Applied Engineering Mathematics, 1st Edition By Brian Vick Contains Chapter 3-11 SOLUTION MANUAL

Chapter 3

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

b

c

d

𝑑𝑑𝑑𝑑 1 = − 𝜃𝜃 𝑑𝑑𝑑𝑑 𝜏𝜏 𝜃𝜃 = 0 , 𝑡𝑡 = 0 2 𝑑𝑑 𝑥𝑥 𝑑𝑑𝑑𝑑 𝑚𝑚 2 + 𝑐𝑐 + 𝑘𝑘 ∙ 𝑥𝑥 = 0 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑥𝑥 = 0 , = 0 , 𝑡𝑡 = 0 𝑑𝑑𝑑𝑑 𝜕𝜕 2 𝑇𝑇 𝜕𝜕 2 𝑇𝑇 + =0 𝜕𝜕𝜕𝜕 2 𝜕𝜕𝜕𝜕 2 𝑇𝑇 = 0 on all boundaries 𝜕𝜕𝜕𝜕 𝜕𝜕 2 𝑇𝑇 = 𝛼𝛼 2 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 = 0 , 𝑥𝑥 = 0 𝜕𝜕𝜕𝜕 𝑇𝑇 = 0 , 𝑥𝑥 = 𝐿𝐿 𝑇𝑇 = 0 , 𝑡𝑡 = 0

Solution All these cases have zero forcing functions, thus they all remain in equilibrium. a) 𝜃𝜃 = 0 b) 𝑥𝑥 = 0 c) 𝑇𝑇 = 0 d) 𝑇𝑇 = 0

1

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? a b

c

d

e

f

Solution a b c d e f

𝑑𝑑𝑑𝑑 1 = − 𝜃𝜃 + 𝑆𝑆 𝑑𝑑𝑑𝑑 𝜏𝜏 𝑑𝑑𝑑𝑑 1 = − 𝜃𝜃 + 𝑆𝑆 𝑑𝑑𝑑𝑑 𝜏𝜏 𝜃𝜃 = 𝜃𝜃0 , 𝑡𝑡 = 0 𝜃𝜃 = 𝜃𝜃1 , 𝑡𝑡 = 𝑡𝑡1 2 𝑑𝑑 𝑥𝑥 𝑑𝑑𝑑𝑑 𝑚𝑚 2 + 𝑐𝑐 + 𝑘𝑘 ∙ 𝑥𝑥 = 𝑓𝑓(𝑡𝑡) 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑥𝑥 = 𝑥𝑥0 , 𝑡𝑡 = 0 2 𝑑𝑑 𝑥𝑥 𝑑𝑑𝑑𝑑 𝑚𝑚 2 + 𝑐𝑐 + 𝑘𝑘 ∙ 𝑥𝑥 = 𝑓𝑓(𝑡𝑡) 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑥𝑥 = 𝑥𝑥0 , 𝑡𝑡 = 0 𝑥𝑥 = 𝑥𝑥1 , 𝑡𝑡 = 𝑡𝑡1 2 𝜕𝜕𝜕𝜕 𝜕𝜕 𝑇𝑇 𝑔𝑔(𝑥𝑥, 𝑡𝑡) = 𝛼𝛼 2 + , 0 < 𝑥𝑥 < 𝐿𝐿 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜌𝜌𝜌𝜌 𝜕𝜕𝜕𝜕 −𝑘𝑘 = 𝑞𝑞0" (𝑡𝑡) , 𝑥𝑥 = 0 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕 2 𝑇𝑇 𝑔𝑔(𝑥𝑥, 𝑡𝑡) = 𝛼𝛼 2 + , 0 < 𝑥𝑥 < 𝐿𝐿 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜌𝜌𝜌𝜌 𝜕𝜕𝜕𝜕 −𝑘𝑘 = 𝑞𝑞0" (𝑡𝑡) , 𝑥𝑥 = 0 𝜕𝜕𝜕𝜕 𝑇𝑇 = 𝑇𝑇0 (𝑡𝑡) , 𝑥𝑥 = 0 𝑇𝑇 = 𝑇𝑇0 , 𝑡𝑡 = 0

Requires an initial condition 𝜃𝜃 = 𝜃𝜃0 , 𝑡𝑡 = 0 Too many initial conditions. Remove 𝜃𝜃 = 𝜃𝜃1 , 𝑡𝑡 = 𝑡𝑡1 𝑑𝑑𝑑𝑑

Requires a second initial condition 𝑑𝑑𝑑𝑑 = 𝑣𝑣0 , 𝑡𝑡 = 0 Remove 𝑥𝑥 = 𝑥𝑥1 , 𝑡𝑡 = 𝑡𝑡1

𝑑𝑑𝑑𝑑

Requires a second initial condition 𝑑𝑑𝑑𝑑 = 𝑣𝑣0 , 𝑡𝑡 = 0

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.

2

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.

x2

x1 b1

b2 m1

f1

k1

m2

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 𝑚𝑚1 𝑥𝑥̈ 1 = � 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓

𝑚𝑚1 𝑥𝑥̈ 1 = −𝑏𝑏1 𝑥𝑥̇ 1 − 𝑏𝑏2 (𝑥𝑥̇ 1 − 𝑥𝑥̇ 2 ) − 𝑘𝑘1 𝑥𝑥1 − 𝑘𝑘2 (𝑥𝑥1 − 𝑥𝑥2 ) + 𝑓𝑓1 𝑚𝑚1 𝑥𝑥̈ 1 + 𝑏𝑏1 𝑥𝑥̇ 1 + 𝑏𝑏2 (𝑥𝑥̇ 1 − 𝑥𝑥̇ 2 ) + 𝑘𝑘1 𝑥𝑥1 + 𝑘𝑘2 (𝑥𝑥1 − 𝑥𝑥2 ) = 𝑓𝑓1 Equation of Motion for Mass-2 𝑚𝑚2 𝑥𝑥̈ 2 = � 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓

Initial conditions at t=0

𝑚𝑚2 𝑥𝑥̈ 2 = −𝑏𝑏2 (𝑥𝑥̇ 2 − 𝑥𝑥̇ 1 ) − 𝑘𝑘2 (𝑥𝑥2 − 𝑥𝑥1 ) + 𝑓𝑓2 𝑚𝑚2 𝑥𝑥̈ 2 + 𝑏𝑏2 (𝑥𝑥̇ 2 − 𝑥𝑥̇ 1 ) + 𝑘𝑘2 (𝑥𝑥2 − 𝑥𝑥1 ) = 𝑓𝑓2 𝑥𝑥1 = 𝑥𝑥1,0 𝑥𝑥2 = 𝑥𝑥2,0 𝑥𝑥̇ 1 = 𝑣𝑣1,0 𝑥𝑥̇ 2 = 𝑣𝑣2,0

3

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). x2

x1

f1

The initial conditions are

xn

fn

f2

•

𝑥𝑥1,0 𝑣𝑣1,0 𝑥𝑥1 (0) ⎡ 𝑥𝑥• 1 (0) ⎤ 𝑥𝑥2,0 𝑣𝑣2,0 𝑥𝑥 (0) ⎢ ⎥ � 2 � = � ⋮ � and ⎢ 𝑥𝑥2 (0) ⎥ = � ⋮ � ⋮ ⎢• ⋮ ⎥ 𝑥𝑥𝑛𝑛,0 𝑣𝑣𝑛𝑛,0 𝑥𝑥𝑛𝑛 (0) ⎣𝑥𝑥𝑛𝑛 (0)⎦ Each mass has an individual external force applied, 𝑓𝑓𝑖𝑖 (𝑡𝑡). 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: 𝑚𝑚𝑥𝑥̈ 1 + 𝑏𝑏𝑥𝑥̇ 1 + 𝑏𝑏(𝑥𝑥̇ 1 − 𝑥𝑥̇ 2 ) + 𝑘𝑘𝑥𝑥1 + 𝑘𝑘(𝑥𝑥1 − 𝑥𝑥2 ) = 𝑓𝑓1

Mass-i, i=2,…n-1 𝑚𝑚𝑥𝑥̈ 𝑖𝑖 + 𝑏𝑏(𝑥𝑥̇ 𝑖𝑖 − 𝑥𝑥̇ 𝑖𝑖 −1 ) + 𝑏𝑏(𝑥𝑥̇ 𝑖𝑖 − 𝑥𝑥̇ 𝑖𝑖+1 ) + 𝑘𝑘(𝑥𝑥𝑖𝑖 − 𝑥𝑥𝑖𝑖−1 ) + 𝑘𝑘(𝑥𝑥𝑖𝑖 − 𝑥𝑥𝑖𝑖+1 ) = 𝑓𝑓𝑖𝑖 Mass-n: 𝑚𝑚𝑥𝑥̈ 𝑛𝑛 + 𝑏𝑏𝑥𝑥̇ 𝑛𝑛 + 𝑏𝑏(𝑥𝑥̇ 𝑛𝑛 − 𝑥𝑥̇ 𝑛𝑛−1 ) + 𝑘𝑘𝑥𝑥𝑛𝑛 + 𝑘𝑘(𝑥𝑥𝑛𝑛 − 𝑥𝑥𝑛𝑛−1 ) = 𝑓𝑓𝑛𝑛

4

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), ρ = 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 sinθ can be replaced with tanθ, where θ 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. 𝜕𝜕 2 𝑢𝑢 𝑚𝑚𝑚𝑚 = 𝜌𝜌𝐴𝐴𝑐𝑐 ∆𝑥𝑥 2 = 𝐹𝐹𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 + 𝐹𝐹𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 + 𝐹𝐹𝑎𝑎𝑎𝑎𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 𝜕𝜕𝜕𝜕 Force due to Tension 𝐹𝐹𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 = −𝑇𝑇 ∙ 𝑠𝑠𝑠𝑠𝑠𝑠�𝜃𝜃(𝑥𝑥)� + 𝑇𝑇 ∙ 𝑠𝑠𝑠𝑠𝑠𝑠�𝜃𝜃(𝑥𝑥 + ∆𝑥𝑥)� Use the small angle approximation 5

𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕 2 𝑢𝑢 𝜕𝜕𝜕𝜕 𝜕𝜕 𝜕𝜕𝜕𝜕 𝑠𝑠𝑠𝑠𝑠𝑠�𝜃𝜃(𝑥𝑥 + ∆𝑥𝑥)� ≅ 𝑡𝑡𝑡𝑡𝑡𝑡�𝜃𝜃(𝑥𝑥 + ∆𝑥𝑥)� = + � � ∆𝑥𝑥 = + ∆𝑥𝑥 𝜕𝜕𝑥𝑥 𝜕𝜕𝜕𝜕 2 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 The net force due to tension becomes 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕 2 𝑢𝑢 𝜕𝜕 2 𝑢𝑢 𝐹𝐹𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 = −𝑇𝑇 + 𝑇𝑇 � + 2 ∆𝑥𝑥� = 𝑇𝑇 2 ∆𝑥𝑥 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 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 𝜕𝜕𝜕𝜕 𝐹𝐹𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 = −𝑏𝑏 ∙ ∆𝑥𝑥 𝜕𝜕𝜕𝜕 where b is the damping coefficient per length. 𝑠𝑠𝑠𝑠𝑠𝑠�𝜃𝜃(𝑥𝑥)� ≅ 𝑡𝑡𝑡𝑡𝑡𝑡�𝜃𝜃(𝑥𝑥)� =

Applied Force The applied force is modeled in terms of a force per unit length along the string, thus 𝐹𝐹𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎 = 𝐹𝐹(𝑥𝑥, 𝑡𝑡) ∙ ∆𝑥𝑥 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 ∆𝑥𝑥. 𝜕𝜕 2 𝑢𝑢 𝜕𝜕𝑢𝑢 𝜕𝜕 2 𝑢𝑢 𝜌𝜌𝐴𝐴𝑐𝑐 2 = −𝑏𝑏 + 𝑇𝑇 2 + 𝐹𝐹 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 Divide by 𝜌𝜌𝐴𝐴𝑐𝑐 to get the standard form of the wave equation. 𝜕𝜕 2 𝑢𝑢 𝑏𝑏 𝜕𝜕𝜕𝜕 𝜕𝜕 2 𝑢𝑢 𝐹𝐹(𝑥𝑥, 𝑡𝑡) 2 = − + 𝐶𝐶 + 𝜕𝜕𝜕𝜕 2 𝜌𝜌𝐴𝐴𝑐𝑐 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 2 𝜌𝜌𝐴𝐴𝑐𝑐 b) Mathematical Model Equation of Motion (zero damping) Boundary Conditions Initial Conditions

c) Zero Forcing Functions

𝑚𝑚 𝑇𝑇 𝐶𝐶 = 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 � � = � 𝑠𝑠 𝜌𝜌𝐴𝐴𝑐𝑐

𝜕𝜕 2 𝑢𝑢 𝜕𝜕 2 𝑢𝑢 𝐹𝐹(𝑥𝑥, 𝑡𝑡) 𝑇𝑇 2 = 𝐶𝐶 + , 𝐶𝐶 = � 2 2 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜌𝜌𝐴𝐴𝑐𝑐 𝜌𝜌𝐴𝐴𝑐𝑐 𝑢𝑢 = 0 , 𝑥𝑥 = 0 𝑢𝑢 = 0 , 𝑥𝑥 = 𝐿𝐿 𝑢𝑢 = 𝑢𝑢0 (𝑥𝑥) 𝜕𝜕𝜕𝜕 � 𝑡𝑡 = 0 = 𝑣𝑣0 (𝑥𝑥) 𝜕𝜕𝜕𝜕

𝑢𝑢(𝑥𝑥, 𝑡𝑡) = 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).

6

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

y-direction BC’s

Initial Condition

𝜌𝜌𝜌𝜌

𝜕𝜕𝜕𝜕 𝜕𝜕 2 𝑇𝑇 𝜕𝜕 2 𝑇𝑇 = 𝑘𝑘 � 2 + 2 � + 𝑔𝑔(𝑥𝑥, 𝑦𝑦, 𝑡𝑡) 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 or 2 𝜕𝜕𝜕𝜕 𝜕𝜕 𝑇𝑇 𝜕𝜕 2 𝑇𝑇 𝑔𝑔 = 𝛼𝛼 � 2 + 2 � + 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜌𝜌𝜌𝜌 𝜕𝜕𝜕𝜕 −𝑘𝑘 = 𝑞𝑞𝑠𝑠" (𝑦𝑦, 𝑡𝑡) , 𝑥𝑥 = 0 𝜕𝜕𝜕𝜕

𝜕𝜕𝜕𝜕 = ℎ(𝑇𝑇∞ − 𝑇𝑇) , 𝑥𝑥 = 𝐿𝐿𝑥𝑥 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 =0 , 𝑦𝑦 = 0 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝑘𝑘 = ℎ(𝑇𝑇∞ − 𝑇𝑇) , 𝑦𝑦 = 𝐿𝐿𝑦𝑦 𝜕𝜕𝜕𝜕 𝑇𝑇 = 𝑇𝑇0 , 𝑡𝑡 = 0 𝑘𝑘

7

(b) 1D Model: y-Direction Only

Energy Equation

𝜕𝜕𝜕𝜕 𝜕𝜕 2 𝑇𝑇 𝑞𝑞𝑠𝑠" ℎ 𝜌𝜌𝜌𝜌 = 𝑘𝑘 2 + 𝑔𝑔 + − (𝑇𝑇 − 𝑇𝑇∞ ) 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝐿𝐿𝑥𝑥 𝐿𝐿𝑥𝑥 2

y-direction BC’s

Initial Condition

or

𝜕𝜕𝜕𝜕 𝜕𝜕 𝑇𝑇 𝑔𝑔 1 " 1 = 𝛼𝛼 2 + + 𝑞𝑞𝑠𝑠 (𝑦𝑦, 𝑡𝑡) − ℎ(𝑇𝑇 − 𝑇𝑇∞ ) 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜌𝜌𝜌𝜌 𝜌𝜌𝜌𝜌𝐿𝐿𝑥𝑥 𝜌𝜌𝜌𝜌𝐿𝐿𝑥𝑥 𝜕𝜕𝜕𝜕 =0 , 𝑦𝑦 = 0 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝑘𝑘 = ℎ(𝑇𝑇∞ − 𝑇𝑇) , 𝑦𝑦 = 𝐿𝐿𝑦𝑦 𝜕𝜕𝜕𝜕 𝑇𝑇 = 𝑇𝑇0 , 𝑡𝑡 = 0

(c) 1D Model: x-Direction Only

𝜕𝜕𝜕𝜕 𝜕𝜕 2 𝑇𝑇 ℎ = 𝑘𝑘 2 + 𝑔𝑔 − (𝑇𝑇 − 𝑇𝑇∞ ) 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝐿𝐿𝑦𝑦 or 2 𝜕𝜕𝜕𝜕 𝜕𝜕 𝑇𝑇 𝑔𝑔 1 = 𝛼𝛼 2 + − ℎ(𝑇𝑇 − 𝑇𝑇∞ ) 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜌𝜌𝜌𝜌 𝜌𝜌𝜌𝜌𝐿𝐿𝑦𝑦 𝜕𝜕𝜕𝜕 −𝑘𝑘 = 𝑞𝑞𝑠𝑠" (𝑡𝑡) , 𝑥𝑥 = 0 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝑘𝑘 = ℎ(𝑇𝑇∞ − 𝑇𝑇) , 𝑥𝑥 = 𝐿𝐿𝑥𝑥 𝜕𝜕𝜕𝜕 𝑇𝑇 = 𝑇𝑇0 , 𝑡𝑡 = 0 𝜌𝜌𝜌𝜌

Energy Equation

x-direction BC’s

Initial Condition

(d) 0-D Model (Lumped Capacity Approximation)

Energy Equation

Initial Condition

𝐿𝐿𝑥𝑥 + 𝐿𝐿𝑦𝑦 𝑑𝑑𝑑𝑑 𝑞𝑞𝑠𝑠" (𝑇𝑇 − 𝑇𝑇∞ ) + = 𝑔𝑔 − ℎ 𝑑𝑑𝑑𝑑 𝐿𝐿𝑥𝑥 𝐿𝐿𝑦𝑦 𝐿𝐿𝑥𝑥 or 𝐿𝐿𝑥𝑥 +𝐿𝐿𝑦𝑦 𝑑𝑑𝑑𝑑 𝑔𝑔 𝑞𝑞𝑠𝑠" (𝑇𝑇 − 𝑇𝑇∞ ) + = −ℎ 𝑑𝑑𝑑𝑑 𝜌𝜌𝜌𝜌 𝜌𝜌𝜌𝜌𝜌𝜌𝑥𝑥 𝐿𝐿𝑦𝑦 𝜌𝜌𝜌𝜌𝜌𝜌𝑥𝑥 𝑇𝑇 = 𝑇𝑇0 , 𝑡𝑡 = 0

𝜌𝜌𝜌𝜌

A steady state temperature is only possible if g and qs " are independent of time. Setting 𝜕𝜕𝑇𝑇/𝜕𝜕𝜕𝜕 = 0 gives 𝐿𝐿𝑥𝑥 𝐿𝐿𝑦𝑦 𝑞𝑞𝑠𝑠" 𝑇𝑇 − 𝑇𝑇∞ = �𝑔𝑔 + � 𝐿𝐿𝑥𝑥 �𝐿𝐿𝑥𝑥 + 𝐿𝐿𝑦𝑦 �ℎ 8

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. 𝑑𝑑𝐸𝐸𝑠𝑠𝑠𝑠 = 𝐸𝐸̇𝑖𝑖𝑖𝑖 − 𝐸𝐸̇𝑜𝑜𝑜𝑜𝑜𝑜 + 𝐸𝐸̇𝑔𝑔 𝑑𝑑𝑑𝑑 𝜕𝜕𝑞𝑞𝑦𝑦 𝜕𝜕𝜕𝜕 𝜕𝜕𝑞𝑞𝑥𝑥 𝜌𝜌𝜌𝜌 ∆𝑥𝑥∆𝑦𝑦 = 𝑞𝑞𝑥𝑥 − �𝑞𝑞𝑥𝑥 + ∆𝑥𝑥� + 𝑞𝑞𝑦𝑦 − �𝑞𝑞𝑦𝑦 + ∆𝑦𝑦� + 𝑔𝑔∆𝑥𝑥∆𝑦𝑦 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 Use Fourier’s Law 𝜕𝜕𝜕𝜕 𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕 𝜕𝜕𝜕𝜕 𝜌𝜌𝜌𝜌 ∆𝑥𝑥∆𝑦𝑦 = − �−𝑘𝑘 ∙ ∆𝑦𝑦 � ∆𝑥𝑥 − �−𝑘𝑘 ∙ ∆𝑥𝑥 � ∆𝑦𝑦 + 𝑔𝑔∆𝑥𝑥∆𝑦𝑦 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 Divide by ∆𝑥𝑥∆𝑦𝑦 𝜕𝜕𝜕𝜕 𝜕𝜕 2 𝑇𝑇 𝜕𝜕 2 𝑇𝑇 𝜌𝜌𝜌𝜌 = 𝑘𝑘 � 2 + 2 � + 𝑔𝑔(𝑥𝑥, 𝑦𝑦, 𝑡𝑡) 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 Divide by 𝜌𝜌𝜌𝜌 𝜕𝜕 2 𝑇𝑇 𝜕𝜕 2 𝑇𝑇 𝑔𝑔 𝜕𝜕𝜕𝜕 = 𝛼𝛼 � 2 + 2 � + 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜌𝜌𝜌𝜌 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. 𝑑𝑑𝐸𝐸𝑠𝑠𝑠𝑠 = 𝐸𝐸̇𝑖𝑖𝑖𝑖 − 𝐸𝐸̇𝑜𝑜𝑜𝑜𝑜𝑜 + 𝐸𝐸̇𝑔𝑔 𝑑𝑑𝑑𝑑 𝜕𝜕𝑞𝑞𝑦𝑦 𝜕𝜕𝜕𝜕 𝜌𝜌𝜌𝜌 𝐿𝐿𝑥𝑥 ∆𝑦𝑦 = 𝑞𝑞𝑦𝑦 − �𝑞𝑞𝑦𝑦 + ∆𝑦𝑦� + 𝑔𝑔𝐿𝐿𝑥𝑥 ∆𝑦𝑦 + 𝑞𝑞𝑠𝑠" ∆𝑦𝑦 − ℎ∆𝑦𝑦(𝑇𝑇 − 𝑇𝑇∞ ) 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 Use Fourier’s Law 9

𝜕𝜕𝜕𝜕 𝜕𝜕 𝜕𝜕𝜕𝜕 𝐿𝐿𝑥𝑥 ∆𝑦𝑦 = − �−𝑘𝑘 ∙ 𝐿𝐿𝑥𝑥 � ∆𝑦𝑦 + 𝑔𝑔𝐿𝐿𝑥𝑥 ∆𝑦𝑦 + 𝑞𝑞𝑠𝑠" ∆𝑦𝑦 − ℎ∆𝑦𝑦(𝑇𝑇 − 𝑇𝑇∞ ) 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 Divide by 𝐿𝐿𝑥𝑥 ∆𝑦𝑦 𝜕𝜕 2 𝑇𝑇 𝑞𝑞𝑠𝑠" ℎ 𝜕𝜕𝜕𝜕 = 𝑘𝑘 2 + 𝑔𝑔 + − (𝑇𝑇 − 𝑇𝑇∞ ) 𝜌𝜌𝜌𝜌 𝜕𝜕𝜕𝜕 𝐿𝐿𝑥𝑥 𝐿𝐿𝑥𝑥 𝜕𝜕𝜕𝜕 Divide by 𝜌𝜌𝜌𝜌 𝜕𝜕𝜕𝜕 𝜕𝜕 2 𝑇𝑇 𝑔𝑔 1 " 1 = 𝛼𝛼 2 + + 𝑞𝑞𝑠𝑠 (𝑦𝑦, 𝑡𝑡) − ℎ(𝑇𝑇 − 𝑇𝑇∞ ) 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜌𝜌𝜌𝜌 𝜌𝜌𝜌𝜌𝐿𝐿𝑥𝑥 𝜌𝜌𝜌𝜌𝐿𝐿𝑥𝑥 Here g must represent the average value of g(x,y,t) over the entire x-direction. 𝜌𝜌𝜌𝜌

𝐿𝐿𝑥𝑥

1 𝑔𝑔 = � 𝑔𝑔(𝑥𝑥, 𝑦𝑦, 𝑡𝑡)𝑑𝑑𝑑𝑑 𝐿𝐿𝑥𝑥 𝑥𝑥=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. 𝑑𝑑𝐸𝐸𝑠𝑠𝑠𝑠 = 𝐸𝐸̇𝑖𝑖𝑖𝑖 − 𝐸𝐸̇𝑜𝑜𝑜𝑜𝑜𝑜 + 𝐸𝐸̇𝑔𝑔 𝑑𝑑𝑑𝑑 𝜕𝜕𝜕𝜕 𝜕𝜕𝑞𝑞𝑥𝑥 𝜌𝜌𝜌𝜌 ∆𝑥𝑥𝐿𝐿𝑦𝑦 = 𝑞𝑞𝑥𝑥 − �𝑞𝑞𝑥𝑥 + ∆𝑥𝑥� + 𝑔𝑔∆𝑥𝑥𝐿𝐿𝑦𝑦 − ℎ∆𝑥𝑥(𝑇𝑇 − 𝑇𝑇∞ ) 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 Use Fourier’s Law 𝜕𝜕𝜕𝜕 𝜕𝜕 𝜕𝜕𝜕𝜕 𝜌𝜌𝜌𝜌 ∆𝑥𝑥𝐿𝐿𝑦𝑦 = − �−𝑘𝑘 ∙ ∆𝑦𝑦 � ∆𝑥𝑥 + 𝑔𝑔∆𝑥𝑥𝐿𝐿𝑦𝑦 − ℎ∆𝑥𝑥(𝑇𝑇 − 𝑇𝑇∞ ) 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 Divide by ∆𝑥𝑥𝐿𝐿𝑦𝑦 𝜕𝜕𝜕𝜕 𝜕𝜕 2 𝑇𝑇 ℎ 𝜌𝜌𝜌𝜌 = 𝑘𝑘 2 + 𝑔𝑔 − (𝑇𝑇 − 𝑇𝑇∞ ) 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝐿𝐿𝑦𝑦 Divide by 𝜌𝜌𝜌𝜌 𝜕𝜕𝜕𝜕 𝜕𝜕 2 𝑇𝑇 𝑔𝑔 1 = 𝛼𝛼 2 + − ℎ(𝑇𝑇 − 𝑇𝑇∞ ) 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜌𝜌𝜌𝜌 𝜌𝜌𝜌𝜌𝐿𝐿𝑦𝑦 10

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

1 𝑔𝑔 = � 𝑔𝑔(𝑥𝑥, 𝑦𝑦, 𝑡𝑡)𝑑𝑑𝑑𝑑 𝐿𝐿𝑦𝑦 𝑦𝑦=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. 𝑑𝑑𝐸𝐸𝑠𝑠𝑠𝑠 = 𝐸𝐸̇𝑖𝑖𝑖𝑖 − 𝐸𝐸̇𝑜𝑜𝑜𝑜𝑜𝑜 + 𝐸𝐸̇𝑔𝑔 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝜌𝜌𝜌𝜌 𝐿𝐿 𝐿𝐿 = 𝑔𝑔𝐿𝐿𝑥𝑥 𝐿𝐿𝑦𝑦 − ℎ𝐿𝐿𝑦𝑦 (𝑇𝑇 − 𝑇𝑇∞ ) − ℎ𝐿𝐿𝑥𝑥 (𝑇𝑇 − 𝑇𝑇∞ ) + 𝑞𝑞𝑠𝑠" 𝐿𝐿𝑦𝑦 𝑑𝑑𝑑𝑑 𝑥𝑥 𝑦𝑦 Divide by 𝐿𝐿𝑥𝑥 𝐿𝐿𝑦𝑦 𝐿𝐿𝑥𝑥 + 𝐿𝐿𝑦𝑦 𝑑𝑑𝑑𝑑 𝑞𝑞𝑠𝑠" (𝑇𝑇 − 𝑇𝑇∞ ) + 𝜌𝜌𝜌𝜌 = 𝑔𝑔 − ℎ 𝑑𝑑𝑑𝑑 𝐿𝐿𝑥𝑥 𝐿𝐿𝑦𝑦 𝐿𝐿𝑥𝑥 Divide by 𝜌𝜌𝜌𝜌 𝐿𝐿𝑥𝑥 +𝐿𝐿𝑦𝑦 𝑑𝑑𝑑𝑑 𝑔𝑔 𝑞𝑞𝑠𝑠" (𝑇𝑇 − 𝑇𝑇∞ ) + = −ℎ 𝑑𝑑𝑑𝑑 𝜌𝜌𝜌𝜌 𝜌𝜌𝜌𝜌𝜌𝜌𝑥𝑥 𝐿𝐿𝑦𝑦 𝜌𝜌𝜌𝜌𝜌𝜌𝑥𝑥 Here g(t) must represent the average value of g(x,y,t) over the entire domain

g (t ) =

Ly

1 Lx g ( x, y, t ) ⋅ dx ⋅ dy .

L x ⋅ L y 0∫ 0∫

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

11

Chapter 4

Calculus Problem 4.1 Determine and plot the derivative of the step function, 0 , 𝑡𝑡 < 0 𝐻𝐻(𝑡𝑡) = � 1 , 𝑡𝑡 ≥ 0 Solution 𝑑𝑑𝑑𝑑(𝑡𝑡) = 𝛿𝛿(𝑡𝑡) = 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 𝑑𝑑𝑑𝑑

x

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

0 , 𝑡𝑡 < 0 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅(𝑡𝑡) = 𝑡𝑡 ∙ 𝐻𝐻(𝑡𝑡) = � 𝑡𝑡 , 𝑡𝑡 ≥ 0 Solution 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑(𝑡𝑡) = 𝐻𝐻(𝑡𝑡) = 𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 𝑑𝑑𝑑𝑑

1

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

Solution

0 , 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃(𝑡𝑡) = 𝐻𝐻(𝑡𝑡) − 𝐻𝐻(𝑡𝑡 − ∆𝑡𝑡) = �1 , 0 ,

𝑡𝑡 < 0 0 ≤ 𝑡𝑡 ≤ ∆𝑡𝑡 𝑡𝑡 > ∆𝑡𝑡

𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑(𝑡𝑡) = 𝛿𝛿(𝑡𝑡) − 𝛿𝛿(𝑡𝑡 − ∆𝑡𝑡) 𝑑𝑑𝑑𝑑 t

x

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

0 , 𝑡𝑡 < −1 𝑡𝑡 + 1 , −1 ≤ 𝑡𝑡 < 0 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑙𝑙𝑒𝑒(𝑡𝑡) = � 1 − 𝑡𝑡 , 0 ≤ 𝑡𝑡 < 1 , 𝑡𝑡 ≥ 1 0 Solution 0 , 𝑡𝑡 < −1 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑(𝑡𝑡) 1 , −1 ≤ 𝑡𝑡 < 0 =� −1 , 0 ≤ 𝑡𝑡 < 1 𝑑𝑑𝑑𝑑 𝑡𝑡 ≥ 1 0 , = �𝐻𝐻(𝑡𝑡 + 1) − 𝐻𝐻(𝑡𝑡)� − �𝐻𝐻(𝑡𝑡) − 𝐻𝐻(𝑡𝑡 − 1)�

2

Problem 4.5 Consider the function 𝑓𝑓(𝑡𝑡) shown.

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

3 a) 𝐼𝐼(𝑡𝑡) = ∫𝑡𝑡 =0 𝑓𝑓(𝑡𝑡𝑜𝑜 )𝑑𝑑𝑡𝑡𝑜𝑜 b) c)

𝑜𝑜

8 𝐼𝐼(𝑡𝑡) = ∫𝑡𝑡 =0 𝑓𝑓(𝑡𝑡𝑜𝑜 )𝑑𝑑𝑡𝑡𝑜𝑜 𝑜𝑜 𝑡𝑡 𝐼𝐼(𝑡𝑡) = ∫𝑡𝑡 =0 𝑓𝑓(𝑡𝑡𝑜𝑜 )𝑑𝑑𝑡𝑡𝑜𝑜 𝑜𝑜

Solution 3 3 (a) 𝐼𝐼(𝑡𝑡) = ∫𝑡𝑡 =0 𝑓𝑓(𝑡𝑡𝑜𝑜 )𝑑𝑑𝑡𝑡𝑜𝑜 = 6 ∫𝑡𝑡 =0 𝑑𝑑𝑡𝑡𝑜𝑜 = 18 𝑜𝑜

𝑜𝑜

8 4 8 (b) 𝐼𝐼(𝑡𝑡) = ∫𝑡𝑡 =0 𝑓𝑓(𝑡𝑡𝑜𝑜 )𝑑𝑑𝑡𝑡𝑜𝑜 = 6 ∫𝑡𝑡 =0 𝑑𝑑𝑡𝑡𝑜𝑜 + 0 ∫𝑡𝑡 =4 𝑑𝑑𝑡𝑡𝑜𝑜 = 24 𝑜𝑜 𝑜𝑜 𝑜𝑜

6𝑡𝑡 , 𝑡𝑡 < 4 24 , 𝑡𝑡 ≥ 4 The integrand is a two-part function; thus the integral is best evaluated in the corresponding twoparts. (c) 𝐼𝐼(𝑡𝑡) = �

3

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 𝑓𝑓(𝑡𝑡) is a two-part function, the integral I(t) is best evaluated as a two-part function. 𝑡𝑡 a) 𝐼𝐼(𝑡𝑡) = ∫𝑡𝑡 =0 𝑓𝑓(𝑡𝑡𝑜𝑜 )𝑑𝑑𝑡𝑡𝑜𝑜 𝑜𝑜

𝑡𝑡

b) 𝐼𝐼(𝑡𝑡) = ∫𝑡𝑡 =0 𝑓𝑓(𝑡𝑡𝑜𝑜 )𝑒𝑒 −(𝑡𝑡−𝑡𝑡𝑜𝑜 )/𝜏𝜏 𝑑𝑑𝑡𝑡𝑜𝑜 𝑜𝑜

Solution (a) The integrand is defined in two distinct parts, thus think of the integral that way. 𝑡𝑡 𝑡𝑡 𝑡𝑡 < 𝑡𝑡1 : 𝐼𝐼(𝑡𝑡) = ∫𝑡𝑡 =0 𝑓𝑓(𝑡𝑡𝑜𝑜 )𝑑𝑑𝑡𝑡𝑜𝑜 = 𝑓𝑓1 ∫𝑡𝑡 =0 𝑑𝑑𝑡𝑡𝑜𝑜 = 𝑓𝑓1 𝑡𝑡 𝑜𝑜

𝑜𝑜

𝑡𝑡 𝑡𝑡 𝑡𝑡 𝑡𝑡 ≥ 𝑡𝑡1 : 𝐼𝐼(𝑡𝑡) = ∫𝑡𝑡 =0 𝑓𝑓(𝑡𝑡𝑜𝑜 )𝑑𝑑𝑡𝑡𝑜𝑜 = 𝑓𝑓1 ∫𝑡𝑡 1=0 𝑑𝑑𝑡𝑡𝑜𝑜 + 0 ∫𝑡𝑡 =4 𝑑𝑑𝑡𝑡𝑜𝑜 = 𝑓𝑓1 𝑡𝑡1 𝑜𝑜 𝑜𝑜 𝑜𝑜

Thus

𝑓𝑓 𝑡𝑡 , 𝑡𝑡 < 𝑡𝑡1 𝐼𝐼(𝑡𝑡) = � 1 𝑓𝑓1 𝑡𝑡1 , 𝑡𝑡 ≥ 𝑡𝑡1 (b) Again consider the integral in two parts since 𝑓𝑓(𝑡𝑡) is two distinct parts. 𝑡𝑡 < 𝑡𝑡1 : 𝐼𝐼(𝑡𝑡) = 𝑓𝑓1 �

𝑡𝑡 ≥ 𝑡𝑡1 : 𝐼𝐼(𝑡𝑡) = 𝑓𝑓1 �

Thus

𝑡𝑡1

𝑡𝑡

𝑡𝑡 =𝑡𝑡

𝑒𝑒 −(𝑡𝑡−𝑡𝑡𝑜𝑜 )⁄𝜏𝜏 𝑑𝑑𝑡𝑡𝑜𝑜 = 𝑓𝑓1 𝜏𝜏 ⋅ 𝑒𝑒 −(𝑡𝑡−𝑡𝑡𝑜𝑜 )⁄𝜏𝜏 |𝑡𝑡𝑜𝑜𝑜𝑜 =0 = 𝑓𝑓1 𝜏𝜏 ⋅ �1 − 𝑒𝑒 −𝑡𝑡⁄𝜏𝜏 �

𝑡𝑡𝑜𝑜 =0

𝑡𝑡 =𝑡𝑡

𝑒𝑒 −(𝑡𝑡−𝑡𝑡𝑜𝑜 )⁄𝜏𝜏 𝑑𝑑𝑡𝑡𝑜𝑜 + 0 = 𝑓𝑓1 𝜏𝜏 ⋅ 𝑒𝑒 −(𝑡𝑡−𝑡𝑡𝑜𝑜 )⁄𝜏𝜏 |𝑡𝑡𝑜𝑜𝑜𝑜 =01 = 𝑓𝑓1 𝜏𝜏 ⋅ �𝑒𝑒 −(𝑡𝑡−𝑡𝑡1 )⁄𝜏𝜏 − 𝑒𝑒 −𝑡𝑡⁄𝜏𝜏 �

𝑡𝑡𝑜𝑜 =0

𝐼𝐼(𝑡𝑡) = �

𝑓𝑓1 𝜏𝜏�1 − 𝑒𝑒 −𝑡𝑡/𝜏𝜏 �

𝑓𝑓1 𝜏𝜏�𝑒𝑒

−(𝑡𝑡−𝑡𝑡1 )/𝜏𝜏

− 𝑒𝑒

−𝑡𝑡/𝜏𝜏

,

� ,

𝑡𝑡 < 𝑡𝑡1 𝑡𝑡 ≥ 𝑡𝑡1

4

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

𝑡𝑡

a) Determine the symbolic solution and plot the function 𝐹𝐹1 (𝑡𝑡) = ∫𝑡𝑡 =0 𝑆𝑆(𝑡𝑡𝑜𝑜 )𝑑𝑑𝑡𝑡𝑜𝑜 . Examine the 𝑜𝑜

limit as ∆t→ 0. 𝑡𝑡 b) Determine the symbolic solution and sketch the function 𝐹𝐹2 (𝑡𝑡) = ∫𝑡𝑡 =0 𝑆𝑆(𝑡𝑡𝑜𝑜 )𝑒𝑒 −(𝑡𝑡−𝑡𝑡𝑜𝑜 )/𝜏𝜏 𝑑𝑑𝑡𝑡𝑜𝑜 . 𝑜𝑜

Examine the limit as ∆t→0. c) Evaluate the function F2 from part (b) using numerical integration. Use t1 = 2, ∆t=1, and τ=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) 0 𝑡𝑡 − 𝑡𝑡1 𝑡𝑡 − 𝑡𝑡1 𝐹𝐹1 (𝑡𝑡) = � � �𝐻𝐻(𝑡𝑡 − 𝑡𝑡1 ) − 𝐻𝐻(𝑡𝑡 − 𝑡𝑡1 − ∆𝑡𝑡)� + 𝐻𝐻(𝑡𝑡 − 𝑡𝑡1 − ∆𝑡𝑡) = � ∆𝑡𝑡 ∆𝑡𝑡 1

Here H(t) is the unit step function.

In the limit as ∆t→0, 𝐹𝐹1 (𝑡𝑡) = 𝐻𝐻(𝑡𝑡 − 𝑡𝑡1 ). (b) 𝐹𝐹2 (𝑡𝑡) =

⎧ ⎪

0 −(𝑡𝑡−𝑡𝑡1 ) 𝜏𝜏 �1 − 𝑒𝑒 𝜏𝜏 � ∆𝑡𝑡

,

,

0 ≤ 𝑡𝑡 < 𝑡𝑡1

,

𝑡𝑡 ≥ 𝑡𝑡1 + ∆𝑡𝑡

, 𝑡𝑡1 ≤ 𝑡𝑡 < 𝑡𝑡1 + ∆𝑡𝑡

0 ≤ 𝑡𝑡 < 𝑡𝑡1

, 𝑡𝑡1 ≤ 𝑡𝑡 < 𝑡𝑡1 + ∆𝑡𝑡

⎨ 𝜏𝜏 (𝑡𝑡−𝑡𝑡 −∆𝑡𝑡) (𝑡𝑡−𝑡𝑡1 ) − ⎪ �𝑒𝑒 − 1𝜏𝜏 𝜏𝜏 � , − 𝑒𝑒 𝑡𝑡 ≥ 𝑡𝑡1 + ∆𝑡𝑡 ⎩∆𝑡𝑡 −(𝑡𝑡−𝑡𝑡1 ) −(𝑡𝑡−𝑡𝑡1 −∆𝑡𝑡) −(𝑡𝑡−𝑡𝑡1 ) 𝜏𝜏 𝜏𝜏 𝜏𝜏 = �1 − 𝑒𝑒 𝜏𝜏 � �𝐻𝐻(𝑡𝑡 − 𝑡𝑡1 ) − 𝐻𝐻(𝑡𝑡 − 𝑡𝑡1 − ∆𝑡𝑡)� + �𝑒𝑒 − 𝑒𝑒 𝜏𝜏 � 𝐻𝐻(𝑡𝑡 − 𝑡𝑡1 − ∆𝑡𝑡) ∆𝑡𝑡 ∆𝑡𝑡

5

Note the limit as ∆t becomes small. lim 𝐹𝐹2 (𝑡𝑡) = 𝑒𝑒 −(𝑡𝑡−𝑡𝑡1 )/𝜏𝜏 𝐻𝐻(𝑡𝑡 − 𝑡𝑡1 ) ∆𝑡𝑡→0

The integrand for part (b) is shown for 𝑡𝑡 > 𝑡𝑡1 + ∆𝑡𝑡.

(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

Problem 4.8 Consider the function f(t) shown.

𝑡𝑡

Determine an expression for the integral 𝐼𝐼(𝑡𝑡) = ∫𝑡𝑡 =0 𝑓𝑓(𝑡𝑡𝑜𝑜 )𝑑𝑑𝑡𝑡𝑜𝑜 over the entire range 0 ≤ t ≤ 3. 𝑜𝑜 Solution 10𝑡𝑡 , 0 < 𝑡𝑡 < 1 𝐼𝐼(𝑡𝑡) = �10 − 5(𝑡𝑡 − 1) , 1 < 𝑡𝑡 < 2 5 , 𝑡𝑡 > 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(𝑡𝑡) b) 𝛿𝛿(𝑥𝑥)𝛿𝛿(𝑡𝑡) c) H(t) = Step function d) Ramp(x) = Ramp function Solution 1 a) 𝑡𝑡 1

1

b) �𝑥𝑥� � 𝑡𝑡 � c) Dimensionless d) x Problem 4.10 Evaluate the following integrals. H is the step function and 𝛿𝛿(𝑥𝑥) is the delta function.

3 a) ∫𝑥𝑥=0 𝐻𝐻(𝑥𝑥 − 1)𝑑𝑑𝑑𝑑 3

b) ∫𝑥𝑥=2 𝐻𝐻(𝑥𝑥 − 1)𝑑𝑑𝑑𝑑 3

c) ∫𝑥𝑥=0 𝛿𝛿(𝑥𝑥 − 2)𝑑𝑑𝑑𝑑 3

d) ∫𝑥𝑥=0 𝛿𝛿(𝑥𝑥 − 5)𝑑𝑑𝑑𝑑 3

e) ∫𝑥𝑥=0 𝑓𝑓(𝑥𝑥)𝛿𝛿(𝑥𝑥 − 2)𝑑𝑑𝑑𝑑 3

f) ∫𝑥𝑥=0 𝑓𝑓(𝑥𝑥)𝛿𝛿(𝑥𝑥 − 5)𝑑𝑑𝑑𝑑

Solution a) 2; (b) 0; (c) 1; (d) 0 (e) 𝑓𝑓(2) (f) 0 7

Problem 4.11 Evaluate the following integrals symbolically. Also sketch the function I(x), clearly showing the locations x=a and x=b. 𝑏𝑏 a) 𝑓𝑓(𝑥𝑥) = ∫𝑥𝑥 =𝑎𝑎 𝛿𝛿(𝑥𝑥 − 𝑥𝑥𝑜𝑜 )𝑑𝑑𝑥𝑥𝑜𝑜 b) c) d) e)

𝑜𝑜

𝑏𝑏 𝑓𝑓(𝑥𝑥) = ∫𝑥𝑥 =𝑎𝑎 𝐻𝐻(𝑥𝑥 − 𝑥𝑥𝑜𝑜 )𝑑𝑑𝑥𝑥𝑜𝑜 𝑜𝑜 𝑏𝑏 𝑓𝑓(𝑥𝑥) = ∫𝑥𝑥 =𝑎𝑎(𝑥𝑥 − 𝑥𝑥𝑜𝑜 )𝑑𝑑𝑥𝑥𝑜𝑜 𝑜𝑜 𝑡𝑡 𝑓𝑓1 (𝑡𝑡, 𝑡𝑡1 ) = ∫𝑡𝑡 =0 𝛿𝛿(𝑡𝑡𝑜𝑜 − 𝑡𝑡1 )𝑑𝑑𝑡𝑡𝑜𝑜 𝑜𝑜 𝑡𝑡 𝑓𝑓2 (𝑡𝑡, 𝑡𝑡1 ) = ∫𝑡𝑡 =0 𝑓𝑓1 (𝑡𝑡𝑜𝑜 , 𝑡𝑡1 )𝑑𝑑𝑡𝑡𝑜𝑜 where 𝑓𝑓1 is the function determined in part (d). 𝑜𝑜

Solution a) Assume a < b.

0 , 𝑥𝑥 < 𝑎𝑎 𝑓𝑓(𝑥𝑥) = � 𝛿𝛿(𝑥𝑥 − 𝑥𝑥𝑜𝑜 )𝑑𝑑𝑥𝑥𝑜𝑜 = 𝐻𝐻(𝑥𝑥 − 𝑎𝑎) − 𝐻𝐻(𝑥𝑥 − 𝑏𝑏) = �1 , 𝑎𝑎 < 𝑥𝑥 < 𝑏𝑏 𝑥𝑥𝑜𝑜 =𝑎𝑎 0 , 𝑥𝑥 > 𝑏𝑏 0 , 𝑥𝑥 < 𝑎𝑎 𝑏𝑏 b) 𝑓𝑓(𝑥𝑥) = ∫𝑥𝑥 =𝑎𝑎 𝐻𝐻(𝑥𝑥 − 𝑥𝑥𝑜𝑜 )𝑑𝑑𝑥𝑥𝑜𝑜 = �𝑥𝑥 − 𝑎𝑎 , 𝑎𝑎 < 𝑥𝑥 < 𝑏𝑏 𝑜𝑜 𝑏𝑏 − 𝑎𝑎 , 𝑥𝑥 > 𝑏𝑏 (𝑥𝑥−𝑎𝑎)2 (𝑥𝑥−𝑏𝑏)2 𝑏𝑏 c) 𝑓𝑓(𝑥𝑥) = ∫𝑥𝑥 =𝑎𝑎(𝑥𝑥 − 𝑥𝑥𝑜𝑜 )𝑑𝑑𝑥𝑥𝑜𝑜 = 2 − 2 𝑜𝑜 d) 𝑓𝑓1 (𝑡𝑡, 𝑡𝑡1 ) = 𝐻𝐻(𝑡𝑡 − 𝑡𝑡1 ) 𝑡𝑡 𝑡𝑡 e) 𝑓𝑓2 (𝑡𝑡, 𝑡𝑡1 ) = ∫𝑡𝑡 =0 𝑓𝑓1 (𝑡𝑡𝑜𝑜 , 𝑡𝑡1 )𝑑𝑑𝑡𝑡𝑜𝑜 = ∫𝑡𝑡 =0 𝐻𝐻(𝑡𝑡𝑜𝑜 − 𝑡𝑡1 )𝑑𝑑𝑡𝑡𝑜𝑜 = (𝑡𝑡 − 𝑡𝑡1 )𝐻𝐻(𝑡𝑡 − 𝑡𝑡1 ) 𝑏𝑏

𝑜𝑜

𝑜𝑜

Problem 4.12 2 2 𝑧𝑧 𝑑𝑑 𝑑𝑑 The error function is erf(𝑧𝑧) = ∫𝑥𝑥=0 𝑒𝑒 −𝑥𝑥 𝑑𝑑𝑑𝑑. Determine 𝑑𝑑𝑑𝑑 𝑒𝑒𝑒𝑒𝑒𝑒(𝑧𝑧) and 𝑑𝑑𝑑𝑑 𝑒𝑒𝑒𝑒𝑒𝑒(𝑧𝑧 2 ). √𝜋𝜋

Solution Using Leibnitz Rule. We can deduce the following 𝑑𝑑 2 𝑒𝑒𝑒𝑒𝑒𝑒(𝑧𝑧) = exp(−𝑧𝑧 2 ) 𝑑𝑑𝑑𝑑 √𝜋𝜋 𝑑𝑑 2 𝑒𝑒𝑒𝑒𝑒𝑒(𝑧𝑧 2 ) = exp(−𝑧𝑧 4 )(2𝑧𝑧) 𝑑𝑑𝑑𝑑 √𝜋𝜋

8

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

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

𝑔𝑔(𝑡𝑡) = � ℎ(𝜏𝜏)𝑓𝑓(𝑡𝑡 − 𝜏𝜏)𝑑𝑑𝑑𝑑 0

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

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

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

t>4

0 , 𝑡𝑡 < 1 ⎧4(𝑡𝑡 − 1) , 1 < 𝑡𝑡 < 2 ⎪ 4 , 2 < 𝑡𝑡 < 3 𝑔𝑔(𝑡𝑡) = ⎨4(4 − 𝑡𝑡) , 3 < 𝑡𝑡 < 4 ⎪ ⎩ 0 , 𝑡𝑡 > 4 = 4(𝑡𝑡 − 1)(𝐻𝐻(𝑡𝑡 − 1) − 𝐻𝐻(𝑡𝑡 − 2)) + 4(𝐻𝐻(𝑡𝑡 − 2) − 𝐻𝐻(𝑡𝑡 − 3)) + 4(4 − 𝑡𝑡)(𝐻𝐻(𝑡𝑡 − 3) − 𝐻𝐻(𝑡𝑡 − 4))

9

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

Problem 4.15: Basics of Numerical Integration 4 Consider the integral 𝐼𝐼 = ∫𝑥𝑥=0(1 − 𝑒𝑒 −2𝑥𝑥 )𝑑𝑑𝑑𝑑. Evaluate this integral using the methods listed.

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

Solution 𝑏𝑏

% Error 0

𝑏𝑏

𝑒𝑒 −2𝑥𝑥 𝑒𝑒 −2𝑏𝑏 𝑒𝑒 −2𝑎𝑎 𝐼𝐼 = � (1 − 𝑒𝑒 = �𝑥𝑥 − � = �𝑏𝑏 − � − �𝑎𝑎 − � −2 𝑎𝑎 −2 −2 𝑥𝑥=𝑎𝑎 For a = 0 and b = 4, the integral evaluates to 𝐼𝐼 = 3.5002. −2𝑥𝑥 )𝑑𝑑𝑑𝑑

Rule Approx Error -------------------------------Exact 3.500 0.00 Trap, n=1 1.999 42.88 Trap, n=2 2.963 15.35 Trap, n=4 3.344 4.47 Simp 1/3, n=2 3.284 6.17 Simp 1/3, n=4 3.471 0.84 Simp 3/8, n=3 3.388 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

2 x

4

0

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

0.5

0

0 0

0.5

0 0

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

2 x

4

0

2 x

4

11

Problem 4.16: Erf a) Use the trapezoid rule to evaluate erf(1) for a sequence of increasing n (decreasing h): 𝑛𝑛 = 2𝑝𝑝 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

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

b) Iexact = Iapprox =

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) 𝑑𝑑 = (12 − 0) � 2 � = 150 𝑚𝑚 0+36

36+25

(b) 𝑑𝑑 = (6) � 2 � + (6) � (c) 𝑑𝑑 = ∑𝑛𝑛𝑖𝑖=1 ℎ𝑖𝑖

𝑣𝑣𝑖𝑖−1 +𝑣𝑣𝑖𝑖 2

2

� = 291 𝑚𝑚

= 263 𝑚𝑚

The data plot using all the data is shown.

12

Problem 4.18: Ellipse 𝜋𝜋/2 The perimeter P of an ellipse is given by 𝑃𝑃 = 4𝑎𝑎 ∫0 √1 − 𝑘𝑘 2 𝑠𝑠𝑠𝑠𝑠𝑠2 𝜃𝜃 𝑑𝑑𝑑𝑑 where a = major axis, b= minor axis, and 𝑘𝑘 = √𝑎𝑎2 + 𝑏𝑏 2 /𝑎𝑎. 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 perimeter of the following ellipses. 𝑥𝑥 2

𝑦𝑦 2

𝑥𝑥 2

𝑦𝑦 2

𝑥𝑥 2

𝑦𝑦 2

𝑎𝑎) 52 + 22 = 1 , 𝑏𝑏) 42 + 72 = 1 , 𝑐𝑐) 1002 + 22 = 1 Solution

*** circular limit *** a = 1.0 , b = 1.0 P = 6.3 *** Case a *** a = 5.0 , b = P = 23.0

2.0

*** Case b *** a = 7.0 , b = P = 35.2

4.0

*** 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

Problem 4.19 Consider the function F(x) shown.

6

We wish to compute the integral 𝐼𝐼 = ∫0 𝐹𝐹(𝑥𝑥)𝑑𝑑𝑑𝑑. 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) 𝐼𝐼 = 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

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 𝐿𝐿 The surface area can be determined from 𝑆𝑆 = 𝜋𝜋 ∫0 𝑑𝑑(𝑧𝑧) ∙ 𝑑𝑑𝑑𝑑 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 𝑛𝑛−1

ℎ 𝑆𝑆 ≅ 𝜋𝜋 �𝑓𝑓(𝑥𝑥0 ) + 2 � 𝑓𝑓(𝑥𝑥𝑖𝑖 ) + 𝑓𝑓(𝑥𝑥𝑛𝑛 )� 2 𝑖𝑖=1

b) Equation

1.5 (0 + 2 ∗ 2.9 + 2 ∗ 4.8 + 2 ∗ 5.8 + 6.2) = 𝜋𝜋 2 = 79.22 𝑖𝑖𝑖𝑖2 (4.79) 𝑛𝑛−1

𝑛𝑛−2

𝑖𝑖=1,3,…

𝑖𝑖=2,4,…

gives

ℎ 𝑆𝑆 ≅ 𝜋𝜋 �𝑓𝑓(𝑥𝑥0 ) + 4 � 𝑓𝑓(𝑥𝑥𝑖𝑖 ) + 2 � 𝑓𝑓(𝑥𝑥𝑖𝑖 ) + 𝑓𝑓(𝑥𝑥𝑛𝑛 )� 3 1.5 (0 + 4 ∗ 2.9 + 2 ∗ 4.8 + 4 ∗ 5.8 + 6.2) 3 = 79.48 𝑖𝑖𝑖𝑖2

= 𝜋𝜋

15

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, 𝑓𝑓(𝑧𝑧) (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 2𝑧𝑧 𝑧𝑧 𝑓𝑓(𝑧𝑧) = 200 � � 𝑒𝑒 − 𝐻𝐻 (𝑙𝑙𝑙𝑙/𝑓𝑓𝑓𝑓) 5 + 𝑧𝑧 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 𝐻𝐻 𝐹𝐹 = ∫0 𝑓𝑓(𝑧𝑧) ∙ 𝑑𝑑𝑑𝑑.

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. 𝐻𝐻 The moment caused by the variable wind force is 𝑀𝑀 = ∫0 𝑧𝑧 ∙ 𝑓𝑓(𝑧𝑧) ∙ 𝑑𝑑𝑑𝑑. 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 𝑓𝑓(𝑧𝑧)and 𝑧𝑧𝑧𝑧(𝑧𝑧)show that these numerical results are at least approximately correct. The graph of 𝑓𝑓(𝑧𝑧)shows an average height of about 45 or 50 lb/ft. Thus the force should be approximately F ≈ (45)(25) = 1250 lb.

16

80

2000 60

Force, F (lb)

f(z) lb/ft

1500 40

20

1000

500

0 0

5

10

15

20

0

25

10

15

20

25

30

35

40

30

35

40

z (ft)

Height, H (ft) 800

15000

10000

400

Tension, T (lb)

z*f(z) lb

600

200

0 0

5

10

15

20

5000

25

0

z (ft)

10

15

20

25

Height, H (ft)

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 𝑝𝑝(𝑧𝑧) = 𝜌𝜌𝜌𝜌(𝐷𝐷 − 𝑧𝑧) where 𝑝𝑝(𝑧𝑧)is pressure in N/m2 exerted at an elevation z meters above the reservoir bottom; ρ is density of water, which for this problem is assumed to be a constant 103 kg/m3; 𝑔𝑔 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 𝑓𝑓𝑡𝑡 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 𝐷𝐷 𝑓𝑓𝑡𝑡 = ∫0 𝜌𝜌𝜌𝜌 ∙ (𝐷𝐷 − 𝑧𝑧)𝑤𝑤(𝑧𝑧) ∙ 𝑑𝑑𝑑𝑑. where w(z) is width of the dam face (m) at elevation z. The line of action can also be obtained by evaluating 𝐷𝐷 ∫0 𝑧𝑧𝑧𝑧𝑧𝑧 ∙ (𝐷𝐷 − 𝑧𝑧)𝑤𝑤(𝑧𝑧) ∙ 𝑑𝑑𝑑𝑑 𝑑𝑑 = 𝐷𝐷 ∫0 𝜌𝜌𝜌𝜌 ∙ (𝐷𝐷 − 𝑧𝑧)𝑤𝑤(𝑧𝑧) ∙ 𝑑𝑑𝑑𝑑 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 𝑓𝑓𝑡𝑡 and d. Test your function using the specific data given.

Solution

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

17

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: 𝐻𝐻

𝐹𝐹𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 = � 𝐹𝐹(𝑧𝑧)𝑑𝑑𝑑𝑑 0

The line of action can also be determined by integration: 𝐻𝐻 ∫0 𝑧𝑧 𝐹𝐹(𝑧𝑧)𝑑𝑑𝑑𝑑 𝑑𝑑 = 𝐻𝐻 ∫0 𝐹𝐹(𝑧𝑧)𝑑𝑑𝑑𝑑 Solution

Wind force on a skyscraper

F (N/m)

4000

2000

0

0

50

100 150 height, z (m)

200

250

50

100 150 height, z (m)

200

250

5

z*F (N)

10

x 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

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 𝑐𝑐1 𝐸𝐸𝜆𝜆,𝑏𝑏 = 𝑐𝑐 𝜆𝜆5 �𝑒𝑒𝑒𝑒𝑒𝑒 � 2 � − 1� 𝜆𝜆𝜆𝜆 Eλ ,b = emissive power (W/m2/µm), λ = wavelength (µm), T = absolute temperature (K) c1 = 3.7418 × 108 W*µm4/m2, c2 = 1.4388 × 104 µm*K The fraction of energy emitted in the wavelength band 0 ≤ 𝜆𝜆 ≤ 𝜆𝜆∗ is

𝜆𝜆∗ 𝑇𝑇

𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵𝐵 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅 𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹𝐹: 𝐹𝐹0→𝜆𝜆∗ (𝜆𝜆∗ 𝑇𝑇) = � 0

𝐸𝐸𝜆𝜆,𝑏𝑏 𝑑𝑑(𝜆𝜆𝜆𝜆) 𝜎𝜎𝑇𝑇 5

σ =Stefan-Boltzmann constant = 5.6696 × 10-8 W/m2 K4 The integrand that defines F0→λ*depends on the product λT, not on λ and T separately. (a) Plot 𝐸𝐸𝜆𝜆,𝑏𝑏 vs λ 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 𝐹𝐹0→𝜆𝜆∗ as a function of the product λT. Please use this function throughout the remainder of this problem. (c) Plot 𝐹𝐹0→𝜆𝜆∗ in the range λT = 0 to 20000 µm K. (d) The range of visible light for a typical human is 0.4 to 0.7 µm. 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 = Eλ,b E E Total emissivity = ε = = 4 Eb σT ∞

E = � ελ Eλ,b dλ 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

(a) Planck Distribution 7

8

3

50 100 300 1000 2000 5800

6

10

4

10

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

lb

2

lb

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

10

10

0

10

x 10

2.5

2

1.5

1

0.5

-2

10

0 -4

10

-1

10

0

1

10 10 Wavelength, lam (micro meter)

2

0

10

20

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

80

90

100

10

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 𝜆𝜆∗ 𝑇𝑇

𝜆𝜆∗ 𝑇𝑇

𝐸𝐸𝜆𝜆,𝑏𝑏 𝑐𝑐1 𝑑𝑑(𝜆𝜆𝜆𝜆) = � 𝑑𝑑(𝜆𝜆𝜆𝜆) 𝐹𝐹0→𝜆𝜆∗ (𝜆𝜆∗ 𝑇𝑇) = � 5 𝜎𝜎𝑇𝑇 5 �exp � 𝑐𝑐2 � − 1� 𝜎𝜎(𝜆𝜆𝜆𝜆) 0 𝜆𝜆𝜆𝜆=0 𝜆𝜆𝜆𝜆 ∗ 𝜆𝜆 𝑇𝑇

𝑐𝑐1 𝑑𝑑(𝑧𝑧) 5 �exp �𝑐𝑐2 � − 1� 𝜎𝜎𝜎𝜎 𝑧𝑧=0 𝑧𝑧 This integral cannot be evaluated analytically and thus must be evaluated numerically. 𝐹𝐹0→𝜆𝜆∗ (𝜆𝜆∗ 𝑇𝑇) = �

(c) Blackbody Radiation Function

20

(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 (λ = 0.4 to 0.7) is

(e) Total Emissivity

𝐹𝐹0→𝜆𝜆∗ (0.7 ∗ 5800) − 𝐹𝐹0→𝜆𝜆∗ (0.4 ∗ 5800) = 0.491 − 0.123 = 0.368 = 36.8%

21

Problem 4.25: Pipe Flow The volume flow rate of a fluid flowing in a round pipe of radius r0 is 𝑟𝑟0

The velocity can be approximated by

𝑄𝑄 = � 𝑣𝑣(𝑟𝑟)2𝜋𝜋𝜋𝜋 𝑑𝑑𝑑𝑑 0

𝑣𝑣(𝑟𝑟) = 2 �1 −

𝑟𝑟 1/6 � 𝑟𝑟0

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 < 𝑟𝑟0 < 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

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. � = [𝐀𝐀, 𝐛𝐛]. b) Determine the rank of the augmented matrix, 𝐀𝐀 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

Case a b

c

d e f g

h

� � �

A

b

Rank(A)

~ Rank( A )

Consistent ?

Solution, x

1 0 � 1 1

2 � � 1

2

2

Yes

1

2

No

1 1 � 0 0

2 � � 1

2  − 1  

0 � � 0

1

1

Yes

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

1

1

Yes

2

2

Yes

2

3

No

1 1 � 0 0

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

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

Problem 5.2: Equation of a Plane Consider the equation of a plane written as 𝑐𝑐1 𝑥𝑥 + 𝑐𝑐2 𝑦𝑦 + 𝑐𝑐3 = 𝑧𝑧. a) Given three known points (𝑥𝑥1 , 𝑦𝑦1 , 𝑧𝑧1 ), (𝑥𝑥2 , 𝑦𝑦2 , 𝑧𝑧2 ), and (𝑥𝑥3 , 𝑦𝑦3 , 𝑧𝑧3 ), derive the system of equations that determine c1, c2, and c3. b) Write a function ‘eq_plane’ that determines the vector of coefficients 𝐜𝐜 = [𝑐𝑐1 , 𝑐𝑐2 , 𝑐𝑐3 ] 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: 𝑐𝑐1 𝑥𝑥1 + 𝑐𝑐2 𝑦𝑦1 + 𝑐𝑐3 = 𝑧𝑧1 𝑐𝑐1 𝑥𝑥2 + 𝑐𝑐2 𝑦𝑦2 + 𝑐𝑐3 = 𝑧𝑧2 𝑐𝑐1 𝑥𝑥3 + 𝑐𝑐2 𝑦𝑦3 + 𝑐𝑐3 = 𝑧𝑧3 This gives 3 equations for the 3 ci values. In matrix form: 𝑧𝑧1 𝑥𝑥1 𝑦𝑦1 1 𝑐𝑐1 𝑧𝑧 𝑐𝑐 𝑥𝑥 𝑦𝑦 1 � 2 � � 2� = � 2� 2 𝑧𝑧3 𝑥𝑥3 𝑦𝑦3 1 𝑐𝑐3 (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 0.4

0.6 0.2

0.8 x

1

0

y

2

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)

b) Solving simultaneously gives

𝑆𝑆1 �𝐹𝐹𝐹𝐹1 𝐶𝐶𝐶𝐶1

𝑆𝑆2 𝐹𝐹𝐹𝐹2 𝐶𝐶𝐶𝐶2

𝑉𝑉𝑠𝑠 𝑆𝑆3 𝑉𝑉1 𝐹𝐹𝐹𝐹3 � �𝑉𝑉2 � = �𝑉𝑉𝑓𝑓𝑓𝑓 � 𝑉𝑉𝑐𝑐𝑐𝑐 𝐶𝐶𝐶𝐶3 𝑉𝑉3

𝑉𝑉1 8000 𝑉𝑉 � 2 � = �4000� 𝑉𝑉3 8000

3

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

200

Case 3: Effect of h’ Effect of h

100 100

h=0

180 90

80

80

120 100

70

60 T (deg C)

140

T (deg C)

T (deg C)

160

40

60 50

h=10

40

80

20

30

60 20

40

0

2

4

6 x (m)

8

10

0

h=100

0

0.2

0.6

0.4 x (m)

0.8

1

10 Ta 0

0

0.1

0.2

0.3

0.4

0.5 x (m)

0.6

0.7

0.8

1

0.9

4

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 𝐹𝐹1 𝑦𝑦𝑖𝑖−1 + 𝐹𝐹2 𝑥𝑥𝑖𝑖+1 = 𝐹𝐹1 𝑦𝑦𝑖𝑖 + 𝐹𝐹21 𝑥𝑥𝑖𝑖 , 𝑖𝑖 = 2,3, … , 𝑛𝑛 − 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 𝑥𝑥𝑖𝑖 𝐾𝐾 = 𝑦𝑦𝑖𝑖 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 𝑥𝑥𝑖𝑖 = Ky𝑖𝑖 , the mass balance at a typical interior stage is

𝐹𝐹1 𝑦𝑦𝑖𝑖−1 − (𝐹𝐹1 + 𝐹𝐹2 𝐾𝐾)𝑦𝑦𝑖𝑖 + 𝐹𝐹2 Ky𝒊𝒊+𝟏𝟏 = 0, i = 2, 3, … , n-1 The mass balances at the inlet and exit are 𝐹𝐹1 𝑦𝑦in − (𝐹𝐹1 + 𝐹𝐹2 𝐾𝐾)𝑦𝑦1 + 𝐹𝐹2 Ky2 = 0 , i = 1 𝐹𝐹1 𝑦𝑦𝑛𝑛−1 − (𝐹𝐹1 + 𝐹𝐹2 𝐾𝐾)𝑦𝑦𝑛𝑛 + 𝐹𝐹2 𝑥𝑥in = 0 , i = n

This produces a simultaneous set of n linear algebraic equations for the unknown concentrations.

5

(𝐹𝐹 + 𝐹𝐹2 𝐾𝐾) −𝐹𝐹2 𝐾𝐾 0 ⎡ 1 (𝐹𝐹 + 𝐹𝐹 𝐾𝐾) −𝐹𝐹 −𝐹𝐹 1 1 2 2 𝐾𝐾 ⎢ ⋱ ⋱ ⎢ ⎢ 0 0 −𝐹𝐹1 ⎢ ⎢ 0 0 0 ⎢ ⎣ 0 0 0

c) yout = 2.6703e-006 xout = 0.049999

0 0 0 0 ⋱ (𝐹𝐹1 + 𝐹𝐹2 𝐾𝐾) −𝐹𝐹2 𝐾𝐾 ⋱ ⋱ 0 −𝐹𝐹1 0 0

0 0

0 0

𝑦𝑦 ⎤ ⎡ 1 ⎤ ⎡𝐹𝐹1 𝑦𝑦in ⎤ 𝑦𝑦 ⎥⎢ 2 ⎥ ⎢ 0 ⎥ ⎥⎢ ⋮ ⎥ ⎢ ⋮ ⎥ ⎥ ⎢ 𝑦𝑦𝑖𝑖 ⎥ = ⎢ 0 ⎥ 0 0 ⎥⎢ ⋮ ⎥ ⋱ ⎢ ⋮ ⎥ ⎥ 𝑦𝑦 (𝐹𝐹1 + 𝐹𝐹2 𝐾𝐾) −𝐹𝐹2 𝐾𝐾 ⎥ ⎢ 𝑛𝑛−1 ⎥ ⎢ 0 ⎥ ⎣ 𝑦𝑦𝑛𝑛 ⎦ ⎣𝐹𝐹2 𝑥𝑥in ⎦ (𝐹𝐹 −𝐹𝐹1 1 + 𝐹𝐹2 𝐾𝐾)⎦

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

Weight Fractions

0.06 y x

0.04

0.02

0

1

1.5

2

2.5

3 Stage #

3.5

4

4.5

5

0

Weight Fractions

10

y x

-2

10

-4

10

-6

10

1

1.5

2

2.5

3 Stage #

3.5

4

4.5

5

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

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: 𝑓𝑓 𝑔𝑔1 ℎ1 ⋅ 0 𝑟𝑟1 ⎡ 1 ⎤ 𝑥𝑥1 ⎡ ⎤ ⎡ 𝑒𝑒 𝑓𝑓 𝑔𝑔 ℎ 2 2 2 𝑟𝑟2 ⎤ ⎢ 2 ⎥ 𝑥𝑥2 ⎢ ⎥ ⎢ 𝑔𝑔3 ℎ3 𝑟𝑟3 ⎥ ⎢𝑑𝑑3 𝑒𝑒3 𝑓𝑓3 ⎥ 𝑥𝑥3 ⎢ ⋅ ⋅ ⋅ ⋅ ⋅ ⋅ ⎥⎢ ⋅ ⎥ ⎢ ⋅ ⎥ ⎢ ⎥ ⎢ 𝑥𝑥𝑖𝑖 ⎥ ⎢ 𝑟𝑟𝑖𝑖 ⎥ 𝑑𝑑𝑖𝑖 𝑒𝑒𝑖𝑖 𝑓𝑓𝑖𝑖 𝑔𝑔𝑖𝑖 ℎ𝑖𝑖 ⎢ ⎥ ⎢ ⎥ ⎢⋅ ⎥ 𝑥𝑥𝑖𝑖+1 = 𝑟𝑟𝑖𝑖+1 𝑑𝑑𝑖𝑖+1 𝑒𝑒𝑖𝑖+1 𝑓𝑓𝑖𝑖+1 𝑔𝑔𝑖𝑖+1 ℎ𝑖𝑖+1 ⋅ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ 𝑥𝑥𝑖𝑖+2 ⎥ ⎢ 𝑟𝑟𝑖𝑖+2 ⎥ 𝑑𝑑𝑖𝑖+2 𝑒𝑒𝑖𝑖+2 𝑓𝑓𝑖𝑖+2 𝑔𝑔𝑖𝑖+2 ℎ𝑖𝑖+2 ⎢ ⎥⎢ ⋅ ⎥ ⎢ ⋅ ⎥ ⋅ ⋅ ⋅ ⋅ ⎢ ⎥ 𝑑𝑑𝑛𝑛−1 𝑒𝑒𝑛𝑛−1 𝑓𝑓𝑛𝑛−1 𝑔𝑔𝑛𝑛−1 ⎥ ⎢𝑥𝑥𝑛𝑛−1 ⎥ ⎢𝑟𝑟𝑛𝑛−1 ⎥ ⎢ ⎣ 𝑥𝑥𝑛𝑛 ⎦ ⎣ 𝑟𝑟𝑛𝑛 ⎦ ⋅ 𝑑𝑑𝑛𝑛 𝑒𝑒𝑛𝑛 𝑓𝑓𝑛𝑛 ⎦ ⎣0

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 𝑥𝑥1 5 ⎡−2 9 −4 −1 0 ⎤ ⎡𝑥𝑥 ⎤ ⎡ ⎤ ⎢ ⎥ ⎢𝑥𝑥2 ⎥ ⎢2⎥ −1 −3 7 −1 −2 ⎢ ⎥ ⎢ 3 ⎥ = ⎢1⎥ ⎢ 0 −4 −2 12 −5⎥ ⎢𝑥𝑥4 ⎥ ⎢1⎥ ⎣0 0 −7 −3 15 ⎦ ⎣𝑥𝑥5 ⎦ ⎣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

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

0

hi gi+1 fi+2 …

0 0

…

0 hi+1 gi+2 …

0 0 0

0 hi+2 … fn-2 0 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. 𝑟𝑟𝑛𝑛 𝑥𝑥𝑛𝑛 = 𝑓𝑓𝑛𝑛 𝑟𝑟𝑛𝑛−1 − 𝑔𝑔𝑛𝑛−1 𝑥𝑥𝑛𝑛 𝑥𝑥𝑛𝑛−1 = 𝑓𝑓𝑛𝑛−1 For i = n-2 : -1 : 1 𝑟𝑟𝑖𝑖 − 𝑔𝑔𝑖𝑖 𝑥𝑥𝑖𝑖+1 − ℎ𝑖𝑖 𝑥𝑥𝑖𝑖+2 𝑥𝑥𝑖𝑖 = 𝑓𝑓𝑖𝑖 Case 1 Case 2 9

1.4

18

x 10

16

1.35

14

1.3 12

1.25 x(i)

x(i)

10

1.2

8

1.15

6 4

1.1 2

1.05

1

1.5

Case 1 xpenta =

2

2.5

1.0825

3 i

3.5

4

4.5

5

0

0

2

6

4

8

i

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

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 (W) (°C) (°C/W) 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

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

70 60 50 40 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

Effect of R4 = R5 on Tc for various Qc (W) values 450 0 5 10 15 20

400

Chip Temperature, Tc (C)

350 300 250 200 150 100 50 0

0

10

20

30 40 50 60 70 80 Thermal Resistance, R4=R5 (deg C/W)

90

100

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

11

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

R1 +

R2

i1

R4

i2

R3

+

R5

v2

v1 -

i4

i3

i5

-

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, 𝑖𝑖1 𝑣𝑣1 𝑅𝑅1 0 0 𝑅𝑅4 0 ⎡𝑖𝑖 ⎤ ⎡ 0 ⎤ ⎡ 0 𝑅𝑅 ⎤ 0 −𝑅𝑅 𝑅𝑅 2 4 5 ⎥ ⎢ 2⎥ ⎢ ⎢ ⎥ 0 0 𝑅𝑅 0 −𝑅𝑅 3 5 ⎥ ⎢𝑖𝑖3 ⎥ = ⎢−𝑣𝑣2 ⎥ ⎢ ⎢ 1 −1 0 −1 0 ⎥ ⎢𝑖𝑖4 ⎥ ⎢ 0 ⎥ ⎣0 1 −1 0 −1 ⎦ ⎣𝑖𝑖5 ⎦ ⎣ 0 ⎦ b) i(mA) 0.8299 0.1909 -0.1162 0.6390 0.3071

12

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

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, 𝐹𝐹𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 = 𝑘𝑘𝑘𝑘. k3

k2

k1

M1

k4

M2

M3

x1+L1 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

𝑥𝑥1 + 𝑥𝑥2 + 𝑥𝑥3 + 𝑥𝑥4 = 𝐿𝐿𝑇𝑇 − 𝐿𝐿1 − 𝐿𝐿2 − 𝐿𝐿3 − 𝐿𝐿4 − 3𝑊𝑊

Mass-1

−𝑘𝑘1 𝑥𝑥1 + 𝑘𝑘2 𝑥𝑥2 = 0

Mass-2

−𝑘𝑘2 𝑥𝑥2 + 𝑘𝑘3 𝑥𝑥3 = 0

Mass-3 In matrix form

1 1 −𝑘𝑘 𝑘𝑘2 � 1 0 −𝑘𝑘2 0 0 (a) Numerical Results

−𝑘𝑘3 𝑥𝑥3 + 𝑘𝑘4 𝑥𝑥4 = 0 1 0 𝑘𝑘3 −𝑘𝑘3

1 𝑥𝑥1 𝐿𝐿𝑇𝑇 − 𝐿𝐿1 − 𝐿𝐿2 − 𝐿𝐿3 − 𝐿𝐿4 − 3𝑊𝑊 𝑥𝑥 0 2 0 � �� � = � 0 𝑥𝑥3 0 𝑘𝑘4 𝑥𝑥4 0

Case 1

Case 2

x

x

0.85 0.85 0.85 0.85

1.632 0.816 0.544 0.408 14

Chapter 6

Non-Linear Algebra: Root Finding Problem 6.1: Graphical Root Finding Consider the equation 𝑒𝑒 −𝑥𝑥 = 𝑥𝑥. Estimate the real root of this equation in the following ways. (a) Graphically by plotting the functions 𝑒𝑒 −𝑥𝑥 and x. (b) Graphically by plotting the function 𝑓𝑓(𝑥𝑥) = 𝑒𝑒 −𝑥𝑥 − 𝑥𝑥. (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 𝑒𝑒 −𝑥𝑥 and x.

These graphs show that the root is approximately xr ≈ 0.57. (b) Graphical by plot of 𝑓𝑓(𝑥𝑥) = 𝑒𝑒 −𝑥𝑥 − 𝑥𝑥.

(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

(d) False Position

Iteration 1 2 3

𝑥𝑥𝑢𝑢 − 𝑥𝑥𝑙𝑙 𝑥𝑥𝑟𝑟 = 𝑥𝑥𝑢𝑢 − 𝑓𝑓(𝑥𝑥𝑢𝑢 ) � � 𝑓𝑓(𝑥𝑥𝑢𝑢 ) − 𝑓𝑓(𝑥𝑥𝑙𝑙 ) 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

𝑓𝑓(𝑥𝑥) = 𝑒𝑒 −𝑥𝑥 − 𝑥𝑥 𝑓𝑓′(𝑥𝑥) = −𝑒𝑒 −𝑥𝑥 − 1

𝑥𝑥𝑖𝑖+1 = 𝑥𝑥𝑖𝑖 −

𝑓𝑓(𝑥𝑥𝑖𝑖 ) 𝑒𝑒 −𝑥𝑥𝑖𝑖 − 𝑥𝑥𝑖𝑖 = 𝑥𝑥 − 𝑖𝑖 𝑓𝑓 ′ (𝑥𝑥𝑖𝑖 ) −𝑒𝑒 −𝑥𝑥𝑖𝑖 − 1

Starting from x0 = 2, we get the following iterations.

Iteration 0 1 2 3

𝑥𝑥𝑖𝑖 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

Problem 6.2: Exploring Root Finding Determine the positive real root of the equation ln(𝑥𝑥 2 ) = 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

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 " " " − 𝑞𝑞𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 − 𝑞𝑞𝑟𝑟𝑟𝑟𝑟𝑟 0 = 𝑞𝑞𝑠𝑠𝑠𝑠𝑠𝑠 0 = 𝛼𝛼𝑠𝑠 ∙ 𝐺𝐺𝑠𝑠 − ℎ(𝑇𝑇 − 𝑇𝑇∞ ) − 𝜀𝜀 ∙ 𝜎𝜎 ∙ 𝑇𝑇 4 2 When Gs = 1200 W/m , the plate temperature is 314.7 K (b) Effect of Solar Flux, Gs 340 330

T (K)

320 310 300 290 280 270

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

Problem 6.4 The Redlich-Kwong equation of state is given by 𝑅𝑅𝑅𝑅 𝑎𝑎 𝑝𝑝 = − 𝑣𝑣 − 𝑏𝑏 𝑣𝑣(𝑣𝑣 + 𝑏𝑏)√𝑇𝑇 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 𝑅𝑅 2 𝑇𝑇𝑐𝑐2.5 𝑇𝑇𝑐𝑐 𝑎𝑎 = 0.427 , 𝑏𝑏 = 0.0866𝑅𝑅 𝑝𝑝𝑐𝑐 𝑝𝑝𝑐𝑐 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

8

x 10

6 4 root 2

f(v)

0 -2 -4 -6 -8 -10

0

0.5

1

1.5

2

2.5

3

v (m3/kg)

-3

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

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 (3𝑅𝑅 − ℎ) 𝑉𝑉 = 𝜋𝜋ℎ2 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 (3𝑅𝑅−ℎ)

(a) 𝑓𝑓(ℎ) = 𝜋𝜋ℎ2 − 𝑉𝑉 3 2 (6𝑅𝑅ℎ − 3ℎ ) 𝑑𝑑𝑑𝑑 = 𝜋𝜋 = 𝜋𝜋(2𝑅𝑅ℎ − ℎ2 ) 𝑑𝑑ℎ 3 (3𝑅𝑅 − ℎ𝑖𝑖 ) 𝜋𝜋ℎ𝑖𝑖 2 − 𝑉𝑉 𝑓𝑓(ℎ𝑖𝑖 ) 3 ℎ𝑖𝑖+1 = ℎ𝑖𝑖 − ′ = ℎ𝑖𝑖 − 𝑓𝑓 (ℎ𝑖𝑖 ) 𝜋𝜋�2𝑅𝑅ℎ𝑖𝑖 − ℎ𝑖𝑖 2 � (b) 0 < ℎ < 2𝑅𝑅 (c) ℎ1 = 2.376

6

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, 𝑆𝑆𝑜𝑜𝑜𝑜𝑜𝑜 . The empirical formula for the optimal spacing is 𝑆𝑆𝑜𝑜𝑜𝑜𝑜𝑜 2 + 𝑆𝑆𝑜𝑜𝑜𝑜𝑜𝑜 /𝐷𝐷 𝐻𝐻 1/3 −1/4 � � = 2.75 � � 𝑅𝑅𝑅𝑅 𝐷𝐷 �1 + 𝑆𝑆 /𝐷𝐷�2/3 𝐷𝐷 𝑜𝑜𝑜𝑜𝑜𝑜

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 𝑆𝑆𝑜𝑜𝑜𝑜𝑜𝑜 /𝐷𝐷 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 H/D = 5 H/D = 10 H/D = 15 H/D = 20

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

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 𝑅𝑅𝑅𝑅𝐷𝐷 for various values of the relative roughness ε/D. Variable definitions are: 𝑑𝑑𝑑𝑑 2 𝑓𝑓 = 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 = − 𝐷𝐷/(𝜌𝜌 𝑢𝑢𝑚𝑚 /2) 𝑑𝑑𝑑𝑑 P = pressure (N/m2), x = distance along pipe (m), ε = roughness (m), D = pipe diameter (m), 𝑅𝑅𝑅𝑅𝐷𝐷 = 𝑢𝑢𝑚𝑚 𝐷𝐷/𝜈𝜈, 𝑢𝑢𝑚𝑚 = mean fluid velocity (m/s), ν = viscosity (m2/s). A correlation between f, 𝑅𝑅𝑅𝑅𝐷𝐷 and ε/D was developed by Colebrook for fully developed turbulent pipe flow 1 𝜖𝜖/𝐷𝐷 2.51 = −2𝑙𝑙𝑙𝑙𝑙𝑙10 � + � 3.7 𝑅𝑅𝑅𝑅𝐷𝐷 �𝑓𝑓 �𝑓𝑓 This equation is valid for 𝑅𝑅𝑅𝑅𝐷𝐷 > 3000. For 𝑅𝑅𝑅𝑅𝐷𝐷 < 3000, the flow is laminar flow and the friction factor for all ε/D values is 𝑓𝑓 = 64/𝑅𝑅𝑅𝑅𝐷𝐷 . a) Write a function to compute the friction factor as a function of ε/D and 𝑅𝑅𝑅𝑅𝐷𝐷 . For 𝑅𝑅𝑅𝑅𝐷𝐷 >3000, the Colebrook correlation should be used. An initial guess at the root can be obtained from 6.9 𝜖𝜖/𝐷𝐷 1.11 𝑓𝑓 = �1.8𝑙𝑙𝑙𝑙𝑙𝑙10 � +� � �� 𝑅𝑅𝑅𝑅𝐷𝐷 3.7

−2

For 𝑅𝑅𝑅𝑅𝐷𝐷 <3000, the laminar flow formula, 𝑓𝑓 = 64/𝑅𝑅𝑅𝑅𝐷𝐷 , 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 ε/D values shown in the figure.

Solution -1

10

friction factor, f

0 1e-005 5e-005 0.0001 0.0002 0.0004 0.0006 0.0008 0.001 0.002 0.004 0.008 0.01 0.015 0.02 0.03 0.04 0.05

-2

10

3

10

4

10

6

5

10

10

7

10

8

10

ReD = u*D/vis

8

Problem 6.8: Enzyme Kinetics The Michaelis-Menten model describes the kinetics of enzyme mediated reactions: 𝑆𝑆 𝑑𝑑𝑑𝑑 = −𝑣𝑣𝑚𝑚 𝑘𝑘𝑠𝑠 + 𝑆𝑆 𝑑𝑑𝑑𝑑

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 𝑆𝑆 = 𝑆𝑆0 − 𝑣𝑣𝑚𝑚 𝑡𝑡 + 𝑘𝑘𝑠𝑠 𝑙𝑙𝑙𝑙(𝑆𝑆0 /𝑆𝑆)

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

Problem 6.9: Basics a) Determine all the roots of 𝑓𝑓(𝑥𝑥) = −14 − 20𝑥𝑥 + 19𝑥𝑥 2 − 3𝑥𝑥 3 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 𝑥𝑥𝑙𝑙 = −1 and 𝑥𝑥𝑢𝑢 = 0, and a stopping criteria of |𝑒𝑒𝑎𝑎 | ≤ 1%.

Solution a) Graphical 30

1.5 20

1 0.5

f(x)

f(x)

10

0

0

-0.5 -10

-1 -20 -1

0

1

-0.6

5

4

3

2 x

-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

10

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

Problem 6.10: Fixed Points of an ODE Consider the first order ODE: 𝑑𝑑𝑑𝑑/𝑑𝑑𝑑𝑑 = 𝑒𝑒 𝑥𝑥 − 𝑐𝑐𝑐𝑐𝑐𝑐(𝑥𝑥). 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 𝑥𝑥0 = 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) 𝑓𝑓(𝑥𝑥) = 𝑒𝑒 𝑥𝑥 − 𝑐𝑐𝑐𝑐𝑐𝑐(𝑥𝑥) 𝑑𝑑𝑑𝑑 = 𝑒𝑒 𝑥𝑥 + 𝑠𝑠𝑠𝑠𝑠𝑠(𝑥𝑥) 𝑑𝑑𝑑𝑑 𝑓𝑓(𝑥𝑥𝑖𝑖 ) 𝑒𝑒 𝑥𝑥𝑖𝑖 − 𝑐𝑐𝑐𝑐𝑐𝑐(𝑥𝑥𝑖𝑖 ) 𝑥𝑥𝑖𝑖+1 = 𝑥𝑥𝑖𝑖 − ′ = 𝑥𝑥𝑖𝑖 − 𝑥𝑥 𝑓𝑓 (𝑥𝑥𝑖𝑖 ) 𝑒𝑒 𝑖𝑖 + 𝑠𝑠𝑠𝑠𝑠𝑠(𝑥𝑥𝑖𝑖 ) d) 𝑥𝑥 = 0

12

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 (𝑥𝑥, 𝑦𝑦) coordinates displayed in the figure.

The trajectory can be derived as 𝑔𝑔 𝑦𝑦 = 𝑦𝑦0 + 𝑥𝑥 tan(𝜃𝜃0 ) − 2𝑣𝑣2 cos(𝜃𝜃 )2 𝑥𝑥 2 .

0

0

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 𝑦𝑦(𝑡𝑡) 𝑣𝑣𝑣𝑣 𝑥𝑥(𝑡𝑡) 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. 𝑑𝑑 2 𝑥𝑥 𝑚𝑚𝑚𝑚𝑥𝑥 = ∑𝐹𝐹𝑥𝑥 𝑚𝑚 2 = 0 𝑑𝑑𝑡𝑡 Integrate twice and use initial conditions to get 𝑥𝑥(𝑡𝑡) = 𝑣𝑣0 ∗ Cos[𝜃𝜃0 ] ∗ 𝑡𝑡 Now in the y-direction 𝑑𝑑2 𝑦𝑦 𝑚𝑚 2 = −𝑚𝑚𝑚𝑚 𝑑𝑑𝑡𝑡 Integrate twice and use initial conditions to get 𝑡𝑡 2 𝑦𝑦(𝑡𝑡) = −𝑔𝑔 ∗ + 𝑣𝑣0 ∗ Sin[𝜃𝜃0 ] ∗ 𝑡𝑡 + 𝑦𝑦0 2 Combine these to get 𝑔𝑔 𝑦𝑦 = 𝑦𝑦0 + 𝑥𝑥 tan(𝜃𝜃0 ) − 𝑥𝑥 2 2 2𝑣𝑣0 cos(𝜃𝜃0 )2 (c)

13

Trajectory of a Projectile 50 15 30 45 60 75

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 θ0. 𝑔𝑔 𝑓𝑓(𝜃𝜃0 ) = 𝑦𝑦 − 𝑦𝑦0 + 𝑥𝑥 tan(𝜃𝜃0 ) − 𝑥𝑥 2 = 0 2 2 ) 2𝑣𝑣0 cos(𝜃𝜃0 (e) Numerical Results

90 25 35 45

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.

14

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 (𝑥𝑥0 , 𝑦𝑦0 ) = (0,0). a) 𝑓𝑓(𝑥𝑥, 𝑦𝑦) = 2𝑥𝑥 − 𝑦𝑦 = 0 𝑔𝑔(𝑥𝑥, 𝑦𝑦) = 𝑥𝑥 + 𝑦𝑦 − 3 = 0 b) 𝑓𝑓(𝑥𝑥, 𝑦𝑦) = 𝑦𝑦 − 2 = 0 𝑔𝑔(𝑥𝑥, 𝑦𝑦) = 𝑥𝑥 2 − 𝑦𝑦 = 0 Solution Use Equations (6.35) and (6.36)

𝑥𝑥𝑖𝑖+1 = 𝑥𝑥𝑖𝑖 − (a)

(b)

𝑓𝑓𝑖𝑖

𝜕𝜕𝑔𝑔𝑖𝑖 𝜕𝜕𝑓𝑓 𝜕𝜕𝑓𝑓 𝜕𝜕𝑔𝑔 − 𝑔𝑔𝑖𝑖 𝑖𝑖 𝑔𝑔𝑖𝑖 𝑖𝑖 − 𝑓𝑓𝑖𝑖 𝑖𝑖 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 , 𝑦𝑦𝑖𝑖+1 = 𝑦𝑦𝑖𝑖 − 𝐽𝐽𝑖𝑖 𝐽𝐽𝑖𝑖

𝐽𝐽 = Jacobian =

𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 − 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕

𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 =2, = −1 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 =1, =1 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 𝐽𝐽 = Jacobian = 3 (0)(1) − (−3)(−1) (−3)(2) − (0)(1) 𝑥𝑥1 = 0 − = 1, 𝑦𝑦1 = 0 − =2 3 3 𝜕𝜕𝜕𝜕 =0, 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 = 2𝑥𝑥 , 𝜕𝜕𝜕𝜕

𝜕𝜕𝜕𝜕 =1 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 = −1 𝜕𝜕𝜕𝜕

𝐽𝐽 = Jacobian = −2𝑥𝑥

Cannot start using an initial guess of 𝑥𝑥0 = 0 since the Jacobian is zero and you get division by zero.

15

Problem 6.13: Graphical Solutions Consider the function 𝑓𝑓(𝑥𝑥) = 𝑎𝑎𝑎𝑎 + 𝑏𝑏 − sin(𝑥𝑥)

a) Graphically show the roots of 𝑓𝑓(𝑥𝑥) = 0 when a = b = 0. What are the roots? b) Graphically show the roots 𝑓𝑓(𝑥𝑥) = 0 when a = 0 for various values of b. Show the roots by plotting 𝑓𝑓(𝑥𝑥) vs x. On a separate plot, show the roots by plotting both 𝑎𝑎𝑎𝑎 + 𝑏𝑏 𝑎𝑎𝑎𝑎𝑎𝑎 sin(𝑥𝑥) 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 𝑓𝑓(𝑥𝑥) = 0 for • a = 0.1 and b = 0 with an initial guess of 𝑥𝑥0 = 4.

• a = 0.5 and b = 1 with an initial guess of 𝑥𝑥0 = 4.

• a = 0.5 and b = 1 with an initial guess of 𝑥𝑥0 = −4.

Solution a) Roots are located at 𝑛𝑛𝑛𝑛.

b)

c)

d) 2.852, Trouble converging, −2.754

16

Problem 6.14: Simple Interest The formula for simple interest is

𝑖𝑖(1 + 𝑖𝑖)𝑛𝑛 𝐴𝐴 = 𝑃𝑃 (1 + 𝑖𝑖)𝑛𝑛 − 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.

Solution a) Need to solve the following for i. 𝑖𝑖(1 + 𝑖𝑖)𝑛𝑛 𝑓𝑓(𝑖𝑖) = 𝑃𝑃 − 𝐴𝐴 = 0 (1 + 𝑖𝑖)𝑛𝑛 − 1

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

17

Problem 6.15: Non-Linear Springs Consider the spring and block system shown with non-linear springs k3

k2

k1 M1

M2

k4 M3

x1+L1 LT

Non-linear springs can be described by 𝐹𝐹𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 = 𝑘𝑘𝑘𝑘 + 𝛾𝛾𝑥𝑥 3 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., 𝐹𝐹𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 = 𝑘𝑘𝑘𝑘. 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, 𝛾𝛾𝑖𝑖 , 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 𝛾𝛾𝑖𝑖 = 0 𝑁𝑁/𝑚𝑚3

Solution (a) Mathematical Model Total length Mass-1 Mass-2 Mass-3

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 𝑁𝑁/𝑚𝑚3

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 𝑁𝑁/𝑚𝑚3

𝑥𝑥1 + 𝑥𝑥2 + 𝑥𝑥3 + 𝑥𝑥4 = 𝐿𝐿𝑇𝑇 − 𝐿𝐿1 − 𝐿𝐿2 − 𝐿𝐿3 − 𝐿𝐿4 − 3𝑊𝑊 −(𝑘𝑘1 𝑥𝑥1 + 𝛾𝛾1 𝑥𝑥13 ) + (𝑘𝑘2 𝑥𝑥2 + 𝛾𝛾2 𝑥𝑥23 ) = 0

−(𝑘𝑘2 𝑥𝑥2 + 𝛾𝛾2 𝑥𝑥23 ) + (𝑘𝑘3 𝑥𝑥3 + 𝛾𝛾3 𝑥𝑥33 ) = 0 −(𝑘𝑘3 𝑥𝑥3 + 𝛾𝛾3 𝑥𝑥33 ) + (𝑘𝑘4 𝑥𝑥4 + 𝛾𝛾4 𝑥𝑥43 ) = 0

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

18

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 𝑑𝑑𝑑𝑑 = 𝑏𝑏 𝑑𝑑𝑑𝑑 Draw sketches for 𝑏𝑏 < 0, 𝑏𝑏 = 0, 𝑎𝑎𝑎𝑎𝑎𝑎 𝑏𝑏 > 0.

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

1

Problem 7.2 𝑑𝑑𝑑𝑑 = −𝜃𝜃 𝑑𝑑𝑑𝑑

Solution 𝜃𝜃 ∗ = 0: stable

Problem 7.3 𝑑𝑑𝑑𝑑 = −𝑎𝑎 ∙ 𝜃𝜃 𝑑𝑑𝑡𝑡

Draw sketches corresponding to a zero, low, and high value of the parameter a.

Solution 𝜃𝜃 ∗ = 0: stable for a >=0

2

Problem 7.4 𝑑𝑑𝑑𝑑 = −𝑎𝑎 ∙ 𝜃𝜃 + 𝑏𝑏 𝑑𝑑𝑑𝑑

Draw sketches corresponding to zero, low and high values of the parameter b.

Solution 𝜃𝜃 ∗ = 𝑏𝑏/𝑎𝑎: Stable

Problem 7.5 𝑑𝑑𝑑𝑑 = 𝑟𝑟 ∙ 𝜃𝜃(1 − 𝜃𝜃) 𝑑𝑑𝑑𝑑

r >0. Draw anticipated solution (θ(t) vs. t) starting from several initial conditions.

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

3

Problem 7.6 Consider the ODE

𝑑𝑑𝑑𝑑 = 𝑟𝑟 ∙ 𝜃𝜃(1 − 𝜃𝜃) + 𝑆𝑆 𝑑𝑑𝑑𝑑

r is a positive constant. a) Find all fixed points and classify their stability (stable or unstable). Are there any critical values of S? 𝑑𝑑𝑑𝑑 b) Plot the phase portraits ( 𝑑𝑑𝑑𝑑 vs. 𝜃𝜃). c) Plot the anticipated solution (𝜃𝜃(𝑡𝑡) vs. t). Draw sketches starting from different initial positions.

Solution S Range 𝑆𝑆 > −

𝑟𝑟 4

Fixed Points 1 𝜃𝜃1∗ = �1 − �1 + 4𝑆𝑆/𝑟𝑟� 2 1 𝜃𝜃2∗ = �1 + �1 + 4𝑆𝑆/𝑟𝑟� 2 1 𝜃𝜃1∗ = 𝜃𝜃2∗ = 2 No real fixed points

Stability of Fixed Points 𝜃𝜃1∗ is Unstable 𝜃𝜃2∗ is Stable

Critical S neutrally stable 𝑆𝑆 = 𝑆𝑆𝑐𝑐 = −𝑟𝑟/4 𝑟𝑟 NA 𝑆𝑆 < − 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

Problem 7.7 Consider the linear first order ODE

𝑑𝑑𝑑𝑑 1 + (𝜃𝜃 − 𝜃𝜃∞ ) = 𝑆𝑆𝑐𝑐 𝑑𝑑𝑑𝑑 𝜏𝜏 𝜃𝜃 = 𝜃𝜃0 , 𝑡𝑡 = 0 The parameters τ, θ∞, and Sc are constants. Assume τ >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 θ∞=0 with a pulsed, time dependent source 𝑆𝑆(𝑡𝑡) = 𝑆𝑆𝑐𝑐 (𝐻𝐻(𝑡𝑡) − 𝐻𝐻(𝑡𝑡 − 𝑡𝑡1 )) Solution

Phase Portrait dθ/dt

θ*

θ

Fixed Point 𝜃𝜃 = 𝜃𝜃∞ + 𝜏𝜏𝑆𝑆𝑐𝑐

This is the only fixed point and it is stable.

Solution Sketches θ

θ*

t

5

Problem 7.8: Tumor Growth The growth of cancerous tumors can be modeled by the Gompertz law 𝑑𝑑𝑑𝑑 = −𝑎𝑎 ∙ 𝑁𝑁 ∙ 𝑙𝑙𝑙𝑙(𝑏𝑏𝑏𝑏) 𝑑𝑑𝑑𝑑 where 𝑁𝑁(𝑡𝑡) 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 𝑁𝑁(𝑡𝑡) 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

Problem 7.9 Consider the model chemical reaction

𝑘𝑘1

�⎯⎯� 𝐴𝐴 + 𝑋𝑋 2𝑋𝑋 �� 𝑘𝑘−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 𝑑𝑑𝑑𝑑 = 𝑘𝑘1 𝑎𝑎𝑎𝑎 − 𝑘𝑘−1 𝑥𝑥 2 𝑑𝑑𝑑𝑑 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 𝑘𝑘1 𝑎𝑎𝑎𝑎 − 𝑘𝑘−1 𝑥𝑥 2 = 0 𝑥𝑥1∗ = 0 : unstable 𝑥𝑥2∗ = 𝑘𝑘1 𝑎𝑎/𝑘𝑘−1 : stable c) Solution Sketches

7

Problem 7.10: Chemical Kinetics Consider the chemical reaction system

𝑘𝑘1

�⎯⎯� 𝐴𝐴 + 𝑋𝑋 2𝑋𝑋 �� 𝑘𝑘−1 𝑘𝑘2

𝑋𝑋 + 𝐵𝐵 �⎯� 𝐶𝐶 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 𝑑𝑑𝑑𝑑/𝑑𝑑𝑑𝑑 = 𝑐𝑐1 𝑥𝑥 − 𝑐𝑐2 𝑥𝑥 2 , 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 𝑘𝑘2 𝑏𝑏 > 𝑘𝑘1 𝑎𝑎, 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

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

𝑑𝑑𝑑𝑑 = 𝑐𝑐1 𝑥𝑥 − 𝑐𝑐2 𝑥𝑥 2 𝑑𝑑𝑑𝑑 𝑐𝑐1 = 𝑘𝑘1 𝑎𝑎 − 𝑘𝑘2 𝑏𝑏, 𝑐𝑐2 = 𝑘𝑘−1

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 𝑘𝑘1 𝑎𝑎 < 𝑘𝑘2 𝑏𝑏 (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

Problem 7.11: Allee Effect For certain species of organisms, the effective growth rate 𝑁𝑁̇/𝑁𝑁 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. 𝑁𝑁̇

a) Show that 𝑁𝑁 = 𝑟𝑟 − 𝑎𝑎(𝑁𝑁 − 𝑏𝑏)2 provides an example of Allee effect, if r, a, and b satisfy certain constraints, to be determined. b) Find all the fixed points of the system and classify their stability. c) Plot the solutions 𝑁𝑁(𝑡𝑡) for different initial conditions. d) Compare the solutions 𝑁𝑁(𝑡𝑡) to those found for the logistic equation. What are the qualitative differences, if any?

Solution b) Fixed Points

c) Numerical Solution

𝑁𝑁 ∗ = 0 𝑁𝑁 ∗ = 𝑏𝑏 ± �𝑟𝑟/𝑎𝑎

Unstable Stable

d) Comparison with Logistic Equation

9

Problem 7.12: Slope Fields Consider an autonomous first order ODE: 𝑑𝑑𝑑𝑑 = 𝑓𝑓(𝑥𝑥) 𝑑𝑑𝑑𝑑 The slope field is shown.

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

Solution a) This vector field seems to be generated from a function such as 𝑓𝑓(𝑥𝑥, 𝑡𝑡) = (𝑥𝑥 + 1)(𝑥𝑥 − 1) b) Solution

d) Phase portrait

10

Chapter 8

Laplace Transforms

Problem 8.1: Linear First Order ODE Consider the linear first- order ODE where τ is constant. 𝑑𝑑𝑑𝑑 1 + 𝜃𝜃(𝑡𝑡) = 𝑓𝑓(𝑡𝑡) 𝑑𝑑𝑑𝑑 𝜏𝜏 1.

2.

3.

𝜃𝜃 = 𝜃𝜃0 , 𝑡𝑡 = 0

Determine the analytical solution using the Laplace transform method for 𝑓𝑓(𝑡𝑡) = 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 𝜃𝜃0 = 0 where fc is a constant and H is the unit step function. Plot each case. a.

Step input 𝑓𝑓(𝑡𝑡) = 𝑓𝑓𝑐𝑐 𝐻𝐻(𝑡𝑡 − 𝑡𝑡0 ) b.

Pulsed input 𝑓𝑓(𝑡𝑡) = 𝑓𝑓𝑐𝑐 �𝐻𝐻(𝑡𝑡 − 𝑡𝑡1 ) − 𝐻𝐻(𝑡𝑡 − 𝑡𝑡2 )� c.

Impulse or Delta function 𝑓𝑓(𝑡𝑡) = 𝑓𝑓𝑐𝑐 𝛿𝛿(𝑡𝑡 − 𝑡𝑡1 ) d.

Harmonic forcing 𝑓𝑓(𝑡𝑡) = 𝑓𝑓𝑐𝑐 𝑠𝑠𝑠𝑠𝑠𝑠(𝜔𝜔𝜔𝜔)

Solution

1) Effect of θ0 Only 𝜃𝜃(𝑡𝑡) = 𝜃𝜃0 𝑒𝑒 −𝑡𝑡⁄𝜏𝜏

t 2

1

1

2

3

4

t

1

2

1

2) Transfer Function ~

𝜃𝜃(𝑠𝑠)

1 𝐺𝐺 (𝑠𝑠) = ~ = 1 𝑓𝑓 (𝑠𝑠) 𝑠𝑠 + 𝜏𝜏

3a) Step Input

𝜃𝜃𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 (𝑡𝑡) = 𝑓𝑓𝑐𝑐 𝜏𝜏(1 − 𝑒𝑒𝑒𝑒𝑒𝑒(− t

1.0

𝑡𝑡 − 𝑡𝑡0 ))𝐻𝐻(𝑡𝑡 − 𝑡𝑡0 ) 𝜏𝜏

0.8 0.6 0.4 0.2

0

1

2

3

4

5

6

t

3b) Pulse 𝜃𝜃𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 (𝑡𝑡) = 𝑓𝑓𝑐𝑐 𝜏𝜏(1 − 𝑒𝑒𝑒𝑒𝑒𝑒(− t

𝑡𝑡 − 𝑡𝑡1 𝑡𝑡 − 𝑡𝑡2 ))𝐻𝐻(𝑡𝑡 − 𝑡𝑡1 ) − 𝑓𝑓𝑐𝑐 𝜏𝜏(1 − 𝑒𝑒𝑒𝑒𝑒𝑒(− ))𝐻𝐻(𝑡𝑡 − 𝑡𝑡2 ) 𝜏𝜏 𝜏𝜏

1.5

1.0

0.5

0

2

4

6

3c) Impulse or Delta function t 1.0

𝜃𝜃𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 (𝑡𝑡) = 𝑓𝑓𝑐𝑐 𝑒𝑒𝑒𝑒𝑒𝑒(−

8

t

𝑡𝑡 − 𝑡𝑡0 )𝐻𝐻(𝑡𝑡 − 𝑡𝑡0 ) 𝜏𝜏

0.8

0.6

0.4

0.2

0

1

2

3

4

5

6

t

2

3d) Harmonic forcing 𝑡𝑡

𝑒𝑒 −𝜏𝜏 𝜏𝜏 2 −𝜏𝜏 2 𝜔𝜔𝜔𝜔𝜔𝜔𝜔𝜔[𝑡𝑡𝑡𝑡] + 𝜏𝜏𝜏𝜏𝜏𝜏𝜏𝜏[𝑡𝑡𝑡𝑡] 𝜃𝜃(𝑡𝑡) = 𝑓𝑓𝑐𝑐 𝜔𝜔( + ) 1 + 𝜏𝜏 2 𝜔𝜔 2 𝜔𝜔(1 + 𝜏𝜏 2 𝜔𝜔 2 ) t

1.0

0.5

5

10

15

20

t

0.5

1.0

3

Problem 8.2 Consider the linear second order ODE where k is constant. 𝑥𝑥̈ + 𝑘𝑘𝑘𝑘(𝑡𝑡) = 𝑓𝑓(𝑡𝑡) 𝑥𝑥(0) = 𝑥𝑥0 𝑥𝑥̇ (0) = 𝑣𝑣0 1. Determine the analytical solution using the Laplace transform method when 𝑓𝑓(𝑡𝑡) = 0. Create a meaningful plot of this solution.

2. Determine the transfer function 𝐺𝐺(𝑠𝑠).

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, 𝑓𝑓(𝑡𝑡) = 𝑓𝑓𝑐𝑐 𝐻𝐻(𝑡𝑡 − 𝑡𝑡0 ) b. Pulsed input, 𝑓𝑓(𝑡𝑡) = 𝑓𝑓𝑐𝑐 �𝐻𝐻(𝑡𝑡 − 𝑡𝑡1 ) − 𝐻𝐻(𝑡𝑡 − 𝑡𝑡2 )� c. Impulse or Delta function, 𝑓𝑓(𝑡𝑡) = 𝑓𝑓𝑐𝑐 𝛿𝛿(𝑡𝑡 − 𝑡𝑡1 ) d. Harmonic forcing, 𝑓𝑓(𝑡𝑡) = 𝑓𝑓𝑐𝑐 𝑠𝑠𝑠𝑠𝑠𝑠(𝜔𝜔𝜔𝜔) Solution

𝑘𝑘

𝑚𝑚

𝑘𝑘

1) 𝑥𝑥 (𝑡𝑡) = 𝑥𝑥0 cos(�𝑚𝑚 𝑡𝑡) + 𝑣𝑣0 � 𝑘𝑘 sin(�𝑚𝑚 𝑡𝑡) 𝑋𝑋(𝑠𝑠)

1

2) 𝐺𝐺 (𝑠𝑠) = 𝐹𝐹(𝑠𝑠) = 𝑚𝑚𝑠𝑠2 +𝑘𝑘 3a) 𝑥𝑥(𝑡𝑡) = 𝑓𝑓𝑐𝑐

(1−Cos[ 𝑘𝑘

1

√𝑘𝑘𝑡𝑡 ]) √𝑚𝑚

3b) 𝑥𝑥(𝑡𝑡) = 𝑓𝑓𝑐𝑐 𝑘𝑘 (1 − Cos[ 3c) 𝑥𝑥(𝑡𝑡) = 𝑓𝑓𝑐𝑐

3d) 𝑥𝑥(𝑡𝑡) = 𝑓𝑓𝑐𝑐

1

√𝑘𝑘√𝑚𝑚

Sin[

1

√𝑘𝑘𝑡𝑡 √𝑘𝑘(𝑡𝑡−Δt) ] + (−1 + Cos[ ])H[𝑡𝑡 − Δt]) 𝑚𝑚 √ √𝑚𝑚

√𝑘𝑘𝑡𝑡 ] √𝑚𝑚

√𝑘𝑘(𝑘𝑘−𝑚𝑚𝑤𝑤 2 )

(−√𝑚𝑚𝑤𝑤Sin[

√𝑘𝑘𝑡𝑡 ] + √𝑘𝑘Sin[𝑡𝑡𝑡𝑡]) √𝑚𝑚

4

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. 𝑑𝑑𝑑𝑑 = 𝑦𝑦 ∙ 𝑡𝑡 2 − 𝑎𝑎 ∙ 𝑦𝑦 𝑑𝑑𝑑𝑑 𝑦𝑦(0) = 𝑦𝑦0 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. 𝑑𝑑𝑑𝑑 = (𝑡𝑡 2 − 𝑎𝑎)𝑑𝑑𝑡𝑡 𝑦𝑦 This equation can now be integrated directly. 𝑑𝑑𝑑𝑑 � = �(𝑡𝑡 2 − 𝑎𝑎)𝑑𝑑𝑑𝑑 𝑦𝑦 𝑡𝑡 3

ln(𝑦𝑦) = 3 − 𝑎𝑎𝑎𝑎+c 𝑡𝑡 3 𝑦𝑦(𝑡𝑡) = exp � − 𝑎𝑎𝑎𝑎 + 𝑐𝑐� 3 The unknown constant is determined by forcing the solution to satisfy the initial condition. 𝑦𝑦0 = exp(𝑐𝑐) Thus 𝑡𝑡 3 𝑦𝑦(𝑡𝑡) = 𝑦𝑦0 exp � − 𝑎𝑎𝑎𝑎� 3

1

2 exact Euler,h=0.5 Euler,h=0.25 Heun,h=0.5

1.8 1.6 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

Problem 9.2: Euler’s Method Consider the differential equation

𝑑𝑑𝑑𝑑 = −𝑦𝑦 + 2 𝑑𝑑𝑑𝑑 a) Sketch the phase diagram (𝑑𝑑𝑑𝑑⁄𝑑𝑑𝑑𝑑 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 ≤ 𝑡𝑡 ≤ 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

b) Manual Calculations

i 1 2 3 4

𝑡𝑡𝑖𝑖 0 t1+h = 0.5 t2+h = 1.0 t3+h = 1.5

y

0

t

𝑦𝑦𝑖𝑖+1 = 𝑦𝑦𝑖𝑖 + ℎ ∙ 𝑓𝑓(𝑡𝑡𝑖𝑖 , 𝑦𝑦𝑖𝑖 ) = 𝑦𝑦𝑖𝑖 + ℎ ∙ (−𝑦𝑦𝑖𝑖 + 2)

𝑦𝑦𝑖𝑖 0 𝑦𝑦2 = 𝑦𝑦1 + ℎ ∙ (−𝑦𝑦1 + 2) = 0 + 0.5(0 + 2) = 1 𝑦𝑦3 = 𝑦𝑦2 + ℎ ∙ (−𝑦𝑦2 + 2) = 1 + 0.5(−1 + 2) = 1.5 𝑦𝑦4 = 𝑦𝑦3 + ℎ ∙ (−𝑦𝑦3 + 2) = 1.5 + 0.5(−1.5 + 2) = 1.75

3

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

3

2

5

4

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

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 𝑑𝑑𝑑𝑑 𝜌𝜌𝜌𝜌𝜌𝜌 = −ℎ𝐴𝐴𝑠𝑠 (𝑇𝑇 − 𝑇𝑇∞ ) + 𝑔𝑔𝑔𝑔 𝑑𝑑𝑑𝑑 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, ρ = 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, ρ, c, h

⇒

Response t, T

⇒

Note that the steady state temperature can be determined form 𝑑𝑑𝑑𝑑 𝜌𝜌𝜌𝜌𝜌𝜌 = −ℎ𝐴𝐴𝑠𝑠 (𝑇𝑇 − 𝑇𝑇∞ ) + 𝑔𝑔𝑔𝑔 = 0 𝑑𝑑𝑑𝑑 𝑔𝑔𝑔𝑔 𝑇𝑇 ∗ = 𝑇𝑇∞ + ℎ𝐴𝐴𝑠𝑠 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

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 𝑔𝑔𝑔𝑔 𝑇𝑇 ∗ = 𝑇𝑇∞ + = 65°𝐶𝐶 ℎ𝐴𝐴𝑠𝑠 Also, the solution is a linear function of the heat source. Thus, doubling g doubles the temperature rise above T∞.

6

Problem 9.4: Population Model The earth’s population can be estimated using the following simplified model 𝑑𝑑𝑑𝑑 = 𝑎𝑎𝑎𝑎 − 𝑏𝑏𝑃𝑃2 𝑑𝑑𝑑𝑑 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, 𝑑𝑑𝑑𝑑⁄𝑑𝑑𝑑𝑑 vs. P. Using the phase plot, sketch the anticipated population history, 𝑃𝑃(𝑡𝑡) 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 𝑑𝑑𝑑𝑑 Steady state occurs when 𝑑𝑑𝑑𝑑 = 0. Thus P = 0 and 𝑃𝑃 = 𝑎𝑎/𝑏𝑏 are possible steady states. The stable steady state is 𝑃𝑃 = 𝑎𝑎/𝑏𝑏 b) Phase Plot and Anticipated Solution dP/dt

0

a/b

P

7

d) 9

10

x 10

9

Earths Population

8 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 Effect of Birth Rate Parameter, a

9

14

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

12

x 10

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

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. 𝑑𝑑𝑑𝑑 𝑑𝑑2 𝑥𝑥 + 𝑘𝑘 ∙ 𝑥𝑥 = 0 𝑚𝑚 2 + 𝑐𝑐 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑥𝑥 = 𝑥𝑥0 𝑑𝑑𝑑𝑑 � 𝑡𝑡 = 0 = 𝑣𝑣0 𝑑𝑑𝑑𝑑 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

4

6

8

10

12

14

16

18

20

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

Problem 9.6: Linear Oscillator with Forcing Consider the forced spring-mass-damper system, described by the following differential equation and initial conditions. 𝑑𝑑𝑑𝑑 𝑑𝑑 2 𝑥𝑥 + 𝑘𝑘 ∙ 𝑥𝑥 = 𝐹𝐹(𝑡𝑡) 𝑚𝑚 2 + 𝑐𝑐 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑥𝑥 = 𝑥𝑥0 , 𝑣𝑣 = 𝑣𝑣0 , 𝑤𝑤ℎ𝑒𝑒𝑒𝑒 𝑡𝑡 = 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) 𝐹𝐹(𝑡𝑡) = 𝐹𝐹0 F0 = 10 N 𝐹𝐹(𝑡𝑡) = 𝐹𝐹0 sin(𝜔𝜔𝜔𝜔) F0 = 5 N/kg, 𝜔𝜔 = 1 rad/s 𝐹𝐹(𝑡𝑡) = 𝐹𝐹0 (𝐻𝐻(𝑡𝑡) − 𝐻𝐻(𝑡𝑡 − 𝑡𝑡𝑜𝑜𝑜𝑜 )) 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

10

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

Problem 9.7: Pendulum The oscillations of an undamped pendulum can be simulated with the following nonlinear equation 𝑑𝑑2 𝜃𝜃 𝑔𝑔 + sin(𝜃𝜃) = 0 𝑑𝑑𝑑𝑑 2 𝐿𝐿 where θ = 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 𝑑𝑑 2 𝜃𝜃 𝑔𝑔 + 𝜃𝜃 = 0 𝑑𝑑𝑑𝑑 2 𝐿𝐿 a) Reformulate these equations as two first order equations for θ and 𝑣𝑣 = 𝑑𝑑𝑑𝑑/𝑑𝑑𝑑𝑑. 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 θ and v as functions of time for both the linear and nonlinear models with L = 0.6 m and initial conditions θ0 = π/8 and v0 = 0. Plot the linear and nonlinear solutions on the same graph. Also plot results for θ0 = π/2, θ0 = 0.99π, and θ0 = π.

Solution a) First Order Equations In order to express as two first order equations, define y1 = θ y2 = v Nonlinear Model 𝑑𝑑𝑦𝑦1 = 𝑦𝑦2 𝑑𝑑𝑡𝑡 𝑑𝑑𝑦𝑦2 𝑔𝑔 = − sin(𝑦𝑦1 ) 𝑑𝑑𝑡𝑡 𝐿𝐿

Linear Model 𝑑𝑑𝑦𝑦1 = 𝑦𝑦2 𝑑𝑑𝑡𝑡 𝑑𝑑𝑦𝑦2 𝑔𝑔 = − 𝑦𝑦1 𝑑𝑑𝑡𝑡 𝐿𝐿

d) Numerical Results

theta0/pi = 0.5 2

0.2

1

Theta (rad)

Theta (rad)

theta0/pi = 0.125 0.4

0 -0.2 -0.4

0

1

2

3

4

5

6

7

8

9

0

1

2

3

4

5

6

7

8

9

10

10

Nonlinear Linear

1

Velocity (rad/s)

Velocity (rad/s)

-1 -2

10

2

0 -1 -2

0

0 -5 -10

0

1

2

3

4

5

Time (s)

6

7

8

9

10

Nonlinear Linear

5

0

1

2

3

4

5

6

7

8

9

10

Time (s)

12

theta0/pi = 1 4

2

2

Theta (rad)

Theta (rad)

theta0/pi = 0.99 4

0 -2 -4

0

1

2

3

4

5

6

7

8

9

0

1

2

3

4

5

6

7

8

9

10

20

Nonlinear Linear

10

Velocity (rad/s)

Velocity (rad/s)

-2 -4

10

20

0 -10 -20

0

0

1

2

3

4

5

Time (s)

6

7

8

9

10

Nonlinear Linear

10 0 -10 -20

0

1

2

3

4

5

6

7

8

9

10

Time (s)

Comments The linear model is more accurate for small angles. However, even at θ0 = π/8, a gradual phase shift between the linear and nonlinear models can be observed. At θ0 = π/2 (vertical starting position), the two models are quite different. At the top position, the two models are completely different.

13

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: 𝑑𝑑𝑑𝑑 𝑑𝑑 2 𝑦𝑦 2) + 𝜇𝜇(1 − 𝑦𝑦 + 𝑦𝑦 = 0 𝑑𝑑𝑑𝑑 2 𝑑𝑑𝑑𝑑 a) Reformulate these equations as two first order equations for y1 = y and 𝑦𝑦2 = 𝑑𝑑𝑑𝑑⁄𝑑𝑑𝑑𝑑. 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 𝑑𝑑𝑑𝑑

a) Using the definitions 𝑦𝑦1 = 𝑦𝑦 and 𝑦𝑦2 = 𝑑𝑑𝑡𝑡 , the original second order system can be decomposed into a set of first order equations. 𝑑𝑑𝑦𝑦2 𝑑𝑑𝑦𝑦1 = 𝑦𝑦2 , = −𝜇𝜇(1 − 𝑦𝑦1 2 )𝑦𝑦2 − 𝑦𝑦1 𝑑𝑑𝑡𝑡 𝑑𝑑𝑡𝑡 c) d) Numerical Results

14

Problem 9.9: Sky Diver The free-fall velocity of a parachutist can be estimated using the force balance 𝑐𝑐𝑑𝑑 𝑑𝑑𝑑𝑑 = 𝑔𝑔 − 𝑣𝑣 2 𝑚𝑚 𝑑𝑑𝑑𝑑 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 𝑔𝑔𝑔𝑔

The steady state occurs when the velocity no longer changes with time: 𝑣𝑣 ∗ = � 𝑐𝑐 Terminal velocity, chute closed = 56.0 m/s Terminal velocity, chute open = 12.5 m/s

𝑑𝑑

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

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 𝑑𝑑2 𝑥𝑥1 𝑚𝑚1 2 = −𝑘𝑘1 (𝑥𝑥1 − 𝐿𝐿1 ) + 𝑘𝑘2 (𝑥𝑥2 − 𝑥𝑥1 − 𝑤𝑤1 − 𝐿𝐿2 ) 𝑑𝑑𝑑𝑑 𝑑𝑑 2 𝑥𝑥2 𝑚𝑚2 = −𝑘𝑘2 (𝑥𝑥2 − 𝑥𝑥1 − 𝑤𝑤1 − 𝐿𝐿2 ) 𝑑𝑑𝑑𝑑 2 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 𝑥𝑥1,0 , 𝑥𝑥2,0 𝑣𝑣1,0 , 𝑣𝑣2,0

⇒

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 = −𝑘𝑘1 (𝑥𝑥1∗ − 𝐿𝐿1 ) + 𝑘𝑘2 (𝑥𝑥2∗ − 𝑥𝑥1∗ − 𝑤𝑤1 − 𝐿𝐿2 ) 0 = −𝑘𝑘2 (𝑥𝑥2∗ − 𝑥𝑥1∗ − 𝑤𝑤1 − 𝐿𝐿2 )

Arrange as a 2-by-2 matrix 𝑘𝑘 + 𝑘𝑘2 −𝑘𝑘2 𝑥𝑥1 𝑘𝑘 𝐿𝐿 − 𝑘𝑘2 (𝑤𝑤1 + 𝐿𝐿2 ) � 1 � �𝑥𝑥 � = � 1 1 � −𝑘𝑘2 𝑘𝑘2 𝑘𝑘2 (𝑤𝑤1 + 𝐿𝐿2 ) 2 Solve this 2-by-2 system using Cramer’s rule. 𝑥𝑥 ∗ 𝐿𝐿1 � 1∗ � = � � 𝐿𝐿1 + 𝑤𝑤1 + 𝐿𝐿2 𝑥𝑥2 The fixed points correspond to the upstretched positions of the springs.

16

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

10

x2

5

1

2

x , x (m)

15

0 -5

2

0

4

6

8

10

12

14

16

20

18

Velocities of a Coupled Spring/Mass System v1

5

v2

0

1

2

v , v (m)

10

-5

2

0

4

6

8

10 t (sec)

12

14

16

18

5

6

20

Phase Plot of a Coupled Spring/Mass System 13 12 11 10

2

x (m)

-10

9 8 7 6 5 -2

-1

0

1

2 x 1 (m)

3

4

17

Problem 9.11: An Epidemic The following ODE’s have been proposed as a model of an epidemic: 𝑑𝑑𝑑𝑑 = −𝑎𝑎𝑎𝑎𝑎𝑎 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 = 𝑎𝑎𝑎𝑎𝐼𝐼 − 𝑟𝑟𝑟𝑟 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 = 𝑟𝑟𝑟𝑟 𝑑𝑑𝑑𝑑 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 ρR, where ρ = 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

b) Improved Model With loss of immunity and reinfection, the equations become 𝑑𝑑𝑑𝑑 = −𝑎𝑎𝑎𝑎𝑎𝑎 + 𝜌𝜌𝜌𝜌 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 = 𝑎𝑎𝑎𝑎𝑎𝑎 − 𝑟𝑟𝑟𝑟 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 = 𝑟𝑟𝑟𝑟 − 𝜌𝜌𝜌𝜌 𝑑𝑑𝑑𝑑 In matrix form, the system of equations can be written as −𝑎𝑎𝑎𝑎𝑎𝑎 + 𝜌𝜌𝜌𝜌 𝑑𝑑 𝑆𝑆 � 𝐼𝐼 � = � 𝑎𝑎𝑎𝑎𝑎𝑎 − 𝑟𝑟𝑟𝑟 � 𝑑𝑑𝑑𝑑 𝑟𝑟𝑟𝑟 − 𝜌𝜌𝜌𝜌 𝑅𝑅 In order to input these equations systematically into the computer, redefine the variables as follows. 𝑦𝑦1 𝑆𝑆 𝑦𝑦 � 𝐼𝐼 � = � 2 � 𝑦𝑦3 𝑅𝑅 𝑦𝑦 −𝑎𝑎𝑦𝑦1 𝑦𝑦2 + 𝜌𝜌𝑦𝑦3 𝑑𝑑 1 �𝑦𝑦2 � = � 𝑎𝑎𝑦𝑦1 𝑦𝑦2 − 𝑟𝑟𝑦𝑦2 � 𝑑𝑑𝑑𝑑 𝑦𝑦 𝑟𝑟𝑦𝑦 − 𝜌𝜌𝑦𝑦 3

2

3

19

Chapter 10

First Order Ordinary Differential Equations Problem 10.1 The functions 𝜓𝜓1 (𝑡𝑡)and 𝜓𝜓2 (𝑡𝑡) are the solutions of the following simple, elementary problems. 𝜓𝜓1 (𝑡𝑡) 𝑑𝑑𝜓𝜓1 1 = − 𝜓𝜓1 + 1 𝑑𝑑𝑑𝑑 𝜏𝜏 𝜓𝜓1 = 0 , 𝑡𝑡 = 0

𝜓𝜓2 (𝑡𝑡) 𝑑𝑑𝜓𝜓2 1 = − 𝜓𝜓2 𝑑𝑑𝑑𝑑 𝜏𝜏 𝜓𝜓2 = 1 , 𝑡𝑡 = 0

Source, 𝑆𝑆(𝑡𝑡)

IC, 𝜃𝜃0

Express the solutions to the following problems in terms of the ψ1 and ψ2 functions. Be specific about the arguments. Do not actually solve for ψ1 and ψ2. A few cases are given as examples in order to clarify the objectives. 𝑑𝑑𝑑𝑑 1 = − 𝜃𝜃 + 𝑆𝑆(𝑡𝑡) 𝑑𝑑𝑑𝑑 𝜏𝜏 𝜃𝜃 = 𝜃𝜃0 , 𝑡𝑡 = 0 Case

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

0 8 𝐻𝐻((𝑡𝑡 − 𝑡𝑡𝑜𝑜 ) Sc 0 3 Sc 𝐼𝐼(𝑡𝑡, 𝑡𝑡𝑜𝑜 , ∆𝑡𝑡) 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅(𝑡𝑡, 𝑡𝑡𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 , 𝑡𝑡𝑜𝑜𝑜𝑜 ) Arbitrary 𝑆𝑆(𝑡𝑡)

4 0 0 𝜃𝜃0 7 𝜃𝜃0 0 0 0

Solution in terms of 𝜓𝜓1 (𝑡𝑡) and 𝜓𝜓2 (𝑡𝑡) 4 ∙ 𝜓𝜓2 (𝑡𝑡) 8 ∙ 𝜓𝜓1 (𝑡𝑡 − 𝑡𝑡𝑜𝑜 )

H(t) is the step function. 𝐻𝐻((𝑡𝑡 − 𝑡𝑡𝑜𝑜 ) − 𝐻𝐻((𝑡𝑡 − 𝑡𝑡𝑜𝑜 − ∆𝑡𝑡) 𝐼𝐼(𝑡𝑡, 𝑡𝑡𝑜𝑜 , ∆𝑡𝑡) = 𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 = ∆𝑡𝑡 𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅(𝑡𝑡, 𝑡𝑡𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 , 𝑡𝑡𝑜𝑜𝑜𝑜 ) = 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝

1

Solution Case 1 2 3 4 5 6 7

Source, 𝑆𝑆(𝑡𝑡) Sc 0 3 Sc 𝐼𝐼(𝑡𝑡, 𝑡𝑡𝑜𝑜 , 𝛥𝛥𝛥𝛥)

IC, 𝜃𝜃0 0 𝜃𝜃0 7 𝜃𝜃0 0

Arbitrary 𝑆𝑆(𝑡𝑡)

0

RepPulse(𝑡𝑡, 𝑡𝑡cycle , 𝑡𝑡on )

0

Solution in terms of 𝜓𝜓1 (𝑡𝑡)and 𝜓𝜓2 (𝑡𝑡) 𝜃𝜃(𝑡𝑡) = 𝑆𝑆𝑐𝑐 ⋅ 𝜓𝜓1 (𝑡𝑡) 𝜃𝜃(𝑡𝑡) = 𝜃𝜃0 ⋅ 𝜓𝜓2 (𝑡𝑡) 𝜃𝜃(𝑡𝑡) = 3 ⋅ 𝜓𝜓1 (𝑡𝑡) + 7 ⋅ 𝜓𝜓2 (𝑡𝑡) 𝜃𝜃(𝑡𝑡) = 𝑆𝑆𝑐𝑐 ⋅ 𝜓𝜓1 (𝑡𝑡) + 𝜃𝜃0 ⋅ 𝜓𝜓2 (𝑡𝑡) 1 𝜃𝜃(𝑡𝑡) = (𝜓𝜓1 (𝑡𝑡 − 𝑡𝑡𝑜𝑜 ) − 𝜓𝜓1 (𝑡𝑡 − 𝑡𝑡𝑜𝑜 − 𝛥𝛥𝛥𝛥))𝜓𝜓𝑝𝑝 (𝑡𝑡, 𝑡𝑡𝑜𝑜 , 𝛥𝛥𝛥𝛥) 𝛥𝛥𝛥𝛥 𝜃𝜃(𝑡𝑡) = � 𝜓𝜓𝑝𝑝 (𝑡𝑡, (𝑛𝑛 − 1)𝑡𝑡cycle , 𝑡𝑡on ) 𝑛𝑛=1,2,... 𝑁𝑁

𝜃𝜃(𝑡𝑡) ≃ � 𝑆𝑆𝑖𝑖 𝜓𝜓𝑝𝑝 (𝑡𝑡, 𝑡𝑡𝑖𝑖 , 𝛥𝛥𝛥𝛥) 𝛥𝛥𝛥𝛥 𝑖𝑖=1

𝑡𝑡𝑖𝑖 = (𝑖𝑖 − 1)𝛥𝛥𝛥𝛥 , 𝑡𝑡 = 𝑁𝑁 ⋅ 𝛥𝛥𝛥𝛥, 𝑆𝑆𝑖𝑖 = 𝑆𝑆(𝑡𝑡𝑖𝑖 )

2

Problem 10.2 Consider the nonlinear problem

1 𝑑𝑑𝑑𝑑 = − 𝜃𝜃 2 𝜏𝜏 𝑑𝑑𝑑𝑑 𝜃𝜃 = 𝜃𝜃0 , 𝑡𝑡 = 0 Can the solution be expressed in terms of the ψ1 and ψ2 functions defined Problem 10.1? If not, what is the difficulty?

Solution Try 𝜃𝜃(𝑡𝑡) = 𝑐𝑐 ⋅ 𝜓𝜓2 (𝑡𝑡),

𝑑𝑑𝑑𝑑2 1 2 2 + 𝑐𝑐 𝜓𝜓2 = 0 dt 𝜏𝜏 𝑐𝑐 ⋅ 𝜓𝜓2 = 2 , 𝑡𝑡 = 0 It is impossible to express the solution in terms of 𝜓𝜓1 (𝑡𝑡)and 𝜓𝜓2 (𝑡𝑡). The nonlinearity couples the coefficients. 𝑐𝑐

3

Problem 10.3 A basic population dynamics model is 𝑑𝑑𝑑𝑑 = 𝑎𝑎 𝑃𝑃 − 𝑏𝑏 𝑃𝑃2 + 𝑆𝑆 𝑑𝑑𝑑𝑑 𝑃𝑃 = 𝑃𝑃0 , 𝑡𝑡 = 0 Can the general solution can be expressed as the superposition of simpler solutions? 𝑃𝑃(𝑡𝑡) = 𝑃𝑃1 (𝑡𝑡) + 𝑃𝑃2 (𝑡𝑡) 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. 𝑑𝑑(𝑃𝑃1 + 𝑃𝑃2 ) = 𝑎𝑎(𝑃𝑃1 + 𝑃𝑃2 ) − 𝑏𝑏 (𝑃𝑃1 + 𝑃𝑃2 )2 + 𝑆𝑆 𝑑𝑑𝑑𝑑 = 𝑎𝑎(𝑃𝑃1 + 𝑃𝑃2 ) − 𝑏𝑏 �𝑃𝑃1 2 + 2𝑃𝑃1 𝑃𝑃2 + 𝑃𝑃2 2 � + 𝑆𝑆 The squared term 2𝑃𝑃1 𝑃𝑃2 makes the equation impossible to separate algebraically into individual problems with 𝑃𝑃1 only and 𝑃𝑃2 only.

4

Problem 10.4 Consider the thermal system described by the equations 𝑑𝑑𝑑𝑑 = −ℎ𝐴𝐴𝑠𝑠 (𝑇𝑇 − 𝑇𝑇∞ ) + 𝑔𝑔𝑔𝑔 𝜌𝜌𝜌𝜌𝜌𝜌 𝑑𝑑𝑑𝑑 𝑇𝑇 = 𝑇𝑇0 , 𝑡𝑡 = 0 Show that the general solution can be expressed as the superposition of the simpler solutions 𝑇𝑇(𝑡𝑡) = 𝑇𝑇1 (𝑡𝑡) + 𝑇𝑇2 (𝑡𝑡) 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 𝑑𝑑(𝑇𝑇1 + 𝑇𝑇2 ) = −𝐴𝐴𝑠𝑠 ℎ(𝑇𝑇1 + 𝑇𝑇2 − 𝑇𝑇∞ ) + 𝑔𝑔 𝑉𝑉 𝜌𝜌cV

dt

𝑇𝑇1 + 𝑇𝑇2 = 𝑇𝑇0 , 𝑡𝑡 = 0

Break this problem into sub-problems. 𝑑𝑑𝑇𝑇1 𝑑𝑑𝑇𝑇2 �𝜌𝜌cV + 𝐴𝐴𝑠𝑠 ℎ(𝑇𝑇1 − 𝑇𝑇∞ )� + �𝜌𝜌cV + 𝐴𝐴𝑠𝑠 ℎ(𝑇𝑇2 ) − 𝑔𝑔 𝑉𝑉� = 0 ����������������� ������������������� 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 0

𝑇𝑇1 + 𝑇𝑇⏟2 = 𝑇𝑇0 , 𝑡𝑡 = 0

x1(t) Problem 𝑑𝑑𝑇𝑇1 𝜌𝜌cV = −𝐴𝐴𝑠𝑠 ℎ(𝑇𝑇1 − 𝑇𝑇∞ ) 𝑑𝑑𝑑𝑑 𝑇𝑇1 = 𝑇𝑇0 , 𝑡𝑡 = 0

0

0

x2(t) Problem 𝑑𝑑𝑇𝑇2 𝜌𝜌cV = −𝐴𝐴𝑠𝑠 ℎ(𝑇𝑇2 ) + gV 𝑑𝑑𝑑𝑑 𝑇𝑇2 = 0 , 𝑡𝑡 = 0

5

Problem 10.5 A lumped thermal system with radiation and heat generation is described by the equations 𝑑𝑑𝑑𝑑 4 ) = −𝐴𝐴𝑠𝑠 𝜀𝜀𝜀𝜀(𝑇𝑇 4 − 𝑇𝑇𝑠𝑠𝑠𝑠𝑠𝑠 + 𝑔𝑔(𝑡𝑡)𝑉𝑉 𝜌𝜌𝑐𝑐𝑐𝑐 𝑑𝑑𝑑𝑑 𝑇𝑇 = 𝑇𝑇0 , 𝑡𝑡 = 0 Can the general solution can be decomposed as the superposition of the simpler solutions 𝑇𝑇(𝑡𝑡) = 𝑇𝑇1 (𝑡𝑡) + 𝑇𝑇2 (𝑡𝑡) 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 𝑑𝑑(𝑇𝑇1 + 𝑇𝑇2 ) 4 𝜌𝜌cV = −𝐴𝐴𝑠𝑠 εσ((𝑇𝑇1 + 𝑇𝑇2 )4 − 𝑇𝑇sur ) + 𝑔𝑔 𝑉𝑉

dt

The 4th order term makes the equation impossible to separate algebraically into individual problems for 𝑇𝑇1 only and 𝑇𝑇2 only.

6

Problem 10.6 Consider the linear first order ODE where τ is constant. 𝑑𝑑𝑑𝑑 1 = − 𝜃𝜃 + 𝑆𝑆(𝑡𝑡) 𝑑𝑑𝑑𝑑 𝜏𝜏 𝜃𝜃 = 𝜃𝜃0 , 𝑡𝑡 = 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 𝜃𝜃0 = 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 θ0 and source S(t). d) Determine the analytical solution for S(t)=Sc=constant and initial condition, θ0, using the superposition principle demonstrated in part (c) and the results found in parts (a) and (b). e) Consider the case where θ0 =0 with a finite pulse source, defined by 𝑆𝑆 𝑆𝑆(𝑡𝑡) = ∆𝑡𝑡𝑐𝑐 (𝐻𝐻(𝑡𝑡 − 𝑡𝑡1 ) − 𝐻𝐻(𝑡𝑡 − 𝑡𝑡1 − ∆𝑡𝑡)) 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 𝑆𝑆(𝑡𝑡) = 0.

𝜃𝜃(𝑡𝑡) = 𝜃𝜃0 𝑒𝑒 −𝑡𝑡⁄𝜏𝜏

b) 𝑆𝑆(𝑡𝑡) = 𝑆𝑆𝑐𝑐 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 with 𝜃𝜃0 = 0 𝜃𝜃(𝑡𝑡) = 𝑆𝑆𝑐𝑐 𝜏𝜏�1 − 𝑒𝑒 −𝑡𝑡⁄𝜏𝜏 �

c) Linearity 𝜃𝜃(𝑡𝑡) = 𝜃𝜃ic (𝑡𝑡) + 𝜃𝜃𝑠𝑠 (𝑡𝑡) is allowable for linear problems.

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d) Combined Effects of θ0 and Sc 𝜃𝜃(𝑡𝑡) = 𝜃𝜃0 𝑒𝑒 −𝑡𝑡⁄𝜏𝜏 + 𝑆𝑆𝑐𝑐 𝜏𝜏�1 − 𝑒𝑒 −𝑡𝑡⁄𝜏𝜏 �

e) Pulsed Source 𝜃𝜃(𝑡𝑡) =

⎧ ⎪

0

𝑆𝑆𝑐𝑐 𝜏𝜏�1 − 𝑒𝑒 −(𝑡𝑡−𝑡𝑡𝑜𝑜 )⁄𝜏𝜏 � 𝛥𝛥𝛥𝛥

, ,

⎨ 𝑆𝑆𝑐𝑐 ⎪ 𝜏𝜏�𝑒𝑒 −(𝑡𝑡−𝑡𝑡𝑜𝑜 −𝛥𝛥𝛥𝛥)⁄𝜏𝜏 − 𝑒𝑒 −(𝑡𝑡−𝑡𝑡𝑜𝑜 )⁄𝜏𝜏 � , ⎩𝛥𝛥𝛥𝛥 0 , 𝑡𝑡 < 𝑡𝑡𝑜𝑜 In the limit as ∆t → 0, 𝜃𝜃(𝑡𝑡) = � 𝑆𝑆𝑐𝑐 𝑒𝑒 −(𝑡𝑡−𝑡𝑡𝑜𝑜 )⁄𝜏𝜏 , 𝑡𝑡 ≥ 𝑡𝑡𝑜𝑜

𝑡𝑡 < 𝑡𝑡𝑜𝑜

𝑡𝑡𝑜𝑜 𝑡𝑡 ≤ 𝑡𝑡𝑜𝑜 + 𝛥𝛥𝛥𝛥 𝑡𝑡 > 𝑡𝑡𝑜𝑜 + 𝛥𝛥𝛥𝛥

8

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

T∞ = 30°C h

L Chip

δ

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

𝜌𝜌cV

𝐸𝐸˙st = 𝑃𝑃 − 𝑞𝑞conv

𝑑𝑑𝑑𝑑 = 𝑃𝑃 − hA𝑠𝑠 (𝑇𝑇 − 𝑇𝑇∞ ) 𝑑𝑑𝑑𝑑

At steady-state, energy storage is zero, thus 𝑃𝑃 = 𝑞𝑞conv = ℎ 𝐴𝐴𝑠𝑠 (𝑇𝑇ss − 𝑇𝑇∞ ) 𝑃𝑃 0.5 𝑇𝑇ss = 𝑇𝑇∞ + 2 = 30 + = 155°C (40)(0.01)2 hL Part 2.

Power On

Power Off

Tss T(t)

T∞ Time, t

t2

An energy balance during time t2 gives

𝑑𝑑𝑑𝑑 = −hA𝑠𝑠 (𝑇𝑇 − 𝑇𝑇∞ ) 𝑑𝑑𝑑𝑑 Solve this differential equation subject to the initial condition T = Tss when t = 0; 𝑇𝑇 − 𝑇𝑇∞ 𝑡𝑡 = exp �− � 𝑇𝑇ss − 𝑇𝑇∞ 𝜏𝜏 𝜌𝜌𝜌𝜌𝜌𝜌 where 𝜏𝜏 = ℎ𝐴𝐴𝑠𝑠 Substitute numbers 𝑡𝑡2 = 563 sec = 9.4 min 𝜌𝜌cV

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Part 3.

Tfail Power On

No Coolant

Tss T(t)

T∞ t3

Time, t

An energy balance during time t3 gives, 𝐸𝐸˙st = 𝑃𝑃 𝑑𝑑𝑑𝑑 𝜌𝜌cV = 𝑃𝑃 𝑑𝑑𝑑𝑑

Solve subject to the initial condition, T = Tss when t = 0.

Substitute numbers,

𝑇𝑇(𝑡𝑡) = 𝑇𝑇ss +

𝑃𝑃 𝑡𝑡 𝜌𝜌cV

𝑡𝑡3 = 203 sec = 3.4 min

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Problem 10.8 Consider a thin plate of area A and thickness L with material properties k, c, and ρ. 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 𝑑𝑑𝑑𝑑 = − 𝜃𝜃 + 𝑆𝑆 𝜏𝜏 𝑑𝑑𝑑𝑑 𝜃𝜃 = 𝜃𝜃0 , 𝑡𝑡 = 0 What are the temperature rise, θ(t), initial temperature, 𝜃𝜃0 , time constant, τ, and the source term, S? d) Using the model from part (c), solve for the transient temperature solution, 𝜃𝜃(𝑡𝑡) when S(t) is a constant. Plot your solution. e) Using the model from part (c), solve for the transient temperature solution, 𝜃𝜃(𝑡𝑡) when S(t) is a pulse, 𝑆𝑆(𝑡𝑡) = 𝑆𝑆0 (𝐻𝐻(𝑡𝑡) − 𝐻𝐻(𝑡𝑡 − ∆𝑡𝑡)) 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 𝜃𝜃0 only. Do not simply examine the mathematical solution to verify that this works (zero credit for this approach). Instead, try the superposition 𝜃𝜃(𝑡𝑡) = 𝜃𝜃𝑠𝑠 (𝑡𝑡) + 𝜃𝜃𝑖𝑖𝑖𝑖 (𝑡𝑡) directly in the governing equations. What equations govern the functions 𝜃𝜃𝑠𝑠 and 𝜃𝜃𝑖𝑖𝑖𝑖 ?

Solution a) Mathematical Model

𝑑𝑑𝑑𝑑 = −ℎ𝐴𝐴(𝑇𝑇 − 𝑇𝑇∞ ) + 𝑔𝑔(𝑡𝑡)𝑉𝑉 + 𝑞𝑞𝑠𝑠" (𝑡𝑡)𝐴𝐴 𝑑𝑑𝑑𝑑 or 𝑑𝑑𝑑𝑑 ℎ 1 1 " (𝑇𝑇 − 𝑇𝑇∞ ) + 𝑔𝑔(𝑡𝑡) + =− 𝑞𝑞 (𝑡𝑡) 𝑑𝑑𝑑𝑑 𝜌𝜌𝜌𝜌𝜌𝜌 𝜌𝜌𝜌𝜌 𝜌𝜌𝜌𝜌𝜌𝜌 𝑠𝑠 𝜌𝜌𝜌𝜌𝜌𝜌

𝑇𝑇 = 𝑇𝑇0 , 𝑡𝑡 = 0

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b) Sketches At steady-state

𝑔𝑔𝑔𝑔 + 𝑞𝑞𝑠𝑠" 𝐴𝐴 𝑇𝑇𝑠𝑠𝑠𝑠 = 𝑇𝑇∞ + ℎ𝐴𝐴

c) Equivalent Form

𝑑𝑑𝑑𝑑 1 = − 𝜃𝜃 + 𝑆𝑆 𝑑𝑑𝑑𝑑 𝜏𝜏 𝜃𝜃 = 𝜃𝜃0 , 𝑡𝑡 = 0

𝜏𝜏 =

d) Solution for constant S

where 𝜃𝜃(𝑡𝑡) = 𝑇𝑇(𝑡𝑡) − 𝑇𝑇∞ , 𝜃𝜃0 = 𝑇𝑇0 − 𝑇𝑇∞

𝜌𝜌𝜌𝜌𝜌𝜌 𝜌𝜌𝜌𝜌𝜌𝜌 𝑔𝑔(𝑡𝑡) 1 " = , 𝑆𝑆(𝑡𝑡) = + 𝑞𝑞 (𝑡𝑡) ℎ ∙ 𝐴𝐴 ℎ 𝜌𝜌𝜌𝜌 𝜌𝜌𝜌𝜌𝜌𝜌 𝑠𝑠

𝜃𝜃(𝑡𝑡) =

𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐

��𝑒𝑒�−𝑡𝑡/𝜏𝜏 �� 𝜃𝜃 0 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠

𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠

���� ���) + �� 𝑆𝑆 ∙ � 𝜏𝜏(1 − 𝑒𝑒�−𝑡𝑡/𝜏𝜏 𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡

���� ������� −𝑡𝑡/𝜏𝜏 𝜃𝜃(𝑡𝑡) = 𝑆𝑆� ∙ 𝜏𝜏 + (𝜃𝜃 0 − 𝑆𝑆 ∙ 𝜏𝜏)𝑒𝑒

e) Pulsed Solution Define 𝜓𝜓𝑐𝑐 (𝑡𝑡) as the response to a constant source of unit magnitude (S = 1) with zero initial condition ( 𝜃𝜃0 = 0). For this case the previous solution reduces to 0 , 𝑡𝑡 < 0 𝜓𝜓𝑐𝑐 (𝑡𝑡) = 𝜏𝜏�1 − 𝑒𝑒 −𝑡𝑡/𝜏𝜏 �𝐻𝐻(𝑡𝑡) = � −𝑡𝑡/𝜏𝜏 𝜏𝜏�1 − 𝑒𝑒 � , 𝑡𝑡 ≥ 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. 𝜃𝜃(𝑡𝑡) = 𝑆𝑆0 �𝜓𝜓𝑐𝑐 (𝑡𝑡) − 𝜓𝜓𝑐𝑐 (𝑡𝑡 − 𝛥𝛥𝛥𝛥)� 𝜓𝜓𝑐𝑐 (𝑡𝑡) , 0 < 𝑡𝑡 < 𝛥𝛥𝛥𝛥 𝜃𝜃(𝑡𝑡) = 𝑆𝑆0 � 𝜓𝜓𝑐𝑐 (𝑡𝑡) − 𝜓𝜓𝑐𝑐 (𝑡𝑡 − 𝛥𝛥𝛥𝛥) , 𝑡𝑡 > 𝛥𝛥𝛥𝛥 −𝑡𝑡/𝜏𝜏 𝜏𝜏�1 − 𝑒𝑒 � , 0 < 𝑡𝑡 < 𝛥𝛥𝛥𝛥 𝜃𝜃(𝑡𝑡) = 𝑆𝑆0 � −(𝑡𝑡−𝛥𝛥𝛥𝛥)/𝜏𝜏 𝜏𝜏�𝑒𝑒 − 𝑒𝑒 −𝑡𝑡/𝜏𝜏 � , 𝑡𝑡 > 𝛥𝛥𝛥𝛥 Alternately, the solution can be obtained by carefully evaluating the resulting integral using an integrating factor.

13

f) Superposition of Elementary Solutions 𝜃𝜃𝑠𝑠 Problem 𝑑𝑑𝜃𝜃𝑠𝑠 1 = − 𝜃𝜃𝑠𝑠 + 𝑆𝑆 𝑑𝑑𝑑𝑑 𝜏𝜏 𝜃𝜃𝑠𝑠 = 0 , 𝑡𝑡 = 0

𝜃𝜃𝑖𝑖𝑖𝑖 Problem 𝑑𝑑𝜃𝜃𝑖𝑖𝑖𝑖 1 = − 𝜃𝜃𝑖𝑖𝑖𝑖 𝑑𝑑𝑑𝑑 𝜏𝜏 𝜃𝜃𝑖𝑖𝑖𝑖 = 𝜃𝜃0 , 𝑡𝑡 = 0

14

Problem 10.9 Consider a large, thin plate of area A and thickness L with material properties k, c, and ρ. 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, 𝑔𝑔 (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, 𝑇𝑇0 = (𝑇𝑇∞1 + 𝑇𝑇∞2 )/2 and 𝑇𝑇∞1 > 𝑇𝑇∞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

𝑑𝑑𝑑𝑑 = −ℎ1 𝐴𝐴(𝑇𝑇 − 𝑇𝑇∞,1 ) − ℎ2 𝐴𝐴(𝑇𝑇 − 𝑇𝑇∞,2 ) + gV 𝑑𝑑𝑑𝑑 b) Steady State Temperature ℎ1 𝑇𝑇∞,1 + ℎ2 𝑇𝑇∞,2 + gL 𝑇𝑇ss = ℎ1 + ℎ2 c) General Solution 𝑇𝑇(𝑡𝑡) = 𝑇𝑇0 𝑒𝑒 −𝑡𝑡⁄𝜏𝜏 + 𝑇𝑇ss �1 − 𝑒𝑒 −𝑡𝑡⁄𝜏𝜏 � 𝜌𝜌cL 𝜌𝜌cV = 𝜏𝜏 = (ℎ1 + ℎ2 )𝐴𝐴 (ℎ1 + ℎ2 ) d) Nondimensional Form 𝜌𝜌cV

Option 3 𝑑𝑑𝑇𝑇 𝑑𝑑𝑇𝑇 𝑑𝑑𝑇𝑇 + + + + + = −𝑇𝑇 + 𝑆𝑆 = −(𝑇𝑇 − 1) + 𝑆𝑆 = −𝑇𝑇 + 𝑑𝑑𝑡𝑡 + 𝑑𝑑𝑡𝑡 + 𝑑𝑑𝑡𝑡 + 𝑇𝑇 + = 1 , 𝑡𝑡 + = 0 𝑇𝑇 + = 0 , 𝑡𝑡 + = 0 𝑇𝑇 + = 1 , 𝑡𝑡 + = 0 𝑇𝑇 − 𝑇𝑇eq + 𝑡𝑡 𝑇𝑇 − 𝑇𝑇0 𝑡𝑡 𝑇𝑇 − 𝑇𝑇eq − 𝜏𝜏𝜏𝜏 + 𝑡𝑡 𝑇𝑇 + = , 𝑡𝑡 + = 𝑇𝑇 + = , 𝑡𝑡 = 𝑇𝑇 + = , 𝑡𝑡 = 𝑇𝑇eq − 𝑇𝑇0 𝜏𝜏 𝑇𝑇0 − 𝑇𝑇eq 𝜏𝜏 𝑇𝑇0 − 𝑇𝑇eq − 𝜏𝜏𝜏𝜏 𝜏𝜏 𝜏𝜏 𝜏𝜏 𝜏𝜏 𝑆𝑆 + = 𝑆𝑆 𝑆𝑆 + = 𝑆𝑆 𝑆𝑆 + = 𝑆𝑆 𝑇𝑇eq − 𝑇𝑇0 𝑇𝑇0 − 𝑇𝑇eq 𝑇𝑇0 − 𝑇𝑇eq − 𝜏𝜏𝜏𝜏 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. +

Option 1

+

Option 2

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Problem 10.10 A long thin copper wire of diameter D and length L has an electrical resistance per length of wire R’e (Ω/m), density ρ, specific heat c , and total emissivity ε. 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 Wire

I

Power supply

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

• Negligible conduction from the ends or stand. a) Derive the mathematical model governing the transient temperature of the wire. b) For the case where 𝑇𝑇0 = 𝑇𝑇∞ = 𝑇𝑇𝑠𝑠𝑠𝑠𝑠𝑠 , 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 𝑇𝑇0 = 𝑇𝑇∞ = 𝑇𝑇𝑠𝑠𝑠𝑠𝑠𝑠 = 300 𝐾𝐾 D = diameter = 0.001 m , L = length = 0.4 m ρ = density = 8933 kg/m3 , k = 400 W/m K, c = specific heat = 385 J/kg K ε = emissivity = 0.9 I=6 Amps, ρe = electrical resistivity = 80 µΩ-cm Solution (a) Energy Equation 𝜌𝜌𝜌𝜌𝜌𝜌

𝑑𝑑𝑑𝑑 = 𝐼𝐼 2 𝑅𝑅𝑒𝑒 − 𝐴𝐴𝑠𝑠 ℎ(𝑇𝑇 − 𝑇𝑇∞ ) − 𝐴𝐴𝑠𝑠 𝜀𝜀𝜀𝜀�𝑇𝑇 4 − 𝑇𝑇𝑠𝑠𝑠𝑠𝑠𝑠 4 � 𝑑𝑑𝑑𝑑 𝑇𝑇 = 𝑇𝑇0 , 𝑡𝑡 = 0 𝜋𝜋𝐷𝐷2 𝐴𝐴𝑠𝑠 = 𝜋𝜋𝜋𝜋𝜋𝜋, 𝑉𝑉 = 𝐿𝐿 4

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(b) Anticipated Solution free convection T(t)

mild forced

strong forced Ti Time, t

(c) Transient Linear solution neglecting radiation 𝜃𝜃(𝑡𝑡) = 𝜃𝜃0 𝑒𝑒 −𝑡𝑡/𝜏𝜏 + 𝑆𝑆 ∙ 𝜏𝜏�1 − 𝑒𝑒 −𝑡𝑡/𝜏𝜏 � In terms of the original physical variables the solution is 𝐼𝐼 2 𝑅𝑅𝑒𝑒 𝑇𝑇(𝑡𝑡) − 𝑇𝑇∞ = (𝑇𝑇0 − 𝑇𝑇∞ )𝑒𝑒 −𝑡𝑡/𝜏𝜏 + �1 − 𝑒𝑒 −𝑡𝑡/𝜏𝜏 � 𝐴𝐴𝑠𝑠 ℎ Steady State Solution 𝐼𝐼 2 𝑅𝑅𝑒𝑒 𝑇𝑇𝑠𝑠𝑠𝑠 = 𝑇𝑇∞ + 𝐴𝐴𝑠𝑠 ℎ (d) Numerical Solution Effect of Convection Coefficient

Convection and Radiation Effects

17

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 𝑞𝑞𝑖𝑖" 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

q i" ri

ro

ρ = 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 𝑞𝑞𝑖𝑖" = 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 𝜌𝜌𝜌𝜌𝜌𝜌 where

𝑑𝑑𝑑𝑑 = −ℎ𝐴𝐴𝑜𝑜 (𝑇𝑇 − 𝑇𝑇∞ ) + 𝑞𝑞𝑖𝑖" 𝐴𝐴𝑖𝑖 𝑑𝑑𝑑𝑑 𝑇𝑇 = 𝑇𝑇𝑖𝑖 , 𝑡𝑡 = 0 𝑉𝑉 =

b) At steady-state

4 𝜋𝜋(𝑟𝑟𝑜𝑜3 − 𝑟𝑟𝑖𝑖3 ) 3

𝐴𝐴𝑖𝑖 = 4𝜋𝜋𝑟𝑟𝑖𝑖2 , 𝐴𝐴𝑜𝑜 = 4𝜋𝜋𝑟𝑟𝑜𝑜2 𝑇𝑇𝑠𝑠𝑠𝑠 = 𝑇𝑇∞ +

𝑞𝑞𝑖𝑖" 𝐴𝐴𝑖𝑖 ℎ𝐴𝐴𝑜𝑜

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Problem 10.12 Consider the system 𝑑𝑑𝑑𝑑/𝑑𝑑𝑑𝑑 = 𝑟𝑟 ∙ 𝑥𝑥 − 𝑥𝑥 3 . For 𝑟𝑟 < 0, 𝑟𝑟 = 0, 𝑎𝑎𝑎𝑎𝑎𝑎 𝑟𝑟 > 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 𝑥𝑥 ∗ = 0 and ± √𝑟𝑟 are found from setting 𝑑𝑑𝑑𝑑/𝑑𝑑𝑑𝑑 = 0. Stability is ascertained 𝑑𝑑𝑑𝑑(𝑥𝑥 ∗ ) from the sign of = 𝑟𝑟 − 3𝑥𝑥 ∗ 2 . The bifurcation point is rc = 0 since real fixed points 𝑥𝑥 ∗ = 𝑑𝑑𝑑𝑑

±√𝑟𝑟 exist only for r ≥ 0 . The findings are: • 𝑟𝑟 < 𝑟𝑟𝑐𝑐 : one fixed point, stable at x* = 0 • 𝑟𝑟 = 𝑟𝑟𝑐𝑐 = 0: one fixed point, stable at x* = 0, but weakly stable • 𝑟𝑟 > 𝑟𝑟𝑐𝑐 : 3 fixed points, unstable at x* = 0, stable at 𝑥𝑥 ∗ = −√𝑟𝑟, and stable at 𝑥𝑥 ∗ = √𝑟𝑟

The following phase portraits and solutions with fixed points and bifurcation diagram reveal the dynamics of this system.

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Problem 10.13: Bifurcations Consider the following first order ODEs. 𝑑𝑑𝑑𝑑 𝑟𝑟 Case 1: 𝑑𝑑𝑑𝑑 = 𝑥𝑥 �1 + 1+𝑥𝑥 2 � 𝑑𝑑𝑑𝑑

𝑥𝑥

Case 2: 𝑑𝑑𝑑𝑑 = −𝑟𝑟𝑟𝑟 + 1+𝑥𝑥 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 𝑟𝑟𝑐𝑐 = −1. Subcritical pitchfork bifurcation. 𝑥𝑥1∗ = 0

𝑥𝑥2∗ = +�−(1 + 𝑟𝑟) 𝑥𝑥3∗ = −�−(1 + 𝑟𝑟)

𝒓𝒓 < 𝒓𝒓𝒄𝒄 stable unstable unstable

𝒓𝒓 > 𝒓𝒓𝒄𝒄 unstable DNE DNE

20

Case 2 A bifurcation occurs at 𝑟𝑟𝑐𝑐 = 1. Transcritical bifurcation. 𝑥𝑥1∗ = 0

𝑥𝑥2∗ =

1 −1 𝑟𝑟

𝒓𝒓 < 𝒓𝒓𝒄𝒄 unstable stable

𝒓𝒓 > 𝒓𝒓𝒄𝒄 stable unstable

21

Problem 10.14: Transient Parameter Consider a first order ODE of the form

𝑑𝑑𝑑𝑑 = 𝑓𝑓(𝑥𝑥, 𝑟𝑟) 𝑑𝑑𝑑𝑑 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

22

Problem 10.15: Improved Laser An improved model of a laser is

𝑑𝑑𝑑𝑑 = 𝐺𝐺𝐺𝐺𝐺𝐺 − 𝑘𝑘𝑘𝑘 𝑑𝑑𝑑𝑑

𝑑𝑑𝑑𝑑 = −𝐺𝐺𝐺𝐺𝐺𝐺 − 𝑓𝑓𝑓𝑓 + 𝑝𝑝 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 a) Assuming 𝑑𝑑𝑑𝑑 ≈ 0, derive a differential equation for n. b) Show that n* = 0 becomes unstable for p > pc. Determine pc. c) What type of bifurcation occurs at pc. 𝑑𝑑𝑑𝑑 d) For what range of parameters is it valid to assume that 𝑑𝑑𝑑𝑑 ≈ 0?

Solution a) Adiabatic elimination 𝑑𝑑𝑑𝑑 𝑝𝑝 Assume 𝑑𝑑𝑑𝑑 ≈ 0. Then, 𝑁𝑁 = 𝐺𝐺𝐺𝐺+𝑓𝑓. The differential equation for n becomes: 𝑑𝑑𝑑𝑑 𝑝𝑝 𝑝𝑝 = 𝐺𝐺𝐺𝐺 � � − 𝑘𝑘𝑘𝑘 = 𝑛𝑛 � � − 𝑘𝑘𝑘𝑘 𝑑𝑑𝑑𝑑 𝐺𝐺𝐺𝐺 + 𝑓𝑓 𝑛𝑛 + 𝑓𝑓/𝐺𝐺

b) Fixed Points 𝑑𝑑𝑑𝑑 The fixed points are the values n* where 𝑑𝑑𝑑𝑑 = 0. 𝑝𝑝 𝑛𝑛∗ � ∗ � − 𝑘𝑘𝑛𝑛∗ = 0 𝑛𝑛 + 𝑓𝑓/𝐺𝐺 We find 2 possible fixed points: 𝑛𝑛1∗ = 0 𝑝𝑝 𝑓𝑓 𝑛𝑛2∗ = − 𝑘𝑘 𝐺𝐺 Stability Analysis of the fixed Points 𝑑𝑑𝑛𝑛̇ 1 𝑛𝑛 = 𝑝𝑝 − 𝑝𝑝 2 − 𝑘𝑘 𝑓𝑓 𝑑𝑑𝑑𝑑 𝑓𝑓 𝑛𝑛 + 𝐺𝐺 �𝑛𝑛 + 𝐺𝐺 � Stability of n1*=0 𝑑𝑑𝑛𝑛̇ 𝐺𝐺 � � = 𝑝𝑝 − 𝑘𝑘 𝑑𝑑𝑑𝑑 𝑛𝑛1 =0 𝑓𝑓 We see a critical value of p, 𝑓𝑓 𝑝𝑝𝑐𝑐 = 𝑘𝑘 𝐺𝐺 𝑝𝑝 < 𝑝𝑝𝑐𝑐 : Stable 𝑝𝑝 > 𝑝𝑝𝑐𝑐 : Unstable Stability of n2* 𝑝𝑝 𝑓𝑓 − 𝐺𝐺 𝑑𝑑𝑛𝑛̇ 1 𝑘𝑘 2 𝑝𝑝 𝑓𝑓 𝑘𝑘 � � = 𝑝𝑝 𝑝𝑝 − 𝑝𝑝 − 𝑘𝑘 = − � − � 𝑝𝑝 2 𝑑𝑑𝑑𝑑 𝑛𝑛2 𝑝𝑝 𝑘𝑘 𝐺𝐺 � � 𝑘𝑘 𝑘𝑘 𝑓𝑓 We see a critical value, 𝑝𝑝𝑐𝑐 = 𝑘𝑘 𝐺𝐺, again dictates stability.

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c) Type of Bifurcation

𝑝𝑝 < 𝑝𝑝𝑐𝑐 : Unstable 𝑝𝑝 > 𝑝𝑝𝑐𝑐 : Stable

We have a transcritical bifurcation. d) Validity of the Adiabatic Elimination Assumption Consider two linear, uncoupled differential equations: 𝑑𝑑𝑥𝑥1 = −𝑎𝑎1 𝑥𝑥 + 𝑆𝑆1 𝑑𝑑𝑑𝑑 𝑑𝑑𝑥𝑥2 = −𝑎𝑎2 𝑥𝑥 + 𝑆𝑆2 𝑑𝑑𝑑𝑑 These have solutions 𝑆𝑆1 𝑥𝑥1 = 𝑥𝑥10 𝑒𝑒 −𝑎𝑎1 𝑡𝑡 + (1 − 𝑒𝑒 −𝑎𝑎1 𝑡𝑡 ) 𝑎𝑎1 𝑆𝑆2 𝑥𝑥2 = 𝑥𝑥20 𝑒𝑒 −𝑎𝑎2 𝑡𝑡 + (1 − 𝑒𝑒 −𝑎𝑎2 𝑡𝑡 ) 𝑎𝑎2 If a2 >> a1, then x2 will relax or approach steady state much faster than x1. Under such dx conditions, we could assume dt2 ≈ 0. Our equations are both nonlinear and coupled, however, let’s try to apply the same line of reasoning. 𝑑𝑑𝑑𝑑 = (𝐺𝐺𝐺𝐺 − 𝑘𝑘)𝑛𝑛 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 = −(𝐺𝐺𝐺𝐺 − 𝑓𝑓)𝑁𝑁 + 𝑝𝑝 𝑑𝑑𝑑𝑑 Our logic would dictate that 𝐺𝐺𝐺𝐺 + 𝑓𝑓 ≫ 𝑘𝑘 − 𝐺𝐺𝐺𝐺. The trouble is that n and N are variables.

However, at fixed point-1, we have 𝑝𝑝 𝑛𝑛1∗ = 0, 𝑁𝑁1∗ = 𝑓𝑓 Substitute to find 𝑝𝑝 𝑓𝑓>>𝑘𝑘 − 𝐺𝐺 𝑓𝑓 𝑓𝑓 𝑓𝑓𝑓𝑓 𝑝𝑝>> �1 − � 𝑘𝑘 𝐺𝐺 Similarly, fixed point-2, 𝑝𝑝 𝑓𝑓 𝑘𝑘 𝑛𝑛2∗ = − , 𝑁𝑁2∗ = 𝑘𝑘 𝐺𝐺 𝐺𝐺 Substitute and we find, 𝐺𝐺𝐺𝐺 >>0 𝑘𝑘 Thus, the criteria for validation of the model are 𝑓𝑓 𝑓𝑓𝑓𝑓 𝐺𝐺𝐺𝐺 𝑝𝑝>> �1 − 𝑘𝑘� 𝐺𝐺 and 𝑘𝑘 >>0

24

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 𝑘𝑘3 𝑔𝑔2 𝑑𝑑𝑑𝑑 = 𝑘𝑘1 𝑠𝑠𝑜𝑜 − 𝑘𝑘2 𝑔𝑔 + 𝑘𝑘4 + 𝑔𝑔2 𝑑𝑑𝑑𝑑

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 𝑑𝑑𝑑𝑑 𝑥𝑥 2 = 𝑠𝑠 − 𝑟𝑟𝑟𝑟 + 𝑑𝑑𝑑𝑑 1 + 𝑥𝑥 2 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. τ behavior.

Solution (a) Dimensionless Form

𝑑𝑑𝑑𝑑 𝑥𝑥 2 = 𝑠𝑠 − 𝑟𝑟𝑟𝑟 + 𝑑𝑑𝑑𝑑 1 + 𝑥𝑥 2 𝑔𝑔 𝑘𝑘3 , 𝜏𝜏 = 𝑡𝑡 𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉𝑉: 𝑥𝑥 = 𝑘𝑘4 𝑘𝑘4 𝑘𝑘1 𝑘𝑘2 𝑘𝑘4 𝑃𝑃𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎𝑎: 𝑠𝑠 = 𝑠𝑠𝑜𝑜 , 𝑟𝑟 = 𝑘𝑘3 𝑘𝑘3 (b) Fixed Points for s=0 Fixed points for s = 0 x*=0 1 − √1 − 4𝑟𝑟 2 2𝑟𝑟 1 + √1 − 4𝑟𝑟 2 𝑥𝑥 ∗ = 2𝑟𝑟 𝑥𝑥 ∗ =

0 ≤ 𝑟𝑟 < 𝑟𝑟𝑐𝑐 Stable Unstable

𝑟𝑟 > 𝑟𝑟𝑐𝑐 Stable Does not exist

Stable

Does not exist

25



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

s 0

dx d

r 0.4

0.2 0.1

0.5

1.0

1.5

2.0

2.5

3.0

x

0.5 0.6

0.1 0.2

(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

0.05

1.0

1.5

x 2.0

0.04 0.06 0.08

0.10

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

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

High

Low

0.02 0.00 0.0

Bistable 0.1

0.2

0.3

0.4

0.5

0.6

r

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

1

0.07 0.072 0.074 50

100

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.

27

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. 𝑑𝑑𝑑𝑑 = −𝑟𝑟𝑟𝑟 + 𝑆𝑆(𝑡𝑡) 𝑑𝑑𝑑𝑑 The stocking function is modeled as a periodic sequence of regularly spaced pulse functions as shown where Nst = number of fished per stocking 𝑡𝑡𝑠𝑠𝑠𝑠 = 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 𝑁𝑁𝑖𝑖 (𝑡𝑡, Δ𝑡𝑡) 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 𝑁𝑁𝑖𝑖 (𝑡𝑡, Δ𝑡𝑡). d) Determine the limit as Δ𝑡𝑡 → 0. e) Plot your solution from part (d). Use parameter values that make sense.

Solution (a) Stocking Function 𝑆𝑆[𝑡𝑡, 𝑡𝑡𝑡𝑡𝑡𝑡, 𝛥𝛥𝛥𝛥, 𝑁𝑁𝑁𝑁𝑁𝑁] =

or

𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖

𝑁𝑁𝑁𝑁𝑁𝑁 H[Δt − Mod[𝑡𝑡, tst]] ∆𝑡𝑡

𝑁𝑁𝑁𝑁𝑁𝑁 𝑆𝑆[𝑡𝑡, 𝑡𝑡𝑡𝑡𝑡𝑡, 𝛥𝛥𝛥𝛥, 𝑁𝑁𝑁𝑁𝑁𝑁] = � (𝐻𝐻[𝑡𝑡 − 𝑡𝑡𝑡𝑡𝑡𝑡(𝑖𝑖 − 1)] − 𝐻𝐻[𝑡𝑡 − 𝑡𝑡𝑡𝑡𝑡𝑡(𝑖𝑖 − 1) − ∆𝑡𝑡]) ∆𝑡𝑡

(b) Unit Response

(c) Pulsed Response

𝑖𝑖=1

ψcon[𝑡𝑡, 𝑟𝑟] =

ψpulse[𝑡𝑡, 𝑡𝑡𝑡𝑡, ∆𝑡𝑡, 𝑟𝑟] =

1 ∗ (1 − Exp[−𝑟𝑟 ∗ 𝑡𝑡]) ∗ 𝐻𝐻[𝑡𝑡] 𝑟𝑟

1 (ψcon[𝑡𝑡 − ti, 𝑟𝑟] − ψcon[𝑡𝑡 − ti − Δt, 𝑟𝑟]) Δt 28

(d) Total Solution

∞

Nfish[𝑡𝑡, 𝑡𝑡𝑡𝑡𝑡𝑡, 𝛥𝛥𝛥𝛥, 𝑁𝑁𝑁𝑁𝑁𝑁, 𝑟𝑟] = Nst � ψpulse[𝑡𝑡, (𝑖𝑖 − 1)tst, Δt, 𝑟𝑟]

(e) Limit as Δt → 0

𝑖𝑖=1

ψpulse[𝑡𝑡, 𝑡𝑡𝑡𝑡, 0, 𝑟𝑟] = Exp[−𝑟𝑟 ∗ (𝑡𝑡 − ti)] ∗ 𝐻𝐻[𝑡𝑡 − ti]

(f) Numerical results for r = 0.1 and r = 0.25 are displayed.

29

Problem 10.18: A Model of a Fishery The equation 𝑁𝑁 𝑑𝑑𝑑𝑑 = 𝑟𝑟𝑟𝑟 �1 − � − 𝐻𝐻 𝐾𝐾 𝑑𝑑𝑑𝑑 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 𝑑𝑑𝑑𝑑 = 𝑥𝑥(1 − 𝑥𝑥) − ℎ 𝑑𝑑𝑑𝑑 What are the dimensionless variables x, τ, 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. τ 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 𝑥𝑥1∗ = �1 − √1 − 4ℎ�/2 𝑥𝑥2∗ = �1 + √1 − 4ℎ�/2

𝑑𝑑𝑑𝑑 = 𝑥𝑥(1 − 𝑥𝑥) − ℎ 𝑑𝑑𝑑𝑑 𝑁𝑁 𝐻𝐻 𝑥𝑥 = , 𝜏𝜏 = 𝑟𝑟𝑟𝑟 , ℎ = 𝐾𝐾 𝑟𝑟𝑟𝑟 𝟎𝟎 ≤ 𝒉𝒉 ≤ 𝒉𝒉𝒄𝒄 unstable stable

𝒉𝒉 > 𝒉𝒉𝒄𝒄 DNE DNE

A bifurcation occurs at 𝒉𝒉𝒄𝒄 = 𝟏𝟏/𝟒𝟒. This is a saddle bifurcation.

30

(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: 𝑁𝑁 2 𝐻𝐻(𝑁𝑁) = 𝐻𝐻𝑚𝑚𝑚𝑚𝑚𝑚 2 𝐴𝐴 + 𝑁𝑁 2

31

Problem 10.19: Improved Model of a Fishery The equation 𝑁𝑁 𝑁𝑁 𝑑𝑑𝑑𝑑 = 𝑟𝑟𝑟𝑟 �1 − � − 𝐻𝐻 𝐾𝐾 𝐴𝐴 + 𝑁𝑁 𝑑𝑑𝑑𝑑

provides a model of a fishery. Here 𝑁𝑁(𝑡𝑡) is the number of fish. In the absence of fishing, the 𝑁𝑁 population is assumed to grow logistically. The effects of fishing are modeled by the term 𝐻𝐻 𝐴𝐴+𝑁𝑁, where 𝐻𝐻 > 0 and 𝐴𝐴 > 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 𝑑𝑑𝑑𝑑 𝑥𝑥 = 𝑥𝑥(1 − 𝑥𝑥) − ℎ 𝑑𝑑𝑑𝑑 𝑎𝑎 + 𝑥𝑥 What are the dimensionless quantities x, τ, 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 𝑑𝑑𝑑𝑑 𝑥𝑥 = 𝑥𝑥(1 − 𝑥𝑥) − ℎ 𝑑𝑑𝑑𝑑 𝑎𝑎 + 𝑥𝑥 𝑁𝑁 𝐴𝐴 𝐻𝐻 𝑥𝑥 = , 𝜏𝜏 = 𝑟𝑟𝑟𝑟, 𝑎𝑎 = , ℎ= 𝐾𝐾 𝐾𝐾 𝑟𝑟𝑟𝑟

(b): Fixed Points

𝑥𝑥1∗ = 0

1 �(1 − 𝑎𝑎) − �(1 − 𝑎𝑎)2 − 4(ℎ − 𝑎𝑎)� 2 1 𝑥𝑥+∗ = �(1 − 𝑎𝑎) + �(1 − 𝑎𝑎)2 − 4(ℎ − 𝑎𝑎)� 2 𝑥𝑥−∗ =

Bifurcations

1

Stability of the Fixed Points 𝑥𝑥1∗ = 0 𝑥𝑥−∗ = 0 𝑥𝑥+∗ = 0

ℎ𝑐𝑐 = (𝑎𝑎 + 1)2 : saddle bifurcation.

4 ℎ𝑐𝑐 = 𝑎𝑎 : transcritical bifurcation.

ℎ < 𝑎𝑎

Unstable 0 < ℎ < 𝑎𝑎 Unstable Stable

𝑎𝑎 < ℎ < ℎ𝑐𝑐 a<1 a>1 Unstable Unstable Stable Unstable

ℎ > 𝑎𝑎 Stable

ℎ > ℎ𝑐𝑐 DNE DNE

32

(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

33

34

Chapter 11

Second Order Ordinary Differential Equations Linear Systems Problem 11.1.1 The motion of a linear damped harmonic oscillator is described by 𝑑𝑑2 𝑥𝑥 𝑑𝑑𝑑𝑑 𝑚𝑚 2 + 𝑏𝑏 + 𝑘𝑘 ∙ 𝑥𝑥 = 0 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑥𝑥 = 𝑥𝑥0 , = 𝑣𝑣0 , 𝑡𝑡 = 0 𝑑𝑑𝑑𝑑 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 𝑑𝑑𝑑𝑑 = 𝑣𝑣 𝑑𝑑𝑑𝑑 𝑘𝑘 𝑏𝑏 𝑑𝑑𝑑𝑑 = − 𝑥𝑥 − 𝑣𝑣 𝑑𝑑𝑑𝑑 𝑚𝑚 𝑚𝑚 In vector form 𝑑𝑑𝑑𝑑 0 1 𝑑𝑑𝑑𝑑 𝑘𝑘 𝑏𝑏 � �𝑥𝑥 � � �=� 𝑣𝑣 𝑑𝑑𝑑𝑑 − − 𝑚𝑚 𝑚𝑚 𝑑𝑑𝑑𝑑 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. −𝜆𝜆 1 𝑏𝑏 𝑘𝑘 𝑏𝑏 𝑘𝑘 𝑘𝑘 𝑏𝑏 𝑑𝑑𝑑𝑑𝑑𝑑 � � = −𝜆𝜆(− − 𝜆𝜆) + = 𝜆𝜆2 + 𝜆𝜆 + = 0 �− − 𝜆𝜆� − 𝑚𝑚 𝑚𝑚 𝑚𝑚 𝑚𝑚 𝑚𝑚 𝑚𝑚 The eigenvalues are

where

𝜏𝜏 ± √𝜏𝜏 2 − 4𝛥𝛥 1 𝑏𝑏 𝑏𝑏 2 𝑘𝑘 � 𝜆𝜆 = = �− ± � � − 4 � 2 2 𝑚𝑚 𝑚𝑚 𝑚𝑚

1

𝛥𝛥 =

The system behavior is indicated by

𝑘𝑘 𝑏𝑏 > 0 , 𝜏𝜏 = − ≤ 0 𝑚𝑚 𝑚𝑚

𝑏𝑏 2 𝑘𝑘 𝜏𝜏 2 − 4𝛥𝛥 = � � − 4 𝑚𝑚 𝑚𝑚 The eigenvalues always have a negative real part (τ < 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. 𝑏𝑏

2

𝑘𝑘

� 𝑚𝑚cr � − 4 𝑚𝑚 = 0

𝑏𝑏cr = 2√𝑚𝑚𝑚𝑚 At this point it seems natural to define a natural frequency, defined as 𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛𝑛 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 = 𝜔𝜔𝑛𝑛 = � 𝑘𝑘 𝑏𝑏cr = 2� = 2𝜔𝜔𝑛𝑛 𝑚𝑚 𝑚𝑚

𝑘𝑘 𝑚𝑚

It now also seems natural to express b in terms of ωn and bcr. 𝑏𝑏 𝑏𝑏cr 𝑏𝑏 𝑏𝑏 = = 2𝜔𝜔𝑛𝑛 = 2𝜔𝜔𝑛𝑛 𝜁𝜁 𝑚𝑚 𝑚𝑚 𝑏𝑏cr 𝑏𝑏cr 𝑏𝑏 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 = 𝜁𝜁 = 𝑏𝑏 cr

The eigenvalues

𝜆𝜆 = 𝜔𝜔𝑛𝑛 �−𝜁𝜁 ± �𝜁𝜁 2 − 1� 𝛥𝛥 = 𝜆𝜆1 𝜆𝜆2 = 𝜔𝜔𝑛𝑛2 > 0 𝜏𝜏 = 𝜆𝜆1 + 𝜆𝜆2 = −2𝜔𝜔𝑛𝑛 𝜁𝜁 ≤ 0 𝜏𝜏 2 − 4𝛥𝛥 = 4𝜔𝜔𝑛𝑛2 (𝜁𝜁 2 − 1) The solution classification is summarized in the following table.

Condition b = 0 or 𝜁𝜁 = 0

b < bcr or 0 < 𝜁𝜁 < 1 b = bcr or 𝜁𝜁 = 1 b > bcr or 𝜁𝜁 > 0

Eigenvalues λ1 and λ2 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

2

Phase Portraits: 𝜁𝜁 = b/bcr

3

Problem 11.1.2 Consider the following system of linear ODE’s. 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 = 𝑎𝑎𝑎𝑎 + 𝑏𝑏 , = 𝑥𝑥 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 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

𝑎𝑎𝑦𝑦 ∗ + 𝑏𝑏 = 0 𝑥𝑥 ∗ = 0

Solve to get

b) Eigenvalues and eigenvectors

𝑥𝑥 ∗ = 0 𝑦𝑦 ∗ = −𝑏𝑏/𝑎𝑎

(𝐀𝐀 − 𝜆𝜆 ∙ 𝐈𝐈)𝐯𝐯 = 𝟎𝟎 0 −𝜆𝜆 𝑎𝑎 𝑣𝑣1 � � �𝑣𝑣 � = � � 0 1 −𝜆𝜆 2 −𝜆𝜆 𝑎𝑎 𝑑𝑑𝑑𝑑𝑑𝑑(𝑨𝑨 − 𝜆𝜆 ∙ 𝑰𝑰) = 𝑑𝑑𝑑𝑑𝑑𝑑 � � = 𝜆𝜆2 − 𝑎𝑎 = 0 1 −𝜆𝜆 𝜆𝜆1 = √𝑎𝑎 , 𝜆𝜆2 = −√𝑎𝑎 Eigenvectors for 𝝀𝝀𝟏𝟏 = √𝒂𝒂: The eigenvectors are determined from 𝑣𝑣1 𝑣𝑣1 −𝜆𝜆 𝑎𝑎 − 𝑎𝑎 𝑎𝑎 0 � 1 � �𝑣𝑣 � = � √ � �𝑣𝑣 � = � � 1 −𝜆𝜆1 2 0 2 1 −√𝑎𝑎 This equation must be singular, thus 𝑣𝑣1 = √𝑎𝑎 ∙ 𝑣𝑣2 Eigenvectors for 𝝀𝝀𝟐𝟐 = −√𝒂𝒂: The eigenvectors are determined from 𝜆𝜆 𝑎𝑎 𝑣𝑣1 𝑎𝑎 𝑎𝑎 𝑣𝑣1 0 � 2 � �𝑣𝑣 � = �√ � �𝑣𝑣 � = � � 1 𝜆𝜆2 2 0 1 √𝑎𝑎 2 This equation must be singular, thus 𝑣𝑣1 = −√𝑎𝑎 ∙ 𝑣𝑣2 c) Condition 𝑎𝑎 > 0 𝑎𝑎 < 0

Eigenvalues λ1 and λ2 One positive, one negative Complex conjugates, Purely imaginary

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

Stability is not affected by b.

4

Problem 11.1.3 Consider the system of linear ODE’s

𝑑𝑑𝑑𝑑 = 𝑎𝑎 ∙ 𝑥𝑥 + 𝑏𝑏 ∙ 𝑦𝑦 + 𝑝𝑝 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 = 𝑐𝑐 ∙ 𝑥𝑥 + 𝑑𝑑 ∙ 𝑦𝑦 + 𝑞𝑞 𝑑𝑑𝑑𝑑 This system is non-homogeneous due to the presence of the terms p and q. a) Determine the fixed points (𝑥𝑥 ∗ , 𝑦𝑦 ∗ ). 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 (𝑥𝑥 ∗ , 𝑦𝑦 ∗ ) = (1,0). Sketch the phase portrait.

Solution a) Fixed points

Solve using Cramer’s rule.

𝑎𝑎 ∙ 𝑥𝑥 ∗ + 𝑏𝑏 ∙ 𝑦𝑦 ∗ + 𝑝𝑝 = 0 𝑐𝑐 ∙ 𝑥𝑥 ∗ + 𝑑𝑑 ∙ 𝑦𝑦 ∗ + 𝑞𝑞 = 0 𝑝𝑝 𝑎𝑎 𝑏𝑏 𝑥𝑥 ∗ � � � ∗ � = �𝑞𝑞 � 𝑦𝑦 𝑐𝑐 𝑑𝑑 𝑥𝑥 ∗ =

b) Variable change

𝑦𝑦 ∗ =

𝑝𝑝𝑝𝑝 − 𝑞𝑞𝑞𝑞 𝑝𝑝𝑝𝑝 − 𝑞𝑞𝑞𝑞 = 𝑑𝑑𝑑𝑑𝑑𝑑[𝐴𝐴] 𝑎𝑎𝑎𝑎 − 𝑏𝑏𝑏𝑏 𝑞𝑞𝑞𝑞 − 𝑝𝑝𝑝𝑝 𝑞𝑞𝑞𝑞 − 𝑝𝑝𝑝𝑝 = 𝑑𝑑𝑑𝑑𝑑𝑑[𝐴𝐴] 𝑎𝑎𝑎𝑎 − 𝑏𝑏𝑏𝑏

𝑥𝑥 + = 𝑥𝑥 − 𝑥𝑥 ∗ 𝑦𝑦 + = 𝑦𝑦 − 𝑦𝑦 ∗ 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.

5

Problem 11.1.4 Consider a series LRC circuit equation: 𝑑𝑑2 𝐼𝐼 𝑑𝑑𝑑𝑑 1 𝐿𝐿 2 + 𝑅𝑅 + ∙ 𝐼𝐼 = 0 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝐶𝐶 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: 𝑥𝑥1 = 𝐼𝐼 𝑑𝑑𝑥𝑥1 𝑑𝑑𝑑𝑑 𝑥𝑥2 = = 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 The second order system then decomposes as two first order ODE’s 𝑑𝑑𝑥𝑥1 = 𝑥𝑥2 𝑑𝑑𝑑𝑑 𝑑𝑑𝑥𝑥2 1 𝑅𝑅 =− 𝑥𝑥1 − 𝑥𝑥2 𝑑𝑑𝑑𝑑 𝐿𝐿𝐿𝐿 𝐿𝐿 In matrix form 0 1 𝑥𝑥 𝑥𝑥˙ 𝑅𝑅 � � 1 � � 1� = � 1 𝑥𝑥2 𝑥𝑥˙2 − − 𝐿𝐿𝐿𝐿 𝐿𝐿 b) Analysis of the fixed point (𝑥𝑥1 , 𝑥𝑥2 ) = (0,0) is the only fixed point. Since the system in linear, the eigenvalues will indicate the global behavior.

1 𝑅𝑅 𝜏𝜏 = − , 𝛥𝛥 = 𝐿𝐿 𝐿𝐿𝐿𝐿 2 𝑅𝑅 1 1 disc = 𝜏𝜏 2 − 4𝛥𝛥 = � � − 4 = 2 (CR2 − 4𝐿𝐿) 𝐿𝐿 𝐿𝐿𝐿𝐿 𝐿𝐿 𝐶𝐶 𝜏𝜏 + √𝜏𝜏 2 − 4𝛥𝛥 𝜏𝜏 − √𝜏𝜏 2 − 4𝛥𝛥 , 𝜆𝜆2 = 2 2 R > 0: 𝜏𝜏 < 0, 𝛥𝛥 > 0, the origin is stable.

R = 0: 𝜏𝜏 = 0, 𝛥𝛥 > 0, the origin is neutrally stable (center). 𝜆𝜆1 =

c) Classification of the fixed point 𝐶𝐶𝑅𝑅 2 − 4𝐿𝐿 < 0 𝐶𝐶𝑅𝑅 2 − 4𝐿𝐿 = 0 𝐶𝐶𝑅𝑅 2 − 4𝐿𝐿 < 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.

6

Problem 11.1.5 Consider the two-dimensional system of equations 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 = −𝑎𝑎(𝑥𝑥 − 𝑦𝑦) , = 𝑎𝑎(𝑥𝑥 − 𝑦𝑦) 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑥𝑥 = 𝑥𝑥0 𝑦𝑦 = 𝑦𝑦 � 𝑡𝑡 = 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.

0

Solution a) Matrix form

b) Eigenvalues

𝑨𝑨 = �

𝑑𝑑𝑑𝑑𝑑𝑑(𝑨𝑨 − 𝜆𝜆 ∙ 𝑰𝑰) = 𝑑𝑑𝑑𝑑𝑑𝑑 �

−𝑎𝑎 𝑎𝑎

𝑑𝑑𝒙𝒙 = 𝑨𝑨 ∙ 𝒙𝒙 𝑑𝑑𝑡𝑡

𝒙𝒙 = 𝒙𝒙𝟎𝟎 , 𝑡𝑡 = 0 𝑥𝑥 𝑥𝑥0 𝑎𝑎 � , 𝒙𝒙 = �𝑦𝑦� , 𝒙𝒙𝟎𝟎 = �𝑦𝑦 � −𝑎𝑎 0

−𝑎𝑎 − 𝜆𝜆 𝑎𝑎

𝑎𝑎 � = (−𝑎𝑎 − 𝜆𝜆)2 − 𝑎𝑎2 = 𝜆𝜆2 + 2𝜆𝜆𝜆𝜆 = 0 −𝑎𝑎 − 𝜆𝜆 𝜆𝜆1 = 0 𝜆𝜆2 = −2𝑎𝑎

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

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

7

e) Numerical Solutions

8

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 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 = 𝐽𝐽 , = −𝑅𝑅 + 𝐽𝐽 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 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. 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 = 𝑎𝑎 ∙ 𝐽𝐽 , = 𝑏𝑏 ∙ 𝑅𝑅 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 Case 3: Fire and Water 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 = 𝑎𝑎 ∙ 𝑅𝑅 + 𝑏𝑏 ∙ 𝐽𝐽 , = −𝑏𝑏 ∙ 𝑅𝑅 − 𝑎𝑎 ∙ 𝐽𝐽 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 Case 1 Solution: Jerry Springer a) Stability Analysis Eigenvalues

𝐴𝐴 = �

0 1 � −1 1

τ = 1, ∆ =1, D = τ2 - 4∆ = -3 𝜏𝜏 + √𝜏𝜏 2 − 4𝛥𝛥 1 √3 𝜆𝜆1 = = + 𝑖𝑖 2 2 2 2 𝜏𝜏 − √𝜏𝜏 − 4𝛥𝛥 1 √3 𝜆𝜆2 = = − 𝑖𝑖 2 2 2 Unstable Spiral b) Transient Behavior of Romeo and Juliet

9

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

0 𝐽𝐽 = � 𝑏𝑏

𝑎𝑎 � 0

𝜏𝜏 = 0 𝜆𝜆 = −√𝑎𝑎𝑎𝑎 � ∆= −𝑎𝑎𝑎𝑎 � → � 1 � 𝜆𝜆2 = √𝑎𝑎𝑎𝑎 𝜏𝜏 2 − 4∆= 4𝑎𝑎𝑎𝑎 The eigenvalues indicate 3 basic types of behavior.

Behavior 1 Behavior 2 Behavior 3

Sign of 𝑎𝑎 ∙ 𝑏𝑏 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 (𝐴𝐴 − 𝜆𝜆𝜆𝜆)𝑣𝑣 = 0. The coefficients are determined by the initial conditions.

Eigenvalues

Eigenvectors 𝑎𝑎 𝑣𝑣1 = �−�𝑏𝑏� 1 𝑎𝑎 𝑣𝑣2 = ��𝑏𝑏� 1

𝜆𝜆1 = −√𝑎𝑎𝑎𝑎 𝜆𝜆2 = √𝑎𝑎𝑎𝑎

Coefficients √𝑎𝑎𝐽𝐽0 − √𝑏𝑏𝑅𝑅0 𝑐𝑐1 = 2√𝑎𝑎

𝑐𝑐2 =

√𝑎𝑎𝐽𝐽0 + √𝑏𝑏𝑅𝑅0 2√𝑎𝑎

⎡�√𝑏𝑏𝑅𝑅0 + √𝑎𝑎𝐽𝐽0 � 𝑒𝑒 √𝑎𝑎⋅𝑏𝑏⋅𝑡𝑡 + �√𝑏𝑏𝑅𝑅0 − √𝑎𝑎𝐽𝐽0 � 𝑒𝑒 −√𝑎𝑎⋅𝑏𝑏⋅𝑡𝑡 ⎤ 𝑅𝑅(𝑡𝑡) ⎢ ⎥ 2√𝑏𝑏 2√𝑏𝑏 � � = 𝑐𝑐1 𝑒𝑒 𝜆𝜆1 𝑡𝑡 𝑣𝑣1 + 𝑐𝑐2 𝑒𝑒 𝜆𝜆2 𝑡𝑡 𝑣𝑣2 = ⎢ ⎥ 𝐽𝐽(𝑡𝑡) ⎢�√𝑏𝑏𝑅𝑅0 + √𝑎𝑎𝐽𝐽0 � 𝑒𝑒 √𝑎𝑎⋅𝑏𝑏⋅𝑡𝑡 − �√𝑏𝑏𝑅𝑅0 − √𝑎𝑎𝐽𝐽0 � 𝑒𝑒 −√𝑎𝑎⋅𝑏𝑏⋅𝑡𝑡 ⎥ ⎣ ⎦ 2√𝑎𝑎 2√𝑎𝑎 The special case of Sign(𝑎𝑎 ⋅ 𝑏𝑏) = 0requires L’Hospital’s rule or a separate analysis. 𝑅𝑅(𝑡𝑡) 𝑅𝑅0 �=� � , for a = 0 𝐽𝐽 + bR 0 𝑡𝑡 𝐽𝐽(𝑡𝑡) 0 𝑅𝑅(𝑡𝑡) 𝑅𝑅 + aJ0 𝑡𝑡 � �=� 0 � , for b = 0 𝐽𝐽0 𝐽𝐽(𝑡𝑡)

�

Special Case of R0 = 1, J0 = 0

1

�𝑒𝑒 √𝑎𝑎⋅𝑏𝑏⋅𝑡𝑡 + 𝑒𝑒 −√𝑎𝑎⋅𝑏𝑏⋅𝑡𝑡 � 2 𝑅𝑅(𝑡𝑡) � � = �1 𝑏𝑏 � 𝐽𝐽(𝑡𝑡) � �𝑒𝑒 √𝑎𝑎⋅𝑏𝑏⋅𝑡𝑡 − 𝑒𝑒 −√𝑎𝑎⋅𝑏𝑏⋅𝑡𝑡 �

(c) Phase Diagrams and Solutions

2

𝑎𝑎

10

J

1

1

Behavior 1: Saddle, 𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆(𝑎𝑎 ∙ 𝑏𝑏) > 0 R 0

2

1

4

6

8

t

1

0

1

R, J

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

Behavior 2: Center, 𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆(𝑎𝑎 ∙ 𝑏𝑏) < 0

J

1

R, J 1

0 5

1

1

0 R

10

15

20

25

30

t

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, 𝑆𝑆𝑆𝑆𝑆𝑆𝑆𝑆(𝑎𝑎 ∙ 𝑏𝑏) = 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)

𝑎𝑎 𝑏𝑏 𝐴𝐴 = � � −𝑏𝑏 −𝑎𝑎 τ = 0, ∆ = b2 - a2, D = τ2 - 4∆ = -4 ( b2 - a2 )

b2 - a2

sign ∆

sign D

+ -

+ -

+

Stability of fixed point (0,0) Center Saddle

11

Problem 11.1.7 Suppose we have a system of three linear, first order, ordinary differential equations. The eigenvalues and eigenvectors are 𝜆𝜆1 = 2 + 𝑖𝑖 𝜆𝜆2 = 2 − 𝑖𝑖 𝜆𝜆3 = −1

𝑉𝑉�1 = [1,0,0] 𝑉𝑉�2 = [0,1,0] 𝑉𝑉�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.

12

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 (𝑥𝑥0 , 𝑦𝑦0 ) = (1,1).

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

13

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

System 1 𝑑𝑑𝑑𝑑 = 𝑥𝑥 − 𝑦𝑦 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 = 𝑥𝑥 2 − 4 𝑑𝑑𝑑𝑑

System 2 𝑑𝑑𝑑𝑑 = 𝑦𝑦 3 − 4𝑥𝑥 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 = 𝑦𝑦 3 − 𝑦𝑦 − 3𝑥𝑥 𝑑𝑑𝑑𝑑

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 𝑥𝑥0 = 1, 𝑦𝑦0 = 0 when 𝑡𝑡 = 0.

Solution System 1 Fixed Points

Stability of Fixed Points The Jacobian is:

(𝑥𝑥1 , 𝑦𝑦1 ) = (−2, −2) (𝑥𝑥2 , 𝑦𝑦2 ) = (2, 2) 1 𝐴𝐴 = � 2𝑥𝑥

−1 � 0

For (𝑥𝑥1 , 𝑦𝑦1 ) = (−2, −2) 1 −1 𝐴𝐴(−2,−2) = � � , 𝜏𝜏 = 1 , 𝛥𝛥 = −4 ⇒ Saddle −4 0 For (𝑥𝑥2 , 𝑦𝑦2 ) = (2, 2) 1 −1 𝐴𝐴(2,2) = � � , 𝜏𝜏 = 1 , 𝛥𝛥 = 4 , disc = -15 ⇒ Unstable Spiral 4 0 Phase Portrait

14

Solution System 2 Fixed Points

Stability of Fixed Points The Jacobian is:

(𝑥𝑥1 , 𝑦𝑦1 ) = (0,0) (𝑥𝑥2 , 𝑦𝑦2 ) = (−2, −2) (𝑥𝑥3 , 𝑦𝑦3 ) = (2, 2)

𝜕𝜕𝜕𝜕 ⎡ 𝜕𝜕𝑥𝑥 𝐴𝐴 = ⎢⎢ 𝜕𝜕𝜕𝜕 ⎢ ⎣ 𝜕𝜕𝜕𝜕

𝜕𝜕𝜕𝜕 ⎤ 𝜕𝜕𝜕𝜕 ⎥ −4 3𝑦𝑦 2 = � � 𝜕𝜕𝜕𝜕⎥ −3 3𝑦𝑦 2 − 1 ⎥ 𝜕𝜕𝜕𝜕⎦ (𝑥𝑥1 , 𝑦𝑦1 ) = (0,0)

−4 0 𝐴𝐴 = � �⇒ {𝜏𝜏 = −5| {𝛥𝛥 = 4| ⇒ Stable spiral −3 −1 (𝑥𝑥2 , 𝑦𝑦2 ) = (−2, −2) −4 12 𝐴𝐴 = � �⇒ {𝜏𝜏 = 7| ⇒ Saddle point −3 11 (𝑥𝑥3 , 𝑦𝑦3 ) = (2, 2) −4 12 𝐴𝐴 = � �⇒ {𝜏𝜏 = 7| ⇒ Saddle point −3 11 (b) x = y If x = y, then dx = 𝑥𝑥 3 − 4𝑥𝑥 dt . dy 3 = 𝑥𝑥 − 4𝑥𝑥 dt The rate of change is the same for x and y, thus the trajectory stays on the line x = y. (c) 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 − = (𝑦𝑦 3 − 4𝑥𝑥) − (𝑦𝑦 3 − 𝑦𝑦 − 3𝑥𝑥) 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑑𝑑(𝑥𝑥 − 𝑦𝑦) = −(𝑥𝑥 − 𝑦𝑦) 𝑑𝑑𝑑𝑑 Thus 𝑥𝑥 − 𝑦𝑦 = 𝑒𝑒 −𝑡𝑡 Phase Portrait

15

Problem 11.2.3 T

g L

θ 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, θ(t). b) Change variables in order to express the equation of motion in the form 𝑑𝑑2 𝜃𝜃 + 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 = 𝛾𝛾 𝑑𝑑𝑑𝑑 2 What are the variables τ and γ? c) Find all the equilibrium points and classify them as γ 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 𝑑𝑑2 𝜃𝜃 𝐼𝐼 2 = � moments dt I = mass moment of inertia = L2 m (for a concentrated mass). Application to the pendulum gives 𝑑𝑑 2 𝜃𝜃 2 𝐿𝐿 𝑚𝑚 2 = −𝑚𝑚𝑚𝑚 ∙ sin(𝜃𝜃) + 𝑇𝑇 dt b) Variable Change Define the natural frequency and dimensionless time as 𝑔𝑔 𝜔𝜔𝑛𝑛 = � , 𝜏𝜏 = 𝜔𝜔𝑛𝑛 𝑡𝑡 𝐿𝐿 Substitute 𝑑𝑑2 𝜃𝜃 + sin(𝜃𝜃) = 𝛾𝛾 𝑑𝑑𝑑𝑑 2 𝑇𝑇 𝑇𝑇 𝛾𝛾 = = 2 2 mL 𝜔𝜔𝑛𝑛 mgL Decompose the system into coupled first order equations. 𝑑𝑑𝑑𝑑 = 𝜔𝜔 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 = −sin(𝜃𝜃) + 𝛾𝛾 𝑑𝑑𝑑𝑑 c) Analysis of Fixed Points Equilibrium requires v* = 0

16

sin(𝜃𝜃 ∗) = 𝛾𝛾 or arcsin(γ) = θ* Note that the formal ‘arcsin’ function on most computers is usually restricted between -π/2 and π/2. This equilibrium condition can be visualized graphically by plotting sin(θ) and γ 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 





2



-0.5 -1 -1.5

To perform a linearized stability analysis, determine the Jacobian 0 1 𝐴𝐴¯ = � � −cos(𝜃𝜃) 0 The fixed points are classified according to ranges of the driving torque parameter. |γ| <1 1) θ1* = arcsin(γ) trace(A) = 0 , det(A) = cos(θ1*) > 0 → Center 2) θ2* = π - arcsin(γ) trace(A) = 0 , det(A) = cos(θ1*) < 0 → Saddle point |γ| =1 Here θ1* = θ2* = π/2 and the fixed point becomes precarious. |γ| >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. 𝑑𝑑𝑑𝑑 𝑑𝑑 2 𝜃𝜃 𝑑𝑑 1 𝑑𝑑𝑑𝑑 2 � 2 + sin(𝜃𝜃) − 𝛾𝛾� = � � � − cos(𝜃𝜃) − 𝜃𝜃 𝛾𝛾� = 0 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 2 𝑑𝑑𝑑𝑑 Thus a conserved quantity does exist 1 𝑑𝑑𝑑𝑑 2 � � − cos(𝜃𝜃) − 𝜃𝜃 𝛾𝛾 = 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 2 𝑑𝑑𝑑𝑑 This is a statement of conservation of energy: Kinetic energy + Potential energy – Work added = constant

17

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

2 3 4 5 6

Fig 2: γ = 0.5



4  3  2  



2 3 4 5 6



-2

-4

Fig 3: γ = 1

f) Small oscillation approximation Small oscillations can only occur for | γ | < 1. The angle must be close to the center fixed point θ1*. Define a small perturbation about θ1* 𝛹𝛹 = 𝜃𝜃 − 𝜃𝜃1∗ Substitute into our equation of motion in dimensionless form 𝑑𝑑 2 𝛹𝛹 + sin(𝛹𝛹 + 𝜃𝜃1∗ ) = 𝛾𝛾 𝑑𝑑𝑑𝑑 2 𝑑𝑑2 𝛹𝛹 + sin(𝛹𝛹)cos(𝜃𝜃1∗ ) + sin(𝜃𝜃1∗ )cos(𝛹𝛹) = 𝛾𝛾 𝑑𝑑𝑑𝑑 2 𝑑𝑑 2 𝛹𝛹 + sin(𝛹𝛹)�1 − 𝛾𝛾 2 + 𝛾𝛾cos(𝛹𝛹) = 𝛾𝛾 𝑑𝑑𝑑𝑑 2 Now use the small angle approximations cos(𝛹𝛹) ≃ 1 sin(𝛹𝛹) ≃ 𝛹𝛹 2 𝑑𝑑 𝛹𝛹 + 𝛹𝛹�1 − 𝛾𝛾 2 = 0 𝑑𝑑𝑑𝑑 2 The dimensionless natural frequency of this system is 𝜔𝜔𝑛𝑛 𝛺𝛺𝑛𝑛 = = (1 − 𝛾𝛾 2 )1⁄4 �g/L Thus, we see an approximate frequency for small oscillations about the centers emerge 𝑔𝑔 𝜔𝜔𝑛𝑛 = � (1 − 𝛾𝛾 2 )1⁄4 𝐿𝐿

18

Problem 11.2.4: Bifurcations Consider the system

𝑑𝑑𝑑𝑑 = 𝑦𝑦 − 2𝑥𝑥 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 = 𝜇𝜇 + 𝑥𝑥 2 − 𝑦𝑦 𝑑𝑑𝑑𝑑 a) Determine all the fixed points. What is the critical value µc at which a bifurcation occurs? b) Classify the stability of the fixed points. c) Plot a phase portrait for µ < µc and one for µ > µc.

Solution a) Fixed points occur when

𝑦𝑦 ∗ − 2𝑥𝑥 ∗ = 0 𝜇𝜇 + 𝑥𝑥 ∗ 2 − 𝑦𝑦 ∗ = 0

Solve to get

𝑥𝑥1∗ = 1 + �1 − 𝜇𝜇, 𝑦𝑦1∗ = 2𝑥𝑥1∗ 𝑥𝑥2∗ = 1 − �1 − 𝜇𝜇, 𝑦𝑦2∗ = 2𝑥𝑥2∗ The critical value is 𝜇𝜇𝑐𝑐 = 1. No real fixed points exist for 𝜇𝜇 > 𝜇𝜇𝑐𝑐 . b) Stability of fixed points

At 𝑥𝑥1∗ = 1 + �1 − 𝜇𝜇

𝑑𝑑𝑑𝑑 = 𝑦𝑦 − 2𝑥𝑥 = 𝑓𝑓(𝑥𝑥, 𝑦𝑦) 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 = 𝜇𝜇 + 𝑥𝑥 2 − 𝑦𝑦 = 𝑔𝑔(𝑥𝑥, 𝑦𝑦) 𝑑𝑑𝑑𝑑 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 ⎡ ⎤ 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕 ⎥ −2 1 ⎢ Jacobian = 𝐉𝐉 = ⎢ =� � 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕⎥ 2𝑥𝑥 −1 ⎢ ⎥ ⎣ 𝜕𝜕𝜕𝜕 𝜕𝜕𝜕𝜕⎦

−2 1 −2 1 �=� � ∗ 2𝑥𝑥1 −1 2�1 + �1 − 𝜇𝜇� −1 The eigenvalues are determined from −2 − 𝜆𝜆 1 𝑑𝑑𝑑𝑑𝑑𝑑 � � = (−2 − 𝜆𝜆)(−1 − 𝜆𝜆) − 2�1 + �1 − 𝜇𝜇� 2�1 + �1 − 𝜇𝜇� −1 − 𝜆𝜆 = 3𝜆𝜆 + 𝜆𝜆2 − 2�1 − 𝜇𝜇 = 0 𝐉𝐉𝟏𝟏 = �

𝜆𝜆 =

This is a saddle point for 𝜇𝜇 > 𝜇𝜇𝑐𝑐 = 1.

−1 ± �1 + 24�1 − 𝜇𝜇

Similarly for at 𝑥𝑥2∗ = 1 − �1 − 𝜇𝜇 we find

6

19

𝜆𝜆 =

This is a stable spiral for 𝜇𝜇 > 𝜇𝜇𝑐𝑐 = 1.

−1 ± �1 − 24�1 − 𝜇𝜇 6

c) Phase Portraits

20

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. 𝑑𝑑𝑑𝑑 1 𝑑𝑑 2 𝑋𝑋 = −𝐵𝐵 ∙ − 𝑋𝑋 �1 − � 2 𝑑𝑑𝜏𝜏 𝑑𝑑𝑑𝑑 √𝐴𝐴2 + 𝑋𝑋 2 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 𝑚𝑚

𝑑𝑑2 𝑥𝑥 𝑑𝑑𝑑𝑑 𝐿𝐿0 = −𝑏𝑏 ∙ − 𝑘𝑘𝑘𝑘 �1 − � 2 𝑑𝑑𝑑𝑑 𝑑𝑑𝑡𝑡 √𝑎𝑎2 + 𝑥𝑥 2 𝑑𝑑𝑑𝑑 𝑥𝑥 = 𝑥𝑥0 , = 𝑣𝑣0 , 𝑡𝑡 = 0 𝑑𝑑𝑑𝑑 21

⇒

Forcing Functions x0, v0 b) Dimensionless Form

⇒

System m, b, k, L0, a

Response x(t), v(t)

𝑑𝑑 2 𝑋𝑋 𝑑𝑑𝑑𝑑 1 = −𝐵𝐵 ∙ − 𝑋𝑋 �1 − � 𝑑𝑑𝑑𝑑 2 𝑑𝑑𝜏𝜏 √𝐴𝐴2 + 𝑋𝑋 2 𝑥𝑥

𝑘𝑘

Variables: 𝑋𝑋 = 𝐿𝐿 , 𝜏𝜏 = 𝑡𝑡�𝑚𝑚 0

𝑎𝑎

Parameters: 𝐴𝐴 = 𝐿𝐿 , 𝐵𝐵 = 0

𝑏𝑏

√𝑚𝑚𝑚𝑚

Decomposition into a system of first order ODE’s 𝑑𝑑𝑑𝑑 = 𝑉𝑉 𝑑𝑑𝜏𝜏 𝑑𝑑𝑑𝑑 1 = −𝐵𝐵 ∙ 𝑉𝑉 − 𝑋𝑋 �1 − � 𝑑𝑑𝜏𝜏 √𝐴𝐴2 + 𝑋𝑋 2 c) Fixed Points Real Variables (𝑥𝑥 ∗ , 𝑣𝑣 ∗ ) = (0,0)

(𝑥𝑥 ∗ , 𝑣𝑣 ∗ ) = �±�𝐿𝐿20 − 𝑎𝑎2 , 0� , 𝑎𝑎 ≤ 𝐿𝐿0

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

(𝑋𝑋 ∗ , 𝑉𝑉 ∗ ) = �±�1 − 𝐴𝐴2 , 0� , 𝐴𝐴 ≤ 1

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

𝑑𝑑𝑑𝑑 1 = 𝑋𝑋 �1 − � 𝑑𝑑𝜏𝜏 √𝐴𝐴2 + 𝑋𝑋 2 d.1) When is the highly damped approximation valid? 𝑏𝑏 2 𝑏𝑏 ≫ 1 𝑜𝑜𝑜𝑜 𝑚𝑚 ≪ 𝑜𝑜𝑜𝑜 𝑏𝑏 ≫ √𝑚𝑚𝑚𝑚 𝐵𝐵 = 𝑘𝑘 √𝑚𝑚𝑚𝑚 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. 𝐵𝐵 ∙

∗

𝑋𝑋 = 0 𝑋𝑋 = ±�1 − 𝐴𝐴2 ∗

A>1 Stable DNE

A=1 Neutrally stable Neutrally stable

A<1 Unstable Stable

22

d.3) Bifurcation Diagram: Supercritical Pitchfork





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



r  rc

r  rc

dx dt

dx dt

1

x

1



1

x

1



1

xt

xt

xt

1

1

1

0

0

0

1 0

2



r  rc dx dt

4

6

8

10

t

1 0

2

4

6

8

10

t

1 0

x

1

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 B=0: Center 1

𝐵𝐵2 + 4 �𝐴𝐴 − 1� < 0: Stable spiral ±�1 − 𝐴𝐴2

1

𝐵𝐵2 + 4 �𝐴𝐴 − 1� > 0: Stable node DNE

A=1

A<1

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

Saddle

Non-isolated fixed point: B=0: Center Solution shows a center for B=0 𝐵𝐵2 − 4(1 − 𝐴𝐴2 ) < 0: Stable spiral and a stable spiral for B>0 𝐵𝐵2 − 4(1 − 𝐴𝐴2 ) > 0: Stable node

23

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)

24

Phase Portraits and Solutions for A<1

Fig 3: Stable fixed point at (±√𝟏𝟏 − 𝑨𝑨𝟐𝟐 ,0) B=0 (center), B=0.1 (stable spiral), B=1 (stable node)

Note that there are

6 different qualitative behaviors.

25

f) Tilted Wire Equation of Motion 𝑚𝑚

Dimensionless Form

𝑑𝑑𝑑𝑑 1 𝑑𝑑 2 𝑥𝑥 = −𝑏𝑏 ∙ − 𝑘𝑘𝑘𝑘 �1 − � + 𝑚𝑚𝑚𝑚 ∙ 𝑠𝑠𝑠𝑠𝑠𝑠(𝜃𝜃) 2 𝑑𝑑𝑡𝑡 𝑑𝑑𝑑𝑑 √𝑎𝑎2 + 𝑥𝑥 2

𝑑𝑑 2 𝑋𝑋 𝑑𝑑𝑑𝑑 1 = −𝐵𝐵 ∙ − 𝑋𝑋 �1 − � + 𝛤𝛤 ∙ 𝑠𝑠𝑠𝑠𝑠𝑠(𝜃𝜃) 𝑑𝑑𝑑𝑑 2 𝑑𝑑𝜏𝜏 √𝐴𝐴2 + 𝑋𝑋 2 𝑥𝑥

𝑘𝑘

Variables: 𝑋𝑋 = 𝐿𝐿 , 𝜏𝜏 = 𝑡𝑡�𝑚𝑚 𝑎𝑎

0

Parameters: 𝐴𝐴 = 𝐿𝐿 , 𝐵𝐵 = 0

𝑚𝑚∙𝑔𝑔

A new dimensionless parameter has emerged: 𝛤𝛤 = 𝑘𝑘∙𝐿𝐿 Fixed Points The fixed points come from the roots of −𝑋𝑋 ∗ �1 −

0

𝑏𝑏

√𝑚𝑚𝑚𝑚

𝑚𝑚∙𝑔𝑔

, 𝛤𝛤 = 𝑘𝑘∙𝐿𝐿

0

1

� + 𝛤𝛤 ∙ 𝑠𝑠𝑠𝑠𝑠𝑠(𝜃𝜃) = 0 �𝐴𝐴2 + 𝑋𝑋 ∗ 2 Actually, the term 𝛤𝛤 ∙ 𝑠𝑠𝑠𝑠𝑠𝑠(𝜃𝜃) can be considered as a single push.

1 � + 𝐹𝐹 = 0 −𝑋𝑋 ∗ �1 − �𝐴𝐴2 + 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.

X 1

1

A

Bifurcation diagram is for F=0.1

26

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 𝑏𝑏 𝑑𝑑𝑥𝑥/𝑑𝑑𝑑𝑑. a) Derive Newton’s law for the motion for the bead. b) Transform the equation of motion into the following dimensionless form. 𝑑𝑑 2 𝑋𝑋 𝑑𝑑𝑑𝑑 1 = −𝐵𝐵 − 𝑋𝑋 �1 − � + 𝐹𝐹 2 𝑑𝑑𝜏𝜏 𝑑𝑑𝑑𝑑 √𝐴𝐴2 + 𝑋𝑋 2 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

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

𝑑𝑑 2 𝑥𝑥 𝑑𝑑𝑑𝑑 𝐿𝐿0 𝑚𝑚 2 = −𝑏𝑏 ∙ − 𝑘𝑘𝑘𝑘 �1 − � + 𝑓𝑓 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 √𝑎𝑎2 + 𝑥𝑥 2 ⇒

⇒

System m, b, k, L0, a

Response x(t), v(t)

𝑑𝑑 2 𝑋𝑋 𝑑𝑑𝑑𝑑 1 = −𝐵𝐵 ∙ − 𝑋𝑋 �1 − � + 𝐹𝐹 2 𝑑𝑑𝑑𝑑 𝑑𝑑𝜏𝜏 √𝐴𝐴2 + 𝑋𝑋 2 𝑥𝑥

𝑘𝑘

Variables: 𝑋𝑋 = 𝐿𝐿 , 𝜏𝜏 = 𝑡𝑡�𝑚𝑚 𝑎𝑎

0

Parameters: 𝐴𝐴 = 𝐿𝐿 , 𝐵𝐵 = 0

𝑏𝑏

√𝑚𝑚𝑚𝑚

𝑓𝑓

, 𝐹𝐹 = 𝑘𝑘∙𝐿𝐿

Decomposition into a system of first order ODE’s 𝑑𝑑𝑑𝑑 = 𝑉𝑉 𝑑𝑑𝜏𝜏 𝑑𝑑𝑑𝑑 1 = −𝐵𝐵 ∙ 𝑉𝑉 − 𝑋𝑋 �1 − � + 𝐹𝐹 𝑑𝑑𝜏𝜏 √𝐴𝐴2 + 𝑋𝑋 2 c) Zero Forcing

0

(c.1) Fixed Points: F = 0 27

Real Variables (𝑥𝑥 ∗ , 𝑣𝑣 ∗ ) (0,0)

(±�𝐿𝐿20 − 𝑎𝑎2 , 0)

A pitchfork bifurcation occurs for a = ac = L0

A<1

A=1

A>1

Dimensionless Variables (𝑋𝑋 ∗ , 𝑉𝑉 ∗ ) (0,0) �±�1 − 𝐴𝐴2 , 0� , 𝐴𝐴 ≤ 1 𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜

A pitchfork bifurcation occurs for A = Ac = 1

Table 1: Stability of the Fixed Points: F = 0 (𝑋𝑋 ∗ , 𝑉𝑉 ∗ ) = (0,0) (𝑋𝑋 ∗ , 𝑉𝑉 ∗ ) = �±�1 − 𝐴𝐴2 , 0� Saddle

B=0: Center 𝐵𝐵 < �4(1 − 𝐴𝐴2 ): Stable spiral 𝐵𝐵 > �4(1 − 𝐴𝐴2 ): Stable node Non-isolated fixed point: Solution shows Non-isolated fixed point: Solution shows a center for B=0 and a stable spiral for a center for B=0 and a stable spiral for B>0 B>0 B=0: Center Does not exist 1 𝐵𝐵 < �4 �1 − 𝐴𝐴�: Stable spiral 1

𝐵𝐵 > �4 �1 − 𝐴𝐴�: Stable node

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

1

A

 1

28

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

29

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

30

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

1

� + 𝐹𝐹 = 0 �𝐴𝐴2 + 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 𝜕𝜕𝜕𝜕 =0 𝜕𝜕𝜕𝜕 𝑊𝑊 = 0 𝜕𝜕𝜕𝜕

Solving 𝜕𝜕𝜕𝜕 = 0 gives (see Mathematica code for details) Substituting into 𝑊𝑊 = 0then gives

𝑋𝑋 = −�𝐴𝐴4⁄3 − 𝐴𝐴2

𝐹𝐹𝐹𝐹(𝐴𝐴) = (1 − 𝐴𝐴2⁄3 )3⁄2

31

Fc 1.0

0.8

0.6

0.4

0.2

0.2

0.4

0.6

0.8

1.0

1.2

A

1.4

Table 1: Stability of the Fixed Points for F > 0 (𝑋𝑋 ∗ , 𝑉𝑉 ∗ ) = (𝑋𝑋−∗ , 0) B = 0: Center A<1 0<F<Fc B > 0: Stable node or spiral DNE A<1

F>Fc A>1 All F

DNE

(𝑋𝑋 ∗ , 𝑉𝑉 ∗ ) = (𝑋𝑋0∗ , 0) Saddle

(𝑋𝑋 ∗ , 𝑉𝑉 ∗ ) = (𝑋𝑋+∗ , 0) B = 0: Center B > 0: Stable node or spiral

DNE

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

DNE

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

As depicted in the following bifurcation diagrams, 𝑋𝑋−∗ is the left most point 𝑋𝑋0∗ is the middle point and 𝑋𝑋+∗ is the right most point. Note that for F = 0, 𝑋𝑋−∗ = −�1 − 𝐴𝐴2 𝑋𝑋+∗ = �1 − 𝐴𝐴2 𝑋𝑋0∗ = 0 The points 𝑋𝑋−∗ 𝑎𝑎𝑎𝑎𝑎𝑎 𝑋𝑋0∗ disappear in a saddle bifurcation as when F becomes greater that Fc.

(d.3) Bifurcation Diagrams

32

(d.4)

33

Problem 11.2.7 The Kermack-McKendrick (1927) model of an epidemic is 𝑑𝑑𝑑𝑑 = −𝑘𝑘𝑘𝑘𝑘𝑘 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 = 𝑘𝑘𝑘𝑘𝑘𝑘 − 𝑟𝑟𝑟𝑟 𝑑𝑑𝑑𝑑 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*

b) Vector Field y

𝛥𝛥 = 0 ⇒ Non-isolated fixed point 𝜏𝜏 = kx − 𝑙𝑙 k1 , l1

2 1.5 1 0.5

0.5

c) Conserved Quantity

1

1.5

2

x

𝑑𝑑𝑑𝑑 𝑙𝑙 = −1 + 𝑑𝑑𝑑𝑑 kx 𝑙𝑙 𝑦𝑦 + 𝑥𝑥 − 𝑘𝑘 ℓ𝑛𝑛(𝑥𝑥) = 𝐶𝐶 = conserved quantity

34

d) Phase Portrait: 𝑘𝑘 = ℓ = 1

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

𝑙𝑙 𝑘𝑘

35

Problem 11.2.8: Predator Prey Model Considered the system

𝑑𝑑𝑑𝑑 = 𝑥𝑥(𝑥𝑥(1 − 𝑥𝑥) − 𝑦𝑦) 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃: = 𝑦𝑦(𝑥𝑥 − 𝑎𝑎) 𝑑𝑑𝑑𝑑 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. 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃:

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) �𝑎𝑎, 𝑎𝑎(1 − 𝑎𝑎)�

Eigenvalues 𝜆𝜆1 = 0 𝜆𝜆2 = −𝑎𝑎 𝜆𝜆1 = −1 𝜆𝜆2 = −𝑎𝑎 + 1

1 𝜆𝜆1,2 = �𝑎𝑎(1 − 2𝑎𝑎) ± 𝑎𝑎 �4𝑎𝑎2 − 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

36

(c) Phase plot for a > 1 0.5

Wolves

0.4

0.3

0.2

0.1

0.0 0.0

0.5

1.5

1.0

Bunnies (d) Examination of the eigenvalues of the fixed point �𝑎𝑎, 𝑎𝑎(1 − 𝑎𝑎)� 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.

  0.4

0.3

0.2

 0.2



Im 

Re , Im 

3 2

1 2

0.2

a

0.1

1  1.0

 0.4

 0.8

 0.6

 0.4

 0.2

 0.1  0.2

0.6 



Re 

 0.3

 0.8

Re - solid, Im - Dashed

(e) According to Strogatz, p. 251, 𝜔𝜔 = 𝐼𝐼𝐼𝐼(𝜆𝜆) evaluated at a = ac =1/2. Thus √2 𝜔𝜔 = 𝐼𝐼𝐼𝐼(𝜆𝜆) = = 0.353 𝑟𝑟𝑟𝑟𝑟𝑟/𝑠𝑠 4 𝜔𝜔 𝑓𝑓 = = 0.056 𝐻𝐻𝐻𝐻 2𝜋𝜋

37

(f) Phase portraits for the different behaviors. a0.5

Wolves

Wolves

a0.25

0.5

1

0

0.5

Bunnies

a0.7

a0.866025

1

Wolves

Bunnies

Wolves

0

0.5

0

1

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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Problem 11.2.9 Consider the predator-prey model.

𝑑𝑑𝑑𝑑 𝑦𝑦 = 𝑥𝑥 �𝑏𝑏 − 𝑥𝑥 − � 𝑑𝑑𝑑𝑑 1 + 𝑥𝑥 𝑥𝑥 𝑑𝑑𝑑𝑑 = 𝑦𝑦 � − 𝑎𝑎𝑎𝑎� 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃: 1 + 𝑥𝑥 𝑑𝑑𝑑𝑑 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(𝑏𝑏 − 2) 𝑎𝑎 = 𝑎𝑎𝑐𝑐 = 2 𝑏𝑏 (𝑏𝑏 + 2) and b > 2. (Hint: A necessary condition for a Hopf bifurcation to occur is 𝜏𝜏 = 0, where τ is the trace of the Jacobian matrix at the fixed point. Show that 𝜏𝜏 = 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. 𝑃𝑃𝑃𝑃𝑃𝑃𝑃𝑃:

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 𝑦𝑦 𝑥𝑥 intersection of 𝑏𝑏 − 𝑥𝑥 − 1+𝑥𝑥 = 0 and 1+𝑥𝑥 − 𝑎𝑎𝑎𝑎 = 0 that has no exact algebraic expression. From 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.

39

Part c After following the suggestions, I was able to show that 4(𝑏𝑏 − 2) 𝑎𝑎 = 𝑎𝑎𝑐𝑐 = 2 𝑏𝑏 (𝑏𝑏 + 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

40

Problem 11.2.10 In Problem 10.15 an improved model of a laser was introduced. 𝑑𝑑𝑑𝑑 = 𝐺𝐺𝐺𝐺𝐺𝐺 − 𝑘𝑘𝑘𝑘 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 = −𝐺𝐺𝐺𝐺𝐺𝐺 − 𝑓𝑓𝑓𝑓 + 𝑝𝑝 𝑑𝑑𝑑𝑑 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

𝑑𝑑𝑑𝑑 = 𝑥𝑥𝑥𝑥 − 𝑥𝑥 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸 𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴𝐴: = −𝑥𝑥𝑥𝑥 − 𝐹𝐹𝐹𝐹 + 𝑆𝑆 𝑑𝑑𝑑𝑑 Dependent variables: 𝑥𝑥 = 𝑛𝑛 𝐺𝐺 , 𝑦𝑦 = 𝑁𝑁 𝐺𝐺 𝑘𝑘 𝑘𝑘 Independent variable: 𝜏𝜏 = 𝑡𝑡𝑡𝑡 Parameters: 𝐹𝐹 = 𝑓𝑓, 𝑆𝑆 = 𝑝𝑝 𝐺𝐺 𝑘𝑘 𝑘𝑘 2 𝑃𝑃ℎ𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜:

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

𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿: 𝐹𝐹 > 𝑆𝑆

Stable node does not exist

𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿𝐿: 𝑆𝑆 > 𝐹𝐹

𝑆𝑆 2 − 4𝑆𝑆 + 4𝐹𝐹 < 0 Saddle Stable spiral

𝑆𝑆 2 − 4𝑆𝑆 + 4𝐹𝐹 > 0 Saddle Stable node

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

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

41

c) Stability Diagram

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

42

Case 1: 𝑺𝑺 < 𝑭𝑭 𝑆𝑆 Stable node at �0, 𝐹𝐹�

43

Case 2: 𝑺𝑺 > 𝑭𝑭 and 𝑺𝑺𝟐𝟐 − 𝟒𝟒𝑺𝑺 + 𝟒𝟒𝑭𝑭 < 𝟎𝟎 Stable spiral at (𝑆𝑆 − 𝐹𝐹, 1) Saddle at (0, 𝑆𝑆/𝐹𝐹)

44

Case 3: 𝑺𝑺 > 𝑭𝑭 and 𝑺𝑺𝟐𝟐 − 𝟒𝟒𝑺𝑺 + 𝟒𝟒𝑭𝑭 < 𝟎𝟎 Stable node at (𝑆𝑆 − 𝐹𝐹, 1) Saddle at (0, 𝑆𝑆/𝐹𝐹)

45

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 𝑑𝑑𝑛𝑛1 = 𝐺𝐺1 𝑁𝑁𝑛𝑛1 − 𝑘𝑘1 𝑛𝑛1 𝑑𝑑𝑑𝑑 𝑑𝑑𝑛𝑛2 = 𝐺𝐺2 𝑁𝑁𝑛𝑛2 − 𝑘𝑘2 𝑛𝑛2 𝑑𝑑𝑑𝑑 𝑁𝑁(𝑡𝑡) = 𝑁𝑁0 − 𝛼𝛼1 𝑛𝑛1 − 𝛼𝛼2 𝑛𝑛2 is number of excited atoms n1 and n2 are the number of photons. The parameters 𝐺𝐺1 , 𝐺𝐺2 , 𝑘𝑘1 , 𝐺𝐺𝐺𝐺2 , 𝛼𝛼1 , 𝛼𝛼2 , 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 𝑑𝑑𝑛𝑛1 𝑘𝑘1 = 𝐺𝐺1 𝑁𝑁 𝑛𝑛1 − 𝑘𝑘1 𝑛𝑛1 = 𝐺𝐺1 (𝑁𝑁0 − 𝛼𝛼1 𝑛𝑛1 − 𝛼𝛼2 𝑛𝑛2 ) 𝑛𝑛1 − 𝑘𝑘1 𝑛𝑛1 = 𝐺𝐺1 𝑛𝑛1 �𝑁𝑁0 − − 𝛼𝛼1 𝑛𝑛1 − 𝛼𝛼2 𝑛𝑛2 � 𝑑𝑑𝑑𝑑 𝐺𝐺1 𝑑𝑑𝑛𝑛2 𝑘𝑘2 = 𝐺𝐺2 𝑁𝑁 𝑛𝑛2 − 𝑘𝑘2 𝑛𝑛2 = 𝐺𝐺2 (𝑁𝑁0 − 𝛼𝛼1 𝑛𝑛1 − 𝛼𝛼2 𝑛𝑛2 ) 𝑛𝑛2 − 𝑘𝑘2 𝑛𝑛2 = 𝐺𝐺2 𝑛𝑛2 �𝑁𝑁0 − − 𝛼𝛼1 𝑛𝑛1 − 𝛼𝛼2 𝑛𝑛2 � 𝑑𝑑𝑑𝑑 𝐺𝐺2 Fixed Points It would help to first draw the nullclines. 𝑛𝑛1 = 0 𝑛𝑛2 = 0 𝑘𝑘1 𝑁𝑁0 − − 𝛼𝛼1 𝑛𝑛1 − 𝛼𝛼2 𝑛𝑛2 = 0 𝐺𝐺1 𝑘𝑘1 𝑁𝑁0 − − 𝛼𝛼1 𝑛𝑛1 − 𝛼𝛼2 𝑛𝑛2 = 0 𝐺𝐺1

𝑑𝑑𝑛𝑛

𝑑𝑑𝑛𝑛

Note that 𝑛𝑛1 = 0 always makes 1 = 0 and 𝑛𝑛2 = 0 always makes 2 = 0. Thus (0,0) is a fixed 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 point. We identify another fixed point when 𝑛𝑛1 = 0 if we also require 𝑘𝑘2 1 𝑘𝑘2 𝑁𝑁0 − − 𝛼𝛼1 0 − 𝛼𝛼2 𝑛𝑛2 = 0 ⇒ 𝑛𝑛2 = �𝑁𝑁0 − � 𝐺𝐺2 𝛼𝛼2 𝐺𝐺2

46

𝑘𝑘

This fixed point only exists when 𝑁𝑁0 − 𝐺𝐺2 > 0. Similarly, another fixed point can be found at 2 1 𝑘𝑘1 � �𝑁𝑁0 − � , 0� 𝛼𝛼1 𝐺𝐺1 𝑘𝑘1 This fixed point exists only when 𝑁𝑁0 − 𝐺𝐺 > 0. Another possibility is for 1 𝑘𝑘1 𝑁𝑁0 − − 𝛼𝛼1 𝑛𝑛1 − 𝛼𝛼2 𝑛𝑛2 = 0 𝐺𝐺1 𝑘𝑘2 𝑁𝑁0 − − 𝛼𝛼1 𝑛𝑛1 − 𝛼𝛼2 𝑛𝑛2 = 0 𝐺𝐺2 These lines have the same slope so they can be true simultaneously only if 𝑘𝑘1 𝑘𝑘2 = 𝐺𝐺1 𝐺𝐺2 In summary, our fixed point possibilities are (n1*, n2* ) (0 , 0) 1 𝑘𝑘2 �0, �𝑁𝑁0 − �� 𝛼𝛼2 𝐺𝐺2 1 𝑘𝑘1 � �𝑁𝑁0 − � , 0� 𝛼𝛼1 𝐺𝐺1 All points along the line 𝑘𝑘1 𝑁𝑁0 − − 𝛼𝛼1 𝑛𝑛1 − 𝛼𝛼2 𝑛𝑛2 = 0 𝐺𝐺1

Analysis of Fixed Points The Jacobian matrix is 𝑘𝑘1 ⎡𝐺𝐺1 �𝑁𝑁0 − − 𝛼𝛼1 𝑛𝑛1 − 𝛼𝛼2 𝑛𝑛2 � − 𝛼𝛼1 𝐺𝐺1 𝑛𝑛1 𝐺𝐺1 𝐴𝐴 = ⎢ ⎢ −𝛼𝛼1 𝐺𝐺2 𝑛𝑛2 ⎣ Analysis of Fixed Point (n1*, n2*) = (0 , 0)

Restriction 𝑁𝑁0 −

𝑘𝑘2 >0 𝐺𝐺2

𝑘𝑘1 >0 𝐺𝐺1 𝑘𝑘1 𝑘𝑘2 = 𝐺𝐺1 𝐺𝐺2

𝑁𝑁0 −

⎤ ⎥ 𝑘𝑘2 𝐺𝐺2 �𝑁𝑁0 − − 𝛼𝛼1 𝑛𝑛1 − 𝛼𝛼2 𝑛𝑛2 � − 𝛼𝛼2 𝐺𝐺2 𝑛𝑛2 ⎥ ⎦ 𝐺𝐺2 −𝛼𝛼2 𝐺𝐺1 𝑛𝑛1

𝑘𝑘1 ⎡𝐺𝐺1 �𝑁𝑁0 − � ⎤ 0 𝐺𝐺1 ⎢ ⎥ 𝐴𝐴(0,0) = 𝑘𝑘2 ⎥ ⎢ 0 𝐺𝐺2 �𝑁𝑁0 − � ⎣ 𝐺𝐺2 ⎦ 𝑘𝑘1 𝑘𝑘2 𝜏𝜏 = 𝐺𝐺1 �𝑁𝑁0 − � + 𝐺𝐺2 �𝑁𝑁0 − � 𝐺𝐺1 𝐺𝐺2 𝑘𝑘1 𝑘𝑘2 𝛥𝛥 = 𝐺𝐺1 �𝑁𝑁0 − � 𝐺𝐺2 �𝑁𝑁0 − � 𝐺𝐺1 𝐺𝐺2 𝑘𝑘

𝑘𝑘

Stability depends on the signs of �𝑁𝑁0 − 𝐺𝐺1 � and�𝑁𝑁0 − 𝐺𝐺2 �. A table is useful for summarizing the possibilities.

1

2

47

𝑘𝑘

Sign of �𝑁𝑁0 − 𝐺𝐺1 � + + -

𝑘𝑘

Sign of �𝑁𝑁0 − 𝐺𝐺2 �

1

+ + -

Sign of τ

Sign of ∆

Stability of (0, 0)

+ ? ?

-

+ +

Unstable node Saddle point Saddle point Stable node

𝑘𝑘

𝑘𝑘

2

The only stable or attracting combination is �𝑁𝑁0 − 𝐺𝐺1 �and �𝑁𝑁0 − 𝐺𝐺2 �. Note that the 1

2

discriminate, τ2 – 4 ∆ > 0 for all cases. The reasoning is that we have the form τ = a + b, ∆ = a*b τ2 – 4 ∆ = (a-b)2 > 0 Thus, no spirals exist.

1

𝑘𝑘

Analysis of Fixed Point (n1*, n2*) =(0, 𝛼𝛼 �𝑁𝑁0 − 𝐺𝐺2 �) 2

⎡𝐺𝐺 ��𝑁𝑁 − 𝑘𝑘1 � − �𝑁𝑁 − 𝑘𝑘2 �� 1 0 0 𝐺𝐺1 𝐺𝐺2 ⎢ 𝐴𝐴 = ⎢ ⎢ 1 𝑘𝑘2 ⎢ −𝛼𝛼1 𝐺𝐺2 �𝑁𝑁0 − � ⎣ 𝐺𝐺2 𝛼𝛼2

2

⎤ ⎡ 𝐺𝐺 �− 𝑘𝑘1 + 𝑘𝑘2 � ⎤ 1 0 0 𝐺𝐺1 𝐺𝐺2 ⎥ ⎢ ⎥ ⎥=⎢ ⎥ ⎥ 1 𝑘𝑘 𝑘𝑘 2 2 ⎢ ⎥ −𝛼𝛼2 𝐺𝐺2 𝑛𝑛2 ⎥ ⎣−𝛼𝛼1 𝐺𝐺2 𝛼𝛼 �𝑁𝑁0 − 𝐺𝐺 � −𝐺𝐺2 �𝑁𝑁0 − 𝐺𝐺 �⎦ 2 2 2 ⎦ 𝑘𝑘1 𝑘𝑘2 𝑘𝑘2 𝜏𝜏 = 𝐺𝐺1 �− + � − 𝐺𝐺2 �𝑁𝑁0 − � 𝐺𝐺1 𝐺𝐺2 𝐺𝐺2 𝑘𝑘1 𝑘𝑘2 𝑘𝑘2 𝛥𝛥 = −𝐺𝐺1 �− + � 𝐺𝐺2 �𝑁𝑁0 − � 𝐺𝐺1 𝐺𝐺2 𝐺𝐺2 𝑘𝑘2 Since this fixed point only exists if �𝑁𝑁0 − 𝐺𝐺 � > 0, stability depends only on the sign of 𝑘𝑘

2

𝑘𝑘

�− 𝐺𝐺1 + 𝐺𝐺2 �.

1

2

𝑘𝑘

𝑘𝑘

Sign of �− 𝐺𝐺1 + 𝐺𝐺2 � + -

1

2

Sign of τ

Sign of ∆

?

-

+

𝐺𝐺 𝑁𝑁 −𝑘𝑘

Stability of (0, 2𝐺𝐺 0𝛼𝛼 2 ) 2 2

Saddle point Stable node

The only stable or attracting combination is 𝑘𝑘 𝑘𝑘 𝑘𝑘 𝑘𝑘 (𝐺𝐺2 𝑁𝑁0 − 𝑘𝑘2 ) > 0 and �− 1 + 2 � < 0 or 2 < 1 𝐺𝐺 𝐺𝐺 𝐺𝐺 𝐺𝐺 1

𝑘𝑘1

1

Analysis of Fixed Point (n1*, n2*) = (𝛼𝛼 �𝑁𝑁0 − 𝐺𝐺 �,0) 1

2

2

1

1

𝑘𝑘1 ⎡ −𝐺𝐺1 �𝑁𝑁0 − � ⎤ ⎡−𝐺𝐺 �𝑁𝑁 − 𝑘𝑘1 � −𝛼𝛼 𝐺𝐺 1 �𝑁𝑁 − 𝑘𝑘1 �⎤ −𝛼𝛼2 𝐺𝐺1 𝑛𝑛1 2 1 𝐺𝐺 1 𝛼𝛼1 0 𝐺𝐺1 ⎥ ⎢ ⎥ ⎢ 1 0 𝐺𝐺1 𝐴𝐴 = ⎢ = 𝑘𝑘2 𝑘𝑘1 ⎥ 𝑘𝑘1 𝑘𝑘2 ⎥ ⎢ 0 𝐺𝐺2 ��𝑁𝑁0 − � − �𝑁𝑁0 − �� ⎥ ⎢ 0 𝐺𝐺2 � − � 𝐺𝐺2 𝐺𝐺1 ⎦ ⎣ ⎦ 𝐺𝐺1 𝐺𝐺2 ⎣

48

𝑘𝑘1 𝑘𝑘1 𝑘𝑘2 � + 𝐺𝐺2 � − � 𝐺𝐺1 𝐺𝐺1 𝐺𝐺2 𝑘𝑘1 𝑘𝑘1 𝑘𝑘2 𝛥𝛥 = −𝐺𝐺1 �𝑁𝑁0 − � 𝐺𝐺2 � − � 𝐺𝐺1 𝐺𝐺1 𝐺𝐺2

𝜏𝜏 = −𝐺𝐺1 �𝑁𝑁0 − 𝑘𝑘

𝑘𝑘

𝑘𝑘

Stability depends on the signs of �𝐺𝐺1 − 𝐺𝐺2 �since�𝑁𝑁0 − 𝐺𝐺1 �>0 for this fixed point to exist. 𝑘𝑘

𝑘𝑘

1

Sign of �𝐺𝐺1 − 𝐺𝐺2 � + -

1

2

2

1

Sign of τ

Sign of ∆

?

-

+

𝐺𝐺 𝑁𝑁 −𝑘𝑘

Stability of (0, 2𝐺𝐺 0𝛼𝛼 2 ) 2 2

Saddle point Stable node

The only stable or attracting combination is 𝑘𝑘 𝑘𝑘 𝑘𝑘 𝑘𝑘 𝑘𝑘 �𝑁𝑁0 − 𝐺𝐺1 �>0 and �𝐺𝐺1 − 𝐺𝐺2 �<0 or 𝐺𝐺1 < 𝐺𝐺2 1

𝒌𝒌

1

2

1

2

Analysis of Fixed Points Along Line 𝑵𝑵𝟎𝟎 − 𝑮𝑮𝟏𝟏 − 𝜶𝜶𝟏𝟏 𝒏𝒏𝟏𝟏 − 𝜶𝜶𝟐𝟐 𝒏𝒏𝟐𝟐 = 𝟎𝟎 𝑘𝑘1

𝟏𝟏

𝑘𝑘

This case is restricted to the case where �𝐺𝐺 − 𝐺𝐺2 �.

1 2 𝑘𝑘1 ⎡𝐺𝐺1 �𝑁𝑁0 − − 𝛼𝛼1 𝑛𝑛1 − 𝛼𝛼2 𝑛𝑛2 � − 𝛼𝛼1 𝐺𝐺1 𝑛𝑛1 ⎤ −𝛼𝛼2 𝐺𝐺1 𝑛𝑛1 𝐺𝐺1 ⎢ ⎥ 𝐴𝐴 = 𝑘𝑘2 ⎢ −𝛼𝛼1 𝐺𝐺2 𝑛𝑛2 𝐺𝐺2 �𝑁𝑁0 − − 𝛼𝛼1 𝑛𝑛1 − 𝛼𝛼2 𝑛𝑛2 � − 𝛼𝛼2 𝐺𝐺2 𝑛𝑛2 ⎥ ⎣ ⎦ 𝐺𝐺2 −𝛼𝛼1 𝐺𝐺1 𝑛𝑛1 −𝛼𝛼2 𝐺𝐺1 𝑛𝑛1 =� � −𝛼𝛼1 𝐺𝐺2 𝑛𝑛2 −𝛼𝛼2 𝐺𝐺2 𝑛𝑛2 𝜏𝜏 = −𝛼𝛼1 𝐺𝐺1 𝑛𝑛1 − 𝛼𝛼2 𝐺𝐺2 𝑛𝑛2 𝛥𝛥 = (𝛼𝛼1 𝐺𝐺1 𝑛𝑛1 ) (𝛼𝛼2 𝐺𝐺2 𝑛𝑛2 ) − (𝛼𝛼1 𝐺𝐺2 𝑛𝑛2 )(𝛼𝛼2 𝐺𝐺1 𝑛𝑛1 ) = 0 Note that τ <0 and ∆ = 0. This is a line of non-isolated fixed points. They are stable or attracting.

49

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 𝑘𝑘 a =𝑁𝑁0 − 𝐺𝐺1 -

1

sign of 𝑘𝑘 b =𝑁𝑁0 − 𝐺𝐺2 -

Long-term Behavior of (n1, n2)

2

(0, 0) 1

2a 2b

+ -

+

3a

+

+

3b

+

+

4

+

+

𝑘𝑘

(𝛼𝛼 �𝑁𝑁0 − 𝐺𝐺1 �,0)

1

�0,

1

1

1 𝑘𝑘2 �𝑁𝑁0 − �� 𝛼𝛼2 𝐺𝐺2 𝑘𝑘

𝐺𝐺

𝐺𝐺

(𝛼𝛼 �𝑁𝑁0 − 𝐺𝐺1 �,0) if 𝑘𝑘1 > 𝑘𝑘2 1

1

1

1

𝑘𝑘

𝐺𝐺

2

𝐺𝐺

�0, 𝛼𝛼 �𝑁𝑁0 − 𝐺𝐺2 �� if 𝑘𝑘1 < 𝑘𝑘2 2

2

1

2

(n1, n2) on the line 𝑘𝑘1 𝑁𝑁0 − − 𝛼𝛼1 𝑛𝑛1 − 𝛼𝛼2 𝑛𝑛2 = 0 𝐺𝐺1 𝐺𝐺 𝐺𝐺 If 𝑘𝑘1 = 𝑘𝑘2 1

2

Case 4 is the most interesting since both modes of the laser can be active simultaneously, depending on the initial conditions.

50

Case 1 𝑘𝑘 𝑘𝑘 In the following graphs, (a,b) = �𝑁𝑁0 − 𝐺𝐺1 , 𝑁𝑁0 − 𝐺𝐺2 � 1

2

Case 2a

Case 2b

51

Case 3a

Case 4

52

Problem 11.2.12: Budworms vs. Forest S = Average size of trees E = Energy reserve; a measure of health B = constant budworm population 𝑆𝑆 𝐾𝐾𝐸𝐸 𝑑𝑑𝑑𝑑 = 𝑟𝑟𝑠𝑠 𝑆𝑆 �1 − � 𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇𝑇: 𝐾𝐾𝑠𝑠 𝐸𝐸 𝑑𝑑𝑑𝑑 𝑑𝑑𝑑𝑑 𝐸𝐸 𝐵𝐵 𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸𝐸: = 𝑟𝑟𝐸𝐸 𝐸𝐸 �1 − � − 𝑃𝑃 𝑑𝑑𝑑𝑑 𝐾𝐾𝐸𝐸 𝑆𝑆 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. 𝑠𝑠 =

𝑆𝑆 𝐸𝐸 , 𝑒𝑒 = , 𝜏𝜏 = 𝑟𝑟𝑠𝑠 𝑡𝑡 𝐾𝐾𝑠𝑠 𝐾𝐾𝐸𝐸

53

Step 5: Substitute and identify dimensionless parameters groups which naturally emerge. ds 𝑑𝑑𝑑𝑑 𝐾𝐾𝑠𝑠 𝑠𝑠 𝐾𝐾𝐸𝐸 𝐾𝐾𝑠𝑠 = 𝑟𝑟𝑠𝑠 𝐾𝐾𝑠𝑠 𝑠𝑠 �1 − � 𝑑𝑑𝑑𝑑 dt 𝐾𝐾𝑠𝑠 𝐾𝐾𝐸𝐸 𝑒𝑒 𝐾𝐾𝐸𝐸 𝑒𝑒 𝐵𝐵 de 𝑑𝑑𝑑𝑑 𝐾𝐾𝐸𝐸 = 𝑟𝑟𝐸𝐸 𝐾𝐾𝐸𝐸 𝑒𝑒 �1 − � − 𝑃𝑃 𝐾𝐾𝑠𝑠 𝑠𝑠 𝑑𝑑𝑑𝑑 dt 𝐾𝐾𝐸𝐸 𝑑𝑑𝑑𝑑 since dt = 𝑟𝑟𝑠𝑠 𝑑𝑑𝑑𝑑 𝑠𝑠 = 𝑠𝑠 �1 − � 𝑑𝑑𝜏𝜏 𝑒𝑒 𝑑𝑑𝑑𝑑 𝑟𝑟𝐸𝐸 𝑃𝑃 𝐵𝐵 1 = 𝑒𝑒(1 − 𝑒𝑒) − 𝑑𝑑𝜏𝜏 𝑟𝑟𝑠𝑠 𝑟𝑟𝑠𝑠 𝐾𝐾𝐸𝐸 𝐾𝐾𝑠𝑠 𝑠𝑠 Two dimensionless groups have emerged: 𝑃𝑃 𝑟𝑟𝐸𝐸 𝑅𝑅 = , 𝑏𝑏 = 𝐵𝐵 𝑟𝑟𝑠𝑠 𝐾𝐾𝐸𝐸 𝐾𝐾𝑠𝑠 𝑟𝑟𝑠𝑠 Our dimensionless mathematical model is: 𝑠𝑠 𝑑𝑑𝑑𝑑 = 𝑠𝑠 �1 − � 𝑒𝑒 𝑑𝑑𝜏𝜏 𝑑𝑑𝑑𝑑 𝑏𝑏 = 𝑅𝑅 𝑒𝑒(1 − 𝑒𝑒) − 𝑑𝑑𝜏𝜏 𝑠𝑠 𝑆𝑆 𝐸𝐸 𝑠𝑠 = , 𝑒𝑒 = , 𝜏𝜏 = 𝑟𝑟𝑠𝑠 𝑡𝑡 𝐾𝐾𝑠𝑠 𝐾𝐾𝐸𝐸 𝑟𝑟𝐸𝐸 𝑃𝑃 𝑅𝑅 = , 𝑏𝑏 = 𝐵𝐵 𝑟𝑟𝑠𝑠 𝑟𝑟𝑠𝑠 𝐾𝐾𝐸𝐸 𝐾𝐾𝑠𝑠 c) Nullclines ds 𝑠𝑠 = 𝑠𝑠 �1 − � = 0 → 𝑠𝑠 = 𝑒𝑒 𝑑𝑑𝑑𝑑 𝑒𝑒 𝑏𝑏 𝑏𝑏 1 de = 𝑅𝑅 𝑒𝑒(1 − 𝑒𝑒) − = 0 → 𝑠𝑠 = 𝑑𝑑𝑑𝑑 𝑠𝑠 𝑅𝑅 𝑒𝑒(1 − 𝑒𝑒) 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

Determination of Critical b/R The critical value is the point at which the two nullclines touch tangentially.

Equate nullclines: Equate tangents of nullclines:

𝑏𝑏

1

𝑏𝑏

𝑠𝑠 = 𝑒𝑒 = 𝑅𝑅 𝑒𝑒(1−𝑒𝑒) ⇒ 𝑒𝑒 2 (1 − 𝑒𝑒) = 𝑅𝑅

𝑑𝑑𝑒𝑒 𝑑𝑑 𝑏𝑏 1 = � � 𝑑𝑑𝑒𝑒 𝑑𝑑𝑒𝑒 𝑅𝑅 𝑒𝑒(1 − 𝑒𝑒) 𝑏𝑏 −1 + 2𝑒𝑒 � 1= � 𝑅𝑅 (1 − 𝑒𝑒)2 𝑒𝑒 2 Solve these two equations simultaneously for e and b/R. 2 𝑒𝑒cr = 𝑠𝑠cr = 3 𝑏𝑏 2 2 2 4 � � = � � �1 − � = = 0.148 𝑅𝑅 cr 3 3 27 In real variables, the critical value of B is: 𝑃𝑃 𝑃𝑃 𝑏𝑏 𝐵𝐵 𝑟𝑟𝑠𝑠 𝐾𝐾𝐸𝐸 𝐾𝐾𝑠𝑠 = = 𝐵𝐵 𝑟𝑟 𝐸𝐸 𝑅𝑅 𝑟𝑟𝐸𝐸 𝐾𝐾𝐸𝐸 𝐾𝐾𝑠𝑠 𝑟𝑟𝑠𝑠 𝑏𝑏 𝑟𝑟𝐸𝐸 𝐾𝐾𝐸𝐸 𝐾𝐾𝑠𝑠 4 𝑟𝑟𝐸𝐸 𝐾𝐾𝐸𝐸 𝐾𝐾𝑠𝑠 𝐵𝐵cr = � � = 𝑅𝑅 cr 𝑃𝑃 27 𝑃𝑃 d) Phase Portraits b/R < (b/R)cr

b/R > (b/R)cr

55

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, 𝑅𝑅, 𝑚𝑚 • Angular velocity of (again, very dumb) duck, 𝜔𝜔, 𝑟𝑟𝑟𝑟𝑟𝑟/𝑠𝑠 • Speed of (dumb) dog, 𝑉𝑉𝑑𝑑𝑑𝑑𝑑𝑑 , 𝑚𝑚/𝑠𝑠 Solution Equation of Motion for the (very dumb) duck: 𝑅𝑅 cos(𝜔𝜔𝜔𝜔 + 𝜃𝜃0 ) 𝑥𝑥⃗𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 = � � 𝑅𝑅 sin(𝜔𝜔𝑡𝑡 + 𝜃𝜃0 ) Equation of Motion for the (not quite as dumb) dog: 𝑥𝑥⃗𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 − 𝑥𝑥⃗𝑑𝑑𝑑𝑑𝑑𝑑 𝑥𝑥⃗̇𝑑𝑑𝑑𝑑𝑑𝑑 = 𝑉𝑉𝑑𝑑𝑑𝑑𝑑𝑑 � � �𝑥𝑥⃗𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 − 𝑥𝑥⃗𝑑𝑑𝑑𝑑𝑑𝑑 � ����������� Initial condition for dog:

𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢 𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 𝑑𝑑𝑑𝑑𝑑𝑑 𝑡𝑡𝑡𝑡 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑

𝑥𝑥⃗𝑑𝑑𝑑𝑑𝑑𝑑 (𝑡𝑡 = 0) = 𝑥𝑥⃗𝑑𝑑𝑑𝑑𝑔𝑔0

Nondimensionalize Select two reference parameters (two independent dimensions, length and time) Parameter Variable Units Pond Radius 𝑅𝑅 𝑚𝑚 (Dumb) Duck Velocity 𝑉𝑉𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 = 𝑅𝑅𝑅𝑅 𝑚𝑚/𝑠𝑠 Develop Reference Length

Length Time

Develop Reference Time

Length Time

𝑏𝑏 𝐿𝐿𝑐𝑐ℎ𝑎𝑎𝑎𝑎 = 𝑅𝑅 𝑎𝑎 𝑉𝑉𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 𝑚𝑚 𝑏𝑏 𝑚𝑚 = 𝑚𝑚𝑎𝑎 � � 𝑠𝑠

{𝑎𝑎, 𝑏𝑏} = {1,0} 𝐿𝐿𝑐𝑐ℎ𝑎𝑎𝑎𝑎 = 𝑅𝑅 𝑏𝑏 𝜏𝜏 = 𝑅𝑅 𝑎𝑎 𝑉𝑉𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 𝑚𝑚 𝑏𝑏 𝑠𝑠 = 𝑚𝑚𝑎𝑎 � � 𝑠𝑠

𝑎𝑎 + 𝑏𝑏 = 1 −𝑏𝑏 = 0

𝑎𝑎 + 𝑏𝑏 = 0 −𝑏𝑏 = 1

56

Dimensionless variables: Time

{𝑎𝑎, 𝑏𝑏} = {1, −1} 𝑅𝑅 1 𝜏𝜏 = = 𝑉𝑉𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 𝜔𝜔

Duck position Dog position

Non-dimensional equations of motion: Duck:

Dog:

𝑥𝑥⃗𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 + = 𝑥𝑥⃗𝑑𝑑𝑑𝑑𝑑𝑑𝑘𝑘 = 𝑅𝑅

𝑡𝑡 = 𝜔𝜔𝜔𝜔 𝜏𝜏 𝑥𝑥⃗𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 + 𝑥𝑥⃗𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 = 𝑅𝑅 𝑥𝑥 ⃗ 𝑑𝑑𝑑𝑑𝑑𝑑 + 𝑥𝑥⃗𝑑𝑑𝑑𝑑𝑑𝑑 = 𝑅𝑅 𝑡𝑡 + =

𝑡𝑡 + 𝑅𝑅 cos �𝜔𝜔 �𝑡𝑡 = � + 𝜃𝜃0 � 𝜔𝜔 � � 𝑡𝑡 + 𝑅𝑅 sin �𝜔𝜔 �𝑡𝑡 = 𝜔𝜔 � + 𝜃𝜃0 � 𝑅𝑅

+ + − 𝑥𝑥⃗𝑑𝑑𝑑𝑑𝑑𝑑 𝑉𝑉𝑑𝑑𝑑𝑑𝑑𝑑 𝑥𝑥⃗𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 𝑥𝑥⃗̇𝑑𝑑𝑑𝑑𝑑𝑑 + = 𝑥𝑥⃗̇𝑑𝑑𝑑𝑑𝑑𝑑 = � + �, + 𝜔𝜔𝜔𝜔 𝜔𝜔𝜔𝜔 ����������� �𝑥𝑥⃗𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 − 𝑥𝑥⃗𝑑𝑑𝑑𝑑𝑑𝑑 � 𝑢𝑢𝑢𝑢𝑢𝑢𝑢𝑢 𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 𝑑𝑑𝑑𝑑𝑑𝑑 𝑡𝑡𝑡𝑡 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑 𝑉𝑉𝑑𝑑𝑑𝑑𝑑𝑑

cos(𝑡𝑡 + + 𝜃𝜃0 ) =� � sin(𝑡𝑡 + + 𝜃𝜃0 )

+ (𝑡𝑡 = 0) = 𝑥𝑥⃗𝑑𝑑𝑑𝑑𝑑𝑑

𝑥𝑥⃗𝑑𝑑𝑑𝑑𝑔𝑔0 + = 𝑥𝑥⃗𝑑𝑑𝑑𝑑𝑔𝑔 0 𝑅𝑅

Thus, the only parameter to vary for the dog is 𝜔𝜔𝜔𝜔 , which, for the current purposes, will be referenced as 𝑓𝑓𝑑𝑑𝑑𝑑𝑑𝑑 .

57

Solutions for various 𝒇𝒇𝒅𝒅𝒅𝒅𝒅𝒅 ’s:

f=0.6

f=0.4

dog duck

1

0.5

y position, y/R

0.5

y position, y/R

dog duck

1

0

0

-0.5

-0.5

-1

-1

-1

-0.5

0 x position, x/R

0.5

-1

1

-0.5

0.5

1

f=1

f=0.8 dog duck

1

dog duck

1

0.5

y position, y/R

0.5

y position, y/R

0 x position, x/R

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