if $u(x,y) $ satisfies $u_{xx} + u_{yy} = 0$ then $v=u(\frac{x}{x^2+y^2},\frac{y}{x^2+y^2})$ satisfies $v_{xx} + v_{yy} = 0$
I tried using the chain rule and it didn't work for me.. I think I might have missed something maybe.. Could anyone show how to properly write $v_{xx} $ and $v_{yy} $ ?
Thanks.
EDIT : my question is also about generally differentiating such function, like is that okay to do :
let $t =\frac{x}{x^2+y^2}$ and $s=\frac{y}{x^2+y^2}$
$v_{x } = \frac{\partial u}{\partial t} \frac{\partial t}{\partial x} + \frac{\partial u}{\partial s} \frac{\partial s}{\partial x} $
or if $v$ is just substituting any $x$ in $u$ for $\frac{x}{x^2+y^2} $ and any $y$ in $u$ for $\frac{y}{x^2+y^2}$ then maybe $v_{x } = \frac{\partial u}{\partial x} \frac{\partial t}{\partial x} + \frac{\partial u}{\partial y} \frac{\partial s}{\partial x} $
Let $g:\Bbb R^2\setminus\{(0,0)\}\to\Bbb R^2,\ (x,y)\mapsto (\frac{x}{x^2+y^2},\frac{y}{x^2+y^2})$. Now consider the polar coordinates $(r,\varphi)$ instead of the cartesian coordinates $$(x,y)=(r\cos(\varphi),r\sin(\varphi)).$$ Using these, $g$ has the following simple form $$g(r,\varphi)=\left(\frac{r\cos(\varphi)}{(r\cos(\varphi))^2+(r\sin(\varphi))^2},\frac{r\sin(\varphi)}{(r\cos(\varphi))^2+(r\sin(\varphi))^2}\right)=\left(\frac{\cos(\varphi)}{r},\frac{\sin(\varphi)}{r}\right).$$ Expressing the result also in polar coordinates (whence expressing $g$ completely in polar coordinates) gives $$g(r,\varphi)=\left(\frac 1r,\varphi\right).$$ The Laplace operator in polar coordinates is given by (see Wikipedia) \begin{align} [2]:\Delta f(r,\varphi):=\frac{\partial^2f}{\partial r^2}(r,\varphi)+\frac1r\frac{\partial f}{\partial r}(r,\varphi)+\frac{1}{r^2}\frac{\partial^2f}{\partial \varphi^2}(r,\varphi) \end{align} which gives zero by our assumption (your $u$ is my $f$; sorry for that). We want to show that $\Delta(f\circ g)=0$. Thus, we have to show $$\frac{\partial^2(f\circ g)}{\partial r^2}+\frac1r\frac{\partial (f\circ g)}{\partial r}+\frac{1}{r^2}\frac{\partial^2(f\circ g)}{\partial \varphi^2}=0.$$ Using $\frac{\partial (f\circ g)}{\partial r}=\frac{\partial f}{\partial r}\frac{\partial g_1}{\partial r}+\frac{\partial f}{\partial\varphi}\frac{\partial g_2}{\partial\varphi}$ where $g=(g_1,g_2)$ and similar for $\frac{\partial (f\circ g)}{\partial\varphi}$ we obtain $$\frac{\partial (f\circ g)}{\partial r}(r,\varphi)=-\frac1{r^2}\frac{\partial f}{\partial r}(\frac1r,\varphi)\\ \frac{\partial^2 (f\circ g)}{\partial r^2}(r,\varphi)=\frac1{r^4}\frac{\partial^2f}{\partial r^2}(\frac1r,\varphi)+\frac2{r^3}\frac{\partial f}{\partial r}(\frac1r,\varphi)\\ \frac{\partial (f\circ g)}{\partial\varphi}(r,\varphi)=\frac{\partial f}{\partial \varphi}(r,\varphi).$$ So we have to prove that $$\frac1{r^4}\frac{\partial^2f}{\partial r^2}(\frac1r,\varphi)+\frac2{r^3}\frac{\partial f}{\partial r}(\frac1r,\varphi)+\frac1r\left(-\frac1{r^2}\frac{\partial f}{\partial r}(r,\varphi)\right)+\frac1{r^2}\frac{\partial^2 f}{\partial \varphi}\left(\frac1r,\varphi\right)=0.$$ Multiplying by $r^4$ gives the equivalence to $$\frac{\partial^2f}{\partial r^2}(\frac1r,\varphi)+r\frac{\partial f}{\partial r}(\frac1r,\varphi)+r^2\frac{\partial^2 f}{\partial \varphi}\left(\frac1r,\varphi\right)=0$$ which is $$\frac{\partial^2f}{\partial r^2}(\frac1r,\varphi)+\frac1{\frac1r}\frac{\partial f}{\partial r}(\frac1r,\varphi)+\frac1{\frac1{r^2}}\frac{\partial^2 f}{\partial \varphi}\left(\frac1r,\varphi\right)=0$$ and thus true by [2].