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Consider $f(x,y,t)$ whose spatial second order derivatives evolve over time based on the following pair of PDEs:

$$\begin{align} \frac{\partial^3}{\partial x^2 \partial t}f\,&=\,\frac{1}{2}\left(\frac{\partial^2}{\partial y^2}f-\frac{\partial^2}{\partial x^2}f\right)\\ \\ \frac{\partial^3}{\partial y^2 \partial t}f\,&=\,\frac{1}{2}\left(\frac{\partial^2}{\partial x^2}f-\frac{\partial^2}{\partial y^2}f\right) \end{align}$$

The process is subject to initial condition $f(x,y,t=0)=f_0(x,y)$ for some given $f_0(x,y)$.

Is there an analytical expression for the solution $f$ in terms of $f_0$?

Help would be greatly appreciated.

Golabi

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    $\begingroup$ You have two equations for the evolution of a single unknown. How is this possible? If you combine the two equations you obtain $\partial_t \nabla^2 f= 0$ and therefore $\nabla^2 f = \nabla^2 f_0$. Is this useful? $\endgroup$ – Dmoreno Apr 12 '17 at 17:06
  • $\begingroup$ Thanks! You are right. Thank you also for the tip that if I combine the two equations, I can constrain the solution space of $f$ by requiring its Laplacian to match that of $f_0$ for all time $t$. $\endgroup$ – Golabi Apr 12 '17 at 17:18
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Write $v = f_{xx}$ and $w = f_{yy}$, your equation simplifies to the pair of ODE

$$ \partial_t (v + w) = 0, \qquad \partial_t(v-w) = - (v-w) $$

So you have that

$$ f_{xx}(t,x,y) = \frac12 \left[ f_{xx}(0,x,y)(1 + e^{-t}) + f_{yy}(0,x,y)(1 - e^{-t}) \right] $$

and

$$ f_{yy}(t,x,y) = \frac12 \left[ f_{xx}(0,x,y)(1 - e^{-t}) + f_{yy}(0,x,y)(1 + e^{-t}) \right] $$

Now, these would be the solutions assuming that a solution exists. However, necessarily you need to have

$$ f_{xxyy} = f_{yyxx} $$

which will require, by our explicit formula, that

$$ f_{xxxx}(0,x,y) = f_{yyyy}(0,x,y) $$

so you are not allowed to prescribe data freely for this equation. (I am not sure if there are other constraints also.)

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