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As the title suggests, I would like to find the eigenfunctions, $y_n(x)$ of a self-adjoint linear differential operator $\dfrac{d^2y}{dx^2}$ that satisfy the boundary conditions $y_n(0)=y_n(\pi)=0$. Thereafter, I would like to construct its Green's function, $G(x, \xi).$.

I know that it is common to define the eigenvalue of a linear operator by: $$\mathcal{Ly_n=\lambda_n\rho y_n}.$$ I was given that the required eigenfunctions satisfying the boundary conditions to be $$y_n(x)=\sqrt{\dfrac{2}{\pi}}\sin (nx).$$

Can anyone explain to me how this was derived or the intuition in approaching this problem. Thereafter, how should I proceed to construct its Green's functions? Any help would be appreciated.

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The problem in solving the eigenfunction equation is that solutions are unique only up to non-zero multiplicative constants. That is, any non-trivial solution of $$ y''=\lambda y , \;\;\; y(0)=y(\pi)=0 $$ may be scaled by any non-zero constant, and the new function is also a solution. Define $y_l$ and $y_r$ to be solutions with $$ y_l(0)=0,\;y_l'(0)=1,\;\;\;\;\; y_r(\pi)=0,\;y_r'(\pi)=1 $$ These solutions are uniquely given by $$ y_l(x)=\frac{\sin(\sqrt{\lambda}x)}{\sqrt{\lambda}} \\ y_r(x)=\frac{\sin(\sqrt{\lambda}(x-\pi))}{\sqrt{\lambda}}. $$ $\lambda$ is an eigenvalue iff these solutions are linearly dependent, which is the case iff the Wronskian of these solutions is $0$: $$ w(\lambda) = y_ly_r'-y_l'y_r \\ =\frac{1}{\sqrt{\lambda}}\{\sin(\sqrt{\lambda}x)\cos(\sqrt{\lambda}(x-\pi))-\cos(\sqrt{\lambda}x)\sin(\sqrt{\lambda}(x-\pi))\} \\ =\frac{\sin(\sqrt{\lambda}\pi)}{\sqrt{\lambda}} $$ $\lambda=0$ is not a solution because the limit as $\lambda\rightarrow 0$ is $\pi$. The eigenvalues are $\lambda_n=n^2$ for $n=1,2,3,\cdots$. The eigenfunctions are $B_n\sin(nx)$. The unique solution of $y''-\lambda y=g$ subject to $y(0)=(\pi)=0$ and $\lambda\ne \lambda_n$ is $$ y(x) = \frac{1}{w(\lambda)}\left[y_r(x)\int_{0}^{x}g(t)y_l(t)dt-y_l(x)\int_{x}^{1}g(t)y_r(t)dt\right] $$ The Green function is $$ \frac{1}{w(\lambda)}(y_r(x)y_l(t)_{t<x}-y_l(x)y_r(t)_{t>x}) $$ You are interested in the case $\lambda=0$.

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  • $\begingroup$ Thank you for the detailed explanation! $\endgroup$ – Cleytus Jul 11 '18 at 13:21
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You find the general solution to $y''=λy$ and try to satisfy the boundary conditions. For positive $λ$ and almost all negative ones the only solution should be the zero solution. Cases with non-trivial solutions are the eigenvalues/eigenfunctions.

For the Green function compute IVP solutions $y_1,y_2$ with $y_1(0)=0$, $y_1'(0)=1$ and $y_2(\pi)=0$, $y_2'(\pi)=-1$ and then consider $$ G(s,t)=\frac{y_1(\min(s,t))y_2(\max(s,t))}{W(s)} $$ as candidate for the Green function.

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