Polar coordinate

Question

Let $f(x,y)$ be a differntiable function in $\mathbb{R}^2$ so that $f_x(x,y)y=f_y(x,y)x$ for all $(x,y)\in\mathbb{R}^2$.

Find $g(r)$ so that $g(\sqrt{x^2+y^2})=f(x,y)$ and $g$ is differentiable in $[0,\infty)$.

Answer 1

Set $h(r,\theta):=f(r\cos\theta,r\sin\theta)$. By composition, this is differentiable on $\mathbb{R}^2$ and by the chain rule $$ \frac{\partial h}{\partial \theta}(r,\theta)=\frac{\partial f}{\partial x}(r\cos\theta,r\sin\theta)\cdot(-r\sin\theta)+\frac{\partial f}{\partial y}(r\cos\theta,r\sin\theta)\cdot(r\cos\theta). $$ By assumption, this yields $$ \frac{\partial h}{\partial \theta}(r,\theta)=0\qquad\forall r,\theta. $$ So $h$ does not depend on $\theta$, by the mean value theorem. Now set $g(r):=h(r,0)=f(r,0)$. This is differentiable on $\mathbb{R}$ by composition. And by the above, $$ g(r)=h(r,\theta)=f(r\cos\theta,r\sin\theta)\qquad\forall r,\theta. $$ Applying this to the polar coordinates of $(x,y)$, we get $$ g\left(\sqrt{x^2+y^2}\right)=f(x,y)\qquad \forall x,y. $$ Since $g$ is differentiable on $\mathbb{R}$, it is a fortiori differentiable on $(0,+\infty)$.