Ordinary differential equation eigenvalues/eigenfuntions

Question

Consider the DE $$y''+\lambda y=0$$ where $\lambda$ is a constant

subject to the boundary conditions $$y(0)=0$$ and $$y(a)=0$$ where $a$ is a positive constant

I want to find the eigenvalues and eigenfunctions related to this problem

My attempt:

The auxiliary equation is $$m^2-\lambda=0\implies m=\sqrt{-\lambda}$$

Thus we have two cases

Case (i) $$\lambda<0$$ in which we have real different roots so let $$\lambda=-k^2\implies m=\pm k$$ thus the solution is $$y=Ae^{kx}+Be^{-kx}$$

Now when $y(0)=0$ we have that $$0=Ae^{k(0)}+Be^{-k(0)}\implies -B=A$$ since $k\ne 0$

Now when $y(a)=0$ we have that $$0=Ae^{ka}+Be^{-ka}\implies \frac{Ae^{ka}}{e^{-ka}}=-B$$ since $a\ne 0$ thus there are no eigenvalues for when $\lambda <0$

Can anyone tell me if i have made a mistake somewhere in this, because i'm unsure what to do from here, i know the next case to check will be when $\lambda>0$ which would give complex roots, but i would like to know if my approach up to now is correct, thanks for taking the time to read this, and any help would be appreciated also i know that when i multiply through by $$q^*=\frac{1}{a_2}e^{\int\frac{a_1}{a_2}}dx=1$$ thus in normal ?self-adjoint form.

Answer 1

Yes, your approach so far is correct. One should also check (preferably first, so you don't forget!) the $\lambda=0$ case: here, it is easy to check that the boundary conditions can't be satisfied.

One can actually extract the general solution from your analysis. In order for there to be a nonzero solution, the system of equations $$ \begin{pmatrix} 1 & 1 \\ e^{ka} & e^{-ka} \end{pmatrix} \begin{pmatrix} A \\ B \end{pmatrix} = 0 $$ must be degenerate (else the only solution is zero). Hence the determinant of the matrix must vanish; this is $$ e^{-ka}-e^{ka}=0, $$ or $e^{2ka}=1$. Recalling Euler's formula, this occurs if and only if $2ka=2n\pi i $ for some integer $n$. Hence $k=\frac{n\pi i}{a}$. The condition at zero gives $y=2A\sinh{kx}$, and $\sinh{iz} = i\sin{z}$, so the eigenfunctions are proportional to $\sin{(n\pi x/a)}$, for $n \in \{ 1,2,3,\dotsc\}$. Certainly this looks plausible, since this satisfies the equation and boundary conditions.

Answer 2

I'll suggest a general method that separates the equations at the endpoints. This does not directly answer your question, but the method is worth knowing.

For a solution $y$ of the second order equation $y''+\lambda y$, there is no non-trivial solution with $y(0)=0=y'(0)$. So, by scaling $y$ if necessary, it must be a solution of $$ y''+\lambda y = 0, \;\; y(0)=0,\;\; y'(0)=1, $$ which has unique solution $$ y_{\lambda}(x) = \frac{\sin(\sqrt{\lambda}x)}{\sqrt{\lambda}}. $$ This version of the solution works for $\lambda=0$ by taking a limit: $$ y_{0}=\lim_{\lambda\rightarrow 0}y_{\lambda}(x) = x. $$ (This technique eliminates special cases because the limits always exist.)

Any non-trivial solution of the solution where $y(0)=0=y(a)$ must be a non-zero constant multiple of $y_{\lambda}$, which means that the only values of $\lambda$ for which there are non-trivial solutions must satisfy $y_{\lambda}(a)=0$. $\lambda=0$ does not work, which is easily checked because $y_0(x)=x$ does not vanish at $a$. The non-zero $\lambda$ must satisfy $$ \sin(\sqrt{\lambda}a) = 0 \\ \implies \sqrt{\lambda}a = \pm \pi,\pm 2\pi,\pm 3\pi,\cdots \\ \implies \lambda = \frac{n^2\pi^2}{a^2},\;\; n=1,2,3,\cdots. $$