Abstract problem involving implicit function theorem

Question

Let $W = \{(x,y) \in \mathbb{R}^2 ; x,y>0\}$ and $z:W\rightarrow \mathbb{R}$ a positive $\mathcal{C}^1$ function. Let $F:\mathbb{R}^2\rightarrow\mathbb{R}$ also $\mathcal{C}^1$ function so that $F \left(x+\frac{z(x,y)}{y}, y+ \frac{z(x,y)}{x} \right) =0$ for any $(x,y) \in W$. And $\nabla F\neq 0$ in the entire plane.

I need to prove that $$x\frac{ \partial z }{\partial x} + y\frac{ \partial z }{\partial y} = z-xy$$

I tried to expressing $y$ as a function of $x$ : $y=g(x)$ where $\frac{\partial F}{\partial y} \neq 0$. so $g'(x)= -\left[\frac{\partial F}{\partial y} (x,g(x))\right]^{-1}\frac{\partial F}{\partial x}(x,g(x))$ but I don't think it's going anywhere.

Answer 1

The function $$\Phi(x,y):=F\left(x+{z(x,y)\over y}, y+{z(x,y)\over x}\right)$$is $\equiv0$ on $W$. It follows that $$\eqalign{ \Phi_x&=F_{.1}\left(1+{z_x\over y}\right)+F_{.2}\left({z_x\over x}-{z\over x^2}\right)\equiv0\cr \Phi_y&=F_{.1}\left({z_y\over y}-{z\over y^2}\right)+F_{.2}\left(1+{z_y\over x}\right)\equiv0\cr}\tag{1}$$ in $W$. Since by assumption $\bigl(F_{.1}(u,v), F_{.2}(u,v)\bigr)=\nabla F(u,v)\ne(0,0)$ for all $(u,v)$ it follows that the determinant of the homogeneous system $(1)$ for the unknowns $F_{.1}$, $F_{.2}$ has to vanish identically: $$\left(1+{z_x\over y}\right)\left(1+{z_y\over x}\right)-\left({z_x\over x}-{z\over x^2}\right)\left({z_y\over y}-{z\over y^2}\right)=0\ .$$ Clearing denominators we get $$x^2y^2-z^2+(x^2y+xz)z_x+(xy^2+zy)z_y=0\ .$$ Here you can factor out $xy+z>0$ to obtain the stated relation.

Answer 2

We can set $F(u,v)=0$ and because $\nabla F\ne 0$, by the implicit function theorem there exists $f$ such that:

$$u=f(v);\;\text{and then}\;x+\frac{z}{y}=f(y+\frac{z}{x})$$

We can now derive with respect to $x$ and respecto to $y$. As long $f=f(v)$, $f'$ stands for $\mathrm df/\mathrm dv$.

$$\left\{\begin{array}{1} 1+\dfrac{z_x}{y}=\left(\dfrac{z_x}{x}-\dfrac{z}{x^2}\right)f'\\ \dfrac{z_y}{y}-\dfrac{z}{y^2}=\left(1+\dfrac{z_y}{x}\right)f' \end{array}\right.$$

$$\left\{\begin{array}{1} 1+\dfrac{z_x}{y}=f'\dfrac{z_x}{x}-f'\dfrac{z}{x^2} \\ \dfrac{z_y}{y}-\dfrac{z}{y^2}=f'+f'\dfrac{z_y}{x} \end{array}\right.$$

$$\left\{\begin{array}{1} \dfrac{z_x}{y}-f'\dfrac{z_x}{x}=-1-f'\dfrac{z}{x^2} \\ \dfrac{z_y}{y}-f'\dfrac{z_y}{x}=f'+\dfrac{z}{y^2} \end{array}\right.$$

$$\left\{\begin{array}{1} z_x\left(1/y-f'/x\right)=-1-f'\dfrac{z}{x^2} \\ z_y\left(1/y-f'/x\right)=f'+\dfrac{z}{y^2} \end{array}\right.$$

$$\left\{\begin{array}{1} z_x=\dfrac{-1-f'z/x^2}{1/y-f'/x}\\ z_y=\dfrac{f'+z/y^2}{1/y-f'/x} \end{array}\right.$$

We can now try to write $xz_x+yz_y$ with the expressions we've got:

$$xz_x+yz_y=x\dfrac{-1-f'z/x^2}{1/y-f'/x}+y\dfrac{f'+z/y^2}{1/y-f'/x}=$$

$$=\dfrac{-x-f'zx/x^2+yf'-yz/y^2}{1/y-f'/x}=\dfrac{z(1/y-f'/x)-xy(f'/x-1/y)}{1/y-f'/x}=$$

$$=(z-xy)(1/y-f'/x)/(1/y-f'/x)=z-xy$$

So is:

$$xz_x+yz_y=z-xy$$

Furthermore, solving this pde we find that $F \left(x+\frac{z(x,y)}{y}, y+ \frac{z(x,y)}{x} \right) =0$ is its the general solution.