Solutions to Assignment 7

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1 MTHE 237 Fall 215 Solutions to Assignment 7 Problem 1 Show that the Laplace transform of cos(αt) satisfies L{cosαt = s s 2 +α 2 L(cos αt) e st cos(αt)dt A s α e st sin(αt)dt e stsin(αt) α { e stsin(αt) A + e st cos(αt) s s α α 2 A α 2e st cos(αt)dt e stsin(αt) A α + s { α 2 e st cos(αt) A s e st cos(αt)dt e stsin(αt) A α + lim s A {e st α 2 cos(αt) A se st cos(αt)dt = s α 2 s2 α 2L(cosαt) (1) Hence, L(cosαt)(1+ s2 α 2) = s α 2, and the result follows by dividing both sides by (1+ s2 α 2 ). Problem 2 Use the Laplace transform method to obtain the solution to the following: y 2y +4y =, y() = 2,y () = 1 Taking the Laplace transform of y: leads to: L{y (s) = sl{y y() L{y (s) = s 2 L{y sy() y () L{y(s 2 2s+4) = 2y()+y () Hence,

2 L{y(s) = 2 4(s 2) s 2 2s+4 = s 1 (s 1) (s 1) 2 +3 By observing that s 1 term gives rise to a sine and a cosine, and a shift in the s domain is equivalent to multiplication by an exponential in the original function, we obtain, y(t) = 2e t cos( 3t) 1 3 e t sin( 3t) Problem 3 Suppose that g(t) = t f(τ)dτ. If G(s) and F(s) are the Laplace transforms of g(t) and f(t), show that G(s) = F(s)/s It could be noticed that g(t) = t f(τ)dτ implies that g (t) = f(t) with g() =. Hence, F(s) = sg(s). Problem 4 Suppose that exists for s >. F(s) = L{f(t), a) Show that if c > then, for s >, L{f(ct) = 1 c F(s c ) b) Show that if k > then, for s >, L 1 {F(ks) = 1 k f(t k ) b) Show that if a,b > are constants, then, for s >, L 1 {F(as+b) = 1 a e bt/a f( t a )

3 a) In the following, let u = ct: L(f(ct))(s) e st f(ct)dt /c e s(u/c)f(u)du c /c e s/cu f(u) du c = 1 c F(s c ) (2) b) This part follows from the result above, that is the Laplace transform of 1 k f( t k ) equals k kf(ks) = F(ks) c) Consider the Laplace transform L( 1 a e bt/a f(t/a))(s) We know the following: L{( 1 a e bt/a f(t/a))(s) = 1 a lim e st ( 1 a e bt/a f(t/a)) e (s+b/a)t f(t/a) Now, defining a variable u = t/a, and applying the steps in part a, it follows that L{( 1 a e bt/a f(t/a))(s) = 1 a lim e (as+b)u f(u)adu = F(as+b) (3) Problem 5 Compute the inverse Laplace transform of where the inverse is a continuous function. s 2 +9s+2 (s 1) 2 (s+3), Hint: Use partial fraction expansion and the properties of the derivative of a Laplace transform. We express and find a = 2,b = 3,c = 2 The inverse transform leads to: s 2 +9s+2 (s 1) 2 (s+3) = a (s 1) 2 + b s 1 + c s+3, 2e t +3te t 2e 3t Problem 6 The transfer function, H(s), of a linear system is defined as the ratio of the Laplace transform of the output y(t) to the Laplace transform of the input function g(t), under the assumption that all the initial conditions are set to zero. If the linear system is governed by a differential equation:

4 Verify that the transfer function is given by: ay +by +cy = g(t) H(s) = 1 as 2 +bs+c Now suppose the Laplace transform of g(t) satisfies G(s) = 1. In this case, you can see that y(t) is the inverse Laplace of the transfer function, which we denote by h(t). But, G(s) = 1 is the Laplace transform of the impulse. This is why h(t) is called the impulse response of the system. Taking the Laplace transform of both sides, and observing that the initial conditions are set to zero, we obtain L(ay +by +cy) = L(g) and Hence, if G(s) = 1, we obtain: L(y) = H(s) = L(g) (as 2 +bs+c). 1 (as 2 +bs+c) H(s) is the transfer function; it multiplies the Laplace transform of the input to generate the Laplace transform of the output at a given value of s. Problem 7 Recall that a differential equation that entails more than one independent variable is called a partial differential equation. An important example is the heat equation u t = u k 2 x2, (4) where u is the dependent function and x,t are the independent variables. Here, k is a constant. This equation models the variation of temperature u with position x and time t in a heated rod where it is assumed that at a given cross-section at length x from one end of the rod the temperature is a constant. Suppose that the rod has a finite length L, extending from x = to x = L. Its temperature function u(x,t) solves the heat equation. Suppose that the initial temperature profile is given by There are also fixed temperatures at the two ends of the rod so that u(x,) = sin( π x) (5) L u(,t) = u(l,t) =. (6) Using the separation of variables technique, obtain a solution to this partial differential equation. Recall that this technique seeks a solution of the form u(x,t) = X(x)T(t), where X only depends on X and T only depends on T. Assuming such a separated structure in u(x,t) = X(x)T(t) holds, (4) reduces to: X (2) X = T(1) kt,

5 Now, suppose that both of these terms are equal to a constant λ. Of course, there may be multiple such λ values, and we will need to account for all such λ values. First, consider the term: X (2) X +λ = We know from our earlier analysis that the general solution for such a problem will be given by either an exponential or a harmonic family, depending on whether λ or λ <. The conditions u(,t) = u(l,t) = actually rule out the case with λ for a non-zero solution, and it turns out that the general solution will write as: X n (x) = sin( λ x), where λ { n2 π 2 L 2,n Z +. Now with this set of λ values, we can solve the second equation: which leads to the solution: T (1) kt = n2 π 2 L 2, T n (t) = K n e n2 π 2 k L 2 t where K n is a constant depending on n. Thus, a candidate solution will be K n e n2 π 2 k L 2 t sin( nπ L x) Using the principle of superposition, we can take a linear combination so that any such linear combination u(x,t) = n K n e n2 π 2 k L 2 t sin( nπ L x) is a solution to the homogenous problem. Now, if we also satisfy the initial condition: u(x,) = f(x,) = n K n sin( nπ L x), it follows that K n = 1 for n = and elsewhere. Thus, u(x,t) = e π2 k L 2 t sin( π L x) * * * Have a good, well-deserved winter break * * *

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