How to solve Sturm-Liouville problem $y'' + \lambda y = 0$ with unknown initial conditions?

Question

I am trying to solve the following Sturm-Liouville problem:

$$ \begin{cases} y'' + \lambda y = 0 \\ y(x_0) = 0 = y(x_1) \end{cases} $$

In the interesting case where $\lambda > 0$ I get the following equations:

$$ \begin{cases} A \sin(\sqrt{\lambda}x_0) + B \cos(\sqrt{\lambda}x_0) = 0 \\ A \sin(\sqrt{\lambda}x_1) + B \cos(\sqrt{\lambda}x_1) = 0 \end{cases} $$

The main difficulty here is discussing this system. Apparently the solution should be $$y_n(x) = B_n \sin\Big(\frac{n \pi x}{x_1-x_0}\Big).$$

Answer 1

Define $u(x) = y(x + x_0)$, then$$ \begin{cases} u'' + λu = 0\\ u(0) = u(x_1 - x_0) = 0 \end{cases} $$ which reduces the original problem to an easier form.

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

Write your solution as $\ds{a\sin\pars{\root{\lambda}\bracks{x - x_{0}}}}$ which already satisfies the boundary condition at $\ds{x = x_{0}}$.

Then, \begin{align} 0 & = \mrm{y}\pars{x_{1}} = a\sin\pars{\root{\lambda}\bracks{x_{1} - x_{0}}} \stackrel{\large\mrm{non\ trivial\ solution} \atop \large a\ \not=\ 0}{\implies} \root{\lambda}\bracks{x_{1} - x_{0}} = n\pi\,,\qquad n \in \mathbb{N}_{\ \geq\ 1} \\[5mm] & \implies \root{\lambda} = n\,{\pi \over x_{1} - x_{0}} \implies \bbx{B_{n}\sin\pars{n\,{\pi \over x_{1} - x_{0}}\bracks{x - x_{0}}}} \end{align}