How to solve $(f_x)^2+(f_y)^2=4(1-f(x,y))(f(x,y))^2$?

Question

Let $f:\Bbb R^2\to \Bbb R$ such that $$(f_x)^2+(f_y)^2=4\Big(1-f(x,y)\Big)\Big(f(x,y)\Big)^2,\qquad 0<f(x,y)<1.$$ then which functions satisfy the above property?

Answer 1

$$(f_x)^2+(f_y)^2=4\Big(1-f(x,y)\Big)\Big(f(x,y)\Big)^2$$ Looking for particular solutions on the form $f(x+y)$ and on the form $f(x-y)$ , the differential equation is common : $$2(f')=4(1-f)f^2$$ The solutions of this separable ODE are : $$f(X)=\left(\cosh\left(\frac{X+c}{\sqrt{2}}\right)\right)^{-2}$$ where $\quad X=x+y\quad$ or $\quad X=x-y$

$$f(x,y)=\left(\cosh\left(\frac{x\pm y+c}{\sqrt{2}}\right)\right)^{-2}$$

HINT :

With the same method, look for more general solutions on the form $f(x,y)=f(ax+by)$ and find : $$f(x,y)=\left(\cosh\left(\frac{ax+by+c}{\sqrt{a^2+b^2}}\right)\right)^{-2}$$

Answer 2

Hint:

Let $u=\tanh^{-1}\sqrt{1-f}$ with reference to http://www.wolframalpha.com/input/?i=int1%2F(x(1-x)%5E(1%2F2)),

Then $u_x=-\dfrac{f_x}{2f\sqrt{1-f}}$ and $u_y=-\dfrac{f_y}{2f\sqrt{1-f}}$

$\therefore(-2f\sqrt{1-f}~u_x)^2+(-2f\sqrt{1-f}~u_y)^2=4(1-f)f^2$

$4f^2(1-f)u_x^2+4f^2(1-f)u_y^2=4(1-f)f^2$

$u_x^2+u_y^2=1$

$u_y=\mp\sqrt{1-u_x^2}$

$u_{xy}=\pm\dfrac{u_xu_{xx}}{\sqrt{1-u_x^2}}$

Let $v=u_x$ ,

Then $v_y=\pm\dfrac{vv_x}{\sqrt{1-v^2}}$ with $v(x,0)=f_x(x)$

Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:

$\dfrac{dy}{dt}=1$ , letting $y(0)=0$ , we have $y=t$

$\dfrac{dv}{dt}=0$ , letting $v(0)=v_0$ , we have $v=v_0$

$\dfrac{dx}{dt}=\mp\dfrac{v}{\sqrt{1-v^2}}=\mp\dfrac{v_0}{\sqrt{1-v_0^2}}$ , letting $x(0)=f(v_0)$ , we have $x=f(v_0)\mp\dfrac{v_0t}{\sqrt{1-v_0^2}}=f(v)\mp\dfrac{vy}{\sqrt{1-v^2}}$ i.e. $v=F\left(x\pm\dfrac{vy}{\sqrt{1-v^2}}\right)$