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 $$\begin{array}{c}{y}^{″}(x)=f(x);0<x<\pi \\ y(0)=y(\pi )=0.\end{array}$$

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 $\nu =n=1,2,3,\dots$ Therefore we look for solutions $$y=\sum _{n=1}^{\mathrm{\infty }}{c}_n\sin nx.$$ 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 $$y^{″}(x)=\sum _{n=1}^{\mathrm{\infty }}-n^2{c}_n\sin nx.$$ 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 $$f(x)=\sum _{n=1}^{\mathrm{\infty }}{b}_n\sin nx,$$ where $$b_n=\frac{2}{\pi }{\int }_0^{\pi }f(x)\sin nxdx.$$

Equating the coefficients of $\sin nx$ gives $$-n^2{c}_n=b_n=\frac{2}{\pi }{\int }_0^{\pi }f(x)\sin nxdx$$ so that the coefficients are $$c_n=-\frac{b_n}{n^2}=-\frac{2}{n^2\pi }{\int }_0^{\pi }f(x)\sin nxdx.$$

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 $$y=\frac{1}{6}(-{\pi }^{2}x+x^3).$$ 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 $$\begin{array}{c}{y}^{″}(x)+cy=f(x);0<x<\pi \\ y(0)=y(\pi )=0\end{array}$$ 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 $$(c-n^2)c_n=\frac{2}{\pi }{\int }_0^{\pi }f(x)\sin nxdx$$ 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: $$c_n=\frac{2}{\pi (c-n^2)}{\int }_0^{\pi }f(x)\sin nxdx.$$ 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

1. 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?
2. Modify the above examples to satisfy Neumann boundary conditions, $y^{\prime }(0)=y^{\prime }(\pi )=0$. 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.

My solution (pdf).