Decouple Two Partial Differential Equations
Sophia Terry 
I am solving a wave propagation problem and after lots of derivations I ended up with the following differential equations. Any idea how to decouple this system so I can get two decoupled equations of the following forms: $\frac{\partial^{2} \Phi}{\partial t^{2}} = $ function of $(\Phi)$, and $\frac{\partial^{2} \Psi}{\partial t^{2}} = $ function of $(\Psi)$.
Equation (1):$$\rho \frac{\partial}{\partial x} \left(\frac{\partial^{2} \Phi}{\partial t^{2}}\right) + \rho \frac{\partial}{\partial z} \left(\frac{\partial^{2} \Psi}{\partial t^{2}}\right) = (\lambda+2\mu)\frac{\partial}{\partial x} (\bigtriangledown^{2} \Phi) + \mu \frac{\partial}{\partial z} (\bigtriangledown^{2} \Psi) + \eta \frac{\partial}{\partial t} \left[2 \frac{\partial}{\partial x} \left(\frac{\partial^{2} \Phi}{\partial z^{2}}\right) + \frac{\partial}{\partial z} \left(\frac{\partial^{2} \Psi}{\partial z^{2}} - \frac{\partial^{2} \Psi}{\partial x^{2}}\right)\right]$$
Equation (2):$$\rho \frac{\partial}{\partial z} \left(\frac{\partial^{2} \Phi}{\partial t^{2}}\right) - \rho \frac{\partial}{\partial x} \left(\frac{\partial^{2} \Psi}{\partial t^{2}}\right) = (\lambda+2\mu)\frac{\partial}{\partial z} (\bigtriangledown^{2} \Phi) - \mu \frac{\partial}{\partial x} (\bigtriangledown^{2} \Psi)$$
$\endgroup$ 02 Answers
$\begingroup$Starting from the answer of @Cosmas Zachos, one can rewrite the two equations compactly as:
$$D_1\Phi=L_1\Psi~~(1)\\D_2\Phi=L_2\Psi~~ (2)$$
with the definitions :
$$D_1=\rho {\partial_x} \partial^{2}_t - (\lambda+2\mu)\partial _x \nabla^{2} -2\eta {\partial_ t} {\partial_ x} \partial^{2}_z \\L_1= - \rho \partial _z \partial^{2}_z + \mu {\partial _z} \nabla^{2} + \eta {\partial _t} {\partial _z} \left({\partial _z^{2}} - \partial^{2}_x \right )\\D_2=\rho {\partial_ z} \partial^{2} _t - (\lambda+2\mu){\partial_ z} \nabla^{2}\\L_2=\rho {\partial _x} \partial^{2}_t - \mu {\partial _x} \nabla^{2} $$
However these operators are mutually commutative. Thus we can eliminate one of the two functions each time to obtain that both functions satisfy the same 7th order partial differential equation defined by the following operator:
$$(D_1L_2-L_1D_2)\pmatrix{\Phi\\ \Psi}=0$$
I don't believe you can rewrite these equations as the form mentioned above, since that would imply some kind of order reduction of the original equation, and that could only happen if the equation was an exact derivative of some sort, which doesn't seem to be true here.
$\endgroup$ $\begingroup$I doubt what you are asking for is feasible, but, as an extended comment, I'd recommend rewriting your equations in a friendlier fashion,
$$\Bigl (\rho {\partial_x} \partial^{2}_t - (\lambda+2\mu)\partial _x \nabla^{2} -2\eta {\partial_ t} {\partial_ x} \partial^{2}_z \Bigr )~\Phi=\Bigl ( - \rho \partial _z \partial^{2}_z + \mu {\partial _z} \nabla^{2} + \eta {\partial _t} {\partial _z} \left({\partial _z^{2}} - \partial^{2}_x \right ) \Bigr ) ~\Psi \tag{1}$$ and$$ \Bigl ( \rho {\partial_ z} \partial^{2} _t - (\lambda+2\mu){\partial_ z} \nabla^{2} \Bigr )\Phi = \Bigl ( \rho {\partial _x} \partial^{2}_t - \mu {\partial _x} \nabla^{2}\Bigr ) \Psi ~~. \tag{2}$$
Judicious $\pm$ linear combinations might lead to more attractive or evocative or factorable expressions, but...
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