# Supplement to Section 10.4 of Boyce & Diprima

## Solving linear boundary value problems using Fourier series

### Example 1

Let’s try to solve the simple inhomogeneous boundary value problem $$ \begin{gathered} y’’(x) = f(x); 0< x< \pi \\\ y(0) = y(\pi) = 0. \end{gathered} $$

First of all, let’s find the eigenfunctions. These solve $y’’(x) = \lambda y(x)$. We know from section 10.2 of Boyce that the eigenvalues are negative, so let $\lambda = -\nu^2$. The general solution is $y = a \sin{\nu x} + b \cos{\nu x}$. The boundary conditions then require $b=0$ and $\nu = n=1,2,3,\ldots$ Therefore we look for solutions $$ y = \sum_{n=1}^\infty c_n \sin{n x}. $$ So far everything we are doing is straight out of section 10.2. If this part has been difficult, go back there and review.

Then $$ y’’(x) = \sum_{n=1}^\infty -n^2 c_n \sin{n x}. $$ Now, following section 10.3 of Boyce, we can expand $f(x)$ in Fourier sine series on $0\le x\le \pi$, so that we must use $L=\pi$ in our computations. The formulas give $$ f(x)= \sum_{n=1}^\infty b_n \sin{n x}, $$ where $$ b_n = \frac{2}{\pi}\int_0^{\pi} f(x) \sin{nx}\ dx. $$

Equating the coefficients of $\sin{nx}$ gives $$ -n^2 c_n = b_n = \frac{2}{\pi}\int_0^{\pi} f(x) \sin{nx}\ dx $$ so that the coefficients are $$ c_n = -\frac{b_n}{n^2}=-\frac{2}{n^2\pi}\int_0^\pi f(x) \sin{nx} \ dx. $$

If we specify the right-hand side, we may arrive at a closed form solution. For example, we follow Example 1 on page 484, letting $f(x)=x$ and taking $L=\pi$, then $$ b_n = \frac{ 2(-1)^{n+1}}{n} $$ and $$ c_n = -\frac{2}{n^2} \frac{ (-1)^{n+1}}{n} = \frac{2(-1)^n}{n^3}. $$ In fact, we can solve this problem exactly using undetermined coefficients and the ideas in the first set of supplementary notes. The exact solution is $$ y = \frac{1}{6}\left(-\pi^2 x + x^3\right). $$ The coefficients $c_n$ match those of the Fourier sine coefficients of the odd extension of this function.

### Example 2

Modify the problem slightly to $$ \begin{gathered} y’’(x) + c y = f(x); 0< x< \pi \\\ y(0) = y(\pi) = 0 \end{gathered} $$ Once again, we find the eigenfunctions are $y_n = \sin{nx}$, but the eigenvalues are $\lambda_n = c-n^2$. Plugging in the same expansion as before, we now find that $$ \left(c-n^2\right) c_n = \frac{2}{\pi}\int_0^\pi f(x) \sin{nx} \ dx $$ So if $c=m^2$ for any integer $m$, this equation clearly has no solution. However, as long as this condition is not satisfied, we can solve for the coefficients: $$ c_n = \frac{2}{\pi \left(c-n^2\right)}\int_0^\pi f(x) \sin{nx} \ dx. $$ Note that if $c=n^2$ for any positive integer $n$ then the problem has no solution (at least not in the form of a Fourier sine series).

## Exercises

- Find the Fourier sine-coefficients if we use $f(x)=x$ and $c=-1$ in Example 2. Find the exact solution using the method of undetermined coefficients. Suppose $c=+1$. Then the problem can’t be solved unless $b_1=0$. Why not?
- Modify the above examples to satisfy Neumann boundary conditions, $y’(0)=y’(\pi)=0$. This will give a Fourier cosine series, and will correspond to an even extension. You will find that Example 1 does not yield a solvable problem unless the term $a_0=0$.

## Solutions to Exercises

Solutions will appear here after the due date.