TEST 3 MA-4451

SOLUTIONS

1. The Fourier cosine transform will be:

2.a. Using separation of variables we get:

and so the two ODEs are:

For positive eigenvalues we have:

We remark that C=0 since otherwise Y would be unbounded on the interval y >0. Also, there are no negative or positive eigenvalues (you should check that these would render an unbounded X). From here the solution is of the form:

From the condition for y=0 we get that:

and thus A and B are the Fourier cosine and sine transform of the function:

From page 231 or from the notes the transforms are:

By substituting in A and B the solution can be explicitly computed.

2.b. By taking the Fourier transform of the PDE with x as variable of integration:

This is a second order ODE, with solutions of the form:

Since the solution u has to be bounded for y > 0, it means that A = 0 for positive and B = 0 for negative . Therefore:

or equivalently:

By transforming the boundary condition we get (see page 231 or notes):

To determine u we have to compute the inverse transform:

Remark next that the above integral can be reduced to the solution of 2.a. by substituting in the formula for the complex exponential.

3. Using separation of variables we have:

Thus:

From here and the boundary conditions we get the following problems:

Hence we obtain the following eigenfunctions and eigenvalues:

The solution of the PDE can be expressed as a double Fourier sum:

(see table 1 page 248 for the coefficients). The coefficients can be explicitly computed. Remark that the the integral in (8) is zero and thus the solution is of the form: