# Solutions to Sample Midterm 2 Math 121, Fall 2004

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1 Solutions to Sample Midterm Math, Fall 4. Use Fourier series to find the solution u(x, y) of the following boundary value problem for Laplace s equation in the semi-infinite strip < x <, y > : u x + u y =, u(, y) = u(, y) =, u(x, ) =, u(x, y) as y. The separated solutions of Laplace s equation that satisfy the boundary conditions at x =, and as y are sin(nπx)e nπy, where n is a positive integer. We therefore look for a solution of the form u(x, y) = b n sin(nπx)e nπy. n= Imposing the boundary condition at y =, we obtain so b n sin(nπx) =, n= b n = sin(nπx) dx [ ] cos(nπx) = nπ [ ( ) n+ = + ] nπ nπ = 4/(nπ) for n odd, for n even.

2 The solution is therefore u(x, y) = 4 sin(πx)e πy + π 3 sin(3πx)e 3πy + } 5 sin(5πx)e 5πy +....

3 . Use Fourier series to find the solution u(x, t) of the following initialboundary value problem for the wave equation in < x < and t > : u t u x =, u u (, t) = (, t) =, x x u(x, ) =, u (x, ) = x. t The separated solutions of the wave equation that are zero at t = and satisfy the boundary conditions at x =, are t and cos(nπx) sin(nπt), where n =,,.... We therefore look for a solution of the form u(x, t) = a t + a n cos(nπx) sin(nπt). n= Differentiating this series with respect to t, we find that u t (x, t) = a + nπa n cos(nπx) cos(nπt). n= Imposing the initial condition for u/ t at t =, we get the Fourier cosine expansion: a + nπa n cos(nπx) = x. Hence, for n we have nπa n = n= x cos(nπx) dx [ x sin(nπx) = + cos(nπx) nπ (nπ) [ ( ) n = (nπ) ] (nπ) 4/(nπ) for n odd, = for n even, ]

4 and For n =, we get a n = 4/(nπ) 3 for n odd, for n even. a = [ = =. x x dx Hence, the solution is u(x, t) = t 4 π 3 cos(πx) cos(3πx) cos(5πx) +... }. ]

5 3. Use Fourier transforms to solve the following initial value problem for u(x, t) in < x <, t > : u t = u 4 x, 4 u(x, ) = f(x). Write the solution for u(x, t) as a convolution, but do not compute any inverse transforms explicitly. How smooth is the solution for t >? Let û(k, t) = u(x, t)e ikx dx be the Fourier transform of u with respect to x. Then, taking the Fourier transform of the initial value problem, we get û t = ( ik)4 û, û(k, ) = f(k), where f is the Fourier transform of f. It follows that which has the solution û t = k4 û, û(k, ) = f(k), û(k, t) = f(k)e k4t. According to the convolution theorem, if f, g have Fourier transforms f, ĝ respectively then f ĝ is the Fourier transform of f g. It follows that u(x, t) = G(x y, t)f(y) dy where Ĝ(k, t) = e k4t. The solution is smooth (infinitely differentiable with respect to x) for t > since its Fourier transform decays exponentially quickly as k (assuming, for example, that f(k) is a bounded function of k).

6 4. (a) Give the formulas for the Fourier transform f(k) of a function f(x) and the inverse Fourier transform. (b) Compute the Fourier transform of e x. (c) State Parseval s theorem, and use it to evaluate ( + k ) dk. (a) A function f(x) and its Fourier transform f(k) are related by f(k) = f(x) = f(x)e ikx dx, f(k)e ikx dk, (b) If f(x) = e x, then using x for x, x = x for x, and changing x x in the integral for < x <, we find that f(k) = = = e x e ikx dx, e ( ik)x dx + e ( ik)x + e (+ik)x} dx = [ e ( ik)x ik + e (+ik)x + ik = [ ik + ] + ik = π + k. } e (+ik)x dx ]

7 (c) Parseval s theorem states that f(k) dk = For f(x) = e x, we compute that f(x) dx = f(x) dx. e x dx = [ e x] =. It follows from Parsevals theorem and (b) that so π ( + k ) dk =, ( + k ) dk = π 4. Remark. The integral in (c) can also be evaluated directly by use of the substitution k = tan θ, which gives ( + k ) dk = = = = π/ π/ π/ π/ = π = π 4, ( + tan θ) sec θ dθ sec 4 θ sec θ dθ sec θ dθ cos θ dθ which verifies Parseval s theorem explicitly in this case.

8 5. Use Laplace transforms to solve the following initial value problem: y + y + y =, y(t) =, y () =. Let Y (p) be the Laplace transform of y(t). Then, taking the Laplace transform of the ODE and using the initial conditions, we get that Solving for Y, we get Y (p) = p Y + py + Y = p. p + p + + p(p + p + ). We have p + p + = (p + ) +, so (from L3 of the table) [ ] L = e t sin t. p + p + Also p(p + p + ) = [ ] p p + p + p + = [ p p + (p + ) + (p + ) + ]. So (from L, L3, L4) we have [ ] L = [ e t cos t e t sin t ] p(p + p + ) Hence, combining these inverse transforms, we get y(t) = [ e t cos t + e t sin t ].

9 6. (a) Say what jump conditions the solution of y(t) of the following initial value problem satisfies at t =, and find the solution directly (do not use Laplace transforms): y 4y = δ(t), y(t) = for t <. (b) Write the solution of the following initial value problem, where f(t) is an arbitrary function, as a convolution (you don t need to derive your answer): y 4y = f(t), y() = y () =. (a) The derivative of y has a jump discontinuity of size one at t =. The solution is therefore y+ (t) for t, y(t) = for t <, where y + 4y + = for t >, y + () =, y +() =. The general solution of the ODE is y + (t) = a cosh t + b sinh t, and the initial conditions imply that a = and b = /. Hence, y(t) = sinh t for t. (b) The solution is y(t) = t sinh (t s)f(s) ds.

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