Sturm-Louiville eigenvalues

Question

I am given this Sturm-Louiville problem: $$\frac{d^2y}{dx^2}+\lambda y=0$$

With boundary conditions: $y'(0) = 0, y(2) = 0$, where $n = 1, 2, 3, ...$

I am now trying to find the eigenvalues.

I tried using the general solution:

$y = A \cos(px) + B\sin(px)$, where $λ = -p^2 < 0$

I differentiated this and substituted the boundary conditions $y'(0) = 0$ to obtain that $B = 0$.

From this, I used the boundary condition $y(2) = 0$ to obtain:

$A\cos(2p) = 0.$

From this, I arrived at $p = \frac{n \pi}{4}$.

I was hoping for some kind of confirmation if I am on the right track...I would also like some guidance of where to go from this to obtain the eigenvalues.

I tried to find the eigenvalues by simply finding $-p^2$, but I found that this was wrong.

Thank you for your time, I appreciate any help.

Answer 1

$$\frac{d^2y}{dx^2}+\lambda y=0\tag1$$

For $\lambda =0$, the general solution is $y(x)=Ax+B$, where $A, B$ are arbitrary constants..

The given conditions are $y'(0) = 0, y(2) = 0$ gives $A=B=0$

Hence we get $y(x)=0$, which is the trivial solution.

For $\lambda\lt 0$, say $\lambda=-p^2$

then the general solution is $y(x)=Ae^{px}+Be^{-px}$, where $A, B$ are arbitrary constants..

The given conditions are $y'(0) = 0, y(2) = 0$ gives

$pA-pB=0\implies A-B=0$

$Ae^{2p}+Be^{-2p}=0$ which gives $A=B=0$

Hence we again get $y(x)=0$, which is the trivial solution.

So we discard these two cases and take $\lambda\gt0$.

Let $\lambda=p^2 \gt 0$

The general solution is $y(x)=A \cos(px) + B \sin(px) $, where $A, B$ are arbitrary constants.

Now the given conditions are $y'(0) = 0, y(2) = 0$

$y'(0) = 0\implies B=0$ and $y(2) = 0\implies A\cos(2p)=0$

If we take $A=0$, then we again get the trivial equation, so we take $A \ne 0$. This gives

$~\cos(2p)=0\implies 2p=(2n-1)\frac{\pi}{2}~~~~~\text{where $~n=1,~2,~3,~\cdots~$ }$ $~\implies ~p=(2n-1)\frac{\pi}{4}~~~~~\text{where $~n=1,~2,~3,~\cdots~$ }$

So the eigen value of the given differential equation $(1)$ is $$~\lambda=p^2=~(2n-1)^2\frac{\pi^2}{16}~~~~~\text{where $~n=1,~2,~3,~\cdots~$ }~$$

and the corresponding eigen functions are $$~y(x)~=~A~\cos\left\{(2n-1)\frac{\pi}{4}~x\right\}\qquad \text{where $~n=1,~2,~3,~\cdots~~~$and $~a~$ is a constant}~.$$

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You can also take the following

$~\cos(2p)=0\implies 2p=(2n+1)\frac{\pi}{2}~~~~~\text{where $~n=0,~1,~2,~3,~\cdots~$ }$ $~\implies ~p=(2n+1)\frac{\pi}{4}~~~~~\text{where $~n=0,~1,~2,~3,~\cdots~$ }$

So the eigen value of the given differential equation $(1)$ is $$~\lambda=p^2=~(2n+1)^2\frac{\pi^2}{16}~~~~~\text{where $~n=0,~1,~2,~3,~\cdots~$ }~$$

and the corresponding eigen functions are $$~y(x)~=~A~\cos\left\{(2n+1)\frac{\pi}{4}~x\right\}\qquad \text{where $~n=0,~1,~2,~3,~\cdots~~~$and $~a~$ is a constant}~.$$

Note: In both the two cases the values of $~\lambda~$ are same, in the first one $~n~$ does not take the value $~0~$, and in the second case it takes.

Answer 2

Any eigenfunction must be a non-zero constant multiple of the solution of $$ y''+\lambda y =0,\;\; y(0)=1,\; y'(0)=0, $$ which has unique solution $$ y(x) = \cos(\sqrt{\lambda}x). $$ Then, in order to satisfy $y(2)=0$, $\lambda$ must satisfy $$ \cos(2\sqrt{\lambda})=0. $$ The general solution is $$ 2\sqrt{\lambda} = (n+1/2)\pi ,\;\; n=0,1,2,3,\cdots, \\ \lambda = (n+1/2)^2\pi^2/4,\;\;\; n=0,1,2,3,\cdots. $$ The corresponding eigenfunctions are $$ y_n(x) = \cos((n+1/2)\pi x/2),\;\; n=0,1,2,3,\cdots. $$

Answer 3

The characteristic equation is

$$z^2+\lambda=0$$ and there are three regimes:

• $\lambda > 0\to y=A\cos\sqrt\lambda x+B\sin\sqrt\lambda x$,

• $\lambda = 0\to y=Ax+B$,

• $\lambda < 0\to y=Ae^{\sqrt{-\lambda}x}+Be^{-\sqrt{-\lambda }x}$.

By plugging the given limit conditions (for generality, $l=2$),

• $\lambda > 0\to 0=B,0=A\cos \sqrt\lambda l+B\sin \sqrt\lambda l$,

• $\lambda = 0\to 0=A,0=Al+B$,

• $\lambda < 0\to 0=A-B,0=Ae^{\sqrt{-\lambda}l}+Be^{-\sqrt{-\lambda}l}$

or

• $\lambda > 0\to \cos \sqrt\lambda l=0$,

• $\lambda = 0\to 1=0$, which is not possible,

• $\lambda < 0\to \cosh\sqrt{-\lambda}l=0$, which is not possible.

So for $\lambda>0$,

$$\sqrt\lambda l=(2k+1)\frac\pi2$$ or $$\lambda=\frac{(2k+1)^2\pi^2}{4l^2}$$ must hold for non-trivial solutions to exist.

The first Eigenfunctions: