$F(x,y)=f(x)g(y)$, $F(r\cos\theta,r\sin\theta)=h(\theta)$, find $F(x,y)$

Question

My progress: $$F_x(x,y)=f'(x)g(y)$$ $$F_y(x,y)=f(x)g'(y)$$ also, $$F(r\cos\theta,r\sin\theta)=h(\theta)$$ taking partial derivative with respect to $r$ and using chain rule: $$F_x(x,y)\cos\theta+F_y(x,y)\sin\theta=0$$ $$F_x(x,y)r\cos\theta+F_y(x,y)r\sin\theta=0$$ $$F_x(x,y)x+F_y(x,y)y=0$$ $$f'(x)g(y)x+f(x)g'(y)y=0$$ And......I got stuck. if only the positive sign were negative.....then that would be easy. Any help would be appreciated!

Answer 1

I don't know if I am missing something but this looks trivial. Put $r=0$ to see that $h(\theta)=F(0,0)$. So $h$ is constant and this means $F$ is a constant.

Answer 2

Write your last equation in the form ${x f'(x)\over f(x)}+{y\, g'(y)\over g(y)}=0$. This implies that $$-{x f'(x)\over f(x)}={y\, g'(y)\over g(y)}=\lambda$$ for some real (or complex) constant $\lambda$. The ODE $$g'(y)={\lambda\over y}g(y)$$ has the solutions $$g(y)=C' y^\lambda\qquad(y>0)\ ,$$ and similarly you obtain $$f(x)=C'' x^{-\lambda}\qquad(x>0)\ .$$ It follows that in the first quadrant $I:=\bigl\{(x,y)\bigm|x>0, \ y>0\bigr\}$ you have $$F(x,y)=C\left({y\over x}\right)^\lambda\qquad\bigl((x,y)\in I\bigr)\ .$$ This $F$ is obviously constant on rays emanating from the origin.

One can also produce an $F$ with a larger domain. E.g., when $\lambda>0$ you can put $$F(x,y):=\left|{y\over x}\right|^\lambda\ .$$ This is defined on all of ${\mathbb R}^2\setminus\{y$-axis$\}$.