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The original equation is $$u_{xx}-u_{yy}+2u_x+2u_y=0$$ What I did is change variables $$v=x-y,w=x+y$$ This way I get $4u_{vw}+4u_w=0$.

However, if I do instead $$\partial_x+\partial_y=\partial_v , \partial_x-\partial_y=\partial_w$$ I get the equation $u_{vw}+2u_v=0$.

Is this the same result? How do I get to the final expression of u?

Michael
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  • Consider the new differential equation as a first-order PDE for $u_v$ (or $u_w$ depending which of the two you use). You must have made an error somewhere since your transformations in either case are the same but with $u,v$ swapped. – jcneek Sep 13 '22 at 21:17
  • Only one of them can be right. I'm sure you can redo your workings and check which is right. You should get the same equation in both cases. – jcneek Sep 13 '22 at 21:20
  • That is exactly what I'm asking.. after repeating quite some times I keep on getting the same thing. – Michael Sep 13 '22 at 21:24
  • Show your workings and I'll tell you where the mistake is. The first looks right to me – jcneek Sep 13 '22 at 21:27
  • The second is that $u_{xx}-u_{yy}=u_{vw}$ by multiplying and $u_x=\frac 12 (u_v+u_w), u_y=\frac 12 (u_v-u_w)$ – Michael Sep 13 '22 at 21:33

1 Answers1

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The first method is correct, giving the p.d.e. $$u_{vw} + u_w = 0 .$$

Hint In the new variable $r := u_w$, the p.d.e. is $$r_v + r = 0 .$$

The p.d.e. in $r$ has solutions $$u_w = r = f(w) e^{-v}$$ for differentiable $f$, and integrating again gives $$u(v, w) = F(w) e^{-v} + G(v) .$$ for twice-differentiable $F, G$. In terms of $x, y$, $$\boxed{u(x, y) = F(x + y) e^{-(x - y)} + G(x - y)} .$$

Travis Willse
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  • But what is wrong with the second method? A bit more details are in the comments. – Michael Sep 13 '22 at 21:44
  • Why is $v,w$ swapped in the final answer? – Michael Sep 13 '22 at 22:00
  • Your formulae for converting from $\partial_x, \partial_y$ to $\partial_v, \partial_w$ appear incorrect. To answer your second question, the switch was just a transcription error. I've fixed the issue. – Travis Willse Sep 14 '22 at 00:06