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:


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.


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?

  • $\begingroup$ Something of the form $\sin(x) + \cos(x)$ satisfy the BC. $\endgroup$ – Piotr Benedysiuk Jan 6 '17 at 19:04
  • $\begingroup$ @PiotrBenedysiuk That works, but another solution is needed for a counter-example since the Wronskian of two linearly dependent solutions is uniformly 0 and will satisfy selfadjoint condition. $\endgroup$ – Jared Goguen Jan 6 '17 at 19:11
  • $\begingroup$ so you just want to check if all the functions with the boundary conditions satisfy $u(\pi)v'(\pi)-u(0)v'(0)=0$ ? $\endgroup$ – Onir Jan 6 '17 at 19:29
  • $\begingroup$ @JorgeFernándezHidalgo Yes, unless I made a mistake in the derivation of that condition. $\endgroup$ – Jared Goguen Jan 6 '17 at 19:30
  • $\begingroup$ but then it is clearly false right? $\endgroup$ – Onir Jan 6 '17 at 20:02

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.

  • $\begingroup$ hopefully I didn't make a dumb mistake. $\endgroup$ – Onir Jan 6 '17 at 19:07
  • $\begingroup$ I'm not sure what this answer is trying to get at. Also, should the upper limit of the first integral be $\pi$? $\endgroup$ – Jared Goguen Jan 6 '17 at 19:12
  • $\begingroup$ oh yeah, my bad, $\endgroup$ – Onir Jan 6 '17 at 19:13
  • $\begingroup$ Where does the first integral come from? How do we know it is equal to zero? Where does the conclusion come from? How do we know that there are functions that do not satisfy that equation? $\endgroup$ – Jared Goguen Jan 6 '17 at 19:20

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.


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