# Symmetric boundary conditions and self-adjoint boundary value problem

I am given the following boundary value problem: \begin{align} -y''&=\lambda{y},\hspace{0.5cm} a 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?

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

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 $$(\ast\ast)$$.