A friend is studying some Finsler metrics and she has come upon the following partial differential equation:
$$\begin{cases} y^i\frac{\partial b(x,y)}{\partial x^i}=0\\ y^i\frac{\partial b(x,y)}{\partial y^i}=1\\ \end{cases},$$ where $b:\mathbb{R}^{2n}\to\mathbb{R},\ x=(x^1,\dots,x^n)\in\mathbb{R}^n, \ y=(y^1,\dots,y^n)\in\mathbb{R}^n.$
I have managed to find a family of solutions: $$b_k(x,y)=\sqrt{|y|^2-k(|x|^2|y|^2-\langle x,y\rangle^2)},\ k\in\mathbb{R}.$$ but I don't see how to solve the equation in general.
Change the $y$ variables to spherical coordinates $(r,\theta)=(r,\theta_1,\ldots,\theta_{n-1})$. Then $$1=y^i\frac{\partial b(x,y)}{\partial y^i}=r\frac{\partial b(x,r,\theta)}{\partial r}$$ Hence $b(x,r,\theta)=\log r+F(x,\theta)$. By the first PDE, $F$ must satisfy: $$0=y^i\frac{\partial b(x,y)}{\partial x^i}=y^i(\theta)\frac{\partial F(x,\theta)}{\partial x^i}$$ Thus it is of the form $F(x,\theta)=F(x^iv_i(\theta),\theta)$ where $v$ can be functions of $\theta$ which are orthogonal to $y(\theta)$: $y^i(\theta)v_i(\theta)=0$ for all $\theta$. For example when $n=2$ and $(y^1,y^2)=(r\cos\theta,r\sin\theta)$: $$b(x,y)=\log r+F(x^2\cos\theta-x^1\sin\theta,\theta)$$
As you can see, the above solution does not cover the solutions you have found. I believe the solutions you suggested do not satisfy the second PDE but rather: $$y^i\frac{\partial b(x,y)}{\partial y^i}=b(x,y)$$