Homework #5 Solutions

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1 Homework #5 Solutions Math 123: Mathematical Modeling, Spring 2019 Instructor: Dr. Doreen De Leon 1. Exercise Stefan-Boltzmann s Law of Radiation states that the temperature change dt/ of a body at T degrees elvin is proportional to E 4 T 4 ; this means that dt/ = ke 4 T 4. Here, E refers to the constant absolute temperature of the surroundings measured in degrees elvin, and k is a constant. a Rewrite the differential equation dt/ = ke 4 T 4 by introducing a nondimensional temperaturet and nondimensional time t. b Determine the solution T of the nondimensional differential equation by applying the method of separation of variables. Hint: use the following integral a and C refer to any constants: dx a 4 x 4 = 1 4a 3 ln a + x + 1 x a x 2a arctan + C. 3 a c Explain how the dependence of T on t can be graphically illustrated for a given value of the initial nondimensional temperature T 0. a Dividing both sides by E 4 gives 1 d T E E 3 = k 1 4 T, E so defining T = T gives a nondimensional temperature temperature. The differential equation is thus E 1 dt E 3 = k1 T 4. Noting that the constant k has dimension of temp 3 T 1, if we divide both sides by k, we have 1 dt ke 3 = 1 T 4. So, defining t = tk E 3 gives a nondimensional time, and the resulting differential equation is dt = 1 T 4. 1

2 b Using the given hint, we may solve the differential equation as follows. dt = 1 T ln T T 2 arctan T + C = t + C ln T T 2 arctan T = t + c. c Given a particular T 0, we may solve for the constant c. Then, using computational software, we can obtain a graph of T versus t. 2. Exercise A modification of the logistic model is given by the model of Schaefer = 1 1 E. τ The model, which was developed for the simulation of the development of fish populations, is equivalent to the logistic model for E = 0, where L = = 0 is assumed for simplicity. The last term E takes into account human predation that reduces the rate of population growth. It is reasonable to consider this term to be proportional to : the effect of predation will increase with population density. The variables, E, 1/τ, and τ are assumed to be non-negative and constant. a Write the model in the form of the logistic model the structure of this rewritten model will be equal to the logistic model but the parameters are different. b Calculate the solution of this rewritten model by taking reference to the solution of the logistic model. c Explain the effect of a nonzero E on the population dynamics in comparison to the logistic model. a The equation may be written = 1 τ = 1 Eτ τ 1 Eτ, and so, by factoring 1 Eτ out of the portion in parentheses gives 1, or = 1 τ 1 Eτ 1 Eτ 1. 1 Eτ b Using the solution to the logistic equation from the textbook, we replace τ by τ and by 1 Eτ to obtain 1 Eτ t =. 1 1 Eτ 0 e 1 Eτ t t τ 0 2

3 c A nonzero E will cause the population to grow more slowly, or even have negative growth, when the population size is sufficiently small. In addition, the population will not stabilize at the carrying capacity as t. 3. Exercise The model of Gompertz, which is used for the modeling of the growth of a tumor, is given by the equation = 1 τ ln. Here, the variables and τ are assumed to be nonnegative and constant. a Calculate the solution of this differential equation. Hint: use the variable transformation y =. The hint in the textbook is incorrect. b Calculate the equilibrium solution. c Show for 0 < < that the change / determined by the Gompertz model is larger than / = 1 //τ as given by the logistic model. Hint: you may wish to show the validity and make use of fx = 1 + ln x x 0 for x 0. a er the hint, let y =. Then we obtain dy = dy 1 = t ln = 1 τ ln 1 y y = 1 y ln y. τ ] ] 1 Solving gives dy 1 y ln y = τ ln ln y = t τ + C. ln y = ce t τ y = e ce t τ b The equilibrium solution is found by solving 1 τ ln = 0, giving = 0 and =. = e ce t τ. 3

4 c For this part, we need to show that 1 τ ln > 1 τ 1, or, multiplying through by τ and using the fact that ln ln > 1 for 0 < <. This is equivalent to showing that 1 + ln < 0. = ln, Let x = /, and define fx = 1 + ln x x. Then f x = 1 1 > 0 for all x 0 < x < 1. Therefore, fx is an increasing function on the interval 0, 1. Since f1 = 0, we have that for 0 < x < 1, fx < 0. When x = 1, =, and we have that for 0 < <, 1 τ ln > 1 1. τ 4. Exercise Electrical vibrations in electric circuits can be described by the following differential equation for the charge Qt measured in coulombs C. L d2 Q 2 + RdQ + 1 C Q = 0. Here, R refers to the resistance, which is measured in ohms Ω, L refers to the inductance, which is measured in henrys H = Ωs, and C refers to the capacitance, which is measured in farads F = s/ω. The variables L, R, and C are non-negative and assumed to be constant. An impressed voltage, which can involve a forcing on the right-hand side of the equation, is not considered. The equation considered represents irchhoff s Second Law. a Assume that L = 2 H, R = 20 Ω, and C = 0.01 F. Find Qt for the case that Q0 = 5 C and dq/0 = 0. b Assume that L = 2 H and C = 0.01 F. Which R value is required so that the circuit is critically damped? c Find the charge Qt for the case of a critical damping considered in b Assume that Q0 = 5 C and dq/0 = 0. a The equation is thus 2Q + 20Q + 100Q = 0, Q0 = 5, Q 0 = 0. 4

5 The characteristic equation is 2λ λ = 0, or λ λ + 50 = 0. The roots of this equation are given by λ = 10 ± = 5 ± 5i. Therefore, the general solution is Qt = c 1 e 5t cos5t + c 2 e 5t sin5t. Since Q0 = 5, we see that c 1 = 5, giving Then Qt = 5e 5t cos5t + c 2 e 5t sin5t. Q t = 25e 5t cos5t + sin5t + 5c 2 e 5t sin5t + cos5t. Therefore, Q 0 = 0 gives c 2 = 0, or c 2 = 5. Therefore, the solution is b In this case, the differential equation is and the characteristic equation is for which the roots are given by Qt = 5e 5t cos5t + 5e 5t sin5t C. 2Q + RQ + 100Q = 0, 2λ 2 + Rλ = 0, λ = R ± R The circuit is critically damped if R = 0, or R 2 = 800. So, if R = 20 2 Ω, the circuit is critically damped. c We need to solve 2Q Q + 100Q = 0, Q0 = 5, Q 0 = 0. From part b, we know that the double root of the equation is λ = 1 4 R = 5 2, giving the general solution Qt = c 1 e 5 2t + c 2 te 5 2t. 5

6 Since Q0 = 5, we see that c 1 = 5, giving Therefore, Qt = 5e 5 2t + c 2 te 5 2t. Q t = 25 2e 5 2t + c 2 e 5 2t 5 2c 2 te 5 2t. Since Q 0 = 0, we see that c 2 = 0, or c 2 = Therefore, the charge in the circuit satisfies Qt = 5e 5 2t te 5 2t C. 5. Compare the period of the nonlinear pendulum with that of the linear pendulum when θ0 = 4, 10, 20,..., 85 degrees and θ 0 = 0. The table below demonstrates the results obtained using Maple. θ0 nonlinear pendulum linear pendulum As we can see, since , the period of the nonlinear pendulum is larger than that of the linear pendulum for all of the measured angles. The relative difference is small for small values of θ0, but increases as θ0 increases. Note that this is not unexpected, because the linear approximation is only a good approximation for small values of θ0. 6. Using the equation for the pendulum obtained via Newton s second law: mx = mg sinθe θ + mg cosθe r T e r, 6

7 derive the equation of motion for a pendulum system where the rod has been replaced by a spring with natural equilibrium length L 0. Hint: The force exerted by the spring is given by kl L 0 e r. By Newton s second law, we have mx = mg sinθe θ + mg cosθe r kl L 0 e r. Note that the length of the pendulum is no longer constant, i.e., L = Lt, so we need to determine x. Since x = Le r, we have that Then x = d L e r + L de r x = L e r + L de r. = L e r + L de r + de L r + e r Ld2 2 = L e r + 2L de r + Ld2 e r 2. Using the definitions of e r and e θ and the equations derived in class, we may write this as x = d2 L e 2 r + 2 dl dθ e d 2 2 θ dθ θ + L e 2 θ e r. Applying this to our equation from Newton s second law gives d 2 L m e 2 r + 2 dl ] dθ e d 2 2 θ dθ θ + L e 2 θ e r = mg sinθe θ +mg cosθe r kl L 0 e r. Writing this componentwise gives the system m 2 dl ] dθ + θ Ld2 = mg sin θ 2 ] d 2 2 L dθ m L = mg cos θ kl L 2 0, or L d2 θ + 2dL dθ = g sin θ 2 d 2 2 L dθ L = g cos θ k 2 m L L 0. 7

Differential Equations: Homework 8

Differential Equations: Homework 8 Alvin Lin January 08 - May 08 Section.6 Exercise Find a general solution to the differential equation using the method of variation of parameters. y + y = tan(t) r +