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.