Sturm-Liouville Problems

Section 7.4 Sturm-Liouville Problems

We now turn to a class of differential equations that arise in solving partial differential equations. This class is called Sturm-Liouville problems and they satisfy boundary conditions instead of the initial conditions that we saw in the previous section.

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Definition 7.4.1.

Consider the differential equation

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\begin{equation*} [r(x) y']' + [p(x) + \lambda q(x)] y = 0 \end{equation*}

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for $r(x), p(x) \in \mathcal{C}^{(1)[a,b]}\text{,}$ $q(x)\in \mathcal{C}^{(0)}[a,b]\text{,}$ and $\lambda \in \mathbb{R}\text{.}$ The differential equation is also subject to the boundary conditions:

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\begin{align*} a_1 y(a) + a_2 y'(a) \amp = 0\\ b_1 y(b) + b_2 y'(b) \amp = 0 \end{align*}

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such that both $a_1$ and $a_2$ cannot be zero as well as both $b_1$ and $b_2\text{.}$ The differential equation with these boundary conditions are called a Sturm-Liouville Problem. The solution is called an eigenfunction of the problem and the values of $\lambda$ are called the eigenvalues.

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The following example shows how to solve a Sturm-Liouville problem, that is, find the eigenvalues and eigenfunctions of the problem.

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Example 7.4.2.

Find the eigenvalue and eigenfunctions of the Sturm-Liouville problem:

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\begin{align*} y''+\lambda y \amp = 0 \amp y(0) \amp = y(L) = 0 \end{align*}

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Solution.

First note that $r(x)=q(x) \equiv 1$ and also $a_1=b_1=1$ and $a_2=b_2=0\text{,}$ which satisfies the conditions on the boundaries. The characteristic equation for this problem is:

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\begin{equation*} r^2+\lambda = 0 \end{equation*}

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which has the solutions $r = \pm \sqrt{-\lambda}\text{.}$ The form of the equation depends on $\lambda\text{.}$ If $\lambda=0\text{,}$ we get:

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\begin{equation} y = A + B x\tag{7.4.1} \end{equation}

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if $\lambda < 0\text{,}$ then the solution is

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\begin{equation} y = c_1 e^{\sqrt{-\lambda} x} + c_2 e^{-\sqrt{-\lambda} x}\tag{7.4.2} \end{equation}

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and if $\lambda > 0\text{,}$ then we get

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\begin{equation} y = C_1 \cos \sqrt{\lambda} x + C_2 \sin \sqrt{\lambda} x\tag{7.4.3} \end{equation}

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Next, we apply the boundary conditions on all three solutions. Recall that $y(0)=y(L)=0\text{.}$ If $\lambda = 0\text{,}$ substituting the boundary conditions into (7.4.1),

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\begin{align*} A+ B(0)\amp = 0 \amp\amp \text{so $A=0$}\\ A+B(L) \amp = 0 \amp\amp \text{so $B=0$} \end{align*}

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so the only solution to (7.4.1) is the trivial solution $y\equiv 0\text{.}$ If $\lambda > 0\text{,}$ then substituting the boundary conditions into (7.4.2) results in

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\begin{align*} c_1 (1) + c_2(1) \amp = 0 \\ c_1(e^{\sqrt{-\lambda} L} + c_2 e^{-\sqrt{-\lambda} L}) \amp = 0 \end{align*}

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From the first equation, $c_1=-c_2$ and substituting this into the 2nd equation

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\begin{align*} c_1 e^{\sqrt{-\lambda} L} - c_1 e^{-\sqrt{-\lambda} L} \amp = 0 \\ c_1 e^{\sqrt{-\lambda}L} \bigl( 1- e^{-2\sqrt{-\lambda} L}) \amp = 0 \end{align*}

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$c_1=0$ is a solution, the second term is never zero and the third term is only zero if $L=0\text{,}$ which is not true or $\lambda=0\text{,}$ which is also not true, since this case is $\lambda > 0\text{.}$ Therefore again, the only solution to (7.4.2) is the trivial solution. If $\lambda < 0$ then substituting the boundary conditions into (7.4.3) results in

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\begin{align*} C_1 (1) + C_2 (0) \amp = 0 \amp\amp \text{so $C_1=0$} \\ (0) \cos \sqrt{\lambda} L + C_2 \sin \sqrt{\lambda} L = 0 \end{align*}

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the second states that either $C_2=0\text{,}$ again the trivial solution or

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\begin{equation*} \sin \sqrt{\lambda} L = 0 \end{equation*}

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which occurs if $\sqrt{\lambda} L = n \pi$ for $n=0,\pm 1, \pm 2, \pm 3, \ldots\text{.}$ Or

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\begin{equation} \lambda = \frac{n^2 \pi^2}{L^2}\tag{7.4.4} \end{equation}

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We now check which values of $n$ result in valid values of $\lambda\text{.}$ When $n=0\text{,}$ we get $\lambda=0$ again, which violates $\lambda \lt; 0$ and for $n$ both plus and minus the same number, we get the same eigenvalue, so we will discard the negative values of $n$ and just include $n=1,2,3, \ldots.$ The eigenvalues of this problem are those in (7.4.4) for $n=1,2,3,\ldots$and the eigenfunctions of this problem are:

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\begin{equation*} \phi_n(x) = \sin \frac{n\pi x}{L} \end{equation*}

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It may have appears that we were lucky that there was a solution to the Sturm-Liouville problem in the above example. However, this is not the case and any Sturm-Liouville problem has a solution as the following theorem shows.

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Theorem 7.4.3.

Let $\lambda_n$ and $\phi_n(x)$ be any eigenvalue and eigenfunction of the Sturm-Liouville problem.

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The eigenvalue $\lambda_n$ is real.
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There are an infinite number of eigenvalues that can be ordered $\lambda_1 < \lambda_2 < \lambda_3 < \cdots $ and for each eigenvalue, there is only one eigenfunction.
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Eigenfunctions $\phi_n$ and $\phi_m$ with $\lambda_n \neq \lambda_m$ satisfy $\langle \phi_n , \phi_m \rangle = 0\text{.}$
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Let $f$ and $f'$ be piecewise continuous functions on $[a,b]\text{.}$ If
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\begin{equation*} a_n = \frac{\langle f,\phi_n \rangle }{ \langle \phi_n, \phi_n \rangle}, \end{equation*}

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then the series:
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\begin{equation*} \sum_{n=1}^{\infty}a_n \phi_n(x) \end{equation*}

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converges to $f$ if $f$ is continuous at $x$ and to the value $(f(x+)+f(x-))/2$ if $f$ is discontinuous at $x$ for each point in $[a,b]\text{.}$
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Example 7.4.4.

Let $f(x)=x$ on $[0,L]\text{.}$ Find the series expansion listed in the theorem corresponding to the Sturm-Liouville problem $y''+\lambda y= 0\text{.}$

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Solution.

\begin{align*} a_n \amp = \frac{ \langle f,\phi_n \rangle}{ \langle \phi_n, \phi_n \rangle} \\ \amp = \frac{1}{L/2} \int_{0}^L f(x) \sin \frac{n\pi x}{L} \, dx \end{align*}

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which is the Fourier sine series.

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Example 7.4.5.

Find the solution of the Sturm-Liouville problem

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\begin{align*} y'' + \lambda y \amp= 0 \amp y'(0) = 0, \amp\amp y'(L) \amp = 0 \end{align*}

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Solution.

Since this the same differential equation as in Example 7.4.2, we note that when $\lambda > 0\text{,}$ there was no solution and the same is true here. In the case of $\lambda=0\text{,}$ the solution is

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\begin{equation*} y = A+Bx \end{equation*}

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and the derivative is needed as well,

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\begin{equation*} y' = B \end{equation*}

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and then applying the boundary condition $y'(0)=y'(L)=0\text{,}$ implies that $B=0\text{,}$ however, $A$ is not determined and $y=A$ is a solution. Next, we turn to $\lambda > 0\text{,}$ with the solution,

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\begin{equation*} y = C_1 \cos \sqrt{\lambda} x + C_2 \sin \sqrt{\lambda} x \end{equation*}

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and again, we need the derivative,

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\begin{equation*} y' = - C_1 \sqrt{\lambda} \sin \sqrt{\lambda} x + C_2 \sqrt{\lambda} \cos \sqrt{\lambda} x \end{equation*}

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Applying the boundary condition, $y'(0)=0\text{,}$ results in

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\begin{equation*} y'(0) = C_2 \sqrt{\lambda} = 0 \end{equation*}

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which implies that $C_2=0\text{.}$ Applying the boundary condition $y'(L)=0$ results in

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\begin{equation*} y'(L) = -C_1 \sqrt{\lambda} \sin \sqrt{\lambda} L = 0 \end{equation*}

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and if $C_1=0\text{,}$ this results in the trivial solution, $\lambda$ cannot be zero, however

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\begin{equation*} \sin \sqrt{\lambda} L = 0 \end{equation*}

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when $\sqrt{\lambda} L = n \pi$ or $\lambda = n^2 \pi^2/L^2\text{.}$ The eigenvalues and eigenfunctions of this problem then are $\lambda_0 =0$ and $\phi_0=1\text{,}$ and

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\begin{align*} \lambda_n \amp = \frac{n^2 \pi^2}{L^2} \amp \phi_n \amp = \cos \frac{n \pi x}{L} \end{align*}

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There are other Sturm-Liouville problems that arise commonly and we will see others later in this chapter and solve them as they arise. We will use these solutions that we just found in solving the PDEs that we derived above.

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