Solve the Sturm-Liouville eigenvalue problem

Question

Consider the Sturm-Liouville eigenvalue problem on 0≤≤2,

y'' + y = 0 , y(0) = 0 ,y(2)=y'(2).

1. Assume the infinite set of eigenvalues are ordered so that 0<1<2<…. Let =|0| and give an implicit equation for of the form =()
2. Consider the eigenvalue $_$ as →∞ Determine that $_$ ∼ $(2+1)^2$ in this limit, and give an equation for

My attempt:

1. Solved the differential equation with the boundary conditions and ended up with = $\tan^2(2\sqrt{}))$, hence we have that p = $\tan^2(2\sqrt{p}))$ for first case.
2. Same solution for this part = $\tan^2(2\sqrt{}))$, then used that $_$ ∼ $(2+1)^2$ hence , $\tan^2(2\sqrt{}))$ ∼ $(2+1)^2$ , hence ∼ $\tan^2(2\sqrt{}))/(2+1)^2$ , hence ended up with = $\tan^2(2\sqrt{(2+1)^2}))/(2+1)^2$ , hence = $\tan^2(2\sqrt{}(2+1))/(2+1)^2$

The solutions for diff equation is correct, but I was told that my the equations I found are not correct , Do you guys have any advice? Where was incorrect and how the correct will look like ?

Answer 1

I think I did this right but got something a little different.

$y''=-\lambda y$. $r^2=-\lambda\implies r=\pm\sqrt{|\lambda|}$ or $r=\pm i\sqrt{|\lambda|}$

TRIG SOLUTIONS

$y=c_1\sin (rx)+c_2\cos( rx)$

$y(0)=0\implies c_1=0$

$y(2)=c_2\cos(2r). y'(2)=-c_2r\sin(2r)$

So $1=-r\tan{2r}\implies 1=|\lambda|\tan^2(2\sqrt{|\lambda|})$.

Since for increasing $\lambda$, $1/|\lambda|\to 0\implies \tan^2(2\sqrt{|\lambda|})=0$

$\implies \sqrt{|\lambda|}\approx(2p+1)\pi/2\implies |\lambda|\approx(2p+1)^2\pi^2/4$ for integer $p$.

$\alpha=\pi^2/4$

Also note: $D^2y=-\lambda y\implies D^2 (Dy)=-\lambda(Dy)$. So given a solution to the eigenvalue problem, the derivative is also a solution. Has bearing on the use of Ladder Operators to produce new solutions to an SL problem.

EXP SOLUTIONS

$y=c_3e^{rx}+c_4e^{-rx}$

$y(0)=0\implies c_3=-c_4$

$y(2)=c_3e^{2r}-c_3e^{-2r}$. $y'(2)=c_3re^{2r}+c_3re^{-2r}$

$y(2)=y'(2)\implies e^{2r}-e^{-2r}=re^{2r}+re^{-2r}\implies e^{4r}-1=re^{4r}+r$

$e^{4r}(1-r)=1+r\implies e^{4r}=\frac{1+r}{1-r}$

Only two solutions one for when $r=0$ and another negative value.

Answer 2

Start by solving $$ y''+\lambda y = 0,\;\; y(0)=0,\;y'(0)=1. $$ You can add the normalization $y'(0)=1$ because that just adjusts the scale of the eigenfunction. And it creates an eigenfunction that depends holomorphically on $\lambda$. The solution of this equation is $$ y_{\lambda}(x)=\frac{\sin(\sqrt{\lambda}x)}{\sqrt{\lambda}}. $$ For $\lambda=0$, this reduces to the limiting case as $\lambda\rightarrow 0$, which is $$ y_0 = x. $$ Then the condition at $x=2$ becomes an equation for the zeros of an entire function of $\lambda$: $$ y(2)-y'(2)=0\\ \frac{\sin(2\sqrt{\lambda})}{\sqrt{\lambda}}-2\cos(2\sqrt{\lambda})=0 $$ This may also be written as $$ \tan(2\sqrt{\lambda})=2\sqrt{\lambda}. $$ $\lambda=0$ is an eigenvalue with solution $$ y_0(x)=x. $$