Determine whether a given linear differential operator is selfadjoint

Question

The problem is to determine whether the given differential operator $L[y]$, whose domain consists of all functions that have continuous second derivatives on the interval $[0,\pi]$ and satisfy the given boundary conditions, is selfadjoint.

$$L[y]=y''+\lambda y;\;\;\;\;y(0)+y'(\pi)=0,\;\;\;\;y'(0)+y(\pi)=0$$

For an operator to be selfadjoint over an interval, it must satisfy the equation $(u,L[v])=(L[u],v)$ for all $u,v$ in the domain, where the inner product over the interval $[a,b]$ is defined as:

$$(f,g)=\int_a^b f(x)g(x)dx$$

Applying integration-by-parts twice to the integral for $(u,L[v])$ goes as follows:

$$ (u,L[v])= \int_0^\pi u(v''+\lambda v)dx = \int_0^\pi uv''\;dx + \int_0^\pi \lambda v\;dx = uv'|_0^\pi - \int_0^\pi u'v'\;dx + \int_0^\pi \lambda v\;dx $$

$$ = uv'|_0^\pi - u'v|_0^\pi + \int_0^\pi u''v\;dx + \int_0^\pi \lambda v\;dx = (uv'-u'v)|_0^\pi + (L[u],v) $$

Thus, the operator is selfadjoint iff $(uv'-u'v)|_0^\pi=0$. The boundary conditions can be used to show that $uv'|_0^\pi=-u'v|_0^\pi$, leading to the following condition for selfadjointness:

$$uv'|_0^\pi=0$$

This seems promising, but I'm not sure where to go from here. I can't even think of functions that satisfy the boundary conditions, which makes finding a counter-example difficult.

Edit:

It looks like the following is a counter-example:

$$u(x)=2\cos{2x}-\sin{2x},\;\;\;\;v(x)=\cos{x}+\sin{x}$$ $$ uv'|_0^\pi = (2\cos{2x}-\sin{2x})(\cos{x}-\sin{x})|_0^\pi = -4$$

Unfortunately, I don't find this particularly enlightening. Is there any way to show that the operator is not selfadjoint without a counter-example?

Answer 1

Take a polynomial $u$ such that $u(0)=1,u'(0)=1,u(\pi)=-1,u'(\pi)=-1$ and another polynomial $v$ such that $v(0)=1,v'(0)=2,v(\pi)=-2,v'(\pi)=-1$.

We therefore have that $u(\pi)v'(\pi)-u(0)v'(0)=1-2=-1$. And clearly both functions satisfy the conditions.

Answer 2

The condition that $uv′|_0^\pi=0$ can be rewritten using the boundary equations as shown below.

$$uv′|_0^\pi=u(\pi)v'(\pi)-u(0)v'(0)=u(\pi)v'(\pi)-u'(\pi)v(\pi)$$ $$=W[u,v](\pi)=0$$

If the Wronskian vanishes at a single point on the interval, it must be zero across the interval, and $u$ and $v$ are linearly dependent. Thus, if we can show that there exist two linearly independent solutions that satisfy the boundary conditions, the operator is not selfadjoint.

These solutions could be the polynomials described in the other answer, or the solutions in the provided counter-example.