Solve a partial diferential equation $u_x^2+u_y^2=u^2$

Question

Solve the equation

$$u_x^2+u_y^2=u^2, \\ u(x,0)=1$$

Is there anybody who can help me?

Answer 1

Hint: Try to find a solution of the form $u(x,y)=f(x)g(y)$.

Answer 2

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 $u(x,y)=e^{\pm y}$