Lots of information is available about Poisson's PDE $\operatorname{div}(\operatorname{grad}(u(\vec{x}))))=f(\vec{x})$. However it is hard to find information about the more generalized case \begin{eqnarray} \operatorname{div}(\Lambda(\vec{x})\operatorname{grad}(u(\vec{x})))-a(\vec{x})u(\vec{x})&=&f(\vec{x}) \end{eqnarray} where $\Lambda(\vec{x})$ is a real and symmetric matrix and $a(\vec{x})$ a variable coefficient. I have seen some recent posts at Mathoverflow about this PDE type - however my questions are going somewhat beyond. So id like to ask (please kindly review) weather i am right / not right with the following four statements:

1.) In order to augment Green's 2nd identity we may examine the expression $I:=\langle v,Lu\rangle _{L2}-\langle Lv,u\rangle _{L2}$ where $L\omega :=\operatorname{div}(\Lambda \operatorname{grad}(\omega))-a\omega$ and obtain: \begin{eqnarray} I&=&\iint_{\Omega } v[\operatorname{div}(\Lambda \operatorname{grad}(u))-au]-u[\operatorname{div}(\Lambda \operatorname{grad}(v))-av]d\vec{v}\\&=&\int_{\partial \Omega} v\Lambda \operatorname{grad}(u)-u\Lambda \operatorname{grad}(v)d\vec{S} \end{eqnarray} The necessary calculation steps are $\operatorname{div}(f\vec{g})=\operatorname{grad}(f)\vec{g}+f\operatorname{div}(\vec{g})$ and $\Lambda \vec{a} \cdot \vec{b}=\vec{a} \cdot \Lambda \vec{b}$ since $\Lambda$ is symmetric (hence is self-adjoint).

2.) The differential operator $L$ becomes self-adjoint (i.e. the integral I vanishes identical zero) if either $u,v=0$ or $\Lambda \operatorname{grad}(u) \cdot d\vec{S}=0$, $\Lambda \operatorname{grad}(v) \cdot d\vec{S}=0$, i.e. the boundary conditions are homogeneous. Note: To be homogeneous the product $\Lambda \operatorname{grad}(\omega )$ must be orthogonal to the boundary which is different in case of Poisson's equation.

3.) Green's function is symmetric (reciprocal) in case of homogeneous boundary conditions. We only need to plug into 1.) two different point sources $LG(\vec{x},\vec{\zeta})=\delta(\vec{x}-\vec{\zeta})$, $LG(\vec{x},\vec{\xi})=\delta(\vec{x}-\vec{\xi})$ including their solutions $u:=G(\vec{x},\vec{\zeta })$ and $v:=G(\vec{x},\vec{\xi })$ respectively. The integral I yields the reciprocity theorem
\begin{eqnarray}
G(\vec{\xi},\vec{\zeta})&=&G(\vec{\zeta},\vec{\xi})
\end{eqnarray}

4.) Green's 2nd identity yields into Kirchhoff's equation. Here we ask for the solution u of the PDE $Lu=f$ when a spatial excitation f is given inside the domain $\Omega $ and a Dirichlet condition h:=u is given at the border $\partial \Omega$. Now use Green's identity from 1.) and let $v:=G(\vec{x},\vec{\zeta})$ and its derivative $LG(\vec{x},\vec{\zeta})=\delta(\vec{x}-\vec{\zeta})$. Let $Lu=f$. If we request from v=G that it solves the PDE including homogeneous Dirichlet boundary conditions - i.e. v=G=0 at the edge $\partial \Omega$ - and let $u=h$ at the border $\partial \Omega$ we obtain the solution inside the domain $\Omega$ as:
\begin{eqnarray}
u(\vec{\zeta)}&=&\iint_{\Omega} f(\vec{x})G(\vec{x},\vec{\zeta})d\vec{v}+\int_{\partial \Omega} h(\vec{x})[\Lambda(\vec{x})\operatorname{grad}_{\vec{x}}G(\vec{x},\vec{\zeta})] d\vec{S}
\end{eqnarray}

Does anyone know some useful books about these issues. Maybe especially suitable for engineers?

Best regards and thanks in advance chloros2

`\operatorname{grad}`

to get $\operatorname{grad}$ instead of $grad$, etc. (4)`\langle \rangle`

look much better than`< >`

: $\langle u,v \rangle$ versus $<u,v>$. $\endgroup$ – Nate Eldredge Mar 3 '18 at 18:51