence of positive real numbers converging to -kX:>. After finding the = 1,2, ... and after a little algebra, we find eigenvalues

A,,

F,,, n

F,, + cosh F,, F,, = sin F,, + sinh F,,

E,, = cos

and thus we can find D,, =-F,,, C,, =-E,,. Knowing C ,D,,,E,,,F,, (a certain number of them anyway using a computer) we can find the eigenfunctions 11

X,,

(x) = C,, sin( F,,x) +D,, cos( F,,x )+ E sinh( F,,x )+ F,,cosh( F,,x) 11

Using ,111 we now solve the ODE in T to get

T,,(t) = a cos( l,,t )+ b,, sin( l,,t) 11

and so the fundamental solutions of the problem are "'

u(x,t)= ~) (x,t) 11

n= I

=

f [C,, sin( F,,x) +D,, cos( F,,x) +E,, sinh( F,,x) +F,, cosh(F,,x)] n=I

x [ a11 cos( A,,t) + b11 sin ( -t,,t)]

120

Solution Manual

The final job is to find the coefficients an and bn so that u(x,t) satisfies the initial conditions. Plugging u(x,O) = f(x),u, (x,O) =g(t) into the initial conditions gives <X)

u(x,0)= ~)11 X 11 (x)=f(x) 00

u, (x,0)= LA,,anxll(x)=g(x) n=l

We now use the fact that the functions X 11 (x) are orthogonal (the Sturm Liouville theory of boundary-value problems of ordinary differential equations extends to 4th order equations as well as 2 nd order). In other words

1oX (x)X,, (x)dx= 0 1

{

111

m-:t- n

-:t-0 m=n

and so

+fg(x),l'n(x)dx

an = """'"''' fl o- ,.,- - - JoX,;(x )cl,

_ f>tx)X,.(x )dx b 11- =-,,- - -

f X,;(x )dx 0

2. What is the solution to the simply supported beam (at both ends) with initial conditions ICs:

{

u(x,O)=sin(Jrx)

u, (x,O) =sin( n-x)

Solution: The idea is to determine the coefficients C,D,E, F and frequencies

Ji in the general form

X(x)=Csin( fix)+ Dcos( fix)+ Esinh( fix)+ Fcosh( fix) that will satisfy the four boundary conditions for a simply supported bean; that is

21. THE VIBRATING BEAM

121

X(O)=O X"(O) = 0 X(l)=O X"(l) = 0 After a brief examination we see the only function having all four of these properties is X 11 (x)=sin(mrx), n=l,2, ...

Hence the solution has the general form

u(x,t) =

f sin( mrx )[a sin[(mzf t] + b cos[ ( n.nf t]] 11

11

n= I

where we are left to determine the coefficients a,,,b11 • So, plugging this expression in the ICs gives us 00

u(x,0)= ~),,sin(mrx)=sin(K x) 00

u, (x,O) =

2> a sin(nJrx) =sin(Kx) 2

11

n= I

which after a quick examination, we see b1 =I, a1 = 1 / Jr 2 , all other coefficients zero. Hence the vibrations of this beam are

which says the beam has a profile of sin ( JZ"X) and vibrates with a frequency of Jr 2 /21r =;r/2 oscillations per second and period 2/ Jr seconds.

3. What is the solution ofa simply supported beam with ICs ICs:

0 sx sl

122

Solution Manual

Solution: In Problem 2 we found the possible shapes of a simply supported beam to be X(x)=sin(mrx ), n=l,2, ... ( we don't allow n = 0 since that gives X ( x) = 0) and so the vibrations have the form

u(x,t) =

f sin(mrx )[ an cos[ ( nnf t] + b,, sin[ (mz-}2 t]] n=I

where the coefficients a,,,bn are determined from the TCs. Plugging u in the !Cs gives

u( x,O) =fa,, sin( mrx) =I -x 2 JCs:

{

11=1

"' u,(x,0)= z )nJr 2 sin(mrx)=O n= I

which yields bn = 0 for all n = 1,2, ... and a11 are found by multiplying each side of the respective equation by sin ( mJrx) and integrating each side of the equation from O to I, giving

t (I

-x 2 )sin( mJrx )dx = a,,,t sin 2 ( mTCx )dx

or 2(m 2 Jr 2 +4) 3

m Jr

2 mJr

3

m=2,4, ...

Hence, the solution

u(x,t)=

fa cos[(n.7r}2t]sin(mrx) 11

n= I

The first term in the series is

m=l,3, ...

123

21. THE VIBRATING BEAM

firstterm = 0.89cos(JZ" 2 t)sin(7l"x) which is an approximation of the vibration of the beam. Note that this first term gives us an approximation of frequency ( 2/ 7l" hertz) and period (2/ 7l" second) .

4. Let the left end ( x = 0) of a beam be rigidly fastened to a wall and let the right end ( x = l) be simply fastened according to the BCs shown in Figure 2 l.3 in the text. Solve the beam problem with these BCs and tell how to find the natural frequencies of vibration of the beam. Knowing the natural frequencies of the beam is important to know since various kinds of inputs of the same frequency give rise to resonance.

Sol,ution: The natural frequencies are the eigenvalues of the Sturm-Liouville problem ODE:

BCs:

The general solution of the differential equation is

X(x) = Csin( /X,x )+ Dcos( /X,x )+ Esinh( Jix )+ Fcosh( /X,x) Plugging this function in the BCs gives

X(0)=D+F =0 X'(0) = c/X, + E/X, = 0 X(l) = Csin( JX,)+ Dcos( /X,)+ Esinh( /X,)+ Fcosh(

../X) = 0

X"(l) =-C,1,sin( /X,)-D,1,co s( /X,)+£,1,sin h( JX,)+F ,1,cosh( /X,)= 0

124

Solution Manual

The first two equations give D = - E,C = - F. Plugging these values in the last two equations gives [ -sin (.Ji)+ cosh (

.Ji)]F + [-cos( .Ji)+ sinh (.Ji)] E = 0

[ ,t sin (.Ji)+ ,t cosh (.Ji)] F + [ ,t cos (.Ji)+ ,t sinh (.Ji)] E = 0

There will be nonzero solutions for A,B,C,D unless the values of ;i, are chosen so that the determinant of the coefficient matrix is zero, i.e.,

.Ji) -cos( .Ji)+ sinh (.Ji) =0 Asin( .Ji)+ ,1,cosh( .Ji) Acos( .Ji)+ Asinh ( .Ji) -sin (.Ji)+ cosh (

These equations can be solved on a computer getting eigenvalues n = 1,2, ... which represent the frequency of vibrations of the bean. We can with a little algebraic work solve the above equations for E,F getting

A,

E,, = cos( A)+ cosh ( A)

F,, =-sin(F,)-s inh(A)

and hence D,, =-E11 ,C11 =-F,,. We now have the solution in the form "' u(x,t)= L:U,, (x,t) n=I

"'

= z:;xn(x) T,, (t) 11=1

t,

= {[-cos( .fix)+ sinh( .fix)] En +[-sin (.fix)+ cosh ( jix)] F,,} [ an cos( A,,t) + bnsin ( l,,t)] We substitute this series in the IC to find the a11 ,b as we did in Problem 1.

Lesson 22: Dimensionless Problems

l. Transform the vibrating string problem 0 < x < L, 0 < t < oo

with independent variables x,t to dimensionless form PDE: urr = u~~ BC: { IC ·.

0 <,; < 1, 0 < r < oo

u(O, T) = 0 u(l, ,) = 0

O<r<oo

{u(. ,;, 0) =sin( 1r,;) +-21 sin (37r,;) u,(,;,0)=0

0 '.5,; '.5 I

in dimensionless variables ,;, r.

Solution: Making the change of variables

,; = !_ or

X

L

we have

125

=l,;

126

Solution Manual

Plugging the above derivatives in the PDE u,, = a 2uxx yields the dimensionless form

(;)\rr

=a 2

(±J ⇒ urr u;~

=u;;

The BCs and ICs transform in the same way by simply substituting for x and t. Doing this we get the PDE with the same dependent variable u but new dimensionless independent variables t;, r : BC:

=0 { u(O,r) ( )

IC:

{

u l,r =0

O<r<oo

u( t;,O) = sin ( 1r;) + ½sin (h;) u,(;,0)=0

Note: In the above BCs and ICs we simply replace x,t by ;, r although t; ranges from O to I , and although r ranges from O to oo the same as t , they do not change at the same rate.

22. DIMENSIONLESS PROBLEMS

127

2. Find the dimensionless formation of the problem 0 < X < L, 0 < t < 00

BC:

IC:

Solution: In this problem we want to transform both the dependent variable u and the independent variables x,t to dimensionless form. We do this by transforming the dependent variable by 1

U(x,t) =-u(x,t)

r;

and the independent variables by

Note that the coefficient of t, which is ( a/ L )2, has units of I /time, hence the new time -r is dimensionless. Taking the required derivatives with the help of the functional diagram in Figure 22.1

u

u

T

X

Figure 22.1: Functional diagram for computing partial derivatives

128

Solution Manual

we find

u,

=r;u, =r;[~~ ~; ]= r;[(~J u,]

u, = I'iUx =Ti[~~ 1]= Ti[(±)u,; ] u,,

=r;u" =r;[! (u,)]=r;[!(±)u,]=r;[±u"'.;] =r;[(±)' u,,]

Plugging these derivatives in the PDE u, == a 2 u,x, we find

and after cancellation, we get the pure dimensionless PDE U r ==U,.

0<-r<oo ,.; O<J:'<J, ':,

Hence, we arrive at the new dimensionless IBVP PDE: Ur= U.g~

0 < t < I, 0 < r < oo

BC: {U(O,r)=l U(l,,)=0

O<r<oo

IC:

U(t,O)=Ti!T2

O~t::;J

Note that the new space variable ; doesn't have units of inches, feet, cm, meters, .... ; it has no units, it's just a variable. The same holds for the time variable r; it doesn't have units of seconds, hours, days, ... it is just a time variable with no units.

22. DIMENSIONLESS PROBLEMS

3. Transform the IBVP

0 < X < L, 0 < I < 00 BC: { IC:

u(O,t)=J; O<t<oo u(L,t) =I;

u(x,O)=sin(:-)

to

O<x<L, O<t<oo

PDE: U, =a 2U_u BC: {U(O,t)=O

O<t<oo

U(L,t)=l

IC:

U(x,0)= sin(Jrx/L)-7; T2 -7;

Solution: Making the substitution

u(x,t)=I; +(I;-I;)U(x,t) we have

u(O,t)=?; +(T2 -T)U(O,t) u(1,t)=7; +(T2 -T)U(1,t) u(x,0)=7; +(I;-T)U(x,O) and plugging these values in the IBVP

129

130

Solution Manual

PDE: uI =a 2u xx BC: { IC:

0 < X < L, 0 < t < 00

u(O,t) = 7; u(L,t)=T;

O<t<oo

u(x,0)=sin(7)

yields PDE: u / = a 2U,x BC: { lC:

O< X < L, 0 < t < 00

r; +(r;-r;)u(o,t)=r;

7; +(7;-I;)U(l,t)=Tz

0 <t <OO

r;+(7;-7;)U(x,O)=sin(7)

Osxsl

or simplifying, we have the desired IBVP PDE: v i = a 2U,x

O< X < L, 0 < t < 00

BC: {U(O,t)=O

O<t<OO

U(L,t) = I IC:

U(x,0)= sin(,irx/ L)-7;

:r; - r;

4. Can you think of a physical reason why the new time variable r = ct would eliminate the parameter c 2 in the wave equation 2

Utt =C Uxx

Remember what a means in terms of the velocity of the wave. Intuition plays a big role in finding the most desirable new coordinates.

Solution: With respect to time scale t velocity of the wave is c and so with respect to the time scale r = ct the velocity of the wave is 1.

22. DIMENSIONLESS PROBLEMS

5. How would you pick a new space variable ~ so that the equation

u, +vux =0 is transformed to dimensionless form

u, +u~ =0

Solution: Trying the new variable ~ = kx, we have

and so ifwe pick k = 1 / v (i.e., ~ = x Iv ), we get the equation

131

Lesson 23: Classification of PDEs (Canonical Form of the Hyperbolic Equation) 1. State whether the following PDEs are hyperbolic, parabolic, or elliptic. a) u,.T - uxy = 0 b) Utt = u,a + ux + hu c) uxx +3u.Y.Y =sinx

d) uxx+uyy=f(x,t) 1 l e) Urr+-ur+2Uoo=f(r,0) r r

Solution: The general second-order linear PDE in two independent variables IS

where A,B,C,D,E,F,G are functions of x,y (where of course constants are also functions of x,y ). The equation is classified as elliptic, parabolic, or hyperbolic according to B 2 - 4AC < 0 ⇒ equation is elliptic B 2 - 4AC = 0 ⇒ equation is parabolic B 2 - 4AC > 0 ⇒ equation is hyperbolic a) A= 1,B =-1,C = 0 ⇒ B 2 -4AC = l ⇒ hyperbolic b) A=l,B=O,C=-1 ⇒ B 2 -4AC=4 ⇒ hyperbolic c) A= 1,B = O,C =3 ⇒ B 2 -4AC =-12 ⇒ elliptic d) A= 1,B = O,C = 1 ⇒ B 2 -4AC = -4 ⇒ elliptic e) A= 1,8 = O,C =I/ r 2 ⇒ B 2 -4AC =-41 r 2 ⇒ elliptic

132

23 . CLA SS IFICAT IO N O F IJDE S

133

2. Verify equations (23.4),(23.5) and (23.6) in the text.

Solution: Use the diagram in Figure 23.2

u

y

Figure 23.2 : Functional diagram

to compute the required (messy) partial derivatives like you learned how to do in multivariable calculus.

3. Verify that the equation

3uxx + 7uxy + 2u»' = 0 is hyperbolic for all x,y and find the new characteristic coordinates.

Solution: We have

and so the equation is hyperbolic for all x,y. The characteristic equations are

134

Solution Manual

dy B- ✓B 2 -4AC 7 - ✓49-24 1 = = = dx 2A 6 3 dy =B+ ✓B 2 -4AC = 7+ ✓49-24 = 2 dx 2A 6 Hence

I y=-x+c, 3 y= 2x+c2 and so the characteristic coordinates <;,'7 for the canonical form are

I 3 l]=y-2x

<; = y--x

4. Continue Problem 3 by finding the canonical form of

3uxx + 7uxy + 2uY.Y =0 which has the form

Solution: We see that A=3,B=7,C=2,D=E=F= G=0 and so the characteristic equations are

dy B- ✓8 2 -4AC 7 - ✓49-24 I = = =3 dx 2A 6 dy =B+ ✓B 2 -4AC = 7+ ✓49-24 = 2 dx 2A 6 Hence

1 y= 3x+c 1

y=2x+c 2

23. CLASSIFICATION OF PDES

We now find the new canonical coordinates l;,17 by solving the above equations for Ci, c2 , or I l;=y--x 3 17 = y-2x We now find the canonical equation

where

A= 31;; + 1!;xi;,. + 21;>~ B = 6/;x17., + 7(/;x17,, + /;,'lx) + 4{y17y E = 317; + 717,17-" + 217~ B = 31;~, + 1!;_'!}. + 21;,, E = 317xx + 717'!}. + 217,.v F=F=O G=G=O and computing the derivatives ;=

':, X

= -1• 13' ' :;:, _\ ' = 1' 17X = -2 ' 17y = l

we find

A =0' B=-25/3 ' C=O' D=E=F=G=G=O Hence, the canonical equation is

25 3 '

- - u 0 ,1 =0

or simply

135

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Solution Manual

5. Continue Problem 4 by finding the alternate canonical form

Solution: Given a hyperbolic PDE in canonical form u ,gq = <1>( ~,17,u,u,g ,uq), it can be written in an alternate canonical form

using the transformation

and with the help of the diagram in Figure 23.1 we compute the following partial derivatives U,g = Uaa,; + Up/J,g = Ua + Up Ury = Uaap + Up/J,1 = Ua - Up

=Uaa -Upp

" p

Figure 23.1: Diagram illustrating functional dependence

23. CLASSIFICATIO N OF PDES

In Problem 4 we had u qlJ = 0 and since u aa - Upp = U,:1/ the alternate canonical form is

6. Find the new characteristic coordinates for uxx + 4u»' = 0

Solve the transformed equation in the new coordinates and then transform back to solve the original problem in the original coordinates.

Solution: We see that A= 1,B = O,C =4,D=E = F= G = 0 and so the characteristic equations are . dy B- ✓B 2 - 4AC O-.J0-16 -=------=----=-2z 2 2A dx dy = B+ ✓B 2 - 4AC = O + ~ =2i

dx

2A

2

Hence

y=-2ix+c1 y = 2ix+c 2 and so the new coordinates <;,17 for the canonical form are

;=y+2ix 1J = y- 2ix The original equation is elliptic, hence it has complex characteristic coordinates.

137

Lesson 24: The Wave Equation in Two and Three Dimensions (Free Space) 1. Show that in one dimension, we can find the solution of PDE: u/1 = c2ux, - 00 < X < oo, 0 < t < 00

{u(x,O)=¢(x) -oo<x<oo u,(x,O) =0

ICs:

by differentiating (with respect to t) the solution of PDE: u11 =c2u_""

{

ICs:

- oo < x < oo, 0 < t < oo

u(x,O) =0 u,(x,0)=¢(x)

-oo<x<oo

Solution: The solution of the initial-value problem when the initial conditions are

ICs:

{ u(x,O) = 0 u, (x,O) = ¢(x)

-oo<x<oo

is

1 u(x,t)=-

f x+CI

2C x -cl

¢(s)dx

lfwe differentiate this function with respect to t using the Leibniz rule

:!_f g(,J ¢( s )ds =¢[ g(t) ]g'(t)-¢[/(t )]f'(t) dt 1(,)

we get

du= I [ ¢(x+ct)-(x+ct) d d ] -¢(x-ct)-(x-ct) dt

2c dt 1 = -[¢(x +ct)+ ¢(x-ct)] 2

138

dt

24. THE WAVE EQ\)ATION IN TWO AND THREE DIMENSION S

139

which is the solution when the initial conditions are

{

ICs:

u(x,0)=_¢(x) u, (x,0)-0

-oo<x<oo

2. Apply the result from Problem I to find the solution of

- oo < x < oo, 0 < t < oo

PDE: u11 = uxx

{

ICs:

u(x,O) =x u, (x,O) = 0

-oo<x<oo

Solution: We start with the solution of the initial-value problem I

PDE:

u 11

ICs:

{

=u xx

- oo < x < oo, 0 < t < oo

u(x,O) = 0 u, (x,O) = x

-oo<x<oo

which is 1 f x+I

u(x,t)=-2

¢(s)ds

X- 1

1 f x+ I sds 2 x- r

=-

=xt Now ifwe differentiate this solution with respect tot we get x. Now ifwe find the solution of the initial-value problem with initial conditions

140

Solution Manual

{

ICs:

u(x ,O)= x

u,(x, 0) =0

-oo< x <co

we get the same result

I u(x,t) =-[¢(x + t) + ¢(x -t )] 2

= _!_[ (X + f) + ( X - t)] 2

=X

3. Illustrate by picture and words the spherical wave solution of the threedimensional problem

u(x,y,z,0)= 0

lCs:

{

1 x 2 + y 2 + z 2 $: I u, (x,y,z,O) = { 0 elsewhere

Solution: Since the solution of the above problem is

u(x,y,z,t)=t iji one can see that the initial disturbance region (which is the unit sphere) gives rise to an outwardly propagating annulus. Each point ( x,y,z) outside the unit sphere will suddenly experience a disturbance and will be disturbed for a length of time after which u( x,y, z ,t) = 0.

24. THE WAVE EQ1JATION IN TWO AND THREE DIMENSIONS

141

trailing edge of disturbance

leading edge of disturbance

initial disturbance

Figure 24.1: Propagating disturbance

4. (Worked in conjunction with Problem 3) What is the two-dimensional solution of the analogous cylindrical problem

u(x,y,0)=0 ICs:

{

_ 1 x 2 + y2 sl ) { 0 elsewhere

u, x,y, 0 (

Solution: For this problem each point ( x,y) outside the unit circle will suddenly experience a shock wave and as time gets large the solution u( x,y,t) gradually goes to zero. The solution would look more or less like a circular wave after dropping a pebble in a lake.

142

Solution Manual

5. (Worked in conjunction with Problem 3) What is the one-dimensional solution of the plane wave equation PDE:

U11

ICs:

{

= C 2 U:cx - 00 < X < 00 u(x,0)=0 u, ( x,O ) -

1 !xi::; 1 { 0 elsewhere

Solution: We have seen in the lesson that the solution will be

0

u(x,t) =

l +x+ ct 2c 1

- oo < x < -ct - I

- ct - 1 ::; x < -ct + l - ct + 1 ::; x < ct - I

C

1-x+ct 2c 0

ct - I ::; x < ct + I ct+ l::; x < oo

This solution is illustrated in Figure 18.4 in the text.

6. What is the physical interpretation of why Huygen's principle does not hold in two dimensions with respect to t?

Solution: Imagine a single initial disturbance at the origin (0,0). Since Huygen's principle does not hold in two dimensions this means that after a given length of time any other point ( x, y) will suddenly experience a shock due to this disturbance and then gradually die out as time gets large. The reason the shock gradually goes to zero is because the origin in two dimensions can physically be interpreted as an infinite line perpendicular to the .xy plane and the points in the .xy plane are in reality being disturbed by all the points along this line. As time increases all points along this line will

24. THE WAVE EQVATION IN TWO AND THREE DIMENSION S

143

be disturbing the point ( x,y) and as time increases it is the points further away that are doing the disturbing.

7. Use Leibniz's rule !!_f g(,) l (1l

dt

F( c;,t)dc; = f g(t) oF (c;,t )dt + g'( x)F[g(t ),t ]- F[f (t ),t ]J'(t) 1( ot 1)

to differentiate the integral

1 f rp(c;)dc; u(x,t)=2c x- c/ x+CI

with respect to t.

Solution: Leibniz's rule

!!__f g(,J F( c;,t )de;= f g(i) oF (c;,t )dt + g'(x )F[g(t ),t ]-F[f (t ),t] dx 1(

1)

1(1)

ar

tells us how to differentiate an integral with respect to t where the limits and the integrand can depend on t. Direct application of the formula gives

a

a

+ ct)-rp(x-ct )-(x -ct) -d f x+cl rp( s )ds =rp( c +t)-(x ot ot dt x- c, = c[rp(x +ct)+ rp(x-ct)]

Lesson 25: The Finite Fourier Transforms (Sine and Cosine Transforms) I. Solve the diffusion following diffusion problem with insulated boundaries, given by PDE: ult =uxx BCs: {

0 < X < 1, 0 < t < 00

ux(O,t)=O ux(I,t)=O

O<t< oo

ICs:

Solution: Applying the cosine transform C,, = C[ u], we get d~,, + (mr) 2 C,,(t) = 0

1 n = 0,1 C,,(0)= { 0 1/2 Solving these equations, we get

n=2,4,5, .. .

n=3 n=O

C,,(t) = 0

n =2

Hence, we have u( x,t) = 1+ e-,T't cos( nx) + .!.e-(3,,-J' i cos(3nx) 2

144

25. THE FINITE FOURJER TRANSFORMS

145

2. Solve the general IBVP 0 < X < I, 0 < t < 00 BCs: { ICs:

u(O,t) =0

O<t<oo

u(I,t) =0

u(x,O) =0

Solution: Taking the sine transform Sn= S[ u] of the lBVP, we get ODE: IC:

dS" +[(mra) 2 +b]Sn=F,,(t) dt S,,(0)=0

Solving this ODE by finding an integrating factor, we get

Hence, the solution is "'

u(x,t)= I:S,, (t)sin(mrx) n= I

3. Derive the basic laws for the sine and cosine transforms. Can you see why it would be hard to solve a differential equation that contained the term u, ?

S [u ] = dS,, [u] dt I n S [u ] = d 2S,, [ u] dtl 11 n

S[ u.u ] = -[ mrLJ2 S[ u] + 2; 2.ir [ u(O,t) + (-1r1 u(L,O)] C[ux,· ] = -[ n.irL J2 C[ u ]- ~ [ ux ( 0,t) + (-lr1 ux (L,O)]

146

Solution Manual

Solution: The sine and cosine transforms ofa function u(x,t)are defined by

J

2 OLu(x,t)sm • (L nrc x) dx Sn[U ] =z

2 fOLu(x,t)cos ( L nl[x) dx C,,[u]=z a) The law

S [u ] = dS u] dt 11 [

11

I

which says

L2 f 1, u,(x,t)sin (nrcx) L dx= dtd {2LJ Lu(x,t)sin (nrcx) L dx } O

O

is a restatement of the following general law of calculus

-d f bf(t,s )ds = f bof -(t,s )ds dt a a ot which allows one to differentiate (as long as the function is differentiable of course) inside the integral. The law is called Leibniz's rule in calculus. b) To prove

we must show 2 2

L2fL u (x,t)sm• (nl[x) L dx= dd t {2fL L u(x,t)sm• (nrcx) L dx } 0

11

0

Again, this statement is a special case of Leibniz's rule that lets one differentiate a function inside an integral.

c) To prove

S[uxx ]= - [nrcL]2 S[u]+ 2;:[u(O,t)+(-1t1 u(L,O)]

147

25. THE FINITE FOURIER TRANSFORMS

we must show

lfyou look carefully you will see that this law is a glorified version of the integration by parts law

r

J: U(x )V'(x )dx =[U(b )V(b )-U(a )v(a )]- V(x )U'(x )dx except in our case we have functions of two variables not one. So letting

and using the integration by parts fommla, we get

( )I

( )

nrcx 2 nrc L . mrx 2 . mrx 2 L zfo uxx(x,r)sm T dx=zu_,(x,t)sm T OL -z ( T fo u,(x,t)cos T c1x

( )

J

2nrc)rL L dx Jou,(x,t)cos (nrcx) =- ( 7 Doing this again, we let

nrcx) , dV=u_, U =cos ( L we get 2mr

-( f3

/ I

)fa u, (x,t)cos ( L )dx=-( f3 )[u(x,t)cos( L )l-( I})( L )fa u(x,t)sm ( L )ctx I

d) The law

mrx

2mr

mrx

2n,r

n,r

1

.

n,rx

148

Solution Manual

is another fun integration by parts we leave for the reader.

4. What is the finite sine transform of

f( x) = sin( ll'X) + .!.sin(37l'x) 2

Graph the sine transform with L = l.

Solution: When the function is a sum of sine functions, we can simply pick the coefficients, that is 1 n=l { S = 1/2 n=3 11 0 all other n's

5. What is the finite cosine transform of

f(x)=x, Osxsl What would the graph of the inverse transform look like for all values of x?

You know it reproduces J(x)=x, Osxsl but what would it look like outside the interval 0, 1l?

r

Solution: The finite cosine transform of the function f( x) = x, 0 s x s l of a single variable gives the coefficients in the Fourier cosine series of the function. Finding it we get

149

25. THE FINITE FOURIER TRANSFORMS

1

C,, [/] =I f1, /( x )cos( n,cx )dx = 2 f1 xcos( n,cx )dx =

Jo

LJo

n=O

2 -4

n = 1,3,

n 2TC 2

n=2,4, ...

0

Hence the cosine series for f (x) = x, 0::;; x::;; I is

x = _!_ _ __i._[cos(,cx) + .!.cos(3,cx)+ _.!._cos(5,cx)+ · .. ] 2

9

TC 2

25

'

0 :5: x :::::1

Figures 25. l a) and 25.1 b) shows graphs of the function /( x) = x, 0::;; x::;; I and the first two and three terms of the Fourier cosine series. You can see how accurate the series approximates the function with only a couple of terms.

0

Figure 25.1 a): Graphs of f(x)=x and _!__~cos(u) 2

1f

150

Solution Manual

0 Figure 25.1 b): Graphs of/ ( x) = x and

½-:

2

cos( nx) +¾cos(37Tx)

6. Solve the IBVP PDE: u,=ux,+sin(3.7ix)

u(O,t)=O

BCs: { ( ) u l,t =0 ICs:

0<x<l, O<t<oo

O<t<oo

u(x,O) =sin(1rx)

0 ::;x::; l

Solution: Taking the sine transform Sn= S[ u], we get

{l

dS +(mr) 2 S = dt n 0

ODE:

_n

IC:

S ( O) = { "

l n=l 0 n = 2,3,4, ...

Writing these equations separately, we have

n=3 n :;c 3

25. THE FINITE FOURIER TRANSFORMS

ODE: n= l

IC: n=3

dS, +,r 2 S =0 ⇒ S,(t)=e-lT', ' dt S,(O)=l

ODE: IC: ODE:

n# 1,3

IC:

dS3 + (3nf S3 = I dt S3 (0)=0

⇒ S3

-(3irJ', ) 1 ( ( ) t = - -, 1-e (Jn}

dSn + (n1[) 2 S = 0 ⇒ S,,(t)=O n dt Sn(O)=O

Hence, we have

u(x,t)=

151

"' (t)sin(nn) Is 11

n= I

= e-ir', sin( l[X) + (J~ )2 (I -e- (Jlf)' , )sin (hx) You can check to see that this function satisfies the IBVP.

Lesson 26: Superposition (The Backbone of Linear Systems) 1. Show that if u, and u2 are solutions to the following problems Pi and Pi PDE: u, =u.o:+ sin(ll"x)

u(O,t)=O u(l,t)=O

O<t<oo

u(x,O) = 0

0$x$l

BCs: { ICs:

PDE: u, =u_u

Pi

0 < X < 1, Q < t < 00

BCs: { ICs:

u(O,t) = 0 u(l,t) = 0

0 < X < 1, Q < t < 00

O<t<oo

u(x,O)=sin(2nx)

O::;;x::;;1

respectively, then u1 +u2 satisfies the following problem P PDE: u, = uxx + sin ( nx)

p

BCs: { ICs:

u(O,t) = 0 u(l,t)=O

0 <x <1, O<t<oo

O<t<oo

u( x,O) = sin (27rx)

0$x$1

Solution: Since u1 satisfies Pi and u2 satisfies Pi, we have 0 < X < l, 0 < t < 00

152

153

26. SUPERPOSIT ION

Pi

o 2u2 =-ox 2 at

OU2

0 < x < I, O<t <oo

PDE:

-

BCs:

{

TCs:

u2 (x,O) = sin(2nx)

U 2 (O,t) = 0

u2 (1 ,t)=0

O<t <oo o ::; x ::; J

lfwe now add the respective equations in the problems Pi and Pi , we get that u 1 + u2 satisfies problem P:

a

PDE:

2 . ( ) o (u1+u2) +sin Jr X = (u1 +u2) ax 2 ot

BCs:

{(u 1 +uJ(O,t)=O

TCs :

(u 1 +u 2 )(x,0)=sin (2u)

0 < X < I, 0 < t < 00

(u 1 +u 2 )(1,t)=0

O:s;x:s; I

2. Solve the following lBVP by superposition. Each sub-problem can be solved in any manner you wish.

PDE: u11 = uxx + sin (31rx) BCs: {

ICs:

u(O,t) = 0 u(l,t)=O

{

O<x <I, O<t< oo

O<t <oo

u(x,O) =~in(1rx) u, (x,0)-0

0 :s;x ::; l

Solution: The solution with homogeneous PDE and nonhomogeneous IC can be solved by separation of variables getting u1 (x,t) =cos(1rt)sin ( 1rx ) The solution to the problem with nonhomogeneous PDE and homogeneous IC can be solved by the finite sine transform, getting

154

Solution Manual

1 - 2 cos(3m )- l]sin (3;rx) u2 ( x,t) = [ - (3;r) Hence, the solution of the original problem is the sum u(x,t) =u, ( x ,t) + u 2 ( x, t)

=cos(m)sin(JZ'x ) - [ -1- 2 cos(3m) + l]sin(3;rx) (3JZ')

3. Suppose u1 and u2 are solutions to the following equations. For which equation is u1 + u2 a solution? a) u, = uxx b) u, =u.« +e' -I c) u,=euxx

d) U1 =Uxx +u 2

What conclusions can you reach from your answers?

Solution: a) b) c) d)

u 1 + u 2 is a solution since the equation is linear and homogeneous. u1 + u 2 is not a solution since the equation is not homogeneous. u 1 + u 2 is a solution since the equation is linear and homogeneous. u1 + u2 is not a solution since the equation is not linear.

26. SUPERPOSIT ION

155

4. Find four initial-value problems whose solutions sum to the solution of the following problem: PDE: u, = u.<-< + f(x,t)

0 <x<l, O<t< oo

BCs:

{u( O,t)

= g 1 (t) u ( 1, t) = g 2( t)

O<t<oo

ICs:

u(x,O) = ¢(x)

0$;x;S;l

Solution: In each of the four sub-problems let one of the right-hand sides be non-zero, and the others zero. The sum of the four solutions will then be the solution of the given problem.

5. Solve the following initial-value problem. Can you verify your answer?

ODE:

dU11 (t) (n7r) 2U = F (t ) II II dt +

IC:

Un(O) =O

Solution: This is a standard ordinary differential equation with initial conditions. One finds the general solution to the differential equation by multiplying each side of the equation by the integrating factor method you learned in ordinary differential equations class. Doing thi s we get the general solution of the initial-value problem f1 (mr)'s F,, ( s )ds U (t) = e - (mr)' r Joe 11 6. Suppose u, and u2 both satisfy the following linear homogeneou s BCs. Does the sum u, + u2 satisfy the boundary conditions?

Solution: Yes, this can be verified by direct substitution.

Lesson 27: First-Order Equations (Method of Characteristics) 1. Solve the following simple convection problem. What does the solution look like?

PDE: ux + u, = 0 - 00 < X < oo, 0 < t < 00 IC:

u(x,O)=cosx -oo<x<oo

Solution: Using the method of characteristics we let dx=l dt=l ds ' ds which yield x(s)=s+c1 t(s) = S + C2 Using IC x(O) = T, t(0) =0 we get c1 =r, c2 = 0 and so x=s+r t=s Ifwe eliminate s from the previous equations, we get the characteristic curves in the xt plane as a function of parameter r, which are x-t=T (-<XJ<r<oo) which is a family of straight lines in the xt.

Now, we can write the PDE and IC in terms of s and r as ODE: IC:

du =0 0 < s < oo ds u( r,0) = cosr

which has the solution 156

27. FIRST-ORDE R EQ!)ATIONS

157

u(s,r)=sinr or in terms of the original coordinates

u( x,t) = cos( x-t) Note that this solution is a moving cosine curve moving to the right with velocity one.

2. Solve the following convection problem. Note that the IC begins at t = l.

What do the characteristics look like? Check your solution.

PDE: xux +tu,+ 2u =0 IC:

-oo < x <oo, l < t <oo

u(x,l)=sinx -oo<x<oo

Solution: Using the method of characteristics we let

dt dx -=x -=t ' ds ds which yield

We now use the IC (note time starts at t = 1 in this problem) x(O)=r,t(O )=l we get c1 =r, c2 =1 and so

x= ,es t = es We now eliminate the variable s to get the characteristic curves, which are curves in the xt plane with parameter r, and in this case they are

158

Solution Manual

X

-=r (-oo<r<oo) t The PDE and IC can now be written as

du ODE: -+2u=O ds IC: u(r,0)=sinr

0<s<oo

which has the solution

u ( s, r) = (sin r) e-2s or substituting in X

r=-, s=lnt t we have u(s,r)=(sinr) e - is = sin(

7

I • =fsm

}

- 2(111,)

(x)t

3. Solve the following convection problem in higher dimensions where a,b,c,d are given constants.

PDE: aux+buy +cu,+du=O IC:

-oo<x<oo, -oo<y<oo, O<t<oo

(x'+JI") u(x,y,O)=e-oo<x<oo, -oo<y<oo

Solution: Using the method of characteristics we let

dx = a dy = b dt = c ds ' ds ' ds which yield

159

27. FIRST-ORDER EQVATIONS

x(s)=as+c 1 y(s)=bs+c 2 t(s)=s+c 3 Using IC

x =as+ r 1 y = bs + r 2 t=s Now, we can write the PDE and IC in terms of s and r as

du ODE: -+du=O ds

O<s< oo

u( r,0) = e-(ri'+r?)

IC: which has the solution

u ( s, r) = sin r or in terms of the original coordinates

u ( x,t ) =e

-(r,'+ri)

=ds

e

now using the transformations

r 1 =x-at r 2 =y-bt s =t we have

u ( x,t ) =e

-(rf+rn

e

-ds

-[(x-ar)'+(y-bt )2]

=e

e

- dt

160

Solution Manual

4. Solve the following convection problem. Check your answer.

PDE: ux + u, +tu= 0 IC:

-oo < x < oo, 0 < t < oo

u(x,O)=F(x) -oo<x<oo

Solu tion: Using the method of characteristics we let

which yield x(s)=s+c 1 t(s)=s+c 2 Using IC x(O)=,, t(O)=O we get c1 =,, c2 =0 and so x=s+r t=s If we eliminate s from the previous equations, we get the characteristic curves in the xt plane as a function of parameter r, which are x-t=r (-oo<r<oo) which is a family of straight lines in the xt.

Now, we can write the PDE and IC in terms of s and r as du ODE: -+su=O ds IC: u(r,0)=F(r)

O<s<oo

The differential equation can be solved by separation of variables, writing

27. FIRST-ORDER EQ\)ATIONS

161

1 2 +c ⇒ u=ee -s'/2 = Ce-s'/2 ds ⇒ lnu=--s du -=-s 2 u C

where C is an arbitrary (positive) constant. Plugging in the IC gives C = F( r) and so the solution of the ODE and IC is

u( T,s) = F( T )e-s'/i or in terms of the original coordinates

u( x,t) = F( x -t )e-" 12 5. It is possible to specify the solution u on a curve other than the usual initial time line t = 0. In fact, the differential equation doesn't have to involve the time variable at all (maybe u depends only on space variables).

Solve the general first-order equation where F( x,y) is a given function.

PDE: u,. + 2u.v + 2u = 0 IC:

- oo < x < oo, - oo < y < oo

u(x,y) = F(x,y) on the cw-ve C: y = x

Solution: Using the method of characteristics we let

dx=l dy=2 ' ds ds which yield

x(s)=s+c 1 y(s)=2s+c 2 Since the IC is along the cw-ve y = x, the IC for our equation in s is

x(0)=,,y(0)=,

162

Solution Manual

from which we get c1 = r, c2 = r, hence x=s+, y= 2s+ T

Solving these equations for the new variables s and , , we have s=y-x

, = x- s = x -(y-x) = 2x - y which, for each value of , , gives characteristic curves in the xy plane as the family of straight lines with slope 2x.

Now, we can write the original PDE and IC in terms of s and , as

du ODE: -+2u=O -oo<s<oo ds IC: u(,,,)=F(,,,) which has the solution

u(s,r)=F(r,,)e_2, and so plugging in the values of s,, in terms of x,y yields

u( x,y) = F(2x - y,2x- y )e- z(v-x) Note that this function satisfies the IC and PDE. For example, if the initial condition were F( x,y) = x 2 + y 2 on the line y = x, then the solution is

u(x,y)=[(2x-y) 2 +(2x-y) 2 -- 2(2x -y)2 e - i(y-x)

Je-

- 00

2(y-x)

< X < oo, -00 < y < 00

Lesson 28: Nonlinear First-Order Equations (Conservative Equations) 1. Solve the following initial-value problem. Draw the solution for different values of time. What is your interpretation of the solution? What is the relationship between the flux and density in the problem?

PDE: u, + uux =0 IC:

- r:t:) < X < oo, - 00 < y < r:t:)

o x<O u(x,0)= {x x~O

Solution: We first find the characteristic curves (curves where u is a constant) starting at an arbitrary point ( x0, 0), which are x = Xo + g ( u ( Xo, 0)) t = x0 + u(x0 ,0)t

x0 < 0 x0 { - x0 + x0t x0 ~ 0

or solving for t when x0 ~ 0 gives

which is the family of straight lines drawn in Figure 28.1.

0

Figure 28.1: Characteristic lines

163

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Solution Manual

Here fx =uux and so the flux f ( u) as a function of density u is

I f(u)=-u 2 .

2

2. Solve the following initial-value problem. What is the flux-density relationship? Would you expect the solution to behave as it does? Compare the solution to the solution in Problem 1.

PDE: u, +u 2ux =0 IC:

-oo<x<oo, O<t< oo

u(x,O)={O x<O x x~O

Solution: The implicit solution is U

= <p( X - g( U )t)

where

o x<O ¢(x)= {x x~O g(u) = u2 Hence, for x -u 2t < 0 or u 2 > x I t we have u( x,t) = 0 , and for x - u 2t ~ 0 or u 2 $ x I t we have the implicit solution

We can also find the characteristic curves (curves where u is a constant) starting at an arbitrary point ( x0 , 0), which are

28. NONLINEAR FIRST- ORDER EQ!)ATIONS

X

165

= X 0 + g( u(x0 ,0))t = x0 + u 2 ( x0 ,0)t {

x0 < 0

x0

= x0 + xit x0 ~ 0 which is the family of straight lines drawn in Figure 28.2.

Figure 28.2: Characteristic lines

3. Suppose a non-viscous liquid is traveling through a pipe and suppose the liquid leaks through the walls of the pipe according to the law F ( u) =ku (g/cm sec) where u is the density of the liquid, the flux of the liquid is F(u)=u, and the initial density of the liquid is u(x,0)=¢(x). What is the density u ( x, t) of the liquid for any x, t? What is your interpretation of the solution?

Solution: The (non) conservation problem is

But df du

fx=-u,,=u,, and so the problem becomes

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Solution Manual

PDE: u, + u x + ku = 0 IC:

u(x,O) = ¢(x)

- oo < x < oo, 0 < t < oo -OO<X<OO

But this problem is linear so we use the method of characteristics to convert the partial differential equation into an ordinary differential equation along the characteristics. We do this by letting dt ds dx

-=l ⇒ t=s+c 1

- = I ⇒ x=s+c2

ds

Letting the initial conditions be x( 0) = ,, t( 0) = 0, we find c1 = 0, c 2 = r and so t = s, x = s +,(ors= t,, = x-t ). Hence, we now have the ordinary differential equation along the characteristics du -+ku=O ds u(0)=¢(,)

which has the solution

or in terms of x,t

u( x,t) = ¢( x-t )e-k' The solution u( x,t) moves to the right with constant velocity 1 and damps exponentially with damping rate k > 0.

4. What is the solution of the non-conservation equation if the loss of the liquid to the outside is F( x) = 1/x, the flux of the liquid is /( u) = ku, and initial density of the liquid is u( x, 0) = <7J( x)? Does your solution check?

Does it make sense physically?

Solution: The initial-value problem is

28. NONLINEAR FIRST-ORDER EQl}ATIONS

U1

I

+ Ux + - = 0

167

- 00 < X < oo, 0 < t < 00

X

u(x,0)=¢(x)

-oo<x<oo

But this problem is linear so we use the method of characteristics to convert the PDE into an ODE along the characteristics. We do this by letting

dt ds dx -=l ⇒ x=s+c 2 ds

-=1 ⇒ t=s+c 1

Letting !Cs x(0) = T, t(0)=0, we find c1 =0, c2 = T and sot = s, x= s + T (or s = t, T = x - t ). Hence, we now have the ODE along the characteristics

du +-1-=0 ds s+T u(0)=¢(r) The ODE has the general solution

u ( s, T) = - In js + rj + c From the JC

u(0, T) =-lnjrj +c =¢( T) we get

c=¢(r)+ln(r) hence

u ( s, T) =- ln js + rj + ¢ ( T) + In ( T) = QJ( T) + lnls: Tl

or in terms of x,t, we have

I

x-ti ~ u(x,t) = ¢(x-t)+ I n

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Solution Manual

5. Verify that

u(x,t)=¢[x-g(u)t ] is an implicit solution of the nonlinear initial-value problem.

PDE: u, + g(u)ux = 0 IC:

u(x,0)=¢(x)

-oo < X <oo, 0 <t <oo -oo<x<oo

Solution: Given

u=¢[x-g(u)t] differentiate each side of the equation with respect to t and x getting

u, =¢'(x-g(u)t)(-g( u)-guu,t) Ux == ¢'(x- g( U )t )(1- g,,uxt) and substituting into we get

u, + g( u )u_, = ¢'(x- g( u )t )(-g(u )- guu, t) + g(u )[¢'(x- g(u )t )(1- g,,u,,t )] =-¢'( X- g( U )t )[( U + gitx )gJ] 1

In order that this equation hold, it must be true that

Lesson 29: Systems of PDEs

1. Write the system of PDEs in the following equations

in the form

where A,B and C are 3 x 3 matrices.

Solution: -]

0

-c 0

2

2. Find the eigenvalues and eigenvectors of the matrix

Solution: The eigenvalues are the roots of

169

l

0

~l

0

-aJ

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Solution Manual

IA - Ml=

I'~

,l

I~ ,ii=

=(1-J) 2 -4 = /4 2 -2,l-3 =(.-l-3)(.-l+l) =0 Hence, the eigenvalues are -1, = 3, Ai = - I . To find the eigenvector

associated with A, = 3, we find all nonzero solutions of

which gives

Doing the same thing with the eigenvalue Ai = -1 we get the eigenvector j\ associated with Ai = -1, which is

29. SYSTEMS OF PDES

171

3. Using the eigenvalues and eigenvectors of Problem 2, find the general solution of the following system. Hint: First write the system in matrix form u, +Au ==0.

Solution: Writing the system in matrix form u, + Aux =0, we have

In Problem 2 we found the eigenvectors Xp l\ of Ai =3, Ai == -1, respectively and so we transform u to new dependent variable v via the transformation

which transforms the original system

to the uncoupled system

which is

Written out, these equations and their solutions are

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Solution Manual

We can now find the solution vector u of the original system by plugging the vector v back in the equations u = Pv, or

or in terms of the two components of the solution

u1 (x,t) = ¢(x-3t) + ¢(x + t) U2 ( X,f)

= 2¢( X -3t )- 2¢( X + t)

4. Given the 2 x 2 matrix equation

u=Pv defined in Problem 3, write the equation in scalar form to verify

ou = Pov ot

or

ou =Pov

ox

Solution: We saw in Problem 3 that

ox

29. SYSTEMS OF PDES

and so the system u = Pv can be written as

or in scalar form U1 = vi +v2

u2 = 2v1 -2v2 and ifwe differentiate these equations with respect to t we have OU1

-

OV1

OV2

=-+ at at ot

OU2

ot

= 2 OV1- 2 OV2 a,

or in matrix form

or

To prove the second equation

simply replace t by x in the previous proof.

ot

173

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Solution Manual

5. Show that the functions u1 (x,t) = 2¢(x-4t)-21f1 (x + 4t) u2 (x,t) =¢(x -4t) + lfl(x + 4t)

satisfy the PDEs

Solution: Differentiating, we find au1 = 2¢'(x - 4t)( - 4)- 21f1'(x+ 4t )( 4) = -8¢'(x - 4t ) - 81f/(x+ 4t) at au ~ = 2¢'(x-4t )(l)-2f//'(x+ 4t )(1) = 2¢'(x- 4t)-2lfl'(x + 4t)

au1

-

at

= ¢' ( X - 41) ( -4) + lf/ 1( X + 4t) ( 4) = -4¢' ( X - 4t) + 41jf' ( X + 41)

au

- 2 = ¢' ( X - 4t) ( J) + lf/ 1( X + 4t) ( 1) = ¢' ( X - 4t) + If/ ( X + 4t) ax

Careful examination of these equations verifies the desired PDEs.

Lesson 30: The Vibrating Drumhead (Wave Equation in Polar Coordinates) I. Substituting u(r,0,t) = U(r,0)T(t) into the wave equation

in order to separate it into the two equations 1

1

r

r

2

U,, +-U, + 2 U00 +JU= 0 T' + li, 2c2T = 0

Solution: Plugging

u(r,0,t) = U (r ,0)T(t) into I JIJ ) u11 = c 2 ( u,, + -I u,+-u/ r2 r

we get

Dividing by U(r,0)T(t) gives

Setting each side of this equation to a constant, say - ,1, 2 (for convenience) , we have the desired result 2 I 1 U,, +-U, + 2 U00 + AU = 0

r

r

T' +

c T =0

Ji, 2 2

175

176

Solution Manual

2. Substitute U(r,0) = R(r )0( 0) into the following Helmoltz BVP. Why do we choose the separation constant to be 0,1,2, .. which we call n2 l 1 2 PDE: Urr +-Ur+ 2 V 08 +..1 U = 0 r r BC: U(l,0)=O to get

Solution: Direct substitution (similar to Problem I).

3. Solve 1 I PDE: U11=Urr+-ur+ 2u00 O<r<l, O<0<21C, O<t<oo r r BC: u(l,0,t )=O O< 0 < 21t, 0 < t < oo, IC:

{

u(r,0,O)=1-r 2 u,(r,0,O)=O

0 :<,; r < l, 0 5, 0 < 21r

Solution: Since the BC and ICs do not depend on 0, then neither will the solution, hence we seek a solution of the form

u(r,t) = R(r )T(t)

Plugging this in the PDE (neglecting the term u08 ), we get

T"R =TR"+_!_ TR' r

30. THE VIBRATING DRUMHEAD

177

and dividing by RT gives T"

R"

IR'

T

R

r R

-=-+-Setting each side of this equation to - . f we arrive the ODE

and an eigenvalue/eigenvector problem

This problem is called a Sturm-Liouville problem and its solutions are called eigenvalues and the possible values of A. are called the eigenvalues of the Sturm-Liouville system. In this problem there are an infinite number of functions that satisfy both the ODE and BCs, and they are

R,,,(r)=J0 (Av,,,r), m=0,1,2, ... , where the first four eigenvalues J

are

Av1

2.40

✓ i.o ,,,

L

Av2 _J 5.52 A.o3 L 8.65

Av4 J 11.79

and the co1Tesponding eigenfunctions.

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Solution Manual

'~ J0 (2.4r)

0

1 Figure 30.1 a): First eigenfunction J 0 ( 2.4r)

Figure 30.1b): Second eigenfunction J 0 ( 5.2r)

30. THE VIBRATING DRUMHEAD

179

J 0 (8.65r)

Figure 30. lc): Third eigenfunction J 0 (8.65r)

We now solve the ODE

getting

hence we have a sum of product solutions "'

u(r,t) = L:0",J0 ( k0,,,r) cos( k0mt) m= I

where the coefficients a,, are determined so u(r,t) satisfies the ICs "'

u(r,0)= LaJ0 (k 011 r) =l-r 2 O:S:r:S:l, O:S:0 < 2;,r n= I

We find the coefficients by multiplying each side of the equation by rJ 0 ( k0,,,r) and integrating each side of the equation with respect to r on the interval [ 0, 1] , and using the orthogonality of the eigenfunctions

(where J 1(r) is the first-order Bessel function , we get

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Solution Manual

2fr( 1- r )J0 ( k r )dr =~o _ _ _ _ _ __ J 2(kOm ) 2

a 111

0111

I

The integrals for a n can be evaluated on a computer. The Mathematica statement for finding a, is lntegrate[2 * r* (1 -r* r )* Bessel[0,2.4r),{r,O,I }]! Bessel [1, 2.4) /\ 2 and gives the values a, □ 1.1, l'2 □ -0.34, l'3 □ 0.045. After finding the coefficients (or 3 or 4) the solution is then ~

u(r,t) = Ia,,,J0 (k0 ,,,r) cos(k0 ,,,1) m=I

1 ' l. l J 0 (2.4r )cos (2.41 ) -0.34 J O ( 5.52r )cos (5 .521) +0.045 J 0 (8.65r )cos (8 .651)

4. Solve PDE: u11 =v' 2u BC:

u(l,0,t)=0

IC:

{

O< r < l, 0 < 0 < 21r, 0 < t < oo Os0<21r,O<t<oo,

u(r,0,0) = J 0 (2.4r) ( )_ 0 s rs 1, 0 s 0 < 21r u, r,0,0 -0

Solution: u(r,t) = J 0 ( 2.4r )cos( 2.4t)

5. Solve the following IBVP. What is the highest frequency of vibration?

PDE: Utt= v' 2 u

O< r < 1, 0 < 0 < 27l', 0 < t < 00

BC:

u(l,0,t )=O

IC:

{u(r,0,0) = J 0 (2.4r )-½J0 (8.65r) + ¾J0 (l 4.93r)

O < 0 < 21r, 0 < t < oo,

u,(r,0,0)=0

30. THE VIBRATING DRUMHEAD

181

Solution: The general solution of the vibrating circular drumhead that is independent of 0 is 00

u ( r,t) = La,,,J0( k 0111 r) cos( k0 ,,,t) m= I

The general idea is that we expand the IC u(r,0,0) = f(r) as

then simply add the factor cos( k0i) in each term, getting the solution

In this problem the IC is already expanded as a series of Bessel functions l l u(r,0,0) = J 0 (2.4r )- -J0 (8.65r) +-J0 (14.93r) 4 2 and so the solution is simply I u(r,0) = J 0 (2.4r )cos(2.4t ) - -J0 (8.65r )cos(8.65t) 2 1 +-J0 (14.93r )cos(l4.93t) 4

6. Graph the following Bessel functions. a) J 0( 5.52r) b) J 0 (14.93r)

Solution: The following graphs were done on Mathematia. Instructions for graphing the functions are given below the graphs. You could also graph them by going to the webpage www.wolframalpha.com and entering the commands under the graphs.

182

Solution Manual

1.0 0.8

0.6 0.4

02

02

0.4

0.6

0.8

.0

02 0.4

Figure 30.3: Plot [BesselJ [ 0, 5. 52r], { r, 0, l)] 1.0 08

04

0.2 04

Figure 30.4: Plot[BesselJ[0,14.93r],{r,0,1}]

1.0

Lesson 31: The Laplacian (an Intuitive Description) 1. What does the wave equation in spherical coordinates u11 = c

1 2 + 2u;; 2[ u,, +-u, --u; + +cot¢ r

r

r

2

2

1

.

2

r sm r/J

u00

]

reduce to if the solution u depends only on rand t?

Solution: The general wave equation in (both forms) spherical coordinates is u11

1 u00 ] 1-. -(sinr/Ju; ) + u,)' +-r smrp =c [-;-(r ; r Sin rp r 2

=c

2

2-

2

cot¢ I 2[ u,, +-u, 2 +,u,,,; + --u;+ r

r

r-

2

.

I

2 • 2

r sm r/J

2

u00

]

Hence, if u depends only on t and r , all partial derivatives of u with respect to 0 and r/J are zero, and so the equation reduces to

A good exercise is to work out the derivatives in the first above form of the wave equation to show that it is the same as the second form.

2. What does the wave equation u11 =c 2V 2u in polar coordinates reduce to if the solution u only depends only on r and t ?

Solution: The wave equation in polar coordinates is

183

184

Solution Manual

so if u depends only on t and r , we have u()(} = O, and so the equation reduces to

Note that this equation is almost the same as the spherical equation in Problem l that only depends on r ( coefficient of I I r in this problem for polar coordinates, compared with a coefficient of 2 / r for spherical coordinates), but this small difference results in a big difference in the solution. The solution of the wave equation in spherical coordinates (three dimensions) that only depends on r is u(r) = 1/ r , whereas the solution of the wave equation in polar coordinates (two dimensions) that only depends on r is u(r)=ln(l ! r).

3. What is Laplace's equation V 2u = 0 in polar coordinates if the solution u only depends only on r? What are the solutions of this equation? These solutions are called the circularly symmetric potentials in two dimensions.

Solution: In general, Laplace's equation in polar coordinates is

If the function u only depends on r and is independent on 0, then we have u (](J =0 and so the PDE becomes the ODE

which, although it is a second-order equation, it is a first-order equation in the derivative U ( t) =du/ dr , giving us the first-order ODE

Separating variables, we have

31. THE LAPLACIAN

185

or lnU(r)=-1 .nr+c=ln(; )+c Solving for U(r) gives

where we simply replace e c by c (you can verify that c I r satisfies the ODE for any real constant c and not just positive constants e c ) . Now, since U ( t) = du/ dr , we integrate to get the general solution

where c 1 and c 2 are arbitrary constants. This solution is called the logarithmic potential and is often written in equivalent form

since In ( 1/ r) = - In r.

4. What is Laplace's equation 'v 2 u =0 in spherical coordinates if the solution u only depends only on r ? What are the solutions of this equation? These solutions are called the spherically symmetric potentials in three dimensions.

Solution: Laplace's equation in spherical coordinates can be expressed either in the more compact form (so-called divergence form)

or in the more expanded form cot¢ I 2 urr +-u,. + 2 u¢¢ + -2-u¢ +

r

r

r

I 2 . 2 u(i(} = 0 r sm ¢

186

Solution Manual

lfthe solution u only depends on r, the partial derivatives with respect to 0 and ¢ are zero, and hence Laplace's equation reduces to the ODE

which can easily be solved by first letting U(t) = du/ dr, after which we get

Separating variables, we have

or

lnU =-2lnr + c = lnr- 2 +c = tn(r\ )+c Solving for U (r), we get

U(r)= c2 r and since U(t)=du/dr we integrate this equation to get

where c1 and c 2 are arbitrary constants. Note the similarity between the Laplacian in three dimensions and the Laplacian in two dimensions that only depends on the distance r from the origin.

• • d 2u2- +--= l du 0 ⇒ u ( r ) =c +c In ( -I ) 2 d11nens1ons= 1 2

dr

r dr

r

• • d 2 u2- +--= 2 du 0 ⇒ u ( r ) =c +c (I) 3 d11nens1ons= 1 2 -

dr

r dr

r

187

31. THE LAPLACIAN

5. What can you say about the (hannonic) surface u( x,t) = xy ?

Solution: By quick examination, you will discover that u(x,t) = xy satisfies u""' =u», =0 and hence satisfies Laplace's equation

Such functions are called harmonic functions. Although there are many functions that satisfy the Laplace's equation, they have important properties in common, one being the 'mean value property' which states that if you take a sphere (or circle in two dimensions) surrounding a point, then the average value of u on the sphere (or circle) is the value of u at the center of the sphere (or circle). Harmonic functions are important in physics, engineering, ....

If you study complex variables, you will discover that the real u and complex v parts ofan analytic function (i.e., differentiable) satisfy Laplace's equation. A few examples are Z2

=( X + ty• ) 2 =X 2 - y 2 + I• ( 2xy ) '--.,....---'

e cr. = e a(x+iy ) = e a.' e '0 >' = e a.' cos ay + i e m sin ay '-----v---'

-

'----v-----'

. 0 ,z-:f. O I = - I e-mo = - I cosn0 - -1 smn

Zn

r 11

r"

r"

'-----v---'

lnz =ln(re' 0 ) = lnr +i0 u

~

188

Solution Manual

6. Transform the two-dimensional Laplacian

to the polar coordinate form 2

1 r

1 r

V U=u,, +-u, + 2 u(}(J =0 by the transformation

x=rcos0 y=rsin0

Solution: To change Laplace's equation in Cartesian coordinates to Laplace's equation in polar coordinates, we make the transformation

x = rcos0 y = rsin0 The following diagram may be helpful in computing the partial derivatives required to make this transformation.

X

y

r

{}

Figure 31.1 : Diagram for computing partial derivatives of u, u, and u0

189

3l. THE LAPLACIAN

First observe

x, =cos0, x0 =-rsin0 Y, =sin 0, y 0 =rcos0 We begin by taking first partial derivatives u, =uxx, +uyyr =u,(cos0)+uy(sin0 )

u0 =uxx0 +u.vy0 =u,(-rsin0)+uArc os0) We now continue on and compute the second derivatives, still using the diagram in Figure 3 I .1 and the fact that u xy = u yx . The second partial derivatives take a little more thought, but we have (watch carefully)

but

(!

u, )cos0 = uxx cos 2 0 + ux.v sin 0cos0

~cos0=0

ar

( ~u ar ),)sin 0 = uX),sin 0cos0 + u".. sin 0 2

~sin0=0

ar

and so

urr =uxx cos 2 0 + 2uxy sin 0cos0 + uyy sin 2 0

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Solution Manual

Also

a

Uoo =-uu

a0

= a~[u, (-rsin0)+u\' (rcos0)]

:i

=

u,(-rsin0)] +

:i

u,(rcos0)]

=[( : 0 u_, )(-rsin0) + u.,( : 0 (-rsin0)) ] + [ ( : 0 u, j(rcos0) + ur( : 0 (rcos0))] = [u,, (r 2 sin 2 0) + u,,, (r 2 sin 0cos0 )- ux(rcos0)]

+ [ u"' (-r 2 sin0cos0) + u»· (r 2 cos 2 0) ]- uY(rsin0) Plugging u, ,u,., ,u/)/) in Laplace's equation in polar coordinates, we see u,.,. + ~u, + ,\ u00 =[uu cos 2 0 + u>Y sin 2 0 + 2u,,,(sin0)( cos0)] + ~[ u, (cos0) + uY(sin0)]

1 [[llu (r 2 sin 2 0) + u,y (r 2 sin 0cos0 )-ux(rcos0)]

+?

+ [ u.,Y (-r 2 sin0cos0) + u>Y (r 2 cos 2 0) ]- ur (rsin 0)

=u,, + u», (after considerable algebra) There is another way to do this problem. We could solve for r,0 in the transformation

x= rcos0 y = rsin0 getting

r =.Jx 2 + y2 B=tan- 1 ( ~ )

l

31. THE LAPLACIAN

191

and then compute the derivatives u xx and u YY in terms of r,0 and plug them into uxx + uy v = 0 and arrive at Laplace's equation in polar coordinates; but this is more difficult since the partial derivatives of r and 0 with respect to x and y are more messy.

Note: If you like computing complicated partial derivatives, try changing the three-dimensional Laplace's equation in Cartesian coordinates

to Laplace's equation in spherical Laplace's equation

by means of the change of variables

x = rsin¢ cos0 y=rsin¢sin0 z = rcos¢ Real fun!

Lesson 32: General Nature of Boundary Value Problems I. Based on intuition, can you find the solution of the following Dirichlet problem?

1 1 PDE: urr +-u,+ 2 u00 =0 , 0<r<I r r BC: u(l,0) =sin0 0 $. 0 < 21r

Solution: lfwe imagine Laplace's equation as a crude model for a rubber sheet, imagine stretching the rubber sheet around the outside (r =1) in the shape of u( 1,0) =sin 0. If you make the observation that sin 0 =-sin ( 0 + 1r) , then the rubber sheet is stretched in the opposite direction on the opposite side of the origin; i.e., 180 degrees apart. For that reason, one might be tempted to think the rubber sheet is linear in r and maybe the solution is

u(r,0) =rsin 0 This function clearly satisfies the BC, and ifwe plug it in the PDE, we find

2. Does the following Neumann problem have a solution inside the circle?

l

1

PDE: u,, +-u, + 2u(}() =0, 0 < r < I r r BC: Ur (1,0) =sin 2 0 0 $. 0 < 21r

Solution: The net flux across the circle must satisfy

f 2"

net flux across the circle= Jo u, (1, 0)d0 =0 and in this problem we have 192

32. GENERAL NATURE OF BOUNDARY VALUE PROBLEMS

193

i 2,r

net flux across the circle =Jo sin 2 0 d0 > 0

and so there is no solution since the total flux (integral of the normal derivative on the boundary) across the boundary for the Neumann problem must be zero. The proof that the total flux is zero comes from Gauss' theorem in multivariable calculus that if R is a bounded region of 91 2 (it also holds in three dimensions but we state the result in two dimensions) with boundary oR and if F is a vector field defined over V, then the integral of the divergence of F over R is the line integral of the normal component of F on the boundary curve oR. Mathematically stated

Now, if the vector field Fis the gradient vector, i.e. , F =Vu and the region is a circle of radius 1, then v rrvu = v' 2u and Gauss' theorem reduces to

But, if Laplace's equation holds, the integrand on the left is zero, hence the flux integral on the right is zero.

3. Give an intuitive description of the solution of the following boundary value for different values of h. l 1 PDE: u,, + -u,. + 2u00 = 0, 0 < r < 1 r r BC: u,(1,0)+h[u(l,0)-sin0]=0

Os;0<2ll'

Solution: When h = 0 the BC is the Neumann BC u, (1,0) = 0. The larger h becomes the more weight is placed on making the BC look like the Dirichlet BC u (1, 0) = sin 0. You can think of h as a parameter going from a

194

Solution Manual

Neumann problem to a Dirichlet problem. For 'medium sized' h , the boundary condition acts like a thermostat adding or subtracting heat depending on whether u(1,0) > sin0 or u(l,0) > sin0.

4. What is the boundary value problem one must solve to find the steadystate solution of 0 < X < 1, 0 < t < 00 PDE: u,, =Uxx -ut +u

u(0,t)=0 BCs: { ( ) 0<t<OO u l,t =0 ICs:

{u( x,O) = sin (3JZ"x) O::;x<l u (x.0) = 0 1

Solution: We suspect there will be a steady-state solution due to the damping term -u,. Setting u, = u11 = 0 in the PDE and along with the BC, we get

d 2u ODE: - , + u = 0 dx BC:

{

0 < x <1

u(O) = 0 u(l) = 0

The general solution of the ODE is

u(x) = c1 sinx+ c2 cosx where c, and c2 are arbitrary constants. Plugging this general solution in the BC gives c 2 =0

c, sin(l) = O ⇒ c, = 0 Hence, the steady state is u ( x) = 0, 0 ::; x ::;; I . If you remember from your ODEs the y' term in the damped harmonic oscillator equation

32. GENERAL NATURE OF BOUNDARY VALUE PROBLEMS

195

y"+y'+y=O is what causes the solutions to go to zero, the same principle in PDEs. The first-time derivative in the somewhat similar PDE

is what causes u( x,t) ➔ 0 as t ➔ oo.

5. Given the physical interpretation of the Laplacian, what is the general nature of the solution of the following Helmholtz BVP?

BC:

u(l,0) =0

Solution: If you first look at the one-dimensional Helmholtz BVP, which is ODE: d 2u =-,1,1u dx 2

{u(O) = 0 • u(l) =0

BC·

the general solution of the ODE is u( x) = c1 sin Jx + c2 coslx

and plugging this solution in the BC gives c2 =0

c1 sinl =0 We can pick c1 = O wi II give us the zero solution u ( x) =0 , which we knew from the start, but we want to find nonzero solutions and so we set sin l =0

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Solution Manual

which yields values (eigenvalues) ~ = 1r,2tr,31t, ... ,mr, ... and so we find an infinite number of solutions of this one-dimensional Helmholtz problem, namely un (x) = csin( mrx ), n = 1,2, ... where c is any constant. Figure 32.1 shows the eigenfunctions u1 ( x ),u2 ( x),u3 ( x),u4 (x).

1.0

0,5

.0

-

sm,- x1

-

sm,2~x1

-0.5

-1.0

-sm,4:,x,

Figure 32. l: Solutions of the I-dimensional Helmholtz eigenvalue problem

We imagine now what happens in higher dimensions. In two dimensions we can interpret Laplace's equation as roughly describing a stretched membrane (in the absence of gravity), where the height of every point on the membrane is the average height of values on the membrane on a small circle around the point. The Helmholtz equation is different. Whenever the height u of the membrane is positive the Laplacian is negative, and whenever u < 0 then V 2u > 0, which means when the membrane is sagging then the overall shape of the membrane is concave up, and when u > 0 the membrane is concave down. There will be an infinite number of these possible shapes, and they constitute the various shapes of a vibrating drumhead in the same way as the eigenfunctions in Figure 32.1 form the basic shapes of a vibrating string.

32. GENERAL NATURE OF BOUNDARY VALUE PROBLEMS

197

6. What is the physical interpretation of the following mixed boundary-value problem inside the given boundary conditions that describe a square?

PDE: uxx +u», =0

0 < x < I, 0 < y < I

uY (x,O) = h[ u(x,0)-2]

BCs: .

u(x,1)=1

O<x<I

0<x<l

u.,(O,y)=O O<y<I ux(l,y)=O O<y<I

Solution: The function u( x,t) might define the temperature of a square plate, insulated on the left and right due to the boundary conditions

so that heat only flows across the top and bottom sides. The temperature on the top side of the square is kept at u ( x, 1) =l, hence heat flows in or out across that side depending on whether the temperature near that boundary is less or greater than I, respectively (a very strong control system). On the bottom of the square, heat can enter or leave, depending on whether the temperature at that boundary is less than or greater than 2. If the temperature along the bottom edge is 1 (degree) then uy( x, 0) < 0 and heat will enter through the boundary (remember heat goes from hot to cold), but, if the temperature at the bottom edge ( when y =1) is positive, then uY( x, 0) > 0 and heat will leave the square. The constant h is proportional to this process. The solution to this problem won't be exactly 2 on the bottom edge but the larger the h the closer it will be.

Lesson 33: Interior Dirichlet Problem for a Circle 1. Carry out the details or the separation of variables to the following interior Dirichlet problem.

1 1 PDE: urr +-u, + 2 u1){) =0 0 < r < 1, 0 < 0 < 2,r BC:

r r u(l,0) = g( 0) 0 ~ 0 < 2,r

Solution: We look for a solution of the form u(r,0) = R(r )0( 0) , and plug this into the PDE, getting R"0+~R'0+2-R0" = 0 r r2 Now, if we divide each side of the equation by R0, multiply by r 2 , we arrive at , R" R' 0" ,-- -+r-=-R R 0 Now, since the left-hand side of this equation depends only on r and the right-hand side only on 0 , and since r and 0 are independent, each side must be equal to a constant, which we call ,1, and after minimal algebra, gives the two ODEs r 2R" +rR' - ,1R = 0 0" +A.-0=0 The ODE in 0 has the general solution

cIe.JJ. 11 + c2e-.JJ.o

l. < 0

0(0)= { a+b0 c1 cos ( .J::i

l.=O

0) + c sin ( .J::i 0) 2

198

l. > 0

33 . INTERIOR DIRICHLET PROBLEM FOR A CIRCLE

199

Since the function 0( 0) must be periodic in 0 with period 2TC, the constant

A can be any of the values (called eigenvalues) .JT,, =n, n = 0,1,2, ... which give acceptable solutions 0n ( 0) = C 1 cos( n0) + c2 sin( n0), n = 0,1,2, ... where c"c2 are arbitrary constants. If we now solve the DE in R with acceptable values of .JT,, = n we find

l

a+ blnr

R(r) = C rn + C ,--n 2

I

A,11 =0

.JT,, = n, n = 1,2, ...

The only solutions R(r) that do not blow up at the origin (r =0) are the constant solution a and r ;. . We have now found an infinite number of solutions of the desired form R(r )0( 0) and they are u,, (r,0) = R,, (r )0n ( 0)= r" [ a,, cos( n0) + b,, sin ( n0)]

for n = 0,1,2, ... (note that n = 0 includes the constant solution. But since the PDE is linear and homogeneous we know that the sums of solutions of this type are solutions and hence sums of the form u(r,0)= fr"[ ancos(n0)+b,,sin(n0 )] n=O

also satisfies the PDE. We now want to find the an,b,, so the BC is satisfied.

Plugging the solution in the BC gives u(l,0)=

L"' a,,cos(n0)+b,,sin(n0)=g(0) n=O

We now multiply both sides of this equation by cos(m0) and integrate each side of the equation with respect to 0 from Oto 2TC, and using the orthogonal properties of sines and cosines

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Solution Manual

m n 1o sin ( n0)sin ( m0)d0 ={O;rm=n =t=

2"

m n 1o cos ( n0) cos ( m0) d 0 = {O;r m=n =t=

2,r

r2"

Jo sin(n0)cos(n0)d0=O n=O,1,2, ... we get or

f"cos(m0)g(0)d0 =-Jo 1 cos(m0)g(0)d0, n=l,2, ...

a,,, = 0

2"

2 ,,.

r cos m0)d0

Jo

2(

1r

We find the coefficients b,, by multiplying each side of the above equation by sin( n0) and integrating, arriving at r2"

J, sin ( n0) g( 0)d0

b,, = 0

2 ,,.

fo sin 2 (n0)d0

I 2ir • =-fo sm(n0)g(0)d0, n=0,1,2, ...

Jr

33. INTER.JOR DIRICHLET PROBLEM FOR A CIRCLE

201

2. What is the solution of the interior Dirichlet problem

BC:

u(l,0)=g(0 ) 0~0<21(

for the following BCs? What do the solutions look like? a) u{I,0)=1+s in0+l.cos0 2 b) u(l,0)=2 c) u{l,0)=sin0 d) u{l,0)=sin{ 30)

Solution:

a) From a physical point of view a stretched rubber sheet is fixed on the boundary at

u(I,0) =I+ sin0 + _!_cos0 2 We can think of decomposing this problem into three parts, each one with one of the terms on the boundary. A value of u(l,0) = I on the boundary gives the solution of u ( x, 0) = 1 inside the circle (clearly any constant satisfies Laplace's equation since all terms are derivatives).

The BCs of sin0 and cos0 give rise to the stretched rubber sheet being linear in r . The conclusion is that the solution to the given BC is

u(r,0) = 1+ rsin0 + l.rcos0 2 You can verify that this function satisfies both the PDE and BC. b)

u(r,0)=2

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Solution Manual

c) u(r,0)= rsin0

d) This solution is more complicated. lfyou envision a stretched rubber sheet being fixed at u(l,0)=sin(30 ) on the boundary, the r dependence is more complicated. It is not linear in r as it was for sin 0 and cos0. You might be tempted to guess that if u(r,0) = r 3 sin(30) is the solution of Laplace's equation with BC u(l ,0) = sin (30) , and that u(l,0) =sin( 0) ⇒ u(r,0) = rsin 0 u(I,0) = sin(20) ⇒ u(r,0) = r 2 sin0 u(l,0) = sin(30) ⇒ u(r,0) = r 3 sin0

If you guessed this, you would be right. You can also verify that u(r,0) = r 3 sin(30) satisfies Laplace's equation.

3. What is the solution of the following interior Dirichlet problem when the radius of the circle is R = 2?

1 1 PDE: urr +-u,. + , u(l(J = 0 0 < r < 2, 0 < 0 < 21r r rBC: u(2,0)=g(0) O~0<21t

Solution: For an integral type solution, we simply plug in R = 2 in the general Poisson integral formula getting

l

R2 - r 2 u r,0 = - J J 2,r [ g OJ dw ( ) 21t O R 2 +2rRcos(i;-w )+r 2 ()

l

4-r 2 g OJ dw 21t O 4+4rcos(i;-w )+r 2 ( )

1 r2,r [ =-J,

There is also a Fourier series type solution. We have solved this Dirichlet problem for O< r < 1, but we can modify the solution for O< r < 2 by making a change of variable i; = r I 2, where now

33 . INTERIOR DIRICHLE T PROBLEM FORA CIRCLE

203

We can now use the solution with radius 1 in tenns of u( ; ,0) and then plug in ; = r I 2 in this solution. The solution for 1 1 PDE: u,;,; + J u,; + ; 2 u00 = 0

0 <; < l , 0 < 0 < 2;r

u(1 ,0) = g ( 0) 0 ~ 0 < 21r

BC:

u( ;, 0) =

is

~> f' "'

11

sin(n:1r0)

n== I

where fl

J, sin(mr0)g (0)d0 an = 0

1

Jo sin n;r0)d0 2(

J 1

= 2 sin ( n:1r0)g( 0)d0

°

Hence, the solution in terms of r,0 is

where the coefficients a,, are the same values as given above.

4. What is the solution of the following interior Dirichlet problem with radius R =2 iftheBCa re u(2,0)=si n(20)?

I 1 PDE: u" + - u, + 2uM = 0 0 < r < 2 , 0 < 0 < 21r r r O~0<2;r n(20) BC: u(2,0)=si

Solution: We essentially solved this problem in Problem 3, but we will do it again here in slightly more detail. We can transform this problem to a problem on a circle ofradius 1, solve the problem with radius 1, and then transform back to this problem . We do this by introducing a new variable

204

Solution Manual

Then since U , =Urr, =2u . ⇒ u -

,

,

I

r

=-U 2 ,,

the new transformed problem becomes I 1 1 PDE: -u,, +-u, + -2 u()(/=O 4 " 4,; ' 4,;

BC:

O<,;<I, O<0<21r

11(1,0)=sin(20 ) Q5,_0<21r

or I I PDE: U,. + -u , + - 2 U M = 0 _, ,; ' ,; "''

BC:

0 < J: < 1 ' 0 < 0 < 21' ':,

u(l,0)=sin(20 ) Q5,_0<21r

The transformed problem has the so lution

and so 2

u(r,0) = !:_sin(20) 4

You can verify this function satisfies the Dirichlet problem on a circle of radius R = 4. In a nutshell we went from a circle of radius l to radius 2 by replacing r in the solution by r/2.

5. Solve the following interior Dirichlet problem

1 I PDE: u,, +-u, + 2u00 =0

BC:

0 < r < 2 , 0 < 0 < 21r

r

r

u ( 2,0 ) = {

sin 0

0 5,_ 0 < 1r

0

1r 5,_0 < 2;.

33. INTERIOR DIRICHLET PROBLEM FOR A CIRCLE

205

Solution: We have seen that in polar coordinates, Laplace's equation has solutions of the form

u(r,0) =fr" [ a,, cos( nB) + b,, sin( n0)] 11; 1

To find the coefficients a,,, b,,, we plug this solution in the BC, getting

To find the a,, we multiply both sides of the equation by cos( m0) and integrate each side of the equation from Oto 2,r, getting

i

iff

2"'a'" 0 cos 2 (m0)d0= 0 sin0cos(m0)d 0

or

r

2ff

{O m = l,),S, ••• = ( 1 )( I )' m =0,2,4, ••• 2,r am= 2'" fu cos 2 ( m0)d0 2 m 1 _ m2 sin0 cos(m0) d0

0

(these integrals were carried out on Mathematica; the reader should have access to a variety of computer algebra systems). We find the coefficients b,,, in a similar way.

6. What does the Poisson kernel R 2 -r2

Poisson kernel = - - - - - - - - -2 R 2 -2rRcos(0-a )+r look like as a function of a, 0 ~a< 2,r for r =3R 0 =!!_ 2 4'

Solution: Picking R = 1 we graph the following Poisson kernel using Mathematica.

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Solution Manual

R 2 -r 2

Poisson kernel = - - - - - - - - -

R2 -2rRcos(0-a )+r 2 1-(9/ 16)

= I - I. 5 cos

(

%-a ) ( ) , + l:

Poisson function

I

0

1f :

-

2

Figure 33 .1: Poisson's function

a

0 :5 a :5 2Jr

33. INTERIOR DIRICHLET PROBLEM FOR A CIRCLE

7. If V(r,0) is any solution of the PDE (not including the BC) 1

1

PDE: V,.,+-V,.+ 2 V60 = f(r,0) r r

O<r<l, 0<0<21r

and if W(r,0) is a solution of

l

1

PDE: W,,+-W,+ 2 Ww =O O<r<l , O<0<21r r r BC: W(l,0) = V(l,0) 0 s 0 < 2tr then the solution u(r,0) of 1 1 PDE: u,, +-u, + 2 u08 =f(r,0) r r 0 s 0 < 21r 0 = 0) BC: u ( l,

O<r<l, O<0<2tr

is

u(r,0) = V(r,0)-W(r,0 ) Note: The note in the book was stated more generally, but we state a restrictive version here so the proof is more concrete.

Solution: Plugging

u(r,0) = V(r,0)-W(r,0 ) into

BC: gives

u(l,0)=O Os0 <21r

207

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Solution Manual

ii (V-W) l o(V-W) 1 o 2 (V-W) PDE: ~ - ~ + ~ - ~ + or 2 r r2 00 2

or

=[ 02v + .!_ ov + _.!__ a2 v ]-[ 02w + .!_ ow + _.!__ 02w ] or 2

r or

r 2 00 2

or 2

r or

r 2 00 2

= f (r ,0) - 0 = f (r ,0) V (1,0)-W (1,0)= V(I,0)- V (1,0) = 0

BC:

8. If V(r,0) is any function that satisfies the BC V(l,0) = f( 0)for a given f( 0) , and if the function W(r,0) satisfies the boundary-value problem I

1

I

1

r

r

r

r

PDE: W,,. +-W,. + 2 W00 = V,., +-V,. + 2 V00 BC:

O<r <l, O<0<21r

W(l,0) = 0 0 5' 0 < 2;r

then u(r,0) = V(r,0)-W( r,0)

satisfies l l PDE: u,.,. +-u,.+ 2 u/J0 =O r

BC:

r

O<r<I, O<0<21r

u(l,0) = /( 0) 0 5' 0 < 2;r

Note: The note in the book was stated more generally, but we state a more restrictive version here so the proof if more concrete.

Solution: We are given that W(r ,0) satisfies the boundary-value problem

BC:

W(l ,0)=0 O5'0<21r

where V(r,0) is an arbitrary (differentiable) function that satisfies V(l,0) == f (0) . From these equations, we have

33. INTERIOR DIRICHLET PROBLEM FOR A CIRCLE

PDE: BC:

ci (V - w) + I a(V - w) + _!_ a2( V - w) == 0 or 2

r

or

r2

80 2

V(I,0)-W (1,0) == f (0)- 0 == f (0)

which proves the desired result.

209

Lesson 34: The Dirichlet Problem in an Annulus 1. Solve the Dirichlet problem in an annulus I I r r u(l,0) =cos0

PDE:

urr +-Ur + 2U00

BCs:

{

=0

1< r < 2 ' 0 < 0 < 2,r

u(2,r)=sin0

Solution: Due to the BCs we know the general fonn of the Dirichlet problem inside an annulus has the general form

f[

u(r,0) =

(a,,r" + b,,r - )cos( n0) + ( c,,r " + d ,,r - )sin(n0)] 11

11

n=I

It is clear the only nonzero coefficients are a 1A,c1, and d 1 • To find these coefficients, we choose them so u satisfies the BCs, which result in the following four equations, the first two in a 1A, the second two in ci,d1 :

c1 +d1 =-I 12" sin(0)cos(0)d0=O ,r 0

Solving these equations results in

I

4

2

2

a=-- b =- c =- d =-1 3' I 3 ' I 3' I 3

Hence, we have the solution 210

34. THE DIRICHLET PROBLEM IN AN ANNULUS

2 11

.

2] sm0 4] cos0+ [2r - -u(r,0)= [- -r +3

3

3r

3r

2. What is the solution to the exterior Dirichlet problem 1 1 PDE: u" + -u, + 2u/}(J = 0 1 < r < oo , 0 < 0 < 21r r r BC: u(l,0) = g( 0) 0 s 0 < 2tr for the following BCs: a) g(0)=1 b) g(0)=1+cos(30) c) g( 0) = sin 0 + cos(30) l O~B <tr

d)

g( 0 )= { 0 1rs0<21r

Solution: The general solution of Laplace's equation in an annulus is

u(r,0) = a0 + b0 lnr +

f [( a,,r'' + bnr -" )cos(n0) + (c/ + d,,r -")sin(n0)] n= I

hence we simply assign the coefficients a11 ,b11 , n=O,1,2,... to match the BCs. Doing this, we find a)

u(r,0) = l

I b) u(r,0) =I+ 1 cos(30) r·

c) u(r,0) = .!.. sin0 + ~cos(30) r

r

d) For the exterior Dirichlet problem, the form of the solution is

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Solution Manual

"'

u(r,0) = 1>_,, [a,, cos( n0) + b,, sin ( n0)] n=O

=a0 +![a, cos( 0) + b, sin( 0)] + ~[ a2 cos(20) + b2 sin(20)] + ••• r

r

Plugging this series in the BC gives

{I

O~0 < n u(1,0)=~[a,,cos(n0)+b,,sin (n0)]= 0 n~ 0 < 21( 00

To find the coefficients a,, we multiply both sides of this equation by cos( m0) and integrate each side of the equation with respect to 0 from Oto 2,r, getting 2,r

1[

a111 f cos (m0)d0= f cos(m0)d0 2

0

0

which gives

fcos(m0)d0 1[

a,,,= 2 ~

f cos

=0 2(

m0)d0

0

Computing b111 by the same strategy, we find 1[

fsin(m0)d0 ={O m=O,2,4, •••

=

0 Ill

2;,r

f sin 2 ( m0)d0

2 13 5 -m m = ' ' , •••

0

Hence, the solution is

L -nrl ,, sin(n0) oo

u(r,0)=2

n=I.L.

= 2[!sin(0)+~sin(30)+~si n(50)+ · ··] r 3r" Sr

34. THE DIRICHLET PROBLEM IN AN ANNULUS

213

for l :;:; r < oo, 0 :;:; 0 < 2,r .

3. The solution of the exterior Neumann problem I l PDE: urr +-u,+ 2 u{J(J =O r r

BC:

l<r<oo, O<0<2,r

au or

-(1,0)=g(0) O:;;0<2,r

has the same form as the exterior Dirichlet problem, that is

u(r,0) =i>-"[ a,,cos(n0) + b,.sin(n0)] n=I

where the constants a,,,b,, are chosen so that u(r,0) satisfies the BC. Find the solution of the above Neumann problem with BC (1,0) = sin0 au or

Solution: Computing the derivative (with respect to r) of the general form, we have

ur (r ,0) = f-nr -(n+i)[ a,, cos( n0) + b,, sin ( n0)] 11=1

and plugging this into the BC we get

ur(l,0)=-fn [ a,, cos(n0)+b,, sin(n0)]=g(0) ,r:;;: J

We find the coefficients a,,,b,, in the usual manner. To find a,, we multiply each side of this equation by cos(m0) and integrate each side of the equation from Oto 2JC, getting r 2r

r 2r

-ma., = Jo g(0)cos(m0)d0= Jo sin(m0)cos(m0)d0=O

214

Solution Manual

or a m = 0 . To find b we multiply by sin ( m0) and integrate, getting 11

or b"' =

=

fo2'' sin(m0)g(0)d0

fo

m 2 " sin 2 ( m0)

J:"sin(m0)sin(0)d0 m

r·

sin 2( m0)

m :;t I m=l

Hence, the solution

u(r,0) = _ _!_sin ( 0) r

Checking the solution we find it satisfies the BC

u,.(r,0)= r\ sin(0) ⇒ ur (1,0)=sin(0)

215

34. THE DIRICHLET PROBLEM IN AN ANNULUS

4. Substitute the general solution

u(r,0) = a0 + b0 lnr +

f [(a/'+ b,,r-")cos(n0)+( c/ +d,,r-")sin(n0)] n,,,J

in the BC

u(R,,0) =g 1( 0) u(R2 ,0)=g2 (0) and integrating with respect to 0 gives the following equations from which we can find a,,,b,,, n=O,1,2, ... ,

Solution: To find a0 ,b0 we evaluate the BCs u(r,0) = g( 0)at r = R,,Ri getting

f [(G"R;' + b,, R,-" )cos( n0) +( cnR:' + dA-" )sin (n0)] g, (0) u( R ,0) G + b ln R + f [(G R; + b,. R;" )cos( n0) +( c,,R~' + d, R;" )sin ( n0)] g 2(0)

u( R,,0) = G0 + b0 In R, +

=

n= I

2

=

0

0

2

11

11 = 1

=

216

Solution Manual

then integrating each equation with respect to 0 from O to 21r, and using the fact that 21r

f sin(n0)d0=O, n=l,2, ...

0 2,r

f cos( n0)d0 = 0, n = 1,2, ...

0

we get

To find the coefficients a,,, b" we evaluate the BCs u(r,0) = g( 0) at r = R1,R2 , getting the equations u( R, ,0) = a0 +b0 In R, +

I[( a,,R;' +h,,R,-" )cos(n0)+( c,, R;' +d,A" )sin(n0)] = g 1(0) 11 =1

u( R1 ,0) = a0 + b0 In R2 +

f [(a,,R;' + h,,R;" )cos( n0) + (c,,R;' + d,,R;" )sin (n0) ] g 0) =

2 (

n=I

Now multiply each equation by cos( m0) and integrate each equation with respect to 0 from O to 21r, and using the orthogonality properties 2,r

f sin(m0)cos(n0)d0=O, n=l,2, ...

0

2I cos( m0)cos( n0)d0 = {: : : : we (simple exercise) arrive at the equations

34. THE DIRICHLET PROBLEM IN AN ANNULUS

217

Finally, to find the coefficients c,,,d,, we do the same thing as we did when we found a,,,b,, except now we multiply the equations by sin(m0). We leave this simple exercise for the reader.

Lesson 35: Laplace's Equation in Spherical Coordinates (Spherical Harmonics) 1. Find two linear independent solutions of Euler's

d 2R dR r 2 - 2 +2r--n(n+1 )R=0 dr dr of the form R(r) = r" .

Solution: Plugging in Euler's equation gives

or r 2 [ a(a-l)r(a-2)]+2r[ ar(a-i)]-n(n+ l)r" =0

a( a - l)r" + 2ar" -n( n + l)r" = 0 r" [ a ( a - I) + 2a - n ( n + 1)] =0 a 2 +a-n(n+1)= 0 This last equation is a quadratic equation in a, whose solutions can easily be found to be a= n, -( n + 1) . Hence, we have found two Iinearly independent solutions of Euler's equation

which yields the general solution

218

35. LAPLACE 'S EQ!)ATION IN SPHERICAL COORDINAT ES

219

2. Make the change of variables x == cos¢ to change the Legendre's equation in independent variable r/> !!_[sin¢ dcf>]+n(n+l )sin¢<D==O d¢ d¢

O<tp<tr

to the following one in independent variable x

dcf> cf> -2x-+n(n+ d 2( 1-x2) 1)<D==0, -l<x<l 2 dx dx

Solution: First realize that Legendre's equation

can be rewritten in the alternate form 2 sin¢ d ; +cos¢ d<P +n(n + l)sin¢ <P == 0 d</J d¢

by simple differentiation. Now, by making the change of variables x == cos¢ we see

220

Solution Manual

Hence, Legendre's equation . ,;. d 2© ,;. d<fJ ( ) . ,;.

0 sm'l'2 +cos'l'-+n n+l sm'I' <fJ = d</J

d</J

is transformed to 2

. ,;. d © ,;. d© ( ) . ,;. m. sm'l'2 +cos'l'-+n n+ I sm'I' ',I-'

d</J

d</J

=sin¢[sin 2 ¢:;-cos¢: ]+ cos¢[-sin¢ :] + n(n+ !)sin¢©

) ✓1- x 2 <fJ

2

= sin 3 ¢ d ; - 2sin ¢cos¢ def)+ n( n + 1

dx r:--:;-

d"

<fJ ,.---i d© r:--:; =-v I - x- ( I - x 2)d - 2x-v 1- x - - + n( n + I)v1I - x 2 ([) 2

d<jJ2

dx

2 = P [ ( 1 - x 2)d ; -2xd(f) +n(n+I)©] d</J dx

35. LAPLACE'S EQVATION IN SPHERICAL COORDINATES

Hence Legendre's equation

in ¢ holds, ifand only if Legendre's equation

(1-x2)-d -<P2 x d<!J - + n(n+ l)<t> =0 2

dr/)2

dx

in x holds.

3. Verify Rodrigues' formula

1 d" [

- - X -I ) 2"n!dx" ( n( X ) =

p

2

"]

n=0,1,2, ...

for the Legendre polynomials Po (x ), Pi (x ), Pi (x ), and~ ( x) .

Solution: By direct computation

221

222

Solution Manual

legendre polynomials

1

0.5

3: rf

0

-0.5

·1 ·1

-0.5

0

0.5

1

Figure 35. l : Graphs of some Legendre polynomials

4. Solve the interior Dirichlet problem PDE: (r 2u, ) +-.-1-[sin¢u; ] + + u0(1 =0 ' sin¢ ; sm ¢ BC: u(I,¢) =cos(3¢) 0 -5, ¢ < 1r

O<r<l

Hint: Use a trig formula to write cos(3¢) in terms of cos¢,cos 2 ¢,cos3 ¢ and then use them in the expansion cos(3¢) =a0 f>o (cos¢)+ aJ~ (cos¢)+· · ·

Solution: Since the BC only depends on the co-latitude ¢ , the solution has the general form 00

u( r ,¢) = ~>nr nP,, ( cos¢) n=O

so the idea is to expand the BC

35. LAPLACE'S EQ1JATION IN SPHERICAL COORDINATES

223

"'

u( 1,¢) = La,,P,, ( cos¢) n=O

as a series of Legendre polynomials, and then slip the factor r" in the nth term of this series. For our BC we use the trig identity cos(3¢) =4cos 3 ¢- 3cos¢

=( ~ )(½J[ 5cos ¢-3cos¢]-}cos¢ 3

3

8 5

=- ~ (cos¢)- -5 ~ (cos¢) and so the solution is

u(r,¢) = ~r 3 ~ ( cos¢)-ir?i ( cos¢) 5 5 = ~r 3 [ 5cos3 ¢- 3cos¢ J-}rcos¢

=4r 3 cos 3 ¢-.!.( 12r 3 - 3r )cos¢ 5

5. Solve the Dirichlet problem 0 <r <l

PDE: BC:

Solution: We saw in the book the solution is of Laplace's equation in spherical coordinates that does not depend on the longitude 0, but only on r and the co-latitude (fJ is ,x;

u(r,¢) = La/'P,,( cos¢) n=O

where the coefficients a,, are chosen so u satisfies the BC. In other words, it satisfies the equation

224

Solution Manual

00

u(I,¢)= z>nP,,(cos¢)= n=I

{

1 0~¢<tr/2 I -1 tr 2 ~ tp < tr

To find the coefficients a,, we multiply both sides of this equation by Pm(cos¢)sin¢ and integrate with respect to¢ from Oto tr, getting

:i:a"r P,, (cos¢)?,,. (cos¢)sin¢ d¢ = ru(l,¢)P,,, (cos¢)sin¢ d¢ n=O

o

0

and using the orthogonality property of the Legendre polynomials 0 m:i:-n foir P,,,(cos¢)P,,(c os¢)sin¢d¢= tP,,,(x)P,,(x)dx= { _2_ m=n

2m+ 1 we have (-

2 -)a,,, =

2m+ 1

=

r°

u(l,¢) Pm(cos¢)sin¢d ¢

i P,,.(cos¢)sin¢d¢- f" P,,.(cos¢)sin¢d¢ ,r/2 0

,r/2

Io

m=0,2,4, .. .

=12f~P, (x)dx 11

m=l,3,5, ...

Solving for a,,,, we find a,,, =0, m=0,2,4, ... and

t

a,,,= (2m+ I) P,,,(x) dx =(2m + 1)[

( ) ( )]' m = 1,3, ••• 2r 1- m [' m + 3 2 2

where f'( ) is the gamma function. Although we could evaluate the gamma function on Mathematica or even at wolframalpha.com, the above integral is best left for Mathematica or Maple to evaluate numerically, and doing this using Mathematica commands

35. LAPLACE'S EQ1JATION IN SPHERICAL COORDINATES

225

3*Nintegrate[LegendreP [ l ,x], {x, 0, l }] 5*Nlntegrate[LegendreP[3,.x], {x,0, 1}] gives a1 = 3 / 2, ~ = - 7 / 8 and so the first few terms of the solution are '°

u(r,¢) = 1>nrn P,, (cos¢) n;O

7 □ -r Pi (cos¢)--r 3~ ( cos¢) 3

8

2

= l..rcos¢ - '!_r 3 [5cos 3 ¢ - 3cos¢] 2

8

2

3 35 3 cos¢ 3 ] cos¢--r 21r 3 +-r =[ -

16

16

2

Note that this solution says that 5 5 u(l,0) =-,u(l,n/2) = 0, u(l,;r) = -8 8 which means the temperature on the surface of the sphere will be zero at the equator ¢ =;r/2, 5/8 at the north pole, and -5/8 at the south pole. You can play around with this solution by yourself going from the center of the sphere to the boundary ( r going from 0 to I) at ¢ = ;r/ 4 (45 degrees north latitude) to see how the temperature changes. In fact, we will do it getting the curve 1.2 1.0 .B

.6 .4 .2

0.2

0.4

0.6

O.B

I .0

Figure 35.2: Temperature u(r,;r/4) for O~r ~ 1

226

Solution Manual

6. Solve the exterior Dirichlet problem PDE: (r 2u,) +-.-1-[ sin¢u;] + + uM = 0 r

BC:

Stntp

Sill

¢

u(I,¢) = 1+ cos¢

tp

I< r < oo

0-::; ¢ < 1r

Solution: Since the BC is independent of 0, the solution has the form

u(r,¢) =

b L"' -;;'±iPn (cos¢) r n=O

We first find the Legendre polynomials

Po( cos¢)= 1 Pi (cos¢)= cos¢

Pi (cos¢) = ½[3 cos 2 ¢ - I] P; (cos¢) = ½ [5cos 3 ¢ - 3 cos ¢]

Plugging the general form of the solution in the BC we have 00

u(I,¢) = Lb11 P,1 (cos¢)= 1+ cos¢ n:O

which requires

n=O b,,

={i = 2,3,=l... n

n

Hence, the solution

35. LAPLACE'S EQVATION IN SPHERICAL COORDINATES

I

= Po (cos¢) + - p (cos¢) r

I

I

r

r

=-+ 2 cos¢

227

Lesson 36: A Nonhomogeneous Dirichlet Problem (Green's Function) 1. Find the potential due to a point source in three dimensions.

Solution: The total outward flux across the boundary of the sphere ofradius r is outward flux == -4n-r 2u,. and since the total outward flux is equal to the total charge inside the sphere, we set this value equal to q , getting

-4n-r 2 u == q ⇒ u == __ q_ ,. ' 4n-r 2 and solving for u(r) yields

u(r) ==_!L 4n-r

2. Find Green's function G( x,y,.;,JJ) for Laplace's equation in the upper half plane y ;c: 0. In other words, find the potential in the upper half plane at the point ( x,y) (zero on the boundary y == 0) due to a point charge at ( .;,JJ).

Hint: Jfwe place a negative charge at Q == ( ,;,-17) then it is clear that the potential field on the line y = 0 due to the two charges at Q == ( ,;,17) and Q = ( ;,-17) is zero. Hence, Green's function would be the result due to these two charges.

Solution: Using the strategy hinted at in the problem, we place a negative charge at Q=(;,-17) which along with the positive charge at Q == ( ,;,17) results in a zero charge on the x-axis. Adding the logarithmic potential due to the positive charge at Q = (i;, 17) and the negative charge at Q = (i;, -17), the charge at ( x, y), or Green's function, will be 1 ln(R) 1 In(R)-G(x,y,i;,17)=21r

1 2n-

21r

= - ln(!!_ )

R

228

36. A NONHOMOGENEOUS DIRICHLET PROBLEM

229

which satisfies Laplace's equ~tion for all ( x,y) in the upper half plane except at Q=(;,77),and R,R are the distances from (x,y) to (;,77) and (;,-77), respectively, i.e., R=.j(x-;) 2 +(y-77) 2

R= ✓( X - ;)2 + (y + '7 )2 Note: Observe that the Green's function is zero for y = 0 as desired since 1 ln(l?)

1 ln(R)-G(x,0,;,77)=27[ 27[ = 417[ {tn[(x-;) 2 +(-77)2]-tn[(x-;) 2 +(11)2]}=0 3. Use the results of Problem 2 to find the solution u( x,y) of Poisson's equation PDE: uxx +uw =-p -oo<x<oo, O<y<oo u(x,0)=0

BC:

-oo<x<oo

Solution: Green's f01m of the solution of the Dirichlet problem

PDE: u_o: +u-'J' =f(x,y) BC:

u(x,0)=0

-oo<x<oo, O<y<oo

-oo<x<oo

is

We saw in Problem 2 that G(x,y,;,lJ)=-1 1n(i!_)=-l 1n[(x-;):+(y-77):) (x-;) +(y+77) 47[ R 27[ where

230

Solution Manual

R= ✓(x-i;)2 +(y-17)2

R = ✓(x-,;) 2 +(y + 17 )2 Hence, the solution with f(x,y)=-pis

This integral will give on the potential at a point ( x,y) in the upper half plane. However, there are problems with this integral for two reasons. First, the integrand is infinite when ( x,y) = (c;, 77) and, secondly, the domain is infinite. What we can do, however, to find u ( x, y) is not to integrate on a small circle around ( x,y) and take smaller and smaller circles getting the limiting value, which will exist. Second, do not integrate over the entire half plane but on a large rectangle, talcing larger and larger rectangles to determining the limiting value. In other words, this integral formula is generally useful for finding the potential u( x,y) at a few points ( x,y). If one wants to find u ( x, y) over a wide range of values, it would be better to construct a grid and use some numerical method like we will study shortly.

4. How would you construct Green's function for the first quadrant x~O, y~O?

Solution: Place positive charges at (c;,17) and (-i;,-17) and negative charges (-f,17) and ( c;,-1]) so Green's function will be

where Ri,R2 ,R;;, and R4 are the distances as illustrated in Figure 36.1.

36. A NONHOMOGENEOUS DIRICHLET PROBLEM

23 1

y

(i:,y)

Figure 36.1: Positive charges at ( ,;,17 ),(-,;,-77), negatives at ( ,;,-17 ),(-,;,77)

5. An alternative approach to solving Poisson's equation that sometimes works is the following. Suppose you want to find the solution of PDE: BC:

1

1

u,, +-u, + 2 u 00 =1

r r u(l,0)=1 O~0<21l"

0< r < 1

You start by finding any solution of

by trying a solution of the form U(r,0) = Ar 2 into the PDE and solving for A . After finding this solution, let u = W + U .

Solution: Trying

up(r) =Ar 2 and substituting this quantity into the PDE, we find

232

Solution Manual

and substituting this quantity into the partial differential equation, we find

Ifwe now let

1 2 u=w+-r 4 we see thatwsatisfies the boundary-value problem with homogeneous partial differential equation

BC:

u(l,0)=-_!_+sin0 4

which has the solution

w(r,0) = _ _!_ + rsin0 4

Hence

O~0<21r

Lesson 37: Numeric Solution (Elliptic Problems) 1. Derive the following approximation equation for the second derivative.

f"(x)

=~[J(x+h )-2/(x )+ f (x-h )] 2

Solution: Writing the first three terms of the Taylor series for f(x+h) and f(x-h) we have h2 f ( X + h) 0 / ( X) + hf' ( X) + 2 .f" ( X)

f(x-h) □ f(x)-hf'(x)+! i_2 .f"(x) Adding these equations, then solving for f"(x), gives the desired result.

2. Carry out the computation for two iterations in the following Dirichlet problem using the Liebmann iterative process.

PDE: uxx + u>Y = 0 0 < x < l, 0 < y < l BC:

U= {

O on the top and sides of the square .

sm(.n-x) O~x~l

Solution: Start by making an n x n grid in the unit square

At the grid points on the top and sides set u i.J = 0 and on the top, evaluate them at the appropriate value of sin ( .n-x). Then initialize the value of u;_1 at all the interior grid points to the average value of all u ,,J on the boundary.

After then evaluate u, _1 =

l( u, _ 1 + u,+1.J + u, _ + u,, + 1_

_

1

233

1

1

1

) ,

i,J = l, ... ,n

234

Solution Manual

Then repeat this process again until you get a desired convergence of the interior values of uI .) . You will discover that the numbers u1,/ after a few iterations. But realize that they are not converging to the solution of the BV problem but to the solution of the difference equation, which is close to the solution of the BV problem. If you want a better approximation you should choose the grid size h smaller. A good strategy would be to pick a couple of different grid sizes and see if your procedure converges to the same numbers.

3. What are the algebraic equations that must be solved when you use a finite-difference approximation to solve the following Poisson equation inside a square?

PDE: ux, +u»' = f(x,t) 0< x <1, 0 < y <1 BC:

u(x,y)=g(x,y) on the boundary

Solution: Using the finite-difference approximations for the derivatives, we have

-

1 ( u/,J+ I -2uI ,) +uI .J- 1 ) uxx □ h2

i = 2, ... ,n -1, j = 2, ... ,n-1 i = 2, ... ,n-1, j = 2, ... ,n -1

Plugging this into the PDE, we have

1(

) 1(

)

-h2 u1,j+ I - 2u I.J . + u1,J . - 1 +h2 - u1+ 1. J - 2u1,/ + u1- l,J. = J, _J

and solving for u1,/ gives 1 h -[u I + U I + U . I+ U . I]--/, .

4 2

U1,/ = 4

1+ ,j

I- .j

l, j+

1.J-

1,/

When the grid point (i,J) is adjacent to a boundary (i.e., if any of i = 2,i = n -1, j = 2, j = n - I), then the corresponding value of g . should 1./

replace the corresponding value of u;_1 .

37. NUMERIC SOLUTION

235

4. What are the algebraic equations that must be solved when you use a finite-difference approximation to solve the following Poisson equation inside a square?

PDE: uxx +u.w +2u=O O<x<l, O<y<l BC:

u(x,y )= g(x,y) on the boundary

Solution: Using the finite-difference approximations for the derivatives, we have

I u>)' □ h2 ( u1+ 1,1 - 2u1_1 + ui-1_1 ) , i = 2, ... ,n -1, j = 2, ..., n-1 Plugging this into the PDE, we have

-h 12 ( u1.1+1 - 2u1,1 + u1.1-I) + -h 12 ( u1+l .1 - 2u1,1 + u1- l ,1 ) + 2u1,1 = 0 and solving for u 1,l gives

. +u1.1 - I ui+ l.1 +uH.1 ) u1.1 =-,--...,...

2( h -2 1- 2) ( u1.1+1 When the grid point (i,J) is adjacent to a boundary (i.e., if any of i = 2, i = n - 1, j = 2, j = n - 1 ), then the corresponding value of g 1.1 should replace the c01Tesponding value of u .i .

1

5. How would you solve the following Neumann problem inside the square by the finite-difference method?

PDE: uxx +u.w =O O<x<l,O<y< l

u ~~, y) = 0 on the top, bottom, and left-hand side { uu BCs: o::;y::;1 ax(I,y)=g(y)

236

Solution Manual

Solution: The finite-difference formula for the PDE is

uI,] .=_!__[u.

4 I+ 1,J.+u/- 1,J. +uI,].+ ,+ul ,J.- 1] for the interior grid points, and for the boundary points, we have u1,1 = 0

(bottom side)

un,J =0 ( top side)

=0 (left side) U; ,n =ui ,n- l + hg(y;) (right side) u;, 1

,Y 0

0

0

0

9

0

0

I

0

g(y.

Ui+l,j

0 ~ 'i,j-1

u,,j Ul,j+I-

...

i 0

uH,j

g(y

0

g(y,

0

g(y, ~

0

0

0

0

0

0

0

X

j Figure 37 .1: Typical grid construction for Laplace's equation inside a square.

For grid points (i,n-1), i =1,2, ... ,n on the vertical line next to the right boundary grid points (i,n ),i = 1,2, ... ,n, we use the formula

ui,n =ui,n-i +hg(y;) to estimate the values of u(I,y;), i=l,2, ... ,n, where h is the grid size. This will satisfy the derivative BC ux (1,y) = g(y ), 0 ~ y ~ I at the right side of the square.

37. NUMERIC SOLUTION

237

6. Write a flow diagram to solve the following Dirichlet problem inside a square with an arbitrary number of grid lines.

PDE: uxx +uw = f(x,y) 0 <x < 1, 0 <y < l BC:

u(x,y)=g(x,y) ontheboundaty

Solution: We saw in Problem 3 the finite-difference approximations for the derivatives are u _rx

□ ; 2 ( u i.J+I - 2u,,1 + u, _1_1)

u YY

□ ; 2 ( u ,+i,, - 2u,,1 + u, - 1.1 )

Plugging this into the PDE, we have

and solving for u ; ,). gives h2 1 ] +u +u +u uI..,) =-[u.

1 1 1 1 4-J'i ,J ,Jl + ,J l ,J 4 I+ ,J I-

This is the equation that is used to approximate the solution of the problem.

We leave the details of a flow diagram for the reader.

Lesson 38: An Explicit Finite-Difference Method l. Find the finite-difference approximation of the following heat conduction problem for t = 0.005, 0.010, 0.0015 by the explicit method. Let h =LlX = 0.1. Plot the solution at x = 0, 0.1, 0.2, 0.3, 0.4, ... ,0.9,1 for t=0.015.

PDE: u, =ux., BCs:

{

0<x<l, O<t<oo

u(O,t) =0 u(l,t) =0

O<t<oo

fC: u(x,O)=sin(.1rx)

0:s;x:s;1

Solution: Student project

2. Solve PDE: u, =Uxx BCs:

{

O<x<l, O<t<oo

u(O,t)=O u(l,t)=O

O<t<oo

by separation of variables, then evaluate the analytical solution at the same grid points where you carried out the finite-difference approximation in Problem I.

Solution: We find the analytic solution by seeking product solutions

u(x,t) = X(x)T(t) Plugging this equation in the PDE gives the two ODEs

T' +(mr}1T=0 X" + ( mr ) 2 X = 0 238

n=l,2, ...

38. AN EXPLICIT FINITE-DIFFERfNCE METHOD

239

which has solutions

T (t) = Cne -(mr)' ,

n =1,2, ...

11

xn ( x ) = ancos( n0) + b,, sin( n0) By superposition we have the series solution 2

"'

"'

u( x, t) = L /n(t)X (x) = }:e- (mr) ' [ Ancos(n0)+ B,, sin(n0)] 11

n=I

~ l

The last step is to find the coefficients An,Bn so that u(x,t) satisfies the IC u(x,O)=sin(;,rx )

Os x s l

It is easy to see that all coefficients must be zero, except B1 = I . Hence, we have the solution

We leave it to the reader to compare the numeric solutions found in Problem I with these values.

3. Write a flow diagram to carry out the computations of the following hyperbolic problem.

PDE: Utt =Uxx

u(O,t)=g,(t)

BCs:

{

lCs:

{

Solution: Student project

0 < x < I, 0 < t < ro

u(I,t)=g2 (t) u(x,O)=1(x) U1( X, 0) = If/ ( X)

O<t< ro

Osxsl

240

Solution Manual

4. Find the finite-difference approximation of the following heat conduction problem for t = 0.005, 0.0 I 0, 0.0015 by the explicit method. Let h = Llx = 0.1. Plot the solution at x = 0, 0.1, 0.2, 0.3, 0.4, ... , 0.9, 1 for

t=0.015.

PDE: u, = u.a

O < x < 1, 0 < t < oo

BCs· { u(O,t)=O

• u,. (l,t)=-[u(l,t)-1] IC: u(x,0)=sin(ll'x)

Solution: Student project

0sxsl

0<t<oo

Lesson 39: An Implicit Finite-Difference Method (Crank-Nicolson Method) 1. Show that the implicit Crank-Nicolson method applied to the JBVP PDE: u, =u_"' 0 < x < 1, 0 < t < oo BC: {

IC:

u(0,t)=0 u(l,t) = 0

0<t <oo

u(x,0)=0 0:S:x:S:l

gives rise to the formula -,1,ru,+1_1 • 1 +(1+2r,1,)u1+ 1,1 -,1,ru,+1,1 _ 1 =r(l-,1,)u,,1 +1 +[t-2r(l-,,1,)]u,_ 1 +r(l-,1,)u,_1 _1

Solution: Direct computation by plugging in the proper finite-difference fonnulas for u, and u xx as illustrated in the book. The BC and IC should be replaced by

=0 u, ,n =0 u ,.1

i =1,2, ... ,m

BCs:

{

IC:

u,,1 = I j =1,2, ... ,n-1

2. How would you solve the following problem by the implicit finitedifference method?

PDE: u, =u.o: - ux

0 < x < 1, 0 < t < oo

u(0,t) = l

BCs:

{

ICs:

u(x,0) =0

ux (I,t) + u(l,t) = g(t)

0<t<oo

0 :S: x :::;1

Solution: The finite-difference approximation for the PDE is the same as in the lesson. The finite-difference approximations to the BCs should be 241

242

Solution Manual

u1,n -u.,,n- 1 + U = g h 1,n

1

where g =g(ik). For each value of time we solve n -1 equations for 1

The BC at x = 1 will give you an extra equation.

3. How would you solve the following problem by the implicit finitedifference method?

PDE: U1 =U.TX + U

0 < x < I, 0 < t < oo

BCs:

{

u(O,t)=1 u(l,t)=O

O<t<oo

ICs:

u(x,0)=1

O:s;x:s;I

Solution: This problem is almost identical to the one in the lesson. We let you carry out the computations.

4. What is the molecular form of the following equation ifwe pick,1, =1?

-Jru,+l.i+J+(1 + 2rJ)u,+1•1 -Jru,+1•1_1 = r(l--i)u,_1 +1 +[1-2r(l-J)Juy + r {1-A )u,.1 _ 1

Solution: The molecular form is drawn in Figure 39.1.

39 . AN IMPLICIT FINITE- DIFFERENCE METHOD

1--8 Figure 39.1: Molecular form of implicit method

5. Write a flow diagram to approximate the solution of the following heat problem.

PDE: ut =uxx 0 < X < 1, 0 < I < 00 BCs:

{

u(O,t) = 1 u(l,t) = 0

O<t<oo

ICs:

u(x,O) =1

0 :s;x:s;l

Solution: Direct application of the formula from the lesson.

6. Solve the following system of equations

Solution: Student project using the iterative method.

243

Lesson 40: Analytic versus Numeric Solutions

1. How would you construct an experiment to estimate the parameter a in the problem PDE: U1 = a 2Uxx u(O,t)=l

BCs:

{

ICs:

u(x,0) = 1

u(l,t)=0

0 < X < J, 0 < t < 00

0<t<oo O~x~l

using the analytical solution

Solution: Writing the first two terms of the analytical solution gives

Hence, when x =0.5 we have 2 • ( 1r) I - (3nn) 2, sm • ( -3,r)] u (0 .5 t ) D -2 [ e- (nn) t sm - +-e ' tr 2 3 2

_2 [ e- (nn) 2 1 --e 1 - (3,ra)',] 1r 3 We can now perform an experiment where we observe values of u at x = 0.5 for different values of time. We can then do a least squares analysis to find the value of a that best fits the above function to the data points.

244

245

40. ANALYTIC VERSUS NUMERIC SOLUTIONS

2. Least-squares approximation minimizes the sum of squares (SS) between a curve and the data points. For example, the least squares line y( x) =a+ bx that approximates the data points (Xi,Y, ),(x,, y , ), ...,( x",y" ), would be the line that minimizes

y

X

Figure 40.1: Least squares diagram

Find the constants a,b in terms of the data points that minimizes SS.

Solution: We differentiate S = S( a,b) which is a function of a and b with respect to a and b, and set the derivatives to zero, getting

as

n

as

n

-=2~::[Y; -(a+bx;)](-1)=0 1=1 oa - = 2L[Y; -(a+bx;)](-x;)= 0 i=l ob or

or in matrix form, we have

i=l

i=l

i=l

i=l

i= I

i=l

246

Solution Manual

from which we can (easily but tedious) solve for a, b.

3. An important problem in biochemistry is the determination of the molecular weight of the macromolecule myoglobin. One approach is to place a certain blood solution in an ultracentrifuge and spin it for a given length of time. The equation that describes the concentration of the liquid in the centrifuge is known as Lamm's equation

ou(r,t) 18( au - - Dr--sr.o 2) u r or or at where

r = distance from the center of the centrifuge D = diffusion coefficient (depends on the weight of myoglobin) s =sedimentation coefficient (computed experimentally) w =angular velocity of the centrifuge (known)

u(r,t) =concentration of the medium in the centrifuge The approach in finding the molecular weight of myoglobin is to find the steady-state solution u( r,oo) ofLamm's equation by letting u, = 0 and solving the ODE (which is not difficult) and found to be 2u ( r )) =Drdu d 2 u2 + ( D-sm 2) -=0 du -d ( Dr--sr.o dr dr dr dr getting

J )(D-s•i )/D U(r)= ( -;.

0<r<oo How would you use this analytical solution to design an experiment to estimate the molecular weight a of myoglobin?

40. ANALYTIC VERSUS NUMERIC SOLUTIONS

247

Solution: One designs an experiment where we measure the concentration of myoglobin at different values of rafter the substance has come to steady state.

We can then do a least squares fit of this data on the above analytical solution to estimate D, assuming we know s and co.

Lesson 41: Classification of PDEs (Parabolic and Elliptic Equations) I. Which of the following parabolic and elliptic equations are in canonical form ? a) u, =ux, - hu

b) u.,:v +uxx +3u =sinx c) uxx +3u.w =0 d) uxx = sinx

Solution: The general second order, linear PDE in two independent variables is Aux,. + Bu_,y + Cu_w +Dux + Eu-" + Fu = G where A,B,C, D, E, F,G are functions (which of course includes constants) of x,y. If B2 - 4AC < 0 ⇒ PDE is elliptic B2 - 4AC = 0 ⇒ PDE is parabolic B 2 - 4AC > 0 ⇒ PDE is hyperbolic We also saw that the canonical form for each type of equation is

canonical forms for a hyperbolic equation

f u,;; ~u,,q = Y.1(,;,17,u,u, ,uq)

lu.;,, -<l>(,;,17,u,u;, uq)

canonical parabolic equation u,1,1 = <t>( ,;,17,u,u, ,u,1 ) canonical elliptic equation u;; +u,1,1 = <t>(,;,17,u,u; ,uq)

Identifying these coefficients, we have the following classification:

248

41. CLASSIFICATION OF PDES

249

a) u, = u_r.x - hu ⇒ B 2 - 4AC = 0 ⇒ parabolic, canonical b) u,y +uxx +3u =sinx ⇒ B 2 -4AC = I >0 ⇒ hyperbolic, not canonical c)

u u + 3u,:v = 0 ⇒ B 2 -

4 AC= -12 < 0 ⇒ ellliptic, not canonical

d) u"'= sinx ⇒ B 2 -4AC=O ⇒ parabolic,canonical You should realize that the notation for the two independent variables, whether x,y or ; ,17, is irrelevant.

2. Transform the following parabolic equation into canonical form.

Solution: In this case we have

and so there is only one characteristic equation dy = B- ✓B 2 -4AC = 2- ✓2 2 -4(1)(1) = I 2(1) 2A dx

which gives the single characteristic equation

Solving for c1 gives c 1 = y-x and so ; = y- x (the idea is to pick the new coordinate on constant curves or lines in this case). The second coordinate '7 is arbitrary as long as the coordinate lines I]= constant are not parallel to those of;= y- x =constant, so we choose something simple, like 17 = y, which gives us the new coordinate system i;=y-x

1J = y

250

Solution Manual

We could have just as well chosen 77 = x, 77 = x + y, 77 = y- 2x, ... but not 77 =2y - 2x . We now use the diagram in Figure 41.1

u

,, X

y

Figure 41.1 : Relational diagram useful for computing derivatives

to assist in finding the following partial derivatives:

ux = u, c;x + u,,77x = -u, uY = u, c;Y + u,,77Y =u, + u,, uxx =(ux)x =(-3u,

t =- (u,,c;x +u,,,17x) =u,,

uYY =(uY)Y =(u, + u,, )Y = u,,c;Y + u,,,77y + u,,,77y + u,,,,77y = u?? + 2u,,, + u,,,, uxy =(ux)Y =-(u, \ =-(u,,c;Y +u,,,77Y )=-(u,, +u,,, ) then plugging them into the PDE

uxx +2uxy +uYY +u = 2 gives or

u,,,, +u = 2 which is in canonical form u,,,, = <1>( c;, 17,u,u,, u,, ) of a parabolic equation.

41. CLASSIFICATION OF PDES

3. Transform the following elliptic equation into canonical fonn.

Solution: In this case we have A=l,B=0,C=2 ⇒ B 2 -4AC=-8<0 so the equation is elliptic (for x -:t:- 0) and the characteristic equations are

dy = B- ✓B 2 -4AC = o-,Jo 2 -4(1)(2) = o- ,Jo 2 -4(1)(2) =-i✓2 2(1) 2(1) 2A dx dy = B+ ✓B 2 -4AC = B+ ✓B 2 -4AC = O+,Jo 2 -4(1)(2) =i✓2 2(1) 2A 2A dx Solving these equations gives 2x+c 1 y=-i✓

2x+c 2 y=i✓

and solving for the constants ci,c2 gives c1 = y+i✓2x

c2 =y-i✓2x Hence, we have the new coordinates

!; = y+i✓2x 1J = y-i✓2x

or which we prefer, real coordinates

251

252

Solution Manual

a=;+17=y

2

/3 =; -17 = ✓2x 2i

We now use the diagram in Figure 41.2

u

p

a

X

y

Figure 41.2: Functional diagram useful for computing partial derivatives

to assist in computing the derivatives ux =uaax +up/3., = ✓ 2up Uy= Uaay + Up/Jy = Ua

and plugging them in the PDE

and using the fact thatx = f3 I ✓ 2, we get the new PDE in ; and 77 as

41. C LASS IFI CATION OF POES

✓2 2 _ } - P /4 +-/J Up - -e 2

Uaa + Upp

4

2

which is in canonical form u,, + u,,q = <t>( ,;,77,u,u, ,uq) of an elliptic equation. The notation of the independent variables a,/3 or ,;, 77 is irrelevant.

253

Lesson 42: Monte Carlo Methods 1. Write a computer program to estimate the value of the following integral and make a graph of the approximation as a function of the number of random tosses.

fl .

I= Joe•mx dx

Solution: The graph of the integrand e•inx is drawn in Figure 42.l. .2 .0 1.8

0 .6

0 .8

1.0

Figure 42. l: Graph of esinx, 0::; x::; 1

The integrand is positive and has its value between 1 and e•in(t) □ 2.32, and so we generate a uniformly distributed random sample of points (Ri, 2.321?:i), where Ri,Ri are random numbers between O and 1, hence the points ( R1, 2.321?:i) lie in the rectangle

R ={( x,y): 0::; x::; 1, 0::; y::; 2.32} (like tossing darts randomly inside the rectangle). We then compute the fraction J of the points ( R1, 2.321?:i) that fall below the curve y = esinx , from the rule if 2.321?:i < e•inR, ⇒ random point (R1,2.321?:i) lies below the curve e•inx if 2.321?.i 2". e•inR, ⇒ random point (Ri,2.321?.i) lies on or above the curve e•in x

254

255

42. MONTE CARLO METHODS

J

by the area of the rectangle R, which is 2.32, which We then multiply will be an approximation of the integral. That is

t

esinx

dx O 2.32]

If you carry out this simulation on a computer for say n = l 00 tosses, you will find roughly 70% of the tosses fall under the curve, hence □ 0.70, and multiplying this value by the area of the rectangle, which is 2.3, you will estimate the integral to be 0.7 x 2.32 = 1.62 (the correct value to six places is 1.63187). A flow diagram for this process is essentially like the flow diagram in Problem 3.

J

2. Generate a sequence of random numbers using the algorithm

Solution: The equation says: I . Take a random number r,,, the first one being r0 = 0

2. Compute 3rn + 4 3. Divide by 7 and take the remainder, that will be r,,+1 4. Repeat the process Carrying out these computations, we will generate the sequence 0,4,2,3,6,1,0, ... ➔ numbers repeat

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Solution Manual

3. Write a flow diagram and computer program to estimate the following triple integral.

2 1= i lilil e--(x +y +=') dxdydz 0 0 0

Solution:

Set/ : )

SdSUM = O

Generate R,. R,.R,.

a.

0>et- :zero and 1)

SUM = SUM+l l = U?

/

(N = umberoftoues)

B ~

- ~

-+

I

l = l+l

~-----~

INIBGRAL = SUM/N

Figure 42.2: Flow diagram for approximating a triple integral

257

42. MONTE CARLO METHODS

4. How would you generate a sequence of random points inside the following triangle?

y

·1~· X

l Figure 42.3: Generating random points inside a triangle

Solution: Generate random numbers ~,ls between Oand 1. If

we keep the point, otherwise generate another pair and repeat the process.

Continue this process to get a random sample in the triangle.

5. How would you generate a random sample from the following statistical distribution?

1/2 1/4

1/4

_ _ _ _...__ _ _...__ _- ' - - - - - - - x

0

1

2

Figure 42.4: Sampling from a binomial distribution

258

Solution Manual

Solution: Generate a random number R between 0 and 1 and then compute 0 0 :5:R :5:0.25 { X = I 0.25 < R :5: 0.75 2 0.75 < R:::; 1 The random variable X has the desired statistical (binomial) distribution.

6. The Buffon needle problem states that if a needle of a given length, say 1 inch, is tossed randomly on a flag (or any ruled surface) where the width between the stripes is the same as the length of the needle, i.e., 1 inch, then the probability the needle crosses a stripe is 2/ TC. How would you devise a game to estimate the value of rr?

Solution: This is a problem in what is called geometric probability. In this casewegeneratetworandomnumbers r,0 where 0:5:r:5:l and O5,05,TC/2.

If

or 1 . 0 >1 y 2 =r+-sm 2

then the needle crosses a line, otherwise it does not. Perfonn this experiment n times and take the average p of the number of times the needle crosses a line, and then estimate rr from the formula 2 -2 L.I p- ⇒ 1( [] -=1( p

Figure 42.5 illustrates the problem.

259

42. MO N TE CARLO M ETHO DS

line 1 . {J Y2 = r+ - SID 2

1 . {J 2

Y1 = r - - SIO

line Figure 42.5: Diagram of the Buffon needle problem

Try writing your own program with a different number of tosses and see

how close to tr you come. In fact, make a graph of your approximation to tr as a function of n, the number of tosses.

Lesson 43: Monte Carlo Solution of Partial Differential Equations 1. Write a flow diagram for tour du wino that finds the approximate solution to the following BV problem at some interior grid point. Let the number of horizontal and vertical grid points be arbitrary.

PDE: uxx + u_,,, = 0, 0 < x < l, 0 < y < 1 BC:

u(x,y)=g(x,y)

ontheboundary

Solution: Hint: One idea would be to set up an n x n grid matrix G = (g, _1 ), i, j = 1, 2, ... ,n where increasing i is in the increasing y direction (when i = I we have y = 0 and when i = n we have y = 1; also when j = I we have x = 0 and when j = n we have x =I). We also define another n x n array P =(p,_1), i, j = l,2, .. .,n , where initially all elements are set to zero, later the first and last rows, first and last columns, will contain the fraction of times the random walks ends at the given boundary. Finally, we set up a matrix G where the first and last rows, and the first and last columns, are the BCs g(x,y). We leave the details ofa program up to a good programmer.

2. Write a computer program to carry out the flow diagram from Problem I. PDE: uxx + x 2uw = 0, 0 < x < I, 0 < y < I BC:

u(x,y)=g(x,y)

on the boundary

Solution: The finite-difference approximation to the derivatives in the PDE are 1( u.0- □ h2 u l.,j+I -2u1.J +u1.J-1)

and so the PDE reduces to (letting h = k for simplicity) 260

43. MONTE CARLO SOLUTION OF PARTIAL DIFFERENTIAL E~ATIONS 261

and solving for ul,j. gives u I ,}

= ~ -1- ~ [ u

2 ( x~ + 1)

l 0 J+ I

+ul ,J - 1 +xj2( u i+J.J +u /-l,j. )]

Hence the probability of moving from u, _1 to one of its four neighbors is the coefficient of the respective terms, which if you note are positive and

I 1 x2 x2 1 1 ---+---+ +--1-=---(2x 2 +2)=1 1 2x} + 2 2x} + 2 2x} + 2 2x~ + 2 2x} + 2 sum to one. We leave the problem of writing a computer program to the reader.

3. How would the game of tour du wino be modified to solve PDE: uxx + x 2uw = 0, 0 < x < l, 0 < y < 1 BC: u(x,y)=g(x,y) ontheboundary

Solution: Plugging the finite-difference approximations u xx

□ ~2 ( u ,,1+1 - 2u;,J + u ,.1- 1)

u yy

n ) 2 ( u,+I.J - 2ui,J + u i-1.J )

with h = k, into the PDE, we find (after brief algebra)

262

Solution Manual

and solving for u1,j. gives u1 = ~ 1- ~ [ u. ,J

2( X~ +])

I,J+ I

+u .. +x 2( u 1•.1-I

J

i+ I.J

+u

i - 1. }

.)]

Hence the probability of moving from u;,; to one of its four neighbors is the coefficient of the respective terms, which if you note are positive and

sum to one. We leave the problem of writing a computer program to the reader.

4. Devise a modified tour du wino game that will solve PDE: uxx + UY.Y =0, 0 < x < 1, 0 < y < 1

l

u(x,l) =0

BCs:

(

) = u(x,0) =0 ' u(0,y)=l

U X y

0 < X < }, 0 < y < 1

ux (l,y) = 0

Solution: We construct a grid of points for the unit square in Figure 43.1, which are shown in Figure 43.2. The BCs give rise to the following equations at the boundary grid points u(0,y) = l ⇒ ui.1 = I, i = 1,2, ... ,n ux (l,y) = 0 ⇒ u, ,n -

(left edge)

u;_,,_ , = 0, i = 1,2, ... ,n (right edge)

u(x,0) = 0 ⇒ u1.1 = 0, j = 1,2, ... ,n

(bottom edge)

u(x,I) =0 ⇒ un.J =0, j =1,2, ... ,n

(top edge)

43. MONTE CARLO SOLUTION OF PARTIAL DIFFERENTIAL EQ1JATIONS 263

u(x,1) =0

11--------~

u(O,y)= 1

uxx+ uyy =0

ux(l,y) =0

0.__________ 1 0 u(x,O) =0

....1....._~

X

Figure 43. J: Laplace's equation inside a square

X-+

u 1J =0, j=l,2, ... ,n

O

0

• • • •

y

i u,_ =I, i =1,2,...,n

• •

•••

. ..

1

ui,n -

ui ,n- 1 =

i = 1,2, ... ,n

1 u•.1 =O,j =I, 2, ... , n

Figure 43.2: Grid of points for the unit square

0'

264

Solution Manual

Note that the derivative BC u_,( l,y) = 0 becomes u i,n - U; _,,_ 1 = 0 or u;_,, = U;_,,_1 which means that the wino arriving at the boundary grid point (i,n) has the same reward as it had the previous step at (i,n-1). Hence, this gives us the strategy at boundaries with derivative BCs; when the wino hits such a boundary, like the right edge at x = I in this problem, it simply bounces back to the interior. The wino may bounce back to the boundary, but eventually it will move on to other interior points, eventually arriving at one with fixed BCs. 5. Derive the Monte Carlo game for solving the following parabolic IBVP. PDE: u, == uxx - ux - u

0 < x < l, 0 < y < l

u(O,t) = f(t)

BC: u(x,y)= { ( ) () u l,t = g t IC: u(x,0)=¢(x)

O<t<oo

0:s;x:s;l

Solution: The Crank-Nicolson method gives us the finite-difference equation

where the P,s are numbers computed from the coefficients in the PDE and represent the probabilities of moving to the corresponding point. That is

One can modify the BCs of this problem to solve problems with derivative BCs (i.e., flux across the boundaries).

Lesson 44: Calculus of Variations (Euler-Lagrange Equations) 1. Find the minimizing function y( x) of the functional

J[y] =

s~J1 +(:J dx

among all (differentiable) functions y(x) satisfying the BC y( 0) = l, y(l) = 1. What is your interpretation of your answer? What is the value of J[y]?

Solution: We are given the integrand

F(x,y,y')=~ and so the Euler-Lagrange equation is

y' ] -0 F - -dF - - d- [ -==== dx .v· - dx ~ y which implies

⇒ y(x)=Ax+B f+7=c ⇒ y'= ~=A 2 -vl -c

and applying the BCs y( 0) = 0 and y(l) = l, we have

y(x)=x Since J[y] stands for the length of the curve y(x) between two points, the answer is not surprising since it is a straight line. The value of the functional 2, which is the length of the curve for this minimizing function is J[ x] = ✓ (line).

265

266

Solution Manual

2. A vibrating mass attached to a spring has kinetic energy (KE) and potential energy (PE) of 2

KE =.!.m(dy) PE =.!.ky 2 2 dt ' 2 where

m = mass of the object k = spring constant y =position of the mass Hamilton's principle states that at all times the position y(t) of the mass is such that the following integral

f'' [KE -PE ]dt = f'' [.!.m(dy)_.!.1cy 2 ]dt= min J,, 2 dt 2

J,,

is minimized. Show that if Hamilton's principle holds, then the position y(t) of the mass must satisfy the differential equation d2

m-f+ky=O dt

Solution: Differentiating the Lagrangian

I , 2 - ky2 l=KE-P£ =-my 2 with respect to y and y', we have

8l(x,y,y') =~[.!.my' 2 _.!_ky 2 ]=-ky 8y 8y 2 2 8l(x,y,y') =~[.!.my' 2 _.!_ky 2] = my' 8y' 8y' 2 2 Plugging these values into the Euler-Lagrange equations, we get

44. CALCULUS OF VARIATIONS

267

Lv - ; L,. =-ky-; (my')=-ky-my"=O Hence, it is not surprising that we have the undamped harmonic oscillator

mJl +ky=O. 3. Show that the minimizing function ji ( x) of the functional

among all (smooth) functions satisfying the BC y( 0) = 0, y( ;./2) = 1 is y( x) = sinx and evaluate J[ sinx].

Solution: We differentiate

Plugging these in the Euler-Lagrange equation gives

L,, - ~Ly• =-2y-:(2y')=-2 y-2y"=0 or

y" + y=O which has the general solution

y(x) = c, cosx+c2 sinx Plugging this in the BCs y(O)=O, y(;./2)=1 gives y(x)=sinx. The value of the functional is

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Solution Manual

r1r12

1r12

J [y] = Jo sinx dx =-cosxl 0 = 1

4. Derive the Euler-Lagrange equation

a

a

F - - F - - F =0 II ax 11, cy Uy for the functional J[u]=

ff F(x,y,u,ux,uy)dxdy D

Solution: The strategy of finding the minimizing function u( x,y) is to take the derivative of ¢(&) =

J[ u(x,y )+ &17(x,y )]

with respect to & and set the derivative to zero. That is dr/J( &) =~J[u + &lJ ]I d& d& s~o

= d:

ff F( x,y,u + &17,u, + &lJ,,u,, + &lJ.v )dxdy D

=ff (aF + oF + oF )dxd au '7 aux lJx au,. y l)

=ff

D

'7y

[a~_ j___l a~ J -.£(a~)] 17( x,y )dxdy au ax au, cy GUY

=0 where the last step in the above equations is simply integration by parts, i.e., taking the derivatives off lJx and lJy and putting them on

44. CALCULUS OF VARIATIONS

8F and

8F

au,

ouy

269

respectively. But in order that the above integral is zero for an arbitrary function '7( x,y), we have the other factor in the integrand to be zero. That is

Lesson 45: Variational Methods for Solving PDEs (Method of Ritz) I. What would be the energy functional J[ u] corresponding to the following problem?

PDE: u xx + UY.)' =1

0 < x < 1, 0 < y < l

BC: u(x,y)=O on the boundary

Solution: In Lesson 44 we saw that the Euler-Lagrange equation for the functional

Jf F( x,y,u,ux,uy )dxdy

J[ u] =

D

was

a

a

F - - F - - F =0 II ax U,: 8y u~-

The goal in this problem is to go backward, i.e., find the functional whose Euler-Lagrange equation is the given PDE

There really isn't a simple procedure for finding the energy functional of a PDE, but by trial and error

tfJ

J[u] =

u_; +u~ +2u] dxdy

we get F =u X2 +u y2 +u FII =2 F11_, =2u X F =2u y " i

which gives the Euler-Lagrange equation 270

45. VARIATIONAL METHODS FOR SOLVING PDES

271

F -~F -~F =2-~(2u )-~(2u) U

ox

11 ,

cy OX cy u, =2 - 2( Uxx + U>,,) X

y

=0 or the given Poisson equation u xx +uY>'

=I

2. How could we minimize the following functional by the method of Ritz?

Hint: Since the function y(t) does not have zero BC we introduce the new function z( x) defined by

z( x) = (1-x )y(x) which converts the BCs to z( 0) = z(l) = 0

Solution: We could Jet

z( x) = (1-x )y( x) which gives

z'( x) = -y(x) + (1-x )y'( x) and solve for y and y' , then plug these values in the original functional J[y] getting

J[z]=

s:[(,~J'

+[,:'x + (1:x)' ndx

Since z( 0) = z(l) = 0. After we find z we find y( x) from

272

Solution Manual

-

z

y=-

1-x

The point is we can use the Method of Ritz since it satisfies the BC z(O)=z( l)=O'.

We could also replace the function y( x) in the functional J[y] with the approximation

which results in the functional taking the form

The minimum could then be found by solving for the coefficients ak, k= 1,2, ... ,n from the equations

a¢ = 0, k = 1,2, ... ,n

-

3ak

I3. Write a computer program to carry out the computations in Figure 43.1.

Solution: Student project 4. Show that the Euler-Lagrange equation for the functional

is

Solution: The proof is similar to the proof in Lesson 44 when we found the Euler-Lagrange equation for a function y( x) of one variable. In this case we consider the function of r:: defined by

45. VARIATIONAL METHODS FOR SOLVING PDES

273

where u(x,y) is the minimizing function; i.e., the function that minimizes the functional J [u], and 17( x, y) is a small arbitrary (differentiable) variation around u which is zero on the boundaries of the square. The minimizing function ii can be found by setting the derivative (with respect to & ) of r/J( & ) to zero at 6 = 0 and then solving for ii. Doing this gives drp( c) = ~J[ii + 617 ]I

de

de

=: 6

t

c=O

fJF(x,y,u+B17 ,ii,+B17x, u" +B17,,)]dxdy

= Jo f 1 r1[ 0~77(x)+ a~ 1lx(x)+ a~ 17,.(x)]dxdy Jo au au au .l'

=

)'

•

f f[aO~ U 0

0

_!!__( o~)-!!_[ o~J:11(x,y)dxdy=O dx OUx dy au,

Now, if the above integral is zero for an arbitrary function 77( x,y), the function ii must satisfy

Hence, if u(x,y) minimizes the functional

J[ u] = f~ f~ F( x,y,u,ux, u,,) dxdy then u(x,y) must satisfy the PDE (Euler-Lagrange equation)

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Solution Manual

In the current problem, the functional is

and so

hence the Euler-Lagrange equation is

oF _!!.._( oF J_!!.._[ oF) = _!!.._(luJ-!!.._( lu .)

OU

dx au,

dy

ouy

dx

dy

)

= 2uxx - 2u!Y = 0 or Laplace's equation

5. The following Dirichlet problem can be solved PDE: uxr +u»,=sin(1rx)

0<x<l, 0<y<l

BC: u(x,y) = 0 on the boundary of the square can be solved by the finite sine transform (transforming the x variable) and has the solution

u(x,y)=[ 0.06e"Y +0.04e-"Y - : 2 }in(1rx) How would you find the potential energy of this system?

Solution: The finite sine transform is ODE BCs

n=I n=2,3, ...

45 . VARIATIONAL METHODS FOR SOLVING PDES

275

which has the solution

ul (y) = Ae"Y + Be-n:y --;1[

Un(y)=O, n=2,3, ...

We can now take the inverse sine transform of these functions to get u ( x, y) after which the potential energy is

J[ u] =

t t [u; + u_;, + 2usin(.7rx )]

dxdy

Most likely this integral would have to be evaluated numerically.

Lesson 46: Perturbation Methods for Solving PDEs 1. Substitute

into theBVP 1 1 PDE: u,, +-u,.+ 2 u00 = 0 r r

0 < r < 1 + &sin0

BC: u(1,0)+u,(l,0)(&sin0)+u,,.(l,0)( &s:eJ +···=cos0 to find a sequence of problems .f'o,~,Pi_, ... .

Solution: We use only the first two terms u = u0 + t:it1 in the perturbation series since the calculations become long and tedious, although the ideas are the same. Substituting this sum into the PDE and first two terms of the BC expansion gives PDE: BC:

Now, setting the coefficients of 1 and & equal to each other on each side of the equations, we find the problems Pa and Fi, respectively

276

46. PERTURBATION METHODS FOR SOLVING PDES

277

Note the solution to Po, which is u0 (r,0)==rcos0, is used in problem Pi· So problem Pi becomes

which has the solution

u1 (r,0) == _.!..r 2 sin (20) 2

Hence, the first two terms in the perturbation expansion ( & == 1) are

2. Show that the following nonlinear problem 2 1 l PDE: u" +-u, +2 u00 + u == 0 0 < r < 1 r r BC: u(l,0) == cos0 0 s;, 0 < 2n

discussed in the text gives rise to the sequence of linear problems

Solution: We seek a solution of the form

278

Solution Manual

and plugging this into the PDE and BC gives 1 (Uo +cu1 +c 2U2 )w +& ( llo +cu1 +c 2!12 )2 = 0 ( uo +&u1 +t: 2!12 )" +-;::I (!lo +&u1+c 2ll2 )r + 7 u0 (I. 0) + cu 1 (1,0) + c 2u2 (1.0) = cos0

Now, setting the coefficients of ),&and £ 2 equal to each other on each side of the equations, we find the problems f'o, ~ and Pi, respectively,

Although problems Fi and Pi are linear they are not homogeneous and not trivial. However, they can be solved and, when done, we will have a good analytical approximation

46. PERTURBATION METHODS FOR SOLVING POES

279

3. To check for accuracy in our perturbation approximation, substitute u(r,0) =u0 +u 1

(r4

1) (r

4

r2 )

= r cos0- - - - -'----"-cos( 20) 24 32 in the nonlinear problem

PDE: V 2u + u 2 = 0

0<r <I

BC: u(I,0) = cos0 0 ~ 0 < 21r

Solution: lfwe substitute

into 'v 2 u + u2 and u(l,0), we will see that u(l,0) = 0 and that the function V 2 u + u 2 is very small inside the unit circle O< r < 1. The term r 2 will contain nine terms but can be evaluated nevertheless.

280

Solution Manual

4. Solve the problem

in the boundary perturbation problem discussed in the book to see how well the approximation

1

u(r,0) =u0 (r,0) +-u 1(r,0) 4 satisfies PDE: V 2 u =0

0 <r <I

BC: u(i+¼sin0,0)=co s0 Os0<21r

Solution: We begin by making the transformation u(r,0) = sin0cos0 + U(r,0) to a new function U ( x,0) which will satisfy the new problem

I

PDE: V2U= 2 O<r<1 r BC: U(I,0)=O Os0<21r We now solve this problem by letting

where U11 (r,0) is the general solution of the homogeneous equation V2 U = 0 and UP (r ,0) is any single solution (called a particular solution) of the nonhomogeneous equation V 2U = I/r 2 • A general form of the solution of the homogeneous equation V2 U = 0 (Laplace's equation) is

46. PERTURBAT ION METHODS FOR SOLVING PDES

u h

281

(r,0) = i[ an cos( n0) + b,, sin( n0)] n= I

To find a particular solution, we try a solution of the fonn U" (r,0) = Acos(20) and find A = -1 / 4. Hence, we have

U(r,0) = U"(r ,0) + U" (r,0) ,,,

I

- 1

4

=zI ancos(n0)+ b,,sin(n0)]- -cos(20) If we now plug this value into the BC U ( I, 0) = 0 we get a2 = 1 / 4 and all the other a;, b; are zero. Hence, we have the solution

1 4

I 4

U(r,0) = -r 2 cos(20)--c os(20) or

u(r,0) = sin0cos0 + _!_r 2 cos(20) - _!_cos(20) 4

4

One would now have to use a computer to evaluate this function to see how well it satisfies the BC

Lesson 47: Conformal Mapping Solutions of PD Es

1. Where is the following mapping conformal? z-1] w=u+iv= log [ z+I Convince yourself that the mapping maps the upper half z plane into the strip -oo < u < oo, 0 < v < 1t in the w plane.

Solution: We have

z-1] dw 2 w=log [ -. ⇒ -=-z+I dz z 2 - 1 and so the mapping is conformal except at z = ±1. Now, from the fact that

w = logz = log(re'0 ) = logr + loge' 0 = logr + i01oge =1ogr+i0 =loglzl+iarg(z ) hence, we have z-1] 1z-ll +iarg [z-1] w=log [ - =log--z+I z+I z+I We see that the real part

l

z-11 u=log-z+I can get as close to -oo as we wish by letting z ➔ I and as close to +oo as we wish by letting z ➔ -1. Hence, --oo < u < oo. On the other hand, the complex part 282

47. CO NFORMAL MAPPING SOLUTIONS OF POES

283

w=arg [ -z-1] =arg(z-1)- arg(z+l) z+ 1 Now, if you look carefully at the drawing in Figure 47.1 you will see that for a complex number in the upper half complex plane that the argument (or angle) of z - I (angle between z = I and the positive x axis) will be equal or larger than the argument of z + I (angle between z = - I and the positive x axis). If you look carefully, the maximum this difference will be is when z lies on the positive complex axis and gets close to the origin we see that arg(z-l) ➔ 1r and arg(z+l) ➔ O and so arg( z -1)-arg( z + 1) ➔ 1l" y

z

-1 Figure 4 7. I : Graph showing angles of z + l and z - I

2 2. What is the image of the first quadrant under the mapping w = z ? It form polar in might be helpful to write the complex numbers

So~ution: Write

and so the first quadrant maps into quadrants 1, 2, and 3.

284

Solution Manual

3. Solve the following Dirichlet problem in the first quadrant by means of the conformal mapping w = z 2 .

PDE: fP.u + <p)'Y = 0

¢ ( x,O ) ={ BCs:

¢ ( O,y ) ={

0 < X < OO, 0 < Y < 00 l O<x<l 0 l<x<oo 1 O<y<l Ol<y<oo

Solution: Applying the transformation

w = z 2 = ( x + iy )2 = ( x 2 - y 1) + i ( 2.xy) or in terms of two real transformations

u = x2 - y2 v=2xy The first quadrant is mapped into the upper half plane and the BCs mapped onto the real axis PDE: lp1111 + fPvv = 0

- 00 < U < oo, 0 < V < 00

¢(u,O) =0 -oo < u < -1 BCs: { ¢(u,0)=1 -1:S:u:S:l ¢(u,0)=0

l<u<oo

This problem has the solution ] - I( u ( u,v ) =-tan JT

or in terms of x,y

2v2 ) 2 u+v-1

47. CONFORMA L MAPPING SOLUTIONS OF PDES

285

4xy _1[ - - - 1 u ( x,y ) =-tan 2- - - - ] ( x 2 - y2) + 4x 2y2 - 1 ,r 4. Solve the following mixed Neumann problem inside a 45° wedge.

Hint: The complex function

f(z) = logz = loglzJ+iarg( z) maps the ray 0 = c1 in the z complex plane to the line v = c1 in the w complex plane. It also maps the circle r = c2 in the z plane to the straight line u = logc2 in the w plane, where log means the natural logarithm. The diagrams in the book might be helpful for visualizing this problem.

Solution: Jf we define the transformation that maps z = x + iy to w = u + iv by w = logz = loglzl + iarg( z) or equivalently as the two real transformations u = loglzl = logr

v = arg( z) = tan - i (y / x) we have that the transformation maps the 45 degree wedge in the problem to the region drawn in Figure 47.2.

286

Solution Manual

V

8 = ,r/4

;, =0 ♦

+

-----1,r/4

:r

Figure 47.2: Conformal mapping from a wedge to an infinite strip

It is clear that the transformed problem has the solution

4 (6(u,v)=-v tr

or in terms of the original coordinates x, y we have

A CATALOG OF SELECTED

DOVER BOOKS IN SCIENCE AND MATHEMAT ICS

OJ

CATALOG OF DOVER BOOKS

Mathemati cs-Bestselle rs HANDBOOK OF MATHEMATICAL FUNCTIONS: with Formulas, Graphs, and Mathematical Tables, Edited by Milton Abramowitz and Irene A. Stegtm. A classic resource for working with special functions, standard trig, and exponential logarithmic definitions and extensions, it features 29 sets of tables, some to as high as 20 places. 1046pp.

8 X 10 1/2.

0-486-61272-4 ABSTRACT AND CONCRETE CATEGORIES: The joy of Cats,jiri Adamek, Horst Herrlich, and George E. Strecker. This up-to-date introductory treatment employs category theory to explore the theory of structures. Its unique approach stresses concrete categories and presents a systematic view of factorization structures. Numerous examples.

1990 edition, updated 2004. 528pp. 6 1/8 x 9 1/4.

0-486-46934-4 MATHEMATICS: Its Content, Methods and Meaning, A. D. Aleksandrov, A. N. Kolmogorov, and M. A. Lavrent'ev. Major survey offers comprehensive, coherent discussions of analytic geometry, algebra, differential equations, calculus of variations, functions of a complex variable, prime numbers, linear and non-Euclidean geometry, topology, functional analysis, more. 1963 edition. 1120pp. 5 3/8 x 8 1/2.

0-486-40916-3 INTRODUCTION TO VECTORS AND TENSORS: Second Edition-Two Volumes Bound as One, Ray M. Bowen and C.-C. Wang. Convenient single-volume compilation of two texts offers both introduction and in-depth survey. Geared toward engineering and science students rather than mathematicians, it focuses on physics and engineering applications. 1976 edition. 560pp. 6 1/2 x 9 1/4.

0-486-46914-X AN INTRODUCTION TO ORTHOGONAL POLYNOMIALS, Theodore S. Chihara.

Concise introduction covers general elementary theory, including the representation theorem and distribution functions, continued fractions and chain sequences, the recurrence formula, special functions, and some specific systems. 1978 edition. 272pp. 5 3/8 x 8 1/2.

0-486-47929-3 ADVANCED MATHEMATICS FOR ENGINEERS AND SCIENTISTS, Paul DuChateau. This primary text and supplemental reference focuses on linear algebra, calculus, and ordinary differential equations. Additional topics include partial differential equations and approximation methods. Includes solved problems. 1992 edition. 400pp.

7 1/2 X 9 1/4.

0-486-47930-7 PARTIAL DIFFERENTIAL EQUATIONS FOR SCIENTISTS AND ENGINEERS, StanleyJ. Farlow. Practical text shows how to formulate and solve partial differential equations. Coverage of diffusion-type problems, hyperbolic-type problems, elliptic-type problems, numerical and approximate methods. Solution guide available upon request. 1982 edition. 414pp. 6 1/8 x 9 1/4.

0-486-67620-X VARIATIONAL PRINCIPLES AND FREE-BOUNDAR Y PROBLEMS, Avner Friedman. Advanced graduate-level text examines variational methods in partial differential equations and illustrates their applications to free-boundary problems. Features detailed statements of standard theory of elliptic and parabolic operators. 1982 edition.

720pp. 6 1/8 x 9 1/4.

0-486-47853-X LINEAR ANALYSIS AND REPRESENTATIO N THEORY, Steven A. Gaal. Unified treatment covers topics from the theory of operators and operator algebras on Hilbert spaces; integration and representation theory for topological groups; and the theory of Lie algebras, Lie groups, and transform groups. 1973 edition. 704pp. 6 1/8 x 9 1/4.

0-486-47851-3

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A SURVEY OF INDUSTRIAL MATHEMATICS, Charles R. MacCluer. Students learn how to solve problems they'll encounter in their professional lives with this concise single-volume treatment. It employs MATLAB and other strategies to explore 0-486-47702-9 typical industrial problems. 2000 edition. 384pp. .5 3/8 x 8 1/2.

NUMBER SYSTEMS AND THE FOUNDATIONS OF ANALYSIS, Elliott Mendelson. Geared toward undergmduate and beginning graduate students, this study explores natural numbers, integers, rational numbers, real numbers, and complex numbers. Numerous exercises and appendixes supplement the text. 1973 edition.

0-486-4.5792-3 368pp. 5 3/8 X 8 1/2.

A FIRST LOOK AT NUMERICAL FUNCTIONAL ANALYSIS, W. W. Sawyer.

Text by renowned educator shows how problems in numerical analysis lead to concepts of functional analysis. Topics include Banach and Hilbert spaces, contraction mappings, convergence, differentiation and integration, and Euclidean space. 1978 edj0-486-47882-3 tion. 208pp . .5 3/8 x 8 1/2.

FRACTALS, CHAOS, POWER LAWS: Minutes from an Infinite Paradise, Manfred Schroeder. A fascinating exploration of the connections between chaos theory, physics, biology, and mathematics, this book abounds in award-winning computer graphics, optical illusions, and games that clarify memorable insights into self-similarity. 1992 0-486-47204-3 edition. 448pp. 6 1/8 x 9 1/4.

SET THEORY AND THE CONTINUUM PROBLEM, Raymond M. Smullyan and Melvin Fitting. A lucid, elegant, and complete survey of set theory, this three-part treatment explores axiomatic set theory, the consistency of the continuum hypothesis, and 0-486-47484-4 forcing and independence results. 1996 edition. 336pp. 6 x 9.

DYNAMICAL SYSTEMS, Shlomo Sternberg. A pioneer in the field of dynamical systems discusses one-dimensional dynamics, rufferential equations, random walks, iterated function systems, symbolic dynarnks, and Markov chruns. Supplementary materials include PowerPoint slides and MATLAB exercises. 2010 edition. 272pp.

0-486-4770.5-3 6 1/8 X 9 1/4.

ORDINARY DIFFERENTIAL EQUATIONS, Morris Tenenbaum and Harry Pollard. Skillfully organized introductory text examines origin of differential equations, then defines basic terms and outlines general solution of a differential equation.

Explores integrating factors; dilution and accretion problems; Laplace Transforms; 0-486-64940-7 Newton's Interpolation Formulas, more. 818pp . .5 3/8 x 8 1/2.

MATROID THEORY, D. J. A. Welsh. Text by a noted expert describes standard examples and investigation results, using elementary proofs to develop basic matroid properties before advancing to a more sophlsticated treatment. Includes numerous 0-486-47439-9 exercises. 1976 erution. 448pp . .5 3/8 x 8 1/2.

THE CONCEPT OF A RIEMANN SURFACE, Hermann Wey!. This classic on the general history of functions combines function theory and geometry, forming the basis of the modern approach to analysis, geometry, and topology. 19.5.5 edition. 208pp.

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THE LAPLACE TRANSFORM, David Vernon Widder. This volume focuses on the Laplace and Stieltjes transforms, offering a hlghly theoretical treatment. Topics include fundamental formulas, the moment problem, monotonic functions, and Tauberian 0-486-477.55-X theorems. 1941 erution. 416pp . .5 3/ 8 x 8 1/2.

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Mathematics-A lgebra and Calculus VECTOR CALCULUS, Peter Baxandall and Hans Liebeck. This introductory text offers a rigorous, comprehensive treatment. Classical theorems of vector calculus are amply illustrated with figures, worked examples, physical applications, and exercises with hints and answers. 1986 edition. 560pp. 5 3/8 x 8 1/2.

0-486-46620-5 ADVANCED CALCULUS: An Introduction to Classical Analysis, Louis Brand. A course in analysis that focuses on the functions of a real variable, this text introduces the basic concepts in their simplest setting and illustrates its teachlngs with numerous examples, theorems, and proofs. 1955 edition. 592pp. 5 3/ 8 x 8 1/2. 0-486-44548-8 ADVANCED CALCULUS, Avner Friedman. Intended for students who have already completed a one-year course in elementary calculus, this two-part treatment advances from functions of one variable to those of several variables. Solutions. 1971 edition.

432pp. 5 3/8 X 8 1/2.

0-486-45795-8 METHODS OF MATHEMATICS APPLIED TO CALCULUS, PROBABILITY, AND STATISTICS, Richard W. Hamming. This 4-part treatment begins with algebra and analytic geometry and proceeds to an exploration of the calculus of algebraic functions and transcendental functions and applications. 1985 edition. Includes 310 figures and 18 tables. 880pp. 6 1/2 x 9 1/4.

0-486-43945-3 BASIC ALGEBRA I: Second Edition, Nathan Jacobson. A classic text and standard reference for a generation, this volume covers all undergraduate algebra topics, including groups, rings, modules, Galois theory, polynomials, linear algebra, and associative algebra. 1985 edition. 528pp. 6 1/8 x 9 1/4.

0-486-47189-6 BASIC ALGEBRA II: Second Edition, Nathan Jacobson. This classic text and standard reference comprises all subjects of a first-year graduate-level course, including in-depth coverage of groups and polynomials and extensive use of categories and functors. 1989 edition. 704pp. 6 1/8 x 9 1/4.

0-486-47187-X CALCULUS: An Intuitive and Physical Approach (Second Edition), Morris Kline.

Application-oriented introduction relates the subject as closely as possible to science with explorations of the derivative; differentiation and integration of the powers of x; theorems on differentiation, antidifferentiation; the chain rule; trigonometric functions; more. Examples. 1967 edition. 960pp. 6 1/2 x 9 1/4.

0-486-40453-6 ABSTRACT ALGEBRA AND SOLUTION BY RADICALS, John E. Maxfield and Margaret W. Maxfield. Accessible advanced undergraduate-level text starts with groups, rings, fields, and polynomials and advances to Galois theory, radicals and roots of unity, and solution by radicals. Numerous examples, illustrations, exercises, appendixes. 1971 edition. 224pp. 6 1/8 x 9 1/4.

0-486-47723-1 AN INTRODUCTION TO THE THEORY OF LINEAR SPACES, Georgi E. Shilov. Translated by Richard A. Silverman. Introductory treatment offers a clear exposition of algebra, geometry, and analysis as parts of an integrated whole rather than separate subjects. Numerous examples illustrate many different fields, and problems include hints or answers. 1961 edition. 320pp. 5 3/8 x 8 1/2.

0-486-63070-6 LINEAR ALGEBRA, Georgi E. Shilov. Covers determinants, linear spaces, systems of linear equations, linear functions of a vector argument, coordinate transformations, the canonical form of the matrix of a linear operator, bilinear and quadratic forms, and more. 387pp. 5 3/ 8 x 8 1/2.

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Mathematics-Probab ility and Statistics BASIC PROBABILITY THEORY, Robert B. Ash. This text emphasizes the probabilistic way of thinking, rather than measure-theoretic concepts. Geared toward advanced undergraduates and graduate students, it features solutions to some of the prob0-486-46628-0 lems. 1970 edition. 352pp. 5 3/ 8 x 8 1/2.

PRINCIPLES OF STATISTICS, M. G. Bulmer. Concise description of classical statistics, from basic dice probabilities to modern regression analysis. Equal stress on theory and applications. Moderate difficulty; only basic calculus required. Includes problems 0-486-63760-3 with answers. 252pp. 5 518 x 8 1/4.

OUTLINE OF BASIC STATISTICS: Dictionary and Formulas, John E. Freund and Frank J. Williams. Handy guide includes a 70-page outline of essential statistical formulas covering grouped and ungrouped data, finite populations, probability, and more, plus over 1,000 clear, concise defirutions of statistical terms. 1966 edition. 208pp.

0-486-47769-X 5 3/ 8 x 8 1/2.

GOOD THINKING: The Foundations of Probability and Its Applications, Irving].

Good. This in-depth treatment of probability theory by a famous British statistician explores Keynesian principles and surveys such topics as Bayesian rationality, corroboration, hypothesis testing, and mathematical tools for induction and simplicity.

0-486-47438-0 1983 edition. 352pp. 5 3/ 8 x 8 1/2.

INTRODUCTION TO PROBABILITY THEORY WITH CONTEMPORARY APPLICATIONS, Lester L. Helms. Extensive discussions and clear examples, written in plain language, expose students to the rules and methods of probability. Exercises foster problem-solving skills, and all problems feature step-by-step solutions. 1997 edi0-486-47418-6 tion. 368pp. 6 1/ 2 x 9 1/4.

CHANCE, LUCK, AND STATISTICS, Horace C. Levinson. In simple, non-technical language, this volume explores the fundamentals governing chance and applies them to sports, government, and business. "Clear and lively ... remarkably accurate."

0-486-41997-5 - Scientific Monthly. 384pp. 5 3/ 8 x 8 1/2.

FIFTY CHALLENGING PROBLEMS IN PROBABILITY WITH SOLUTIONS, Frederick Mosteller. Remarkable puzzlers, graded in difficulty, illustrate elementary and advanced aspects of probability. These problems were selected for originality, general interest, or because they demonstrate valuable techniques. Also includes detailed 0-486-65355-2 solutions. 88pp. 5 3/ 8 x 8 1/2.

EXPERIMENTAL STATISTICS, Mary Gibbons Natrella. A handbook for those seeking engineering information and quantitative data for designing, developing, constructing, and testing equipment. Covers the planning of experiments, the analyzing of extreme-value data; and more. 1966 edition. Index. Includes 52 figures and 76 tables.

0-486-43937-2 560pp. 8 3/ 8 X 11.

STOCHASTIC MODELING: Analysis and Simulation, Barry L. Nelson. Coherent introduction to techniques also offers a guide to the mathematical, numerical, and simulation tools of systems analysis. Includes formulation of models, analysis, and in0-486-47770-3 terpretation of results. 1995 edition. 336pp. 6 1/ 8 x 9 1/4.

INTRODUCTION TO BIOSTATISTICS: Second Edition, Robert R. Sokal and F. James Rohlf. Suitable for undergraduates with a minimal background in mathematics, this introduction ranges from descriptive statistics to fundamental distributions and the testing of hypotheses. Includes numerous worked-out problems and examples. 1987 0-486-46961 -l edition. 384pp. 6 1/ 8 x 9 1/4.

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