Formal adjoint ($L^{\ast }$)

The formal adjoint of a differential operator $L$ is characterized by moving $L$ between factors under an integral by integration by parts, while discarding boundary terms for compactly supported functions.

The formal adjoint of an infinite matrix does not specify the domain of the true adjoint. For the weighted difference generator $(Qf{)}_n=4^n(f_{n-1}/2-f_n)$, its formal positive-eigenvalue solution is $h_n=2^{-n}{h}_0{\prod }_{k<n}(1+\alpha 4^{-k})$. This is square summable but does not belong to the true adjoint domain of the maximal operator: testing against $f_n=2^{-n}$ contradicts the adjoint identity. The nonzero limiting boundary term is the obstruction.

// Target: analysis.bigb

For smooth positive $f$, let $L=f^{-1}\nabla \cdot (f\nabla )+k^2$. Its unweighted bilinear transpose is the displayed operator. If $v$ satisfies $\alpha v+\beta {\partial }_{n}v=0$, its transpose boundary condition is $\alpha u+\beta ({\partial }_{n}u-u{\partial }_n\log f)=0$. The resulting Green kernels satisfy $G_L({\mathbf{r}}_2,{\mathbf{r}}_1)=G_{L^{\mathrm{T}}}({\mathbf{r}}_1,{\mathbf{r}}_2)$. Equivalently $G_{L^{\mathrm{T}}}(\mathbf{r},{\mathbf{r}}_2)=f(\mathbf{r})G_L(\mathbf{r},{\mathbf{r}}_2)/f({\mathbf{r}}_2)$, exhibiting weighted acoustic Green-function reciprocity.

For an order-$m$ operator with smooth coefficients on a bounded piecewise smooth domain, integration by parts has no boundary remainder when at each boundary point either the entire boundary jet of $u$ through order $m-1$ vanishes or the entire such jet of $v$ vanishes. Every boundary summand pairs derivatives of the two functions of order at most $m-1$. It is insufficient to require only that, for each same multi-index $\beta$, one of ${\partial }^{\beta }u$ and ${\partial }^{\beta }v$ vanish: for $P={\partial }_x^2$, $u=x(a-x)\chi (y)$ and $v=1$ on a rectangle, all those same-index products vanish on the boundary but $\int (Pu)v=-2a\int \chi \ne 0$.