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How would I determine the characteristic coordinates for $xu_{xx} + u_{yy} = 0$?

This PDE reads $au_{xx} + 2b u_{xy} + cu_{yy} = 0$ with $a=x, b=0, c=1$. The polynomial equation $a\lambda^2 -2b\lambda +c =0 $ implies $\lambda^2 = \frac{-1}{x}$. Since $x<0$, we can take $x=-a$ where $a>0$ and so we get $\lambda^2 = \frac{1}{a}$ and thus $\lambda = dy/dx= \pm \frac{1}{\sqrt{a}}$ Solving this would give me $y = \mp 2\sqrt{a} + c = \mp2\sqrt{-x} +c$, and so $c = y \pm 2\sqrt{-x}$. Finally, $$ξ(x, y) = y+2\sqrt{-x} \qquad\text{and}\qquad η(x, y) = y-2\sqrt{-x}$$

Is this correct?

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  • $\begingroup$ Further reading: p. 161-162 of R. Courant, D. Hilbert (1962) Methods of Mathematical Physics vol. II: "Partial differential equations", Wiley-VCH. doi:10.1002/9783527617234 $\endgroup$ – Harry49 Dec 12 '18 at 18:17
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Try it out, set $u(x,y)=v(ξ,η)=v(y+2\sqrt{−x},y-2\sqrt{−x})$ so that \begin{align} u_x&=-\frac1{\sqrt{-x}}(v_ξ-v_η),& u_y&=v_ξ+v_η\\ u_{xx}&=\frac1{-x}(v_{ξξ}-2v_{ξη}+v_{ηη})+\frac1{2x\sqrt{-x}}(v_ξ-v_η),& u_{yy}&=v_{ξξ}+2v_{ξη}+v_{ηη} \end{align} so that $$ xu_{xx}+u_{yy}=4v_{ξη}+\frac1{2\sqrt{-x}}(v_ξ-v_η) $$ which is likely what you wanted to achieve.

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  • $\begingroup$ Cheers mate, just wanted to confirm $\endgroup$ – pablo_mathscobar Dec 12 '18 at 18:03

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