# How to find Sturm-Liouville problem eigenvalue and function?

So I have the following Sturm-Liouville problem: $$y'' + \lambda y = 0$$ Such that $$\lambda > 0$$ and the initial conditions are as follows: $$y (0) + y'(0) = 0$$ $$y(1) + y'(1) = 0$$

So my attempt at this goes something like this:

I know that the $$\lambda$$ is positive so the solution must be: $$y(t) = A\cos(\sqrt(\lambda)t) + B\sin(\sqrt(\lambda)t)$$

and: $$y'(t) = -A\sqrt{\lambda}\sin(\sqrt{\lambda}t) + B\sqrt{\lambda}\cos(\sqrt{\lambda})t)$$

Such that $$A$$ and $$B$$ are constants.

So I can evaluate the solution at the first initial condition: \begin{align} &=A\cos(\sqrt{\lambda}0) + B\sin(\sqrt{\lambda}0) + -A\sqrt{\lambda})\sin(\sqrt{\lambda}0) + B\sqrt{\lambda}\cos(\sqrt{\lambda}0)\\ &=A + B\sqrt{\lambda} \end{align}

Thus I know: $$A = -B\sqrt{\lambda}$$

Evaluating at the second initial condition: \begin{align} &=A\cos(\sqrt{\lambda}1) + B\sin(\sqrt{\lambda}1) -A\sqrt{\lambda}\sin(\sqrt{\lambda}1) + B\sqrt{\lambda}\cos(\sqrt{\lambda}1)\\ &=A\cos(\sqrt{\lambda}) + B\sin(\sqrt{\lambda}) -A\sqrt{\lambda}\sin(\sqrt{\lambda}) + B\sqrt{\lambda}\cos(\sqrt{\lambda}) \end{align}

I'm not sure where to go from here, I factored the second initial condition by cos and sin but that led me to a trivial solution for A and B like this:

$$(A + B\sqrt{\lambda})\cos(\sqrt{\lambda}) + (B - A\sqrt{\lambda})\sin(\sqrt{\lambda}) = 0$$

Plugging in $$A = -B\sqrt{\lambda}$$.

$$(-B\sqrt{\lambda} + B\sqrt{\lambda})\cos(\sqrt{\lambda}) + (B + B\sqrt{\lambda}\sqrt{\lambda})\sin(\sqrt{\lambda}) = 0$$ $$(B + B\lambda)\sin(\sqrt{\lambda}) = 0$$

So assuming $$B$$ isn't $$0$$, then I know $$\lambda = (n\pi)^2$$ but how do I find the value of B? Any guidance would be greatly appreciated!

Thank you.

• Based on your equation and your boundary conditions, you expect to get a linear subspace of solutions (i.e. any rescaling of a solution should also be a solution). Another way to view this is that you have only two boundary conditions but 3 unknowns (A,B, $\lambda$), so you shouldn't expect to find exact values for all 3. – TM Gallagher Apr 4 at 18:07
• That makes sense. So my eigenfunction can be written as $y(t) = -B(n\pi)cos(n\pi t) + Bsin(n\pi t)$? – Safder Apr 4 at 18:14
• Yes. You could also take $B=1$ and say that $-n\pi\cos(n\pi t) + \sin(n\pi t)$ generates the eigenspace. However you want to represent your solution. – TM Gallagher Apr 4 at 18:43
• @TMGallagher perfect. Thank you! – Safder Apr 4 at 18:44

## 2 Answers

Start by solving $$y''+\lambda y = 0$$ subject to $$y'(0)+y(0)=0 \\ y(0)=1.$$ The second condition is arbitrary; you could take $$y(0)-y'(0)=1$$ for example. The point is that no solution can satisfy $$y'(0)+y(0)=0$$ and $$y(0)=0$$ unless it is identically the $$0$$ solution, which means you can always scale $$y$$ so that $$y'(0)+y(0)=0$$ and $$y(0)=1$$. The solution of this problem is $$y_{\lambda}(x)=-\frac{\sin(\sqrt{\lambda}x)}{\sqrt{\lambda}}+\cos(\sqrt{\lambda}x).$$ Notice that the above is the correct solution at $$\lambda=0$$ when interpreted as a limit as $$\lambda\rightarrow 0$$, where it gives $$-x+1$$. In fact, $$y_{\lambda}$$ is a power series in $$\lambda$$. This will always be the case.

The added condition that $$y'(1)+y(1)=0$$ gives an equation in $$\lambda$$: $$-\cos(\sqrt{\lambda})-\sqrt{\lambda}\sin(\sqrt{\lambda})-\frac{\sin(\sqrt{\lambda})}{\sqrt{\lambda}}+\cos(\sqrt{\lambda}) = 0.$$ The solutions $$\lambda$$ are the zeros of a power series in $$\lambda$$, and $$\lambda=0$$ is not a solution. The solutions $$\lambda$$ must satisfy $$\sqrt{\lambda}\sin(\sqrt{\lambda})+\frac{\sin(\sqrt{\lambda})}{\sqrt{\lambda}}=0.$$ Looks like $$\lambda=0$$ is not a solution, but $$\sqrt{\lambda}=n\pi$$ for $$n=1,2,3,\cdots$$ are solutions. And $$\lambda=-1$$ looks like a solution. $$y_{-1}$$ is an exponential solution $$e^{-x}$$.

$$y'' + \lambda y = 0 \qquad \lambda>0$$ Since $$\lambda>0\quad$$ let $$\quad \lambda=\omega^2$$ . $$y'' + \omega^2 y = 0$$ $$y(x)=c_1\cos(\omega x)+c_2\sin(\omega x)$$ $$y'=-c_1\omega \sin(\omega x)+\omega c_2\cos(\omega x)$$ Conditions :

$$y(0)=c_1$$

$$y(1)=c_1\cos(\omega)+c_2\sin(\omega)$$

$$y'(0)=\omega c_2$$

$$y'(1)= -c_1\omega \sin(\omega)+\omega c_2\cos(\omega)$$

$$\begin{cases} y(0)+y'(0)=c_1+\omega c_2=0 \\ y(1)+y'(1)=c_1\cos(\omega)+c_2\sin(\omega)-c_1\omega \sin(\omega)+\omega c_2\cos(\omega)=0 \end{cases}$$ $$c_1= -\omega c_2$$

$$(-\omega c_2)\cos(\omega)+c_2\sin(\omega)-(-\omega c_2)\omega \sin(\omega)+\omega c_2\cos(\omega)=0$$

A first solution is : $$c_2=0 \quad\implies\quad c_1=0\quad\implies\quad y(x)=0$$ This trivial solution was visible at first sight.

Non trivial case $$\quad c_2\neq 0$$ :

$$-\omega\cos(\omega)+\sin(\omega)+\omega^2\sin(\omega)+\omega\cos(\omega)=0$$

$$(1+\omega^2)\sin(\omega)=0$$

$$\omega=\pm n\pi\quad$$ any integer $$n$$ .

If $$\lambda\neq (n\pi)^2\quad$$the case $$c_2\neq 0$$ is impossible. The only solution of the problem is $$y(x)=0$$.

If $$\lambda= (n\pi)^2\quad$$ They are an infinity of solutions :

$$\omega=\pm n\pi \quad;\quad c_1= \mp n\pi c_2\quad$$ any $$c_2$$ .

$$y(x)=\mp n\pi c_2\cos(\pm n\pi x)+c_2\sin(\pm n\pi x)$$

$$y(x)=\mp n\pi c_2\cos(n\pi x)\pm c_2\sin(n\pi x)$$

$$y(x)=\pm c_2\left(-n\pi\cos(n\pi x)+\sin(n\pi x)\right)$$

Since $$c_2$$ is any positive or negative constant, without loss of generality :

$$y(x)=c_2\left(-n\pi\cos(n\pi x)+\sin(n\pi x)\right)\quad\text{if}\quad \lambda= (n\pi)^2$$

• You are missing the solution $e^{-x}$. – DisintegratingByParts Apr 9 at 6:21
• No. $e^{-x}$ is solution of $y''-y=0$ that is $\lambda=-1$. It is specified that $\lambda>0$ in the wording of the question. – JJacquelin Apr 9 at 6:26