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I am given the following boundary value problem: \begin{align} -y''&=\lambda{y},\hspace{0.5cm} a<x<b\\ y(a) = y(b), \hspace{0.25cm} &y'(a)=2y'(b) \end{align} and I am being asked if it has symmetric boundary conditions and if it is self-adjoint. I understand that the following must hold for it to be self-adjoint: \begin{align} (Ly_1,y_2)=(y_1,Ly_2) \end{align} if that were to hold, does that automatically mean that the boundary conditions are symmetric? And also how would I show that the above is true or false?

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  • $\begingroup$ Try to find $L^*$ first (IBP). Then, find a way to get rid of the boundary terms. $\endgroup$
    – hugo d
    Oct 17, 2021 at 22:44

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In Sturm–Liouville theory you are considering the ODE $$\frac{d}{dx} \bigg ( p(x) \frac{d y}{dx} \bigg) +q(x) y = -\lambda w(x) y. $$ For your particular equation you have $p\equiv w\equiv1$ and $q\equiv0$. You also have $$ Ly(x) = - y''(x).$$ Since your weight function $w$ is equal to 1 everywhere the inner product is $$(u,v) =\int_a^b u(x)v(x)\,dx. \tag{$\ast$}$$ Then $$(Ly_1,y_2) = - \int_a^b y_1''(x)y_2(x)\,dx $$ and $$ (y_1,Ly_2) = -\int_a^by_1(x)y_2''(x)\, dx \tag{$\ast\ast$}. $$ Try use integration by parts and your boundary conditions to establish whether $(\ast)$ equals $(\ast\ast)$.

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