Proof that the operator $\Theta\vec{\omega}=-\mu\Delta\vec{\omega}-(\lambda+\mu)\nabla(\nabla\cdot\vec{\omega})$ is positive and self-adjoint

Question

I have the following partial differential equation (eigenvalue equation) \begin{equation} \Theta \vec{\omega}=\alpha\vec{\omega} \end{equation} where I have defined the linear partial differential operator $\Theta$ as $$\Theta\vec{\omega}=-\mu\Delta\vec{\omega}-(\lambda+\mu)\nabla(\nabla\cdot\vec{\omega}).$$ I would like to show that this operator is positive self-adjoint with respect to the $L^2$ scalar product $(u,v)=\int_B u^*\cdot v~ d^3 r$. In a paper which I found (link, doi) they said it is obvious, however, I would like to prove this explicitly but I struggle a lot. Can anyone give me a hint or some intuition how to prove that? In my case $B$ denotes the body which is a sphere with radius $R$ and $\lambda,\mu$ are the Lamé parameters. This equation one encounters often in linear and isotropic elasticity.

If I could show this, than an immediate consequence would be that the eigenvalues are real and positive as well as the eigenfunctions to different eigenvalues are orthogonal, that is clear an easy to prove if the operator is self-adjoint. Is it also possible to prove that the eigenfunction form a complete set of a basis, so that I can write every vector field defined on $B$ as a linear combination of the eigenfunctions?

Answer 1

The operator is self-adjoint trivially by the spectral theorem. In fact, it can be seen as a real-orthogonal-matrix-valued function $f(-i\nabla)$ of the self-adjoint $-i \nabla$ operator.

If the function $f$ is also positive as a matrix (when considered as a function of a real variable), then the operator is positive.

The resulting operator is however unbounded (if, as I assume, the function is unbounded), and does not have compact resolvent so it does not admit a complete basis of eigenvalues.

Answer 2

I'm going to assume $\mu$ and $\lambda$ are constants, you can certainly trace again all the results below if that is not the case. In general you want to study the product

$$ \langle \Theta {\bf f}| {\bf g}\rangle = \langle {\bf f}| \Theta {\bf g}\rangle \tag{1} $$

let's start by pieces

$\nabla^2$

First note that

$$ (\nabla^2 {\bf f}^*)\cdot{\bf g} = \nabla\cdot [(\nabla {\bf f}){\bf g}] - \nabla {\bf f} : \nabla {\bf g} $$

So that

\begin{eqnarray} \int_\Omega (\nabla^2 {\bf f}^*)\cdot{\bf g} ~{\rm d}^3 {\bf x} &=& \int_\Omega \nabla\cdot [(\nabla {\bf f}^*)\cdot{\bf g}]~ {\rm d}^3 {\bf x} - \int_\Omega\nabla {\bf f}^* : \nabla {\bf g}~{\rm d}^3 {\bf x} \\ &=& \int_{\partial \Omega} [(\nabla {\bf f}^*)\cdot{\bf g}]\cdot {\rm d}^2 {\bf S} - \int_\Omega\nabla {\bf f}^* : \nabla {\bf g}~{\rm d}^3 {\bf x} \\ \end{eqnarray}

The first term vanishes if you assume that both fields vanish at the boundary, or that the flow is normal to the boundary. So you end up with the last term, and you can apply the same trick again, the result is

$$ \int_\Omega (\nabla^2 {\bf f}^*)\cdot{\bf g} ~{\rm d}^3 {\bf x} = \int_\Omega {\bf f}^*\cdot(\nabla^2 {\bf g}) ~{\rm d}^3 {\bf x} \tag{2} $$

$\nabla (\nabla \cdot)$

You can repeat the same exercise here, the general idea is to express the term as $\nabla \cdot (\text{something}) - \text{stuff}$, use the divergence theorem to get rid of the first term and then process the second term to obtain the final result.

$$ \int_\Omega (\nabla \nabla \cdot{\bf f}^*)\cdot{\bf g} ~{\rm d}^3 {\bf x} = \int_\Omega {\bf f}^*\cdot(\nabla \nabla \cdot{\bf g}) ~{\rm d}^3 {\bf x} \tag{3} $$

Putting together $(2)$ and $(3)$ you will reach the conclusion in $(1)$, and that's it