Given: $$\frac{\partial^2u}{\partial x^2} + \frac{\partial^2u}{\partial y^2} = 0$$
And: $$v= u\left(\frac{\ x}{\ x^2+y^2},\frac{\ y}{\ x^2+y^2}\right)$$
Proove that v also satisfies the given equation.
I know this question have been asked at the link below, but I didn't learn yet the polar cordinates method and I am soposed to solve it. Is it possibe without it? I tryed with chain rule but didn't succeed, I guess I'm doing something wrong. Thank you.
@MISC {1451549, TITLE = {Laplace equation of composite function}, AUTHOR = {Timo Junolainen (https://math.stackexchange.com/users/98840/timo-junolainen)}, HOWPUBLISHED = {Mathematics Stack Exchange}, NOTE = {URL:Laplace equation of composite function (version: 2015-09-25)}, EPRINT = {Laplace equation of composite function}, URL = {Laplace equation of composite function} }
We have $v(x,y)=u(X(x,y),Y(x,y))$ with $X=\dfrac{x}{x^2+y^2}$ and $Y=\dfrac{y}{x^2+y^2}$. The hypothesis is $\dfrac{\partial^2 u}{\partial X^2}+\dfrac{\partial^2u}{\partial Y^2}=0$
Bt the chain rule,
$\dfrac{\partial v}{\partial x}=\dfrac{\partial u}{\partial X}\dfrac{\partial X}{\partial x}+\dfrac{\partial u}{\partial Y}\dfrac{\partial Y}{\partial x}=\dfrac{\partial u}{\partial X}\dfrac{y^2-x^2}{(x^2+y^2)^2}+\dfrac{\partial u}{\partial Y}\dfrac{-2xy}{(x^2+y^2)^2}$
For $\dfrac{\partial u}{\partial X}$ we have the same form as before, so is $\dfrac{\partial u}{\partial X}(X(x,y),Y(x,y))$. We have the cumbersome task of differentiating products and composited functions.
$\dfrac{\partial^2v}{\partial x^2}=\left(\dfrac{\partial^2u}{\partial X^2}\dfrac{\partial X}{\partial x}+\dfrac{\partial^2u}{\partial X\partial Y}\dfrac{\partial Y}{\partial x}\right)\dfrac{y^2-x^2}{(x^2+y^2)^2}+\dfrac{\partial u}{\partial X}\dfrac{2x(x^2-3y^2)}{(x^2+y^2)^3}+$
$+\left(\dfrac{\partial^2u}{\partial Y^2}\dfrac{\partial Y}{\partial x}+\dfrac{\partial^2u}{\partial X\partial Y}\dfrac{\partial X}{\partial x}\right)\dfrac{-2xy}{(x^2+y^2)^2}+\dfrac{\partial u}{\partial Y}\dfrac{-2y(y^2-3x^2)}{(x^2+y^2)^3}=$
$=\left(\dfrac{\partial^2u}{\partial X^2}\dfrac{y^2-x^2}{(x^2+y^2)^2}+\dfrac{\partial^2u}{\partial X\partial Y}\dfrac{-2xy}{(x^2+y^2)^2}\right)\dfrac{y^2-x^2}{(x^2+y^2)^2}+\dfrac{\partial u}{\partial X}\dfrac{2x(x^2-3y^2)}{(x^2+y^2)^3}+$
$+\left(\dfrac{\partial^2u}{\partial Y^2}\dfrac{-2xy}{(x^2+y^2)^2}+\dfrac{\partial^2u}{\partial X\partial Y}\dfrac{y^2-x^2}{(x^2+y^2)^2}\right)\dfrac{-2xy}{(x^2+y^2)^2}+\dfrac{\partial u}{\partial Y}\dfrac{-2y(y^2-3x^2)}{(x^2+y^2)^3}$
Surely you are now into the idea that in spherical coordinates the task is lighter and probably you are running to learn how it works... Find $\dfrac{\partial v}{\partial y}$ and $\dfrac{\partial^2 v}{\partial y^2}$ and add the second partials. A lot of terms cancel and the remaining can be arranged to form the laplacian for $u$