From Boundary to righthandside

I have a problem coming from linear elasticity in $(x,y,z)\in\mathbb{R}^2\times \mathbb{R}^+$, $t\in \mathbb R$:

\left\{\begin{aligned}\partial_{tt} \sigma&=A(D_x,D_y,D_z) \sigma\\ \sigma\big|_{t=0}&=0\\ \partial_t\sigma\big|_{t=0}&=0\\ \sigma\big|_{z=0}&=f(x,y,t), \end{aligned}\right.

where $\sigma=(\sigma_{xx},\sigma_{yy},\sigma_{zz},\sigma_{xy},\sigma_{xz},\sigma_{yz})$ is the vector of stresses. $A$ is assumend to be a second order differential Matrix, independent of $x,y,z$, symmetric. The symbol $A(\xi_1,\xi_2,\xi_3)$ is of rank three and non negative definite.

This equation describes the equation of motion for the stresses, if one knows the stresses on the boundary.

Since I would like to apply Fourier Transform, I need to transform this into a whole space problem.

Can anyone help me here? Is it possible to write this somehow like this:

\left\{\begin{aligned}\partial_{tt} \widetilde{\sigma}&=A(D_x,D_y,D_z) \widetilde{\sigma} + \text{something}\\ \widetilde{\sigma}\big|_{t=0}&=0\\ \partial_t\widetilde{\sigma}\big|_{t=0}&=0, \end{aligned}\right.

where $\widetilde\sigma$ is definded on the whole space and $\widetilde\sigma\big|_{z>0}\equiv \sigma$?

If anyone has some literature for me, that would be great too!

thanks for the help

EDIT:

Maybe I should have been more clear about the matrix $A$. It splits as follows:

$$A(D_x,D_y,D_z)= A_0 D_z^2 + A_1(D_x,D_y) D_z+ A_2(D_x,D_y),$$

where $D_x= \sqrt{-1} \partial_x$ and so on... So if i want to extend this antisymmetrix, i will end up with the following equation: \left\{\begin{aligned}\partial_{tt} \widetilde{\sigma}&=[A_0 D_z^2 + sign(z) A_1(D_x,D_y) D_z+ A_2(D_x,D_y)] \widetilde{\sigma} + \text{something}\\ \widetilde{\sigma}\big|_{t=0}&=0\\ \partial_t\widetilde{\sigma}\big|_{t=0}&=0, \end{aligned}\right.

Unfortunately, this wont help very much in terms of Fourier Transform...

• If you are willing to make the equations inhomogeneous, then the boundary values $f(x,y,t)$ plays little role. If you let $\tau$ be a fixed smooth function such that $\tau |_{z = 0} = f$, then $$\partial_{tt} (\sigma - \tau) = A(\sigma - \tau) - (\partial_{tt} \tau - A\tau)$$ so $\sigma - \tau$ solves an inhomogeneous problem with vanishing boundary condition at $z = 0$. For the linear wave equation this type of Dirichlet condition is enforced by antisymmetrizing: enforcing that $\tilde{\sigma}(\cdot,-z) = -\tilde{\sigma}(\cdot,z)$. Have you tried something like that? – Willie Wong Dec 7 '16 at 14:38
• Hey, first of all, thanks for the reply. Please see my response in the edit-section above :). – Martin Dec 7 '16 at 16:13