Solution Manual for Advanced Mathematics for Engineering Students The Essential Toolbox 1st Edition

Solution Manual Brent J. Lewis, E. Nihan Onder and Andrew A. Prudil June 2020

Copyright c 2022 Brent J. Lewis, E. Nihan Onder and Andrew A. Prudil All rights reserved

Problems Chapter 2 2.1 Consider the nonhomogeneous system of first-order, linear differential equations y10 = y2 + cosh t and y20 = y1 , with boundary conditions y1 (0) = 0 and y2 (0) = − 21 . Solve this initial value problem by the following three methods: (a) Matrix methods. Use the method of variation of parameters to determine the particular solution. (b) Laplace transform methods. (c) Convert the two first-order differential equations into a single secondorder differential equation and solve this latter equation.

Solution (a) y10 =0 · y1 + 1 · y2 + cosh t y20 =1 · y1 + 0 · y2 + 0

0 cosh t y1 (0) 0 0 1 y1 y1 + and = = .

Therefore, 0 y2 (0) − 21 y20 1 0 y2 0 1 Homogeneous equation: y 0 = y⇒ 1 0 −λ 1 Characteristic equation: det(A − λI) = = λ2 − 1 = 0.

1 −λ Therefore, λ2 = 1 ⇒ Eigenvalues: λ1 = +1 and λ2 = −1.

The eigenvectors are obtained from: −λx 1 + x2 = 0.

1 For λ1 = 1 ⇒ x1 = x2 and we can take .

1

1 For λ2 = −1 ⇒ x1 = −x2 and we can take . −1 The homogeneous solution is: y (h) = c1 x(1) et + c2 x(2) e−t , where the eigenvec

1 1 tors are: x(1) = and x(2) = .

1 −1 Method of Variation of Parameters for y (p) : t

e e−t cosh t (1) (2) Y = [y y ] = and g = . et −e−t 0

−t −t −t −t 1 1 1 y −y −e −e e e 22 12 Y −1 = =− = . t t t t −y y −e e e −e detY 2 2 21 11 2-1

Check: Y Y

−1

et e−t et −e−t

1 g= 2

e−t e−t et −et

et +e−t

e−t e−t et −et 1 = 4

=

1 0 0 1

−2t 1 + e 2 Y .

1 + e2t 0   −2t e −1

Z t+  1 t 1 + e−2t̂ 1  2t−2  .

Therefore, u(t) = d t̂ =  e −1 4 0 4 1 + e2t̂ t+ 2  −2t  e −1 t

 t+  1 e e−t (p)  2t−2  y =Yu= t −t   e −1 4 e −e t+ 2   t t −t e e e−t e

−t t 1  te − 2 + 2 + te + 2 − 2  1 t cosh t + sinh t = =  . e−t et et e−t  2 t sinh t 4 t −t te − + − te − + 2 2 2 2 −1

1 = 2

The last expression is obtained after factoring. The general solution is thus:

1 t cosh t + sinh t 1 1 (g) t −t y = c1 e + c2 e + t sinh t 1 −1 2

y1 (0) 0 Now, = so that: y2 (0) − 12

c1 + c2 = 0 y1 (0) 1 1 0 0 1 . which yields: = c1 +c2 + = y2 (0) 1 −1 0 − 12 c1 − c2 = − 2 1 1 Thus, c1 = − and c2 = . Therefore: 4 4 1 t 1 −t t sinh t t y1 (t) = − e + e + cosh t + = cosh t 2 2 | 4 {z 4 } 2 sinh t 2 1 t 1 −t t 1 y2 (t) = − e − e + sinh t = [t sinh t − cosh t] [answer] 2 | 4 {z 4 } 2 cosh t =− 2 =−

2-2

(b) Using a Laplace Transform method: sY1 = Y2 + sY2 +

s s2 − 1

⇒ Y2 = sY1 −

1 1 = Y1 ⇒ sY2 = Y1 − 2 2

s s2 − 1

(2.1)

Multiplying the first equation through by (−s): −sY2 = −s2 Y1 + sY2 = Y1 −

1 2

s2 s2 − 1

Adding the above two equations and solving for Y1 : s2 1 1 Y1 = 2 − 2 2 (s − 1) 2 (s − 1) Taking the inverse Laplace transform of Y1 : sinh t + t cosh t 1 t cosh t y1 (t) = L −1 {Y1 } = − sinh t = 2 2 2

[answer].

Rewriting eq. (2.1): sY1 = Y2 + Y1 = sY2 +

s s2 − 1 1 2

Multiplying the second equation through by (−s): sY1 = Y2 +

s s2 − 1

−sY1 = −s2 Y2 −

s 2

Adding the above two equations and solving for Y2 : s s Y2 = 2 − 2 2 (s − 1) 2(s − 1) Taking the inverse Laplace transform of Y2 : t sinh t 1 y2 (t) = L −1 {Y2 } = − cosh t [answer].

2 2 The solutions for y1 (t) and y2 (t) are identical to those in part (a).

2-3

(c) Given y20 = y1 ⇒ y200 = y10 = y2 + cosh | {z }t. The homogeneous equation is =r

therefore: y200 − y20 = 0 ⇒ characteristic equation: λ2 − 1 = 0 so that λ = ±1.

Homogeneous solution (Case I) : y2h = c1 et + c2 e−t . et e−t Particular solution (Variation of Parameters): W = t = −1 − 1 = e −e−t −2 6= 0 (linearly independent). Therefore: Z h Z h y1 r y2 r (p) h h y2 = −y1 dt + y2 dt W W

t

Z Z e−t et + e−t et e + e−t t −t = −e dt + e dt (−2) 2 (−2) 2

Z Z e−t et et e−2t e−t e2t −2t 2t (1 + e ) dt − (1 + e ) dt = = t− − t+ 4 4 4 2 4 2 t −t t −t te e e te t 1 = − − − = sinh t − cosh t 4 8 8 4 2 4 Hence, the general solution is: 1 t (h) (p) y2 (t) = y2 + y2 = c1 et + c2 e−t + sinh t − cosh t 2 4 Using the boundary conditions: y2 (0) = − 12 ⇒ c1 + c2 − 14 = − 21 ⇒ 1 c1 + c2 = − . Given y20 = y1 ⇒ y20 (0) = y1 (0) = 0. Hence y20 (t) = 4 t 1 1 c1 et − c2 e−t + cosh t + sinh t − sinh t ⇒ y20 (0) = c1 − c2 = 0 ⇒ c1 = c2 .

2 2 4 Thus, c1 = c2 = − 18 .

1 et + e−t t 1 t 1 The solution is: y2 (t) = − + sinh t− cosh t = sinh t − cosh t.

4 2 2 4 2 2 | {z } =cosh t

t 1 1 t Also y1 (t) = y20 (t) = cosh t + sinh t − sinh t = cosh t 2 2 2 2

Same answer as given in parts (a) and (b).

2-4

[answer]

2.2 Consider the following problems: (a) State the differential equation for the Sturm-Liouville problem where r(x) = x, p(x) = x−1 , and q(x) = 0. (b) Using the transformation x = et , show that this differential equation d2 y reduces to: 2 + λy = 0. dt (c) Given the boundary conditions, y(1) = 0 and y(e) = 0 for the differential equation in (a), what are the corresponding eigenvalues and eigenfunctions?

Solution (a) The differential equation for the Sturm-Liouville problem is: 1 (r(x)y 0 )0 + (q(x) + λp(x))y = 0 where r(x) = x, p(x) = and q(x) = 0 ⇒ x

dy d x λ λ dx 0 0 (xy ) + y = 0 or + y = 0. [answer] x dx x dx = et or dx = et dt. Substituting this latter relation into dt

t dy d e t λ e dt + t y = 0. the differential equation yields: t e dt e d2 y Therefore, 2 + λy = 0. [answer] dt (b) Let x = et ⇒

(c) For the boundary conditions y(1) = 0 and y(e) = 0, the solution of the differential equation in part (b) is: y(t) = A cos νt + B sin νt where λ = ν 2 .

But x = et ⇒ t = ln x. Thus, the solution to the differential equation in part (a) is: y(x) = A cos(ν ln x) + B sin(ν ln x).

The boundary condition y(1) = 0 ⇒ y(1) = A = 0. The boundary condition y(e) = 0 ⇒ y(e) = B sin ν = 0 ⇒ ν = ±π, ±2π, ±3π... or ν = nπ for n = 1, 2, ....

Eigenfunctions are: yn (x) = sin(nπ ln |x|) and eigenvalues are: λ = ν 2 = (nπ)2 for n = 1, 2, ... [answer]

2-5

2.3 Using the transformation x = cosh t, show that the following differential dy d2 y equation: 2 +coth t −20y = 0 reduces to Legendre’s differential equation dt dt and give the expression for the Legendre Polynomial which is a solution to this transformed equation.

Solution

dx d2 y d dy If x = cosh t ⇒ dx = sinh tdt. Therefore: dt = and 2 = = sinh t dt dt dt

d dy d dy dy d sinh t d2 y dy sinh t = sinh t + · = sinh2 t 2 + cosh t . Thus, dt dx dt dx dx dt dx dx the differential equation becomes: 2 2 dy cosh t dy dy 2 d y 2 d y sinh t 2 +cosh t + − −20y = 0 or sinh t 2 +2 cosh t

dx dx sinh t (dx/sinh t) dx dx 20y = 0. But cosh2 t − sinh2 t = 1 =⇒ sinh2 t = x2 − 1 so that the differential d2 y dy equation reduces to: (1 − x2 ) 2 − 2x + 20y = 0. This equation is in the dx dx form of Legendre’s differential equation where n(n + 1) = 20 =⇒ n = 4 so that P4 (x) is a solution.

2 X (8 − 2m)! n (−1)m = 2, P4 (x) = x4−2m , Hence with M = 2 16m!(4 − m)!(4 − 2m)! m=0 which yields:

(−1)0 (8 − 0)!x4 (−1)1 (8 − 2)!x4−2 (−1)2 (8 − 4)!x4−4 + + 16 · 0!4!4!

16 · 1!(4 − 1)!(4 − 2)! 16 · 2!(4 − 2)!(4 − 4)!

8!

6!

4! = x4 − x2 + 16 · 4!4!

16 · 1!3!2!

16 · 2!2!0!

7 · 6 · 5 4 30 2 3 8 · 7 · 6 · 5 · 4! 4 6 · 5 · 4! 2 4! · 3 x − x + = = x − x + 2 · 4 · 3 · 2 8 8 (8 · 2) · 4! · 4!

4 · 4! · 2!

4 · 4! · 2

P4 (x) =

Therefore: P4 (x) =

1 [35x4 − 30x2 + 3].

8

2-6

[answer]

2.4 Consider the hypergeometric equation: x(1−x)y 00 +[b−(2+b)x]y 0 −by = 0, where b is a constant larger than unity. Using a Frobenius method, show that the series solution about x = 0 is the geometric series such that y1 (x) = 1 + x + x2 + . . . ..

Solution The expanded hypergeometric equation is: xy 00 −x2 y 00 +by 0 −(2+b)xy 0 −by = 0. Therefore, using a Frobenius method: ∞ ∞ ∞ X X X m+r−1 m+r (m + r)(m + r − 1)am x − (m + r)(m + r − 1)am x +b (m +

m=0

r)am xm+r−1 − (2 + b) or∞ X

∞ X

m=0

(m + r)am xm+r − b

m=0

(m+r)[m+r+b−1]am xm+r−1 −

m=0

∞ X

∞ X

m=0

am xm+r = 0

m=0

{(m+r)[m+r+b+1]+b}am xm+r = 0

m=0

The indicial equation for the coefficient of xr−1 is: r(r + b − 1) = 0 ⇒ r = 0 or r = 1 − b.

Letting m = s + 1 in the first series and m = s in the second series, on relabelling: ∞ ∞ X X r+s {(s + r)[s + r + b + 1] + b}as xr+s = 0 (s + r + 1)[s + r + b]as+1 x − s=−1

Choosing r = 0:

∞ X

s=0

s

(s + 1)[s + b]as+1 x −

s=−1

∞ X s=0

{(s)[s + b + 1] + b} as xs = 0 | {z } =(s+1)(s+b)

For s = −1: (−1 + 1) ·(b − 1) = 0 (Nothing new) | {z } =0

For s = 0, 1, 2...: (s + 1) (s + b)as+1 = (s + 1) (s + b)as or as+1 = as .

Therefore a0 = a1 = a2 = ... so that ∞ X 0 y=x am xm = a0 + a1 x + a2 x2 + ... = a0 [1 + x + x2 + x3 + ...]. m=0

Hence: y1 (x) = 1 + x + x2 + x3 + ... (a0 = 1)

2-7

[answer]

2.5 Find a general solution, in terms of Bessel functions, for the differential equation: 4x2 y 00 + 4xy 0 + (x − n2 )y = 0, where n is an integer by using either the substitution or the generalized form for the solution of the Bessel’s Equation. What is the solution that satisfies the condition |y(0)| < ∞?

Solution

dz 1 1 = √ = . Also z 2 = x and z 4 = x2 . dx 2z 2 x 2 00 Using the chain rule for 4x y + 4xy 0 + (x − n2 )y = 0 yields:

dz dy dz d dy dz · · + 4x · + (x − n2 )y = 0 4x2 dz dz dx dx dz dx Therefore substituting in the previous expressions into the differential equation gives:

4z 4 d dy 1 dy 1 · + 4z 2 · + (z 2 − n2 )y = 0 2z dz dz 2z dz 2z

1 dy dy 1 d2 y 3 − 2 + 2z + (z 2 − n2 )y = 0 which reduces to: 2z 2 2z dz 2z dz dz 2 dy dy ⇒ z 2 2 + z + (z 2 − n2 )y = 0 dz dz The solution√to Bessel’s differential equation is: y(z) = c1 Jn (z) + c2 Yn (z) or √ y(x) = c1 Jn ( x) + c2 Yn ( x). [answer]

Let z = x1/2 ⇒

An alternative derivation can be obtained by comparing to the general equation: x2 y 00 + x(a + 2bxr )y 0 + [c + dx2s − b(1 − a − r)xr + b2 x2r ]y = 0 to the 2 x n given problem: x2 y 00 + xy 0 + − y = 0. Hence on inspection: 4 4 a + 2bxr = 1 ⇒ a = 1, b = 0 n2 1 1 n2 x c + dx2s = − + ⇒ c = − , d = , s = .

4 s4 4 4 2 s

2

2 1 1−a 1−1 n2 Therefore ν = −c = 2 + = n (integer). Also s 2 2 4 √ r d 1 =2 = 1 (real). Therefore, case b) applies: s 4√ √ y(x) = c1 Jn ( x) + c2 Yn ( x). [answer] √ For |y(0)| < ∞ ⇒ c2 = 0 and therefore y(x) = c1 Jn ( x).

2-8

[answer]

2.6 Using Jo for a series expansion, show that the coefficients of the FourierBessel series for f (x) = 1 over the interval 0 ≤ x ≤ 1 are given as am = 2/ {αm0 J1 (αm0 )}, where αm0 are the zeros of J0 .

Solution

Z ∞

am J0 (km0 x) where f (x) = Here the series expansion for f (x) is: f (x) = m=1 αmn 1 for 0 ≤ x ≤ 1. Thus, kmn = ⇒ km0 = αm0 (since R = 1). The R Fourier-Bessel coefficients are therefore given as: Z R 2 am = 2 2 xf (x)Jn (kmn x) dx R Jn+1 (αmn ) 0 Z 1 2 = 2 xJ0 (αm0 x) dx J1 (αm0 ) 0 Letting ξ = αm0 x ⇒ x = ξ/αm0 and dx = dξ/αm0 . Also as x → 0 to 1, ξZ → 0 to αm0 . Thus, the Zintegral can be written as: 1 αm0 1 xJ0 (αm0 x) dx = 2 ξJ0 (ξ) dξ αm0 0 0 d ν [ξ Jν (ξ)] = ξ ν Jν−1 (ξ), the integral Integrating the recurrence relation: dξ Z αm0

on the right hand side can be evaluated as (with ν = 1): 0 αm0  ξJ1 (ξ) . Hence, the Fourier-Bessel coefficient becomes: 0

am =

2

2

1

. · 2 · αm0 J1 (αm0 ) = αm0 J1 (αm0 ) J21 (αm0 ) αm0

2-9

[answer]

ξJ0 (ξ) dξ =

2.7 Consider the function:

0 if − 1 ≤ x ≤ 0 f (x) = 2 if 0 ≤ x ≤ 1

Calculate the first two terms of the Fourier-Legendre series of f (x) over the interval –1 ≤ x ≤ 1. What is the value of the complete Fourier-Legendre series of f (x) at x = 0?

Solution ∞ X f (x) = am Pm (x) = a0 Po (x) + a1 P1 (x) + ..., where P0 (x) = 1, P1 (x) = x m=0 Z 2m + 1 1 and am = f (x)Pm (x) dx.

2 −1 Z Z 1 1 1 1 1 f (x)P0 (x) dx = 2 dx = x = 1.

For m = 0: a0 = 2 −1 2 0 0 Z 1 1 3 3 3 For m = 1: a1 = 2 · x dx = x2 = .

2 0 2 0 2 3 Therefore: f (x) = 1 + x + ... [answer] 2 f (x + 0) + f (x − 0) From the Dirichlet condition: at x = 0, therefore 2 0+2 f (x) = = 1 at x = 0. [answer] 2

2-10

2.8 Consider the differential equation: 16x2 y 00 + 3y = 0. (a) Calculate the two basis functions using a Frobenius method and give the general solution. (b) Derive the second basis function from y1 (x) using a method of reduction of order.

Solution (a) Let y =

∞ X

am xm+r , y 0 =

m=0

∞ X

am (m + r)xm+r−1 and y 00 =

m=0

∞ X

am (m +

m=0

4)(m + r − 1)xm+r−2 . Substituting these expressions into the differential equation yields: ∞ ∞ X X 16 am (m + r)(m + r − 1)xm+r + 3 am xm+r = 0 ⇒

m=0 ∞ X

m=0

m=0

am [16(m + r)(m + r − 1) + 3]xm+r = 0.

Equating the smallest power to zero: a0 [16r(r−1)+3] = 0 ⇒ 16r2 −16r+3 = 0. The roots of pthe latter equation are: √ 16 ± 162 − (4 × 16 × 3) 2± 4−3 2±1 3 1 r1 , r2 = = = = or .

2 × 16 4 4 4 4 ∞ 3 3 X 1 as [16(s + )(s − ) + 3]xs+r = 0 Letting m = s with r1 = : 4 s=0 4 4

3 1 For s = 0 : a0 [16( )(− ) + 3] = 0 (already satisfied) 4 4 ∞ X 3 1 as [16(s + )(s − ) + 3] = 0 ⇒ a1 = a2 = a3 = ... = 0 For s = 1, 2, 3 : 4 {z 4 | } s=0 just a number

Therefore y1 (x) = x

r

∞ X

3

am xm = x 4 [a0 ]

m=0

3

With a0 = 1, the first basis function is: y1 (x) = x 4

∞ 1 X 1 3 With r2 = : as [16(s + )(s − ) + 3]xs+r = 0 4 s=0 4 4

2-11

[answer]

1 3 For s = 0 : a0 [16( )(− ) + 3] = 0 (already satisfied) 4 4 ∞ X 3 1 For s = 1, 2, 3 : as [16(s + )(s − ) + 3] = 0 ⇒ a1 = a2 = a3 = ... = 0 4 {z 4 | } s=0 just a number

Therefore y2 (x) = x

1 4

∞ X

1

am xm = x 4 [a0 ]

m=0

1

With a0 = 1, the second basis function is: y2 (x) = x 4

[answer]

(b) Method of reduction of order for second basis function: R 0 0 3 − p dx Given y 00 + y + y = 0 ⇒ p(x) = 0 so that e = e−0 = 1.

16x2 16x2 1 R 1 3 3 Therefore U = 2 e− p dx = 2 = (x− 4 )2 = x− 2 .

R y1 R − 3 y1 1 1 3 Hence, u = U dx = x 2 dx = −2x− 2 so that y2 = uy1 = (−2x− 2 )(x 4 ) = 1 −2x 4 .

1 Therefore y2 = x 4 and the general solution is: 3

1

y = c1 y1 (x) + c2 y2 (x) = c1 x 4 + c2 x 4 .

[answer]

2-12

2.9 Give the Legendre polynomial Pn (x) which is a solution to the differential 1 equation: (1 − x2 ) y 00 − xy 0 + 3y = 0.

2 Solution Compare (1 − x2 ) y 00 − 2xy 0 + 6y = 0 with (1 − x2 ) y 00 − 2xy 0 + n(n + 1)y = 0 ⇒ n(n + 1) = 6 or n = 2. Therefore: 1 P2 (x) = (3x2 − 1) [answer] 2

2-13

2.10 Find a general solution, in terms of Bessel functions, for the differential d2 y dy − (1 + 4x4 ) y = 0 by using the transformation (i) equation: x2 2 + x dx dx z = x2 and (ii) employing the inspection method for the generalized form of Bessel’s equation.

Solution

√ dz (i) Letting z = x2 ⇒ = 2x = 2 z. Using the chain rule: dx

dy dz dz dy dz 2 d x · +x · − (1 + 4x4 ) y = 0 dz dz dx dx dz dx

√ √ dy √ d dy √ 2 z 2 z + z 2 z − (1 + 4z 2 ) y = 0 ⇒z dz dz dz √ √ d2 y dy 1 dy √ z 2+ ⇒ 4z z + 2z − (1 + 4z 2 ) y = 0 dz dz 2 z dz 2 dy dy ⇒ 4z 2 2 + 4z − (1 + 4z 2 ) y = 0, which implies the modified Bessel equadz dz 1 tion (with ν = ): 2" 2 # 2 d y dy 1 z2 2 + z − z2 + y = 0. dz dz 2 Hence, the solution is: y = c1 I1/2 (z) + c2 I−1/2 (z) or y = c1 I1/2 (x2 ) + c2 I−1/2 (x2 ) [answer] dy d2 y + x − (1 + 4x4 ) y = 0 with the general form: x2 y 00 + 2 dx dx x(a + 2bxr )y 0 + [c + dx2s − b(1 − a − r)xr + b2 x2r ]y = 0, which yields: (ii) Compare x2

a + 2bxr = 1 =⇒ a = 1, b = 0 4 c + dx2s = −(1 s + 4x ) =⇒ d = −4, s = 2 and c = −1. √ √ 2 1 1−a 1 d −4 1√ Thus, ν = 0 + 1 = (6= integer) and −c= = = s 2 2 2 2 2 imaginary. Hence, these two conditions imply case c) where Zν ≡ Iν , Z−ν ≡ I−ν . The solution is therefore the same answer as in part (i): y = c1 I1/2 (1 · x2 ) + c2 I−1/2 (1 · x2 )

[answer]

2-14

2.11 The steady-state temperature distribution for a fin of cross-sectional area A, constant perimeter P , constant conductivity k and length L can be determined

from the following differential equation: dT d kA − hP (T − T∞ ) = 0, dx dx where h is the heat transfer coefficient for the fin surrounded by a fluid with a constant temperature T∞ . (a) Is this differential equation homogeneous or nonhomogeneous? With a change of variable θ = T –T∞ , find the general solution for the variable θ(x) by putting m2 = hP/(kA). (b) Find a solution to the boundary value problem for the boundary conditions: θ(0) = θ0 and θ0 (L) = 0.

Solution

dθ dT (a) The differential equation is homogeneous. Let θ = T − T∞ =⇒ = . dx dx Substituting the differential equation becomes:

in these expressions, dθ d2 θ hP d θ=0 kA − hP θ = 0 or 2 − dx dx dx kA Therefore: d2 θ hP − m2 θ = 0 where m2 = .

2 dx kA The general solution is: θ = A sinh mx + B cosh mx [answer] (b) From the boundary conditions θ(0) = θ0 =⇒ θ(0) = 0 + B = θ0 so that B = θ0 . Therefore θ(x) = A sinh mx + θ0 cosh mx =⇒ θ0 (x) = A m cosh mx + θ0 m sinh mx.

Hence, with θ0 (L) = 0 for the second boundary condition: θ0 (L) = A m cosh mL+ sinh mL θ0 m sinh mL = 0 =⇒ A = −θ0 = −θ0 tanh mL. cosh mL The solution is: θ(x) = θ0 [cosh mx − tanh mL sinh mx]

2-15

[answer]

2.12 Given the periodic function f (x) = x for –2 < x < 2 and f (x + 4) = f (x). (a) Sketch this function over the interval –6 ≤ x ≤ 6. (b) Find the Fourier series of this function.

(c) What is the value of f (x) that one would compute from the Fourier series solution at the discontinuity of x = 2.

Solution (a)

(b) Since the function is odd, therefore the Fourier series will involve just a ∞ X nπ bn sin sine series such that: f (x) = x, where p = 2L = 4 =⇒ L = 2.

L n=1 Z nπx 2 L f (x) sin dx = The Fourier coefficients are evaluated as: bn = L 0 L  2 nπx nπx Z x cos sin 2 2 nπx 4   f (x) sin dx =  2 2 − nπ 2  = − cos nπ. Therefore: nπ 2 0 2 nπ 2 0 2 ∞ ∞ X 4 nπx 4X nπx f (x) = cos nπ sin = (−1)n+1 sin . [answer] − nπ 2 π 2 n=1 n=1 (c) From the Dirichlet condition at the point of discontinuity: f (2 + 0) + f (2 − 0) −2 + 2 = = 0. [answer] f (2) = 2 2

2-16

2.13 Given the differential equation: xy 00 + (1 − x)y 0 + ny = 0. (a) Determine one of the basis solutions for this differential equation, when n = 2, using a Frobenius method. (b) The Laguerre Polynomial Ln (x) is in fact a solution of this differential dn equation as given by the Rodrigue’s formula: Ln (x) = ex n (xn e−x ). Using dx this formula, specify the Laguerre polynomial L2 (x). What is the value of the arbitrary constant in part (a) to obtain the Laguerre Polynomial solution L2 (x)?

Solution 00

0

(a) Given xy + (1 − x)y + 2y = 0. Let y = r)xm+r−1 and y 00 =

∞ X

m=0

∞ X

am x

m+r

0

,y =

∞ X

am (m +

m=0

m=0

am (m + r)(m + r − 1)xm+r−2 . Substituting these

expressions into the differential equation yields: ∞ ∞ ∞ X X X (m+r)am xm+r + (m+r)am xm+r−1 − am (m+r)(m+r−1)xm+r−1 +

m=0 ∞ X

2

am xm+r = 0

m=0 ∞ X

⇒

m=0

m=0

2 m+r−1

am (m + r) x

m=0

∞ X

−

(m + r − 2)am xm+r = 0

m=0

Equating the smallest power to zero: r2 = 0. The roots of the latter equation are: r = 0 (double root) Letting m + r − 1 = r + s =⇒ m = s + 1 in the first expression and m = s in the second expression: ∞ ∞ X X 2 s+r as+1 (s + r + 1) x − (s + r − 2)as xs+r = 0

s=−1

With r = 0 :

∞ X

s=−1

s=0

2 s

as+1 (s + 1) x −

∞ X s=0

(s − 2)as xs = 0

For s = −1 : (−1 + 1)2 a0 = 0 (already satisfied)

For s = 0, 1, 2... : (s + 1)2 as+1 − (s − 2)as = 0 =⇒ as+1 = s = 0 : a1 =

(−2)a0 = −2a0 1 2-17

(s − 2)as (s + 1)2

(−1)a1 a0 1 = − (−2a0 ) = 2 2 4 2 0 · a2 s = 2 : a3 = 2 = 0 3 1 · a3 s = 3 : a4 = 2 = 0 and so on a3 = a4 = a5 ... = 0.

4 s = 1 : a2 =

Therefore y(x) =

∞ X

m=0

[answer]

am xm = a0 + a1 x + a2 x2 = 0 =⇒ y(x) = a0 [1 − 2x +

x2 ] 2

(b) For n = 2 =⇒

2 d d d 2 −x 2 −x x x 2 −x −x x d (x e ) = e (x e ) =e (−x e + 2xe )) = L2 (x) = e dx2 dx dx dx + x2 + 2 − 2x ] = 2 − 4x + x2 x e [−2x e−x e−x e−x e−x

Therefore with a0 = 2 =⇒ y(x) = 2 − 4x + x2 (same answer as part (a)). [answer]

2-18

d2 y 2.14 Consider the differential equation: − xy = 0. Using the transfordx2 √ √ 2 mations y = u x and ix3/2 = z, where i = −1, show that this differential 3 equation reduces to the Bessel’s differential equation:

2 d u 1 du 1 +u 1− = 0. +z 2 dz dz (3z)2 What is the solution for u(z) and y(x)?

Solution √ √ du dy 1 d2 y 1 −1/2 du √ d2 u y = u x =⇒ = x + x−1/2 u and x + x 2 + = dx dx 2 dx2 2 dx dx 1 −1/2 du 1 −3/2 x − x u. Hence, the differential equation becomes: 2 dx 4 2 √ du 1 du 1 −3/2 3/2 u − ux x 2+√ − x | {z } = 0. dx dx 4 x | {z } | {z } | {z } IV I III II

√ √ 2 3/2 dz 2 3 Letting z = ix =⇒ = ·i· x1/2 = i x so that dz = i xdx.

3 dx 3 2 Thus, each term can be transformed as:

√ d du √ √ d2 u √ d du dz · =i x x = I: x 2 = x dx dz dx dx √

dx dz

√ du d x √ d du dz √ 1 du d2 u i du d2 u i x + x · =i x √ + ix 2 = − x3/2 2 . dz dx dz dz √ dx dz 2 dz dz 2 x dz

1 du dz x du 1 du du =√ · = i =i .

II : √ √ dz dz x dx x dz dx x

1 1 2i iu III : − x−3/2 u = − u=− .

4 6z 4 3 z 3 z IV : −ux3/2 = −u .

2i

2 i du du iu 3z 3/2 d u The above expressions yield: −x + i − − u = 0.

2 dz dz 2 dz 6z 2i 2i Multiplying by − , the Bessel differential equation is: 3z

d2 u 1 du 1 1 + +u 1− = 0. Thus, ν = and the solution is: 2 2 dz z dz (3z) 3 √ 2 u(z) = AJ1/3 (z) + BJ−1/3 (z) or using y = u x with z = ix3/2 gives: 3

√ y(x) = x AJ1/3 ( 23 ix3/2 ) + BJ−1/3 ( 32 ix3/2 ) [answer] 2-19

2.15 Determine the particular solution for the nonhomogeneous, secondorder, ordinary differential equation: y 00 +y 0 −2y = 4 sin 2x using the following methods: (a) Undetermined Coefficients (b) Variation of Parameters (c) Fourier Transforms (see problem 3.10).

Solution (a) Let yp = K cos wx + M sin wx =⇒ yp0 = −Kw sin wx + M w cos wx and yp00 = −Kw2 cos wx − M w2 sin wx.

Substituting in these expressions into the differential equation yields: −Kw2 cos wx−M w2 sin wx−Kw sin wx+M w cos wx−2K cos wx−2M sin wx = 4 sin 2x =⇒ (−Kw2 + M w − 2K) cos wx + (−M w2 − Kw − 2M ) sin wx = 4 sin 2x On comparing the left and right hand sides of the equation, yields: −Kw2 + M w − 2K = 0 and −M w2 − Kw − 2M = 4 with w = 2. Hence, 3 1 K = − and M = − . Thus, the particular solution is: 5 5 3 1 yp = − cos 2x − sin 2x [answer] 5 5 Z h Z h y2 r y1 r h h (b) yp (x) = −y1 dx + y2 dx where y1 and y2 are solutions of the W W homogeneous equation, r = 4 sin 2x and W = y1 y20 − y10 y2 . The homogeneous equation y 00 + y 0 − 2y = 0 (with a = 1 and b = −2) yields the characteristic equation: λ2 + λ − 2 = 0. Here a2 − 4b = 1 + 8 = 9 > 0 =⇒ 2 implies real roots: √ √ 1 1 λ1 = (−1 + 9) = 1 and λ2 = (−1 − 9) = −2. Therefore y1 = ex and 2 2 y2 = e−2x and W = −2ex e−2x − ex e−2x = −3e−x .

The particular solution is evaluated Z −2x Z from: e 4 sin 2x ex 4 sin 2x −2x yp = −ex dx + e dx −x −x Z−3e Z−3e 4 4 =⇒ yp = ex e−x sin 2x dx − e−2x e2x sin 2x dx 3 3 Z ax e (a sin bx − b cos bx) . Hence: But eax sin bx dx = 2 a + b2 2-20

" # −x

2x (2 sin 2x − 2 cos 2x) e (− sin 2x − 2 cos 2x) 4 −2x e 4 x

e − e yp = 3 5 3 8

4 − sin 2x − 2 cos 2x −2 sin 2x + 2 cos 2x = + 3 5 8 1 1 = [−18 sin 2x − 6 cos 2x] = [−3 sin 2x − cos 2x] 30 5 Thus, one obtains the same solution as in part (a): 3 1 yp = − cos 2x − sin 2x [answer] 5 5

2-21

Problems Chapter 3 3.1 Find L form.

−1

s 2 (s − a2 )2

using the property for differentiation of a trans-

Solution

Z s s ds dF = 2 Let F (s) = =⇒ F (s) = . Let u = s2 − a2 =⇒ 2 2 2 ds (s − a ) Z (s − a2 )2 du 1 1 1 =− 2 . Using Laplace du = 2s ds. Hence, F (s) = =− 2 2 u 2u 2(s − a2 ) 1 sinh at transform tables: f (t) = − .

2 a

dF Using the property for differentiation of a transform L −1 = −tf (t), ds

t sinh at s = L −1 . [answer] 2 2 2 (s − a ) 2a 0

3-1

3.2 Find L form.

cos at − cos bt t

using the property for integration of a trans-

Solution

Z ∞

s f (t) F (u) du.

Given f (t) = cos ξt =⇒ F (s) = 2 = and L 2 s

s + Z ξ∞ Zt ∞ cos at − cos bt udu udu Therefore L = − . Letting z = t Z u2 Z + a2 u2 + b2 s s ∞ ∞ udu 1 1 1 dz 2 2 2 2 u + ξ =⇒ dz = 2u du, = ln z = ln(u + ξ ) . = u2 + ξ 2 2 z 2 2 s s Hence: " # Z ∞ Z ∞ K K udu udu 1 − = lim ln(u2 + a2 ) − ln(u2 + b2 ) 2 + a2 2 + b2 u u 2 K→∞ s s s s

1 = lim ln(K 2 + a2 ) − ln(s2 + a2 ) − ln(K 2 + b2 ) + ln(s2 + b2 ) 2 K→∞ 2

1 K 2 + a2 1 s + b2 = lim ln + ln 2 2 K→∞ K 2 + b2 2 s + a2 {z } |    2  1 1 + (a/K)

   

lim ln

2 K→∞ 1 + (b/K)2 Therefore: L

cos at − cos bt t

1 = ln 2

2

s + b2 . s 2 + a2

3-2

[answer]

3.3 Find L

−1

1 4 s −1

using the convolution theorem.

Solution

1 1 1 −1 −1 L =L 4 s4 − a s 2 − a2 s 2 + a2

1 1 =⇒ f (t) = sinh at If F (s) = 2 2 a s − a 1 1 If G(s) = =⇒ g(t) = sin at s 2 + a2 a Thus, using the convolution theorem: L

−1

Z t 1 {F (s)G(s)} = 2 sinh a(t − u) sin au du a 0

Z t Z t 1 −au −at au at = 2 e e sin au du − e e sin au du 2a 0 0

1 = 4 eat {e−at (−a sin at − a cos at) − 1(0 − a)} − e−at {eat (a sin at − a cos at) − 1(0 − a)} 4a 1 ( − ae−at ( + aeat − a sin at + ( ((at ((at a cos a cos = 4 −a sin at − ( 4a 1 = 3 [sinh at − sin at] 2a

Thus, with a = 1 =⇒ L

−1

1 4 s −1

=

3-3

1 [sinh t − sin at] .

2

[answer]

e−πs s 3.4 Find the inverse Laplace transform L (s2 + 1) shifting theorem, and sketch the resultant function. −1

Solution

e−πs s Using the second shifting theorem, L (s2 + 1)

But cos(t − π) = cos t cos sin π = − cos t. | {zπ} + sin t −1

=−1

Therefore f (t) = −U (t − π) cos t.

[answer]

3-4

using the second

= U (t − π) cos(t − π).

3.5 Find the inverse Laplace transform L for the differentiation of a transform.

−1

1 2 (s − 1)

using the property

Solution Using the property for differentiation of a transform L −1 {F 0 (s)} = −tf (t).

Z dF ds 1 1 s−1 = 2 =⇒ F (s) = = ln .

Therefore, F 0 (s) = 2 ds s −1 s − 1 2 s+ 1

1 s−1 1 e−t − et Hence, f (t) = L −1 {F (s)} = L −1 ln = .

2 s+1 2 t t

e − e−t Thus, −tf (t) = = sinh t.

2

1 −1 The solution is therefore: L = sinh t. [answer] (s2 − 1)

3-5

s Γ(2) using the 3.6 Find the inverse Laplace transform L s2 s2 + 1 convolution theorem where Γ is the gamma function. −1

Solution

L −1

     Γ(2)

    

s = t ∗ cos t  s2 s2 + 1     | {z } | {z }  F (s) G(s) Z t (t − u) cos u du = 0 Z t Z t =t cos u du − u cos u du 0

0

t

t

= t sin u − (cos u + u sin u) 0

Therefore the solution is: L

−1

0

t − [cost + t − 1] = t sin t sin

Γ(2) s2

3-6

s 2 s +1

= 1 − cos t.

[answer]

3.7 Consider the function:

0 if − 1 ≤ x ≤ 0 f (x) = 2 if 0 ≤ x ≤ 1

(a) Sketch this function and write it in terms of Heaviside step functions. (b) What is the Laplace transform of f (x)?

Solution (a)

e−as , therefore: s L {f (x)} = L {2U (x − 0)} − L {2U (x − 1)} . (b) Given that L {U (x − a)} =

Hence, the solution is: e−0·s e−s 2 L {f (x)} = 2 −2 = [1 − e−s ] . s s s

3-7

[answer]

3.8 Find the Fourier Transform F {f (x)} by direct integration where f (x) ex if x < 0 is given by: f (x) = 0 if x > 0.

Solution 1 F (w) = √ 2π

Z ∞

−∞

−iwx

f (x) e

1 dx = √ 2π

Z 0

0

1 ex−iwx ex e−iwx dx = √ 2π 1 − iw −∞ −∞

2

−iwx Note: e−iwx = cos x − | = cos2 wx + sin2 wx = 1 (where by √ i sin x =⇒ |e definition |a + ib| = a2 + b2 ) and lim ex = 0. x→−∞

Thus, lim ex[1−iw] = 0 and the solution is: x→−∞ 1 1 F (w) = √ [answer] 2π 1 − iw

3-8

3.9 Show that the Fourier Sine Transform pair as given in mathematical handbooks (for example, [Spiegel, 1973]): r 2 Γ(a) aπ sin f (x) = xa−1 (0 < a < 1) and F (w) ≡ F {f (x)} = a π w 2 reduces to the following transform pair for an appropriate choice of a: 1 1 f (x) = √ and F (w) ≡ F {f (x)} = √ . x w Solution 1 Let a = 1/2 =⇒ f (x) = √ . Hence F {f (x)} = x √ 1 π Γ( 12 ) = π and sin = √ , therefore: 4 2 r r 2 π 1 1 F {f (x)} = √ = √ [answer] w π 2 w

3-9

r

2 Γ( 12 ) π √ sin . But π w 4

3.10 Determine the particular solution for the non-homogeneous, secondorder, ordinary differential equation: y 00 + y 0 − 2y = 4 sin 2x using a Fourier Transform method (compare your answer to the solution in problem 2.15 (a) and (b)).

Solution Consider the Fourier transform Z ∞ in eq. (3.24) and its derivatives: 1 F (w) = F {f (x)} = √ f (x) e−iwx dx, F (w) = F {f 0 (x)} = iwF {f (x)} 2π −∞ and F (w) = F {f 00 (x)} = −w2 F {f (x)}. Taking the transform of the differential equation yields: −w2 Y + iwY − 2Y = F {4 sin 2x} where Y = F {y(x)}. Solving for Y : 1 Y = F {4 sin 2x} = G(w) F (w) where from the convolu2 −w + iw − 2 tion theorem F (w) is the Fourier transform of f (x) and G(w) is the Z ∞Fourier √ transform of g(x). Thus, H(w) = 2πG(w)F (w) implies h(x) = g(x − −∞ Z ∞ g(y) f (x − y) dy. Taking the inverse transform of Y , the y) f (y) dy = −∞

solution for y(x) is: Z ∞ y(x) = g(y) 4 sin 2(x − y) dy −∞

One needs to find g(x), which is the inverse transform of

1 −w2 + iw − 2

such

that: Z ∞ Z ∞ 1 1 −1 −1 iwx √ e dw = √ eiwx dw.

2 2π −∞ w − iw + 2 2π −∞ (w − 2i)(w + i)

This integral is solved in problem 10.9 using contour integration and the residue theorem: √  − 2π x e for x < 0    3 g(x) = √    − 2π e−2x for x > 0.

3 Thus, inserting g(x) into the solution for y(x): 3-10

Z ∞ 1 −2y 1 y − e 4 sin 2(x − y) dy y(x) = − e 4 sin 2(x − y) dy + 3 3 0 −∞ Letting y → −y in the first integral: Z Z 4 ∞ −y 4 ∞ −2y y(x) = − e sin 2(x + y) dy − e sin 2(x − y) dy 3 0 3 0 Z ∞ Z ∞ 4 4 −y e cos 2y dy − cos 2x e−y sin 2y dy = − sin 2x 3 3 0 Z ∞0 Z ∞ 4 4 e−2y cos 2y dy + cos 2x e−2y sin 2y dy − sin 2x 3 3 0 0 Z ∞ Z ∞ a b e−ay cos by dy = 2 But e−ay sin by dy = 2 and .

2 a +b a + b2 0 0 Z 0

Therefore:

1 4 2 4 2 4 2 4 − cos 2x − sin 2x + cos 2x y(x) = − sin 2x 3 1+4 3 1+4 3 4+4 3 4+4

4 1 2 4 2 2 = − sin 2x + − cos 2x − 3 5 8 3 5 8

4 18 6 4 − sin 2x − cos 2x 3 40 3 40

1 3 The solution is: y(x) = − sin 2x − cos 2x.

5 5

[answer]

This is the same answer as in Problems 2.15(a) and (b).

3-11

2

e−(x/a) √ 3.11 Show that the normal distribution reduces to: f (x) = if it is a π √ centred at the origin where a = 2σ. The maximum of this function occurs 1 at f (0) = √ , where a is the width of the curve at half the maximum value. a π (a) Show that the Fourier transform of this function is: F (w) = F {f (x)} = 1 2 √ e(−w/b) . This function is also a Gaussian function, where b = 2/a is the 2π corresponding width of the transformed curve at half-maximum. (b) Recover f (x) by evaluating the inverse Fourier transform F −1 {F (w)} for the function F (w) given in part (a).

Solution

−(x−µ)2

e 2σ2 √ Consider the normal distribution in eq. (8.1): f (x) = . If this σ 2π√ distribution is centred at the origin (µ = 0) and letting a = 2σ then: 2 e−(x/a) √ . Here the maximum of this Gaussian function occurs at f (x) = a π 1 f (0) = √ . The parameter a is the width of the curve at half-maximum a π value. (a) From eq. (3.24):

Z ∞ 1 1 2 √ e−(x/a) e−iwx dx. f (x) e dx = √ 2π a π !−∞ −∞ 2 Z ∞ − x+ iwa2 2 w2 a2 1 1 a2 √ Completing the square: F (w) = √ e e− 4 dx.

2π a π −∞ !2  iwa2 1 F (w) = F {f (x)} = √ 2π

−

Note: e

x+

2 a2

2 2

− w 4a

e

Z ∞

−iwx

i h 2 2 xiw

2 2 2a w2 a4 − x2 + − w 4a 2 − 4a2

a 2a

=e

e

−

=e

2 2 x2 +xiw− w 4a a2

2

− x2

−xiw

=e a e

3-12

w2 a2

e− 4

Letting u =

2

x+ iwa 2 a

1 1 dx w2 a2 √ e− 4 , yields: F (w) = √ =⇒ du = a 2π a π

Z ∞

2

ae−u du . −∞ | {z√ } =a π

1 2 w2 a2 Therefore: F (w) = √ e− 4 and with b = yields: a 2π 1 −( wb )2 F (w) = √ e . [answer] 2π

Thus, the Fourier transform is also a Gaussian of width b at half-maximum.

2 The width of f (x) and F (w) are inversely proportional (b = ). As such, a a general feature of a Fourier transform is that a “narrow” function f (x)gives a “broad” Fourier transform function F (w) and vice-versa. (b) From eq. Z(3.25): Z ∞ ∞ w2 a2 1 1 1 iwx f (x) = √ F (w) e dw = √ √ e− 4 eiwx dw.

2π −∞ Z2π∞ 2π −∞ 2 x2 wa ix 1 Completing the square: f (x) = e−( 2 − a ) e− a2 dw.

2π −∞ 2 Z ∞ wa ix a e−(x/a) 2 −u2 Letting u = − =⇒ du = dw. Therefore f (x) = e du .

2 a 2 2π −∞ a {z } | 2√ = π a 1 2 √ − x22 πe a and the solution is: Therefore f (x) = 2 π a x2 1 f (x) = √ e− a2 . [answer] a π

3-13

3.12 Given the function:

 1 −x/a e for x > 0    a 1 f (x) = e−|x|/a =  a 1   ex/a for x < 0 a

(a) Show that the Fourier transform of this function f (x) is: F (w) = r

2 1 . π 1 + (wa)2

(b) Recover f (x) by evaluating the inverse Fourier transform F −1 {F (w)} for the function F (w) given in part (a) based on a contour integration method (see also chapter 10).Hint: Consider the contour integral: I 1 eizx

dz, πa2 C z + ai z − ai

where the contours C1 and C2 are given in fig. 3.7(a) and fig. 3.7(b) for (x > 0) and (x < 0), respectively, in the limit that R −→ ∞.

Figure 3.7: Contours for the integral (a) for the case that x > 0, and (b) for the case that x < 0.

Solution

3-14

(a) From eq. (3.24): Z ∞ 1 f (x) e−iwx dx F (w) = F {f (x)} = √ 2π −∞

Z 0 Z ∞ 1 1 −x/a −iwx x/a −iwx e e dx e e dx + =√ 2π a 0 −∞

Z 0 Z ∞ 1 1 x[ a1 −iw] −x[ a1 +iw] =√ e dx + e dx 2π a −∞ 0 # " 0 ∞ 1 1 1 1 1 1 =√ − 1 ex[ a −iw] e−x[ a +iw] 1 ( a + iw) 2π a ( a − iw) −∞ 0

1 1 1 =√ + 2π (1 − iwa) (1 + iwa) r

2 1 Therefore F (w) = . [answer] π 1 + (wa)2 (b) From eq. Z(3.25): Z ∞ 1 ∞ 1 eiwx iwx F (w) e dw = f (x) = √ dw. π −∞ 1 + (wa)2 2π −∞ Z ∞ 1 eiwx Therefore f (x) = 2 dw. πa −∞ (w + i/a)(w − i/a) The poles of the above equation are w = ±i/a. One can then take a contour along the real axis and close by a semi-circle. The semi-circle will close in the upper-half plane or lower-half plane depending on the sign of x as shown in fig. 3.7(a) and fig. 3.7(b) for (x > 0) and (x < 0), respectively. The choice is made so that the integral around the semi-circle vanishes in the limit that the radius R of the semi-circle goes to infinity (that is, R → ∞).

I I 1 eizx eizx 1

dz. dz = Consider the contour integral: π C 1 + z 2 a2 πa2 C z + ai z − ai The exponential factor in the integral eizx = ei x[Re(z) + Im(z)] where Re(z) and Im(z) are the real and imaginary parts of z. This exponential factor can be written using Euler’s identity: eizx = e−xIm(z) eixRe(z) = e−xIm(z)[cos(Re(z))+i sin(Re(z))] Since sin and cos are bounded by +1 and -1, one only needs to concentrate on the factor e−xIm(z) → 0 as Im(z) → ∞ for x > 0, which corresponds to 3-15

the upper-half plane for the contour in this case. Likewise, e−xIM (z) → 0 as Im(z) → −∞ for x < 0, corresponding to the lower-half plane. Thus, there are two contours of interest for the upper- and lower-half planes with the requirement that the integral goes to zero as R → ∞. Hence, using the residue theorem for these two cases yields: I in eq. (10.22) eizx 1 1

dz = 2πa−1 (at z = i/a) × . (i) x > 0: i i π C a2 z + a z − a π The residue for the simple pole at z = i/a is calculated from eq. (10.20): (z − ai )eizx ei(i/a)x e−x/a

a−1 = lim 2 = =

i z→i/a a (z + ) 2ia (z − ai ) a2 2ia a I izx −x/a −x/a

1 e 2πie e

Hence, dz = = .

2 2 π C1 1 + z a 2πia a

Doing the integral explicitly using polar coordinates where z = Reiθ =⇒ dz = iReiθ I dθ: Z Z iθ 1 1 R eiwx 1 π e−x/a eizx eixRe = dz = dw+ iReiθ dθ a π C 1 1 + z 2 a2 π −R 1 + w2 a2 π 0 1 + R2 a2 ei2θ | {z } I

where we have substituted w for z in the first integral since it is a dummy variable. Examining the second integral as R → ∞ yields: Z π ixR cos θ −xR sin θ e e ieiθ dθ. Here the first term in the numerator eixR cos θ is 2 ei2θ (1/R) + Ra 0 bounded and the second term in the numerator e−xR sin θ goes to zero since ieiθ = x > 0 and sin θ is positive over 0 to π. Also the term: lim R→∞ (1/R) + Ra2 ei2θ 0. Hence I → 0 as R → ∞ so that: f (x) = lim

R→∞

Z R

eiwx = 2 2 −R (1 + w a )

Therefore: f (x) =

e−x/a a

Z ∞

eiwx e−x/a = 2 2 a −∞ (1 + w a )

(x > 0).

(x > 0)

(ii)I Similarly for the case x < 0, one uses the contour in the lower-half plane: 1 eizx 1

dz = (−1) 2πb−1 (at z = −i/a) × . i i | {z } π C a2 z + a z − a π clockwise

The residue for the simple pole at z = −i/a is: (z + ai )eizx ei(−i/a)x ex/a

b−1 = lim 2 = = − z→−i/a a 2ia (z + ai )(z − ai ) a2 −2i a 3-16

I −e−x/a 2πi 1 eizx ex/a

From the residue theorem, = . dz = (−1)

2πia π C2 1 + z 2 a2 a

Doing the integral: Z Z iθ ex/a 1 R eiwx eixRe 1 π = dw + iReiθ dθ. Examining the seca π −R 1 + w2 a2 π 2π 1 + R2 a2 ei2θ {z } | II

ond as R → ∞ yields: Z π integral ixR cos θ −xR sin θ e e ieiθ dθ. Similarly the first term in the numerator is bounded 2 i2θ 0 (1/R) + Ra e and the term e−xR sin θ goes to zero since x < 0 and sin θ is negative over 2π ieiθ = 0. Hence II → 0 as R → ∞ to π. Also the term: lim R→∞ (1/R) + Ra2 ei2θ ex/a so that: f (x) = (x < 0) . a Thus from parts (i) and (ii):  1 −x/a e for x > 0    a 1 f (x) = e−|x|/a =  a 1   ex/a for x < 0 a

3-17

[answer]

3.13 Consider the Laplace transform F (s) =

1 . s−2

M in eq.

Rk (3.12) by finding the values of M and k. Hint: Given the relation |a + b| ≤ |a|+|b|, and letting a = z1 −z2 and b = z2 , it follows that |z1 −z2 | ≥ |z1 |−|z2 | 1 1 ≤ . Using this latter relation, with z1 = s = Reiθ and or |z1 − z2 | |z1 | − |z2 | 1 1 z2 = 2, one obtains the required result: . ≤ s−2 R−2 est (b) What is the residue of at the simple pole s = 2? s−2

1 −1 (c) From parts (a) and (b), calculate L using the complex ins−2 version formula. Compare your result to that given in table 3.1 for a = 2. (a) Show that this transform satisfies the condition |F (s)| <

Solution 1 1 1 1 2 = ≤ ≤ < for large iθ iθ s−2 Re − 2 |Re | − |2| R−2 R M enough R (R > 4). Hence, the condition |F (s)| < k is satisfied by k = 1 R and M = 2. [answer]

(a) For s = Reiθ :

(b) The residue at the simple pole s = 2: st e 2t lim (s − 2) [answer] =e

s→2

(s − 2)

(c) From parts (a) and (b), calculate L −1 est F (s) = e2t . [answer]

From table 3.1: If f (t) = eat then L {f (t)} = with a = 2, f (t) = e2t (same answer)

3-18

= sum of residues of

1 s−a

(s > a). Therefore

1 s−2

3.14 Solve the system of chemical kinetic equations in Example 2.4.3 using Laplace transforms.

Solution The system under consideration is: N10 = −.02N1 + .01N2 N20 = .02N1 − .01N2 with initial conditions N1 (0) = 300 and N2 (0) = 0.

Taking Laplace transforms of the system: sN̂1 − 300 = −.02N̂1 + .01N̂2 sN̂2 − 0 = .02N̂1 − .01N̂2 Adding these two equations yields: sN̂1 + sN̂2 = 300 or N̂2 =

300 − sN̂1 . s

Substituting this result into the first equation for N̂2 gives: !

300 − sN̂1 3 sN̂1 − 300 = −.02N̂1 + .01 =⇒ (s + .02 + .01)N̂1 = 300 + s s 300 3 or N̂1 = + . By partial fractions the last term can be (s + .03) s(s + .03) 3 A B written as: = + . Therefore: A(s + .03) + Bs = 3. s(s + .03) s (s + .03) For s = 0, A(.03) = 3 =⇒ A = 100.

For s = −.03, B(−.03) = 3 =⇒ B = −100.

300 100 100 200 100 Hence, N̂1 = + − = + . Taking the (s + .03) s (s + .03) (s + .03) s inverse Laplace transform gives: N1 (t) = 200 −.03t + 100 or:

N1 (t) = 100 1 + 2e−.03t From the original transformed equation sN̂2 = .02N̂1 − .01N̂2 with the pre200 100 vious result: N̂1 = + gives: (s + .03) s 4 2 (s + .01)N̂2 = + . Solving for N̂2 : (s + .03) s 4 2 + .

N̂2 = (s + .01)(s + .03) s(s + .01) | {z } | {z } I

II

3-19

Using partial fractions again: B1 A1 + = 4. Therefore: A1 (s + .03) + B1 (s + .01) = 4.

I: (s + .01) (s + .03) For s = −.03, B1 (−.02) = 4 =⇒ B1 = −200.

For s = −.01, A1 (.02) = 4 =⇒ A1 = 200.

A2 B2 II : + = 2. Therefore: A2 (s + .01) + B2 s = 2. s (s + .01) For s = −.01, B2 = −200.

For s = 0, A2 (.01) = 2 =⇒ A2 = 200.

200 200 200 200 200 200

− + − = − .

(s + .01) (s + .03) s (s + .01) s (s + .03)

Taking the inverse Laplace transform gives: N2 (t) = 200 1 − e−.03t Therefore N̂2 =

The final result is:

N1 (t) = 100 1 + 2e−.03t and

N2 (t) = 200 1 − e−.03t [answer]

Same result as Example 2.4.3 using matrix methods.

3-20

Problems Chapter 4 4.1 Given the following system of equations: 3y + 2x = z + 1 3x + 2z = 8 − 5y 3z − 1 = x − 2y solve for x, y, and z using the following methods: (a) Inverse Method; (b) Cramer’s Rule; (c) Gauss Elimination; (d) Maple software package.

Solution

4-1

4-1 4.1(a)

4-2

4.1(b)

4-2

4-3

4-3 4.1(c)

4-4

4-4 4.1(d)

4-5

4.2 Find the directional derivative of φ = x2 y +xz at (1,2,-1) in the direction → − → − A A = 2î − 2ĵ + k̂. Hint: (i) determine the vector û = , (ii) find ∇φ at |A|

dφ (1,2,-1) and then calculate from eq. (4.10). ds Solution

4-6

4-5 4.2

4-7

→ − → − 4.3 What is the work done if F = xy î − y 2 ĵ and dr = îdx + ĵdy assuming the path is: (a) A straight line where y = 12 x from A = (0, 0) to B = (2, 1)? (b) Parabolic where y = 41 x2 from A = (0, 0) to B = (2, 1)? (c) Comment if the evaluated work is the same or different in (a) and (b).

Solution

4-8

4-6 4.3

4-9

→ − 4.4 Let the force F = (2xy − z 3 ) î + x2 ĵ − (3xz 2 + 1) k̂. → − (a) Show that F is conservative. (b) Find φ(x, y, z) − φ(0, 0, 0). Note that from (a), since the force is conservative, any path can be taken to evaluate the line integral. Therefore, use simple straight line paths along the same directions as the axes starting from the origin (0, 0, 0) to (x, 0, 0), then from (x, 0, 0) to (x, y, 0), and finally up to the final end point from (x, y, 0) to (x, y, z).

Solution

4-10

4-7

4.4

4-11

Problems Chapter 5 5.1 Consider a solid ball of radius R, with a constant thermal diffusivity κ. Initially, the ball has a uniform temperature uo . If this ball is dropped into an ice-water bath of temperature zero, show that the temperature distribution u(r, t) in the ball can be described by the following partial differen∂u = tial equation (PDE) when only radial heat conduction is considered: ∂t

∂u 1 ∂ r2 with the initial condition: u(r, t) = uo , 0 < r < R, t = 0, κ 2 r ∂r ∂r and boundary conditions: u(r, t) is finite at r = 0, t > 0, and u(r, t) = 0, r = R, t > 0. ∂ 2ξ ∂ξ =κ 2 Using the transformation ξ = ur, show that the PDE becomes: ∂t ∂r subject to an initial condition: ξ(r, t) = uo r, 0 < r < R, t = 0 and boundary conditions: ξ(r, t) = 0, r = 0, t > 0, and ξ(r, t) = 0, r = R, t > 0. Solve this transformed problem for ξ(r, t). What is the corresponding solution for u(r, t)?

Solution

5-1

5-1 5.1

5-2

5-2

5-3

5-3

5-4

5.2 A chemical pollutant is released into the atmosphere where it diffuses radially with a constant diffusion coefficient D. It can be assumed that the initial concentration of the pollutant is zero, where the pollutant is a point source of strength S (atom s−1 ). The diffusion equation (DE) for the pollutant concentration t) (atom m−3 ) is given by:

C(r, ∂C 1 ∂ ∂C =D 2 r2 . ∂t r ∂r ∂r The source condition can be represented as a boundary condition at r = 0. ∂C Using the Fick’s First Law of Diffusion, where the flux F = −D , the ∂r release rate of the pollutant (at r = 0) is equal to the source strength such that: lim (4πr2 F ) = S. The other conditions follow for the given physical r→0 problem: C = 0 as r → ∞, t > 0 and C = 0, 0 < r < ∞, t = 0. (a) As proved in problem 5.1, with the transformation ξ = Cr, the DE ∂ 2ξ ∂ξ reduces to: = D 2 . Using this transformation, show that the initial ∂t ∂r and boundary conditions become: ξ(r, t) = 0, 0 < r < R, t = 0 and S , r = 0, t > 0, and ξ(r, t) is finite as r → ∞, t > 0. Using ξ(r, t) = 4πD a Laplace Transform Method, show that the solution, subject to the given

r S erfc √ conditions, is given by ξ(r, t) = where the latter function 4πD 2 Dt is the complementary error function. (b) What is the corresponding solution for C(r, t)? (c) Briefly mention why the Method of Separation of Variables is inappropriate for this given problem. (d) Assuming a diffusivity D = 10−2 m−2 s−1 , and a source strength S = 100 atom s−1 , what is the concentration at a distance of 100 m from the point source at a time of 104 s?

Solution

5-5

5-4 5.2

5-6

5-5

(b)

(c)

(d) (c)

5-7

5.3 Consider the radial diffusion of material in a sphere of radius a, with a concentration distribution C(r, t) and constant diffusivity D. It is assumed that: (i) the concentration at the centre of the sphere is finite for t > 0, (ii) the surface of the sphere is maintained at a zero concentration for t > 0, and (iii) initially there is a uniform concentration Co throughout the sphere 0 < r < a.

Dt C C r r , τ = 2 and u = = x, the transa a Co a Co ∂ 2u ∂u = formed problem is: with conditions: u(x, τ ) = x, 0 < x < 1, τ = ∂τ ∂x2 0, u(x, τ ) = 0, x = 0, τ > 0, and u(x, τ ) = 0, x = 1, τ > 0.

Defining the variables: x =

Using the Laplace Transform Method with respect to the variable τ , such that U (x, s) =√L {u(x, τ )}, show that the transformed solution is: U (x, s) = sinh x s 1 √ . What are the corrsponding solutions for u(x, τ ) and x− s sinh s C(r, t)? (a) The release fraction F (t) can be defined as the total amount of material which has diffused through the surface of the sphere at time Z t divided t

2

J(t)dt

4πa

by the initial amount of material in the sphere: F (t) = 3 aCo

Z t

0

4 3 πa Co 3

=

J(t)dt. The flux of material J(t) is evaluated from the Fick’s law of

0

∂C(r, t) . ∂r r=a Z τ Z τ 3a du Hence, show that: F (τ ) = J(τ )dτ = −3 dτ , and definDCo 0 dx x=1 0 3 dU (x, s) ing the Laplace Transform F (s) = L {F (τ )}, show that: F (s) = − . s dx x=1 diffusion: J(t) = −D

(b) Using the solution for U (x, and the latter expression for F (s) from s), √ coth s 1 part (a), show that: F (s) = 3 − 2 .

3/2 s s (c) An infinite series will eventually result for the release fraction F (t) which is more difficult to evaluate. However, an analytical form for F (t) is possible 5-8

by considering a “short-time” approximation for the condition that: τ << 1.

Show that this condition is equivalent to s >> 1 in Laplace transform space.

Determine an analytic expression for F (t) by applying the condition s >> 1 and taking the inverse transform of this resultant expression.

Solution

5-9

5-6 5.3 (a)

5-10

5-7

5-11

5-8

(a)

5-12

5-9 (b)

(c)

5-13

5-10

5-14

5.4 A standard model for describing the dispersion of chemical pollutants in the atmosphere is the Gaussian plume model. In this model it is assumed that the wind carries the pollutant in the x-direction with an average wind speed u, while the material diffuses in the other directions. In the case of a plane source of pollutant at x = 0, assuming only diffusion in the y-direction (with a diffusivity D), the steady-state transport equation for the concentra∂ 2C ∂C = D 2 . Using the transformation τ = xD/u, this equation tion C is: u ∂x ∂y ∂ 2C ∂C = can be written as: , subject to the conditions: ∂τ ∂y 2 C = Sδ(y), −∞ < y < ∞, τ = 0 ∂C C is finite at y = 0, which implies = 0, y = 0, τ > 0, ∂y C = 0, y = ±∞, τ > 0.

The condition at τ = 0 accounts for the plane source S of material being released into the atmosphere (that is, at x = τ = 0).Z Here δ(y) is a so-called distribution function which has the property that: Z ∞ and Sδ(y) dy = S.

∞

−∞

δ(y − a) f (y) dy = f (a)

−∞

(a) Using a Fourier Cosine Transform method such that Ĉc (w, τ ) = Fc {C(y, τ )}, show that the transformed partial differential equation is: r 2 ∂C ∂ Ĉc (w, τ ) = −w2 Ĉc (w, τ ) − . ∂τ π ∂y y=0 (b) Using the given conditions, and the property of the distribution funcS 2 tion, show that the transformed solution in part (a) is: Ĉ(w, τ ) = √ e−w τ .

2π (c) Taking the inverse Fourier cosine transform of the expression in part S 2 (b), show that the final solution is given by: C(y, τ ) = √ e−y /(4τ ) . Why 4πτ is the Method of Separation of Variables inappropriate for this problem?

5-15

Solution ∂ 2C ∂C = , subject to the conditions: ∂τ ∂y 2 (i) C = Sδ(y), −∞ < y < ∞, τ = 0

(ii) C is finite at y = 0, which implies

∂C = 0, y = 0, τ > 0, ∂y

(iii) C = 0, y = ±∞, τ > 0. (a) Let Ĉc (w, τ ) = Fc {C(y, τ )} =⇒ Fc {f (x)} = −w Fc {f (x)}− 00

Thus, transforming the differential equation gives: r ∂ Ĉc (w, τ ) 2 ∂Cc (w, τ )

= −w2 Ĉc (w, τ ) −

∂τ π | ∂τ {z } y=0

2

r

2 0 f (0). π

[answer]

=0 condition (ii)

∂ Ĉc 2 = −w2 Ĉc =⇒ Ĉc (w, τ ) = A(w)e−w τ . Using condition (i): Ĉc (w, τ = ∂τr Z r Z ∞ 2 ∞ 2 S 0) = S δ(y) cos wy dy = δ(y) cos wy dy. Using the πZ 0 π 2 −∞

(b)

∞

δ(y) f (y) dy = f (0) yields: r r 2 S S Ĉc (w, τ = 0) = . cos(0) = π 2 r 2π r S S −w2 τ Hence, Ĉc (w, τ = 0) = A(w) = . Therefore: Ĉc (w, τ ) = e .

2π 2π [answer] property

−∞

(c) Takingr the inverse transform: τ ) = Fc−1 {Ĉc (w, τ )}. Thus: r C(y,

S −1 −w2 τ S 1 −y2 /4τ √ e F {e . }= C(y, τ ) = 2π c 2τ r2π S −y2 /4τ The final solution is: C(y, τ ) = e . [answer] 4πτ By inspection of the solution, this problem is not separable into Y (y) τ (τ ).

5-16

5.5 The model in problem 5.4 can be generalized further to account for diffusion in both the y and z directions, with a time-dependent diffusiveconvective equation for the concentration of the pollutant in the atmosphere C(x, y, z) (atom m−3 ) and point source of pollutant S (atom s−1 ). If the movement of the effluent in the x-direction from the wind (with a constant speed u m/s) is much greater than that the following isotropic 2of diffusion,

2 ∂ C ∂ C ∂C ∂C =D where D is the + −u diffusion equation applies: 2 2 ∂t ∂y ∂z ∂x diffusion coefficient (m2 s−1 ). For steady-state conditions, and assuming ∂C ∂ 2C ∂ 2C D = 1 m2 s−1 and u = 1 m/s, this equation simplifies to: = + 2. ∂x ∂y 2 ∂z (a) Separating the variables where C(x, y, z) = C1 (x, y)·C2 (x, z), show that ∂ 2 C1 ∂C2 ∂ 2 C2 ∂C1 = and = . two partial differential equations result: ∂x ∂y 2 ∂x ∂z 2 (b) Because of the similarity of these two equations, one need only consider a solution for one (since the other follows by symmetry). The boundary conditions for the given y-region are: C1 (x, y) < M at y = 0, where M is a constant since the C1 concentration must remain finite at y = 0, ∂C1 = 0, x > 0. The second boundary condition which implies that ∂y y=0 C1 (x, y) = 0, y = ∞, x > 0, follows since far away from the source, the concentration must approach zero. Using a Fourier Cosine Transform method, such that c(x, w) = Fc {C 1 (x,2 y)}, show that the solution of this A1 y problem is: C1 (x, y) = √ exp − , where A1 is an arbitrary con4x x stant. definite integral has been used in this derivation: Z ∞ Note the following r 1 π − b2 2 e−aw cos (bw) dw = e 4a .

2 a −∞ (c) The solution for the z-direction follows on replacing the variable y with z in the solution of part (b) because of the symmetry of the probA2 z2 lem: C2 (x, z) = √ exp − . The final solution for C(x, y, z) therefore 4x x

A 1 2 follows as: C(x, y, z) = C1 (x, y) · C2 (x, z) = exp − [y + z 2 ] , where x 4x A = A1 · A2 . The arbitrary constant A follows from a conservation of mass using the source condition for the pollutant at x = 0. Hence, evaluate A by integrating the flux of material [u · C(x, y, z)] (where u = 1 m/s) over all 5-17

spaceZ in the y and z-direction and equating this result to the quantity S: ∞ Z ∞ [1 · C(x, y, z)] dydz = S. Hint: note the definite integral result lim x→ 0 −∞ −∞ r Z ∞ π −aξ 2 e dξ = . This analysis therefore yields the steady-state Gaussian a −∞ plume model for the dispersal of a pollutant with diffusion in the y and z-direction and convective transport by the wind in the x-direction.

Solution

5-18

5-12 5.5(a)

5-19

5-13

5-20

5.6 Consider heat conduction in a semi-infinite slab where the slab is initially (t = 0) at a constant temperature um , and the end of the slab (at x = 0) is maintained at the constant temperature uw . The partial differential equation for the temperature distribution u(x, t) for thermal diffusivity κ is: ∂ 2u ∂u = κ 2, ∂t ∂x

x > 0, t > 0

with u(x, 0) = um , x > 0, u(0, t) = uw , t > 0, |u(x, t)| < M (that is u is finite as x → ∞, t > 0). Using a Laplace Transform Method: √s e − κ x um + . (a) Show that the transformed solution is: U (x, s) = (uw −um ) s s

−a√s a e = erfc √ , (b) Using the inverse Laplace Transform L −1 s 2 t show that the solution is:

x + um . u(x, t) = L −1 {U (x, s)} = (uw − um )erfc √ 2 κt Solution

5-21

5.6

5-14

(a)

(b)

5-22

5.7 A problem of interest to chemical and material engineers is the melting of a material. This problem can be solved as a moving-boundary heat conduction problem as a so-called “Stefan problem,” where the melt front (that is, the liquid/solid interface) progresses with time. Consider a slab of material which is initially at the constant temperature um at t = 0. At time t > 0, the temperature at the wall (x = 0) is raised to a value above the melting temperature of the material, uw . At the liquid/solid interface, x = s(t), the temperature is equal to um . These boundary conditions can written mathematically as: u(0, t) = uw , t > 0, u(x, t)|x=s(t) = um , t > 0.

As shown in problem 5.6, the general solution of the heat conduction equation x , where A and B are constants determined is of the form: A + B erfc √ 2 κt from the boundary conditions. (a) Show that by applying the boundary conditions, the solution of this x √ erf u(x, t) − uw 2 κt = , where s(t) = moving boundary problem is: u − u erf(γ) m w √ 2γ κt. The relation for s(t) indicates the position of the moving liquid front. (b) Assuming that heat transfer occurs in the liquid phase by conduction only, the constant γ can be determined from an energy balance across the du ds interface: −K . Using the definition of the thermal con= τL dx x=s(t) dt du ds ductivity K, show that this energy balance becomes: −κσ =L . dx x=s(t) dt Here τ is the density, κ is the thermal diffusivity, σ is the specific heat for the liquid phase and L is the latent heat of melting/solidification. Show that by applying this condition with the solution for u(x, t) one obtains the followσ(uw − um ) 2 √ ing transcendental equation for γ: γeγ erf(γ) = . As shown in L π problem 9.2, various approximate methods can be used to numerically solve this transcendental equation. (c) To numerically solve for γ in part (b), one requires that the error func5-23

Z γ 2 2 tion erf(γ) = √ e−u du be evaluated. Using a Gaussian integration π 0 rule (with n = 2 terms) (Section 9.4.3), derive an analytic relation for erf(γ).

Solution

5-24

5-15 5.7

(a)

5-25

5-16 (b)

5-26

5-17 (c)

5-27

5.8 Give a brief physical interpretation of the following partial differential equations. Classify each equation as elliptic, hyperbolic or parabolic by comparing it to the general form of a linear partial differential equation with two independent variables. ∂ 2u ∂ 2u + =0 ∂x2 ∂y 2 ∂ 2u ∂u =κ 2 (b) ∂t ∂x 2 ∂ 2u 2∂ u (c) = α ∂t2 ∂x2 Solution (a)

5-28

5-18

5.8 (a)

(b)

(c)

5-29

5.9 Consider the heat conduction in a thin insulated bar of length 3 m where the initial temperature at t = 0 is f (x) = 15 − 10x o C, and the ends of the bar are kept at 0o C. The partial differential equation for the temperature distribution u(x, t) at the distance x and time t in the bar ∂ 2u ∂u = . Here the thermal diffusivity is simply is therefore given by: ∂t ∂x2 2 −1 equal to 1 m s . The boundary conditions for this problem are given mathematically by: u(0, t) = 0, t > 0 u(3, t) = 0, t > 0, and the initial condition is: u(x, 0) = 15 − 10x, for 0 < x < 3, t = 0. Solve this problem using a Separation of Variable Technique.

Solution

5-30

5-19 5.9

5-31

5.10 Solve problem 5.9 using a Laplace Transform Method with the variable transformation U (x, s) = L {u(x, t)}, where: (a) The transformed by:

√solution for U (x, s)√is given 15 − 10x sinh (x s) sinh ((x − 3) s) √ + √ + .

U (x, s) = 15 s s sinh (3 s) s sinh (3 s) (b) By taking the inverse Laplace transform of U (x, s) in part (a), what is the solution for u(x, t)?

Solution

5-32

5-20 5.10

(a)

5-33

5-21

(b)

5-34

5.11 Consider the heat conduction in a thin insulated bar of length 3 m where the initial temperature at t = 0 is 25o C, and the ends of the bar are kept at 10o C at x = 0 and 40o C at x = 3. The partial differential equation for the temperature distribution u(x, t) at the distance x and time ∂ 2u ∂u = . Here the thermal diffusivity is simply t in the bar is given by: ∂t ∂x2 2 −1 equal to 1 m s . The boundary conditions for this problem are given by: u(0, t) = 10, t > 0, u(3, t) = 40, t > 0, and the initial condition is: u(x, 0) = 25, for 0 < x < 3, t = 0. To solve this complicated boundaryvalue problem, assume that u(x, t) = v(x, t) + ψ(x), where ψ(x) is to be suitably determined in part (a). (a) Explain why u(x, t) can be written as two separate functions. Show that the boundary value problem can be rewritten as two separate problems: ∂ 2v ∂v = with v(0, t) = 0, v(3, t) = ψ 00 (x) = 0, ψ(0) = 10, ψ(3) = 40, and ∂t ∂x2 0, v(x, 0) = 15 − 10x. (b) What is the solution for ψ(x)? The solution for ψ(x) is in fact needed to obtain the initial condition for v(x, 0).

(c) The partial differential equation problem for the function v(x, t) is in fact the same boundary-value problem as given in problems 5.9 and 5.10.

Using this solution for v(x, t), give the complete solution for u(x, t). (d) What is the physical significance of the solution for ψ(x)?

Solution

5-35

5-22 5.11

(a) (b)

5-36

5-23 (c)

(d)

5-37

5.12 The slowing down of neutrons in the moderator of a nuclear reactor can be described by the slowing down density q (neutron m−3 s−1 ) in accordance ∂ 2q ∂q = , where τ is the “age” of with the so-called “Fermi age” equation: ∂τ ∂x2 the neutron (in this theory, the units of τ are in m2 ). This equation can be solved assuming a plane source of neutrons S (neutrons m−2 s−1 ) in the y − z plane at the origin of the coordinate system where x is the distance from the plane source. The boundary conditions for this equation consider symmetry ∂q at x = 0 and require that: (i) q remains finite such that: = 0, τ > 0, ∂x x=0 and (ii) q is finite as x → ∞, τ > 0. (a) Using a Fourier cosine transform on defining Q(w, τ ) = Fc {q(x, τ )}, 2 show that this problem yields the transformed solution: Q(w, τ ) = A1 e−w τ , where A1 is an arbitrary constant. (b) Taking the inverse Fourier cosine transform, show that the solution for A 2 q(x, τ ) is: q(x, τ ) = √ e−x /(4τ ) , where A is a constant. τ (c) ZUsing a source condition that follows from the conservation of mass S=

∞

q(x, τ )dx, evaluate the constant A. This source condition replaces

−∞

the initial condition for τ = 0.

Solution

5-38

5-24 5.12

(a)

(b)

5-39

5-25 (c)

5-40

5.13 Consider the solution of D’Alembert’s for the wave equation as mentioned in section 1.1.1. Here a solution to the wave equation: ∂ 2 u(x, t) 1 ∂ 2 u(x, t) = ∂x2 c2 ∂t2 is given by: u(x, t) = F (x + ct) + G(x − ct) where F (x + ct) and G(x − ct) are arbitrary functions. This solution is obtained by changing the variable x and t to w = x + ct and z = x − ct, respectively, considering that u(x, t) = ū(w, z) for the partial differential equation.

Solution

5-41

5.13

5-26

5-42

Problems Chapter 6 6.1 Explain the advantages and disadvantages of the Adam-Moulton, RungeKutta and Euler Methods for the numerical solution of first-order ordinary differential equations.

Solution Euler method is first-order whereas Runge-Kutta and Adam-Moulton method is fourth-order. Therefore the latter methods have significantly better accuracy. Adam-Moulton is faster since it is a predictor-corrector type (two equations) where as Runge-Kutta needs four axillary equations plus the timestep equation. However, because Adam-Moulton is a multi-step method, it requires the Runge-Kutta method to start (for the first three time steps).

The Euler method is a simple one-step (explicit) method.

6-1

dy = f (x, y).

6.2 Consider the first-order ordinary differential equation y 0 = dx Derive the predictor and corrector equations for the Heun’s Method by using a truncated Taylor series for yn+1 .

Solution

6-2

6-2 6.2

6-3

6.3 Given the initial value problem: y 0 = y − y 2 , y(0) = 0.5: (a) Solve the problem exactly using standard analytical methods for firstorder ordinary differential equations. (b) Apply the Runge-Kutta method (of fourth-order), and solve the problem for 0 ≤ x ≤ 1, with a step size of h = 0.2. Compute the error, using the exact values in part (a).

Solution

6-4

6-3 6.3(a)

6-5

6-4

6-6

6-5

6-7

6.4 Using the Maple software package: (a) Consider the square plate in Example 6.2.1 for a mesh spacing of h = 4 cm (that is, h = a/n where a = 12 cm and n = 3). However, the bottom and side boundaries each have a temperature of 100o C and the top boundary is at 0o C. Therefore show that u11 = u21 = 87.5o C and u12 = u22 = 62.5o C. (b) Repeat the calculation for a mesh spacing of h = 3 cm (that is, n = 4) and find the steady-state temperature at the internal mesh points. Hint: The number of mesh points can be reduced if symmetry is considered. (c) The analytical solution is given by:

∞ (2n + 1)π(a − y) 1 sin(2n + 1)πx 400 X sinh u(x, y) = π n=0,1... (2n + 1) sinh(2n + 1)π a a

sin(2n + 1)πy (2n + 1)πx (2n + 1)π(a − x) + + sinh sinh . a a a

Show that this analytical solution yields the values u11 = u21 = 88.1o C and u12 = u22 = 61.9o C for the given mesh in part (a). Compare the numerical results of part (b) with the values obtained from the analytical solution.

Solution

6-8

6-6

6.4

6-9

6-7

6-10

6-8

6-11

6-9

6-12

6-10

6-13

6.5 Write the steady-state heat conduction equation (that is, the Laplace equation) for a Cartesian coordinate system for 2-spatial dimensions. During the winter (in which the air has an ambient temperature of −10o C), a thin car window with dimensions 1 m high by 1.5 m wide is completely covered with ice. A poor quality electric heater is only able to apply heat to the bottom edge of the window at 23o C. The other edges of the window remain at the ambient temperature of the air. Determine if one can see through the middle third of the bottom half of the window if one waits for a long enough time by performing a numerical solution of the steady-state heat conduction equation with stated boundary conditions (that is, calculate the temperature at the mesh points u11 and u21 with a given mesh-size of h = 0.5 m).

Solution

6-14

6-11

6.5

6-15

6.6 Consider the temperature distribution u(x, t) in an insulated bar of ∂ 2u ∂u = κ 2 , sublength L as governed by the heat conduction equation: ∂x ∂ t̃ ∂u ject to the boundary conditions: = 0 t > 0, u(L, t) = 0 t > 0 and ∂x x=0 initial condition: u(x, 0) = uo , 0 < x < L, t = 0. (a) Consider the transformation of the above parabolic heat conduction x u , η = and τ = equation. Using the non-dimensional variables: ν = uo L κt , give the transformed partial differential equation for ν(η, τ )) with the L2 corresponding boundary and initial conditions. (b) Consider the Direct Method for the numerical solution of the parabolic partial differential equation. Show that with κ = 1 and L = 1, with a reflexive boundary condition at i = 0, the following formula results: u0,j+1 = (1 − 2r)u0,j + 2ru1,j , where r = k/h2 .

Solution

6-16

6-12

6.6 (a)

6-17

6-13

6-18

6.7 Explain what methodology could be used to determine if a numerical solution for a given problem is correct and accurate.

Solution Benchmark against an analytical solution (with constant coefficients) or experimental data.

6-19

6.8 Consider the heat conduction in a thin thermally-insulated bar of length L which is initially at a temperature of 0o C. At t > 0, heat is applied to the ends of the bar, which are kept at a constant temperature of 55o C. The temperature distribution is therefore described by the partial differential ∂ 2u ∂u = κ 2 with boundary conditions: u(0, t̃) = u(L, t̃) = 55, t̃ > equation: ∂ x̃ ∂ t̃ 0, and initial condition: u(x̃, 0) = 0, for 0 < x̃ < L, t̃ = 0. (a) What transformation can be applied to the independent variables in ∂ 2u ∂u = order to reduce the problem to: or ut = uxx with the corre∂t ∂x2 sponding boundary and initial conditions: u(0, t) = u(1, t) = 55, t > 0, and u(x, 0) = 0, for 0 < x < 1, t = 0? (b) Using a Crank-Nicolson method, solve the two resulting finite-difference equations for the first time step for the unknowns u11 and u21 with r = k/h2 = 1 and h = 0.2. What is the advantage of the Crank-Nicolson method over the conventional direct method? (c) By transforming the dependent variable, and using the more general solution in Example 5.2.4, give the analytical solution for the problem in part (a).

Solution

6-20

6-15

6.8 (a)

6-21

6-16 (b)

6-22

(c)

6-17

6-23

Problems Chapter 7 7.1 Solve the time-independent Laplace heat conduction equation in example 6.2.1 with the COMSOL software package.

Solution 7.1 See the methodology in example 7.1.1.

7.2 Solve the Poisson equation in example 6.2.2 with the COMSOL software package.

Solution 7.2 See the methodology in example 7.1.2.

7.3 Solve the time-dependent heat conduction equation in example 6.2.3 with the COMSOL software package.

Solution 7.3 See the methodology in example 7.1.3.

7.4 Solve the wave equation in example 6.2.4 with the COMSOL software package.

Solution 7.4 See the methodology in example 7.1.4.

7.5 Solve the first-order ordinary differential equation in example 6.1.4 with the COMSOL software package.

Solution 7.5 See the methodology in example 7.1.5.

7.6 Solve the second-order ordinary differential equation in example 6.1.5 with the COMSOL software package.

Solution 7.6 See the methodology in example 7.1.6.

7-1

7.7 A time-dependent fission product diffusion model in the nuclear fuel matrix can be coupled with a mass balance in the fuel-to-sheath gap and coolant to predict the coolant activity behaviour from a defective fuel element for variable power operation in a nuclear reactor.1 The diffusion equation for the fission product concentration C in an idealized fuel grain sphere of radius a, accounting for radioactive decay λ with a time-dependent diffusion coefficient D and fission product generation rate B (= fission rate F times the fission product yield Y ), is: D(t) ∂ ∂C(r, t) = 2 ∂t r ∂r

r

2 ∂C(r, t)

∂r

− λC(r, t) +

F (t)Y V

(7.19)

Defining the dimensionless variable, x = r/a, and multiplying through by the volume V yields: ∂u(x, t) D0 (t) ∂ = ∂t x2 ∂x

2 ∂u(x, t) x − λu(x, t) + F (t)Y ∂x

(7.20)

where u = CV and D0 = D/a2 . The initial and boundary conditions are given as:

u(x, 0) = 0, 0 < x < 1, t = 0 ∂u = 0, x = 0, t > 0 ∂x u(1, t) = 0 = 0, x = 1, t > 0

(7.21a) (7.21b) (7.21c)

A zero concentration is assumed at time zero. For the boundary conditions, symmetry is assumed at the centre of the sphere (i.e., a zero flux) and a zero concentration at the fuel grain surface with fission product release from the grain surface. The diffusional release to-birth rate (R/B) ratio derived from the Fick’s law of diffusion is:

R 4πa2 ∂CV 3D0 ∂u = −D =− B F Y (4πa3 /3) ∂r r=a F Y ∂x x=1 1

(7.22)

B.J. Lewis, P. K. Chan, A. El-Jaby, F.C. Iglesias and A. Fitchett, ”Fission product release modelling for application of fuel-failure monitoring and detection - An overview,” J. Nucl. Mater. 489 (2017) 64-83.

7-2

Equivalently, the release rate R (atom s−1 ) from the fuel matrix is: R = −3D0

∂u ∂x x=1

(7.23)

Thus, the time-dependent diffusion equation in eq. (7.20) can be solved by numerical methods subject to the conditions in eq. (7.21a) to eq. (7.21c).

The derivative of this solution (at x = 1) is subsequently used in eq. (7.23), which is the source release rate from the fuel matrix into the fuel-to-sheath gap. As such, the mass balance for the number of atoms Ng in the gap is: dNg (t) = R(t) − (λ + ν(t))Ng (t) dt

(7.24a)

Ng (t) = 0,

(7.24b)

with initial condition t = 0.

A first-order loss rate process is assumed from the gap characterized by an escape rate coefficient ν(s−1 ) which can also be considered as a function of time with element deterioration. The mass balance in the coolant is similarly given by: dNc (t) = ν(t)Ng (t) − (λ + β(t))Nc (t) dt

(7.25a)

Nc (t) = 0,

(7.25b)

with initial condition t = 0.

Here a time-dependent coolant purification rate constant β(t) for the coolant is considered. For this problem, the empirical diffusion coefficient D0 (s−1 ) as a function of the element linear power P (kW m−1 ) is derived from historical coolant activity data in the commercial stations as: 0

−12

D = 8.6 × 10

10(0.114·P ) P2

(7.26)

In addition, assuming 180 MeV fission−1 and a fuel stack length of 0.48 m, the fission rate F (fission s−1 ) = 1.489 × 1013 P (kW m−1 ).

7-3

(a) The solution of the coupled eq. (7.20) to eq. (7.26) provides a prediction of both the number of atoms in the gap and coolant as a function of time for a variable fuel element linear rating and coolant purification history. Solve the time-dependent governing transport equations using the commercial COMSOL software package (Version 5.4) employing a finite-element technique following the methodology in section 7.1. For this calculation, evaluate the I-131 activity concentration Ac (t)(µCi kg−1 ) = λNc (t) / (3.7 × 10−4 Bq µCi−1 ) / 244,000 kg in the reactor coolant over an irradiation period of 30 days at a constant linear power rating of P = 40 kW m−1 . For I-131, the following data apply: Y = 0.0288 atom fission−1 and λ = 9.98 × 10−7 s−1 .

One can take ν = 3 × 10−6 s−1 as a typical escape rate coefficient for iodine with a reactor coolant purification constant of β = 5 × 10−5 s−1 .

Solution

7-4

7-1 7-2 7.1 7.7

The governing equations are solved in a one-dimensional space ranging from 0 to 1 with the finite mesh distribution shown in Figure 1, consisting of 3003 degrees of freedom (using a maximum element size of 0.001 and mesh growth rate of 1.1).

(a)

Neumann Boundary Condition (center of grain)

Dirichlet Boundary Condition (surface of grain)

x

Figure 1.

Finite element distribution for radial diffusion in an idealized fuel grain model.

The three-coupled equations are solved by adding the “stabilized convection-diffusion equation” of the form:

d a u c c u Du J a u E u

f

in :

(7.27)

three times into the COMSOL “multiphysics” environment (see Figure 2).

Figure 2.

Comsol multiphysics environment for the problem.

Here Ω represents the subdomains in which the equations are applied. The vector u corresponds to the dependent variables u, Ng and Nc and the independent variables are the time t, and the dimensionless

7-5

7-3 7-2 axial position x. The da term is a time-dependent coefficient. Hence, on inspection of Eqs. (7.20), (7.24a) and (7.25a), the variable coefficients can be defined as given in Table 1. Here Eq. (7.20) has been multiplied through by x2/Dc in order to remove the singularity at the point x = 0 (i.e., for the spherical geometry). Furthermore, the boundary settings are obtained from the general form:

n c u Du J q u

g hT P; h u

r on w:

(7.28)

where w: is the boundary on which the conditions are applied. Here n is a unit normal vector and P is a Lagrange multiplier. For a Neumann-type boundary condition (which is used when the flux on the boundary is known), Eq. (7.28) takes the form:

n c u Du J q u

(7.29a)

g

If the boundary condition is reflexive (where there is a zero flux across the boundary), q and g are set to zero. For a Dirichlet-type condition, Eq. (7.28) takes the form:

h u

(7.29b)

r

For a zero quantity on a given boundary, then h = 1 and r = 0. The given boundary conditions and initial conditions are summarized in Table 2 and labeled on Figure 1. Initial conditions are also presented in Table 2.

Table 1:

Multiphysics Subdomain Quantities for Transport Equations

Dependent Variable

Subdomain Component Quantity

U u

c x2

a x2O/Dc

f F Y x2/ Dc

da x2/ Dc

D

E

J

0

0

0

Ng

0

O +Q

(R/B) FY

1

0

0

0

Nc

0

O +Ep

QNg

1

0

0

0

Table 2:

Initial and Boundary Conditions for the Defected Fuel Model

Dependent Defining Condition Variable Equation Type No. u Ng Nc

(7.21a) (7.21b) (7.21c) (7.24b) (7.25b)

Initial Neumann Dirichlet Initial Initial

Initial Condition Representation

Boundary Condition Representation

u(t0) 0 -

q 0 -

Ng(t0) 0 -

Nc(t0) 0

g 0 -

h 1 -

r 0 -

The source release term from the fuel to the gap, R, in the gap mass balance of Eq. (7.24a) is applied in Table 1 as a coupling variable (see the “f” quantity for Ng) where (R/B) is evaluated via Eq. (7.22). This coupling is obtained by right clicking on “Definitions” under “Component 1” and selecting “Probes” ->

7-6

7-4 7-3 “Global Variable Probe” with the variable coupling for the probe point expression as shown in Figure 2.

The x: Coordinate: 1 is selected as the point at x =1 for the “Domain Point Probe 1”. The source term from the gap to the coolant in the coolant mass balance of Eq. (7.25a) is further given as QNg (see the “f” quantity for Nc in Table 1) for the coupling into the coolant.

The output calculation (see Figure 3) for “Study 1” -> “Step 1: Time Dependent” is set to “Time Unit:” d, “Times:” range (0, 2, 30), and a user controlled “Relative Tolerance:” 0.1. The output coolant activity concentration Ac result for I-131 is shown in Figure 4. .

Figure 3.

Time solver choice and relative error tolerance for the problem.

7-7

7-57-4

Figure 4.

Output results of coolant activity concentration versus time.

7-8

7.8 Find the weak form of ∇ · ∇u + 2u + exp(x) = 0.

Solution

7-9

7.8

7-6

7.

7.

7-10

7.9 Starting with the weak from problem 7.8 use the Galerkin method to find the contribution to the element load vector from each shape function for a 1D linear Lagrange finite-element on the range x1 to x2 .

Solution

7-11

7.97.7.

7-7

7-12

7.10 Evaluate the contributions to the load vector from problem 7.9 for a mesh consisting of 1D linear Lagrange finite-elements with four elements with nodes positioned at x = 0, 1, 1.5, 1.75, and 2.

Solution

7-13

77.10

7-14

7.11 Assemble the global load vector from problem 7.10.

Solution

7-15

77.11

7-16

7.12 Starting from the weak form from problem 7.8 use the Galerkin method to find the contribution to the element stiffness matrix from each shape function for a 1D linear Lagrange finite-element on the range x1 to x2 .

Solution

7-17

7.12

7-10

7-18

7.13 Evaluate the contributions to the stiffness matrix from problem 7.12 for the mesh described in problem 7.10.

Solution

7-19

7-11 7.13

7-20

7.14 Assemble the global stiffness matrix from problem 7.13.

Solution

7-21

7-12 7.14

7-22

7.15 Solve the linear system Ku = F for the unknown nodal values of u using K from problem 7.14 and F from problem 7.11.

7.16 Graph the approximate solution from problem 7.15 for u(x) over the over the range x = 0 to 2. range x = 0 to 2. Also graph du dx 7.17 Repeat the step found in problems 7.8 to 7.16 for the same mesh using second-order Lagrange elements with the internal node placed at the centre of each element. Compare with the approximate solution obtained using linear finite-elements.

7.18 Solve the differential equation in problem 7.8 with the COMSOL software package with a fine mesh and compare to the results obtained in problems 7.16 and 7.17.

Solutions using Maple to Questions 7.9 to 7.18

7-23

7-13 7.

7.1

Question 7.9) Calculate the contributions to the element load vector FE for a general element The shape functions for a linear 1D element are:

(1)

(2) The element contributions are: (3)

(4)

(5)

(6) Question 7.10) Evaluate the contributions form each element using the mesh provided (7) (8) (9) (10) (11) (12) (13) (14) Question 7.11) Assemble the global load vector F. Using the FE1 and FE2 functions defined above which require the location of nodes 1 and 2 for each element

7-24

7-14

(15)

Question 7.12) Calculate the contributions to the element stiffness vector FE for a general element

(16)

(17)

(18)

(19)

(20)

(21)

7-25

7-15

(22) Question 7.13) Evaluate the contribtution on to the stiffness from each element Start by converting the KE expressions to functions of nodal positions to easily evalute on different elements.

(23)

(24)

(25) From element#1 from x=0 to 1 (26) 4 3

(27)

(28) From element#2 from x=1 to 1.5 (29) 2.166666667

(30) (31)

From element#3 from 1.5 to 1.75 (32) 4.083333333

(33) (34)

From element#4 from 1.75 to 2 (35) 4.083333333

7-26

(36)

7-16

(37) Question 7.14) Assemble the global stiffness matrix

(38)

Question 7.15) Solve the linear system Ku=F for unknow nodal values of u

(39)

Question 7.16) Plot the approximate solution of u and du/dx.

Create the vector of shape functions for the mesh using piecewise to restrict them to the appropriate element.

(40)

7-27

7-17

The approximate solution is:

(41)

7-28

7-1

Finally we plot the derivative of the apprximate solution with respect to x.

7-29

7-1

7.17 Question Question7.17 7.16) Repeat the process for quadratic lagrange elements Weak form doesn't change with finite-element order. Therefore this remains the same.

The new shape functions assuming x2 is located halfway between x1 and x3:

(42)

(43)

(44)

7-30

7-20 The new contributions to the load vector are (45)

(46)

(47)

(48)

(49)

(50) For element#1 from 0-1 (51) (52) (53) For element#2 from 1-1.5 (54) (55) (56) For element#3 from 1.5-1.75 (57) (58) (59) For element#4 from 1.75-2 (60) (61)

7-31

7-21 (62) Now the global load vector is

(63)

Now calculate stiffness matrix contribution from an arbitary element. Here are the indefinite integrals: KE2_11

(64)

KE2_12

(65)

KE2_13

7-32

7-22

(66)

KE2_22

(67)

KE2_23

(68)

KE2_33

(69)

Calculate the definite integrals:

(70)

(71)

7-33

7-23

(72)

(73)

(74)

(75)

(76)

7-34

7-24 (77)

Show the subblocks from each element for reference

(78)

(79)

(80)

(81)

Now we combine the subblocks into the global matrix. Start with an empty global matrix, then add the subblocks one at a time.

(82)

7-35

7-25

Now solve the linear system as before.

(83)

Create a vector of the shape functions on each element using piecewise to restrict each shape function to the correct element.

The solution 2nd order approximation for u on this mesh is:

(84)

7-36

7-26

7-37

7-27

Finally, we can plot the derivate with respect to x.

7-38

7-2

7-39

7-40

Question 7.18) Plot of u with fine mesh. 7-2

7-41

Question 7.18) Plot of du/dx with fine mesh. 7-30

Problems Chapter 8 8.1 Given the function, y = ab/c, where a, b and c are independent variables with random errors of sa , sb and sc , show that using the q general error propa(sa )2r + (sb )2r + (sc )2r .

gation formula one obtains the specific result: (sy )r = Solution

s

2 ∂y y = ab/c. In general sy = + + sc . ∂c s 2 as 2 abs 2 ∂y b ∂y a ∂y ab bsa b c + =⇒ Thus, = , = , = − 2 . Hence: sy = + − 2 ∂a c ∂b c ∂c c c c c s

sy b 2 2 a2 2 a2 b 2 2 c 2 = (sy )r = s + s + 4 sc 2 2 . Therefore: y c2 a c2 b c ab r q

2 2 2 sa sb sc (sy )r = + + = (sa )2r + (sb )2r + (sc )2r . [answer] a b c ∂y sa ∂a

2

∂y sb ∂b

2

8.2 What is the expression for calculating the standard error of the intercept (yo ) of a linear regression line (Hint: use the confidence interval relation to determine this expression)?

Solution

r

1 (x̄ − x∗ )2 + . For the stann A2 r x̄2 1 dard error of the interval (that is at x∗ = 0): se 1 + + 2 . Hown rA x̄2 1 + 2 or ever, for the regression limit, omit the unit quantity =⇒ se n A s 2 1 x̄ se 1 + + P 2 . [answer] n xi − nx̄ The prediction interval for y is y ± tse

1+

8.3 For the logarithmic expression y = log x, show that the standard deviation sy is given by sy = 0.434(sx )r .

Solution dy sx . Given y = log x = 0.434 ln x, hence: dx dy d ln x 0.434 sx = 0.434 = . Therefore sy = 0.434 or sy = 0.434 (sx )r . dx dx x x [answer]

y = log x. In general, sy =

8.4 A concentration measurement (in ppm) was made consisting of the following values: 21, 15, 23, 21 and 24. Should the anomalously low value be refused based on a Q critical value at a 90% confidence?

Solution For ppm measurements:

15, 21, 21, 23, 24. |{z}

anomolous

The difference from the next nearest neighbour is: 21 − 15 = 6.

Data spread: 24 − 15 = 9.

6 Therefore Qexp = = 0.67.

9 At 90% confidence for 5 observations Qcrit = 0.64.

Reject if Qexp > Qcrit , that is: 0.67 > 0.64 =⇒ reject data point.

[answer]

8.5 What is the difference between a t distribution and a normal distribution as used for confidence interval prediction?

Solution A t-distribution has a flatter distribution. This implies more area under the tail of the t-distribution. In the limit, as the number of degrees of freedom become infinite, the t-distribution approaches that of the z-distribution.

Problems Chapter 9 9.1 Solve x = cosx by: (a) Fixed-Point Iteration (x0 = 1, 20 steps, 6 significant figures) (b) Newton’s Method (x0 = 1, 6 decimal places). Sketch the function first. (c) Secant Method (x0 = 0.5, x1 = 1, 6 decimal places).

Solution

9-1

9-1 9.1

9-2

9-2

9-3

9-3

9-4

9.2 As shown in the moving-boundary problem 5.7 with a liquid front during the melting of a material, the following transcendental equation arises for the σ(uw − um ) 2 √ . To numerically solve this transcendenvariable γ: γeγ erf(γ) = L π tal equation (if all of the other constants and material property values were known), define the function f (γ) to be used and give the iteration scheme for each of the following methods: (i) fixed point iteration, (ii) Newton’s method and (iii) secant method. For application of the Newton’s method what is the expression for f 0 (x)?

Solution

9-5

9-4 9.2

9-6

9.3 Consider the following tabular data: ln 9.0 = 2.1972, ln 9.5 = 2.2513 and ln 11.0 = 2.3979. (a) Calculate the Lagrange interpolation polynomial p2 (x) and compute approximations of ln 9.4, ln 10, ln 10.5, ln 11.5, ln 12. Evaluate the errors by using exact values to four decimal places. (b) Comment on the effect of extrapolation. (c) Set up Newton’s divided difference formula for the same data, and derive from it p2 (x).

Solution

9-7

9-5 9.3

9-8

9-6

9-9

9-7 (c)

9-10

9.4 Find the cubic spline to the given data with k0 = −2 and k2 = −14.

Data: f0 = f (−2) = 1, f1 = f (0) = 5, f2 = f (2) = 17. (a) What is the advantage of a spline fit compared to a Lagrange polynomial for interpolation?

Solution

9-11

9-8

9.4

(a)

9-12

9.5 Given the definite integral: J =

Z 1 0

dx . Evaluate this integral using: 1 + x2

(a) An exact analytic formula obtained from calculus. (b) Approximate methods: (i) Trapezoidal rule (n = 4); (ii) Simpson’s rule (2m = 4); and (iii) Gaussian integration formula (n = 4). (c) Compute the error bounds for the Trapezoidal rule in part (b). (d) Compare the results obtained in parts (a) and (b).

Solution 9.5 (a) From calculus:

Z 1 0

swer]

1

dx

−1 = tan−1 x = tan−1 1 − tan 0 = 0.785398. [an2 1+x 0

9-13

9-10 (b)

(i)

(ii)

(iii)

9-14

9-11

(c)

9-15

9-12

(d)

9-16

9.6 The probability P that a measurement will fall within t standard devi −(x − µ)2 Z µ+tσ exp 2σ 2 √ ations is given by: P = dx. σ 2π µ−tσ (a) Show how this probability expression reduces to the normal error inteZ t z2 − 1 e 2 dz. gral: P = √ 2π −t (b) Using the normal error integral and Gaussian integration formula (for n = 2), calculate the probability that a measurement will fall within one standard deviation (that is, t = 1). What is the relative error for this estimation?

Solution

9-17

9-13 9.6

(a)

(b)

9-18

Z x

sin u du. Evaluate the quanu 0 tity Si(1) (that is, for x = 1) using a Gaussian integration formula to 4 significant digits using the zeros of the Legendre polynomial P2 (x). Using the actual value from a mathematical handbook (for example, [Spiegel, 1973]), estimate the relative and absolute error for the estimate.

9.7 The Sine Integral is defined as: Si(x) =

Solution

9-19

9-14 9.7

9-20

9.8 For application of the Gauss integration method, show that using eq.

(9.31) the following integral can be transformed as:

Z Z 1 1 1 1 2 −x2 exp − (t + 1) dt. e dx = 2 −1 4 0 .

Solution

9-21

9-15

9.8

9-22

Problems Chapter 10 10.1 Expand the functionf (z) =

1 in a Laurent series, which (z + 1)(z + 3)

is valid for 1 < |z| < 3.

Hint: Use partial fractions and a binomial series expansion: (1 + x)−1 = 1 − x + x2 − x3 ... for x values in the interval of convergence of −1 < x < 1.

Solution

10-1

10-1 10.1

10-2

I

2 dz where C is a circle of radius 1/2 and 2 C z −1 centred at 1, and positively oriented.

10.2 Consider the integral

2

1 1 − . z−1 z+1 (b) Evaluate the integral using Cauchy’s Integral Formula and Cauchy’s Theorem with these partial fractions, respectively. (a) Using a partial fraction method, show that:

Solution

10-3

z2 − 1

=

10-2

10.2

10-4

10.3 The complex inversion formula in√eq. (3.9) with a branch point s = 0

−a s a a e −1 √ = erfc √ . = 1−erf yields [Spiegel, 1971]: f (t) = L s 2 t 2 t This transform is used in Example 5.2.10. √ a 2 (a) Show that L −1 e−a s = √ t−3/2 e−a /(4t) . To find this transform, 2 π use the Laplace transform of a derivative L {f 0 (t)} = sF (s) (that is, for

a −1 √ f (0) = 0), with the previous result: f (t) = L {F (s)} = erfc 2 t √ e−a s where F (s) = . s Solution

10-5

10-3 10.3

(a)

10-6

10.4 Given the function u(x, y) = x2 − y 2 . (a) Prove that this equation satisfies the Laplace’s equation. (b) Find the function v(x, y) such that f (z) = u+iv is an analytic function.

Solution

10-7

10-4 10.4

10-8

10.5 Show that f (z) = sin z is analytic in the following steps: (a) Find the real and imaginary parts of f (z). (b) Show that u(x, y) and v(x, y) from part (a) satisfy the Cauchy-Reimann conditions.

Solution

10-9

10-5 10.5

10-10

10.6 Evaluate the integral

H

C

z 2 dz along the path shown in fig. 10.6.

Figure 10.6: Schematic of a closed semi-circle contour C.

Solution

10-11

10-6

10.6

10-12

H

dz dz where C is a circle of radius b C (z 2 − a2 ) centred at the origin with b > a.

10.7 Evaluate the integral Solution

10-13

10.7

10-7

10-14

R

+∞

dx where the contour runs along (1 + x2 ) −∞ the x -axis from −ρ to +ρ, and then closes by a semi-circle in the upper half plane of radius ρ centred at the origin.

10.8 Evaluate the definite integral

Solution

10-15

10-8

10.8

10-16

10.9 Show that using contour integration and the residue theorem, the def-

R

+∞

inite integral

−eiwx dw is evaluated as: (w2 − iw + 2)

−∞ √  − 2π x e for x < 0    3 g(x) = √   − 2π −2x  e for x > 0.

3 This result is used in the solution of problem 3.10.

Solution

10-17

10.

10-

10.

10.9

10-18

11-1

228 Chapter 11 Problems

CHAPTER 11. ADVANCED CONCEPTS

the “Method of Repeating Variables” is discussed here. While using the 11.1 Determine the variables on which a pendulum’s amplitude (or sometheorem, the following steps are applied times called the pendulum position) depend upon (see fig. 11.2). (Hint: use the Buckingham’s π theorem; the relevant dimensional variables can be • List relevant variables/parameters: the problem, identify listed as: pendulum’s mass, m, length of the rigidClearly rod, l, define pendulum’s period, all the variables that are important, and determine which is the main τ , time, t, gravitational acceleration, g, and initial position of the pendulum, variable of interest (i.e., dependent variable). Indicate the total number φ0 = φ(0)). of independent and dependent variable and parameters, n • Indicate basic dimensions of each parameter: Express fundamental dimensions of each of n variables/parameters, and hence determine the number of basic dimensions, m • Determine the number of Π groups, which is expressed by (n-m) • Choose repeating variables: While choosing the repeating variables, the following considerations need to be taken into account – Never pick the dependent variable

Motion that of a are pendulum.

–Figure Never 11.2: pick variables already dimensionless – Never pick two variables with the same dimensions – Chosen variables must represent all the primary dimensions – Chose common variables as they may appear in each of the Π groups – Chose simple variables over complex ones

Solution

• Form dimensionless Π groups and check that they are actually all dimensionless: Combine repeating parameters into products with each of the remaining parameters, one at a time, to create the Π groups

11.1 Example 6: Determine the variables on which pendelum’s amplittude (or sometimes called as the pendulum’s position) depends.

The motion of the pendulum is confined to a vertical plane, and the position of the pendulum can be indicated by the angle, φ, with the vertical (see Figure 11.2). In this example, the pendulum’s amplitude, φ, is the dependent variable. Independent variables and parameters, with their dimensions in capital letters, are :

11-1

11-2

11.6. NONDIMENSIONALIZATION

229

• (variable) pendulum’s mass, [m], M • (variable) length of the rigid rod, [l], L • (variable) pendulum’s period, [τ ], T • (variable) time, [t], T • (parameter) gravitational acceleration, [g], L·T−2 • (parameter) initial position of the pendulum, [φ0 = φ(0)], -

Figure 11.2: Pendulum.

In order to simplify the terminology, instead of indicating each time “variable” and “parameters” separately, the term “variables” will be used onword to indicate both.

The model is of the form f (φ, m, l, t, τ, g, φ0 ) = 0

(11.9)

This is written as a relation, but not as a function. The following relationship expresses the pendulum’s amplitude, φ, which is dependent variable, in terms of a complete set of independent variables: φ = f (m, l, t, τ, g, φ0 )

(11.10)

In this equation, φ and φ0 are already dimensionless and no need to apply any nondimensionalization; they have already formed the Π groups. There

11-2

11-3

230

CHAPTER 11. ADVANCED CONCEPTS

are (m=) 3 basic physical dimensions in this equation: time, T, mass, M and length, L; and (n=) 5 dimensional variables: t, τ , m, l, and g. Consequently, there will be two [i.e., (=5-3)] more dimensionless Π parameters, besides the Π groups formed for φ and φ0 . As discussed earlier, t and τ cannot be choosen at the same time, because two parameters with the same dimensions should not be picked to form Π groups. Consequently, one of the dimesional variables that appears in the Π group is t and the other one should be τ .

Given that all the rest (i.e., m, l, and g) have basic dimensions, they can be picked as repeating variables. Based on this analysis, the form of the model is Π0 = f (Π1 , Π2 , Π3 )

(11.11)

where • Π0 = φ • Π1 = φ 0 With m, l, and g being the repeating variables, Π2 and Π3 can be formed as Π2 = t a 1 m a 2 l a 3 g a 4

(11.12)

Π3 = τ b 1 m b 2 l b 3 g b 4

(11.13)

For the dimensionless Π2 parameter, the dimensional D matrix, where the rows correspond to the basis dimensions T, M, and L, and the columns to the dimensional variables t, m, l, and g, is given by t m l g T 1 0 0 −2 M 0 1 0 0 L 0 0 1 1 or

⎡

⎤ 1 0 0 −2 D = ⎣0 1 0 0 ⎦ 0 0 1 1

11-3

(11.14)

11-4

11.6. NONDIMENSIONALIZATION

231

The elements of the dimensional D matrix correspond to the powers to which the respective dimensions are to be raised.

For a dimensionless constant Π2 = ta1 ma2 la3 g a4 , a vector in the form of a = [a1 , a2 , a3 , a4 ] is sought so that the matrix-vector product Da equals the zero vector [0,0,0]. In linear algebra, the set of vectors with this property is known as the kernel (or nullspace) of (the linear map represented by) the dimensional matrix. In this particular case its kernel is one-dimensional. The matrix-vector multiplication is shown below ⎡ ⎤ ⎡ ⎤ a1 D11 D12 D13 D14 ⎢ ⎥ ⎣D21 D22 D23 D24 ⎦ ⎢a2 ⎥ = [0, 0, 0] (11.15) ⎣ a3 ⎦ D31 D32 D33 D34 a4 which leads to the following linear equations: a1 − 2a4 = 0 a2 = 0 a3 + a4 = 0

(11.16)

Taking the index of the prime quantity, which is a1 (for the prime quantity t), to be equal to 1 (i.e., a1 = 1), the other indexes become a3 = −1/2 and a4 = 1/2. With a2 = 0, the vector, a, becomes ⎡ ⎤ 1 ⎢ 0 ⎥ ⎥ a=⎢ (11.17) ⎣−1/2⎦ 1/2 Substituting the values of the vector a as exponents in Equation 11.12, Π2 = t1 m0 l−1/2 g 1/2

(11.18)

or in basic dimensions terms Π2 = (T )1 (M )0 (L)−1/2 (L/T 2 )1/2 = 1 Rearranging Equation 11.18, Π2 = t

g l

(11.19)

11-4

11-5

232

CHAPTER 11. ADVANCED CONCEPTS or expressing it in terms of dimensionless time parameter, t∗ , as

g ∗ (11.20) Π2 = t = t l

Given that τ has exactly the same dimension as time, t, the results obtained for Π2 hold true for Π3 , which therefore can be written as Π3 = τ 1 m0 l−1/2 g 1/2 or Π3 = τ

g l

(11.21)

Assuming the zeroes of F (in Equation 11.7) are discrete, it can be stated that τ g/L = Cn , where Cn is the nth zero of the function F. If there is only one zero, then τ g/L = C. It requires more physical insight or an experiment to show that there is indeed only one zero and that the constant is in fact given by C = 2π. If it is the case, Equation 11.21 can be rewritten as Π3 = τ ∗ =

2π τ

g l

(11.22)

where τ ∗ is the dimensionless pendulum period.

Note that the mass, m, is not present in any of the dimensionless quantities φ, φ0 , t∗ and τ ∗ . This implies that pendulum motion is independent from m. This conclusion is deduced directly from the Buckingham π theorem.

Therefore, the movement of the pendulum is given by a relation between φ and φ0 , t∗ and τ ∗ . One should realize that the number of independent variables are reduced from 6 in the dimensional form to 3 in the nondimensional form.

Inserting Πs into Equation 11.11 , the relationship between pendulum’s amplitude and the independent variables, in the form of dimensionless parameters, is given by φ = f (φ0 , t∗ , τ ∗ )

11-5

(11.23)

11-6

11.6. 11.2 NONDIMENSIONALIZATION 243 Apply the nondimensionalization technique to the Navier–Stokes equa-

tion for the x-direction component; the incompressible Navier–Stokes moInsering Equation into by: Equation 11.52, mentum equation11.53 is given ∂u ρ[ ∂t + (u · ∇)u] = −∇P + µ∇2 u + ρg, x0 dξP is the pressure, µ2 is2 the dynamic viscosity, u is the where ρ is the density, = a + (bx0 )ξ + (cx0 )ξ (11.54) t0 dτand g is the gravitational acceleration. flow velocity, t is time • Divide through by the coefficient of the highest order polynomial or derivative term The coefficient of the highest ordered term is in front of the first derivative term. Dividing by this gives dξ at0 + (bt0 )ξ + (cx0 t0 )ξ 2 = dτ x0

(11.55)

• Choose the characteristic unit for each variable to make the coefficients of as many terms as possible to become 1 By choosing the constant term (i.e., the first term in the right hand side of Equation 11.55) and the coefficient in front of ξ (i.e., the second term in the right hand side of 11.55) to become equal to 1, 1 bt0 = 1 ⇒ t0 = ; b

at0 a = 1 ⇒ x0 = x0 b

(11.56)

The coefficient of ξ 2 becomes (i.e., the last term in 11.55) κ = ca/b2 Finally, the nondimensional equation can be expressed as: dξ = 1 + ξ + κξ 2 (11.57) dτ Solution Instead of three undetermined parameters (a, b, c) in the dimensional version (i.e., 11.52), the nondimensional version has only one parameter, κ.

11.2 Example 10: Apply the nondimensionalization technique to the Navier–Stokes equation for the x-direction component.

Besides that nondimensionalization of the Navier–Stokes equation helps to reduce the number of parameters, it helps to gain a greater insight into the relative size of various terms present in the equation. Neglecting the smaller terms against the larger ones allows the simplification of the phenomena. In most of the cases (e.g., without density variations due to concentration and

11-6

11-7

244

CHAPTER 11. ADVANCED CONCEPTS

temperature or with the negligence of the gravity term), the nondimensionalized Navier–Stokes equation depend only on the Reynolds number and hence all physical realizations of the related experiment will have the same value of non-dimensionalized variables for the same Reynolds number.

The incompressible Navier–Stokes momentum equation is written as ρ[

∂u + (u · ∇)u] = −∇P + μ∇2 u + ρg ∂t

(11.58)

where ρ is the density,P is the pressure, μ is the dynamic viscosity, u is the flow velocity, t is time and g is the gravitational acceleration. Writing Equation 11.58 for the x-direction component and dividing by ρ gives ∂ux ∂ux ∂ux ∂ux + ux + uy + uz = ∂t ∂x ∂y ∂x −

1 ∂P ∂ 2 u x ∂ 2 ux ∂ 2 ux + ν( 2 + + ) + gx ρ ∂x ∂x ∂y 2 ∂z 2 (11.59)

Given that the gravitational acceleration vector is (0,-9.81,0) [ i.e., the gravitational acceleration has no impact on the x-direction but on the ymomentum equation], and hence the last term in Equation 11.59 becomes nil.

• Identify all the independent and dependent variables: ux and P are dependent variables; and t, x, y and z, uy , uz , ρ (i.e., in the case of no dependency on concentration and temperature in the x direction) and ν are independent variables or parameters.

• List primary dimensions of all dimensional variables and parameters/constants: [x]=L; [t]=T; [ux ]=L·T−1 ; [uy ]=L·T−1 ; [uz ]=L·T−1 ; [ρ]=M·L−3 ; [ν]=L2 ·T−1 ; [P]=M·L−1 · T−2 Dimensional homogeneity can be verified by checking if each term in Equation 11.59 has a dimension equal to sm2 .

• Select characteristic units (i.e., scaling parameters) to nondimensionalize variables: Nondimensional variables and parameters are: ux ∗ , uy ∗ , uz ∗ , P ∗ , t∗ , x∗ , y ∗ and z ∗

11-7

11-8

11.6. NONDIMENSIONALIZATION

245

Characteristic units which have the same dimensions as the original variables and parameters are: U is the flow velocity in L·T−1 , L is the lenght in L and T is the time in T. • Replace each of dimensional variables with a quantity scaled relative to a characteristic unit (See Equation 11.34 in Example 7 for nondimensionalization of the pressure term): ux = u x ∗ U ;

uy = uy ∗ U ;

uz = u z ∗ U ;

P = P ∗U 2ρ

y = y ∗ L;

z = z ∗ L;

t = t∗ T

x = x∗ L;

Substituting the terms defined in nondimensional form in Equation 11.59 ∗ ∗ ∗ ∂ux ∗ U ∗ ∂ux U ∗ ∂ux U ∗ ∂ux U + u +u + u = U U U x y z ∂t∗ T ∂x∗ L ∂y ∗ L ∂z ∗ L

−

1 ∂P ∗ U 2 ρ ∂ 2 ux ∗ U ∂ 2 ux ∗ U ∂ 2 ux ∗ U + ν( ∗2 2 + + ∗2 2 ) ∗ ∗2 2 ρ ∂x L ∂x L ∂y L ∂z L (11.60)

With T=L/U and dividing Equation 11.60 by U2 /L, ∗ ∗ ∗ ∂ux ∗ ∗ ∂ux ∗ ∂ux ∗ ∂ux + u +u + u = x y z ∂t∗ ∂x∗ ∂y ∗ ∂z ∗

(11.61) −

∗

2

∗

2

∗

2

∗

ν ∂ ux ∂ ux ∂ ux ∂P ( + + + ) ∂x∗ U L ∂x∗2 ∂y ∗2 ∂z ∗2

Rearranging Equation 11.61 for the Reynold’s number defined as U L/ν (where ν = μ/ρ) ∗ ∗ ∗ ∂ux ∗ ∗ ∂ux ∗ ∂ux ∗ ∂ux + u +u + u = x y z ∂t∗ ∂x∗ ∂y ∗ ∂z ∗

(11.62) ∂P ∗ 1 ∂ 2 ux ∗ ∂ 2 ux ∗ ∂ 2 ux ∗ − ( + + + ) ∂x∗ Re ∂x∗2 ∂y ∗2 ∂z ∗2

11-8

11.3 Consider a simplification of the problem in Example 11.2.1 for the pressure drop ∆p in a straight pipe of diameter D. Here, in an experiment, it is observed that with laminar flow conditions in the pipe, the pressure drop is observed to depend on three parameters: the distance l along the pipe between the pressure sampling points, the velocity v of the fluid and the fluid viscosity µ. For this problem, using the Buckingham’s π theorem, determine how the pressure drop varies with the pipe diameter.

Solution

11-9

1111.3

11-10

11-10

11-11

11-11

11-12

11.4 As shown in [Lewis et al., 2017], fuel swelling in nuclear fuel is caused by the presence of fission gases in small bubbles. Consider a gas bubble of radius R embedded in a solid medium with no hydrostatic stress, where the pressure p acting to expand the bubble is balanced by the surface tension γ of the solid (see fig. 11.3). Using the Buckingham’s π theorem, determine the dependence of the pressure on the bubble radius and solid surface tension.

Hint: for this analysis, use the basic physical dimensions for the system of: force (F), length (L) and time (T).

Figure 11.3: Gas-filled bubble in mechanical equilibrium with a solid (adapted from fig. 4.61 in [Lewis et al., 2017]).

Solution

11-13

11-12 11.4

11-14

Problems Chapter 12 12.1 Discuss the finite difference solution(s) to the Navier-Stokes equations.

Solution

12-1

12-1

12.1

Since the Navier-Stokes equations find a wide range of applications in modelling from weather, ocean currents, water flow in a pipe to air flow around a wing, many numerical schemes have been proposed to solve the Navier-Stokes equations.

The following articles have been suggested for discussions on the finite difference schemes for the Navier-Stokes equations with students: x x

x

A.J. Chorin, “Numerical Solution of the Navier-Stokes Equations”, Mathematics of Computation, 22, pp. 745-762, 1968.

M. Yanwen, F. Dexun, T. Kobayashi and N. Taniguchi, “Numerical Solution of the Incompressible Navier-Stokes Equations with an Upwind Compact Difference Scheme”, International Journal for Numerical Methods in Fluids, 50, pp. 509-521, 1999.

B. Seibold1, “A compact and fast Matlab code solving the incompressible Navier-Stokes equations on rectangular domains”, http://math.mit.edu/~gs/cse/, March 31, 2008.

ͳͺͲͺ͸ Ǥ ǡ Course 18.086: Computational Science and Engineering II Massachusetts Institute of TechnologyǤ Computational Science and Engineering ǣȀȀ Ǧ Ǥ Ǥ Ȁ Ǥ ǣȀȀ Ǧ Ǥ Ǥ ȀͳͺͲͺ͸Ǥ ǡ Ǥ Ǧ Ǥ ǡ ǡ ǡ ǡ Ǥ 1

12-2

12.2 Obtain the solution for the KdV equation given in eq. 12.39 by assuming a travelling wave of the form: u(x, t) = f (ξ) where ξ = (x − ct − xo ).

Solve this problem with direct integration by explicitly separating the differentials and using anti-derivatives. Use Maple to solve the resulting integral df √ . that arises in this derivation, f c + 2f Solution

R

12-3

12-2 12.2

12-4

12-3

12-5

∂u ∂u +c = 0, with u = ∂t ∂x u(x, t), can be reproduced with the Lax operator pair:

12.3 Show that the classical advection equation:

L=

∂2 − uI ∂x2

M = −c where I is the identity operator.

Solution

12-6

∂ ∂x

12-4 12.3

12-7

12.4 Classify the following equations as linear or nonlinear: (a) utt = c2 uxx . (b) σ ρ ut = ∇ · (K(u) ∇u) where σ and ρ are constants. (c) u0 = f (u, t).

(d) ∇2 u = λ sin(u) where λ is a constant. ∂u ∂u (e) +u = 0. ∂t ∂x Solution

12-8

12-5 12.4

12-9

12.5 Consider the nonlinear ordinary differential equation: u0 (t) = au(t) + b(u(t))2 , where a and b are constants. The time derivative for this equation can be discretized as: un+1 − un u0 (t) = , ∆t where un and un+1 are the values of the function u(t) at the respective nth and n + 1st mesh points. (a) What is the iteration scheme for this nonlinear equation using the Forward Euler discretization method in eq. 6.4? (b) Consider a Crank-Nicolson discretization method for this ordinary differential equation with a central difference approach given by: h i2 un+1 − un = a un+ 1 + b un+ 1 .

2 2 ∆t What is the resulting iteration scheme if un+ 1 is approximated by a standard 2 1 arithmetic mean: un+ 1 = [un + un+1 ] but linearized with a geometric mean 2 h 2 i2 for the nonlinear term: un+ 1 ∼ un · un+1 ?

2

Solution

12-10

12.5

12-6

12-11

12.6 A well-known model for population growth involves the nonlinear ordinary differential equation: u0 = a(u) u,

u(0) = 1

where the growth term is given by: u r a(u) = ρ 1 − M

and r is a constant. Here the growth rate factor a(u) ∼ ρ to allow the population to grow initially with unlimited access while the population u is small but tails off where a(t) approaches zero as the population u approaches a maximum M that the environment can sustain.

The standard Crank-Nicolson approach for the iteration scheme for this differential equation from section 6.2.2 would be:

1 un+1 − un = a(un )un + a(un+1 )un+1 . ∆t 2

However, this equation is nonlinear. Thus, linearize the iteration scheme using a geometric mean as presented in problem 12.5.

Solution

12-12

12-7 12.6

12-13

Problems Chapter 13 13.1 Solve theZ following integral equation using a Laplace transform method: t

y(t) = sin t +

0

y(τ ) sin(t − τ ) dτ

Solution

13-1

13-1 13.1

13-2

13.2 Consider the Sturm-Liouville problem in Example 13.2.1. This problem can be recast as the second-order ordinary differential equation y 00 (x) = f (x) with boundary conditions y(0) = y(1) = 0 on letting λ = −1, p(x) = 1 and f (x) = y(x) in eq. (13.8). Thus, the solution follows from eq. 13.11 as: Z 1 y(x) = G(x, s) f (s) ds.

0

Here the Green’s function is symmetric (that is, G(x, s) = G(s, x)) and is defined as the negative quantity of eq. (13.10) (which satisfies the boundary conditions):  s(x − 1), 0 ≤ s < x .

G(x, s) =  x(s − 1), x < s ≤ 1

(a) Show that for f (x) = x2 , the solution using the Green’s function forx 3 (x − 1) . malism is: y(x) = 12 (b) Show that:  x − 1, 0 ≤ s < x ∂G(x, s)  =  ∂s x, x < s ≤ 1

so that: (∂G(x, s)/∂s) = x − 1 + U (s − x), where U (s − x) is the unitary step function (or Heaviside function) (see section 3.1.3):

0 if U (s − x) = 1 if

s<x s > x.

(c) Given that the derivative of the Heaviside function is the Dirac delta ∂ 2 G(x, s) function, show that from part (b): = δ(s − x) = δ(x − s). Hence, ∂s2 Z x+ using the property of the Dirac delta function:

x−

δ(s − x)f (s) ds = f (x),

d2 y show that one obtains the original ordinary differential equation: = dx2 Z 1 2 Z 1 ∂ G (x, s) f (s) ds = f (x). Thus, y(x) = G(x, s) f (s) ds is a solution 2 0 ∂x 0 of the nonhomogeneous boundary value problem. Moreover, the Green’s function can be thought of as a response function to a unit impulse at s = x.

13-3

13.2

13-2

Solution

13-4

13.3 Consider Example 5.2.10 for heat conduction in a semi-infinite solid but where the temperature is initially zero in the solid, and at t > 0 a constant temperature of uo is applied and maintained at the face x = 0. The initial and boundary conditions for this problem become: u(x, 0) = 0, u(0, t) = uo and |u(x, t)| < M .

(a) Using a Laplace transform method with L {u(x, t)} = U (x, s), and transformed boundary condition: U (0, s) = L u √{u(0, t)} = uo /s, show that o − s/k x e . Thus, using the inverse the transformed solution is: U (x, s) = s −1 transform from √ problem 10.3, the final solution is: u(x, t) = L {U (x, s)} = uo erfc[x/(2 kt)]. (b) If the boundary condition, u(0, t) = uo is replaced by the more general condition √ u(0, t) = g(t), show that the transformed solution is: U (x, s) = − s/k x G(s) e . √ x 2 (c) From problem 10.3, L −1 {e− s/k x } = √ t−3/2 e−x /(4κt) . Hence, 2 πκ show that using the convolution theorem (see section 3.1.4) for the inverse transformZof part (b), the solution for u(x, t) is given by the integral equation: t x 2 √ u−3/2 e−x /(4κu) g(t − u) du. u(x, t) = 0 2 πκ Solution

13-5

13-3 13.3

13-6

13.4 Prove that eq. (13.16) is a general solution of the Helmholtz equation in eq. (13.14).

Solution

13-7

13.4

13-4

13-8

13.5 Classify each of the following integral equations as a Fredholm or Volterra type integral equation, linear or non-linear, homogeneous or nonhomogeneous, and identify the parameter λ abd the kernel K(x, y): (a) u(x) = x +

R1

xy u(y) dy

0

Rx (b) u(x) = 1 + x2 + (x − y) u(y) dy 0

(c) u(x) = ex +

Rx

y u2 (y) dy

0

R1

(d) u(x) = (x − y)2 u(y) dy 0

(e) u(x) = 1 + Solution

x R1 1 1 dy 4 0 (x + y u(y)

13-9

13-5 13.5

13-10

13.6 Convert the following Volterra integral equation into an initial value Rx problem: u(x) = x + (y − x) u(y) dy.

0

Solution

13-11

13-6 13.6

13-12

13.7 Convert the following initial value problem into a Volterra integral equation: u00 (x) + u(x) = cos x, subject to the initial conditions u(0) = 0 and u0 (0) = 1.

Solution

13-13

13-7 13.7

13-14

13.8 Convert the following boundary value problem into a Fredholm integral equation: y 00 (x)+y(x) = x, for x ∈ (0, 1), subject to the boundary conditions y(0) = 1 and y(1) = 0.

Solution

13-15

13-8

13.8

13-16

13-9

13-17

R1 13.9 Given the Fredholm integral equation: u(x) = x+λ (xy) u(y) dy, solve 0

this equation using (i) the successive approximation method, (ii) a Neumann series and (iii) the resolvent method.

Solution

13-18

13.9

13-10

13-19

13-11

13-20

13-12

13-21

Check: Z 1 λ u(x) = x + λ (xy) u(y) dy = x + x y · y dy. Thus, integrating: 1 − λ/3 0 0 !

1 y3 3 λx 1 λx =x + u(x) = x + 1 − λ/3 3 0 3 − λ 3

λ =x 1 + 3−λ 3x = 3−λ x = 1 − λ/3 R1

Thus, one recovers the solution.

13-22

Rx 13.10 Given the Volterra integral equation: u(x) = 1 + (x − y) u(y) dy, 0

solve this equation using the successive approximation method.

Solution

13-23

13-14 13.10

13-24

Problems Chapter 14 14.1 Consider a bead on a rotating wheel as shown in fig. 14.4. Here F = mv 2 is the centrifugal force, m is the mass of the bead, v is the tangential r velocity at the radius r of the wheel and n is the number of rotations per second such than ω = 2πn.

Figure 14.4: Schematic of a bead on a rotating wheel.

1 (a) Show that the kinetic energy in the problem is: T = mṙ2 . Given that 2 dV mω 2 r2 v = ωr and using eq. (14.11) such that: F = − , show that V = − . dr 2 (b) Using the Euler-Lagrange equation, show that the equation of motion is r̈ = ω 2 r, which has the solution: r(t) = Ae−ωt + Beωt .

Solution

14-1

14-1 14.1

(a)

(b)

14-2

14.2 Consider a pendulum shown in fig. 14.5 of mass m and length l. The angle of the pendulum from the vertical is φ with a velocity of v = l dφ/dt.

Using the Euler-Lagrange equation in eq. (14.8), with the Lagrangian L defined by eq. (14.9) as the difference between the kinetic energy (T = 21 m v 2 ) and the potential energy (V = m g h), show that the equation of motion for the pendulum is given by: d2 φ g sin φ = 0. + dt2 l Hint: the state variable for this problem is the angle φ that the pendulum makes from the vertical.

Figure 14.5: Schematic of the pendulum for an Euler-Lagrange analysis.

Solution

14-3

14-2

14.2

14-4

14.3 Consider a small block of mass m sliding a distance s down a moving wedge of mass M as shown in fig. 14.6. The wedge is also sliding a distance x along the floor. Neglecting frictional effects, calculate the following:

Figure 14.6: Schematic of a sliding block on a moving wedge.

(a) Given that the velocity of the wedge is simply ẋ, show that the x and y components of the velocity of the small block are: vx = ẋ − ṡ cos θ and vy = −ṡ sin θ, respectively.

(b) Show that the total kinetic energy of the combined system is: T = 1 1 M ẋ2 + m(ẋ2 + ṡ2 − 2ẋṡ cos θ). Also show that the potential energy for the 2 2 small block is: V = −m g s sin θ where g is the acceleration due to gravity.

(c) Determine the Euler-Lagrange equations for this problem for the two variables s and x and their derivatives.

(d) Solving the Euler-Lagrange equations in part (c), show that the two accelerations s̈ and ẍ are given by: s̈ =

M (M + m) g sin θ = c1 and ẍ = g sin θ cos θ = c2 , 2 (M + m sin θ) (M + m sin2 θ)

where c1 and c2 are simply constants.

Solution

14-5

14-3 14.3

(a)

(b)

(c)

14-6

14-4

(d)

14-7

14.4 Show that the shortest curve that has an area A below it is a circular arc: (λ x − c)2 + (λ y − d)2 = 1 as shown in fig. 14.7. Here λ is a Lagrange multiplier constant, and c and d are constants R 1 of integration. For this constrained problem: y(0) = a, y(1) = b and A = 0 y(x) dx.

Hint: use the Euler-Lagrange equation in eq. (14.47) with the constraint ϕ = A. From Example 14.1.2, the Lagrangian under consideration is given by eq. (14.17).

Figure 14.7: Schematic of a circular arc for the optimal shape of a curve with an area under it.

Solution

14-8

14-5 14-3

14.4 14.3

14-9

14-4 14-6

14-10

Bibliography [Lewis et al., 2017] Lewis, B. J., Onder, E. N., and Prudil, A. A. (2017).

Fundamentals of Nuclear Engineering. Wiley international editions: Nuclear engineering. Wiley.

[Spiegel, 1971] Spiegel, M. (1971). Advanced Mathematics for Engineers and Scientists. Schaum’s outline series. McGraw-Hill.

[Spiegel, 1973] Spiegel, M. (1973). Mathematical Handbook of Formulas and Tables. Schaum’s outline series. McGraw-Hill.