Answers and Comments on Test 2

1. The answer is 1/(2*sqrt(pi))*exp(-x²/4). Use the formula for exp(-a²*x²) in the table. Set a = 1/2, and then adjust the multiplicative factor. This was very similar to the calculation in the lab on the Friday before the test.

2. We have cos(2x) - sin(3x) = 1/2*(exp(2xi) + exp(-2xi)) + 1/(2i)*(exp(3xi) - exp(-3xi)).

   So, the complex form of the Fourier series is -1/(2i)*exp(-3xi) + 1/2*exp(-2xi) + 1/2*exp(2xi) + 1/(2i)*exp(3xi).

   As I discussed this in class, almost everyone worked far too hard. There was no partial credit for attempting to find the real form -- which was, in fact, given. Watch for another problem like this on the final.

3. The integral is half of the integral from minus infinity to infinity. The integral from minus infinity to infinity is

   2*pi*i*(Res(f(z), z = exp(pi/4*i) + Res(f(z), z=exp(3*pi/4*i)).

   where f(z) = z²/(1 + z⁴). So the given integral is

   1/2*2*pi*i*1/4*(exp(-pi/4*i) + exp(-3/4*i)) = pi*sqrt(2)/4.

   The poles of f are the four 4th roots of -1. We only used the ones in the upper half-plane.

4. The initial condition is given in its half-range sine expansion. So there is no need to calculate Fourier coefficients. The solution is

   u(x,t) = sin(2*pi*x/3)*exp(-64/9*pi^2*t)

   The verification step had to include checking that your solution satisfied the pde.

5. Both of the initial conditions are given in half-range sine expansions. Again, there is no need to calculate Fourier coefficients. The solution is

   y(x,t) = sin(pi*x)*cos(2*pi*t) + 1/(4*pi)*sin(2*pi*x)*sin(4*pi*t)

   Although it was not required, you should have checked your answer to make sure it satisfied the initial conditions and the pde. This is quick and catches most careless errors.

Takehome

1. This requires three steps. Start with the Cauchy Riemann equations:
   du/dx = dv/dy and du/dy = - dv/dx.

   Now take partials as appropriate:

   d²u/dx² = d²v/dx dy and d²u/dy² = -d²v/dy dx.

   Use the fact that the mixed second partials of v are equal to conclude that

   d²u/dx² = -d²u/dy².

   Then it follows that

   d²u/dx² + d²u/dy² = 0.

2. Here, and in Part 3, the functions need to be expressed as functions of z.
   f₁(z) = 3z² and f₂(z) = -iz³.

3. So f(z) = 3z² + iz³.

4. I saw all sorts of responses here. In fact, I did not take off any points for this part. What I was looking for was that we now had a function that satisfied Laplace's equation everywhere and the given boundary condition on the circle, so it must be the solution.

5. In polar coordinates u(r, theta) = Re(f(r*exp(i*theta))). This is the real part of
   3r²exp(2*theta*i) + i*r³exp(3*theta*i)

   or the real part of

   3r²(cos(2*theta) + i*sin(2*theta)) + i*r³(cos(3*theta) + i*sin(3*theta))

   This is just 3r²cos(2*theta) - r³sin(3*theta). We will talk some more about Laplace's equation in the final review.

Last modified: April 13, 1999