Coordinate Transformation to solve Laplace Equation

annakolish

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Nov 26, 2006
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I have the following problem. I got the answer the part (a), but I could not figure the solution for the part (b), please help!
Consider the coordinate transformation \(\displaystyle y = uv, x = (1/2)(u^2 - v^2)\), v\(\displaystyle \geq\)0, u goes from negative infinity to positive infinite
part (a) (I solved this one already) For \(\displaystyle \phi(u,v) = F(x,y)\), show that \(\displaystyle \phi_{uu} + \phi_{vv} = (u^2 + v^2)(F_{xx}+ F_{yy})\).
Part (b)Show tht the parabolas, \(\displaystyle y^2 = 2(x+1/2)\) and \(\displaystyle y^2=-2(x-1/2)\) correspond respectively to v = 1 and u = 1 in (u,v) space, and that the line y=0, v\(\displaystyle \prec\)0 correspond respectively to v = 0 and u = 0 in (u,v) space (these are parabolic cylindrical coordinate)

Did is what I did
basically, I just substitute uv for y and x

there fore, \(\displaystyle y^2 = 2(x+1/2)=u^2v^2=u^2-v^2+1\)
I'm stuck here, I don't see how I can reduce this to v =1;

I have a similar problem with the rest,

I also try this way, \(\displaystyle y^2 = 2(x+1/2) \mbox{, therefore,}\frac{y^2}{x+\frac{1}{2}}=2 \mbox{ or } \frac{y^2}{2x+1}=1, \mbox{ so, in order for v = 1, v has to be equal to }\frac{y^2}{2x+1}
\\\mbox{ Originally, I know that} v = y/u = \frac{y^2}{uy}=\frac{y^2}{u^2v}
\\\mbox{ I'm stuck here also
Some hints would be helpful, thanks}\)
 
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