# Question on coupled partial differential equations.

While doing some mental exercises from elasticity theory, I came across the following equations for the elastic waves in a two-dimensional anisotropic medium:

$$\rho\frac{\partial^{2}}{\partial{t}^{2}}\begin{bmatrix}u_{1}(x_{1}, x_{2}, t) \\ u_{2}(x_{1}, x_{2}, t)\end{bmatrix}=\begin{bmatrix}\alpha\frac{\partial^{2}}{\partial{x_{1}^{2}}}+\beta\frac{\partial^{2}}{\partial{x_{2}^{2}}} & \alpha\frac{\partial^{2}}{\partial{x_{1}}\partial{x_{2}}} \\ \alpha\frac{\partial^{2}}{\partial{x_{1}}\partial{x_{2}}} & \beta\frac{\partial^{2}}{\partial{x_{1}^{2}}}+\alpha\frac{\partial^{2}}{\partial{x_{2}^{2}}}\end{bmatrix}\begin{bmatrix}u_{1}(x_{1}, x_{2}, t) \\ u_{2}(x_{1}, x_{2}, t)\end{bmatrix}$$ It is quite easy to obtain the dispersion relation for the waves by, say doing the Fourier Transform, or bysimply performing the anzatz $u(x_{1}, x_{2}, t)=u_{0}^{1}\cos(\omega{t}-k\cdot{x})+u_{0}^{2}\sin(\omega{t}-k\cdot{x})$. One is then left with the following diagonalization broblem: $$\det\Big[\begin{bmatrix}\alpha{k_{1}^{2}}+\beta{k_{2}^{2}} & \alpha{k_{1}k_{2}} \\ \alpha{k_{1}k_{2}} & \beta{k_{1}^{2}}+\alpha{k_{2}^{2}}\end{bmatrix}-\omega^{2}{I}\Big]=0$$ Which gives two eigenvalues for frequency $\omega$, so using the anzatz above we get $4$ linearly independent solutions as it should have been. My question is weather there is a way to decouple the equations anyhow, so that other methods of solution can be used?

• yes i am sure the operator can be diagonalizable, applying then a change of variable. IF IT can be decoupled, that is the way. – Brethlosze Jun 7 '17 at 13:04
• Ok, I have another question. Is it possible to find the solutions to this equation that have the property $u(x_{1}, L_{2}, t)=u(x_{1}, 0, t)=u(L_{1}, x_{2}, t)=u(0, x_{2}, t)=0$. If so, how? – Kiryl Pesotski Jun 8 '17 at 11:06
• But if there are equal to zero, i dont see a problem with the decoupling and or superposition if applicable – Brethlosze Jun 8 '17 at 11:24
• Can you specify how can you do it? No superposition can work unless you get rid of the off-diagonal terms. I do not see how to get rid of them. – Kiryl Pesotski Jun 8 '17 at 11:46