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I'm trying to get an exact solution to this second order inhomogeneous PDE:

$$ \frac{\partial^2}{\partial{x}^2} y(x, z) - \frac{\partial^2}{\partial z^2} y(x, z)=k^2y(x, z)-\frac{1}{3}e^{4(x-2z)}y(x, z) $$

where $k^2$ is a constant. No boundary conditions. Any ideas? I tried with $t=(x-cz)$ and with variable separation but it isn't the right way.

Yesterday I shared the same question on Mathematics, maybe I'm luckier here

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    $\begingroup$ Transformation as $u=x-2y$ and $v=-2x+y$ reduces this equation to $\frac{\partial^2{y}}{\partial v^2}-\frac{\partial^{2}y}{\partial u^2}=(\frac{1}{9}e^{4u}-\frac{k^2}{3})y$. Now, you can use separation of variables. $\endgroup$
    – Alapan Das
    Jan 14, 2021 at 14:19
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    $\begingroup$ @AlapanDas thank you so much! Instead, I just tried with $u=4x-8z$ and $v=c_1 x+ c_2 z$ and with a canonical transformation and conditions for $c_1$ and $c_2$ I arrive at $$ 48 \frac{\partial ^2 y}{\partial u^2}-3\frac{\partial ^2 y}{\partial v^2}=(-k^2+e^u)y$$ but I think that your method is simpler than mine. Thank you very much! $\endgroup$
    – Gin
    Jan 14, 2021 at 14:46

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I am writing my comment as answer. If we change the variables $x,z \rightarrow u,v$ such that $u=x-2y$ (Just to make the exponential term a single variable function), and $v=ax+by$, then

Then, to eliminate $\frac{\partial^2 y}{\partial u\partial v}$ term we need to choose $\frac{a}{b}=-2$. Choose, $a=-2, b=1$.

Then, the equation becomes $\frac{\partial ^2 y}{\partial v^2}+\frac{k^2}{3}y=\frac{\partial ^2 y}{\partial u^2}+\frac{1}{9}{e^{4u}}y$.

Separating $y=B(v)A(u)$ we get, $B''(v)+(\frac{k^2}{3}-b)B(v)=0$ for some constant which communicates between two sides. And this is easy to solve.

The second equation is $A''(u)+(\frac{1}{9}e^{4u}-b)A(u)=0$.

Now ,again change the variable as $u \rightarrow t=e^{2u}$, then the equation becomes $u^2\bar{A}''+u\bar{A}'+((\frac{t}{6})^2-(\frac{\sqrt{b}}{2})^2)\bar{A}=0$ where, $\bar{A}(t)=A(u)$.

And this is the Bessel equation, hence, $A(u)=c_1J_{\frac{\sqrt{b}}{2}}(\frac{e^{2u}}{6})+c_2J_{-\frac{\sqrt{b}}{2}}(\frac{e^{2u}}{6})$.

($\frac{\sqrt{b}}{2}$ isn't a natural number, in that case Bessel function of second kind should be taken).

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  • $\begingroup$ yes I obtain the same result but with different coefficients ahahah, I have to check it! Btw, helpful, thank you! $\endgroup$
    – Gin
    Jan 14, 2021 at 15:29
  • $\begingroup$ You are welcome. $\endgroup$
    – Alapan Das
    Jan 14, 2021 at 15:31

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