# Necessary and Sufficient Condition for Boundary Conditions to be Self-adjoint

## Theorem

Let $\mathbf p$ be continuously differentiable.

$\map{\mathbf y} a\Big\vert_{x\mathop=a}=\map{\boldsymbol\psi} {\mathbf y}\Big\vert_{x\mathop=a}$
$\displaystyle\forall i,k\in\N:1\le i,k\le N:\frac{\partial p_i \sqbrk{x,\mathbf y,\map{\boldsymbol\psi}{\mathbf y} } } {\partial y_k}\Big\vert_{x\mathop=a}=\frac{\partial p_k\sqbrk{x,\mathbf y,\map{\boldsymbol\psi}{\mathbf y} } } {\partial y_i}\Big\vert_{x\mathop=a}$

## Proof

### Necessary Condition

By assumption the boundary conditions are self-adjoint.

Then exists $\map g {x,\mathbf y}$ such that:

$\map{p_i} {x,\mathbf y,\map{\boldsymbol\psi} {\mathbf y} }=\frac{\partial \map g {x\mathbf y} } {\partial y_i}$

Since $\mathbf p\in C^1$, $g\in C^2$.

Differentiate both sides with respect to $y_k$:

$\frac{\partial\map{p_i} {x,\mathbf y,\map{\boldsymbol\psi} {\mathbf y} } } {\partial y_k}=\dfrac {\partial^2 \map g {x,\mathbf y} } {\partial y_k\partial y_i}$

By the Schwarz-Clairaut Theorem, partial derivatives commute, hence indices can be mutually replaced:

$\dfrac{\partial\map {p_i} {x,\mathbf y,\map{\boldsymbol\psi} {\mathbf y} } } {\partial y_k}=\frac {\partial\map{p_k} {x,\mathbf y,\map{\boldsymbol\psi} {\mathbf y} } } {\partial y_i}$

Fixing $x=a$ provides the result.

$\Box$

### Sufficient condition

By assumption:

$\displaystyle{\frac{\partial p_i} {\partial y_j}\Big\vert_{x\mathop=a}={\frac{\partial p_j} {\partial y_i}\vert_{x\mathop=a}$

Then

$\displaystyle\exists\map g {x,\mathbf y}\in C^2:\frac{\partial p_i}{\partial y_j}\Big\vert_{x\mathop=a}={\frac{\partial p_j} {\partial y_i} }\Big\vert_{x\mathop=a}=\frac{\partial^2 g} {\partial y_i\partial y_j}\Big\vert_{x\mathop=a}$

In other words:

$\displaystyle p_i\Big\vert_{x\mathop=a}=\frac{\partial g} {\partial y_i} \Big\vert_{x\mathop=a}$

Hence, the boundary conditions are self-adjoint.

$\blacksquare$