# How do you solve the following nonhomogeneous eigenvalue problem?

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

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

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

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.

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

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

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

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

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

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

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

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

• Thanks! I don't know if I should make another post for this, but how can I use this then with the superposition principle to obtain solutions for Laplace's equation?
– MNM
Commented Apr 19, 2020 at 18:19
• @M.Navarro it's better to acceot Jan answer and open a new post for that question Commented Apr 19, 2020 at 18:23