Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Chapter 3

Modeling of Physical Processes Problem 3.1: Ground Level Determine the solutions to the following problems. a

b

c

d

𝑑𝑑𝑑 1 = − 𝑑𝑑 𝑑𝑡 𝜏 𝑑𝑑 = 0 , 𝑡 = 0 𝑑2𝑥 𝑑𝑥 𝑚 +𝑐 +𝑘∙𝑥 =0 𝑑𝑡2 𝑑𝑡 𝑑𝑥 𝑥 =0, =0, 𝑡 =0 𝑑𝑡 𝜕2𝑇 𝜕2𝑇 + =0 𝜕𝑥2 𝜕𝑦2 𝑇 = 0 on all boundaries 𝜕𝑇 𝜕2𝑇 =𝛼 𝜕𝑡 𝜕𝑥2 𝜕𝑇 =0 , 𝑥=0 𝜕𝑥 𝑇=0 , 𝑥=𝐿 𝑇 =0, 𝑡 =0

Solution All these cases have zero forcing functions, thus they all remain in equilibrium. a) 𝑑𝑑 = 0 b) 𝑥 = 0 c) 𝑇 = 0 d) 𝑇 = 0

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Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 3.2: Mathematical Models Consider the following systems of equations. In each case identify if the mathematical model is formulated properly in order to allow for the possibility of a unique solution. If the system is not properly formulated, what additions or changes are necessary? 𝑑𝑑𝑑 1 = − 𝑑𝑑 + 𝑆 𝑑𝑡 𝜏 𝑑𝑑𝑑 1 = − 𝑑𝑑 + 𝑆 𝑑𝑡 𝜏 𝑑𝑑 = 𝑑𝑑0 , 𝑡 = 0 𝑑𝑑 = 𝑑𝑑1 , 𝑡 = 𝑡1 2 𝑑𝑥 𝑑𝑥 𝑚 +𝑐 + 𝑘 ∙ 𝑥 = 𝑓(𝑡) 𝑑𝑡2 𝑑𝑡 𝑥 = 𝑥0 , 𝑡 = 0 2 𝑑𝑥 𝑑𝑥 𝑚 +𝑐 + 𝑘 ∙ 𝑥 = 𝑓(𝑡) 𝑑𝑡2 𝑑𝑡 𝑥 = 𝑥0 , 𝑡 = 0 𝑥 = 𝑥1 , 𝑡 = 𝑡1 2 𝜕𝑇 𝜕 𝑇 𝑔(𝑥, 𝑡) =𝛼 + , 0<𝑥<𝐿 𝜕𝑡 𝜕𝑥2 𝜌𝑐 𝜕𝑇 −𝑘 = 𝑞" (𝑡) , 𝑥 = 0 0 𝜕𝑥 𝜕𝑇 𝜕2𝑇 𝑔(𝑥, 𝑡) =𝛼 + , 0<𝑥<𝐿 𝜕𝑡 𝜕𝑥2 𝜌𝑐 𝜕𝑇 −𝑘 = 𝑞" (𝑡) , 𝑥 = 0 0 𝜕𝑥 𝑇 = 𝑇0(𝑡) , 𝑥=0 𝑇 = 𝑇0 , 𝑡 = 0

a b

c

d

e

f

Solution a b c

Requires an initial condition 𝑑𝑑 = 𝑑𝑑0 , 𝑡 = 0 Too many initial conditions. Remove 𝑑𝑑 = 𝑑𝑑1 , 𝑡 = 𝑡1 Requires a second initial condition 𝑑𝑥 = 𝑣 0 , 𝑡 = 0

d

Remove 𝑥 = 𝑥1 , 𝑡 = 𝑡1 Requires a second initial condition 𝑑𝑥 = 𝑣 0 , 𝑡 = 0

𝑑𝑡

𝑑𝑡

e f

Requires a boundary condition at x=L. Requires an initial condition.

Requires a boundary condition at x=L. Over specified at x+L. Remove one of the conditions.

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Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 3.3: Coupled Mass-Spring-Dampers Consider the following system with two sets of mass-spring-dampers. A force f1 is applied directly to mass-1 while a force f2 is applied to mass-2. The spring constants are k1 and k2, and the damping coefficients are b1 and b2. The equilibrium positions with f1=f2=0 correspond to x1=x2=0.

x1

x2 b2 f1

f2 k2

Derive the complete mathematical model of this system.

Solution Draw free body diagrams of the two masses and apply Newton’s 2nd law.

Equation of Motion for Mass-1 𝑚1𝑥1 = � 𝑓𝑜𝑟𝑐𝑒𝑠 𝑚1𝑥1 = −𝑏1𝑥1 − 𝑏2(𝑥1 − 𝑥2) − 𝑘1𝑥1 − 𝑘2(𝑥1 − 𝑥2) + 𝑓1 𝑚1𝑥1 + 𝑏1𝑥1 + 𝑏2(𝑥1 − 𝑥2) + 𝑘1𝑥1 + 𝑘2(𝑥1 − 𝑥2) = 𝑓1 Equation of Motion for Mass-2 𝑚2𝑥2 = � 𝑓𝑜𝑟𝑐𝑒𝑠 𝑚2𝑥2 = −𝑏2(𝑥2 − 𝑥1) − 𝑘2(𝑥2 − 𝑥1) + 𝑓2 𝑚2𝑥2 + 𝑏2(𝑥2 − 𝑥1) + 𝑘2(𝑥2 − 𝑥1) = 𝑓2 Initial conditions at t=0 𝑥1 = 𝑥1,0 𝑥2 = 𝑥2,0 𝑥1 = 𝑣1,0 𝑥2 = 𝑣2,0 3

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 3.4: A Series of Spring-Mass-Dampers A series of coupled mass-spring-dampers consisting of n-masses in series are connected by springs and dampers. The masses m (kg) are identical, the springs are identical with spring constant k (N/m), and the dampers are identical with damping coefficient b (N m/s). x1

x2

f1

The initial conditions are

xn

f2

•

𝑥1,0 𝑣1,0 𝑥1(0) ⎡𝑥•1(0)⎤ 𝑣 𝑥 2,0 2,0 𝑥 (0) � 2⋮ � = � ⋮ � and 𝑥2(0) = � ⋮ � 𝑥 ⎢• ⋮ ⎥ 𝑥 (0) 𝑣𝑛,0 𝑛,0 𝑛 ⎣𝑥𝑛(0)⎦ Each mass has an individual external force applied, 𝑓𝑖(𝑡). The displacements are measured from static equilibrium with no applied forces. Derive the mathematical model of the system for the displacements xi, for i =1 to n.

Solution Using free body diagrams for mass-1, general mass-i, and mass–n, we get the following coupled set of ODEs. Mass-1: 𝑚𝑥1 + 𝑏𝑥1 + 𝑏(𝑥1 − 𝑥2) + 𝑘𝑥1 + 𝑘(𝑥1 − 𝑥2) = 𝑓1 Mass-i, i=2,…n-1 ) 𝑚𝑥𝑖 + 𝑏(𝑥𝑖 − 𝑥𝑖−1 + 𝑏(𝑥𝑖 − 𝑥𝑖+1) + 𝑘(𝑥𝑖 − 𝑥𝑖−1) + 𝑘(𝑥𝑖 − 𝑥𝑖+1) = 𝑓𝑖 Mass-n: 𝑚𝑥𝑛 + 𝑏𝑥𝑛 + 𝑏(𝑥𝑛 − 𝑥𝑛−1) + 𝑘𝑥𝑛 + 𝑘(𝑥𝑛 − 𝑥𝑛−1) = 𝑓𝑛

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Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 3.5: Mathematical Model of a Vibrating String Consider the motion of an elastic string. u 0

x

L

Fixed ends

u(x,t) = transverse displacement (m), T = tension (N),  = density (kg/m3), Ac = cross-sectional area (m2), F(x,t) = distributed load (N/m) Assumptions 1. The motion takes place entirely in one plane, and in this plane each particle moves at right angles to the equilibrium position of the string.

2. The deflection of the string during the motion is so small that the resulting change in length of the string has no effect on the tension T. 3. The string is perfectly flexible, i.e., can transmit force only in the direction of its length.

4. The slope of the deflection curve is small so that sin can be replaced with tan, where  is the inclination angle of the tangent to the deflection curve.

Objectives a) Derive the equation governing the transverse displacement of the string, u(x,t). b) State the complete mathematical formulation of the equations necessary to determine the motion of the string. c) For the special case where the distributed load is zero, F = 0, and the string is initial in equilibrium (the initial displace and velocity is zero), determine the solution u(x,t) for the displacement of the string. d) What is the relationship between this mathematical model and the series of discrete springmass-dampers in Problem 3.4?

Solution a) Equation of Motion Consider a differential length x.

Apply Newton’s law in the transverse direction. 𝜕2𝑢 𝑚𝑎 = 𝜌𝐴𝑐∆𝑥 2 = 𝐹𝑑𝑎𝑚𝑝𝑖𝑛𝑑𝑑 + 𝐹𝑡𝑒𝑛𝑠𝑖𝑜𝑛 + 𝐹𝑎𝑝𝑝𝑙𝑖𝑒𝑑 𝜕𝑡 Force due to Tension 𝐹𝑡𝑒𝑛𝑠𝑖𝑜𝑛 = −𝑇 ∙ 𝑠𝑖𝑛(𝑑𝑑(𝑥)) + 𝑇 ∙ 𝑠𝑖𝑛(𝑑𝑑(𝑥 + ∆𝑥)) Use the small angle approximation

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Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

𝜕𝑢 𝑠𝑖𝑛(𝑑𝑑(𝑥)) ≅ 𝑡𝑎𝑛(𝑑𝑑(𝑥)) = 𝜕𝑥 𝜕𝑢 𝜕 𝜕𝑢 𝜕𝑢 𝜕2𝑢 𝑠𝑖𝑛(𝑑𝑑(𝑥 + ∆𝑥)) ≅ 𝑡𝑎𝑛(𝑑𝑑(𝑥 + ∆𝑥)) = + ( ) ∆𝑥 = + ∆𝑥 𝜕𝑥 𝜕𝑥 𝜕𝑥 𝜕𝑥 𝜕𝑥2 The net force due to tension becomes 𝜕𝑢 𝜕𝑢 𝜕2𝑢 𝜕2𝑢 𝐹𝑡𝑒𝑛𝑠𝑖𝑜𝑛 = −𝑇 + 𝑇 ( + 2 ∆𝑥) = 𝑇 2 ∆𝑥 𝜕𝑥 𝜕𝑥 𝜕𝑥 𝜕𝑥 Force due to Damping The force due to friction is proportional to the speed at which the element is moving, and acts in a direction opposite to the motion of the element, thus 𝜕𝑢 𝐹𝑑𝑎𝑚𝑝𝑖𝑛𝑑𝑑 = −𝑏 ∙ ∆𝑥 𝜕𝑡 where b is the damping coefficient per length.

Applied Force The applied force is modeled in terms of a force per unit length along the string, thus 𝐹𝑎𝑝𝑝𝑙𝑖𝑒𝑑 = 𝐹(𝑥, 𝑡) ∙ ∆𝑥 The applied force can be spatial and time dependent, as would be the case for the playing of a musical instrument. Substitute these force terms into Newton’s law and cancel ∆𝑥. 𝜕2𝑢 𝜕𝑢 𝜕2𝑢 𝜌𝐴𝑐 2 = −𝑏 +𝑇 2 +𝐹 𝜕𝑡 𝜕𝑡 𝜕𝑥 Divide by 𝜌𝐴𝑐 to get the standard form of the wave equation. 𝜕2𝑢 𝑏 𝜕𝑢 𝜕2𝑢 𝐹(𝑥, 𝑡) =− + 𝐶2 + 𝜕𝑡2 𝜌𝐴𝑐 𝜕𝑡 𝜕𝑥2 𝜌𝐴𝑐 𝑚 𝑇 𝐶 = 𝑤𝑎𝑣𝑒 𝑠𝑝𝑒𝑒𝑑 ( ) = � 𝑠 𝜌𝐴𝑐 b) Mathematical Model Equation of Motion (zero damping)

𝜕2𝑢 𝐹(𝑥, 𝑡) 𝑇 𝜕2𝑢 = 𝐶2 + , 𝐶=� 𝜕𝑡2 𝜕𝑥2 𝜌𝐴𝑐 𝜌𝐴𝑐

Boundary Conditions

𝑢=0 , 𝑥=0 𝑢=0 , 𝑥=𝐿 𝑢 = 𝑢0(𝑥) 𝜕𝑢 � 𝑡=0 = 𝑣0(𝑥) 𝜕𝑡

Initial Conditions

c) Zero Forcing Functions 𝑢(𝑥, 𝑡) = 0 A physical response is due to some combination of volumetric forces, surface interactions, and initial disturbances. If all of these are identically zero, the system must to at rest in equilibrium.

Thus, the only possible solution is u =0. This intuitive answer can be checked by substituting back into the mathematical formulation (equation of motion, BC’s, and IC’s).

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Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 3.6: 2-D, 1-D, 0-D Mathematical Models Consider transient heat transfer in a 2-D region (no temperature gradients in the z-direction), initially at temperature T0, with the boundary conditions shown.

a) Formulate the equations governing the temperature distribution, T(x,y,t). b) Assuming that temperature gradients are negligible in the x-direction, formulate the equations governing the temperature distribution. c) Assuming that temperature gradients are negligible in the y-direction, formulate the equations governing the temperature distribution. d) Assuming that temperature gradients are negligible in all directions (lumped capacity approximation), formulate the equations governing the temperature distribution. For this lumped capacity model, determine the steady state temperature if qs” and g are constant.

Solution (a) 2D Model

Energy Equation

x-direction BC’s

𝜌𝑐

𝜕𝑇

𝜕2𝑇 𝜕𝑥2

+

𝜕2𝑇

) + 𝑔(𝑥, 𝑦, 𝑡) 𝜕𝑦2 or 2 𝜕𝑇 𝜕 𝑇 𝜕2𝑇 𝑔 = 𝛼( + )+ 𝜕𝑡 𝜕𝑥2 𝜕𝑦2 𝜌𝑐 𝜕𝑇 −𝑘 = 𝑞"(𝑦, 𝑡) , 𝑥 = 0 𝑠 𝜕𝑥

𝜕𝑡

= 𝑘(

𝜕𝑇 = ℎ(𝑇∞ − 𝑇) , 𝑥 = 𝐿𝑥 𝜕𝑥 𝜕𝑇 =0 , 𝑦=0 𝜕𝑦 𝜕𝑇 𝑘 = ℎ(𝑇∞ − 𝑇) , 𝑦 = 𝐿𝑦 𝜕𝑦 𝑇 = 𝑇0 , 𝑡 = 0 𝑘

y-direction BC’s

Initial Condition

7

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

(b) 1D Model: y-Direction Only 𝜕𝑇

𝜌𝑐

Energy Equation

𝜕𝑡

=𝑘

𝜕2𝑇 𝜕𝑦2

+𝑔+

𝑞"𝑠

−

𝐿𝑥

ℎ 𝐿𝑥

(𝑇 − 𝑇 ) ∞

or 𝜕𝑇 𝜕𝑡

𝜕2𝑇

=𝛼

𝜕𝑦2

𝑔

+

+

1

𝑞"(𝑦, 𝑡) −

1

𝜌𝑐 𝜌𝑐𝐿𝑥 𝑠 𝜌𝑐𝐿𝑥 𝜕𝑇 =0 , 𝑦=0 𝜕𝑦 𝜕𝑇 𝑘 = ℎ(𝑇∞ − 𝑇) , 𝑦 = 𝐿𝑦 𝜕𝑦 𝑇 = 𝑇0 , 𝑡 = 0

y-direction BC’s

Initial Condition

ℎ(𝑇 − 𝑇 ) ∞

(c) 1D Model: x-Direction Only 𝜌𝑐

Energy Equation

𝜕𝑇

=𝑘

𝜕𝑡

𝜕2𝑇 𝜕𝑥2

+𝑔−

ℎ 𝐿𝑦

(𝑇 − 𝑇∞)

or 𝜕𝑇 𝜕𝑡 x-direction BC’s

Initial Condition

𝜕2𝑇

𝑔

1

ℎ(𝑇 − 𝑇∞) 𝜕𝑥2 𝜌𝑐 𝜌𝑐𝐿𝑦 𝜕𝑇 −𝑘 = 𝑞𝑠"(𝑡) , 𝑥 = 0 𝜕𝑥 𝜕𝑇 𝑘 = ℎ(𝑇∞ − 𝑇) , 𝑥 = 𝐿𝑥 𝜕𝑥 𝑇 = 𝑇0 , 𝑡 = 0

=𝛼

+

−

(d) 0-D Model (Lumped Capacity Approximation) 𝜌𝑐 Energy Equation

𝑑𝑇 𝑑𝑡

𝑑𝑇 𝑑𝑡 Initial Condition

=

=𝑔−ℎ 𝑔 𝜌𝑐

−ℎ

𝐿𝑥 + 𝐿𝑦

𝐿𝑥𝐿𝑦 or 𝐿𝑥+𝐿𝑦

(𝑇 − 𝑇 ) + ∞

(𝑇 − 𝑇 ) +

∞ 𝜌𝑐𝐿𝑥𝐿𝑦 𝑇 = 𝑇0 , 𝑡 = 0

𝑞𝑠" 𝐿𝑥

𝑞𝑠" 𝜌𝑐𝐿𝑥

A steady state temperature is only possible if g and qs " are independent of time. Setting 𝜕𝑇/𝜕𝑡 = 0 gives 𝐿𝑥𝐿𝑦 𝑞𝑠" 𝑇 − 𝑇∞ = (𝑔 + 𝐿𝑥 ) (𝐿 + 𝐿 )ℎ 𝑥

𝑦

8

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 3.6 Appendices Appendix (a): 2D Model Our Control Volume (CV) is a differential section in both x and y of size x-by-y. The zdimension can be taken as unity since no changes occur in z.

Apply conservation of energy to the control volume shown. 𝑑𝐸𝑠𝑡 = 𝐸𝑖𝑛 − 𝐸𝑜𝑢𝑡 + 𝐸𝑑𝑑 𝑑𝑡 𝜕𝑞𝑦 𝜕𝑇 𝜕𝑞𝑥 ∆𝑦) + 𝑔∆𝑥∆𝑦 𝜌𝑐 ∆𝑥∆𝑦 = 𝑞𝑥 − (𝑞𝑥 + ∆𝑥) + 𝑞𝑦 − (𝑞𝑦 + 𝜕𝑦 𝜕𝑡 𝜕𝑥 Use Fourier’s Law 𝜕𝑇 𝜕 𝜕𝑇 𝜕 𝜕𝑇 𝜌𝑐 ∆𝑥∆𝑦 = − (−𝑘 ∙ ∆𝑦 ) ∆𝑥 − (−𝑘 ∙ ∆𝑥 ) ∆𝑦 + 𝑔∆𝑥∆𝑦 𝜕𝑡 𝜕𝑥 𝜕𝑥 𝜕𝑦 𝜕𝑦 Divide by ∆𝑥∆𝑦 𝜕𝑇 𝜕2𝑇 𝜕2𝑇 𝜌𝑐 = 𝑘 ( 2 + 2 ) + 𝑔(𝑥, 𝑦, 𝑡) 𝜕𝑡 𝜕𝑥 𝜕𝑦 Divide by 𝜌𝑐 𝜕𝑇 𝜕2𝑇 𝜕2𝑇 𝑔 = 𝛼 ( 2 + 2) + 𝜕𝑡 𝜕𝑥 𝜕𝑦 𝜌𝑐 Appendix (b): 1D Model, y-Direction Only Our Control Volume (CV) is a differential section in y, y-by-Lx. The z-dimension can be taken as unity since no changes occur in z.

qs” y  y

qy+y  gLxy

 h y (T-T∞)

Lx  qy Apply conservation of energy to the control volume shown. 𝑑𝐸𝑠𝑡 = 𝐸𝑖𝑛 − 𝐸𝑜𝑢𝑡 + 𝐸𝑑𝑑 𝑑𝑡 𝜌𝑐 𝜕𝑇 𝐿 ∆𝑦 = 𝑞 − (𝑞 + 𝜕𝑞𝑦 ∆𝑦) + 𝑔𝐿 ∆𝑦 + 𝑞" 𝑦 − ℎ𝑦(𝑇 − 𝑇 ) 𝑦 𝑦 𝑥 𝑠 ∞ 𝜕𝑡 𝑥 𝜕𝑦 Use Fourier’s Law 9

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

𝜌𝑐

𝜕𝑇 𝜕𝑡

𝐿 ∆𝑦 = − 𝑥

𝜕 (−𝑘 ∙ 𝐿 𝜕𝑇

) ∆𝑦 + 𝑔𝐿 ∆𝑦 + 𝑞" 𝑦 − ℎ𝑦(𝑇 − 𝑇 ) 𝑥 𝑠 ∞ 𝑥 𝜕𝑦

𝜕𝑦

Divide by 𝐿𝑥∆𝑦 𝜌𝑐

" ℎ 𝑞𝑠 − (𝑇 − 𝑇∞) =𝑘 2+𝑔+ 𝐿𝑥 𝐿𝑥 𝜕𝑡 𝜕𝑦

𝜕𝑇

𝜕2𝑇

Divide by 𝜌𝑐

𝑔 1 𝜕𝑇 𝜕2𝑇 1 =𝛼 2+ + 𝑞𝑠"(𝑦, 𝑡) − ℎ(𝑇 − 𝑇∞) 𝜕𝑡 𝜕𝑦 𝜌𝑐 𝜌𝑐𝐿𝑥 𝜌𝑐𝐿𝑥 Here g must represent the average value of g(x,y,t) over the entire x-direction. 𝐿𝑥

1 𝑔= � 𝑔(𝑥, 𝑦, 𝑡)𝑑𝑥 𝐿𝑥 𝑥=0

The energy equation can be derived directly from an energy balance on the CV y-by-Lx or by integrating the 2-D energy equation from x = 0 to Lx. Appendix (c): 1D Model: x-Direction Only Our Control Volume (CV) is a differential section in x, x-by-Ly. The z-dimension can be taken as unity since no changes occur in z. hx(T-T∞)  gx Ly qx 

 qx+x Ly

x Apply conservation of energy to the control volume. 𝑑𝐸𝑠𝑡 = 𝐸𝑖𝑛 − 𝐸𝑜𝑢𝑡 + 𝐸𝑑𝑑 𝑑𝑡 𝜕𝑇 𝜕𝑞𝑥 ∆𝑥) + 𝑔∆𝑥𝐿𝑦 − ℎ𝑥(𝑇 − 𝑇∞) 𝜌𝑐 ∆𝑥𝐿𝑦 = 𝑞𝑥 − (𝑞𝑥 + 𝜕𝑡 𝜕𝑥 Use Fourier’s Law 𝜕𝑇 𝜕 𝜕𝑇 𝜌𝑐 ∆𝑥𝐿𝑦 = − (−𝑘 ∙ ∆𝑦 ) ∆𝑥 + 𝑔∆𝑥𝐿𝑦 − ℎ𝑥(𝑇 − 𝑇∞) 𝜕𝑡 𝜕𝑥 𝜕𝑥 Divide by ∆𝑥𝐿𝑦 𝜕𝑇 𝜕2𝑇 ℎ 𝜌𝑐 = 𝑘 2 + 𝑔 − (𝑇 − 𝑇∞) 𝜕𝑡 𝜕𝑥 𝐿𝑦 Divide by 𝜌𝑐 𝜕𝑇 𝜕2𝑇 𝑔 1 =𝛼 2+ − ℎ(𝑇 − 𝑇∞) 𝜕𝑡 𝜕𝑥 𝜌𝑐 𝜌𝑐𝐿𝑦

10

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Here g(x,t) must represent the average value of g(x,y,t) over the entire y-direction. 𝐿𝑦

1 𝑔= � 𝑔(𝑥, 𝑦, 𝑡)𝑑𝑦 𝐿𝑦 𝑦=0

Similarly, qs”(t) represent the average value of qs”(y,t) over the entire y-direction.

Appendix (d): 0-D Model (Lumped Capacity Approximation) The energy equation can be derived a number of ways: • directly from an energy balance on the CV of dimensions Lx-by-Ly • integrating the 2-D energy equation from x = 0 to Lx and from y = 0 to Ly • integrating the 1-D energy equation from x = 0 to Lx Let’s choose a Control Volume (CV) of the entire domain, Lx-by-Ly. h Lx(T-T∞) 

qs ” Ly 

 h Ly(T-T∞)

g Lx Ly Ly Lx

Apply conservation of energy to the control volume. 𝑑𝐸𝑠𝑡 = 𝐸𝑖𝑛 − 𝐸𝑜𝑢𝑡 + 𝐸𝑑𝑑 𝑑𝑡 𝜌𝑐 𝑑𝑇 𝐿 𝐿 = 𝑔𝐿 𝐿 − ℎ𝐿 (𝑇 − 𝑇 ) − ℎ𝐿 (𝑇 − 𝑇 ) + 𝑞" 𝐿 Divide by 𝐿𝑥𝐿𝑦

𝑑𝑡

𝑥 𝑦

𝑥 𝑦

𝑑𝑡

∞

𝑥

𝐿𝑥 + 𝐿𝑦

𝑑𝑇 𝜌𝑐

𝑦

=𝑔−ℎ

𝐿𝑥𝐿𝑦

(𝑇 − 𝑇∞) +

∞

𝑠

𝑦

𝑞𝑠" 𝐿𝑥

Divide by 𝜌𝑐

𝐿𝑥+𝐿𝑦 𝑔 𝑑𝑇 𝑞𝑠" −ℎ = (𝑇 − 𝑇∞) + 𝜌𝑐𝐿𝑥 𝜌𝑐𝐿𝑥𝐿𝑦 𝑑𝑡 𝜌𝑐 Here g(t) must represent the average value of g(x,y,t) over the entire domain 1 Lx Ly g(t) =  g(x, y, t)  dx  dy .

Lx  Ly 0 0 Similarly, qs”(t) represent the average value of qs”(y,t) over the entire y-direction.

11

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Chapter 4

Calculus Problem 4.1 Determine and plot the derivative of the step function, 0 , 𝑡<0 𝐻(𝑡) = � 1 , 𝑡≥0 Solution 𝑑𝐻(𝑡) = 𝛿𝛿(𝑡) = 𝑑𝑒𝑙𝑡𝑎 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑑𝑡

x

Problem 4.2 Determine and plot the derivative of the ramp function, Ramp(t).

0 , 𝑡<0 𝑅𝑎𝑚𝑝 (𝑡) = 𝑡 ∙ 𝐻 (𝑡) = �𝑡 , 𝑡 ≥ 0 Solution 𝑑𝑅𝑎𝑚𝑝(𝑡) = 𝐻(𝑡) = 𝑆𝑡𝑒𝑝 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑑𝑡

1

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 4.3 Determine and plot the derivative of the pulse function.

0 , 𝑃𝑢𝑙𝑠𝑒(𝑡) = 𝐻(𝑡) − 𝐻(𝑡 − ∆𝑡) = �1 , 0 ,

𝑡<0 0 ≤ 𝑡 ≤ ∆𝑡 𝑡 > ∆𝑡

Solution 𝑑𝑃𝑢𝑙𝑠𝑒(𝑡) = 𝛿𝛿(𝑡) − 𝛿𝛿(𝑡 − ∆𝑡) 𝑑𝑡

t

x

Problem 4.4 Determine and plot the derivative of the triangle function.

0 , 𝑡 < −1 𝑡 + 1 , −1 ≤𝑡<0 𝑇𝑟𝑖𝑎𝑛𝑔𝑙𝑒 (𝑡 ) = � 1−𝑡 , 0≤ 𝑡 < 1 , 𝑡≥1 0 Solution 0 , 𝑡 < −1 𝑑𝑇𝑟𝑖𝑎𝑛𝑔𝑙𝑒(𝑡) 1 , −1 ≤ 𝑡 < 0 = �−1 , 0≤𝑡<1 𝑑𝑡 , 𝑡≥1 0 = (𝐻(𝑡 + 1) − 𝐻(𝑡)) − (𝐻(𝑡) − 𝐻(𝑡 − 1))

2

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 4.5 Consider the function 𝑓(𝑡) shown.

Evaluate and sketch the following integrals, I(t).

3 a) 𝐼(𝑡) = ∫ 𝑓(𝑡𝑜 )𝑑𝑡𝑜 𝑡𝑜=0 8

b) 𝐼(𝑡) = ∫

𝑓(𝑡𝑜 )𝑑𝑡𝑜

c) 𝐼(𝑡) = ∫

𝑓(𝑡𝑜 )𝑑𝑡𝑜

𝑡𝑜=0 𝑡 𝑡𝑜=0

Solution 3 (a) 𝐼(𝑡) = ∫

3

8𝑜

4

𝑓(𝑡𝑜 )𝑑𝑡𝑜 = 6 ∫𝑡 =0 𝑑𝑡𝑜 = 18 𝑡 =0

(b) 𝐼(𝑡) = ∫

𝑜

8

𝑓(𝑡𝑜 )𝑑𝑡𝑜 = 6 ∫ 𝑑𝑡𝑜 + 0 ∫𝑡 =4 𝑑𝑡𝑜 = 24 𝑡𝑜=0 𝑡𝑜=0 𝑜 6𝑡 , 𝑡 < 4 (c) 𝐼(𝑡) = � 24 , 𝑡 ≥ 4 The integrand is a two-part function; thus the integral is best evaluated in the corresponding twoparts.

3

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 4.6 Consider the pulse function f(t) shown.

Evaluate and sketch the following integrals. Note that the independent variable t can be less than t1 or greater than t1. Since the integrand 𝑓(𝑡) is a two-part function, the integral I(t) is best evaluated as a two-part function. 𝑡 a) 𝐼(𝑡) = ∫ 𝑓(𝑡𝑜 )𝑑𝑡𝑜 𝑡𝑜=0 𝑡 b) 𝐼(𝑡) = ∫ 𝑓(𝑡𝑜 )𝑒−(𝑡−𝑡𝑜)/𝜏𝜏𝑑𝑡𝑜 𝑡𝑜=0

Solution (a) The integrand is 𝑡defined in two distinct parts, thus think of the integral that way. 𝑡 < 𝑡 : 𝐼(𝑡) = ∫ 𝑓(𝑡 )𝑑𝑡 = 𝑓 ∫ 𝑡 𝑑𝑡 = 𝑓 𝑡 1

𝑡𝑡𝑜=0

𝑡 ≥ 𝑡 : 𝐼(𝑡) = ∫ 1

𝑡𝑜=0

𝑜

𝑜

1 𝑡𝑜=0

𝑜

𝑜

𝑜

1 𝑡𝑜=0

𝑜

1

𝑓(𝑡 )𝑑𝑡 = 𝑓 ∫𝑡1 𝑑𝑡 + 0 ∫ 𝑡

𝑡𝑜=4

𝑑𝑡 = 𝑓 𝑡 𝑜

1 1

Thus 𝑓1𝑡 , 𝑡 < 𝑡1 𝐼(𝑡) = � 𝑓1𝑡1 , 𝑡 ≥ 𝑡1 (b) Again consider the integral in two parts since 𝑓(𝑡) is two distinct parts. 𝑡

𝑡 < 𝑡 : 𝐼(𝑡) = 𝑓 � 1

1

𝑡1

𝑡 ≥ 𝑡 : 𝐼(𝑡) = 𝑓 � 1

1

𝑒

𝑒−(𝑡−𝑡𝑜)⁄𝜏𝜏𝑑𝑡 = 𝑓 𝜏 ⋅ 𝑒−(𝑡−𝑡𝑜)⁄𝜏𝜏|𝑡𝑜=𝑡 = 𝑓 𝜏 ⋅ (1 − 𝑒−𝑡⁄𝜏𝜏)

𝑡𝑜=0 −(𝑡−𝑡𝑜)⁄𝜏𝜏

𝑜

1

1

𝑡𝑜=0

𝑑𝑡 + 0 = 𝑓 𝜏 ⋅ 𝑒−(𝑡−𝑡𝑜)⁄𝜏𝜏|𝑡𝑜=𝑡1 = 𝑓 𝜏 ⋅ (𝑒−(𝑡−𝑡1)⁄𝜏𝜏 − 𝑒−𝑡⁄𝜏𝜏) 𝑜

1

𝑡𝑜=0

𝑡𝑜=0

1

Thus 𝑓1𝜏(1 − 𝑒−𝑡/𝜏𝜏) 𝐼(𝑡) = � 𝑓1 𝜏(𝑒−(𝑡−𝑡1)/𝜏𝜏 − 𝑒−𝑡/𝜏𝜏)

, 𝑡 < 𝑡1 , 𝑡 ≥ 𝑡1

4

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 4.7: Pulse Integral Consider the pulse function S(t).

𝑡

a) Determine the symbolic solution and plot the function 𝐹 (𝑡) = ∫ 1

𝑆(𝑡 )𝑑𝑡 . Examine the 𝑜

𝑡𝑜=0

limit as t→ 0. 𝑡 b) Determine the symbolic solution and sketch the function 𝐹 (𝑡) = ∫ 2

𝑜

𝑆(𝑡 )𝑒−(𝑡−𝑡𝑜)/𝜏𝜏𝑑𝑡 . 𝑜

𝑡𝑜=0

𝑜

Examine the limit as t→0. c) Evaluate the function F2 from part (b) using numerical integration. Use t1 = 2, t=1, and =1.

Plot both the analytical solution and numerical solution on the same graph.

Solution The integrand is defined in 3 distinct parts, thus evaluate the integral that way. (a) 𝑡 − 𝑡1 𝑡 −0𝑡1 𝐹 (𝑡) = ( ) (𝐻(𝑡 − 𝑡 ) − 𝐻(𝑡 − 𝑡 − ∆𝑡)) + 𝐻(𝑡 − 𝑡 − ∆𝑡) = � 1 1 1 1 ∆𝑡 ∆𝑡 1

, 0 ≤ 𝑡 < 𝑡1 , 𝑡 ≤ 𝑡 < 𝑡 + ∆𝑡 1

,

1

𝑡 ≥ 𝑡1 + ∆𝑡

Here H(t) is the unit step function.

In the limit as t→0, 𝐹1(𝑡) = 𝐻(𝑡 − 𝑡1). (b) 0 𝜏 𝐹2(𝑡) =

𝜏 ⎩∆𝑡

=

∆𝑡 (𝑒

,

−(𝑡−𝑡1) (1 − 𝑒 𝜏𝜏 )

(𝑡−𝑡1−∆𝑡) − 𝜏𝜏

−𝑒

∆𝑡

, 𝑡1 ≤ 𝑡 < 𝑡1 + ∆𝑡

(𝑡−𝑡1) − 𝜏𝜏 )

−(𝑡−𝑡1) 𝜏 (1 − 𝑒 𝜏𝜏 ) (𝐻(𝑡 − 𝑡1) − 𝐻(𝑡 − 𝑡1 − ∆𝑡)) +

0 ≤ 𝑡 < 𝑡1

,

𝑡 ≥ 𝑡1 + ∆𝑡

−(𝑡−𝑡1−∆𝑡) 𝜏 𝜏𝜏 (𝑒

∆𝑡

−𝑒

−(𝑡−𝑡1) 𝜏𝜏 ) 𝐻(𝑡 − 𝑡1 − ∆𝑡)

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Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Note the limit as t becomes small. lim 𝐹2(𝑡) = 𝑒−(𝑡−𝑡1)/𝜏𝜏𝐻(𝑡 − 𝑡1) ∆𝑡→0

The integrand for part (b) is shown for 𝑡 > 𝑡1 + ∆𝑡.

(c): Numerical Integration The solution using numerical integration is indistinguishable from the previous analytical solution. The difference between the analytical solution and the numerical solution using “NIntegrate” in Mathematica is shown. The difference is about the round off error using 64 bit arithmetic, as expected.

6

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 4.8 Consider the function f(t) shown.

𝑡

Determine an expression for the integral 𝐼(𝑡) = ∫

𝑡𝑜=0

𝑓(𝑡𝑜 )𝑑𝑡𝑜 over the entire range 0 ≤ t ≤ 3.

Solution 10𝑡 𝐼(𝑡) = �10 − 5(𝑡 − 1) 5

, 0<𝑡<1 , 1<𝑡<2 , 𝑡>2

Problem 4.9 What are the units of the following expressions? In all cases, t is in seconds(s) and x is in meters(m). a) 𝛿𝛿(𝑡) b) 𝛿𝛿(𝑥)𝛿𝛿(𝑡) c) H(t) = Step function d) Ramp(x) = Ramp function Solution a) 1 𝑡

1

1

𝑥

𝑡

b) ( ) ( ) c) Dimensionless d) x Problem 4.10 Evaluate the following integrals. H is the step function and 𝛿𝛿(𝑥) is the delta function. a) ∫3 𝐻(𝑥 − 1)𝑑𝑥 𝑥=0 b) ∫3 𝐻(𝑥 − 1)𝑑𝑥 𝑥=2

c) ∫3 𝛿𝛿(𝑥 − 2)𝑑𝑥 𝑥=0 d) ∫3 𝛿𝛿(𝑥 − 5)𝑑𝑥 𝑥=0

e) ∫3 𝑓(𝑥)𝛿𝛿(𝑥 − 2)𝑑𝑥 𝑥=0 f) ∫3 𝑓(𝑥)𝛿𝛿(𝑥 − 5)𝑑𝑥 𝑥=0

Solution a) 2; (b) 0; (c) 1; (d) 0 (e) 𝑓(2) (f) 0

7

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 4.11 Evaluate the following integrals symbolically. Also sketch the function I(x), clearly showing the locations x=a and x=b. 𝑏 a) 𝑓(𝑥) = ∫ 𝛿𝛿(𝑥 − 𝑥𝑜)𝑑𝑥𝑜 𝑥𝑏𝑜=𝑎 b) 𝑓(𝑥) = ∫ 𝐻(𝑥 − 𝑥 )𝑑𝑥 𝑜

𝑥𝑜=𝑎 𝑏

𝑜

c) 𝑓(𝑥) = ∫ (𝑥 − 𝑥𝑜)𝑑𝑥𝑜 𝑥𝑜=𝑎𝑡 d) 𝑓 (𝑡, 𝑡 ) = ∫ 𝛿𝛿(𝑡 − 𝑡 )𝑑𝑡 1

1

𝑡𝑡𝑜=0

2

1

𝑡𝑜=0 1

e) 𝑓 (𝑡, 𝑡 ) = ∫

1

𝑜

𝑜

𝑓 (𝑡 , 𝑡 )𝑑𝑡 where 𝑓 is the function determined in part (d). 𝑜

1

𝑜

1

Solution a) Assume a < b.

0 , 𝑥<𝑎 )𝑑𝑥 𝑓(𝑥) = � 𝛿𝛿(𝑥 − 𝑥𝑜 𝑜 = 𝐻(𝑥 − 𝑎) − 𝐻(𝑥 − 𝑏) = �1 , 𝑎 < 𝑥 < 𝑏 𝑥𝑜=𝑎 0 , 𝑥>𝑏 0 , 𝑥 < 𝑎 𝑏 b) 𝑓(𝑥) = ∫ 𝐻(𝑥 − 𝑥𝑜 )𝑑𝑥𝑜 = �𝑥 − 𝑎 , 𝑎 < 𝑥 < 𝑏 𝑥𝑜=𝑎 𝑏−𝑎 , 𝑥>𝑏 𝑏 c) 𝑓(𝑥) = ∫ (𝑥 − 𝑥 )𝑑𝑥 = (𝑥−𝑎)2 − (𝑥−𝑏)2 𝑜 𝑜 𝑏

𝑥𝑜=𝑎

2

2

d) 𝑓1(𝑡, 𝑡1) = 𝐻(𝑡 𝑡 − 𝑡1) 𝑡 e) 𝑓 (𝑡, 𝑡 ) = ∫ 𝑓 (𝑡 , 𝑡 )𝑑𝑡 = ∫ 2

1

𝑡𝑜=0 1

𝑜

1

𝑜

𝐻(𝑡 − 𝑡 )𝑑𝑡 = (𝑡 − 𝑡 )𝐻(𝑡 − 𝑡 ) 𝑜

𝑡𝑜=0

Problem 4.12 2 𝑧 The error function is erf(𝑧) = ∫ √𝜋 𝑥=0

1

𝑜

1

1

2

𝑒−𝑥 𝑑𝑥. Determine 𝑑 𝑒𝑟𝑓(𝑧) and 𝑑 𝑒𝑟𝑓(𝑧2). 𝑑𝑧

𝑑𝑧

Solution Using Leibnitz Rule. We can deduce the following 2 𝑑 𝑒𝑟𝑓(𝑧) = exp(−𝑧2) 𝑑𝑧 √𝜋 𝑑 𝑒𝑟𝑓(𝑧2) = 2 exp(−𝑧4)(2𝑧) 𝑑𝑧 √𝜋

8

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 4.13: Convolution Integral Two functions 𝑓(𝑡 )and ℎ(𝑡 )are shown in the figure.

A third function 𝑔(𝑡) is related to the two functions plotted above using the following integral. ∞

𝑔(𝑡) = � ℎ(𝜏)𝑓(𝑡 − 𝜏)𝑑𝜏 0

This integral is the well-known convolution integral, used to determine how a signal 𝑓(𝑡) will be filtered by a system’s impulse response ℎ(𝜏). a) Determine the function 𝑔(𝑡) for all values of t. b) Plot the function 𝑔(𝑡).

Hint: This integral is best undertaken by breaking the functions up into sections.

Solution By varying t, one can think of “dragging” the function 𝑓(𝑡 − 𝜏) across the stationary function ℎ(𝜏). As visualized, there are 5 time ranges with different behavior. t<1 1<t<2 2<t<3 3<t<4

t>4

0 , 𝑡<1 4(𝑡 − 1) , 1 < 𝑡 < 2 , 2<𝑡<3 𝑔(𝑡) = 4 4(4 − 𝑡) , 3 < 𝑡 < 4 ⎩ 0 , 𝑡>4 = 4(𝑡 − 1)(𝐻(𝑡 − 1) − 𝐻(𝑡 − 2)) + 4(𝐻(𝑡 − 2) − 𝐻(𝑡 − 3)) + 4(4 − 𝑡)(𝐻(𝑡 − 3) − 𝐻(𝑡 − 4))

9

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 4.14 Why do we even need numerical integration? Why not simply evaluate all integrals in an exact, symbolic form?

Solution Because many integrals cannot be evaluated analytically.

10

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 4.15: Basics of Numerical Integration 4 Consider the integral 𝐼 = ∫ (1 − 𝑒−2𝑥)𝑑𝑥. Evaluate this integral using the methods listed. 𝑥=0

Rule Exact Trapezoid, n=1 Trapezoid, n=2 Trapezoid, n=4 Simpson’s 1/3, n=2 Simpson’s 1/3, n=4 Simpson’s 3/8, n=3

Numerical Approximation

% Error 0

Solution 𝑏

𝑏

𝑒−2𝑥 𝑒−2𝑎 𝑒−2𝑏 ) = (𝑏 − ) 𝐼 = � (1 − 𝑒−2𝑥)𝑑𝑥 = (𝑥 − ) − (𝑎 − −2 −2 −2 𝑥=𝑎 𝑎

For a = 0 and b = 4, the integral evaluates to 𝐼 = 3.5002.

Rule

Approx

Error

Exact Trap, n=1 Trap, n=2 Trap, n=4 Simp 1/3, n=2 Simp 1/3, n=4 Simp 3/8, n=3

3.500 1.999 2.963 3.344 3.284 3.471 3.388

0.00 42.88 15.35 4.47 6.17 0.84 3.19

The accuracy of the trapezoid rule is of the order h2 while Simpson’s 1/3 is of the order h4. The accuracy versus n is shown in the following graphics.

0.5

0

0.5

0

0.5

0

2 4 x Trap: n=16, err=0.3% 1

0.5

0 0

2 x

4

0.5

0 0

0

2 4 x Trap: n=32, err=0.1% 1

f(x)

2 4 x Trap: n=8, err=1.2% 1

f(x)

0

f(x)

Trap: n=4, err=4.5% 1

f(x)

Trap: n=2, err=15.3% 1

f(x)

f(x)

Trap: n=1, err=42.9% 1

0.5

0 0

2 x

4

0

2 x

4

11

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 4.16: Erf a) Use the trapezoid rule to evaluate erf(1) for a sequence of increasing n (decreasing h): 𝑛 = 2𝑝 for p = 1 to 8. b) Determine the n required to get an absolute error less than 10-7.

Solution a) n 1 2 4 8 16 32 64 128 256

h 1.00000 0.50000 0.25000 0.12500 0.06250 0.03125 0.01563 0.00781 0.00391

b) Iexact = Iapprox =

Iapprox Per Error 0.77174333 8.420 0.82526296 2.069 0.83836778 0.514 0.84161922 0.128 0.84243051 0.032 0.84263323 0.008 0.84268390 0.002 0.84269657 0.001 0.84269974 0.000

0.8427007929 0.8427006930

Required number of panels = 832 Required panel size, h = 0.00120192

Problem 4.17: Distance from Velocity Data The following velocity versus time data is available:

Estimate the distance traveled using the trapezoid rule with: a) n = 1 (one trapezoid) b) n = 2 (two trapezoids) c) all the available data Make sketches of the data and the various approximations.

Solution 0+25 ( (a) 𝑑 = 12 − 0) ( ) = 150 𝑚 2 0+36 36+25 (b) ) + (6) ( 𝑑 = (6) ( ) = 291 𝑚 (c) 𝑑 = ∑𝑛

2 𝑣𝑖−1+𝑣𝑖

𝑖=1 ℎ𝑖

2

2

= 263 𝑚

The data plot using all the data is shown.

12

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 4.18: Ellipse 𝜋/2 The perimeter P of an ellipse is given by 𝑃 = 4𝑎 ∫ √1 − 𝑘2𝑠𝑖𝑛2𝜃𝜃 𝑑𝜃𝜃 where a = major axis, b= 0

minor axis, and 𝑘 = √𝑎2 + 𝑏2/𝑎. Write a function that calculates the perimeter of an ellipse. The input arguments should be a and b and the output argument should be P. Use your function to calculate the perimeter2of the following ellipses. 𝑥2 𝑦2 𝑥 𝑦2 𝑥2 𝑦2 𝑎) + = 1 , 𝑏) + = 1 , 𝑐) + =1 52

22

42

72

1002

22

Solution *** circular limit *** a = 1.0 , b = 1.0 P = 6.3 *** Case a *** a = 5.0 , b = 2.0 P = 23.0 *** Case b *** a = 7.0 , b = 4.0 P = 35.2 *** Case c *** a =100.0 , b = P = 400.4

2.0

Note that the circular limit gives the well-known and correct result. Also, the long ellipse limit gives the expected result of P  4 a.

13

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 4.19 Consider the function F(x) shown.

6

We wish to compute the integral 𝐼 = ∫ 𝐹(𝑥)𝑑𝑥.

0

a) Compute the exact value of the integral. b) Approximate the integral with the trapezoid rule using 1, 2, 3, 6, and 106 trapezoids. c) Approximate the integral with the Simpson’s 1/3 rule using 1 and 2 panels (2 and 4 segments).

Solution a) 𝐼 = 1 b) Number of trapezoids 1 2 3 6 106

Numerical Approximation 0 -3 0 1 1

Number of panels in Simpson’s rule 1 2

Numerical Approximation

c) -4 1

14

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 4.20: Football To estimate the surface area of a football, the diameter of the ball is measured at different points along the ball. z

d

L=12 in

z (in) 0 1.5 3 4.5 6 d (in) 0.0 2.9 4.8 5.8 6.2 𝐿 The surface area can be determined from 𝑆 = 𝜋 ∫ 𝑑(𝑧) ∙ 𝑑𝑧 0

a) Use the composite trapezoid rule with the given data to estimate the surface area of the ball. b) Use the composite Simpson’s rule with the given data to estimate the surface area of the ball.

Solution a) We have n = 5 data points, thus 4 trapezoids. Equation (4.67) gives 𝑛−1

ℎ 𝑆 ≅ 𝜋 (𝑓(𝑥0) + 2 � 𝑓(𝑥𝑖) + 𝑓(𝑥𝑛)) 2 𝑖=1

b) Equation

1.5 (0 + 2 ∗ 2.9 + 2 ∗ 4.8 + 2 ∗ 5.8 + 6.2) =𝜋 2 = 79.22 𝑖𝑛2 (4.79) 𝑛−1

𝑛−2

𝑖=1,3,…

𝑖=2,4,…

gives

ℎ 𝑆 ≅ 𝜋 (𝑓(𝑥0) + 4 � 𝑓(𝑥𝑖) + 2 � 𝑓(𝑥𝑖) + 𝑓(𝑥𝑛)) 3 1.5

(0 + 4 ∗ 2.9 + 2 ∗ 4.8 + 4 ∗ 5.8 + 6.2) 3 = 79.48 𝑖𝑛2 =𝜋

15

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 4.21: Sailboat A cross section of a racing sailboat is shown.

H is the height of the mast and W is the distance to support cable attachment. Wind forces exerted per length of mast, 𝑓(𝑧) (lb/ft), vary as a function of distance above the deck, z (ft). A good approximation of the wind force per length of mast is 2𝑧 𝑧 ) 𝑒− 𝐻 (𝑙𝑏/𝑓𝑡) 𝑓(𝑧) = 200 ( 5+𝑧 Assume that the right support cable is completely slack and the mast joins the deck in a manner that transmits horizontal and vertical forces but no moments. The total force, F (lb), exerted by the wind on the mast is 𝐻 𝐹 = ∫0 𝑓(𝑧) ∙ 𝑑𝑧.

Write a function that computes the total force, F (lb), and the tension in the left mast support cable, T (lb) as function of H (ft) and W (ft). To compute T, sum the moments about the base of the mast. 𝐻 The moment caused by the variable wind force is 𝑀 = ∫0 𝑧 ∙ 𝑓(𝑧) ∙ 𝑑𝑧. Next write a separate file to explore the system parameters. For H =30 ft and W = 3 ft, compute F and T. Also plot F and T versus H in the range 10 ft < H <40 ft while keeping W = 3 ft.

Solution Height, H = 30.0 ft Width, W = 3.0 ft Total force, F = 1480.6 lb Cable tension, T = 6474.4 lb The following plots of 𝑓(𝑧)and 𝑧𝑓(𝑧)show that these numerical results are at least approximately correct. The graph of 𝑓(𝑧)shows an average height of about 45 or 50 lb/ft. Thus the force should be approximately F  (45)(25) = 1250 lb.

16

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

2000

Force, F (lb)

f(z) lb/ft

1500

1000

500

0 10

15

20

25

30

35

40

10

15

20

25

30

35

40

15000

Tension, T (lb)

z*f(z) lb

10000

5000

0

Problem 4.22: Force on a Dam Water exerts pressure on the upstream face of a dam. The pressure increases linearly with depth and can be characterized by 𝑝(𝑧) = 𝜌𝑔(𝐷 − 𝑧) where 𝑝(𝑧)is pressure in N/m2 exerted at an elevation z meters above the reservoir bottom;  is density of water, which for this problem is assumed to be a constant 103 kg/m3; 𝑔 is acceleration due to gravity (9.81 m/s2); and D is elevation (in m) of the water surface above the reservoir bottom. Omitting atmospheric pressure (because it works against both sides of the dam face and essentially cancels out), the total force 𝑓𝑡 can be determined by multiplying pressure times the area of the dam face. Because both pressure and area vary with elevation, the total force is obtained by evaluating 𝐷 𝑓𝑡 = ∫0 𝜌𝑔 ∙ (𝐷 − 𝑧)𝑤(𝑧) ∙ 𝑑𝑧. where w(z) is width of the dam face (m) at elevation z. The line of action can also be obtained by evaluating 𝐷 ∫ 𝑧𝜌𝑔 ∙ (𝐷 − 𝑧)𝑤(𝑧) ∙ 𝑑𝑧 𝑑 = 0𝐷 ∫0 𝜌𝑔 ∙ (𝐷 − 𝑧)𝑤(𝑧) ∙ 𝑑𝑧 The following table contains data of stream width at various elevations.

Elevation, z (m)

0

10

20

30

40

50

60

Width, w (m)

122

130

135

160

175

190

200

Write a function that accepts vectors containing the z locations and the corresponding w values on the dam face. The function should perform numerical integration using the trapezoid rule for discrete data and return the values 𝑓𝑡 and d. Test your function using the specific data given.

Solution Total Force = 2.52706e+009 N Line of Action = 21.43 m

17

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 4.23: Wind Force A wind force distributed against the side of a skyscraper is measured as Height z (m) Force, F(z) (N/m)

0 0

30 340

60 1200

90 1600

120 2700

150 3100

180 3200

210 3500

240 3800

Compute the net force and line of action due to this distributed wind. The total force F exerted on the mast is: 𝐻

𝐹𝑡𝑜𝑡𝑎𝑙 = � 𝐹(𝑧)𝑑𝑧 0

The line of action can also be determined by integration: 𝐻 ∫ 𝑧 𝐹(𝑧)𝑑𝑧 𝑑 = 0𝐻 ∫0 𝐹(𝑧)𝑑𝑧 Solution Wind force on a skyscraper

F (N/m)

4000

2000

0 0 x 10

50

100 150 height, z (m)

200

250

50

100 150 height, z (m)

200

250

5

z*F (N)

10

5

0

0

Total force = 526200 N Line of action = 158.6 m Comments By visually estimating the area under the curve from the graphs, it is clear that the numerical results are of the correct order of magnitude.

18

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 4.24: Blackbody Radiation Function and Emissivity The thermal radiation emitted from an object is a function of its absolute temperature. A blackbody emitter is an ideal surface that emits radiation uniformly in all directions and absorbs all radiation incident on its surface. The variation of the emissive power as a function of wavelength for a blackbody emitter is described by the Planck distribution 𝑐1 𝐸𝜆,𝑏 = 𝑐 𝜆5 (𝑒𝑥𝑝 ( 2 ) − 1) 𝜆𝑇 2 E ,b = emissive power (W/m /m),  = wavelength (m), T = absolute temperature (K) c1 = 3.7418  108 W*m4/m2, c2 = 1.4388  104 m*K The fraction of energy emitted in the wavelength band 0 ≤ 𝜆 ≤ 𝜆∗ is 𝜆∗𝑇

𝐵𝑙𝑎𝑐𝑘𝑏𝑜𝑑𝑦 𝑅𝑎𝑑𝑖𝑎𝑡𝑖𝑜𝑛 𝐹𝑢𝑛𝑐𝑡𝑖𝑜𝑛: 𝐹0→𝜆∗(𝜆∗𝑇) = � 0

𝐸𝜆,𝑏 𝜎𝑇5

𝑑(𝜆𝑇)

 =Stefan-Boltzmann constant = 5.6696  10-8 W/m2 K4 The integrand that defines F0→*depends on the product T, not on  and T separately. (a) Plot 𝐸𝜆,𝑏 vs  on both a linear scale and on the classical log-log scale. On each graph, plot various values of T from 50K to 5800K (surface of the sun). (b) Create a function to evaluate the 𝐹0→𝜆∗ as a function of the product T. Please use this function throughout the remainder of this problem. (c) Plot 𝐹0→𝜆∗in the range T = 0 to 20000 m K. (d) The range of visible light for a typical human is 0.4 to 0.7 m. Determine the fraction of solar radiation in the visible range assuming the sun as a blackbody at 5800K. (e) Emissivity is defined as the actual amount of radiation emitted by a real surface divided by the radiation that a blackbody would emit at the same temperature.

Eλ Spectral emissivity = ελ = E λ,b E E Total emissivity = ε = = 4 σT Eb ∞

E = � ελEλ,bdλ λ=0

Consider surfaces with the following spectral variations of emissivity. For each case, plot the total emissivity as a function of temperature in the range T = 0 K to 5000 K. 1.0 0.8

surface a

0.6

surface b 0.4

surface c 0.2 5

10

15

20

m

For more details on thermal radiation, refer to Introduction to Heat and Mass Transfer, 7th Ed. by Bergman, Lavine, Incropera and Dewitt, Chapter 12 with particular emphasis on Sections 12.412.6.

Solution 19

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

(a) Planck Distribution

lb

lb

Emissive power, E (W/m2 micro meter)

Emissive power, E (W/m 2 micro meter)

3

x 10

7

2.5

2

1.5

1

0.5

0

0

10

20

30 40 50 60 70 Wavelength, lam (micro meter)

80

90

100

On a linear scale, only the highest temperature curve is visible due to the order of magnitude differences in the values. Thus, the log-log graph is always shown. Another interesting graph is shown. It helps in understanding the greenhouse effect.

(b) Blackbody Radiation Function 𝜆∗𝑇

𝐹0→𝜆∗(𝜆∗𝑇) = � 0

𝜆∗𝑇

𝐸𝜆,𝑏

𝑑(𝜆𝑇) = 𝜎𝑇5 𝜆∗𝑇

�

𝑐1

𝑑(𝜆𝑇) 𝑐2 5 (exp ( ) − 1) 𝜎(𝜆𝑇) 𝜆𝑇=0 𝜆𝑇

𝑐1 𝐹0→𝜆∗(𝜆 𝑇) = � 𝑑(𝑧) 𝑐 2 5 𝜎𝑧 (exp ( ) − 1) 𝑧=0 𝑧 This integral cannot be evaluated analytically and thus must be evaluated numerically. ∗

(c) Blackbody Radiation Function

20

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

(d) Fraction in the Visible Spectrum Our function is used to analyze spectral distributions. The fraction of sunlight (T = 5800 K) in the visible spectrum ( = 0.4 to 0.7) is 𝐹0→𝜆∗(0.7 ∗ 5800) − 𝐹0→𝜆∗(0.4 ∗ 5800) = 0.491 − 0.123 = 0.368 = 36.8%

(e) Total Emissivity

21

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 4.25: Pipe Flow The volume flow rate of a fluid flowing in a round pipe of radius r0 is 𝑟0

𝑄𝑄 = � 𝑣(𝑟)2𝜋𝑟 𝑑𝑟 0

The velocity can be approximated by 𝑟 1/6 𝑣(𝑟) = 2 (1 − ) 𝑟0

a) Create a function to compute Q as a function of r0. Evaluate the integral numerically. b) Use your function to plot Q versus r0 in the range 0 < 𝑟0 < 6 cm.

Solution Volume Flow Rate in a Circular Pipe 180

160

3

/s)

140

120

Volume Flow Rate (cm

100

80

60

40

20

0 0

1

2

3

4

5

6

Radius (cm)

22

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Chapter 5

Linear Algebra Problem 5.1: Linear Algebra Basics Consider the special cases of the systems of equations Ax = b listed. For all cases: a) Determine the rank of the coefficient matrix, A. � = [A, 𝐛𝐛]. b) Determine the rank of the augmented matrix, A c) Determine if the system consistent? d) For each consistent system, give the solution.

For cases (a)-(f), sketch the row and column interpretation. Summarize these sketches in a table.

Solution Rank(

~ ) A

Case

A

b

Rank(A)

a

�1 0� 1 1

2 � � 1

2

b

1 1 � � 0 0

2 � � 1

1

�1 1� 0 0

0 � � 0

1

1

Yes

0 � � �1 0� 0 0 0 2 1 1 �1� �1 0� 1 0 1 1 1 1 �−1� �1 0� −1 0 1 4 1 2 0 2 �4� �0 1 3 4� 1 0 1 2 4 −1 1 2 −3 1 �0 1 −2 1 � � 2 � 1 0 1 −1 1

1

1

Yes

2

2

Yes

2

3

No

c

d e f g

h

Consistent ?

Solution, x

2

Yes

2 − 1  

2

No No solutions Infinite solutions: Any combination such that x1 = -x2 Infinite number of solutions: Any x2 with x1=0. 1 1  No solutions

3

3

Yes Infinite number of solutions

2

3

No No solutions

1

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 5.2: Equation of a Plane Consider the equation of a plane written as 𝑐1𝑥 + 𝑐2𝑦 + 𝑐3 = 𝑧. a) Given three known points (𝑥1, 𝑦1, 𝑧1), (𝑥2, 𝑦2, 𝑧2), and (𝑥3, 𝑦3, 𝑧3), derive the system of equations that determine c1, c2, and c3. b) Write a function ‘eq_plane’ that determines the vector of coefficients 𝐜𝐜 = [𝑐1, 𝑐2, 𝑐3] given any three points. c) Test your function by finding the equation passing through the points (1,0,0), (0,1,0), (0,0,1).

What are the z values at (x,y) = (0, 0.5), (0.5, 0), (0.25, 0.25), and (0.5, 0.5)? Plot this plane.

Solution (a) The three points must all satisfy the equation of the plane: 𝑐1𝑥1 + 𝑐2𝑦1 + 𝑐3 = 𝑧1 𝑐1𝑥2 + 𝑐2𝑦2 + 𝑐3 = 𝑧2 𝑐1𝑥3 + 𝑐2𝑦3 + 𝑐3 = 𝑧3 This gives 3 equations for the 3 ci values. In matrix form: 𝑧1 𝑥1 𝑦1 1 𝑐1 �𝑥2 𝑦2 1� �𝑐2� = �𝑧2� 𝑧3 𝑥3 𝑦3 1 𝑐3 (c) c = -1 -1 1 z1 = 0.5000 z2 = 0.5000 z3 = 0.5000 z4 = 0

1

z

0.5 0 -0.5 -1 0

1 0.2

0.8 0.4

0.6 0.6

0.4 0.8 x

0.2 1

0

y

2

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 5.3: Pits A civil engineer involved in a construction process requires a volume Vs of sand, a volume Vfg of fine gravel, and a volume Vcg of course gravel. There are three pits from which these materials can be obtained. The composition of these pits is

Pit 1 Pit 2 Pit 3

Sand Fraction S1 S2 S3

Fine Gravel Fraction FG1 FG2 FG3

Course Gravel Fraction CG1 CG2 CG3

a) Develop the equations needed to determine the volumes V1, V2, and V3 that must be hauled from pits 1, 2, and 3 to exactly meet the construction needs. Put in matrix form. b) For the following special case, solve for the volumes V1, V2, and V3.

Pit 1 Pit 2 Pit 3

Sand Fraction 1 0 0

Fine Gravel Fraction 0 1 0

Course Gravel Fraction 0.25 0.25 0.5

Vs = 5000 m3 , Vfg = 5000 m3 , Vcg = 10000 m3 Solution a) 𝑉𝑠 𝑆1 𝑆2 𝑆3 𝑉1 𝑉 �𝐹𝐺1 𝐹𝐺2 𝐹𝐺3� �𝑉2� = � 𝑓𝑓𝑓� 𝐶𝐺1 𝐶𝐺2 𝐶𝐺3 𝑉3 𝑉𝑐𝑓𝑓 b) Solving simultaneously gives 𝑉1 8000 �𝑉2� = �4000� 𝑉3 8000

3

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 5.4 Heated Rod Consider a heated rod with convection from the sides.

a) Write a function to solve for the temperature along the rod for n equally spaced nodes. The function should accept n, L, T0, TL, h’, and Ta as inputs and output the vectors of the x locations and the computed temperatures. Here L is the rod length and TL is the temperature at x = L (T5=200 in the example). b) Write a separate code to compute and plot the following cases: 1. n = 6, L = 1, T0 = 40, TL = 100, h’=0.01, Ta = 20 2. n = 10, L = 1, T0 = 100, TL = 0, h’=0, Ta = 100 3. n = 100, L = 1, T0 = 100, TL = 100, h’=0, 10, and 100, Ta = 10 (single graph with 3 curves for the 3 h’ values) Solution Case 1

Case 2: h = 0

Case 3: Effect of h’

T (deg C)

T (deg C)

T (deg C)

Effect of h

4

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 5.5: Stage Extraction Process A stage extraction process is depicted.

In such systems, a stream containing a weight fraction yin of a chemical enters at a mass flow rate F1. Simultaneously, a solvent carrying a weight fraction xin of the same chemical enters from the other side at a flow rate F2. A mass balance at a typical interior stage can be represented as 𝐹1𝑦𝑖−1 + 𝐹2𝑥𝑖+1 = 𝐹1𝑦𝑖 + 𝐹21𝑥𝑖 , 𝑖 = 2,3, … , 𝑛 − 1 This mass balance must be modified at the first and last stage. At each stage equilibrium is assumed to be established between xi and yi as 𝑥𝑖 𝐾= 𝑦𝑖 where K is a distribution coefficient. a) Draw a cause-effect diagram b) Write a function to solve for the concentrations along the stage extractor with n stages. c) Write a separate function to compute and plot the concentrations and determine yout and xout for the case F1 = 500 kg/h, yin = 0.1, F2=1000 kg/h, xin = 0, and K = 4.

Solution a) A cause-effect diagram is Forcing Functions xin, yin



System K, n F1, F2



Response {y}

Note that yout = yn and xout =K y1. Also, F1 and F2 could be viewed as forcing functions rather than system parameters. b) After using 𝑥𝑖 = Ky𝑖, the mass balance at a typical interior stage is 𝐹1𝑦𝑖−1 − (𝐹1 + 𝐹2𝐾)𝑦𝑖 + 𝐹2Ky𝑖+1 = 0, i = 2, 3, … , n-1 The mass balances at the inlet and exit are 𝐹1𝑦in − (𝐹1 + 𝐹2𝐾)𝑦1 + 𝐹2Ky2 = 0 , i = 1 𝐹1𝑦𝑛−1 − (𝐹1 + 𝐹2𝐾)𝑦𝑛 + 𝐹2𝑥in = 0 , i = n This produces a simultaneous set of n linear algebraic equations for the unknown concentrations.

5

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

(𝐹1 + 𝐹2𝐾) ⎡ −𝐹1

⎢ ⎣

0

−𝐹2𝐾 (𝐹1 + 𝐹2 𝐾) ⋱ 0

0 0

0 0

0 0 −𝐹2 𝐾 0 ⋱ ⋱ (𝐹 −𝐹1 1 + 𝐹2𝐾) ⋱ 0 0 0 0

0 0

0 0

−𝐹2𝐾 ⋱ −𝐹1 0

0 ⋱ (𝐹1 + 𝐹2𝐾) −𝐹1

c) yout = 2.6703e-006 xout = 0.049999

𝐹1𝑦in 𝑦 ⎤ ⎡ 𝑦1 ⎤ ⎡ 0 ⎤ 2 ⎢ ⋮ ⎥ ⎢ ⋮ ⎥ ⎢ ⎥ ⎢ ⎥ =⎢ 0 ⎥ 𝑦 𝑖 ⎥ 0 ⎢ ⎢ ⋮ ⎥ ⎢ ⋮ ⎥ −𝐹2𝐾 ⎥ ⎢𝑦𝑛−1⎥ ⎢ 0 ⎥ (𝐹1 + 𝐹2𝐾)⎦ ⎣ 𝑦𝑛 ⎦ ⎣𝐹2𝑥in⎦ 0 0

The weight fractions are plotted on both a linear-linear scale and on a log-linear scale.

Weight Fractions

0.06

0.04

0.02

0 1

1.5

2

2.5

3 Stage #

3.5

4

4.5

5

1

1.5

2

2.5

3 Stage #

3.5

4

4.5

5

0

Weight Fractions

10 -2

10 -4

10 -6

10

Comments If the purpose of this device is to remove some chemical from fluid stream 1, the device seems to be a great success.

6

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 5.6: Pentadiagonal Solver A pentadiagonal system of equations is a special system with a bandwidth of 5. An n-by-n pentadiagonal system has the following form: 𝑓1 𝑔1 ℎ1 ⋅ 0 𝑟1 ⎡ ⎤ 𝑥1 ⎡ ⎡ ⎤ 𝑒 𝑓 ℎ 2 2 𝑔2 2 𝑟2 ⎤ ⎥ 𝑥2 ⎢ ⎢ ⎢ ⎥ 𝑟3 ⎥ ℎ3 ⎢𝑑3 𝑒3 𝑓3 𝑔3 ⎥ 𝑥3 ⎢ ⋅ ⎥ ⎢ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⎥⎢ ⋅ ⎥ ⎢ ⎥ ⎥ ⎢⎢ 𝑥𝑖 ⎥⎥ ⎢ 𝑟𝑖 ⎥ 𝑑𝑖 𝑒𝑖 𝑓𝑖 𝑔𝑖 ℎ𝑖 ⎢⎢ ⋅ 𝑑𝑖+1 𝑒𝑖+1 𝑓𝑖+1 𝑔𝑖+1 ℎ𝑖+1 ⋅ ⎥ ⎢ 𝑥 𝑖+1 ⎥ = ⎢𝑟𝑖+1 ⎥ ⎢ ⎥ ⎢𝑥𝑖+2 ⎥ ⎢ 𝑟𝑖+2 ⎥ 𝑑𝑖+2 𝑒𝑖+2 𝑓𝑖+2 𝑔𝑖+2 ℎ𝑖+2 ⎥⎢ ⋅ ⎥ ⎢ ⎢ ⋅ ⎥ ⋅ ⋅ ⋅ ⋅ ⎥ ⎢ ⎢𝑥𝑛−1⎥ ⎢𝑟𝑛−1⎥ 𝑑 𝑒 𝑓 𝑔 ⎢ 𝑛−1 𝑛−1 𝑛−1 𝑛−1⎥ ⎣ 𝑥 𝑛 ⎦ ⎣ 𝑟𝑛 ⎦ 𝑒𝑛 𝑓𝑛 ⎦ ⎣0 ⋅ 𝑑𝑛 a) Write a flowchart or pseudocode describing the logic required to solve this special system of equations. Only the diagonals and RHS should be input. Do not form a full matrix or perform a full matrix solution. b) Write a function to solve this pentadiagonal system. c) Test your function with the following special cases.

Case 1: 8 −2 −1 0 0 𝑥1 5 ⎡−2 9 −4 −1 0 ⎤ 𝑥2 2 −1 −3 7 −1 −2 𝑥3 = 1 ⎢ 0 −4 −2 12 −5⎥ ⎢𝑥4⎥ ⎢1⎥ ⎣0 0 −7 −3 15 ⎦ ⎣𝑥5⎦ ⎣5⎦ Case 2: di’s = -1, ei’s = -3, fi’s = 8, gi’s = -3, hi’s = -1, ri’s = 1. Try n = 10000 and 100000.

Please plot your solutions rather than print out all these numbers. For comparison, you should try to solve this problem using the standard full matrix solver: x = A\b.

Solution Pentadiagonal Matrix Algorithm It might help to visualize this system of equations by using a table with gridlines. f1 e2 d3 0 …

0

g1 f2 e3 …

0

h1 g2 f3 … di 0

0 h2 g3 … ei di+1 0

…

0 h3 … fi ei+1 di+2 …

0 … gi fi+1 ei+2 …

0 …

0

hi gi+1 fi+2 … dn-2 0

0 hi+1 gi+2 … en-2 dn-1 0

0 hi+2 … fn-2 en-1 dn

…

0 gn-2 fn-1 en

hn-2 gn-1 fn

x1 x2

r1 r2

… xi xi+1 xi+2 …

… ri ri+1 ri+2 …

xn-1 xn

=

rn-1 rn

7

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Note that n is the number of equations and i is a dummy index. After the forward elimination of the d and e vectors, we are left with the following. f1 0 0 0

g1 f2 0 …

0

…

h1 g2 f3 …

0 0

…

0 h2 g3 …

0 0 0

0 h3 … fi 0 0 …

0 … gi fi+1 0 …

0

0

hi gi+1 fi+2 …

0 0

…

0

0 hi+1 gi+2 …

0 0 0

…

0

0 hi+2 … fn-2 0 0

gn-2 fn-1 0

hn-2 gn-1 fn

x1 x2

r1 r2

… xi xi+1 xi+2 …

… ri ri+1 ri+2 …

=

xn-1 xn

rn-1 rn

The vectors e and f have been eliminated and the vectors f, g, h, and r have been changed. We can now back substitute. 𝑟𝑛 𝑥𝑛 = 𝑓𝑛 𝑟𝑛−1 − 𝑔𝑛−1𝑥𝑛 𝑥𝑛−1 = 𝑓𝑛−1 For i = n-2 : -1 : 1 𝑟𝑖 − 𝑔𝑖𝑥𝑖+1 − ℎ𝑖𝑥𝑖+2 𝑥𝑖 = 𝑓𝑖 Case 1 Case 2 9

x 10

1.4

18 16

1.35

14

1.3 12

1.25 x(i)

x(i)

10

1.2

8 6

1.15

4

1.1 2

1.05

1

1.5

2

2.5

3 i

3.5

4

4.5

5

0 0

2

4

6

8

i

Case 1 xpenta =

1.0825

1.1759

1.3082

1.1854

10 5

x 10

1.1809

The pentadiagonal solver had no problem with 10000 or 100000 equations. The full matrix solver using the left-divide crashed the computer for 10000 equations.

8

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 5.7 Consider the cooling of an integrated circuit package from Section 5.3.1. The purpose of this exercise is to write a function to solve for the unknown heat flows and temperatures and test this function. You are requested to: a) Write a function called ‘CoolingIC’ that computes the heat flow rates and temperatures in the IC package. The inputs should be: R = vector containing the 5 thermal resistance values, Ta = ambient temperature, and Qc = power dissipated. The function returns two vectors: T = a vector containing the 3 unknown temperatures.

Q = a vector containing the 4 unknown heat flow rates b) Test your ‘CoolingIC’ function. This function assigns the input values, calls on ‘CoolingIC’ to compute the T and Q vectors, and displays the output. Consider the following test cases.

Case R1 R2 R3 R4 R5 Qc Ta (C/W) (W) (C) 1 2 2 2 2 2 0 25 2 2 2 2 2 2 10 25 3 2 0.5 35 0.7 1 10 25 4 1000 0.5 35 0.7 1 10 25 Solution Case 1 --- Inputs --Ta = 25.0 degC Qc = 0.00 W R1 = 2.00 degC/W R2 = 2.00 degC/W R3 = 2.00 degC/W R4 = 2.00 degC/W R5 = 2.00 degC/W

Case 2 --- Inputs --Ta = 25.0 degC Qc =10.00 W R1 = 2.00 degC/W R2 = 2.00 degC/W R3 = 2.00 degC/W R4 = 2.00 degC/W R5 = 2.00 degC/W

Case 3 --- Inputs --Ta = 25.0 degC Qc =10.00 W R1 = 2.00 degC/W R2 = 0.50 degC/W R3 =35.00 degC/W R4 = 0.70 degC/W R5 = 1.00 degC/W

Case 4 --- Inputs --Ta = 25.0 degC Qc =10.00 W R1 =1000. degC/W R2 = 0.50 degC/W R3 =35.00 degC/W R4 = 0.70 degC/W R5 = 1.00 degC/W

--- Outputs --Tc = 25.0 degC Tp = 25.0 degC Tw = 25.0 degC Q1 = 0.00 W Q2 = 0.00 W Q3 = 0.00 W Q4 = 0.00 W

--- Outputs --Tc = 37.5 degC Tp = 30.0 degC Tw = 27.5 degC Q1 = 3.75 W Q2 = 1.25 W Q3 = 6.25 W Q4 = 2.50 W

--- Outputs --Tc = 48.1 degC Tp = 29.5 degC Tw = 28.0 degC Q1 = 9.34 W Q2 = 2.97 W Q3 = 0.66 W Q4 = 6.37 W

--- Outputs --Tc =363.2 degC Tp = 25.2 degC Tw = 25.1 degC Q1 = 0.34 W Q2 = 0.11 W Q3 = 9.66 W Q4 = 0.23 W

Observations: • Case 1: zero input (Qc = 0) produces zero output (Q’s = 0, T’s = Ta) • Case 4: when R1 is large (chip not attached well to the plate), the only cooling mechanism is to the air and the chip overheats

9

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 5.8: Cooling of an IC Package, Parameter Study We will now use the function developed in the previously problem to perform a parameter study on the cooling arrangement of the IC system. The most important engineering result of this analysis is the chip temperature, Tc. This is particularly crucial since electronic chips can fail if overheated. The cooling arrangement must maintain the chip temperature below a critical failure temperature, Tcrit. As electrical circuits become smaller and draw more power, overheating becomes a limiting constraint in their design. The purpose of this exercise is to perform the following parameter studies. a) Study of chip temperature as a function of power. Create a plot of Tc vs Qc with Qc ranging from 0 to 100 W. On a single graph, put curves for R4 = R5 = 0, 2, 4, 6 C/W. Take other parameter values from Case 3 in the previous problem. What can you conclude about Tc vs Qc? If Tcrit = 150C, what is the maximum allowable chip power, Qc, for each case. b) Study of chip temperature as a function air flow (R4 and R5). The resistances R4 and R5 represent the effect of the air cooling; a high resistance corresponds to low air velocity and low resistance corresponds to high air velocity. Assume R4 = R5 and create a plot of Tc vs R4 with R4 = R5 ranging from 0 (hurricane) to 100 C/W (stagnant air). On a single graph, put curves for Qc = 0, 5, 10, 15 W. Take other parameter values from Case 3 in the previous problem. If Tcrit = 150C, what is the allowable range of resistance values for Qc = 10 W. You should write a function to perform these studies. This function should call on the previously developed function from Problem 5.7.

Solution a) Effect of Qc on Tc for various R4 = R5 values 500 0 2 4 6

450

Chip Temperature, Tc (C)

400 350 300 250 200 150 100 50 0 0

10

20

30

40 50 60 70 Power Dissipated, Qc (W)

80

90

100

Maximum Power Dissipation R4=R5(degC/W) Qmax (W) 0.000 66.1 2.000 43.8 4.000 33.9 6.000 28.0 b)

10

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Effect of R4 = R5 on Tc for various Qc (W) values 450 400

Chip Temperature, Tc (C)

350 300 250 200 150 100 50 0 0

10

20

30 40 50 60 70 Thermal Resistance, R4=R5 (deg C/W)

80

90

100

For Qc = 10 W, the allowable range of R4=R5 is 0 to 34.6 degC/W.

11

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 5.9: Voltage Range Consider the circuit shown in the figure.

+

+

-

-

a) If all resistance and voltage values are considered known, derive the equations necessary to determine the 5 unknown currents. b) Compute the values of the currents given the following values of the resistances and the voltages; R1 = 5 kΩ, R2 = 100 kΩ , R3 = 200 kΩ , R4 = 150 kΩ , R5 = 250 kΩ v1 = 100 volts, v2 = 100 volts c) Suppose that each resistor is rated to carry a current of no more than 1 milliampere (=0.001 amperes). Determine the allowable range of positive values for the voltage v2. Use the resistances and v1 values from part b. d) Suppose we want to investigate how the resistance R3 limits the allowable range for v2. Obtain a plot of the allowable limits on v2 as a function of R3 for 150 < R3 < 250 kΩ.

Solution a) Using Kirchhoff’s voltage and current laws, the equations describing the circuit shown in the figure are: -v1 + R1 i1 + R4 i4 = 0 -R4 i4 + R2 i2 + R5 i5 = 0 -R5 i5 + R3 i3 + v2 = 0 i1 = i2 + i4 i2 = i3 + i5 In matrix form, 𝑖1 𝑅1 0 0 𝑅4 0 𝑣1 ⎡ 0 𝑅2 ⎡ ⎤ ⎤ 0 0 −𝑅4 𝑅5 𝑖2 0 0 𝑅3 0 −𝑅5 𝑖3 = −𝑣2 ⎢ 1 −1 0 −1 0 ⎥ ⎢𝑖4⎥ ⎢ 0 ⎥ ⎣0 1 −1 0 −1 ⎦ ⎣𝑖5⎦ ⎣ 0 ⎦ b) i(mA) 0.8299 0.1909 -0.1162 0.6390 0.3071

12

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

c) Resistance, R(3) = 200 Ohm Applied voltage, v1 = 100 V Maximum Current, imax = 0.0010 A Minimum required voltage, v2 = 32 V i(amps) imax-abs(i)(amps) 0.00100 0.00000 0.00037 0.00063 0.00013 0.00087 0.00063 0.00037 0.00023 0.00077 Maximum allowable voltage, v2 = 343 V i(amps) imax-abs(i)(amps) 0.00022 0.00078 -0.00043 0.00057 -0.00100 -0.00000 0.00066 0.00034 0.00057 0.00043 The allowable voltage range is 32 to 343 Volts.

Part (d)

13

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 5.10: Equilibrium Position of a System of Linear Springs and Masses Consider the system of masses and linear springs. The total length of each spring is equal to the unstretched length (Li) plus the stretched length (xi). The width of each block is W and the total length of the system is LT. For a linear spring, the force is directly proportional to elongation, 𝐹𝑠𝑝𝑟𝑖𝑛𝑓𝑓 = 𝑘𝑥. k2

k1

k3

k4

LT

a) Considering all the ki, Li, and W as known, derive the mathematical model for the stretched lengths xi. b) Write a function to solve the system of equations in part (a). c) Use the function developed in part (b) to determine the xi’s for the following cases.

Case 1 W=0.2 m LT=8 m Li=1 m ki=2 N/m

Case 2 W=0.2 m LT=8 m Li=1 m k1=1, k2=2, k3=3, k4=4 N/m.

Solution (a) Mathematical Model Total length

𝑥1 + 𝑥2 + 𝑥3 + 𝑥4 = 𝐿𝑇 − 𝐿1 − 𝐿2 − 𝐿3 − 𝐿4 − 3𝑊𝑊

Mass-1

−𝑘1𝑥1 + 𝑘2𝑥2 = 0

Mass-2

−𝑘2𝑥2 + 𝑘3𝑥3 = 0

Mass-3

−𝑘3𝑥3 + 𝑘4𝑥4 = 0

In matrix form 1 1 −𝑘1 𝑘2 �0 −𝑘2 0 0 (a) Numerical Results Case 1

x

0.85 0.85 0.85 0.85

1 0 𝑘3 −𝑘3

𝑥1 𝐿𝑇 − 𝐿1 − 𝐿2 − 𝐿3 − 𝐿4 − 3𝑊𝑊 𝑥2 0 � � � � = � 0 𝑥3 0 𝑘4 𝑥4 0 1 0

Case 2

1.632 x 0.816 0.544 0.408 14

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Chapter 6

Non-Linear Algebra: Root Finding Problem 6.1: Graphical Root Finding Consider the equation 𝑒−𝑥 = 𝑥. Estimate the real root of this equation in the following ways. (a) Graphically by plotting the functions 𝑒−𝑥and x. (b) Graphically by plotting the function 𝑓(𝑥) = 𝑒−𝑥 − 𝑥. (c) Using three iterations of the bisection method with initial guesses xl = 0 and xu = 2. (d) Using three iterations of the false-position method with initial guesses xl = 0 and xu = 2. (e) Using three iterations of the Newton-Raphson method with initial guess x0 = 2.

Summarize your results for parts (c) and (d) with a table.

Solution (a) Graphical by plot of 𝑒−𝑥and x.

These graphs show that the root is approximately xr  0.57. (b) Graphical by plot of 𝑓(𝑥) = 𝑒−𝑥 − 𝑥.

(c) Bisection Iteration 1 2 3

xl 0 0 0.5

xu 2 1 1

xr 1 0.5 0.75

After 3 iterations of the bisection method, our estimate of the root is xr = 0.75.

1

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

(d) False Position

Iteration 1 2 3

𝑥𝑟 = 𝑥𝑢 − 𝑓(𝑥𝑢 ) (

𝑥𝑢 − 𝑥𝑙 ) 𝑓(𝑥𝑢) − 𝑓(𝑥𝑙 )

xl 0 0 0

xu 2 0.6982 0.5815

xr 0.6982 0.5815 0.5687

After 3 iterations of the false position method, our estimate of the root is xr = 0.5687. This is quite a bit more accurate than the estimate after 3 iterations using the bisection. (e) Newton-Raphson 𝑓(𝑥) = 𝑒−𝑥 − 𝑥 𝑓′(𝑥) = −𝑒−𝑥 − 1 𝑓(𝑥𝑖) 𝑒−𝑥𝑖 − 𝑥𝑖 𝑥𝑖+1 = 𝑥𝑖 − ′ = 𝑥𝑖 − 𝑓 (𝑥𝑖 ) −𝑒−𝑥𝑖 − 1 Starting from x0 = 2, we get the following iterations.

Iteration 0 1 2 3

𝑥𝑖 2.0 0.3576 0.5587 0.5671

The iteration is converging on the correct answer very quickly compared to the bisection or false position method.

2

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 6.2: Exploring Root Finding Determine the positive real root of the equation ln(𝑥2) = 0.7 (a) Graphically. (b) Using three iterations of the bisection method, with initial guesses xl = 0.5 and xu = 2. (c) Using three iterations of the false-position method, with initial guesses xl = 0.5 and xu = 2.

Solution

Bisection Method xl xu 0.5000 2.0000 1.2500 2.0000 1.2500 1.6250 1.2500 1.4375

xr 1.2500 1.6250 1.4375 1.3438

False-Position Method xl xu xr 0.5000 2.0000 1.6287 0.5000 1.6287 1.4970 0.5000 1.4970 1.4484 0.5000 1.4484 1.4302

3

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 6.3 Consider a metal plate exposed to the sun with an insulated (no heat flow) bottom surface, as described in Section 6.2.3. a) Determine the plate temperature, T, on a day when Gs = 900 W/m2 and h = 15 W/m2 K. b) Compute a table of the plate temperature T for values of Gs ranging from 0 to 1200 W/m2 in increments of 100 for h = 15 W/m2 K. Make a plot of T vs. Gs. c) Compute a table of the plate temperature, T, for values of the heat transfer coefficient, h, ranging from 10 to 200 W/m2 K in increments of 10 when Gs = 900 W/m2. Make a plot of T vs. h. As h becomes extremely large, what is the plate temperature?

Solution (a) The plate temperature is determined by an energy balance. In steady-state 0 = 𝑞" − 𝑞" − 𝑞" 𝑠𝑢𝑛

𝑐𝑜𝑛𝑣

𝑟𝑎𝑑

0 = 𝛼𝑠 ∙ 𝐺𝑠 − ℎ(𝑇 − 𝑇∞) − 𝜀𝜀 ∙ 𝜎 ∙ 𝑇4 2 When Gs = 1200 W/m , the plate temperature is 314.7 K (b) Effect of Solar Flux, Gs 340 330

320

T (K)

310 300

290

280

270

0

200

400

600

800

1000

1200

Gs (W/m2)

(c) Effect of Heat Transfer Coefficient, h As h becomes extremely large, the plate temperature approached the air temperature.

320 318 316 314

T (K)

312 310 308 306 304 302 300

0

50

100

150

200

h (W/m2 K)

4

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 6.4 The Redlich-Kwong equation of state is given by 𝑅𝑇 𝑎 − 𝑝= 𝑣 − 𝑏 𝑣(𝑣 + 𝑏)√𝑇 where R = gas constant, T = absolute temperature (K), p = absolute pressure (kPa), and v = specific volume (m3/kg). The parameters a and b are calculated by 𝑅2𝑇𝑐2.5 𝑇𝑐 𝑎 = 0.427 , 𝑏 = 0.0866𝑅 𝑝𝑐 𝑝𝑐 For methane, R = 0.518 kJ/(kgK), pc = 4580 kPa, and Tc = 191 K. As a chemical engineer, you are asked to determine the amount of methane fuel that can be held in a 3-m3 tank at a temperature of -50 degrees Celsius with a pressure of 65,000 kPa. a) Estimate v using the graphical method. b) Use a root locating method to calculate v and then determine the mass of methane contained in the tank. c) Compare the Redlich-Kwong equation of state compare to the ideal gas law?

Solution (a) Graphical Approximation 6

x 10 8 6 4 root 2

f(v)

0 -2 -4 -6 -8 -10

0

0.5

1

1.5

2

2.5

3 -3

v (m3/kg)

x 10

(b) Numerical Solution v_RedKw = 0.00187 m^3/kg v_Ideal = 0.00178 m^3/kg mass = 1602.8 kg (c) Comparison with Ideal Gas Law 0.03 RedKw Ideal 0.025

v (m3/kg)

0.02

0.015

0.01

0.005

0 0

1

2

3

4 5 6 Pressure (kPa)

7

8

9

10 4

x 10

5

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 6.5: Depth of Water in a Tank You are designing a spherical tank to hold water. The volume of liquid in the tank is (3𝑅 − ℎ) 𝑉 = 𝜋ℎ2 3 where V = volume (m3), h = depth of water in the tank (m), and R = tank radius (m). (a) We wish to determine the required height h for given values of R and V. Derive the formula that uses the Newton-Raphson method to determine h. (b) What is the range of h values that would physically make sense for an initial guess? (c) Perform one iteration with your formula with parameters R = 3 m, V = 30 m3. Use an initial guess of h0 = 1 m.

Solution (3𝑅−ℎ) (a) 𝑓(ℎ) = 𝜋ℎ2 −𝑉 3

(6𝑅ℎ − 3ℎ2) 𝑑𝑓 =𝜋 = 𝜋(2𝑅ℎ − ℎ2) 𝑑ℎ 3 (3𝑅 − ℎ𝑖) 𝑓(ℎ𝑖 ) 𝜋ℎ𝑖 2 −𝑉 3 ℎ =ℎ − =ℎ − 𝑖+1

𝑖

𝑓′(ℎ𝑖)

𝑖

𝜋(2𝑅ℎ𝑖 − ℎ𝑖2)

(b) 0 < ℎ < 2𝑅 (c) ℎ1 = 2.376

6

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 6.6: Optimal Fin Spacing Heat sinks are often attached to electronic devices to increase the cooling efficiency and thereby lower the temperature of the device. One common configuration of these heat sinks is an array of pin fins. Given the overall dimensions of a heat sink consisting of pin fins, it is desirable to know the optimal fin spacing, 𝑆𝑜𝑝𝑡. The empirical formula for the optimal spacing is 𝑆𝑜𝑝𝑡 𝐻 1/3 ( ) 2 + 𝑆𝑜𝑝𝑡/𝐷2/3 = 2.75 ( ) 𝑅𝑎−1/4 𝐷 (1 + 𝑆𝑜𝑝𝑡 /𝐷) 𝐷 where D = diameter (m), H = height (m), Ra = Rayleigh number. a) Write a function to compute the optimal spacing given D, H, and Ra. Include your algorithm in the form of a flowchart or pseudocode. b) Use your function to plot 𝑆𝑜𝑝𝑡/𝐷 versus Ra over the range 300 < Ra < 10000 for H/D = 5, 10, 15, 20.

Solution a) Algorithm BEGIN ‘SoptStudy’ INPUT: H, Ra Compute Sopt using ‘Sopt’ for all H, Ra values Plot Sopt vs Ra at each H value END ‘SoptStudy’ BEGIN ‘Sopt’ INPUT: D, H, Ra SoptGuess = D Compute Sopt using a ‘fzero’ OUTPUT Sopt END ‘Sopt’

b) Results 1 0.9 0.8 0.7

Sopt / D

0.6 0.5 0.4 0.3 0.2 0.1 0 0

1000

2000

3000

4000 5000 6000 Rayleigh #, Ra

7000

8000

9000

10000

Comments Since the Rayleigh number represents the strength of the buoyancy, the fins can be packed closer at higher Ra. Also, a larger total height requires fins further apart due to the heating effect between fins.

7

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 6.7: Friction Factor and Moody Diagram The Moody diagram is a classic found in virtually all books on fluid mechanics and heat transfer. It was originally published by L. F. Moody in the Transactions of the ASME in 1944 and is one of those rare diagrams that has passed the test of time and is still used today. This diagram shows the friction factor f as a function of Reynolds number 𝑅𝑒𝐷 for various values of the relative roughness /D. Variable definitions are: 𝑑𝑃 𝑓 = 𝑓𝑟𝑖𝑐𝑡𝑖𝑜𝑛 𝑓𝑎𝑐𝑡𝑜𝑟 = − 𝐷/(𝜌 𝑢2 /2) 𝑚 𝑑𝑥 2 P = pressure (N/m ), x = distance along pipe (m),  = roughness (m), D = pipe diameter (m), 𝑅𝑒𝐷 = 𝑢𝑚𝐷/𝜈𝜈, 𝑢𝑚 = mean fluid velocity (m/s),  = viscosity (m2/s). A correlation between f, 𝑅𝑒𝐷 and /D was developed by Colebrook for fully developed turbulent pipe flow 1 𝜖/𝐷 2.51 ) = −2𝑙𝑜𝑔10 ( + 3.7 �𝑓 𝑅𝑒 𝐷 �𝑓 This equation is valid for 𝑅𝑒𝐷 > 3000. For 𝑅𝑒𝐷 < 3000, the flow is laminar flow and the friction factor for all /D values is 𝑓 = 64/𝑅𝑒𝐷. a) Write a function to compute the friction factor as a function of /D and 𝑅𝑒𝐷. For 𝑅𝑒𝐷>3000, the Colebrook correlation should be used. An initial guess at the root can be obtained from −2

𝜖/𝐷 1.11 +( ) )) 𝑓 = (1.8𝑙𝑜𝑔10 ( 3.7 𝑅𝑒 𝐷 6.9

For 𝑅𝑒𝐷<3000, the laminar flow formula, 𝑓 = 64/𝑅𝑒𝐷, should be used. b) Use your function to plot the friction factor data on a log-log scale, just like the Moody diagram. Use the /D values shown in the figure.

Solution -1

friction factor, f

10

-2

10

3

10

4

10

5

6

10

10

7

10

8

10

ReD = u*D/vis

8

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 6.8: Enzyme Kinetics The Michaelis-Menten model describes the kinetics of enzyme mediated reactions: 𝑑𝑆 𝑆 = −𝑣𝑚 𝑑𝑡 𝑘𝑠 + 𝑆 Where S = substrate concentration (moles/L), vm = maximum uptake rate (moles/L/d), and ks = half saturation constant, which is the substrate level at which uptake level is half of the maximum (moles/L). If the initial substrate level at t = 0 is S0, this differential equation can be solved for 𝑆 = 𝑆0 − 𝑣𝑚𝑡 + 𝑘𝑠𝑙𝑛(𝑆0/𝑆) a) Develop a function to determine S as a function of t, S0, vm, and ks. b) Use your function to plot S versus t for S0 = 0, 10, and 20 moles/L. Use vm = 0.5 moles/L/d and ks = 2 moles/L. Experiment with the maximum time so that a steady state is reached. Put all 3 curves on a single graph.

Solution

9

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 6.9: Basics a) Determine all the roots of 𝑓(𝑥) = −14 − 20𝑥 + 19𝑥2 − 3𝑥3 graphically. b) Determine the first root of the function with the bisection method. c) Determine the first root of the function with the false position method.

For parts (b) and (c), use initial guesses of 𝑥𝑙 = −1 and 𝑥𝑢 = 0, and a stopping criteria of |𝑒𝑎| ≤ 1%.

Solution a) Graphical 30 1.5 1

10

0.5

f(x)

f(x)

20

0

0

-0.5 -10 -1 -20 -1

0

1

2 x

3

4

-0.6

5

-0.55

-0.5

-0.45

-0.4

x

From the graph, three roots located at xr  -0.5, 2.1, and 4.7 can be seen. Once the roots have been spotted, any one of them can be zoomed in on to get a better look. For instance, the graph on the right zooms in on the root at xr  -0.5 and shows a more accurate estimated of xr  -0.47. b) Bisection i 1 2 3 4 5 6 7 8

Bisection Method: tol = 0.0100 xl xu xr -1.0000 0.0000 -0.5000 -0.5000 0.0000 -0.2500 -0.5000 -0.2500 -0.3750 -0.5000 -0.3750 -0.4375 -0.5000 -0.4375 -0.4688 -0.5000 -0.4688 -0.4844 -0.4844 -0.4688 -0.4766 -0.4766 -0.4688 -0.4727

ea 0.0200 1.0000 0.3333 0.1429 0.0667 0.0323 0.0164 0.0083

Iterations using Bisection Method 0

xl,xu,xr

-0.2

-0.4

-0.6

-0.8

-1 1

2

3

4 5 iteration

6

7

8

c) False Position False Position Method: tol = 0.0100

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Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

i 1 2 3 4 5

xl -1.0000 -1.0000 -1.0000 -1.0000 -1.0000

xu 0.0000 -0.3333 -0.4362 -0.4633 -0.4701

xr -0.3333 -0.4362 -0.4633 -0.4701 -0.4717

ea 0.0200 0.2359 0.0584 0.0144 0.0036

Iterations using False Position Method 0

xl,xu,xr

-0.2

-0.4

-0.6

-0.8

-1 1

1.5

2

2.5

3 iteration

3.5

4

4.5

5

Comments Results show that the bisection method converges to xr = -0.4727 with a tolerance of ea<=0.01after 8 iterations while the false position method converges to xr = -0.4717 after only 5 iterations. Thus, the false position method seems more efficient.

11

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 6.10: Fixed Points of an ODE Consider the first order ODE: 𝑑𝑥/𝑑𝑡 = 𝑒𝑥 − 𝑐𝑜𝑠(𝑥). a) Can you find the fixed points (steady state) explicitly? b) Show the location of the fixed points graphically. c) For this problem, what is the iteration formula needed to find the fixed points using the Newton-Raphson method? d) Starting with an initial guess of 𝑥0 = 10, what result would the Newton-Raphson converge to after many iterations?

Solution a) No b)

The fixed points appear to be at x=-1.3 and x=0. c) 𝑓(𝑥) = 𝑒𝑥 − 𝑐𝑜𝑠(𝑥) 𝑑𝑓 = 𝑒𝑥 + 𝑠𝑖𝑛(𝑥) 𝑑𝑥 𝑥𝑖 𝑖) 𝑥𝑖+1 = 𝑥𝑖 − 𝑓(𝑥 = 𝑥𝑖 − 𝑒 𝑥 − 𝑐𝑜𝑠(𝑥𝑖) ′ 𝑓 (𝑥 ) 𝑒 𝑖 + 𝑠𝑖𝑛(𝑥 ) 𝑖

𝑖

d) 𝑥 = 0

12

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 6.11: Projectiles Aerospace engineers sometimes compute the trajectories of projectiles such as rockets. A related problem deals with the trajectory of a thrown baseball. The trajectory of a ball thrown by a fielder is defined by the (𝑥, 𝑦) coordinates displayed in the figure.

The trajectory can be derived as 𝑦 = 𝑦 + 𝑥 tan(𝜃𝜃 ) − 0

0

𝑔𝑔

𝑥2 .

2𝑣02 cos(𝜃0)2

a) Derive the above formula from physical principles. b) Create an animation of y vs x with controls for the parameters. c) Create an animation of a parametric plot of 𝑦(𝑡) 𝑣𝑠 𝑥(𝑡) where t is a parameter, along with the other system parameters. This is a “shooting star” animation. d) Create a function to compute the initial angle 𝜃𝜃0 as a function of the other variables. e) Use your function to study the effect of a baseball thrown by a fielder to home plate. Create a plot of 𝜃𝜃0 (in degrees) versus x. Use an x-range of 1m to 130 m (deep outfield). On a single graph, plot curves corresponding to v0 = 25, 35, and 45 m/s. Note that 45 m/s is about 100 mph which is the limit for a top major league baseball player. Also assume the fielder releases the ball at an elevation of 2 m and the catcher receives it at 0 m.

Solution (a) The stated trajectory equation can be derived from Newton’s Law. 𝑑2𝑥 𝑚𝑎𝑥 = ∑𝐹𝑥𝑚 2 = 0 𝑑𝑡 Integrate twice and use initial conditions to get 𝑥(𝑡) = 𝑣0 ∗ Cos[𝜃𝜃0] ∗ 𝑡 Now in the y-direction 𝑑2𝑦 𝑚 2 = −𝑚𝑔 𝑑𝑡 Integrate twice and use initial conditions to get 𝑡2 𝑦(𝑡) = −𝑔 ∗ + 𝑣0 ∗ Sin[𝜃𝜃0] ∗ 𝑡 + 𝑦0 2 Combine these to get 𝑔 𝑦 = 𝑦 + 𝑥 tan(𝜃𝜃 ) − 𝑥2 2 0 0 2 ) 2𝑣 0cos(𝜃𝜃0 (c)

13

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Trajectory of a Projectile 50 45 40 35

y (m)

30 25 20 15 10 5 0 0

10

20

30

40

50 x (m)

60

70

80

90

100

(d) The solution requires a root finder to solve the following non-linear algebraic equation for 0. 𝑔 𝑓(𝜃𝜃 ) = 𝑦 − 𝑦 + 𝑥 tan(𝜃𝜃 ) − 𝑥2 = 0 0 0 0 2𝑣02 cos(𝜃𝜃0)2 (e) Numerical Results 90 80

Initial Angle, theta0 (deg)

70 60 50 40 30 20 10 0 20

30

40

50

60 70 80 90 Horizontal Distance, x (m)

100

110

120

Comments If the velocity is too low (25 m/s), the fielder is unable to throw more than about 65 m.

Mathematically, there is no root if the desired distance is too large and MATLAB returns an answer of ‘NaN’.

When solutions are possible there are always two complimentary solutions, corresponding to a low arc and a high rainbow arc. These angles are almost symmetric about the 𝜃𝜃0 = 45 deg line. This is explained by examining some trajectories.

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Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 6.12: Simultaneous Equations Consider the following equations. Sketch solutions to each of these equations in the x-y plane and indicate the x, y points that are solutions to both equations. Apply one step of the NewtonRaphson formula using initial guess (𝑥0, 𝑦0) = (0,0). a) 𝑓(𝑥, 𝑦) = 2𝑥 − 𝑦 = 0 𝑔(𝑥, 𝑦) = 𝑥 + 𝑦 − 3 = 0 b) 𝑓(𝑥, 𝑦) = 𝑦 − 2 = 0 𝑔(𝑥, 𝑦) = 𝑥2 − 𝑦 = 0 Solution Use Equations (6.35) and (6.36)

𝜕𝑔𝑖

− 𝑔 𝜕𝑓𝑖 𝜕𝑓𝑖 𝜕𝑔𝑖 𝑔𝑖 𝜕𝑥 − 𝑓𝑖 𝜕𝑥 𝑖 𝜕𝑦 𝑥𝑖+1 = 𝑥𝑖 − , 𝑦𝑖+1 = 𝑦𝑖 − 𝐽𝑖 𝐽𝑖 𝜕𝑓 𝜕𝑔 𝜕𝑔 𝜕𝑓 𝐽 = Jacobian = − 𝜕𝑥 𝜕𝑦 𝜕𝑥 𝜕𝑦 𝑓

𝑖 𝜕𝑦

(a) 𝜕𝑓 𝜕𝑓 =2, = −1 𝜕𝑥 𝜕𝑦 𝜕𝑔 𝜕𝑔 =1 =1, 𝜕𝑥 𝜕𝑦 𝐽 = Jacobian = 3 (−3)(2) − (0)(1) (0)(1) − (−3)(−1) =2 𝑥1 = 0 − = 1, 𝑦1 = 0 − 3 3 (b) 𝜕𝑓 =0, 𝜕𝑥 𝜕𝑔 = 2𝑥 , 𝜕𝑥

𝜕𝑓 𝜕𝑦 𝜕𝑔 𝜕𝑦

=1 = −1

𝐽 = Jacobian = −2𝑥 Cannot start using an initial guess of 𝑥0 = 0 since the Jacobian is zero and you get division by zero.

15

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 6.13: Graphical Solutions Consider the function 𝑓(𝑥) = 𝑎𝑥 + 𝑏 − sin(𝑥) a) Graphically show the roots of 𝑓(𝑥) = 0 when a = b = 0. What are the roots? b) Graphically show the roots 𝑓(𝑥) = 0 when a = 0 for various values of b. Show the roots by plotting 𝑓(𝑥) vs x. On a separate plot, show the roots by plotting both 𝑎𝑥 + 𝑏 𝑎𝑛𝑑 sin(𝑥) vs x. What is the range of b for which roots exist? c) Graphically show the roots when b = 0 for various values of a. Are roots always possible?

Are there any ranges of a for which multiple roots exist? d) Find the root of 𝑓(𝑥) = 0 for • a = 0.1 and b = 0 with an initial guess of 𝑥0 = 4.

• a = 0.5 and b = 1 with an initial guess of 𝑥0 = 4.

• a = 0.5 and b = 1 with an initial guess of 𝑥0 = −4.

Solution a) Roots are located at 𝑛𝜋.

b)

c)

d) 2.852, Trouble converging, −2.754

16

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 6.14: Simple Interest The formula for simple interest is

𝑖(1 + 𝑖)𝑛 (1 + 𝑖)𝑛 − 1 where P = present worth, A = annual payments, n = number of years, and i = interest rate. a) Write a function to compute i as a function of A, P, and n. b) For P = $25,000, A = $5,000, and n = 30 years, determine i. c) With P = $25,000, plot i versus n for A = $1000 to $10000. 𝐴=𝑃

Solution a) Need to solve the following for i. 𝑖(1 + 𝑖)𝑛 𝑓(𝑖) = 𝑃

(1 + 𝑖)𝑛 − 1

−𝐴 = 0

We will need a root finder. b) 𝑖 = 0.199 or 19.1% c) Solution for A = 1000, 5000, and 10000.

17

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 6.15: Non-Linear Springs Consider the spring and block system shown with non-linear springs k2

k1 M1

k3 M2

k4 M3

x1+L1 LT

Non-linear springs can be described by 𝐹𝑠𝑝𝑟𝑖𝑛𝑔𝑔 = 𝑘𝑥 + 𝛾𝛾𝑥3 where x is the displacement of the spring from its equilibrium length and k and 𝛾𝛾 are dependent upon the properties of the spring.

For relatively small value of x, the non-linear term is small and the spring behaves as a linear spring, i.e., 𝐹𝑠𝑝𝑟𝑖𝑛𝑔𝑔 = 𝑘𝑥. If 𝛾𝛾 > 0, the spring is called a hardening spring because it takes more force to cause the same displacement. If 𝛾𝛾 < 0, the spring is called a softening spring meaning that it loses its strength after is stretched or compressed a certain amount. a) Considering all the ki, 𝛾𝛾𝑖, Li, and W as known, derive the mathematical model for the stretched lengths xi. b) Write a function to solve the system of equations in part (a). c) Use the function developed in part (b) to determine the xi’s the following cases.

Case 1 W=0.2 m LT=8 m Li=1 m ki=2 N/m 𝛾𝛾𝑖 = 0 𝑁/𝑚3

Case 2 W=0.2 m LT=8 m Li=1 m ki=2 N/m 𝛾𝛾1 = 0.1, 𝛾𝛾2 = 0.2, 𝛾𝛾3 = 0.3, 𝛾𝛾4 = 0.4 𝑁/𝑚3

Case 3 W=0.2 m LT=8 m Li=1 m k1=1, k2=2, k3=3, k4=4 N/m 𝛾𝛾1 = 0.1, 𝛾𝛾2 = 0.2, 𝛾𝛾3 = 0.3, 𝛾𝛾4 = 0.4 𝑁/𝑚3

Solution (a) Mathematical Model Total length Mass-1

𝑥1 + 𝑥2 + 𝑥3 + 𝑥4 = 𝐿𝑇 − 𝐿1 − 𝐿2 − 𝐿3 − 𝐿4 − 3𝑊𝑊 −(𝑘1𝑥1 + 𝛾𝛾1𝑥3) + (𝑘2𝑥2 + 𝛾𝛾2𝑥3) = 0 1

Mass-2

−(𝑘2𝑥2 + 𝛾𝛾2𝑥3) + (𝑘3𝑥3 + 𝛾𝛾3𝑥3) = 0 2

Mass-3

2

3

−(𝑘3𝑥3 + 𝛾𝛾3𝑥3) + (𝑘4𝑥4 + 𝛾𝛾4𝑥3) = 0 3

4

These non-linear algebra equations must be solved numerically. (c) Results

18

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Chapter 7

Introduction to Ordinary Differential Equations Problems 7.1 to 7.6 For each of the ordinary differential equations, a) Sketch the phase portrait (dx/dt vs. x) b) Find all fixed points and classify their stability (stable or unstable). c) Sketch the anticipated solution (x(t) vs. t) Clearly label your sketches.

Problem 7.1 𝑑𝑑𝑑 =𝑏 𝑑𝑡 Draw sketches for 𝑏 < 0, 𝑏 = 0, 𝑎𝑛𝑑 𝑏 > 0.

Solution No fixed (or steady state) points unless b = 0.

1

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 7.2 𝑑𝑑𝑑 = −𝑑𝑑 𝑑𝑡 Solution 𝑑𝑑∗ = 0: stable

Problem 7.3 𝑑𝑑𝑑 = −𝑎 ∙ 𝑑𝑑 𝑑𝑡 Draw sketches corresponding to a zero, low, and high value of the parameter a.

Solution 𝑑𝑑∗ = 0: stable for a >=0

2

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 7.4 𝑑𝑑𝑑 = −𝑎 ∙ 𝑑𝑑 + 𝑏 𝑑𝑡 Draw sketches corresponding to zero, low and high values of the parameter b.

Solution 𝑑𝑑∗ = 𝑏/𝑎: Stable

Problem 7.5 𝑑𝑑𝑑 = 𝑟 ∙ 𝑑𝑑(1 − 𝑑𝑑) 𝑑𝑡 r >0. Draw anticipated solution ((t) vs. t) starting from several initial conditions.

Solution 𝑑𝑑∗ = 0: Unstable 𝑑𝑑∗ = 1: Stable

3

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 7.6 Consider the ODE 𝑑𝑑𝑑 = 𝑟 ∙ 𝑑𝑑(1 − 𝑑𝑑) + 𝑆 𝑑𝑡

r is a positive constant. a) Find all fixed points and classify their stability (stable or unstable). Are there any critical values of S? b) Plot the phase portraits (𝑑𝜃 vs. 𝑑𝑑). 𝑑𝑡

c) Plot the anticipated solution (𝑑𝑑(𝑡) vs. t). Draw sketches starting from different initial positions.

Solution S Range 𝑆>−

𝑟 4

Fixed Points 1 𝑑𝑑1∗ = (1 − �1 + 4𝑆/𝑟) 2 1 𝑑𝑑2∗ = (1 + �1 + 4𝑆/𝑟) 2 1 𝑑𝑑∗ = 𝑑𝑑∗ = 1 2 2 No real fixed points

Stability of Fixed Points 𝑑𝑑∗ is Unstable 1

𝑑𝑑∗ is Stable 2

Critical S neutrally stable 𝑆 = 𝑆𝑐 = −𝑟/4 𝑟 NA 𝑆<− 4 The distinct qualitative change in the solution as S crosses –r/4 is referred to in dynamics as a bifurcation. We study this behavior in detail later.

4

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 7.7 Consider the linear first order ODE

𝑑𝑑𝑑 1 + (𝑑𝑑 − 𝑑𝑑∞) = 𝑆𝑐 𝑑𝑡 𝜏 𝑑𝑑 = 𝑑𝑑0 , 𝑡 = 0 The parameters , ∞, and Sc are constants. Assume  >0. a) Find and classify the stability of the fixed points. b) Sketch the phase portrait for Sc=0 and Sc>0. c) Sketch the solution based on the phase portrait for different initial conditions. Consider Sc=0 and Sc>0. d) Based on the previous findings, sketch the anticipated solution for ∞=0 with a pulsed, time dependent source 𝑆(𝑡) = 𝑆𝑐(𝐻(𝑡) − 𝐻(𝑡 − 𝑡1)) Solution Phase Portrait d/dt

*



Fixed Point 𝑑𝑑 = 𝑑𝑑∞ + 𝜏𝑆𝑐 This is the only fixed point and it is stable.

Solution Sketches 

*

t

5

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 7.8: Tumor Growth The growth of cancerous tumors can be modeled by the Gompertz law 𝑑𝑁 = −𝑎 ∙ 𝑁 ∙ 𝑙𝑛(𝑏𝑁) 𝑑𝑡 where 𝑁(𝑡) is proportional to the number of cells in the tumor, and a, b > 0 are parameters. a) Interpret a and b biologically. b) Sketch the phase portrait c) Sketch 𝑁(𝑡) for various initial values.

Solution a) Interpretation a represents a growth rate 1/b represents a carrying capacity b) Solution Sketches

Fixed Points: x* = 0 : Unstable x* = 1/b : Stable

6

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 7.9 Consider the model chemical reaction

𝑘

1 �⎯⎯� 𝐴 + 𝑋𝑋 2𝑋𝑋 ➛−

𝑘−1

in which one molecule of X combines with one molecule of A to form two molecules of X. This means that the chemical X stimulates its own production, a process called autocatalysis. This positive feedback process leads to a chain reaction, which eventually is limited by a “back reaction” in which 2X returns to A + X. According to the law of mass action of chemical kinetics, the rate of an elementary reaction is proportional to the product of the concentrations of the reactants. We denote the concentrations by lowercase letters x = [X] and a = [A]. Assume that there’s an enormous surplus of chemical A, so that its concentration a can be regarded as constant. Then the equation for the kinetics of x is 𝑑𝑥 = 𝑘 𝑎𝑥 − 𝑘 2 1 −1 𝑥 𝑑𝑡 where k1 and k-1 are positive parameters called rate constants. a) Sketch the phase portrait. b) Find all the fixed points and classify their stability. c) Sketch the solution based on the phase portrait for different initial conditions Solution a) Phase Portrait

b) Fixed Points 𝑘1𝑎𝑥 − 𝑘−1𝑥2 = 0 𝑥1∗ = 0 : unstable 𝑥2∗ = 𝑘1𝑎/𝑘−1: stable c) Solution Sketches

7

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 7.10: Chemical Kinetics Consider the chemical reaction system 𝑘

1 �⎯⎯� 𝐴 + 𝑋𝑋 2𝑋𝑋 ➛−

𝑘−1 𝑘2

𝑋𝑋 + 𝐵 �⎯� 𝐶 This is a generalization of Problem 7-9. The new feature is that X is used up in the production of C. a) Assuming that both A and B are kept at constant concentrations a and b, show that the law of mass action leads to an equation of the form 𝑑𝑥/𝑑𝑡 = 𝑐1 𝑥 − 𝑐2𝑥2, where x is the concentration of X, and c1 and c2 are constants to be determined. b) Find and classify all the fixed points. Show that x*= 0 is stable when 𝑘2𝑏 > 𝑘1𝑎, and explain why this makes sense chemically. c) Sketch the phase portraits and anticipated solutions for all the qualitatively different possibilities.

Solution a) Governing Equation

𝑑𝑥 = 𝑐 𝑥 − 𝑐

2 1 2𝑥 𝑑𝑡 𝑐1 = 𝑘1𝑎 − 𝑘2𝑏, 𝑐2 = 𝑘−1

b) Analysis of the Fixed Points c1 Range c1 > 0 c1 = 0 c1 < 0

Fixed Points x* = 0 x* = c1/ c2 x* = 0

Stability of Fixed Points Unstable Stable Neutrally stable

x* = 0 x* = c1/ c2

Stable Unstable

x* =0 is the only stable fixed point when 𝑘1𝑎 < 𝑘2𝑏 (c1 < 0) because chemical X is being used up faster during its interaction with B than it is being produced through its interaction with A. c) Phase Portraits and Expected Solutions c1 < 0

c1 = 0

c1 > 0

8

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 7.11: Allee Effect For certain species of organisms, the effective growth rate 𝑁/𝑁 is highest at intermediate N. This the called the Allee effect. For example, imagine that it is too hard to find mates when N is very small, and there is too much competition for food and other resources when N is large. a) Show that 𝑁 = 𝑟 − 𝑎(𝑁 − 𝑏)2 provides an example of Allee effect, if r, a, and b satisfy 𝑁

certain constraints, to be determined. b) Find all the fixed points of the system and classify their stability. c) Plot the solutions 𝑁(𝑡) for different initial conditions. d) Compare the solutions 𝑁(𝑡) to those found for the logistic equation. What are the qualitative differences, if any?

Solution b) Fixed Points 𝑁∗ = 0 𝑁∗ = 𝑏 ± �𝑟/𝑎

Unstable Stable

c) Numerical Solution

d) Comparison with Logistic Equation

9

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 7.12: Slope Fields Consider an autonomous first order ODE: 𝑑𝑥 = 𝑓(𝑥) 𝑑𝑡 The slope field is shown.

a) Propose a possible function 𝑓(𝑥) consistent with the slope field. b) Plot the solution 𝑥(𝑡) versus t starting from x0 = -2, -1, 0, 1, and 2. c) Plot the phase portrait, 𝑑𝑥/𝑑𝑡 versus x.

Solution a) This vector field seems to be generated from a function such as 𝑓(𝑥, 𝑡) = (𝑥 + 1)(𝑥 − 1) b) Solution

d) Phase portrait

10

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Chapter 8

Laplace Transforms

Problem 8.1: Linear First Order ODE Consider the linear first- order ODE where  is constant. 𝑑𝑑𝑑 1 + 𝑑𝑑(𝑡) = 𝑓(𝑡) 𝑑𝑡 𝜏 𝑑𝑑 = 𝑑𝑑0 , 𝑡 = 0 1.

2.

3.

Determine the analytical solution using the Laplace transform method for 𝑓(𝑡) = 0. Create a meaningful plot of this solution.

Determine the transfer function G(s).

Apply the results from part (2) to determine the solution for the following forcing functions with 𝑑𝑑0 = 0 where fc is a constant and H is the unit step function. Plot each case. a.

Step input 𝑓(𝑡) = 𝑓𝑐 𝐻(𝑡 − 𝑡0) b. Pulsed input 𝑓(𝑡) = 𝑓𝑐 (𝐻(𝑡 − 𝑡1) − 𝐻(𝑡 − 𝑡2)) c.

Impulse or Delta function 𝑓(𝑡) = 𝑓𝑐 𝛿𝛿(𝑡 − 𝑡1) d. Harmonic forcing 𝑓(𝑡) = 𝑓𝑐 𝑠𝑖𝑛(𝜔𝑡)

Solution 1) Effect of 0 Only 𝑑𝑑(𝑡) = 𝑑𝑑0𝑒−𝑡⁄𝜏𝜏 t

t

1

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

2) Transfer Function ~

𝑑𝑑(𝑠) 1 𝐺(𝑠) = ~ = 1 𝑓(𝑠) 𝑠 + 𝜏 3a) Step Input 𝑑𝑑𝑠𝑡𝑒𝑝(𝑡) = 𝑓𝑐𝜏(1 − 𝑒𝑥𝑝(−

𝑡 − 𝑡0 ))𝐻(𝑡 − 𝑡0)

𝜏

t

0

1

2

3

4

5

6

t

3b) Pulse 𝑑𝑑𝑝𝑢𝑙𝑠𝑒(𝑡) = 𝑓𝑐𝜏(1 − 𝑒𝑥𝑝(−

𝑡 − 𝑡2

𝑡 − 𝑡1 𝜏

))𝐻(𝑡 − 𝑡1) − 𝑓𝑐𝜏(1 − 𝑒𝑥𝑝(−

𝜏

))𝐻(𝑡 − 𝑡2)

1.5

1.0

0.5

0

2

4

6

3c) Impulse or Delta function 𝑑𝑑𝑠𝑡𝑒𝑝(𝑡) = 𝑓𝑐𝑒𝑥𝑝(−

8

t

𝑡 − 𝑡0 )𝐻(𝑡 − 𝑡0)

𝜏

t 1.0

0.8

0.6

0.4

0.2

0

1

2

3

4

5

6

t

2

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

3d) Harmonic forcing 𝑡

𝑒− 𝜏𝜏𝜏2 𝑑𝑑(𝑡) = 𝑓𝑐𝜔( + 1 + 𝜏2𝜔2

−𝜏 2𝜔𝐶𝑜𝑠[𝑡𝜔] + 𝜏𝑆𝑖𝑛[𝑡𝜔] 𝜔(1 + 𝜏2𝜔2)

)

t

3

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 8.2 Consider the linear second order ODE where k is constant. 𝑥 + 𝑘𝑥(𝑡) = 𝑓(𝑡) 𝑥(0) = 𝑥0 𝑥(0) = 𝑣0 1. Determine the analytical solution using the Laplace transform method when 𝑓(𝑡) = 0. Create a meaningful plot of this solution.

2. Determine the transfer function 𝐺(𝑠).

3. Apply the results from part (2) to determine the solution for the following forcing functions with zero initial conditions. Plot each case. a. Step input, 𝑓(𝑡) = 𝑓𝑐 𝐻(𝑡 − 𝑡0) b. Pulsed input, 𝑓(𝑡) = 𝑓𝑐 (𝐻(𝑡 − 𝑡1) − 𝐻(𝑡 − 𝑡2)) c. Impulse or Delta function, 𝑓(𝑡) = 𝑓𝑐 𝛿𝛿(𝑡 − 𝑡1) d. Harmonic forcing, 𝑓(𝑡) = 𝑓𝑐 𝑠𝑖𝑛(𝜔𝑡) Solution 1) 𝑥(𝑡) = 𝑥 cos(�𝑘 𝑡) + 𝑣 0

2)

𝐺(𝑠) =

=

𝐹(𝑠)

3a) 𝑥(𝑡) = 𝑐𝑓

1

𝑚𝑠2+𝑘 ])

√𝑚

𝑘

1

𝑐 √𝑘√𝑚

3d) 𝑥(𝑡) = 𝑓

√𝑘𝑡

√𝑘𝑡

√𝑘(𝑡−Δt)

√𝑚 √𝑘𝑡

√𝑚

(1 − Cos[

𝑐𝑘

3c) 𝑥(𝑡) = 𝑓

𝑚

1

(1−Cos[

3b) 𝑥(𝑡) = 𝑓

0�𝑘

𝑚

𝑋(𝑠)

𝑚 sin(�𝑘 𝑡)

Sin[

] + (−1 + Cos[

])H[𝑡 − Δt])

]

√𝑚

1

𝑐 √𝑘(𝑘−𝑚𝑤𝑤2)

√𝑘𝑡

(−√𝑚𝑤Sin[

] + √𝑘Sin[𝑡𝑤])

√𝑚

4

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Chapter 9

Numerical Solutions of Ordinary Differential Equations Problem 9.1: Runge-Kutta Basics Solve the following initial value problem over the interval from t = 0 to 2 with a = 1.1 and y0=1.

Display all your results on a single graph. 𝑑𝑦 = 𝑦 ∙ 𝑡2 − 𝑎 ∙ 𝑦 𝑑𝑡 𝑦(0) = 𝑦0 a) Analytically. b) Using Euler’s method with h =0.5 and 0.25. c) Using Heun’s method with h =0.5.

Solution (a) Analytical Solution For this particular case, it is possible to separate the variables. 𝑑𝑦 = (𝑡2 − 𝑎)𝑑𝑡 𝑦 This equation can now be integrated directly. 𝑑𝑦 � = �(𝑡2 − 𝑎)𝑑𝑡 𝑦 𝑡3 ln(𝑦) = − 𝑎𝑡+c 3

𝑡3 − 𝑎𝑡 + 𝑐) 3 The unknown constant is determined by forcing the solution to satisfy the initial condition. 𝑦0 = exp(𝑐) Thus 𝑡3 𝑦(𝑡) = 𝑦0exp ( − 𝑎𝑡) 3 𝑦(𝑡) = exp (

1

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

2 1.8

exact

1.6

Euler,h=0.5 Euler,h=0.25 Heun,h=0.5

1.4

y(t)

1.2 1 0.8 0.6 0.4 0.2 0 0

0.2

0.4

0.6

0.8

1 t

1.2

1.4

1.6

1.8

2

Using the Euler scheme, reducing the step size by a factor of 2 reduces the error by approximately a factor of 2. The Heun method is second order accurate and gives better results at the expense of more computations.

2

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 9.2: Euler’s Method Consider the differential equation 𝑑𝑦 = −𝑦 + 2 𝑑𝑡 a) Sketch the phase diagram (𝑑𝑦⁄𝑑𝑡 vs. y) and the anticipated solution (y(t) vs. t). b) Starting with the initial condition y0 = 0, manually perform three steps of the Euler’s method using a step size of h = 0.5. c) Using the Euler’s method, compute and plot the solution starting from initial conditions: y0 = 2, 0, 2, 4, and 6. Put all the curves on a single graph. Use a time range 0 ≤ 𝑡 ≤ 5. Experiment with the required step size in order to get accurate and smooth looking plots. Report the step size you decided upon.

Solution a) Phase Diagram and Anticipated Solution x* = 2: stable Draw sketches corresponding to zero, low and high values of the parameter S. y(t)

dy/dt

2 0

y

0

t

b) Manual Calculations 𝑦𝑖+1 = 𝑦𝑖 + ℎ ∙ 𝑓(𝑡𝑖, 𝑦𝑖) = 𝑦𝑖 + ℎ ∙ (−𝑦𝑖 + 2) i 1 2 3 4

𝑡𝑖 0 t1+h = 0.5 t2+h = 1.0 t3+h = 1.5

𝑦𝑖 0 (−𝑦 𝑦2 = 𝑦1 + ℎ ∙ 1 + 2) = 0 + 0.5(0 + 2) = 1 𝑦3 = 𝑦2 + ℎ ∙ (−𝑦2 + 2) = 1 + 0.5(−1 + 2) = 1.5 𝑦4 = 𝑦3 + ℎ ∙ (−𝑦3 + 2) = 1.5 + 0.5(−1.5 + 2) = 1.75

3

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

c) Results Effect of Step Size, h 4 2 1 0.5 0.25 0.125 0.0625

3.5 3

y(t)

2.5 2 1.5 1 0.5 0 0

1

2

3

4

5

t

The previous graph shows that there is no noticeable difference in the results using step sizes of 1/64 and 1/128. Thus, a step size of h = 0.01 is plenty small for accurate graphical results and is used to obtain the following results. The results using h = 0.5 match the hand calculations.

Effect of Initial Condition, y0 6 -2 0 2 4 6

5 4

y(t)

3 2 1 0 -1 -2 0

1

2

3

4

5

t

4

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 9.3: Lumped Thermal Mass Consider the cooling of a hot object initially at temperature T0 in cold air at T with heat transfer coefficient h. Assume the lumped thermal capacity model, as described in the previous chapter, is valid. The differential equation is 𝑑𝑇 𝜌𝑐𝑉 = −ℎ𝐴𝑠(𝑇 − 𝑇∞) + 𝑔𝑉 𝑑𝑡 a) Draw the cause/effect diagram for this physical problem. b) Write a function to numerically solve for temperature as a function of time. The system parameters and forcing functions should be inputs and the time and temperature vectors should be outputs. c) Explore the effect of the heat transfer coefficient. Plot temperature vs. time for h = 10 W/m2 C (still day), 25 W/m2 C (typical day) and 75 W/m2 C (hurricane). Put all curves on a single graph. Carry out time long enough such that the temperature begins to level out at steady state.

The other parameters are: V =10-6 m3, As =10-4 m2,  = 1000 kg/m3, c = 500 J/kg C, T = 25C, T0 = 400C, g = 0. d) Explore the effect of the heat source strength. Plot the solution for g = 0, 105, 2x105 and 3x105 W/m3. Put all curves on a single graph. Use parameter values from part (b) with h = 25 W/m2 C. Solution Forcing Functions g, T0, T∞

System V, As, , c, h



Response t, T



Note that the steady state temperature can be determined form 𝑑𝑇 𝜌𝑐𝑉 = −ℎ𝐴𝑠(𝑇 − 𝑇∞) + 𝑔𝑉 = 0 𝑑𝑡 𝑔𝑉 𝑇∗ = 𝑇∞ + ℎ𝐴𝑠 With no heat generation (g = 0), the object will simply end up at room temperature.

Part (c): Effect of the Heat Transfer Coefficient Effect of the Heat Transfer Coefficient, h (W/m2 K) 400 h=10 h=25 h=75

350

Temperature (C)

300 250 200 150 100 50 0 0

200

400

600

800

1000 1200 Time (sec)

1400

1600

1800

2000

The object cools quicker with a higher h, representing a higher wind velocity.

5

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Part (d): Effect of Heat Generation Effect of the Heat Source, g (kW/m3) 160 0 100 200 300

Temperature (C)

140

120

100

80

60

40

20 0

100

200

300

400

500

600

700

800

900

1000

Time (sec)

Note that the steady state temperatures can be verified as being correct. For g = 105 𝑔𝑉 𝑇∗ = 𝑇∞ + = 65°𝐶 ℎ𝐴𝑠 Also, the solution is a linear function of the heat source. Thus, doubling g doubles the temperature rise above T∞.

6

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 9.4: Population Model The earth’s population can be estimated using the following simplified model 𝑑𝑃 = 𝑎𝑃 − 𝑏𝑃2 𝑑𝑡 where a is the birth rate parameter and b is the death rate parameter. a) Determine the steady state population. b) Sketch the phase plot, 𝑑𝑃⁄𝑑𝑡 vs. P. Using the phase plot, sketch the anticipated population history, 𝑃(𝑡) vs. time. Draw a single graph with several curves corresponding to different starting populations. c) When the time variable t is measured in years, experimental evident suggests that the parameters a and b in the population model are approximately a = 0.028 1/year and b = 2.9  10-12 1/(people*year). Starting from a population of approximately P0 = 100 million people in the year 1800, compute and plot the earth’s population as a function of time in years. Run your simulation until the population appears to level off. Using these parameters, how many people will we eventually have on this earth? At what year will we have reached 99% of our maximum population? d) Estimating population is an inexact science. Examine the effect of the birth rate parameter by plotting the earth’s population for a = 0.025, 0.03, and 0.035 with b = 2.910-12. Put all curves on a single graph. e) Examine the effect of the death rate parameter by plotting the earth’s population for b = 2.5  10-12, 3  10-12 and 3.5  10-12 with a = 0.029. Put all curves on a single graph.

Solution a) Steady State Steady state occurs when𝑑𝑃 = 0. Thus P = 0 and 𝑃 = 𝑎/𝑏 are possible steady states. The stable 𝑑𝑡

steady state is 𝑃 = 𝑎/𝑏 b) Phase Plot and Anticipated Solution dP/dt

P 0

a/b

7

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

d) x 10

9

10 9 8

Earths Population

7 6 5 4 3 2 1 0 1800

1850

1900

1950

2000 Year

2050

2100

2150

2200

Maximum Population The earth`s population will reach 9.333 billion people! We will reach 99 percent of the earth`s maximum population in the year 2129. e) Birth Rate Study 9

14

Effect of Birth Rate Parameter, a

x 10

0.025 0.03 0.035

Earths Population

12 10 8 6 4 2 0 1800

1850

1900

1950

2000

2050

2100

2150

2200

2150

2200

Year

f) Death Rate Study Effect of Death Rate Parameter, b

9

x 10 12

Earths Population

10

2.5e-012 3e-012 3.5e-012

8

6

4

2

0 1800

1850

1900

1950

2000

2050

2100

Year

8

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 9.5: Linear Oscillator, Effect of Damping Consider the spring-mass-damper system with zero forcing, described by the following differential equation and initial conditions. 𝑑2𝑥 𝑑𝑥 𝑚 2 +𝑐 +𝑘∙𝑥 = 0 𝑑𝑡 𝑑𝑡 𝑥 = 𝑥0 𝑑𝑥 = 𝑣 � 𝑡 = 0 0 𝑑𝑡 a) Draw the cause/effect diagram for this system. b) Write a function to numerically solve this differential equation. c) Use your function to solve and plot x(t) and v(t) versus t. Use the following parameters: x0 = 1 m, v0 = 0, m = 20 kg, k = 20 N/m, c = 0, 8, and 40 Ns/m. Create separate plots for x(t) and v(t) versus t. Put all three curves for different c values on a single graph (see Fig. 9.7 for an example).

Solution Forcing Functions x0, x0

System m, c, k





Response x(t), v(t)

Effect of Damping 1

x (m)

0.5 0 -0.5 -1 0

2

4

6

8

10

12

14

16

18

20

4

6

8

10

12

14

16

18

20

1 0 8 40

v (m/s)

0.5 0 -0.5 -1 0

2

Time (sec)

9

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 9.6: Linear Oscillator with Forcing Consider the forced spring-mass-damper system, described by the following differential equation and initial conditions. 𝑑2𝑥 𝑑𝑥 𝑚 2 +𝑐 + 𝑘 ∙ 𝑥 = 𝐹(𝑡) 𝑑𝑡 𝑑𝑡 𝑥 = 𝑥0, 𝑣 = 𝑣0, 𝑤ℎ𝑒𝑛 𝑡 = 0 a) Draw the cause/effect diagram for this system. b) Write a function to numerically solve this differential equation. c) Use your function to compute the displacement and velocity for the following special cases.

For each case, create a 3-part graph containing F(t), x(t), and v(t) vs. t. For all cases: m = 20 kg, c = 4 Ns/m, k = 20 N/m, and x0 = v0 =0. See Figure 9.6 for an example.

Constant forcing Harmonic forcing Pulsed forcing

Forcing: F(t) (N) 𝐹(𝑡) = 𝐹0 F0 = 10 N 𝐹(𝑡) = 𝐹0sin(𝜔𝑡) F0 = 5 N/kg, 𝜔 = 1 rad/s 𝐹(𝑡) = 𝐹0(𝐻(𝑡) − 𝐻(𝑡 − 𝑡𝑜𝑛)) F0 = 10 N , ton = 40 sec

Solution Forcing Functions F(t), x0, x0

System m, c, k





Response t, x(t), v(t)

Constant Forcing F (N)

10 5 0 0

5

10

15

20

25

30

35

40

0

5

10

15

20

25

30

35

40

0

5

10

15

20 Time (sec)

25

30

35

40

x (m)

1 0.5 0

v (m/s)

0.5 0 -0.5

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Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

F (N)

Harmonic Forcing 4 2 0 -2 -4 0

5

10

15

20

25

30

35

40

45

50

0

5

10

15

20

25

30

35

40

45

50

0

5

10

15

20

25 30 Time (sec)

35

40

45

50

x (m)

2 0 -2

v (m/s)

2 0 -2

Pulsed Forcing F (N)

5

0 0

10

20

30

40

50

60

70

80

0

10

20

30

40

50

60

70

80

0

10

20

30

40 Time (sec)

50

60

70

80

x (m)

0.5 0 -0.5

v (m/s)

0.5 0 -0.5

11

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 9.7: Pendulum The oscillations of an undamped pendulum can be simulated with the following nonlinear equation 𝑑2𝜃𝜃 𝑔 + sin(𝜃𝜃) = 0 𝑑𝑡2 𝐿 where  = angle of displacement, g = gravitational constant = 9.81 m2/s, and L = pendulum length.

For small angular displacements, sin(𝜃𝜃) ≈ 𝜃𝜃 and the model can be linearized as 𝑑2𝜃𝜃 𝑔 + 𝜃𝜃 = 0 𝑑𝑡2 𝐿 a) Reformulate these equations as two first order equations for  and 𝑣 = 𝑑𝜃𝜃/𝑑𝑡. b) Draw the cause/effect diagram. c) Create a function to compute the solution as a function of the input parameters. d) Using the function you developed, solve for  and v as functions of time for both the linear and nonlinear models with L = 0.6 m and initial conditions 0 = /8 and v0 = 0. Plot the linear and nonlinear solutions on the same graph. Also plot results for 0 = /2, 0 = 0.99, and 0 = .

Solution a) First Order Equations In order to express as two first order equations, define y1 =  y2 = v Nonlinear Model 𝑑𝑦1 = 𝑦2 𝑑𝑡 𝑑𝑦2 𝑔 = − sin(𝑦1) 𝑑𝑡 𝐿

Linear Model 𝑑𝑦1 = 𝑦2 𝑑𝑡 𝑑𝑦2 𝑔 = − 𝑦1 𝑑𝑡 𝐿

d) Numerical Results theta0/pi = 0.5 2

0.2

1

Theta (rad)

Theta (rad)

theta0/pi = 0.125 0.4

0 -0.2

-1 -2

-0.4 0

1

2

3

4

5

6

7

8

9

10

0

1

2

3

4

0

1

2

3

4

5

6

7

8

9

10

5

6

7

8

9

10

10

Velocity (rad/s)

2

Velocity (rad/s)

0

1 0 -1

5 0 -5 -10

-2 0

1

2

3

4

5

Time (s)

6

7

8

9

10

Time (s)

12

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

theta0/pi = 1 4

2

2

Theta (rad)

Theta (rad)

theta0/pi = 0.99 4

0 -2

-4

-4 0

1

2

3

4

5

6

7

8

9

10

0

1

2

3

4

0

1

2

3

4

5

6

7

8

9

10

5

6

7

8

9

10

20

Velocity (rad/s)

20

Velocity (rad/s)

0 -2

10 0 -10

10 0 -10 -20

-20 0

1

2

3

4

5

Time (s)

6

7

8

9

10

Time (s)

Comments The linear model is more accurate for small angles. However, even at 0 = /8, a gradual phase shift between the linear and nonlinear models can be observed. At 0 = /2 (vertical starting position), the two models are quite different. At the top position, the two models are completely different.

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Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 9.8: Van der Pol Equation The van der Pol equation is a model of a nonlinear circuit that arose back in the days of vacuum tubes: 𝑑2𝑦 𝑑𝑦 2) + 𝜇(1 − 𝑦 +𝑦 =0 𝑑𝑡2 𝑑𝑡 a) Reformulate these equations as two first order equations for y1 = y and 𝑦2 = 𝑑𝑦⁄𝑑𝑡. b) Create a function to compute the solution as a function of the input parameters. c) Using your function from part (b), solve for y1 and y2 as functions of time with µ = 1 and initial conditions y1 = 0.1 and y2 = 0. Create the following graphs: • y1 vs. t and y2 vs. t • phase plot: y1 vs. y2 d) Repeat part (c) with µ = 0 and 10.

Solution 𝑑𝑦

a) Using the definitions 𝑦1 = 𝑦 and 𝑦2 = 𝑑𝑡, the original second order system can be decomposed into a set of first order equations. 𝑑𝑦2 𝑑𝑦1 = 𝑦 , = −𝜇(1 − 𝑦 2)𝑦 − 𝑦 1 2 1 2 𝑑𝑡 𝑑𝑡 c) d) Numerical Results

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Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 9.9: Sky Diver The free-fall velocity of a parachutist can be estimated using the force balance 𝑑𝑣 𝑐𝑑 = 𝑔 − 𝑣2 𝑑𝑡 𝑚 where v = velocity (m/s), t = time (s), g = 9.81 m/s2 = acceleration due to gravity, cd = drag coefficient (kg/m), and m = mass (kg). For an 80-kg parachutist, we wish to solve this differential equation given that v = 0 at t = 0. During free fall, cd = 0.25 kg/m. However, at t = topen, the chute opens, whereupon cd = 5 kg/m. a) What would be the steady state velocity if the chute was not opened? What is the steady state velocity after the chute opens? b) Sketch the phase diagram (dv/dt versus v) and the anticipated solution for v(t) vs. t based on the phase diagram. c) Write a function to solve this differential equation numerically from t = 0 to 30 s. Create a plot of v(t) vs. t for topen = 0, 5, 10, and 20 sec, putting all curves on a single plot.

Solution (a) Steady or Terminal Velocity The steady state occurs when the velocity no longer changes with time: 𝑣∗ = �

𝑔𝑔𝑚 𝑐𝑑

Terminal velocity, chute closed = 56.0 m/s Terminal velocity, chute open = 12.5 m/s From the phase diagram, an approximate sketch of the solution can be anticipated, even for this relatively difficult problem. c) Effect of Free Fall Time Effect of topen 60 0 5 10 20

50

Velocity (m/s)

40

30

20

10

0 0

5

10

15 Time (s)

20

25

30

15

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 9.10: Coupled Oscillators Two masses are attached to a wall by linear springs.

Force balances based on Newton’s second law can be written as 𝑑2𝑥1 𝑚1 2 = −𝑘1(𝑥1 − 𝐿1) + 𝑘2(𝑥2 − 𝑥1 − 𝑤1 − 𝐿2) 𝑑𝑡 𝑑2𝑥2 𝑚2 2 = −𝑘2(𝑥2 − 𝑥1 − 𝑤1 − 𝐿2) 𝑑𝑡 where ki = the spring constants (N/m), mi = masses (kg), Li = the length of the unstretched springs (m), and wi = the widths of the masses (m). Initially, the masses are at locations x1,0 and x2,0. a) Draw the cause/effect diagram for this system. b) Determine the steady-state or fixed points. c) Write a function to numerically solve this system of differential equations. d) Compute the positions of the masses as a function of time over the range t=0 to 20 using the following parameters: k1=k2=5, m1=m2=2, w1=w2=5, and L1=L2=2. Set the initial conditions as x1,0=L1 and x2,0= L1+w1+L2+6 with zero velocity. Construct time-series plots of both the displacements and the velocities. Indicate the fixed points on your graph. In addition, produce a phase-plane plot of x1 versus x2.

Solution a) Cause/Effect Forcing Functions 𝑥1,0, 𝑥2,0 𝑣1,0, 𝑣2,0

System m1, m2, k1, k2 L1, L2, w1, w2





Response x1(t), x2(t) v1(t), v2(t)

b) Fixed Points The steady or fixed points are where the time derivatives are zero (no motion, v1 = v2 = 0).

0 = −𝑘1(𝑥∗ − 𝐿1) + 𝑘2(𝑥∗ − 𝑥∗ − 𝑤1 − 𝐿2) 1

2

1

0 = −𝑘2(𝑥∗ − 𝑥∗ − 𝑤1 − 𝐿2) 2

1

Arrange as a 2-by-2 matrix 𝑘1 + 𝑘2 −𝑘2 𝑥1 𝑘1𝐿1 − 𝑘2(𝑤1 + 𝐿2) � � � �𝑥 � = � 𝑘2 (𝑤1 + 𝐿2) −𝑘2 𝑘2 2 Solve this 2-by-2 system using Cramer’s rule. 𝐿1 𝑥∗ � ∗1� = � � 𝑥2 𝐿1 + 𝑤1 + 𝐿2 The fixed points correspond to the upstretched positions of the springs.

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Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

d) Results Displacements of a Coupled Spring/Mass System x1

10

x2

5

1

2

x , x (m)

15

0 -5 0

2

4

6

8

10

12

14

16

18

20

Velocities of a Coupled Spring/Mass System 10 v1 v2

0

1

2

v , v (m)

5

-5 -10 0

2

4

6

8

10 t (sec)

12

14

16

18

5

6

20

Phase Plot of a Coupled Spring/Mass System 13 12 11

x2 (m)

10 9 8 7 6 5 -2

-1

0

1

2 x1 (m)

3

4

17

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 9.11: An Epidemic The following ODE’s have been proposed as a model of an epidemic: 𝑑𝑆 = −𝑎𝑆𝐼 𝑑𝑡 𝑑𝐼 = 𝑎𝑆𝐼 − 𝑟𝐼 𝑑𝑡 𝑑𝑅 = 𝑟𝐼 𝑑𝑡 where S = the number of susceptible individuals I = the number of infected individuals R = the number of recovered individuals a = infection rate r = recovery rate A city initially has 10,000 people, all of whom are susceptible. Then a single infectious individual enters the city at t = 0. Use the following estimates for the parameters: a = 0.002/(person week) and r = 0.15/day. a) Compute the progression of the epidemic. At what time does the number of infected individuals drop back to 10? Create time series plots of all the state variables over a time range from 0 until the number of infected individuals falls below 10. Also, create a 3D phase plot of S versus I versus R. b) Suppose that after recovery, there is a loss of immunity that causes recovered individuals to become susceptible. This reinfection mechanism can be modeled as R, where  = the reinfection rate = 0.03/day. Modify the model to include this mechanism and repeat the computations in (a). c) Suppose that some of the infected people recover and some die. F is the number of dead people.

Also consider that there is an influx of susceptible individuals moving to the city at the rate of Q people/day. Modify the model to include these possibilities and repeat the computations in (a).

Solution a) Basic Model

18

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

b) Improved Model With loss of immunity and reinfection, the equations become 𝑑𝑆 = −𝑎𝑆𝐼 + 𝜌𝑅 𝑑𝑡 𝑑𝐼 = 𝑎𝑆𝐼 − 𝑟𝐼 𝑑𝑡 𝑑𝑅 = 𝑟𝐼 − 𝜌𝑅 𝑑𝑡 In matrix form, the system of equations can be written as −𝑎𝑆𝐼 + 𝜌𝑅 𝑑 𝑆 � 𝐼 � = � 𝑎𝑆𝐼 − 𝑟𝐼 � 𝑑𝑡 𝑅 𝑟𝐼 − 𝜌𝑅 In order to input these equations systematically into the computer, redefine the variables as follows. 𝑦1 𝑆 � 𝐼 � = �𝑦2� 𝑦3 𝑅 𝑦 −𝑎𝑦1𝑦2 + 𝜌𝑦3 𝑑 𝑦1 � 2� = � 𝑎𝑦1𝑦2 − 𝑟𝑦2 � 𝑑𝑡 𝑦3 𝑟𝑦2 − 𝜌𝑦3

19

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Chapter 10

First Order Ordinary Differential Equations Problem 10.1 The functions 𝜓𝜓1(𝑡)and 𝜓𝜓2(𝑡) are the solutions of the following simple, elementary problems. 𝜓𝜓1(𝑡) 𝑑𝜓𝜓1 1 = − 𝜓𝜓1 + 1 𝑑𝑡 𝜏 𝜓𝜓1 = 0 , 𝑡 = 0

𝜓𝜓2(𝑡) 𝑑𝜓𝜓2 1 = − 𝜓𝜓2 𝑑𝑡 𝜏 𝜓𝜓2 = 1 , 𝑡 = 0

Express the solutions to the following problems in terms of the 1 and 2 functions. Be specific about the arguments. Do not actually solve for 1 and 2. A few cases are given as examples in order to clarify the objectives.

1 𝑑𝑑𝑑 = − 𝑑𝑑 + 𝑆(𝑡) 𝑑𝑡 𝜏 𝑑𝑑 = 𝑑𝑑0 , 𝑡 = 0 Case

Source, 𝑆(𝑡)

IC, 𝑑𝑑0

Ex. 1 Ex. 2 1 2 3 4 5 6 7

0 8 𝐻((𝑡 − 𝑡𝑜) Sc 0 3 Sc 𝐼(𝑡, 𝑡𝑜, ∆𝑡) 𝑅𝑒𝑝𝑃𝑢𝑙𝑠𝑒(𝑡, 𝑡𝑐𝑦𝑐𝑙𝑒, 𝑡𝑜𝑛) Arbitrary 𝑆(𝑡)

4 0 0 𝑑𝑑0 7 𝑑𝑑0 0 0 0

Solution in terms of 𝜓𝜓1(𝑡) and 𝜓𝜓2(𝑡) 4 ∙ 𝜓𝜓2(𝑡) 8 ∙ 𝜓𝜓1(𝑡 − 𝑡𝑜)

H(t) is the step function. 𝐻((𝑡 − 𝑡𝑜) − 𝐻((𝑡 − 𝑡𝑜 − ∆𝑡) , ∆𝑡) = 𝑖𝑚𝑝𝑢𝑙𝑠𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 = 𝐼(𝑡, 𝑡𝑜 ∆𝑡 𝑅𝑒𝑝𝑃𝑢𝑙𝑠𝑒(𝑡, 𝑡𝑐𝑦𝑐𝑙𝑒, 𝑡𝑜𝑛) = 𝑟𝑒𝑝𝑒𝑡𝑖𝑡𝑖𝑣𝑒 𝑝𝑢𝑙𝑠𝑒

1

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Solution Case 1 2 3 4 5

Source, 𝑆(𝑡) Sc 0 3 Sc 𝐼(𝑡, 𝑡𝑜, 𝛥𝑡)

IC, 𝑑𝑑0 0 𝑑𝑑0 7 𝑑𝑑0 0

Solution in terms of 𝜓𝜓1(𝑡)and 𝜓𝜓2(𝑡) 𝑑𝑑(𝑡) = 𝑆𝑐 ⋅ 𝜓𝜓1(𝑡) 𝑑𝑑(𝑡) = 𝑑𝑑0 ⋅ 𝜓𝜓2(𝑡) 𝑑𝑑(𝑡) = 3 ⋅ 𝜓𝜓1(𝑡) + 7 ⋅ 𝜓𝜓2(𝑡) 𝑑𝑑(𝑡) = 𝑆𝑐 ⋅ 𝜓𝜓1(𝑡) + 𝑑𝑑0 ⋅ 𝜓𝜓2(𝑡) 1 𝑑𝑑(𝑡) = (𝜓𝜓1(𝑡 − 𝑡𝑜) − 𝜓𝜓1(𝑡 − 𝑡𝑜 − 𝛥𝑡))𝜓𝜓𝑝(𝑡, 𝑡𝑜, 𝛥𝑡 𝛥𝑡

6

RepPulse(𝑡, 𝑡cycle, 𝑡on)

0

𝑑𝑑(𝑡) = � 𝜓𝜓𝑝(𝑡, (𝑛 − 1)𝑡cycle, 𝑡on)

7

Arbitrary 𝑆(𝑡)

0

𝑛=1,2,... 𝑁

𝑑𝑑(𝑡) ≃ � 𝑆𝑖 𝜓𝜓𝑝(𝑡, 𝑡𝑖, 𝛥𝑡) 𝛥𝑡 𝑖=1

𝑡𝑖 = (𝑖 − 1)𝛥𝑡 , 𝑡 = 𝑁 ⋅ 𝛥𝑡, 𝑆𝑖 = 𝑆(𝑡𝑖)

2

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 10.2 Consider the nonlinear problem

1 𝑑𝑑𝑑 = − 𝑑𝑑2 𝑑𝑡 𝜏 𝑑𝑑 = 𝑑𝑑0 , 𝑡 = 0 Can the solution be expressed in terms of the 1 and 2 functions defined Problem 10.1? If not, what is the difficulty?

Solution Try 𝑑𝑑(𝑡) = 𝑐 ⋅ 𝜓𝜓2(𝑡), 1 2 2 + 𝑐 𝜓𝜓2 = 0 dt 𝜏 𝑐 ⋅ 𝜓𝜓2 = 2 , 𝑡 = 0 It is impossible to express the solution in terms of 𝜓𝜓1(𝑡)and 𝜓𝜓2(𝑡). The nonlinearity couples the coefficients. 𝑐 𝑑𝜓𝜓2

3

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 10.3 A basic population dynamics model is 𝑑𝑃

= 𝑎 𝑃 − 𝑏 𝑃2 + 𝑆

𝑑𝑡

𝑃 = 𝑃0 , 𝑡 = 0 Can the general solution can be expressed as the superposition of simpler solutions? 𝑃(𝑡) = 𝑃1(𝑡) + 𝑃2(𝑡) where P1(t) is due to P0 only with S =0 P2(t) is due to S only with P0 List the governing equations for P1(t) and P2(t) if this superposition attempt is successful.

Solution Substitute the proposed solution into the ODE. 𝑑(𝑃1 + 𝑃2) 2 = 𝑎(𝑃1 + 𝑃2) − 𝑏 (𝑃1 + 𝑃2 ) + 𝑆 𝑑𝑡 = 𝑎(𝑃1 + 𝑃2) − 𝑏 (𝑃12 + 2𝑃1𝑃2 + 𝑃22) + 𝑆 The squared term 2𝑃1𝑃2 makes the equation impossible to separate algebraically into individual problems with 𝑃1 only and 𝑃2 only.

4

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 10.4 Consider the thermal system described by the equations 𝑑𝑇 𝜌𝑐𝑉 = −ℎ𝐴𝑠(𝑇 − 𝑇∞) + 𝑔𝑉 𝑑𝑡 𝑇 = 𝑇0 , 𝑡 = 0 Show that the general solution can be expressed as the superposition of the simpler solutions 𝑇(𝑡) = 𝑇1(𝑡) + 𝑇2(𝑡) where T1(t) is due to T0 only with g=0 T2(t) is due to g only with T0=0 List the governing equations for T1(t) and T2(t).

Solution Substitute the proposed superposition 𝑑(𝑇1 + 𝑇2) 𝜌cV = −𝐴𝑠ℎ(𝑇1 + 𝑇2 − 𝑇∞) + 𝑔 𝑉 dt 𝑇1 + 𝑇2 = 𝑇0, 𝑡 = 0 Break this problem into sub-problems. 𝑑𝑇2 𝑑𝑇1 (𝜌cV + 𝐴𝑠ℎ(𝑇1 − 𝑇∞)) + (𝜌cV + 𝐴𝑠ℎ(𝑇2) − 𝑔 𝑉) = 0 � � � � 𝑑_𝑡 𝑑𝑡 0

0

𝑇1 + 𝑇⏟2 = 𝑇0 , 𝑡 = 0 0

x1(t) Problem 𝑑𝑇1 𝜌cV = −𝐴𝑠ℎ(𝑇1 − 𝑇∞) 𝑑𝑡 𝑇1 = 𝑇0 , 𝑡 = 0

x2(t) Problem 𝜌cV

𝑑𝑇2 = −𝐴𝑠ℎ(𝑇2) + gV 𝑑𝑡 𝑇2 = 0 , 𝑡 = 0

5

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 10.5 A lumped thermal system with radiation and heat generation is described by the equations 𝑑𝑇 𝜌𝑐𝑉 = −𝐴 𝜀𝜀𝜎(𝑇4 − 𝑇4 ) + 𝑔(𝑡)𝑉 𝑠𝑢𝑟 𝑠 𝑑𝑡 𝑇 = 𝑇0 , 𝑡 = 0 Can the general solution can be decomposed as the superposition of the simpler solutions 𝑇(𝑡) = 𝑇1(𝑡) + 𝑇2(𝑡) where T1(t) is due to T0 only with g=0 T2(t) is due to g only with T0=0 List the governing equations for T1(t) and T2(t) if this superposition attempt is successful.

Solution Substitute the proposed superposition 𝑑(𝑇1 + 𝑇2) 𝜌cV = −𝐴 εσ((𝑇 + 𝑇 )4 − 𝑇4 ) + 𝑔 𝑉 𝑠 1 2 sur dt th The 4 order term makes the equation impossible to separate algebraically into individual problems for 𝑇1 only and 𝑇2 only.

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Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 10.6 Consider the linear first order ODE where  is constant.

1 𝑑𝑑𝑑 = − 𝑑𝑑 + 𝑆(𝑡) 𝑑𝑡 𝜏 𝑑𝑑 = 𝑑𝑑0 , 𝑡 = 0 a) Determine the analytical solution for S(t) = 0. Create a meaningful plot of this solution. b) Determine the analytical solution for S(t)=Sc=constant with 𝑑𝑑0 = 0. Create a meaningful plot of this solution. c) Show that the general solution to this linear problem can be expressed as the superposition of effects due to the initial condition 0 and source S(t). d) Determine the analytical solution for S(t)=Sc=constant and initial condition, 0, using the superposition principle demonstrated in part (c) and the results found in parts (a) and (b). e) Consider the𝑆case where 0 =0 with a finite pulse source, defined by 𝑆(𝑡) = 𝑐 (𝐻(𝑡 − 𝑡 ) − 𝐻(𝑡 − 𝑡 − ∆𝑡)) ∆𝑡

1

1

Determine the analytical solution. Create a meaningful plot of this solution. Consider the effect of t for a fixed Sc. Examine the limit as t  0.

Solution a) Solution for 𝑆(𝑡) = 0. 𝑑𝑑(𝑡) = 𝑑𝑑0𝑒−𝑡⁄𝜏𝜏

b) 𝑆(𝑡) = 𝑆𝑐 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 with 𝑑𝑑0 = 0 𝑑𝑑(𝑡) = 𝑆𝑐𝜏(1 − 𝑒−𝑡⁄𝜏𝜏)

c) Linearity 𝑑𝑑(𝑡) = 𝑑𝑑ic(𝑡) + 𝑑𝑑𝑠(𝑡) is allowable for linear problems.

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Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

d) Combined Effects of 0 and Sc 𝑑𝑑(𝑡) = 𝑑𝑑0𝑒−𝑡⁄𝜏𝜏 + 𝑆𝑐𝜏(1 − 𝑒−𝑡⁄𝜏𝜏) e) Pulsed Source 𝑆𝑐 𝑑𝑑(𝑡) =

0 𝜏(1 − 𝑒−(𝑡−𝑡𝑜)⁄𝜏𝜏)

𝛥𝑡 𝑆𝑐 𝜏(𝑒−(𝑡−𝑡𝑜−𝛥𝑡)⁄𝜏𝜏 − 𝑒−(𝑡−𝑡𝑜)⁄𝜏𝜏) ⎩𝛥𝑡 0 , 𝑡 < 𝑡𝑜 In the limit as t → 0, 𝑑𝑑(𝑡) = � 𝑆𝑐𝑒−(𝑡−𝑡𝑜)⁄𝜏𝜏 , 𝑡 ≥ 𝑡𝑜

, 𝑡 < 𝑡𝑜 , 𝑡 𝑡 ≤ 𝑡 + 𝛥𝑡 𝑜

,

𝑜

𝑡 > 𝑡𝑜 + 𝛥𝑡

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Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 10.7 A square silicon chip of length L = 10 mm on a side and thickness  = 5 mm is embedded in a well-insulated substrate.

T = 30C h

L Chip



Substrate

The chip draws P = 0.5 W of electrical power and is cooled by convection from the top surface to air at T = 30C with a heat transfer coefficient h = 40 W/m2 K. Assume that radiation effects are negligible and that the lumped capacity approximation is valid. The chip has the following thermal properties:  = 2000 kg/m3, c = 700 J/kg K, k = 150 W/m K 1. The chip is initially at the ambient temperature of T with the power off. At t = 0, the power is switched on. a. Sketch the temperature as a function of time for this process b. Derive the energy equation for this case (symbolic form). c. Determine the eventual steady-state temperature of the chip (numerical value).

2. After the steady condition in Part 1 has been reached, the power is shut off. a. Sketch the temperature history as a function of time for this process, starting from the time the power is initially switched on, all the way to a new steady-state after the power is switched off. b. What does the energy equation reduce to for this case (symbolic form). c. Determine how long it takes for the chip temperature to drop to 35 C after the power is shut off (numerical value).

3. After the steady condition in Part 1 has been reached, the system suffers a loss of coolant, causing the heat transfer from the surface due to convection to become negligibly small. a. Sketch the temperature history as a function of time for the entire process, starting from the time the power is initially switched on, all the way to the time the chip reaches its failure temperature. b. What does the energy equation reduce to for this case (symbolic form)? c. Determine the how long it takes for the chip to reach its failure temperature of 300 C after loss of coolant.

Solution Part 1.

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Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Recall the general statement of ENERGY CONSERVATION 𝐸˙st = 𝐸˙𝑔𝑔 + 𝐸˙in − 𝐸˙out Application to our problem gives,

𝐸˙ s t = 𝑃 − 𝑞conv 𝑑𝑇 𝜌cV = 𝑃 − hA𝑠(𝑇 − 𝑇∞) 𝑑𝑡 At steady-state, energy storage is zero, thus 𝑃 = 𝑞conv = ℎ 𝐴𝑠 (𝑇ss − 𝑇∞) 𝑃 0.5 𝑇ss = 𝑇∞ + 2 = 30 + = 155C (40)(0.01)2 hL Part 2.

Tss T(t)

T An energy balance during time t2 gives

𝑑𝑇

= −hA𝑠(𝑇 − 𝑇∞) 𝑑𝑡 Solve this differential equation subject to the initial condition T = Tss when t = 0; 𝑡 𝑇 − 𝑇∞ = exp �− � 𝑇ss − 𝑇∞ 𝜏 𝜌𝑐𝑉 where 𝜏 = ℎ𝐴𝑠 Substitute numbers 𝑡2 = 563 sec = 9.4 min 𝜌cV

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Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Part 3.

Tfail On

No Coolant

Tss T(t)

T t3

t An energy balance during time t3 gives, 𝐸˙st = 𝑃 𝑑𝑇 𝜌cV =𝑃 𝑑𝑡 Solve subject to the initial condition, T = Tss when t = 0. 𝑃 𝑇(𝑡) = 𝑇ss +

𝜌cV

𝑡

Substitute numbers, 𝑡3 = 203 sec = 3.4 min

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Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 10.8 Consider a thin plate of area A and thickness L with material properties k, c, and . The plate is exposed to convection on the bottom side and to a specified heat flux, qs”(t), on the top side. In addition, the plate is subjected to a volumetric heat source, g(t) (W/m3). The initial temperature is T0. Assume the lumped capacity approximation is valid. qs”(t)

g

L T , h

a) Derive the energy equation. Formulate the complete mathematical model of the system. b) Sketch the solution assuming g and qs” are constant. Consider T0 = T. On a single graph, put curves for a 0, medium, and large value of h. On another graph, show the effect of the volumetric heat source, g. Try to create a meaningful graph that highlight the effect of g only. c) Consider the mathematical model from part (a) in the form 1 𝑑𝑑𝑑 = − 𝑑𝑑 + 𝑆 𝑑𝑡 𝜏 𝑑𝑑 = 𝑑𝑑0 , 𝑡 = 0 What are the temperature rise, (t), initial temperature, 𝑑𝑑0, time constant, , and the source term, S? d) Using the model from part (c), solve for the transient temperature solution, 𝑑𝑑(𝑡) when S(t) is a constant. Plot your solution. e) Using the model from part (c), solve for the transient temperature solution, 𝑑𝑑(𝑡) when S(t) is a pulse, 𝑆(𝑡) = 𝑆0(𝐻(𝑡) − 𝐻(𝑡 − ∆𝑡)) where S0 is a constant and t is the pulse time. Plot your solution. f) Show that due to the linear nature of this problem, the solution can be constructed as a superposition or summation of two simpler solutions, one due to the heat source S(t) only, and another due to the initial condition 𝑑𝑑0 only. Do not simply examine the mathematical solution to verify that this works (zero credit for this approach). Instead, try the superposition 𝑑𝑑(𝑡) = 𝑑𝑑𝑠(𝑡) + 𝑑𝑑𝑖𝑐(𝑡) directly in the governing equations. What equations govern the functions 𝑑𝑑𝑠 and 𝑑𝑑𝑖𝑐?

Solution a) Mathematical Model 𝜌𝑐𝑉

𝑑𝑇 𝑑𝑡

= −ℎ𝐴(𝑇 − 𝑇 ) + 𝑔(𝑡)𝑉 + 𝑞"(𝑡)𝐴 ∞

𝑠

or 𝑑𝑇 = − ℎ (𝑇 − 𝑇 ) + 1 𝑔(𝑡) + 1 𝑞"(𝑡) ∞ 𝑑𝑡 𝜌𝑐𝐿 𝜌𝑐 𝜌𝑐𝐿 𝑠 𝑇 = 𝑇0 , 𝑡 = 0

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Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

b) Sketches At steady-state 𝑇𝑠𝑠 = 𝑇∞ + c) Equivalent Form

𝑔𝑉 + 𝑞"𝑠𝐴 ℎ𝐴

1 𝑑𝑑𝑑 = − 𝑑𝑑 + 𝑆 𝑑𝑡 𝜏 𝑑𝑑 = 𝑑𝑑0 , 𝑡 = 0 where 𝑑𝑑(𝑡) = 𝑇(𝑡) − 𝑇∞ , 𝑑𝑑0 = 𝑇0 − 𝑇∞ 𝜌𝑐𝑉 𝜌𝑐𝐿 𝑔(𝑡) 1 "(𝑡) 𝜏= = , 𝑆(𝑡) = + 𝑞 ℎ∙𝐴 ℎ 𝜌𝑐 𝜌𝑐𝐿 𝑠

d) Solution for constant S 𝑑𝑑(𝑡) =

𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑐𝑜𝑛𝑑𝑖𝑡𝑖𝑜𝑛

𝑠𝑜𝑢𝑟𝑐𝑒

−� 𝑡� /𝜏𝜏 𝑑�0�𝑒�

�� (� +𝑆 ∙� 𝜏� 1� −�𝑒�−�𝑡�/𝜏𝜏�)

𝑠𝑡𝑒𝑎𝑑𝑦

𝑡𝑟𝑎𝑛𝑠𝑖𝑒𝑛𝑡

������−�𝑡� /𝜏𝜏 (𝑑�� 𝑑𝑑(𝑡) = 𝑆� ∙𝜏 + � 0− 𝑆 ∙ 𝜏)𝑒

e) Pulsed Solution Define 𝜓𝜓𝑐(𝑡) as the response to a constant source of unit magnitude (S = 1) with zero initial condition ( 𝑑𝑑0 = 0). For this case the previous solution reduces to , 𝑡 < 0 0 𝜓𝜓 (𝑡) = 𝜏(1 − 𝑒−𝑡/𝜏𝜏)𝐻(𝑡) = � 𝑐 𝜏(1 − 𝑒−𝑡/𝜏𝜏) , 𝑡 ≥ 0 It is best to program this type of function on a computer using the step function rather than an ‘if’ condition. The pulsed heat flux can be considered as an on/off sequence, written in either of the following equivalent forms. 𝑑𝑑(𝑡) = 𝑆0(𝜓𝜓𝑐(𝑡) − 𝜓𝜓𝑐(𝑡 − 𝛥𝑡)) 𝜓𝜓𝑐(𝑡) , 0 < 𝑡 < 𝛥𝑡 𝑑𝑑(𝑡) = 𝑆0 � 𝜓𝜓𝑐(𝑡) − 𝜓𝜓𝑐(𝑡 − 𝛥𝑡) , 𝑡 > 𝛥𝑡 𝜏(1 − 𝑒−𝑡/𝜏𝜏) , 0 < 𝑡 < 𝛥𝑡 𝑑𝑑(𝑡) = 𝑆0 � −(𝑡−𝛥𝑡)/𝜏𝜏 𝜏(𝑒 − 𝑒−𝑡/𝜏𝜏) , 𝑡 > 𝛥𝑡 Alternately, the solution can be obtained by carefully evaluating the resulting integral using an integrating factor.

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Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

f) Superposition of Elementary Solutions 𝑑𝑑𝑠 Problem 𝑑𝑑𝑑𝑠 1 = − 𝑑𝑑𝑠 + 𝑆 𝑑𝑡 𝜏 𝑑𝑑𝑠 = 0 , 𝑡 = 0

𝑑𝑑𝑖𝑐 Problem 𝑑𝑑𝑑𝑖𝑐 1 = − 𝑑𝑑𝑖𝑐 𝑑𝑡 𝜏 𝑑𝑑𝑖𝑐 = 𝑑𝑑0 , 𝑡 = 0

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Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 10.9 Consider a large, thin plate of area A and thickness L with material properties k, c, and . The plate is exposed to convection at different environments on the top and bottom. In addition, the plate is subjected to a constant volumetric heat source, 𝑔 (W/m3). The initial temperature is T0.

Assume the lumped capacity approximation is valid.

T1 , h1

g

L

Insulated sides

T2 , h2 a) Derive the energy equation. b) What is the steady state temperature? c) Solve for the transient temperature solution, T(t). d) Transform the energy equation into dimensionless form. e) Sketch the anticipated solution for ℎ1 ≫ ℎ2, ℎ1 = ℎ2 , and ℎ1 ≪ ℎ2 with g = 0, 𝑇0 = (𝑇∞1 + 𝑇∞2)/2 and 𝑇∞1 > 𝑇∞2. Put all curves on a single graph. f) Show the effect of the volumetric heat source, g. Try to create a meaningful graph that highlights the effect of g only.

Solution a) Energy Equation 𝜌cV

𝑑𝑇

𝑑𝑡 b) Steady State Temperature

= −ℎ1𝐴(𝑇 − 𝑇∞,1) − ℎ2𝐴(𝑇 − 𝑇∞,2) + gV

𝑇ss =

ℎ1𝑇∞,1 + ℎ2𝑇∞,2 + gL ℎ1 + ℎ2

c) General Solution 𝑇(𝑡) = 𝑇0𝑒−𝑡⁄𝜏𝜏 + 𝑇ss(1 − 𝑒−𝑡⁄𝜏𝜏) 𝜌cV = 𝜌cL 𝜏= )𝐴 (ℎ + ℎ ) + ℎ (ℎ1 2 1 2 d) Nondimensional Form Option 1

Option 3 𝑑𝑇+ = −𝑇+ + 𝑆+ = −(𝑇+ − 1) + 𝑆+ = −𝑇+ 𝑑𝑡+ 𝑑𝑡+ 𝑑𝑡+ 𝑇+ = 1 , 𝑡+ = 0 𝑇+ = 0 , 𝑡+ = 0 𝑇+ = 1 , 𝑡+ = 0 𝑇 − 𝑇 𝑡 𝑇 − 𝑇eq 𝑇 − 𝑇eq − 𝜏𝑆 𝑡 0 𝑡 𝑇+ = , 𝑡+ = 𝑇+ = , 𝑡+ = 𝑇+ = , 𝑡+ = 𝑇eq − 𝑇0 𝜏 𝑇0 − 𝑇eq 𝜏 𝑇0 − 𝑇eq − 𝜏𝑆 𝜏 𝜏 𝜏 𝜏 𝑆+ = 𝑆 𝑆+ = 𝑆 𝑆+ = 𝑆 𝑇eq − 𝑇0 𝑇0 − 𝑇eq 𝑇0 − 𝑇eq − 𝜏𝑆 Note that the meaningful interpretation of temperature is actually a temperature difference. It is always easier (and more meaningful) to define a new dependent variable as the rise above a logical ground level. 𝑑𝑇+

Option 2

𝑑𝑇+

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Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 10.10 A long thin copper wire of diameter D and length L has an electrical resistance per length of wire R’e (/m), density , specific heat c , and total emissivity . The wire is initially at steady state at temperature T0. At time t=0 an electric current I (amps) is passed through the wire, causing electrical resistance heating. As the wire temperature rises, heat is dissipated by convection to the air at temperature T with a heat transfer coefficient h and by radiation to the walls at temperature Tsur.

Walls, Tsur Air, T , h Power supply

I

Wire

Assumptions • Constant thermal and electrical properties • Lumped capacity, valid for low Bi. That is, spatially uniform temperature at any time, 𝑇 = 𝑇(𝑡).

• Negligible conduction from the ends or stand. a) Derive the mathematical model governing the transient temperature of the wire. b) For the case where 𝑇0 = 𝑇∞ = 𝑇𝑠𝑢𝑟, sketch the anticipated temperature vs time behavior of the wire. On a single graph sketch curves corresponding to free convection, mild forced convection, and strong forced convection. Clearly label your graph. c) Neglecting radiation affects, derive an analytical solution for the transient temperature of the wire. What is the steady temperature of the wire for this case? d) Create a function to evaluate the solution numerically. Plot temperature vs. time using the following parameters. Put all three curves on a single graph. h values of 5, 25, and 100 W/m2 K 𝑇0 = 𝑇∞ = 𝑇𝑠𝑢𝑟 = 300 𝐾 D = diameter = 0.001 m , L = length = 0.4 m  = density = 8933 kg/m3 , k = 400 W/m K, c = specific heat = 385 J/kg K  = emissivity = 0.9 I=6 Amps, e = electrical resistivity = 80 -cm Solution (a) Energy Equation 𝜌𝑐𝑉

𝑑𝑇 𝑑𝑡

= 𝐼2𝑅 − 𝐴 ℎ(𝑇 − 𝑇 ) − 𝐴 𝜀𝜀𝜎(𝑇4 − 𝑇 𝑒

𝑠

∞

𝑠

𝑠𝑢𝑟

4)

𝑇 = 𝑇0 , 𝑡 = 0 𝜋𝐷2 𝐴𝑠 = 𝜋𝐷𝐿, 𝑉 = 𝐿 4 16

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

(b) Anticipated Solution free convection T(t)

mild forced

strong forced Ti Time, t

(c) Transient Linear solution neglecting radiation 𝑑𝑑(𝑡) = 𝑑𝑑0𝑒−𝑡/𝜏𝜏 + 𝑆 ∙ 𝜏(1 − 𝑒−𝑡/𝜏𝜏) In terms of the original physical variables the solution is 𝐼2𝑅𝑒 𝑇(𝑡) − 𝑇∞ = (𝑇0 − 𝑇∞)𝑒−𝑡/𝜏𝜏 + (1 − 𝑒−𝑡/𝜏𝜏) 𝐴𝑠ℎ Steady State Solution 𝐼2𝑅𝑒 𝑇𝑠𝑠 = 𝑇∞ + 𝐴𝑠 ℎ (d) Numerical Solution Effect of Convection Coefficient

Convection and Radiation Effects

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Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 10.11: Canister Wall A spherical, stainless steel (AISI 302) canister is used to store reacting chemicals that provide for a uniform heat flux 𝑞𝑖" to its inner surface. The canister is suddenly submerged in a liquid bath of temperature T< Ti where Ti is the initial temperature of the canister wall.

Cannister

Liquid T, h

qi" ri

ro

 = 8000 kg/m3 c = 500 J/kg K ri = 0.5 m ro = 0.6 m T = 300 K

a) Assuming negligible temperature gradients in the canister wall and a constant heat flux qi", develop an equation that governs the variation of the wall temperature with time during the transient process. b) Develop an expression for the steady temperature of the wall. c) The convection coefficient depends on the velocity of the fluid and whether or not the wall temperature is large enough to induce boiling in the liquid bath. For the parameters listed below, compute and plot the steady-state temperature as a function of h for the range 100 < h < 10,000 W/m2 K. Include curves for 𝑞𝑖" = 105, 2x105, and 3x105 W/m2. Put all three curves on a single graph. d) Is there any value of h below which operation would be unacceptable?

Solution a) Mathematical Model

𝜌𝑐𝑉 𝑑𝑇 = −ℎ𝐴 (𝑇 − 𝑇 ) + 𝑞"𝐴 𝑜 ∞ 𝑖 𝑖 𝑑𝑡 𝑇 = 𝑇𝑖 , 𝑡 = 0

where 𝑉=

4

𝜋(𝑟3 − 𝑟3)

𝑖 3 𝑜 𝐴𝑖 = 4𝜋𝑟2, 𝐴𝑜 = 4𝜋𝑟2 𝑜

𝑖

b) At steady-state 𝑇𝑠𝑠 = 𝑇∞ +

𝑞𝑖" 𝐴𝑖 ℎ𝐴𝑜

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Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 10.12 Consider the system 𝑑𝑥/𝑑𝑡 = 𝑟 ∙ 𝑥 − 𝑥3. For 𝑟 < 0, 𝑟 = 0, 𝑎𝑛𝑑 𝑟 > 0: a) Use linear stability analysis to classify the fixed points. b) Plot the phase portraits and the anticipated solutions, x(t) versus t.

Solution The fixed points 𝑥∗ = 0 and ± √𝑟 are found from setting 𝑑𝑥/𝑑𝑡 = 0. Stability is ascertained (𝑥∗) from the sign of 𝑑𝑓 = 𝑟 − 3𝑥∗2. The bifurcation point is r c = 0 since real fixed points 𝑥∗ = 𝑑𝑥

±√𝑟 exist only for r  0 . The findings are: • 𝑟 < 𝑟𝑐: one fixed point, stable at x* = 0 • 𝑟 = 𝑟𝑐 = 0: one fixed point, stable at x* = 0, but weakly stable • 𝑟 > 𝑟𝑐: 3 fixed points, unstable at x* = 0, stable at 𝑥∗ = −√𝑟, and stable at 𝑥∗ = √𝑟 The following phase portraits and solutions with fixed points and bifurcation diagram reveal the dynamics of this system.

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Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 10.13: Bifurcations Consider the following first order ODEs. 𝑟 Case 1: 𝑑𝑥 = 𝑥 (1 + ) 𝑑𝑡

1+𝑥2

𝑑𝑡

1+𝑥

𝑥 Case 2: 𝑑𝑥 = −𝑟𝑥 +

a) Find all the fixed points. b) Find any critical values rc at which bifurcations occur. What type of bifurcations occur (saddle-node, transcritical, supercritical pitchfork, or subcritical pitchfork)? c) Classify the stability of the fixed points. Collect your results in a table. d) Sketch the bifurcation diagram: x* versus r. e) Sketch the phase portraits and corresponding solutions for each qualitatively different behavior.

Solution Case 1 A bifurcation occurs at 𝑟𝑐 = −1. Subcritical pitchfork bifurcation. 𝑥1∗ = 0 𝑥2∗ = +�−(1 + 𝑟) 𝑥3∗ = −�−(1 + 𝑟)

𝑟 < 𝑟𝑐 stable unstable unstable

𝑟 > 𝑟𝑐 unstable DNE DNE

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Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Case 2 A bifurcation occurs at 𝑟𝑐 = 1. Transcritical bifurcation. 𝑥1∗ = 0 1 𝑥∗ = − 1 2 𝑟

𝑟 < 𝑟𝑐 unstable stable

𝑟 > 𝑟𝑐 stable unstable

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Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 10.14: Transient Parameter Consider a first order ODE of the form 𝑑𝑥 = 𝑓(𝑥, 𝑟) 𝑑𝑡 where r is a parameter. The phase diagrams for two different values of the parameter r are shown.

Consider a process where the parameter r is at a value r1 for a quite some time and then changes to a new value r2. Sketch the resulting dynamic response of the system (x(t) versus t) starting from initial states x0 = -4, -2, 0, and 2.

Solution

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Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 10.15: Improved Laser An improved model of a laser is 𝑑𝑛 = 𝐺𝑛𝑁 − 𝑘𝑛 𝑑𝑡 𝑑𝑁 = −𝐺𝑛𝑁 − 𝑓𝑁 + 𝑝 𝑑𝑡 a) Assuming 𝑑𝑁 ≈ 0, derive a differential equation for n. 𝑑𝑡

b) Show that n* = 0 becomes unstable for p > pc. Determine pc. c) What type of bifurcation occurs at pc. d) For what range of parameters is it valid to assume that 𝑑𝑁 ≈ 0? 𝑑𝑡

Solution a) Adiabatic elimination Assume 𝑑𝑁 ≈ 0. Then, 𝑁 = 𝑑𝑡

𝑝 𝐺𝑛+𝑓

. The differential equation for n becomes:

𝑝 𝑝 𝑑𝑛 = 𝐺𝑛 ( ) − 𝑘𝑛 = 𝑛 ( ) − 𝑘𝑛 𝑑𝑡 𝐺𝑛 + 𝑓 𝑛 + 𝑓/𝐺 b) Fixed Points The fixed points are the values n* where 𝑑𝑛 = 0. 𝑑𝑡 𝑝 ∗ 𝑛 ( ) − 𝑘𝑛∗ = 0 ∗ 𝑛 + 𝑓/𝐺 We find 2 possible fixed points: 𝑛1∗ = 0 𝑝 𝑓 𝑛∗ = − 2 𝑘 𝐺 Stability Analysis of the fixed Points 1 𝑛 𝑑𝑛 = 𝑝 − 𝑝 𝑑𝑛 2−𝑘 𝑓 𝑓 𝑛+𝐺 (𝑛 + 𝐺) Stability of n1*=0 𝐺 𝑑𝑛 ( ) =𝑝 −𝑘 𝑑𝑛 𝑛1=0 𝑓 We see a critical value of p, 𝑓 𝑝𝑐 = 𝑘 𝐺 𝑝 < 𝑝𝑐: Stable 𝑝 > 𝑝𝑐: Unstable Stability of n2* 𝑝 𝑓 𝑘 − 𝐺 − 𝑘 = − 𝑘2 𝑝 𝑓 𝑑𝑛 1 − 𝑝 ( − ) ( ) =𝑝 𝑝 2 𝑝 𝑑𝑛 𝑛2 𝑝 𝑘 𝐺 ( ) 𝑘 𝑘 𝑓 We see a critical value, 𝑝𝑐 = 𝑘 , again dictates stability. 𝐺

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Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

𝑝 < 𝑝𝑐: Unstable 𝑝 > 𝑝𝑐: Stable c) Type of Bifurcation We have a transcritical bifurcation. d) Validity of the Adiabatic Elimination Assumption Consider two linear, uncoupled differential equations: 𝑑𝑥1 = −𝑎1𝑥 + 𝑆1 𝑑𝑡 𝑑𝑥2 = −𝑎2𝑥 + 𝑆2 𝑑𝑡 These have solutions 𝑆 −𝑎1𝑡 + 1 (1 − 𝑒−𝑎1𝑡) 𝑥1 = 𝑥10 𝑒 𝑎1 𝑆 −𝑎2𝑡 + 2 (1 − 𝑒−𝑎2𝑡) 𝑥2 = 𝑥20 𝑒 𝑎2 If a2 >> a1, then x2 will relax or approach steady state much faster than x1. Under such conditions, we could assume dx2 ≈ 0. Our equations are both nonlinear and coupled, however, dt

let’s try to apply the same line of reasoning. 𝑑𝑛 𝑑𝑁

𝑑𝑡

= (𝐺𝑁 − 𝑘)𝑛

= −(𝐺𝑛 − 𝑓)𝑁 + 𝑝 𝑑𝑡 Our logic would dictate that 𝐺𝑛 + 𝑓 ≫ 𝑘 − 𝐺𝑛. The trouble is that n and N are variables.

However, at fixed point-1, we have 𝑝 𝑛∗ = 0, 𝑁∗ = 1 1 𝑓 Substitute to find 𝑝 𝑓>>𝑘 − 𝐺 𝑓 𝑓 𝑓𝑘 𝑝>> (1 − ) 𝑘 𝐺 Similarly, fixed point-2, 𝑝 𝑓 𝑘 𝑛2∗ = − , 𝑁2∗ = 𝑘 𝐺 𝐺 Substitute and we find, 𝐺𝑝 >>0 𝑘 Thus, the criteria for validation of the model are 𝑓 𝑓𝑘 𝐺𝑝 𝑝>> (1 − ) and >>0 𝑘

𝐺

𝑘

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Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 10.16: Biochemical Switch Zebra stripes and butterfly wing patterns are two of the most spectacular examples of biological pattern formation. Explaining the development of these patterns is one of the outstanding problems of biology. As one ingredient in a model of pattern formation, consider a simple example of a biochemical switch, in which a gene G is activated by a biochemical signal substance S. For example, the gene may normally be inactive but can be “switched on” to produce a pigment or other gene product when the concentration of S exceeds a certain threshold. Let g(t) denote the concentration of the gene product, and assume that the concentration so of S is fixed. The model is 𝑑𝑔 𝑘3𝑔2 = 𝑘1𝑠𝑜 − 𝑘2𝑔 + + 𝑔2 𝑑𝑡 𝑘4 where the k’s are positive constants. The production of g is stimulated by so at a rate k1, and by an autocatalytic or positive feedback process (the nonlinear term). There is also a linear degradation of g at a rate k2. a) Show that the system can be put in the dimensionless form 𝑑𝑥 𝑥2 = 𝑠 − 𝑟𝑥 + 𝑑𝜏 1 + 𝑥2 where r>0 and s  0 are dimensionless groups. What are the dimensionless variables and parameters? b) For s = 0, find and classify all the positive fixed points x*. Is there any critical values rc?

Determine and plot the fixed points for all r and s. c) Find the parametric equations for the bifurcation curves in (r, s) space, and classify the bifurcations that occur. Plot the stability diagram in (r, s). d) Assume that initially there is no gene product, i.e., g(0) = 0, and suppose s is slowly increased from zero (the activating signal is turned on); what happens to g(t)? What happens if s then goes back to zero? Does the gene turn off again? Support your findings using phase plots and numerical solutions of the x vs.  behavior.

Solution (a) Dimensionless Form

𝑑𝑥

𝑥2 = 𝑠 − 𝑟𝑥 + 𝑑𝜏 1 + 𝑥2 𝑔 𝑘3 𝑉𝑎𝑟𝑖𝑎𝑏𝑙𝑒𝑠: 𝑥 = , 𝜏=𝑡 𝑘4 𝑘4 𝑘1 𝑘2𝑘4 , 𝑟= 𝑃𝑎𝑟𝑎𝑚𝑒𝑡𝑒𝑟𝑠: 𝑠 = 𝑠𝑜 𝑘3 𝑘3 (b) Fixed Points for s=0 Fixed points for s = 0 x*=0 𝑥∗ = 𝑥∗ =

1 − √1 − 4𝑟2

0 ≤ 𝑟 < 𝑟𝑐 Stable Unstable

𝑟 > 𝑟𝑐 Stable Does not exist

2𝑟 1 + √1 − 4𝑟2

Stable

Does not exist

2𝑟

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Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

The two positive fixed points only exist if 𝑟 < 𝑟𝑐 = 1/2. The phase plots for various r values and s =0 are shown below.

s ◻0

dx d◻

r 0.4 0.5 0.6

x

(c) Fixed Points Next the more general fixed point diagram in r-s space is shown. The fixed points are the solution of a cubic equation, resulting in a fold and cusp.

(d) Phase Portraits and Bifurcation Diagram It seems that rc = 0.5 is an interesting value for r. In addition, it would be interesting, as suggested in the problem, to start at zero and slowly increase s. A sequence of phase portraits is shown below. dx d◻

s 0.

0.05

0.02 0.5

1.0

1.5

2.0

x

0.04 0.06 0.08

The mathematical explanation behind the biochemical switch is now revealing itself. Starting from an initial state of x0 = 0, as s increases slowly the gene concentration rises slowly to the lower stable fixed point. At around s = 0.068, a saddle node bifurcation occurs as the smaller 26

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

two fixed points disappear. x then shoots to the larger stable fixed point, causing a switchinglike mechanism where x changes rapidly from a low value to a high value. The phase portraits show that the saddle node is the only possible type of bifurcation.

Note a hysteresis effect. If s is slowly decreased the gene level x will not return to the low level, it will stay at the higher activated level.

Next is a stability diagram in r-s space is shown. Derivations are in the Appendix. s 0.12 0.10 0.08 0.06 0.04 0.02 0.00 0.0

r 0.1

0.2

0.3

0.4

0.5

0.6

Based on this stability diagram and previous conclusions from our phase portraits, we can qualitatively sketch the solutions for all (s, r) combinations. (e) General Dynamical Behavior Let’s thus examine the dynamical behavior for r = rc =0.5 as s is increased. r◻0.5 xt

0.064 0.066 0.068 0.07 0.072 0.074 t

Note that a switching phenomenon occurs between s = 0.068 and 0.07. For s =0.07, the x vs t behavior shows a slow rise or bottleneck (the ghost of the annihilated fixed points), followed by a rapid rise to an elevated level.

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Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 10.17: Trout Fishing in Stocked Streams Fishing in stocked trout waters usually involves a sudden addition of fish on stocking days immediately followed by lots of fisherman feverishly trying to catch them. The rate of fish caught is usually proportional to the number of fish still available. Assuming no natural reproduction, the following model is proposed. 𝑑𝑁 = −𝑟𝑁 + 𝑆(𝑡) 𝑑𝑡 The stocking function is modeled as a periodic sequence of regularly spaced pulse functions as shown where Nst = number of fished per stocking 𝑡𝑠𝑡= time between stockings t = time required for each stocking

a) Express the stocking function S(t) in mathematical form. b) Determine the analytical solution for a single stocking event. Name this function 𝑁𝑖(𝑡, Δ𝑡) where i = 1,2,3,.. is the ith stocking event. c) Determine the general solution for a sequence of stocking events in terms of the function 𝑁𝑖(𝑡, Δ𝑡). d) Determine the limit as Δ𝑡 → 0. e) Plot your solution from part (d). Use parameter values that make sense.

Solution (a) Stocking Function 𝑆[𝑡, 𝑡𝑠𝑡, 𝛥𝑡, 𝑁𝑠𝑡] = or

𝑁𝑠𝑡 H[Δt − Mod[𝑡, tst]] ∆𝑡

𝑖𝑚𝑎𝑥

𝑁𝑠𝑡 𝑆[𝑡, 𝑡𝑠𝑡, 𝛥𝑡, 𝑁𝑠𝑡] = � (𝐻[𝑡 − 𝑡𝑠𝑡(𝑖 − 1)] − 𝐻[𝑡 − 𝑡𝑠𝑡(𝑖 − 1) − ∆𝑡]) ∆𝑡 𝑖=1

(b) Unit Response ψcon[𝑡, 𝑟] = (c) Pulsed Response ψpulse[𝑡, 𝑡𝑖, ∆𝑡, 𝑟] =

1 Δt

1 𝑟

∗ (1 − Exp[−𝑟 ∗ 𝑡]) ∗ 𝐻[𝑡]

(ψcon[𝑡 − ti, 𝑟] − ψcon[𝑡 − ti − Δt, 𝑟])

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Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

(d) Total Solution

∞

Nfish[𝑡, 𝑡𝑠𝑡, 𝛥𝑡, 𝑁𝑠𝑡, 𝑟] = Nst � ψpulse[𝑡, (𝑖 − 1)tst, Δt, 𝑟] 𝑖=1

(e) Limit as Δt → 0 ψpulse[𝑡, 𝑡𝑖, 0, 𝑟] = Exp[−𝑟 ∗ (𝑡 − ti)] ∗ 𝐻[𝑡 − ti] (f) Numerical results for r = 0.1 and r = 0.25 are displayed.

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Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 10.18: A Model of a Fishery The equation 𝑁 𝑑𝑁 = 𝑟𝑁 (1 − ) − 𝐻 𝑑𝑡 𝐾 provides a simple model of a fishery. Here N(t) is the number of fish. In the absence of fishing, the population is assumed to grow logistically. The effects of fishing are modeled by the term, – H, which says that fish are caught or “harvested” at a constant rate H > 0, independent of the fish population N. This assumes that the fishermen are not worried about fishing the population dry; they simply catch the same number of fish every day. a) Show that the system can be rewritten in dimensionless form as 𝑑𝑥 = 𝑥(1 − 𝑥) − ℎ 𝑑𝑡 What are the dimensionless variables x, , and h? b) Determine the fixed points. Classify the stability of the fixed points. Show that a bifurcation occurs for a certain critical value hc. What type of bifurcation is this? Plot a bifurcation diagram (x* versus h). c) Plot separate phase portraits for 0 < h < hc, h = hc, and h > hc.. d) Plot the solutions x(t) vs.  for 0 < h < hc, h = hc, and h > hc. Create 3 plots, one corresponding to each range of h. Include several initial conditions on each plot. e) Discuss any unrealistic aspects of this model. Can you suggest an improved model? Sketch the anticipated solution for your improved model.

Solution (a) Dimensionless Form

(b) Fixed Points

𝑑𝑥 = 𝑥(1 − 𝑥) − ℎ 𝑑𝜏 𝑁 𝐻 𝑥 = , 𝜏 = 𝑟𝑡 , ℎ = 𝑟𝐾 𝐾

𝑥1∗ = (1 − √1 − 4ℎ)/2

𝟎𝟎 ≤ ℎ ≤ ℎ𝑐 unstable

ℎ > ℎ𝑐 DNE

𝑥2∗ = (1 + √1 − 4ℎ)/2

stable

DNE

A bifurcation occurs at ℎ𝑐 = 1/𝟒𝟒. This is a saddle bifurcation.

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Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

(c) And (d) Phase Portraits and Solutions

(e) Improved Model This model can predict “negative fish” or perhaps I should say “anti-fish”. Also, the harvesting rate should be related to the number of available fish. The harvesting rate should be zero for zero fish and should reach a saturation limit (how many all-you-can-eat fish buffets can we stand). It would be better to replace H by something like: 𝑁2 𝐻(𝑁) = 𝐻𝑚𝑎𝑥 2 𝐴 + 𝑁2

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Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 10.19: Improved Model of a Fishery The equation 𝑁 𝑑𝑁 𝑁 = 𝑟𝑁 (1 − ) − 𝐻 𝑑𝑡 𝐾 𝐴+𝑁 provides a model of a fishery. Here 𝑁(𝑡) is the number of fish. In the absence of fishing, the 𝑁 population is assumed to grow logistically. The effects of fishing are modeled by the term 𝐻 𝐴+𝑁, where 𝐻 > 0 and 𝐴 > 0, which says that fish are caught or “harvested” at a rate that decreases with the fish population N. This is plausible since it gets harder to catch fish as the population decreases. a) Show that the system can be rewritten in dimensionless form as 𝑑𝑥 𝑥 = 𝑥(1 − 𝑥) − ℎ 𝑑𝜏 𝑎+𝑥 What are the dimensionless quantities x, , a, and h? b) Determine the fixed points. Classify the stability of the fixed points using linear stability analysis. Collect your findings in an organized table. What types of bifurcations occur? c) Plot the stability diagram in (a, h) space. Can hysteresis occur in any of the regions? d) For each qualitatively different behavior plot Solution (a) Dimensionless Form 𝑑𝑥 𝑥 = 𝑥(1 − 𝑥) − ℎ 𝑑𝜏 𝑎+𝑥 𝐴 𝑁 𝐻 𝑥 = , 𝜏 = 𝑟𝑡, 𝑎 = , ℎ= 𝐾 𝐾 𝑟𝐾 (b) : Fixed Points 1

𝑥∗ = −

2 1

𝑥∗ = +

Bifurcations

𝑥1∗ = 0 ((1 − 𝑎) − �(1 − 𝑎)2 − 4(ℎ − 𝑎)) ((1 − 𝑎) + �(1 − 𝑎)2 − 4(ℎ − 𝑎))

2 1

ℎ = (𝑎 + 1)2 : saddle bifurcation. 𝑐

4

ℎ𝑐 = 𝑎 : transcritical bifurcation.

Stability of the Fixed Points 𝑥1∗ = 0 0<ℎ<𝑎 ∗ = 0 𝑥− ∗ = 0 𝑥+

Unstable Stable

ℎ<𝑎

ℎ>𝑎

Unstable

Stable

𝑎 < ℎ < ℎ𝑐 a<1 Unstable Stable

a>1 Unstable Unstable

ℎ > ℎ𝑐 DNE DNE

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Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

(c) Stability Diagram

Note that there are three different qualitative behaviors. Hysteresis is possible. Imagine the variable a<1 and slowly vary h back and forth. (d) Phase Portraits and Solutions

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Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

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Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Chapter 11

Second Order Ordinary Differential Equations Linear Systems Problem 11.1.1 The motion of a linear damped harmonic oscillator is described by 𝑑2𝑥 𝑑𝑥 𝑚 2 +𝑏 +𝑘∙𝑥 =0 𝑑𝑡 𝑑𝑡 𝑑𝑥 𝑥 = 𝑥0 , = 𝑣0, 𝑡 = 0 𝑑𝑡 where b  0 is the damping coefficient. a) Rewrite this equation as simultaneous first order equations. b) Classify the fixed point at the origin and plot the phase portrait. Be sure to show all the various cases that can occur, depending on the relative sizes of the parameters. c) How do your results relate to standard notions of overdamped, critically damped, underdamped, and undamped vibrations?

Solution a) Decomposition into coupled first order equations 𝑑𝑥 =𝑣 𝑑𝑡 𝑘 𝑏 𝑑𝑣 =− 𝑥− 𝑣 𝑑𝑡 𝑚 𝑚 In vector form 𝑑𝑥 0 1 𝑑𝑡 𝑏 � �𝑥 ] � �=� 𝑘 𝑣 − − 𝑑𝑣 𝑚 𝑚 𝑑𝑡 b) Classification of Fixed Point The only fixed point is clearly at (0,0). The character of this system is revealed by examining the eigenvalues, obtained from the characteristic equation. −𝜆 1 𝑏 𝑘 𝑏 𝑘 𝑏 𝑑𝑒𝑡 � 𝑘 � = −𝜆(− − 𝜆) + = 𝜆2 + 𝜆 + = 0 − (− − 𝜆) 𝑚 𝑚 𝑚 𝑚 𝑚 𝑚 The eigenvalues are 𝑏 � 𝑏 2 𝑘 𝜏 ± √𝜏2 − 4𝛥 1 = (− ± ( ) − 4 ) 𝜆= 2 𝑚 2 𝑚 𝑚 where 1

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

𝛥=

𝑘 𝑚

> 0 ,𝜏 = −

The system behavior is indicated by

𝑏 𝑚

≤0

𝑏 2 𝑘 𝜏2 − 4𝛥 = ( ) − 4 𝑚 𝑚 The eigenvalues always have a negative real part ( < 0). However, there is a clear behavior change when the quantity under the square root changes sign. This value of b is designated as bcr. 𝑏cr 2 𝑘 ( ) −4 =0 𝑚

𝑚

𝑏cr = 2√𝑚𝑘 At this point it seems natural to define a natural frequency, defined as 𝑛𝑎𝑡𝑢𝑟𝑎𝑙 𝑓𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦 = 𝜔𝑛 = �

𝑘 𝑚

𝑏cr 𝑘 = 2� = 2𝜔 𝑛 𝑚 𝑚 It now also seems natural to express b in terms of n and bcr. 𝑏 𝑏 𝑏cr 𝑏 = = 2𝜔𝑛 = 2𝜔𝑛𝜁 𝑏cr 𝑚 𝑚 𝑏cr 𝑏 𝑑𝑎𝑚𝑝𝑖𝑛𝑔 𝑟𝑎𝑡𝑖𝑜 = 𝜁 = 𝑏cr

The eigenvalues 𝜆 = 𝜔𝑛 (−𝜁 ± �𝜁2 − 1) 𝛥 = 𝜆1𝜆2 = 𝜔𝑛2 > 0 𝜏 = 𝜆1 + 𝜆2 = −2𝜔𝑛𝜁 ≤ 0 𝜏2 − 4𝛥 = 4𝜔𝑛2(𝜁2 − 1) The solution classification is summarized in the following table.

Condition b = 0 or 𝜁 = 0 b < bcr or 0 < 𝜁 < 1 b = bcr or 𝜁 = 1 b > bcr or 𝜁 > 0

Eigenvalues 1 and 2 Purely imaginary

Stability of x* = [0,0] Center

Standard Notions in Vibrations Undamped

Complex conjugates, negative real parts negative real (repeated) negative real

Stable spiral

Underdamped

star or degenerated node Stable node

Critically damped Overdamped

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Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Phase Portraits: 𝜁 = b/bcr

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Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 11.1.2 Consider the following system of linear ODE’s. 𝑑𝑥 𝑑𝑦 = 𝑎𝑦 + 𝑏 , =𝑥 𝑑𝑡 𝑑𝑡 a) Determine the fixed points. b) Determine the eigenvalues and the corresponding eigenvectors. c) Classify the fixed point for various values of parameters a and b. d) Sketch phase portraits showing all the qualitatively different behaviors. e) Determine the solution x(t) and y(t) for a = 0, x0 = 0, and y0 = 1.

Solution a) Fixed points 𝑎𝑦∗ + 𝑏 = 0 𝑥∗ = 0 Solve to get 𝑥∗ = 0 𝑦∗ = −𝑏/𝑎 b) Eigenvalues and eigenvectors (A − 𝜆 ∙ I)𝐯𝐯 = 𝟎𝟎 −𝜆 𝑎 ] �𝑣1 0 ]=� ] � 1 −𝜆 𝑣2 0 −𝜆 𝑎 𝑑𝑒𝑡(𝐴 − 𝜆 ∙ 𝐼) = 𝑑𝑒𝑡 � ] = 𝜆2 − 𝑎 = 0 1 −𝜆 𝜆1 = √𝑎 , 𝜆2 = −√𝑎 Eigenvectors for 𝝀𝝀1 = √𝑎: The eigenvectors are determined from 𝑣1 −√𝑎 𝑎 0 −𝜆1 𝑎 𝑣1 ] � ]=� ] � ] �𝑣 ] = � 1 0 −√𝑎 𝑣2 1 −𝜆1 2 This equation must be singular, thus 𝑣1 = √𝑎 ∙ 𝑣2 Eigenvectors for 𝝀𝝀𝟐𝟐 = −√𝑎: The eigenvectors are determined from 0 𝜆2 𝑎 𝑣1 √𝑎 𝑎 𝑣1 � ] �𝑣 ] = � ] �𝑣 ] = � ] 1 𝜆2 2 2 0 1 √𝑎 This equation must be singular, thus 𝑣1 = −√𝑎 ∙ 𝑣2 c) Condition 𝑎>0 𝑎<0

Eigenvalues 1 and 2 One positive, one negative Complex conjugates, Purely imaginary

Stability of x* = [0,-b/a] Saddle Center

Stability is not affected by b.

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Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 11.1.3 Consider the system of linear ODE’s 𝑑𝑥 =𝑎∙𝑥+𝑏∙𝑦+𝑝 𝑑𝑡 𝑑𝑦 =𝑐∙𝑥+𝑑∙𝑦+𝑞 𝑑𝑡 This system is non-homogeneous due to the presence of the terms p and q. a) Determine the fixed points (𝑥∗, 𝑦∗). b) What variable change would transform this to a homogeneous system (one with p and q zero)? c) Are the eigenvalues and stability of the system affected by the presence of the nonhomogeneous terms p and q? If your answer is yes, explain how? d) Let’s say that for a certain combination of parameters the homogeneous problem is a center.

With the presence of the non-homogeneous terms p and q the fixed point is determined to be (𝑥∗, 𝑦∗) = (1,0). Sketch the phase portrait.

Solution a) Fixed points 𝑎 ∙ 𝑥∗ + 𝑏 ∙ 𝑦∗ + 𝑝 = 0 𝑐 ∙ 𝑥∗ + 𝑑 ∙ 𝑦∗ + 𝑞 = 0 𝑝 𝑎 𝑏 𝑥∗ � ] � ] = �𝑞] 𝑐 𝑑 𝑦∗ Solve using Cramer’s rule. 𝑥∗ =

𝑝𝑑 − 𝑞𝑏 𝑝𝑑 − 𝑞𝑏 = 𝑑𝑒𝑡[𝐴] 𝑎𝑑 − 𝑏𝑐

𝑦∗ =

𝑞𝑎 − 𝑝𝑐 𝑞𝑎 − 𝑝𝑐 = 𝑑𝑒𝑡[𝐴] 𝑎𝑑 − 𝑏𝑐

b) Variable change 𝑥+ = 𝑥 − 𝑥∗ 𝑦+ = 𝑦 − 𝑦∗ c) No, the presence of the non-homogeneous terms p and q simply shifts the fixed points.

Stability of the shifted points is not affected.

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Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 11.1.4 Consider a series LRC circuit equation: 𝑑2𝐼 𝑑𝐼 1 𝐿 2+𝑅 + ∙𝐼 =0 𝑑𝑡 𝑑𝑡 𝐶 where L, C > 0 and R  0. a) Rewrite the equation as a two-dimensional linear system. b) Show that the origin is asymptotically stable if R > 0 and neutrally stable if R = 0. c) Classify the fixed point at the origin. Plot the phase portrait in all cases.

Solution a) Define: 𝑥1 = 𝐼 𝑑𝑥1 𝑑𝐼 = 𝑥2 = 𝑑𝑡 𝑑𝑡 The second order system then decomposes as two first order ODE’s 𝑑𝑥1 = 𝑥2 𝑑𝑡 𝑑𝑥2 1 𝑅 =− 𝑥1 − 𝑥2 𝑑𝑡 𝐿𝐶 𝐿 In matrix form 0 1 𝑥 𝑥˙1 𝑅� � 1] � ]=� 1 𝑥2 − − 𝑥˙2 𝐿𝐶 𝐿 b) Analysis of the fixed point (𝑥1, 𝑥2) = (0,0) is the only fixed point. Since the system in linear, the eigenvalues will indicate the global behavior. 𝑅 1 𝜏 = − ,𝛥 = 𝐿𝐶 𝐿 2 1 1 𝑅 disc = 𝜏2 − 4𝛥 = ( ) − 4 = 2 (CR2 − 4𝐿) 𝐿 𝐿𝐶 𝐿 𝐶 𝜆1 =

𝜏 + √𝜏2 − 4𝛥 , 𝜆2 =

𝜏 − √𝜏2 − 4𝛥 2

2 R > 0: 𝜏 < 0, 𝛥 > 0, the origin is stable.

R = 0: 𝜏 = 0, 𝛥 > 0, the origin is neutrally stable (center). c) Classification of the fixed point 𝐶𝑅2 − 4𝐿 < 0 𝐶𝑅2 − 4𝐿 = 0 𝐶𝑅2 − 4𝐿 < 0

The origin is a stable spiral (or a center if R = 0). This often called under damped.

The origin is a star or degenerate node. This often called critically damped.

The origin is a stable node. This often called over damped.

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Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 11.1.5 Consider the two-dimensional system of equations 𝑑𝑥 𝑑𝑦 = 𝑎(𝑥 − 𝑦) = −𝑎(𝑥 − 𝑦) , 𝑑𝑡 𝑑𝑡 𝑥=𝑥 0 𝑦 = 𝑦0 � 𝑡 = 0 a) Express the system in matrix form. b) Determine the eigenvalues of this system. c) Classify the behavior of this system. d) Sketch the phase portrait. e) Sketch a typical solution.

Solution a) Matrix form 𝑑𝑥 =𝐴∙𝑥 𝑑𝑡 𝑥 = 𝑥𝟎𝟎 , 𝑡 = 0 𝑥 −𝑎 𝑎 𝑥0 𝐴=� ] , 𝑥 = � ] , 𝑥 = � 𝟎𝟎 𝑦 𝑦0] 𝑎 −𝑎 b) Eigenvalues −𝑎 − 𝜆 𝑑𝑒𝑡(𝐴 − 𝜆 ∙ 𝐼) = 𝑑𝑒𝑡 � 𝑎

𝑎

] = (−𝑎 − 𝜆)2 − 𝑎2 = 𝜆2 + 2𝜆𝑎 = 0 −𝑎 − 𝜆 𝜆1 = 0 𝜆2 = −2𝑎

c) Classification This is a non-isolated fixed point (see Figure 11.3).

For 𝑎 > 0 it is stable For 𝑎 < 0 it is unstable. d) Phase Portrait

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Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

e) Numerical Solutions

8

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 11.1.6 For each of the following cases: a) Classify the stability of the fixed point based on the eigenvalues. Identify all the different possible types of behavior depending on the parameters a and b. b) Determine the analytical solution using eigenvalues/eigenvectors. Carefully examine the case where sign(ab)=0. c) For each different type of behavior identified in part (a), plot the phase diagram. Also plot the corresponding solution for the special case R0 = 1, J0 = 0.

Case 1: Jerry Springer 𝑑𝑅 𝑑𝐽 =𝐽 , = −𝑅 + 𝐽 𝑑𝑡 𝑑𝑡 Case 2: Out of Touch with Their Own Feelings Suppose Romeo and Juliet are out of touch with their own feelings such that they react to each other but not to themselves. 𝑑𝑅 𝑑𝐽 =𝑎∙𝐽 , =𝑏∙𝑅 𝑑𝑡 𝑑𝑡 Case 3: Fire and Water 𝑑𝑅 𝑑𝐽 =𝑎∙𝑅+𝑏∙𝐽 , = −𝑏 ∙ 𝑅 − 𝑎 ∙ 𝐽 𝑑𝑡 𝑑𝑡 Case 1 Solution: Jerry Springer a) Stability Analysis

0 𝐴=� −1

1 ] 1

Eigenvalues  = 1,  =1, D = 2 - 4 = -3 𝜏 + √𝜏2 − 4𝛥 1 √3 𝜆1 = = +𝑖 2 2 2 2 𝜏 − √𝜏 − 4𝛥 1 √3 𝜆2 = = − 𝑖 2 2 2 Unstable Spiral b) Transient Behavior of Romeo and Juliet

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Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Case 2 Solution: Out of Touch (a) Stability Analysis

0 𝐽=� 𝑏

𝑎 ] 0

𝜏=0 𝜆 = −√𝑎𝑏 � � ∆= −𝑎𝑏 � → � 1 𝜆 = √𝑎𝑏 2 𝜏2 − 4∆= 4𝑎𝑏 The eigenvalues indicate 3 basic types of behavior.

Behavior 1 Behavior 2 Behavior 3

Sign of 𝑎 ∙ 𝑏 positive Negative 0

Fixed Point (0,0) (0,0) (0,J0) when a=0 (R0,0) when b=0 (R0,J0) when a=b=0

Classification of fixed point Saddle Center Non-isolated fixed points

(b) General Analytical Solution The eigenvectors come from the non-trivial solutions of (𝐴 − 𝜆𝐼)𝑣 = 0. The coefficients are determined by the initial conditions.

Eigenvalues

Eigenvectors 𝑎 −� 𝑣1 = � 𝑏� 1 𝑎 � 𝑣2 = � 𝑏� 1

𝜆1 = −√𝑎𝑏

𝜆2 = √𝑎𝑏

Coefficients 𝑐 =

√𝑎𝐽0 − √𝑏𝑅0 2√𝑎

𝑐 =

√𝑎𝐽0 + √𝑏𝑅0 2√𝑎

1

2

⎡( √𝑏𝑅0 + √𝑎𝐽0 √𝑎⋅𝑏⋅𝑡 + ( √𝑏𝑅0 − √𝑎𝐽0 ) 𝑒−√𝑎⋅𝑏⋅𝑡⎤ 𝑅(𝑡) )𝑒 ⎢ 2√𝑏 � ] = 𝑐 𝑒𝜆1𝑡𝑣 + 𝑐 𝑒𝜆2𝑡𝑣 = 2√𝑏 ⎥⎥ 1 1 2 2 𝐽(𝑡) ⎢ √𝑏𝑅 + √𝑎𝐽 √𝑏𝑅 − √𝑎𝐽 0 0 0 0 ⎢( ) 𝑒√𝑎⋅𝑏⋅𝑡 − ( ) 𝑒−√𝑎⋅𝑏⋅𝑡⎥ ⎣ 2√𝑎 2√𝑎 ⎦ The special case of Sign(𝑎 ⋅ 𝑏) = 0requires L’Hospital’s rule or a separate analysis. 𝑅(𝑡) 𝑅0 � ]=� ] , for a = 0 𝐽0 + bR0𝑡 𝐽(𝑡) 𝑅0 + aJ0𝑡 �𝑅(𝑡) ]=� ] , for b = 0 𝐽(𝑡) 𝐽0 Special Case of R0 = 1, J0 = 0 1

√𝑎⋅𝑏⋅𝑡 + 𝑒−√𝑎⋅𝑏⋅𝑡) 𝑅(𝑡) 2 (𝑒 � ]=� � 1 𝑏 𝐽(𝑡) � (𝑒√𝑎⋅𝑏⋅𝑡 − 𝑒−√𝑎⋅𝑏⋅𝑡) 2

𝑎

(c) Phase Diagrams and Solutions 10

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Behavior 1: Saddle, 𝑆𝑖𝑔𝑛(𝑎 ∙ 𝑏) > 0

a>0, b>0 is similar to a<0, b<0, except the arrows flow in the opposite direction.

Behavior 2: Center, 𝑆𝑖𝑔𝑛(𝑎 ∙ 𝑏) < 0

J

1

R, J 1

0 5

10

15

20

25

30

t

◻1 ◻1

0 R

1

◻1

a>0, b<0 is similar to a<0, b>0, with rotation in the opposite direction.

Behavior 3: Non-isolated fixed points, 𝑆𝑖𝑔𝑛(𝑎 ∙ 𝑏) = 0 R, J

J

1

0 1

◻1 ◻1

0 R

1 2

4

6

8

t

The case a = 0, b  0 is shown. The case where b = 0, a  0 is similar with lines in the phase plane parallel to R axis.

Case 3 Solution: Fire and Water Analysis of Fixed Point at (0,0)

𝑎 𝑏 𝐴=� ] −𝑏 −𝑎  = 0,  = b2 - a2, D = 2 - 4 = -4 ( b2 - a2 )

b2 - a2

sign 

sign D

+ -

+ -

+

Stability of fixed point (0,0) Center Saddle

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Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 11.1.7 Suppose we have a system of three linear, first order, ordinary differential equations. The eigenvalues and eigenvectors are 𝜆1 = 2 + 𝑖 𝜆2 = 2 − 𝑖 𝜆3 = −1

𝑉�1 = [1,0,0] 𝑉�2 = [0,1,0] 𝑉�3 = [0,0,1]

Sketch the phase portraits in the x-y plane, x-z plane, and the y-z plane.

Solution Unstable spiral in x-y plane.

Saddle in the x-z plane.

Saddle in the y-z plane.

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Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Non-Linear Systems Problem 11.2.1: Solutions from Phase Portraits The phase plot for a second order system is shown.

a) Speculate on the location and stability of the fixed points. b) Plot solutions x(t) and y(t) starting from initial conditions (𝑥0, 𝑦0) = (1,1).

Solution a) There appears to be a saddle at (-2,-2) and a unstable spiral at (2,2). b)

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Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 11.2.2 Consider the following non-linear second order systems.

System 1 𝑑𝑥 =𝑥−𝑦 𝑑𝑡 𝑑𝑦 = 𝑥2 − 4 𝑑𝑡

System 2 𝑑𝑥 = 𝑦3 − 4𝑥 𝑑𝑡 𝑑𝑦 = 𝑦3 − 𝑦 − 3𝑥 𝑑𝑡

a) Determine all the fixed points. b) Classify the stability of the fixed points. c) Sketch the phase portrait. d) Sketch the solution starting from the initial conditions 𝑥0 = 1, 𝑦0 = 0 when 𝑡 = 0.

Solution System 1 Fixed Points (𝑥1, 𝑦1) = (−2, −2) (𝑥2, 𝑦2) = (2, 2) Stability of Fixed Points The Jacobian is: 𝐴=�1 2𝑥

−1 ] 0

For (𝑥1, 𝑦1) = (−2, −2) 1 −1 𝐴(−2,−2) = � ] , 𝜏 = 1 , 𝛥 = −4  Saddle −4 0 For (𝑥2, 𝑦2) = (2, 2) 1 −1 𝐴(2,2) = � ] , 𝜏 = 1 , 𝛥 = 4 , disc = -15  Unstable Spiral 4 0 Phase Portrait

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Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Solution System 2 Fixed Points (𝑥1, 𝑦1) = (0,0) (𝑥2, 𝑦2) = (−2, −2) (𝑥3, 𝑦3) = (2, 2) Stability of Fixed Points The Jacobian is: 𝜕𝑓 𝜕𝑥 𝐴= 𝜕𝑔 ⎣𝜕𝑥

−4 0 𝐴=� ] {𝜏 = −5| −3 −1 −4 12 𝐴=� ] {𝜏 = 7| −3 11 −4 12 𝐴=� ] {𝜏 = 7| −3 11 (b) x = y If x = y, then

𝜕𝑓 −4 3𝑦2 𝜕𝑦 =� ] 𝜕𝑔 −3 3𝑦2 − 1 𝜕𝑦⎦ (𝑥1, 𝑦1) = (0,0) {𝛥 = 4|  Stable spiral (𝑥2, 𝑦2) = (−2, −2)  Saddle point (𝑥3, 𝑦3) = (2, 2)  Saddle point

dx = 𝑥3 − 4𝑥

dt

dy = 𝑥3 − 4𝑥

.

dt

The rate of change is the same for x and y, thus the trajectory stays on the line x = y. (c) 𝑑𝑥 𝑑𝑦 − = (𝑦3 − 4𝑥) − (𝑦3 − 𝑦 − 3𝑥) 𝑑𝑡 𝑑𝑡 𝑑(𝑥 − 𝑦) = −(𝑥 − 𝑦) 𝑑𝑡 Thus 𝑥 − 𝑦 = 𝑒−𝑡 Phase Portrait

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Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 11.2.3 T

g L

 Consider an undamped pendulum of mass m driven by a constant torque, T (N m). a) Derive the equation of motion for the angular position, (t). b) Change variables in order to express the equation of motion in the form 𝑑2𝜃𝜃 + 𝑠𝑖𝑛𝜃𝜃 = 𝛾𝛾 𝑑𝜏2 What are the variables  and ? c) Find all the equilibrium points and classify them as  varies. d) Is the system conservative? If so, find a conserved quantity. e) Using computer generated solutions, plot phase portraits for various values of 𝛾𝛾. f) Find the approximate frequency of small oscillations about any centers in the phase portrait.

Solution a) Equation of motion Newtons’s 2nd law of motion for a rotating system is 𝑑2𝜃𝜃 𝐼 2 = � moments dt 2 I = mass moment of inertia = L m (for a concentrated mass). Application to the pendulum gives 𝑑2𝜃𝜃 𝐿2𝑚 2 = −𝑚𝑔 ∙ sin(𝜃𝜃) + 𝑇 dt b) Variable Change Define the natural frequency and dimensionless time as 𝑔 𝜔𝑛 = � , 𝜏 = 𝜔𝑛𝑡 𝐿 Substitute 𝑑2𝜃𝜃 + sin(𝜃𝜃) = 𝛾𝛾 𝑑𝜏2 𝑇 𝑇 𝛾𝛾 = = 2 2 mL 𝜔𝑛 mgL Decompose the system into coupled first order equations. 𝑑𝜃𝜃 =𝜔 𝑑𝜏 𝑑𝜔 = −sin(𝜃𝜃) + 𝛾𝛾 𝑑𝜏 c) Analysis of Fixed Points Equilibrium requires v* = 0 16

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

sin(𝜃𝜃 ∗) = 𝛾𝛾 or arcsin() = * Note that the formal ‘arcsin’ function on most computers is usually restricted between -/2 and /2. This equilibrium condition can be visualized graphically by plotting sin() and  and noting the intersections. We see that for |𝛾𝛾| < 1 we have two fixed points, while for |𝛾𝛾| > 1 fixed points do not exist.

1.5 1 0.5 ◻

◻◻

◻

2◻

◻

-0.5 -1 -1.5

To perform a linearized stability analysis, determine the Jacobian 0 1 𝐴¯ = � ] −cos(𝜃𝜃) 0 The fixed points are classified according to ranges of the driving torque parameter. || <1 1) 1* = arcsin() trace(A) = 0 , det(A) = cos(1*) > 0 → Center 2) 2* =  - arcsin() trace(A) = 0 , det(A) = cos(1*) < 0 → Saddle point || =1 Here 1* = 2* = /2 and the fixed point becomes precarious. || >1 No fixed points can exist. The driving torque is too strong and the pendulum whirls around. d) Conservative quantity Let’s look for a conservative quantity. Multiply the equation of motion by velocity. 𝑑𝜃𝜃 𝑑2𝜃𝜃 𝑑 1 𝑑𝜃𝜃 2 � � ( ) − cos(𝜃𝜃) − 𝜃𝜃 𝛾𝛾� = 0 + sin(𝜃𝜃) − 𝛾𝛾� = 𝑑𝜏 𝑑𝜏 2 𝑑𝜏 2 𝑑𝜏 Thus a conserved quantity does exist 1 𝑑𝜃𝜃 2 ( ) − cos(𝜃𝜃) − 𝜃𝜃 𝛾𝛾 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 2 𝑑𝜏 This is a statement of conservation of energy: Kinetic energy + Potential energy – Work added = constant

17

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

e) Results Imagine the pendulum at the lowest position ( = 0) with various initial velocities. We can then plot a phase portrait for various values of the driving parameter, . v 6

v 6

v 6

4

4

4

2

2

2

◻4 ◻ ◻3 ◻ ◻2 ◻ ◻◻

◻

2 ◻ 3 ◻ 4 ◻ 5 ◻ 6◻

◻

◻4 ◻ ◻3 ◻ ◻2 ◻ ◻◻

◻

-2

-2

-4

-4

Fig 1:  = 0

2◻ 3◻ 4◻ 5◻ 6◻

Fig 2:  = 0.5

◻ ◻4 ◻ ◻3 ◻ ◻2 ◻ ◻◻

◻

2◻ 3◻ 4◻ 5◻ 6◻

-2

-4

Fig 3:  = 1

f) Small oscillation approximation Small oscillations can only occur for |  | < 1. The angle must be close to the center fixed point 1*. Define a small perturbation about 1* 𝛹 = 𝜃𝜃 − 𝜃𝜃1∗ Substitute into our equation of motion in dimensionless form 𝑑2𝛹 ∗ + sin(𝛹 + 𝜃𝜃 1 ) = 𝛾𝛾 𝑑𝜏2 𝑑2𝛹 + sin(𝛹)cos(𝜃𝜃∗) + sin(𝜃𝜃∗)cos(𝛹) = 𝛾𝛾 1 1 𝑑𝜏2 2 𝑑𝛹 2 + sin(𝛹)�1 − 𝛾𝛾 + 𝛾𝛾cos(𝛹) = 𝛾𝛾 𝑑𝜏2 Now use the small angle approximations cos(𝛹) ≃ 1 sin(𝛹) ≃ 𝛹 𝑑2𝛹 2 + 𝛹�1 − 𝛾𝛾 =0 2 𝑑𝜏this The dimensionless natural frequency of system is 𝜔𝑛 = (1 − 𝛾𝛾2)1⁄4 𝛺𝑛 = �g/L Thus, we see an approximate frequency for small oscillations about the centers emerge 𝑔 2 1⁄4 𝜔𝑛 = � (1 − 𝛾𝛾 ) 𝐿

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Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 11.2.4: Bifurcations Consider the system 𝑑𝑥 = 𝑦 − 2𝑥 𝑑𝑡 𝑑𝑦 = 𝜇 + 𝑥2 − 𝑦 𝑑𝑡 a) Determine all the fixed points. What is the critical value c at which a bifurcation occurs? b) Classify the stability of the fixed points. c) Plot a phase portrait for  < c and one for  > c.

Solution a) Fixed points occur when 𝑦∗ − 2𝑥∗ = 0 𝜇 + 𝑥∗2 − 𝑦∗ = 0 Solve to get

𝑥∗ = 1 + �1 − 𝜇, 𝑦∗ = 2𝑥∗ 1 1 1 𝑥∗ = 1 − �1 − 𝜇, 𝑦∗ = 2𝑥∗ 2

2

2

The critical value is 𝜇𝑐 = 1. No real fixed points exist for 𝜇 > 𝜇𝑐. b) Stability of fixed points 𝑑𝑥 = 𝑦 − 2𝑥 = 𝑓(𝑥, 𝑦) 𝑑𝑡 𝑑𝑦 = 𝜇 + 𝑥2 − 𝑦 = 𝑔(𝑥, 𝑦) 𝑑𝑡 𝜕𝑓 𝜕𝑓 𝜕𝑥 𝜕𝑦 = �−2 1 ] Jacobian = J = 𝜕𝑔 𝜕𝑔 2𝑥 −1 ⎣𝜕𝑥 𝜕𝑦⎦ At 𝑥∗1 = 1 + �1 − 𝜇

−2 J1 = � 2𝑥 ∗ 1

−2 1]=� −1 2(1 + �1 − 𝜇)

1 ] −1

The eigenvalues are determined from −2 − 𝜆 1 𝑑𝑒𝑡 �2(1 + ] = (−2 − 𝜆)(−1 − 𝜆) − 2(1 + �1 − 𝜇) �1 − 𝜇) −1 − 𝜆 = 3𝜆 + 𝜆2 − 2�1 − 𝜇 = 0 −1 ± �1 + 24�1 − 𝜇 𝜆= This is a saddle point for 𝜇 > 𝜇𝑐 = 1.

6

Similarly for at 𝑥∗2 = 1 − �1 − 𝜇 we find

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Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

𝜆= This is a stable spiral for 𝜇 > 𝜇𝑐 = 1.

−1 ± �1 − 24�1 − 𝜇 6

c) Phase Portraits

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Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 11.2.5: Bead on a Wire Consider a bead of mass m constrained to slide along a wire. Suppose that the motion is opposed by a viscous damping force b dx/dt.

x = coordinate along wire, measured from closest point to the spring support m = mass k = spring stiffness L0 = relaxed length a = distance between support point and wire a) Derive Newton’s law for the motion of the bead. b) Transform the equation of motion into the following dimensionless form. 𝑑2𝑋𝑋 𝑑𝑋𝑋 1 = −𝐵 ∙ − 𝑋𝑋 (1 − 2 ) 𝑑𝜏2 𝑑𝜏 √𝐴 + 𝑋𝑋2 Determine these dimensionless variables and parameters. c) Find all possible equilibrium points (fixed points). Identify any bifurcations and determine critical parameter values. d) Consider the highly damped case.

1. Under what conditions can we neglect the acceleration term (second derivative term) compared to the damping term?

2. Classify the stability of all the fixed points for this case.

3. Plot the bifurcation diagram (X* vs A). What kind of bifurcation do we have?

4. Plot phase portraits for A > 1, A = 1, and A < 1. Plot some solutions corresponding to each of these phase portraits. e) Now consider the general case.

1. Classify the stability of all the fixed points.

2. Plot phase portraits A > 1, A = 1, and A < 1. Plot some solutions corresponding to each of these phase portraits. Create a set of plots for B=0 and another set for a B=0.1. f) Now consider an inclined wire, shown on p. 73 in the text.

1. Derive Newton’s law for the motion of the bead.

2. Determine and classify all the fixed points for this case.

Solution a) Equation of Motion

𝑑2𝑥 𝑚

𝑑𝑥 𝐿0 = −𝑏 ∙ − 𝑘𝑥 (1 − ) 𝑑𝑡2 𝑑𝑡 √𝑎2 + 𝑥2 𝑑𝑥 𝑥 = 𝑥0 , = 𝑣0, 𝑡 = 0 𝑑𝑡

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Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11



Forcing Functions x0, v0 b) Dimensionless Form



System m, b, k, L0, a

Response x(t), v(t)

𝑑2𝑋𝑋

𝑑𝑋𝑋 1 − 𝑋𝑋 (1 − 2 ) 𝑑𝜏 𝑑𝜏 √𝐴 + 𝑋𝑋2 𝑥 𝑘 Variables: 𝑋𝑋 = , 𝜏 = 𝑡� 2 = −𝐵 ∙

𝐿0 𝑎

Parameters: 𝐴 =

𝐿0

, 𝐵=

𝑚 𝑏 √𝑚𝑘

Decomposition into a system of first order ODE’s 𝑑𝑋𝑋 =𝑉 𝑑𝜏 𝑑𝑉 1 ) = −𝐵 ∙ 𝑉 − 𝑋𝑋 (1 − 𝑑𝜏 √𝐴2 + 𝑋𝑋2 c) Fixed Points Real Variables (𝑥∗, 𝑣∗) = (0,0)

Dimensionless Variables (𝑋𝑋∗, 𝑉∗) = (0,0)

(𝑥∗, 𝑣∗) = (±�𝐿2 − 𝑎2, 0) , 𝑎 ≤ 𝐿 0

0

(𝑋𝑋∗, 𝑉∗) = (±�1 − 𝐴2, 0) , 𝐴 ≤ 1

A supercritical pitchfork bifurcation occurs for a = ac = L0 or in dimensionless variables when A = Ac = 1. d) Highly Damped Case

1 𝑑𝑋𝑋 ) = 𝑋𝑋 (1 − 𝑑𝜏 √𝐴2 + 𝑋𝑋2 d.1) When is the highly damped approximation valid? 𝑏 𝑏2 𝐵= ≫ 1 𝑜𝑟 𝑚 ≪ 𝑜𝑟 𝑏 ≫ √𝑚𝑘 𝑘 √𝑚𝑘 d.2) Stability of Fixed Points X= 0 is always an equilibrium position. Stability depends on the size of A. A>1: spring is in tension, X = 0 is a stable A<1: spring is in compression, X = 0 is an unstable with a pair of stable points on either side. 𝐵∙

𝑋𝑋∗ = 0 𝑋𝑋∗ = ±�1 − 𝐴2

A>1 Stable DNE

A=1 Neutrally stable Neutrally stable

A<1 Unstable Stable

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Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

d.3) Bifurcation Diagram: Supercritical Pitchfork

d.4) Highly Damped: m = 0 or B>>1

r ◻ rc

r ◻rc

r ◻ rc

dx dt

dx dt

dx dt

x

1

x

1

xt

xt

xt

1

1

1

0

0

0

0

2

4

6

8

10

t

0

2

4

6

8

10

t

x

1

0

2

4

6

8

10

t

Fig. 1: Phase portraits and solutions r = A= a/L0, rc = Ac = 1.

e) General Case e.1) Stability of the Fixed Points X* 0

A>1

A=1

A<1

B=0: Center 1 𝐵2 + 4 ( − 1) < 0: Stable spiral

Non-isolated fixed point: Solution shows a center for B=0 and a stable spiral for B>0

Saddle

Non-isolated fixed point: Solution shows a center for B=0 and a stable spiral for B>0

B=0: Center 𝐵2 − 4(1 − 𝐴2) < 0: Stable spiral 𝐵2 − 4(1 − 𝐴2) > 0: Stable node

𝐴 1

𝐵2 + 4 ( − 1) > 0: Stable node 𝐴

±�1 − 𝐴2

DNE

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Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

e.2) Phase Portraits and Solutions for A>1

Fig 2: Stable fixed point at (0,0),:B=0 (center), B=0.1 (stable spiral), B=1 (stable node)

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Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Phase Portraits and Solutions for A<1

Fig 3: Stable fixed point at (±√1 − 𝐴𝟐𝟐,0) B=0 (center), B=0.1 (stable spiral), B=1 (stable node)

Note that there are

6 different qualitative behaviors.

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Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

f) Tilted Wire Equation of Motion 𝑚 Dimensionless Form

𝑑2𝑥 𝑑𝑡

2 = −𝑏 ∙

𝑑𝑥 1 − 𝑘𝑥 (1 − 2 ) + 𝑚𝑔 ∙ 𝑠𝑖𝑛(𝜃𝜃) 𝑑𝑡 √𝑎 + 𝑥2

𝑑2𝑋𝑋 𝑑𝑋𝑋 1 = −𝐵 ∙ − 𝑋𝑋 (1 − ) + 𝛤 ∙ 𝑠𝑖𝑛(𝜃𝜃) 𝑑𝜏2 𝑑𝜏 √𝐴2 + 𝑋𝑋2 𝑥

Variables: 𝑋𝑋 = , 𝜏 = 𝑡� Parameters: 𝐴 =

𝑎

𝐿0

𝑏

, 𝐵=

𝐿0

A new dimensionless parameter has emerged: 𝛤 =

√𝑚𝑘

𝑚∙𝑔𝑔

𝑘

𝑚

, 𝛤=

𝑚∙𝑔𝑔 𝑘∙𝐿0

𝑘∙𝐿0

Fixed Points The fixed points come from the roots of 1

) + 𝛤 ∙ 𝑠𝑖𝑛 (𝜃𝜃) = 0 �𝐴2 + 𝑋𝑋∗2 Actually, the term 𝛤 ∙ 𝑠𝑖𝑛(𝜃𝜃) can be considered as a single push.

1 −𝑋𝑋∗ (1 − )+𝐹 =0 �𝐴2 + 𝑋𝑋∗2 These fixed points cannot be determined analytically but can be visualized graphically and determined numerically. For instance, the following bifurcation diagram is for F=0.1. −𝑋𝑋∗ (1 −

X 1

1

A

Bifurcation diagram is for F=0.1

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Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 11.2.6 Consider the bead of mass m constrained to slide along a wire as described in Problem 11.2.5. A constant force f is applied and the motion is opposed by a viscous damping force 𝑏 𝑑𝑥/𝑑𝑡. a) Derive Newton’s law for the motion for the bead. b) Transform the equation of motion into the following dimensionless form. 𝑑2𝑋𝑋 𝑑𝑋𝑋 1 = −𝐵 − 𝑋𝑋 (1 − )+𝐹 𝑑𝜏2 𝑑𝜏 √𝐴2 + 𝑋𝑋2 Determine these dimensionless variables and parameters. c) For F=0, 1. Find and classify all possible fixed points.

2. Plot the bifurcation diagram (X* vs A).

3. Plot phase portraits for each qualitatively different behavior. d) For any F, 1. How do you find the fixed points? Create a function to determine them.

2. Classify the fixed points.

3. Plot bifurcation diagrams (X* vs A) for F=0.01, F=0.1, F =0.5, and for F=1. Also plot a 3D bifurcation diagram showing X* as a function of A and F. 4. Plot phase portraits and some solutions for each qualitatively different behavior.

Solution Summary (a) Equation of Motion

𝑑2𝑥 𝑚

𝑑𝑡2

= −𝑏 ∙

𝑑𝑡

𝐿0 − 𝑘𝑥 (1 −



Forcing Functions x0, v0, f (b) Dimensionless Form

𝑑𝑥

𝑑2𝑋𝑋 𝑑𝜏

2 = −𝐵 ∙

)+𝑓 √𝑎2 + 𝑥2 

System m, b, k, L0, a

Response x(t), v(t)

𝑑𝑋𝑋 1 − 𝑋𝑋 (1 − 2 )+𝐹 𝑑𝜏 √𝐴 + 𝑋𝑋2

Variables: 𝑋𝑋 =

𝑥

, 𝜏 = 𝑡�

𝐿0

Parameters: 𝐴 =

𝑎 𝐿0

, 𝐵=

𝑘

𝑚 𝑏 √𝑚𝑘

, 𝐹=

𝑓 𝑘∙𝐿0

Decomposition into a system of first order ODE’s 𝑑𝑋𝑋 =𝑉 𝑑𝜏 𝑑𝑉 1 = −𝐵 ∙ 𝑉 − 𝑋𝑋 (1 − )+𝐹 𝑑𝜏 √𝐴2 + 𝑋𝑋2 c) Zero Forcing (c.1) Fixed Points: F = 0 27

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Real Variables (𝑥∗, 𝑣∗) (0,0)

Dimensionless Variables (𝑋𝑋∗, 𝑉∗) (0,0)

(±�𝐿2 − 𝑎2 , 0)

(±�1 − 𝐴2 , 0) , 𝐴 ≤ 1 𝑜𝑛𝑙𝑦

A pitchfork bifurcation occurs for a = ac = L0

A pitchfork bifurcation occurs for A = Ac = 1

0

Table 1: Stability of the Fixed Points: F = 0 (𝑋𝑋∗, 𝑉∗) = (0,0) (𝑋𝑋∗, 𝑉∗) = (±�1 − 𝐴2 , 0) Saddle

A<1

A=1

A>1

Non-isolated fixed point: Solution shows a center for B=0 and a stable spiral for B>0 B=0: Center 1

𝐵 < �4 (1 − ): Stable spiral

B=0: Center 𝐵 < �4(1 − 𝐴2): Stable spiral 𝐵 > �4(1 − 𝐴2): Stable node Non-isolated fixed point: Solution shows a center for B=0 and a stable spiral for B>0 Does not exist

𝐴

1

𝐵 > �4 (1 − ): Stable node 𝐴

(c.2) Bifurcation Diagram: Supercritical Pitchfork X 1

1

A

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Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

(c.3) Phase Portraits and Solutions: B = 0 (X = solid line, V = dashed line)

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Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Phase Portraits and Solutions: B = 0.2 (X = solid line, V = dashed line)

30

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

(d) Constant Forcing (d.1) Fixed Points The fixed points come from the roots of 𝑊𝑊(𝑋𝑋∗, 𝐴, 𝐹) = −𝑋𝑋∗ (1 −

1

)+𝐹 =0 �𝐴2 + 𝑋𝑋∗2 These fixed points cannot be determined analytically but can be visualized graphically and determined numerically using non-linear algebraic methods such as the bisection or NewtonRaphson methods.

(d.2) Classification of Fixed Points Note that a saddle bifurcation occurs when F reaches some critical value Fc. This value occurs when 𝜕𝑊𝑊 =0 𝜕𝑋𝑋 𝑊𝑊 = 0 Solving 𝛛𝖶 = 0 gives (see Mathematica code for details) 𝛛𝑋

𝑋𝑋 = −�𝐴4⁄3 − 𝐴2 Substituting into 𝑊𝑊 = 0then gives 𝐹𝑐(𝐴) = (1 − 𝐴2⁄3)3⁄2

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Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Fc 1.0

0.8

0.6

0.4

0.2

A 0.2

0.4

0.6

0.8

1.0

1.2

1.4

Table 1: Stability of the Fixed Points for F > 0 (𝑋𝑋∗, 𝑉∗) = (𝑋𝑋∗ , 0) −

A<1 0<F<Fc A<1 F>Fc A>1 All F

(𝑋𝑋∗, 𝑉∗) = (𝑋𝑋∗, 0) 0

(𝑋𝑋∗, 𝑉∗) = (𝑋𝑋∗ , 0) +

B = 0: Center B > 0: Stable node or spiral

Saddle

B = 0: Center B > 0: Stable node or spiral

DNE

DNE

B = 0: Center B > 0: Stable node or spiral

DNE

DNE

B = 0: Center B > 0: Stable node or spiral

As depicted in the following bifurcation diagrams, 𝑋𝑋∗ is the left most point 𝑋𝑋∗ is the middle −

0

∗ point and 𝑋𝑋 + is the right most point. Note that for F = 0, ∗ 2 𝑋𝑋 − = − �1 − 𝐴 𝑋𝑋 + ∗ = �1 − 𝐴2 𝑋𝑋0 ∗ = 0 The points 𝑋𝑋∗ 𝑎𝑛𝑑 𝑋𝑋∗ disappear in a saddle bifurcation as when F becomes greater that Fc.

−

0

(d.3) Bifurcation Diagrams

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Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

(d.4)

33

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 11.2.7 The Kermack-McKendrick (1927) model of an epidemic is 𝑑𝑥 = −𝑘𝑥𝑦 𝑑𝑡 𝑑𝑦 = 𝑘𝑥𝑦 − 𝑟𝑦 𝑑𝑡 x = number of healthy population y = number of sick population r = death rate constant for sick people k = infection rate constant The equation for the deaths plays no role in the model and is omitted. a) Find and classify all the fixed points b) Sketch the nullclines and vector field c) Find a conserved quantity for the system d) Plot the phase portrait. What happens as t →  e) Let (xo,yo) be the initial condition. An epidemic is said to occur if y(t) increases initially.

Under what conditions does an epidemic occur?

Solution a) Fixed Points y* = 0 for any x* 𝛥=0  Non-isolated fixed point 𝜏 = kx − 𝑙

b) Vector Field y

k◻1 , l◻1

2 1.5 1 0.5

0.5

c) Conserved Quantity

1

1.5

2

x

𝑙 𝑑𝑦 = −1 + 𝑑𝑥 kx 𝑙 𝑃𝑛(𝑥) = 𝐶 = conserved quantity 𝑦+𝑥− 𝑘

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Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

d) Phase Portrait: 𝑘 = 𝑃 = 1

As t → , y → 0 and x approaches a point between 0 and l/k. e) Epidemic Conditions 𝑥>

𝑙 𝑘

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Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 11.2.8: Predator Prey Model Considered the system 𝑑𝑥 = 𝑥(𝑥(1 − 𝑥) − 𝑦) 𝑑𝑡 𝑑𝑦 𝑃𝑟𝑒𝑑𝑎𝑡𝑜𝑟: 𝑑𝑡 = 𝑦(𝑥 − 𝑎) where x  0 is the dimensionless population of the prey, y  0 is the dimensionless population of the predator, and a 0 is a control parameter. a) Plot the nullclines in the first quadrant x, y  0. b) Find the fixed points. Classify the stability of these fixed points. c) Plot the phase portrait for a >1, and show that the predators go extinct. d) Show that a Hopf bifurcation occurs for at ac= ½. Is it subcritical or supercritical? e) Estimate the frequency of the limit cycle oscillations for a near the bifurcation. f) Plot the topologically different phase portraits for 0 < a <1. 𝑃𝑟𝑒𝑦:

Solution (a)

y 0.4 0.3 0.2 0.1

0.2

0.4

0.6

0.8

1.0

1.2

x

(b) Fixed Points and Stability Fixed Point (0,0) (1,0) (𝑎, 𝑎(1 − 𝑎))

Eigenvalues 𝜆1 = 0 𝜆2 = −𝑎 𝜆1 = −1 𝜆2 = −𝑎 + 1 1 𝜆1,2 = (𝑎(1 − 2𝑎) ± 𝑎�4𝑎2 − 3) 2

Stability Non-isolated fixed point (stable kind) 0<a<1 saddle a>1 stable node 0<a< ½ unstable spiral a=½ center ½ <a<√3/2 stable spiral a = √3/2 star stable node √3/2<a<1 a>1 does not exist

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Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

(c) Phase plot for a > 1 0.5

Wolves

0.4

0.3

0.2

0.1

0.0 0.0

0.5

1.0

1.5

Bunnies (d) Examination of the eigenvalues of the fixed point (𝑎, 𝑎(1 − 𝑎)) indicate the real part of the eigenvalues changes from positive to negative at a = ac = ½. A plot of the real and imaginary parts of the eigenvalues is shown. Graphical observations indicate a supercritical bifurcation.

Im ◻

Re ◻, Im ◻ 0.4

3 2

1 2

a Re ◻

Re - solid, Im - Dashed

(e) According to Strogatz, p. 251, 𝜔 = 𝐼𝑚(𝜆) evaluated at a = ac =1/2. Thus √2 𝜔 = 𝐼𝑚(𝜆) = = 0.353 𝑟𝑎𝑑/𝑠 4 𝜔 𝑓= = 0.056 𝐻𝑧 2𝜋

37

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

(f) Phase portraits for the different behaviors.

Wolves

a◻0.5

Wolves

a◻0.25

0.5

1

0

0.5

Bunnies

a◻0.7

a◻0.866025

1

Wolves

Bunnies

Wolves

0

0.5

1

0

0.5

Bunnies

a◻0.95

a◻1.1

1

Wolves

Bunnies

Wolves

0

0

0.5

Bunnies

1

0

0.5

1

Bunnies

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Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 11.2.9 Consider the predator-prey model.

𝑦 𝑑𝑥 ) = 𝑥 (𝑏 − 𝑥 − 𝑑𝑡 1+𝑥 𝑥 𝑑𝑦 𝑃𝑟𝑒𝑑𝑎𝑡𝑜𝑟: 𝑑𝑡 = 𝑦 ( − 𝑎𝑦) 1+𝑥 where x, y > 0 are the populations and a, b > 0 are parameters. a) Sketch the nullclines and discuss the bifurcations that occur as b varies. b) Show that a positive fixed point x* > 0, y* > 0 exists for all a, b >0. (Don’t try to find the fixed point explicitly; use a graphical argument instead). c) Show that a Hopf bifurcation occurs at the positive fixed point if 4(𝑏 − 2) 𝑎 = 𝑎𝑐 = 2 𝑏 (𝑏 + 2) and b > 2. (Hint: A necessary condition for a Hopf bifurcation to occur is 𝜏 = 0, where  is the trace of the Jacobian matrix at the fixed point. Show that 𝜏 = 0 if and only if 2x* = b – 2.

Then use the fixed point conditions to express a, in terms of x*. Finally, substitute x* = (b – 2)/2 into the expression for a, and you’re done.) d) Using a computer, check the validity of the expression in (c) and determine whether the bifurcation is subcritical or supercritical. Plot typical phase portraits above and below the Hopf bifurcation. 𝑃𝑟𝑒𝑦:

Solution Parts (a) and (b) From the nullclines, 3 fixed points are apparent: (x*,y*) = (0,0), (b,0), and the other one at the 𝑦 intersection of 𝑏 − 𝑥 − = 0 and 𝑥 − 𝑎𝑦 = 0 that has no exact algebraic expression. From 1+𝑥 1+𝑥 the nullclines only, I find it impossible to anticipate a bifurcation. After creating the vector field along with the nullclines, a bifurcation from a stable spiral (for a > ac) to an unstable spiral (a < ac) can perhaps be anticipated. I was not totally convinced until I plotted phase plots of the solution however.

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Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Part c After following the suggestions, I was able to show that 4(𝑏 − 2) 𝑎 = 𝑎𝑐 = 2 𝑏 (𝑏 + 2) Part d The phase portraits above and below ac clearly show the Hopf bifurcation. Slowly crossing ac indicates a supercritical Hopf bifurcation.

a/ac > 1: Stable spiral

a/ac < 1: Unstable spiral going to a limit cycle

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Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 11.2.10 In Problem 10.15 an improved model of a laser was introduced. 𝑑𝑛 = 𝐺𝑛𝑁 − 𝑘𝑛 𝑑𝑡 𝑑𝑁 = −𝐺𝑛𝑁 − 𝑓𝑁 + 𝑝 𝑑𝑡 N(t) = number of exited atoms n(t) = number of photons in laser field G = gain coefficient for stimulated emission k = decay rate of the photons f = decay rate for spontaneous emission p = pump strength All parameters are positive. a) Non-dimensionalize the system. Use photon parameters as reference quantities. b) Find the fixed points. Classify the fixed points using linear stability analysis. c) Plot the stability diagram for the system. What types of bifurcation occur? d) Plot all the qualitatively different phase portraits and transient responses.

Solution Summary a) Non-dimensional System 𝑑𝑥 = 𝑥𝑦 − 𝑥 𝑑𝜏 𝑑𝑦 𝐸𝑥𝑐𝑖𝑡𝑒𝑑 𝐴𝑡𝑜𝑚𝑠: = −𝑥𝑦 − 𝐹𝑦 + 𝑆 𝑑𝜏 Dependent variables: 𝑥 = 𝑛 𝐺, 𝑦 = 𝑁 𝐺 𝑘 𝑘 Independent variable: 𝜏 = 𝑡𝑘 Parameters: 𝐹 = 𝑓, 𝑆 = 𝑝 𝐺 2 𝑃ℎ𝑜𝑡𝑜𝑛𝑠:

𝑘

𝑘

b) Stability of the Fixed Points (x*, y*) (0, 𝑆/𝐹) (𝑆 − 𝐹, 1)

𝐿𝑎𝑚𝑝: 𝐹 > 𝑆 Stable node does not exist

𝐿𝑎𝑠𝑒𝑟: 𝑆 > 𝐹 𝑆2 − 4𝑆 + 4𝐹 < 0 Saddle Stable spiral

𝑆2 − 4𝑆 + 4𝐹 > 0 Saddle Stable node

This table reveals three distinct behaviors, as derived using linearized stability analysis.

A transcritical bifurcation occurs when S = F. • For S < F, the device acts as a lamp with a stable fixed point (0, 𝑆). There are no atoms in 𝐹 laser field • For S > F, the device acts as a laser with a stable fixed point (𝑆 − 𝐹, 1), which can be either a node or a spiral.

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Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

c) Stability Diagram

d) Phase Portraits and Solutions Representative solutions are shown for each of the three qualitatively different cases.

42

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Case 1: 𝑆 < 𝐹 𝑆 Stable node at (0, 𝐹)

43

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Case 2: 𝑆 > 𝐹 and 𝑆𝟐𝟐 − 𝟒𝟒𝑆 + 𝟒𝟒𝐹 < 𝟎𝟎 Stable spiral at (𝑆 − 𝐹, 1) Saddle at (0, 𝑆/𝐹)

44

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Case 3: 𝑆 > 𝐹 and 𝑆𝟐𝟐 − 𝟒𝟒𝑆 + 𝟒𝟒𝐹 < 𝟎𝟎 Stable node at (𝑆 − 𝐹, 1) Saddle at (0, 𝑆/𝐹)

45

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 11.2.11: Two-Mode Laser A two-mode laser produces two different kinds of photons with numbers n1 and n2. By analogy with the simple laser model, the rate equations are 𝑑𝑛1 = 𝐺1𝑁𝑛1 − 𝑘1𝑛1 𝑑𝑡 𝑑𝑛2 = 𝐺2𝑁𝑛2 − 𝑘2𝑛2 𝑑𝑡 𝑁(𝑡) = 𝑁0 − 𝛼1𝑛1 − 𝛼2𝑛2 is number of excited atoms n1 and n2 are the number of photons. The parameters 𝐺1, 𝐺2, 𝑘1, 𝐺𝑘2, 𝛼1, 𝛼2, and N0 are all positive. a) Find and classify all the fixed points. b) Depending on the values of the parameters, how many qualitatively different phase portraits can occur? For each of the qualitatively different phase portraits, what does the model predict about the long-term behavior?

Solution Mathematical Model 𝑑𝑛1 = 𝐺 𝑁 𝑛 − 𝑘 𝑛 = 𝐺 (𝑁 − 𝛼 𝑛 − 𝛼 𝑛 ) 𝑛 − 𝑘 𝑛 = 𝐺 𝑛 (𝑁 − 𝑘1 − 𝛼 𝑛 − 𝛼 𝑛 ) 1 1 2 2 1 1 1 1 1 0 1 1 2 2 1 1 1 1 1 0 𝐺1 𝑑𝑡 𝑑𝑛2 𝑘 = 𝐺 𝑁 𝑛 − 𝑘 𝑛 = 𝐺 (𝑁 − 𝛼 𝑛 − 𝛼 𝑛 ) 𝑛 − 𝑘 𝑛 = 𝐺 𝑛 (𝑁 − 2 − 𝛼 𝑛 − 𝛼 𝑛 ) 1 1 2 2 2 2 2 2 2 0 1 1 2 2 2 2 2 2 2 0 𝐺2 𝑑𝑡 Fixed Points It would help to first draw the nullclines. 𝑛1 = 0 𝑛2 = 0 𝑘1 𝑁0 − − 𝛼1𝑛1 − 𝛼2𝑛2 = 0 𝐺1 𝑘1 𝑁0 − − 𝛼1𝑛1 − 𝛼2𝑛2 = 0 𝐺1

Note that 𝑛 = 0 always makes 𝑑𝑛1 = 0 and 𝑛 = 0 always makes 𝑑𝑛2 = 0. Thus (0,0) is a fixed 1

𝑑𝑡

2

𝑑𝑡

point. We identify another fixed point when 𝑛1 = 0 if we also require 𝑘2 1 𝑘2 𝑁0 − − 𝛼10 − 𝛼2𝑛2 = 0 ⇒ 𝑛2 = (𝑁0 − ) 𝐺2 𝛼2 𝐺2 46

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

𝑘2

This fixed point only exists when 𝑁0 −

> 0. Similarly, another fixed point can be found at

𝐺2

1

(𝑁0 − 𝑘1 ) , 0) 𝛼1 𝐺1 𝑘1 − > 0.

Another possibility is for This fixed point exists only when 𝑁0 (

𝐺1

𝑘1

𝑁0 −

𝐺1 𝑘2

− 𝛼1𝑛1 − 𝛼2𝑛2 = 0

− 𝛼1𝑛1 − 𝛼2𝑛2 = 0 𝐺2 These lines have the same slope so they can be true simultaneously only if 𝑘1 𝑘2 = 𝐺1 𝐺2 In summary, our fixed point possibilities are 𝑁0 −

(n1*, n2* ) (0 , 0) 1 𝑘2 (0, (𝑁0 − )) 𝛼2 𝐺2 1 𝑘1 ( (𝑁0 − ) , 0) 𝛼1 𝐺1 All points along the line 𝑘1 𝑁0 − − 𝛼1𝑛1 − 𝛼2𝑛2 = 0 𝐺1 Analysis of Fixed Points The Jacobian matrix is ⎡𝐺 (𝑁 − 𝑘1 − 𝛼 𝑛 − 𝛼 𝑛 ) − 𝛼 𝐺 𝑛 1 0 1 1 2 2 1 1 1 𝐺1 𝐴=⎢ ⎢ −𝛼1𝐺2𝑛2 ⎣

Restriction 𝑁0 − 𝑁0 −

𝑘2 𝐺2

>0

𝑘1

>0 𝐺1 𝑘1 𝑘2 = 𝐺1 𝐺2

−𝛼 𝐺 𝑛

2 1 1

𝐺2 (𝑁0 −

⎤

𝑘2

⎥ ⎥

𝐺2

− 𝛼1𝑛1 − 𝛼2𝑛2) − 𝛼2𝐺2𝑛2 ⎦

Analysis of Fixed Point (n1*, n2*) = (0 , 0) ⎡𝐺 (𝑁 − 𝑘1 ) ⎤ 0 1 0 𝐺 1 ⎥ 𝐴(0,0) = ⎢ 𝑘2 ⎥ ⎢ 0 𝐺2 (𝑁0 − ) 𝐺2 ⎦ ⎣ 𝑘1 𝑘2 𝜏 = 𝐺1 (𝑁0 − ) + 𝐺2 (𝑁0 − ) 𝐺1 𝐺2 𝑘1 𝑘2 𝛥 = 𝐺1 (𝑁0 − ) 𝐺2 (𝑁0 − ) 𝐺1 𝐺2 𝑘1 𝑘2 Stability depends on the signs of (𝑁 − ) and(𝑁 − ). A table is useful for summarizing the 0

𝐺1

0

𝐺2

possibilities.

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Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

𝑘

𝑘

Sign of (𝑁 − 1) 0

Sign of (𝑁 − 2 ) 0

𝐺1

+ + -

+ + -

Sign of 

Sign of 

Stability of (0, 0)

+ ? ?

-

+ +

Unstable node Saddle point Saddle point Stable node

𝑘

𝑘

𝐺2

The only stable or attracting combination is (𝑁 − 1)and (𝑁 − 2 ). Note that the 0

0

𝐺1

𝐺2

discriminate,  – 4  > 0 for all cases. The reasoning is that we have the form  = a + b,  = a*b 2 – 4  = (a-b)2 > 0 Thus, no spirals exist.

2

1

Analysis of Fixed Point (n1*, n2*) =(0,

𝑘

(𝑁 − 2)) 0

𝛼2

𝐺2

⎡ 𝐺 (− 𝑘1 + 𝑘2 ) ⎤ 1 0 0 𝐺 𝐺 1 2 ⎥ = ⎢⎢ ⎥ ⎥⎥ ⎢ 1 𝑘 𝑘 −𝛼1𝐺2 (𝑁0 − 2 ) −𝐺2 (𝑁0 − 2 )⎥ −𝛼 𝐺 𝑛 2 2 2 𝛼2 𝐺2 𝐺2 ⎦ ⎦ ⎣ 𝑘1 𝑘2 𝑘2 𝜏 = 𝐺1 (− + ) − 𝐺2 (𝑁0 − ) 𝐺1 𝐺2 𝐺2 𝑘1 𝑘2 𝑘2 𝛥 = −𝐺1 (− + ) 𝐺2 (𝑁0 − ) 𝐺1 𝐺2 𝐺2 𝑘2 Since this fixed point only exists if (𝑁0 − ) > 0, stability depends only on the sign of ⎡𝐺 ((𝑁 − 𝑘1 ) − (𝑁 − 𝑘2 )) 0 0 𝐺2 ⎢ 1 𝐺1 𝐴=⎢ ⎢⎢ −𝛼 𝐺 1 (𝑁 − 𝑘2 ) 1 2 ⎣ 𝛼2 0 𝐺2

𝑘

𝑘

𝐺1

𝐺2

⎤ ⎥⎥

𝐺2

(− 1 + 2). 𝑘

𝑘

𝐺1

𝐺2

Sign of (− 1 + 2)

Sign of 

Sign of 

+ -

?

-

+

𝐺 𝑁 −𝑘2

Stability of (0, 2 0

𝐺2𝛼2

)

Saddle point Stable node

The only stable or attracting combination is 𝑘 𝑘 𝑘 𝑘 (𝐺 𝑁 − 𝑘 ) > 0 and (− 1 + 2) < 0 or 2 < 1 2 0

2

1

𝑘

𝐺1

𝐺2

Analysis of Fixed Point (n1*, n2*) = ( (𝑁 − 1),0) 𝛼1

⎡ −𝐺 (𝑁 − 1

𝐴=

⎢ ⎣

0

0

𝑘1 𝐺1

)

−𝛼 𝐺 𝑛

0

𝐺2

𝐺1

𝐺1

𝑘1 ⎡−𝐺1 (𝑁0 − ) 𝐺1 𝑘2 = ⎢⎢ 𝑘1 − ) − (𝑁 ⎥ − )) 0 𝐺 ((𝑁 2 0 0 𝐺2 𝐺1 ⎦ ⎣ 2 1 1

⎤

𝑘1 (𝑁0 − )⎤ 𝛼1 𝐺 𝑘1 𝑘2 1 ⎥⎥ ( − ) 𝐺 2 𝐺1 𝐺2 ⎦

−𝛼2𝐺1

1

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Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

𝑘1 𝑘2 ) + 𝐺2 ( − ) 𝐺1 𝐺1 𝐺2 𝑘1 𝑘1 𝑘2 𝛥 = −𝐺1 (𝑁0 − ) 𝐺2 ( − ) 𝐺1 𝐺1 𝐺2

𝜏 = −𝐺1 (𝑁0 −

𝑘

𝑘

𝐺1

𝐺2

𝑘1

𝑘

Stability depends on the signs of ( 1 − 2)since(𝑁 − 1)>0 for this fixed point to exist.

Sign of (

𝑘1 𝐺1

−

𝑘2 𝐺2

)

+ -

0

𝐺1

Sign of 

Sign of 

?

-

+

𝐺 𝑁 −𝑘2

Stability of (0, 2 0

𝐺2𝛼2

)

Saddle point Stable node

The only stable or attracting combination is 𝑘 (𝑁0 − 𝐺1)>0 and ( 𝑘1 − 𝑘2 )<0 or 𝑘1 < 𝑘2 𝐺1

1

𝐺2

𝐺1

𝐺2

𝑘

Analysis of Fixed Points Along Line 𝑁 − 1 − 𝛼 𝑛 − 𝛼 𝑛 = 𝟎𝟎 𝟎𝟎 𝐺1 𝑘1 𝑘2

1 1

𝟐𝟐 𝟐𝟐

This case is restricted to the case where ( − ). 𝐺1

𝐺2

⎡𝐺 (𝑁 − 𝑘1 − 𝛼 𝑛 − 𝛼 𝑛 ) − 𝛼 𝐺 𝑛 𝐴=⎢ ⎢ ⎣

1

0

𝐺1

1 1

2 2

−𝛼1𝐺2𝑛2

−𝛼 𝐺 𝑛

2 1 1

1 1 1

𝐺2 (𝑁0 −

⎤

𝑘2

⎥ ⎥

𝐺2

− 𝛼1𝑛1 − 𝛼2𝑛2) − 𝛼2𝐺2𝑛2 ⎦

−𝛼1𝐺1𝑛1 −𝛼2𝐺1𝑛1 =� ] −𝛼1𝐺2𝑛2 −𝛼2𝐺2𝑛2 𝜏 = −𝛼1𝐺1𝑛1 − 𝛼2𝐺2𝑛2 𝛥 = (𝛼1𝐺1𝑛1) (𝛼2𝐺2𝑛2) − (𝛼1𝐺2𝑛2)(𝛼2𝐺1𝑛1) = 0 Note that  <0 and  = 0. This is a line of non-isolated fixed points. They are stable or attracting.

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Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Summary We have only 4 qualitatively different phase portraits, corresponding to combinations of parameters where each of the 3 fixed points takes its turn being stable.

Case 1

sign of 𝑘 a =𝑁 − 1

sign of 𝑘 b =𝑁 − 2

𝐺1

𝐺2

0

-

0

-

Long-term Behavior of (n1, n2) (0, 0) 𝑘

( 1 (𝑁 − 1),0)

2a 2b

+ -

+

3a

+

+

(0,

+

+

0

(0,

1 𝛼2

4

+

+

1

𝐺1

(𝑁0 −

𝑘2

)) 𝛼2 𝐺2 𝑘 𝐺 ( 1 (𝑁 − 1),0) if 𝐺1 > 2 𝛼1

3b

0

𝛼1

𝐺1

𝑘1 𝑘

(𝑁 − 2)) if 𝐺1 < 0

𝐺2

𝑘1

𝑘2 𝐺2 𝑘2

(n1, n2) on the line 𝑘1 𝑁0 − − 𝛼1𝑛1 − 𝛼2𝑛2 = 0 𝐺1 𝐺 If 𝐺1 = 2 𝑘1

𝑘2

Case 4 is the most interesting since both modes of the laser can be active simultaneously, depending on the initial conditions.

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Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Case 1 𝑘 𝑘 In the following graphs, (a,b) = (𝑁 − 1 , 𝑁 − 2) 0

𝐺1

0

𝐺2

Case 2a

Case 2b

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Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Case 3a

Case 4

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Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 11.2.12: Budworms vs. Forest S = Average size of trees E = Energy reserve; a measure of health B = constant budworm population 𝑆 𝐾𝐸 𝑑𝑆 𝑇𝑟𝑒𝑒𝑠: = 𝑟𝑠𝑆 (1 − ) 𝑑𝑡 𝐾𝑠 𝐸 𝑑𝐸 𝐸 𝐵 𝐸𝑛𝑒𝑟𝑔𝑦: = 𝑟𝐸𝐸 (1 − ) − 𝑃 𝑆 𝑑𝑡 𝐾𝐸 a) Interpret the terms in the model biologically. b) Nondimensionalize the system. c) Plot the nullclines. Show fixed points. Identify the type of bifurcation that occurs. d) Plot the phase portraits.

Solution a) Biological Interpretation rs = growth rate for S rE = growth rate for E Ks = carrying capacity for S KE = carrying capacity for E P = rate parameter relating destruction of E to B/S b) Non-dimensional Formulation Step 1: List all variables and parameters along with their units.

Dependent variables: S (size) E (health) Independent variables: t (sec) Parameters: rs (1/sec) rE (1/sec) Ks (size) KE (health) P*B (health*size/sec) Step 2: Count the number of independent dimensions in the problem.

3 independent dimensions: size, health, sec Step 3: Choose independent reference quantities equal to the number of dimensions.

Let’s completely normalize the S equation.

Choose: rs, Ks, KE Step 4: Form dimensionless variables. 𝑠=

𝑆 𝐾𝑠

,𝑒 =

𝐸 𝐾𝐸

, 𝜏 = 𝑟𝑠𝑡 53

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Step 5: Substitute and identify dimensionless parameters groups which naturally emerge. ds 𝑑𝜏 𝐾𝑠𝑠 𝐾𝐸 ) 𝐾𝑠 = 𝑟𝑠𝐾𝑠𝑠 (1 − 𝐾𝑠 𝐾𝐸𝑒 𝑑𝜏 dt de 𝑑𝜏 𝐵 𝐾𝐸𝑒 ) − 𝑃 𝐾𝐸 = 𝑟𝐸𝐾𝐸𝑒 (1 − 𝐾𝐸 𝐾𝑠𝑠 𝑑𝜏 dt 𝑑𝑑𝑑 since = 𝑟 dt

𝑠

𝑠 𝑑𝑠 = 𝑠 (1 − ) 𝑑𝜏 𝑒 𝑑𝑒 𝑟𝐸 𝑃 𝐵 1 = 𝑒(1 − 𝑒) − 𝑑𝜏 𝑟𝑠 𝑟𝑠 𝐾𝐸𝐾𝑠 𝑠 Two dimensionless groups have emerged: 𝑟𝐸 𝑃 𝑅 = ,𝑏 = 𝐵 𝑟𝑠 𝑟𝑠𝐾𝐸𝐾𝑠 Our dimensionless mathematical model is: 𝑠 𝑑𝑠 = 𝑠 (1 − ) 𝑑𝜏 𝑒 𝑑𝑒 𝑏 = 𝑅 𝑒(1 − 𝑒) − 𝑑𝜏 𝑠 𝑆 𝐸 𝑠 = ,𝑒 = , 𝜏 = 𝑟𝑠𝑡 𝐾𝑠 𝐾𝐸 𝑟𝐸 𝑃 𝑅 = ,𝑏 = 𝐵 𝑟𝑠 𝑟𝑠𝐾𝐸𝐾𝑠 c) Nullclines ds 𝑠 = 𝑠 (1 − ) = 0 → 𝑠 = 𝑒 𝑑𝜏 𝑒 de 𝑏 𝑏 1 = 𝑅 𝑒(1 − 𝑒) − = 0 → 𝑠 = 𝑑𝜏 𝑠 𝑅 𝑒(1 − 𝑒) A plot of these two nullclines shows: • b/R sufficiently small: 2 fixed points • b/R sufficiently large: 0 fixed points

We see a Saddle Node bifurcation for a critical value of b/R. 54

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Determination of Critical b/R The critical value is the point at which the two nullclines touch tangentially.

Equate nullclines: 𝑠=𝑒=

𝑏

1

𝑅 𝑒(1−𝑒)

Equate tangents of nullclines:

𝑏 𝑅

𝑑 𝑏 1 ( ) 𝑑𝑒 𝑑𝑒 𝑅 𝑒(1 − 𝑒 ) 𝑏 −1 + 2𝑒 ) 2 2 1= ( 𝑅 (1 − 𝑒) 𝑒 Solve these two equations simultaneously for e and b/R. 2 𝑒cr = 𝑠cr = 3 𝑏 2 2 2 4 ( ) = ( ) (1 − ) = = 0.148 𝑅 cr 3 3 27 In real variables, the critical value of B is: 𝑃 𝐵 𝑃 𝑏 𝑟𝑠𝐾𝐸𝐾𝑠 = 𝐵 = 𝑟𝐸 𝑟𝐸𝐾𝐸𝐾𝑠 𝑅 𝑟𝑠 𝑏 𝑟𝐸𝐾𝐸𝐾𝑠 4 𝑟𝐸𝐾𝐸𝐾𝑠 = 𝐵cr = ( ) 𝑅 cr 𝑃 27 𝑃 d) Phase Portraits b/R < (b/R)cr

𝑑𝑒

 𝑒2(1 − 𝑒) =

=

b/R > (b/R)cr

55

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Problem 11.2.13 A (Dumb) dog, instead of seeking to head off the duck, swims at constant speed in a circular pond straight at the (even dumber) duck. (Very dumb) duck, deciding not to fly, makes no other attempt at escape from the dog beyond swimming (at constant speed) in a circle the radius of the pond.

Parametrically determine the path of the (previously admittedly dumb) dog. Develop a non-dimensionalization that minimizes the number of parameters necessary for characterization of this solution.

Parameters: • Pond radius, 𝑅, 𝑚 • Angular velocity of (again, very dumb) duck, 𝜔, 𝑟𝑎𝑑/𝑠 • Speed of (dumb) dog, 𝑉𝑑𝑜𝑔𝑔, 𝑚/𝑠 Solution Equation of Motion for the (very dumb) duck: 𝑅 cos(𝜔𝑡 + 𝜃𝜃0) ] 𝑥 𝑑𝑢𝑐𝑘 = � 𝑅 sin(𝜔𝑡 + 𝜃𝜃0) Equation of Motion for the (not quite as dumb) dog: 𝑥 𝑑𝑢𝑐𝑘 − 𝑥 𝑑𝑜𝑔𝑔 𝑥 𝑑𝑜𝑔𝑔 = 𝑉𝑑𝑜𝑔𝑔 ( ) � 𝑥 − 𝑥 � 𝑑𝑢 𝑘___ 𝑑𝑜𝑔𝑔 � �_ _ _𝑐_ 𝑢𝑛𝑖𝑡 𝑣𝑒𝑐𝑡𝑜𝑟 𝑝𝑜𝑖𝑛𝑡𝑖𝑛𝑔𝑔 𝑓𝑟𝑜𝑚 𝑑𝑜𝑔𝑔 𝑡𝑜 𝑑𝑢𝑐𝑘

Initial condition for dog: 𝑥 𝑑𝑜𝑔𝑔(𝑡 = 0) = 𝑥 𝑑𝑜𝑔𝑔0 Nondimensionalize Select two reference parameters (two independent dimensions, length and time) Parameter Variable Units Pond Radius 𝑅 𝑚 (Dumb) Duck Velocity 𝑉𝑑𝑢𝑐𝑘 = 𝑅𝜔 𝑚/𝑠 Develop Reference Length

𝑏 𝐿𝑐ℎ𝑎𝑟 = 𝑅𝑎𝑉𝑑𝑢𝑐𝑘 𝑚 𝑏 𝑚 = 𝑚𝑎 ( ) 𝑠

𝑎+𝑏 =1 −𝑏 = 0

Length Time {𝑎, 𝑏} = {1,0} 𝐿𝑐ℎ𝑎𝑟 = 𝑅 Develop Reference Time 𝑏 𝜏 = 𝑅𝑎𝑉𝑑𝑢𝑐𝑘 𝑚 𝑏 𝑠 = 𝑚𝑎 ( ) 𝑠

Length Time

𝑎+𝑏 =0 −𝑏 = 1 56

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

{𝑎, 𝑏} = {1, −1} 𝑅 1 𝜏= = 𝜔 Dimensionless variables: Time

𝑡+ =

Duck position

𝑡 𝜏

= 𝜔𝑡

+ 𝑥𝑑𝑢𝑐𝑘 =

Dog position

+ = 𝑥𝑑𝑜𝑔𝑔

𝑥𝑑𝑢𝑐𝑘 𝑅 𝑥 𝑑𝑜𝑔𝑔 𝑅

Non-dimensional equations of motion: Duck:

Dog:

𝑡+ 𝑅 cos (𝜔 �𝑡 = 𝜔 ] + 𝜃𝜃0) � � + 𝑡 𝑥𝑑𝑢𝑐𝑘 𝑅 sin (𝜔 �𝑡 = 𝜔 ] + 𝜃𝜃0) cos(𝑡+ + 𝜃𝜃0) + =� ] = 𝑥 𝑑𝑢𝑐𝑘 = 𝑅 sin(𝑡+ + 𝜃𝜃0) 𝑅 𝑥 𝑑𝑜𝑔𝑔 𝜔𝑅

=𝑥+ = 𝑑𝑜𝑔𝑔

𝑉𝑑𝑜𝑔𝑔

𝑥 + −𝑥+ 𝑑𝑜𝑔𝑔 ), 𝑑𝑢𝑐𝑘 𝜔𝑅 ( �𝑥 + − 𝑥 + � � 𝑑𝑢_𝑐_ 𝑘 𝑑𝑜𝑔𝑔 �

𝑥 𝑑𝑜𝑔𝑔 0 =𝑥 + 𝑥 + (𝑡 = 0) = 𝑑𝑜𝑔𝑔 𝑑𝑜𝑔𝑔0 𝑅

𝑢𝑛𝑖𝑡 𝑣𝑒𝑐𝑡𝑜𝑟 𝑝𝑜𝑖𝑛𝑡𝑖𝑛𝑔𝑔 𝑓𝑟𝑜𝑚 𝑑𝑜𝑔𝑔 𝑡𝑜 𝑑𝑢𝑐𝑘 𝑉𝑑𝑜𝑔

Thus, the only parameter to vary for the dog is

𝜔𝑅

, which, for the current purposes, will be

referenced as 𝑓𝑑𝑜𝑔𝑔.

57

Solution Manual for Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11

Solutions for various 𝑓𝑑𝑜𝑔’s: f=0.4

f=0.6 dog

dog duck

1

1

0.5

y position, y/R

0.5

y position, y/R

duck

0

0

-0.5

-0.5

-1

-1

-1

-0.5

0 x position, x/R

0.5

-1

1

-0.5

f=0.8

0 x position, x/R

0.5

f=1 dog

1

dog 1

duck

duck

0.5

y position, y/R

0.5

y position, y/R

1

0

0

-0.5

-0.5

-1

-1

-1

-0.5

0 x position, x/R

0.5

1

-1

-0.5

0 x position, x/R

0.5

1

f=1.2 dog duck

1

y position, y/R

0.5

0

-0.5

-1

-1

-0.5

0 x position, x/R

0.5

1

58