Solve for the eigenvalues and eigenfunctions for the equation $y''+\lambda y =0$

Question

The boundary conditions are $$y(-L)=0=y(L)$$ where $L>0$.

I know how to solve these kind of eigenvalue problems on the interval from 0 to L, but I don't know how to approach this set of boundary conditions. Any help? Thanks!

Answer 1

The same way as with boundary $0$ and $L$.

When $\lambda\leq 0$ there are no nontrivial solutions, and for $\lambda=\mu^2$ you get nontrivial solutions only when $\mu=n\pi/(2L)$: either $y=c\sin(\mu x)$ (even $n$) or $y=c\cos(\mu x)$ (odd $n$).

Answer 2

$$y''+\lambda y=0$$

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

The given conditions are $y(-L)=0=y(L) \qquad \text{where} \qquad L\gt0$

So $AL+B=0=-AL+B\implies A=B=0$

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 $y(-L)=0=y(L) \qquad \text{where} \qquad L\gt0$ gives

$Ae^{pL}+Be^{-pL}=0 \qquad \text{and} \qquad Ae^{-pL}+Be^{pL}=0$ 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 $y(-L)=0=y(L) \qquad \text{where} \qquad L\gt0$

$y(L) = 0\implies A \cos(pL) + B \sin(pL)=0\qquad$ and

$y(-L) = 0\implies A \cos(-pL) + B \sin(-pL)=0\implies A \cos(pL) - B \sin(pL)=0$

For non trivial solution if we take $A=0, B \ne 0$, then we get

$B \sin(pL)=0\implies \sin(pL)=0\implies pL=n\pi \implies p=\frac{n\pi}{L}$

$\qquad \text{where}\qquad n=0,\pm1,\pm2, . . . \quad \text{and}\quad L\gt 0$.

If we take $A\ne 0, B = 0$, then we get

$A\cos(pL)=0\implies \cos(pL)=0\implies pL= (2n+1)\frac{\pi}{2}\implies p=(2n+1)\frac{\pi}{2L}$

$\qquad \text{where}\qquad n=0,\pm1,\pm2, . . . \quad \text{and}\quad L\gt 0$.

Again if we take both $A\ne 0, B \ne 0$, then we get

$A\cos(pL) =0= B \sin(pL)=0\implies \cos(pL) =0= \sin(pL)$

There is no $p \quad \text{where}\quad L\gt 0$ for which $\cos(pL) =0= \sin(pL)$ holds. So we again discard this case.

Conclusion:

For $n=0,\pm1,\pm2, . . . \quad \text{and}\quad L\gt 0$

$y(x)=\sin(\frac{n\pi x}{L})$ be the eigenfunction corresponding the eigen value $\frac{n^2\pi^2}{L^2}$

and

$y(x)=\cos((2n+1)\frac{\pi x}{2L})$ be the eigenfunction corresponding the eigen value $(2n+1)^2\frac{\pi^2}{4L^2}$