How do you solve the following nonhomogeneous eigenvalue problem?

Question

Let's say we want to solve this Sturm-Louiville problem:

$$ h^{"}(z) + \lambda h(z) = 0$$

subject to boundary conditions $$ h(0) = 0 \ ; h(5) = 25. $$

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This is my approach:

1. $\lambda = 0$

$$h(z) = c_1z + c_2$$

$$h(0) = c_2 = 0;\ h(5) = 5c_1 = 25 \rightarrow c_1 = 5$$

$$h(z) = 5z $$ so I guess this means that the eigenfunction is $h(z) = z$ and $0$ is an eigenvalue.

2. $\lambda = -\alpha^2 < 0 $

$$h(z) = c_1 \cosh(\alpha z) + c_2 \sinh(\alpha z)$$

$$h(0) = c_1 = 0; \ h(5) = c_2 \sinh(5 \alpha) = 25$$ and I don't know what this means.

3. $\lambda = \alpha^2 > 0$

   $$h(z) = c_1 \cos(\alpha z) + c_2 \sin(\alpha z)$$ $$h(0) = c_1 = 0$$

$$h(5) = c_2 \sin(5 \alpha) = 25$$ which I also don't know what it means.

Answer 1

Your treatment of the case $\lambda = 0$ is correct.

Now suppose $\lambda = - \alpha^2$. Since $\alpha \neq 0$, we get $c_2 = 25 / \sinh(5 \alpha)$ (recall that $\alpha$ is given). Hence $\lambda = - \alpha^2$ is an eigenvalue with

$$h(z) = \frac{25}{\sinh(5\alpha)} \sinh(\alpha z).$$

If $\lambda = \alpha^2$, we get $c_2 \sin(5\alpha) = 25$. Now it is clear that $\alpha \neq k \pi /5$ with $k \in \mathbb{Z}$, hence $\lambda = \alpha^2$ is an eigenvalue for all $\alpha \neq k\pi / 5$ and

$$h(z) = \frac{25}{\sin(5\alpha)} \sin(\alpha z).$$