I have a PDE of the following form, from a physics problem: $$ y \left(\alpha \frac{\partial }{\partial y}+x \frac{\partial^2 }{\partial x \partial y} \right)f(x,y) = \left( z_1 + z_2 x^\alpha y^{-2} \right) f(x,y) $$ $f(x,y)$ is a real-space real-valued function and $z_{1,2},\alpha$ are real numbers, generally irrational. The latter, specifically the $z_2$ coefficient term, seems to make all of the textbook methods (characteristics, Froebnius, Fourier transform) fail. Does any one know weather a method exists to solve this? Apologies if this is a simple question but, well, I am a theoretical physicist and it is not simple for me.
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1$\begingroup$ What do you mean by "solve"? In what region, under what boundary conditions, or you want just one solution or all solutions? And what are your $z$ and $\alpha$? $\endgroup$– Alexandre EremenkoCommented Dec 21, 2015 at 0:20
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1$\begingroup$ Maybe change variables to $\xi = \log x$, $\eta = \log y$ and try Fourier transforms again. $\endgroup$– Igor KhavkineCommented Dec 21, 2015 at 1:08
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$\begingroup$ Looks like a semilinear wave equation, so initial data should be prescribed on one of the two light rays. $\endgroup$– Fan ZhengCommented Dec 21, 2015 at 4:48
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$\begingroup$ Igor, thank you, in fact Froebnius might work with this variable change $\endgroup$– GiorgioCommented Dec 21, 2015 at 14:36
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$\begingroup$ Do you have an initial condition? (and are you sure characteristics don't work?) $\endgroup$– Pietro MajerCommented Dec 21, 2015 at 15:24
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1 Answer
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The Ansatz $$f(x,y)=x^pu(x^\alpha y^{-2}) $$ yields to a linear second order ODE for $u(t)$
$$2t^2 u''+ (4\alpha+2p)tu'+(z_1+z_2t)u=0\ , $$ which can be immediately solved by series in terms of hypergeometric functions.
answered Dec 21, 2015 at 16:29
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2$\begingroup$ Actually Bessel functions: $${t}^{1/2-\alpha-p/2}{{ J}_{\sqrt {4\,{\alpha}^{2}+4\,\alpha\,p+{p}^ {2}-4\,\alpha-2\,p-2\,z_{{1}}+1}}\left(\sqrt {2z_{{2}} t}\right)} $$ and the corresponding expression with $Y$ instead of $J$. $\endgroup$ Commented Dec 21, 2015 at 17:00
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$\begingroup$ Thank you Pietro and Robert. This really helps, sometimes the right trick overcomes months of thinking :(( $\endgroup$– GiorgioCommented Dec 22, 2015 at 21:15