Answers and Comments on Review Questions II

1. • (a) = 2*pi*a0 = 0

   • (b) Since all the a_k coefficients are zero, the function must be odd. Then the integral from 0 to pi is (pi/2)*b₃ = (pi/2)*(1/27) = pi/54.

   • (c) = pi*b₄ = pi/48

   • (d) = pi*a₅ = 0

   • (e) = 0, since the interval of integration is symmetric with respect to 0 and f(x)*cosh(x) is odd.

2. The only d coefficient that is non-zero is d₃. Then
   3*2*d₃ = 2
   and d₃ = 1/3. So u(x,t) = (1/3)*sin(3*x)*sin(6*t).

3. Here L = 4, so it helps to write the initial condition in the form
   u(x,0) = sin(12*(pi/4)*x).
   The corresponding t-function is then
   exp(-12²*9*(pi²/4²)).
   So
   u(x,t) = sin(3*pi*x)*exp(-81*pi²*t).

4. • (a) We look for a solution in the form u(x,t) = F(x)*G(t), where F(0) = F(pi) = 0. From the partial differential equation, we want
     F"(x)/F(x) + F'(x)/F(x) = G"(t)/G(t) = c,
     where c is a constant. Then the F-differential equation becomes
     F" + F' - cF = 0.
     This is a second-order linear homogeneous differential equation. The corresponding characteristic equation is
     lambda² + lambda - c = 0.
     The roots of this are
     -1/2 + or - (1/2)*sqrt(1 + 4*c).
     In order to satisfy the boundary conditions, we need 1 + 4*c to be a 4*negative integer squared, say -n². Then, we may take F(x) = exp(-x/2)*sin(n*x) and satisfy the conditions, F(0) = F(pi) = 0, without F being identically 0.

     Since we only want one solution, we arbritrarily pick n = 1. Then 1 + 4*c = -4 and c = -5/4. Thus, the corresponding t-function is G(t) = exp(-(5/4)*t), and one solution is

     u(x,t) = exp(-(1/2)*x)*sin(x)*exp(-(5/4)*t).

   • (b) For any fixed value of x, the negative exponent in t means that the limit as t approaches infinity of u(x,t) is 0.

5. pi/3

6. 2/5*pi*sqrt(5)

7. • (a) a₀ = the average value = 2/4 = 1/2.

   • (b) All the b_k coefficients are 0 since the function is even.

   • (c) a1 = 2/pi, a₂ = 0, a₃ = -2/(3*pi), a₄ = 0

   • (d) The general form of the Fourier series is
     1/2 + (2/pi)*(cos((pi/2)*x) -1/3*cos((3*pi/2)*x) + 1/5*cos((5*pi/2)*x) - + ...

   • (e) If we evaluate at x = 0, we have
     1/2 + (2/pi)*(1 - 1/3 + 1/5 - 1/7 +-...) = 1. Thus,
     1 - 1/3 + 1/5 - 1/7 +-... = pi/4.

8. (a) For this problem it is most convenient to use the traveling wave solution. At t = 2, the graph of u as a function of x looks like
   

   (b) For t = 2, u at x = -4.5 is halfway up the line segment from 0 to 2, i.e., u(-4.5, 2) = 1.

9. Suppose f is real-valued. Then the Fourier transform of f is
   

   In the expression on the right, the first integral is the real part and the second is the imaginary part.

   (a) If f is even, then f(s)*sin(omega*s) is odd, so the imaginary part of F is zero. Thus, F is real-valued.

   (b) If f is odd, then f(s)*cos(omega*s) is odd, and the real part is zero. Thus, F is purely imaginary.