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.

  • 2
    $\begingroup$ 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. $\endgroup$ – TM Gallagher Apr 4 at 18:07
  • $\begingroup$ That makes sense. So my eigenfunction can be written as $ y(t) = -B(n\pi)cos(n\pi t) + Bsin(n\pi t)$? $\endgroup$ – Safder Apr 4 at 18:14
  • 1
    $\begingroup$ 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. $\endgroup$ – TM Gallagher Apr 4 at 18:43
  • $\begingroup$ @TMGallagher perfect. Thank you! $\endgroup$ – Safder Apr 4 at 18:44

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 :



$y'(0)=\omega c_2$

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

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

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

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 $ :



$\omega=\pm n\pi\quad$ 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 :

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

  • $\begingroup$ You are missing the solution $e^{-x}$. $\endgroup$ – DisintegratingByParts Apr 9 at 6:21
  • $\begingroup$ 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. $\endgroup$ – JJacquelin Apr 9 at 6:26

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