Solve a partial diferential equation $u_x^2+u_y^2=u^2$ Solve the equation
$$u_x^2+u_y^2=u^2,  \\ u(x,0)=1$$
Is there anybody who can help me?  
 A: Hint: Try to find a solution of the form $u(x,y)=f(x)g(y)$.
A: Let $u=e^v$ ,
Then $u_x=e^vv_x$
$u_y=e^vv_y$
$\therefore(e^vv_x)^2+(e^vv_y)^2=(e^v)^2$ with $v(x,0)=0$
$e^{2v}v_x^2+e^{2v}v_y^2=e^{2v}$ with $v(x,0)=0$
$v_x^2+v_y^2=1$ with $v(x,0)=0$
$v_y^2=1-v_x^2$ with $v(x,0)=0$
$v_y=\pm\sqrt{1-v_x^2}$ with $v(x,0)=0$
$v_{xy}=\mp\dfrac{v_xv_{xx}}{\sqrt{1-v_x^2}}$ with $v(x,0)=0$
Let $w=v_x$ ,
Then $w_y=\mp\dfrac{ww_x}{\sqrt{1-w^2}}$ with $w(x,0)=0$
$w_y\pm\dfrac{ww_x}{\sqrt{1-w^2}}=0$ with $w(x,0)=0$
Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$\dfrac{dy}{dt}=1$ , letting $y(0)=0$ , we have $y=t$
$\dfrac{dw}{dt}=0$ , letting $w(0)=w_0$ , we have $w=w_0$
$\dfrac{dx}{dt}=\pm\dfrac{w}{\sqrt{1-w^2}}=\pm\dfrac{w_0}{\sqrt{1-w_0^2}}$ , letting $x(0)=f(w_0)$ , we have $x=\pm\dfrac{w_0t}{\sqrt{1-w_0^2}}+f(w_0)=\pm\dfrac{wy}{\sqrt{1-w^2}}+f(w)$ , i.e. $w=F\left(x\mp\dfrac{wy}{\sqrt{1-w^2}}\right)$
$w(x,0)=0$ :
$F(x)=0$
$\therefore w=0$
$v_x=0$
$v(x,y)=g(y)$
$v_y=g_y(y)$
$\therefore(g_y(y))^2=1$
$g_y(y)=\pm1$
$g(y)=\pm y+C$
$\therefore v(x,y)=\pm y+C$
$v(x,0)=0$ :
$C=0$
$\therefore v(x,y)=\pm y$
Hence $u(x,y)=e^{\pm y}$
