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How can the PDE $y^2u_{xx}+2xyu_{xy}+u_{yy}=0$ be reduced to canonical form in its hyperbolic region, namely $|x|>1,y\neq0$?

I know the required substitution $(\xi(x,y),\eta(x,y))$ should be given by the two solutions of $y^2(f_x)^2+2xyf_xf_y+(f_y)^2=0$, but after substituting $x=\cosh\theta$ I got the solution $f(x,y)=C\exp\Big(\frac{y^2}{2}\pm\cosh^{-1}x-\frac{1}{2}e^{\pm2\cosh^{-1}x}\Big)$. Substituting this sort of expression into the PDE looks way too ghastly for a problem that's supposed to take less than half an hour!

Maybe there's a different way of solving $y^2(f_x)^2+2xyf_xf_y+(f_y)^2=0$? It looks as though you should be able to spot a solution, but I can't quite manage! Or can we change this to get a different type of substitution which will still yield the canonical form?

Many thanks for any help with this!

Harry Macpherson
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2 Answers2

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$\xi$ and $\eta$ are null coordinates. In the $\xi,\eta$ coordinates the $1+1$ linear hyperbolic PDE is always equivalent to $u_{\xi\eta}(\xi,\eta) = 0$. (If you do the explicit change of coordinates, you will end up with a conformal factor $\Omega(\xi,\eta)$ multiplying $u_{\xi\eta}(\xi,\eta)$. But as long as your change of coordinates $(x,y) \to (\xi,\eta)$ is regular, the conformal factor $\Omega \neq 0$ and you can divide it away.)

That is sort of the point of the canonical form. Once you found the correct change of variables, you can just solve the equation $u_{\xi\eta} = 0$ and do the reverse coordinate transform at the end.

Willie Wong
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The procedure to get explicit expressions for the canonical form coordinates $\xi(x,y)$ and $\eta(x,y)$ is given here. I obtain (but I might have made some errors) \begin{align} \xi(x,y) &= y^2 + x^2 + x\sqrt{x^2-1} - \log(x+\sqrt{x^2-1}),\\ \eta(x,y) &= y^2 + \frac{2}{3} (x^2-1)^{\frac{3}{2}}. \end{align}

Frits Veerman
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