4
$\begingroup$

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.

$\endgroup$
  • 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
2
$\begingroup$

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 $e^{-x}$.

$\endgroup$
2
$\begingroup$

$$y'' + \lambda y = 0 \qquad \lambda>0$$ 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 :

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

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.