Characteristics method applied to the PDE $u_x^2 + u_y^2=u$

I am trying to solve: $u_x^2 + u_y^2=u$ with boundary conditions: $u(x,0)=x^2$. Unfortunately it leads to equations that makes no sense (sum of squares is $0$ and all constants are $0$). I would be grateful for a reasonable explanation how to solve that with help of characteristics.

Solution 1:

I don't see how the method of characteristics (as I know) could be applied to an equation that involves the derivatives in a nonlinear way.

Since $u\ge 0$, we can look for $u$ in the form $u=v^2$ which leads to a simplification: by the chain rule, $$u_x^2+u_y^2 = 4v^2(v_x^2+v_y^2)$$ so in terms of $v$ the PDE becomes $$ v_x^2+v_y^2 = \frac14 \tag{1} $$ This is a form of the Eikonal equation, the solution of which is $1/2$ times the distance function from the set $\{v=0\}$.

However, the boundary condition is incompatible with the PDE. Indeed, it asserts that $v(x,0)=x$. But according to (1), $|\nabla v|=1/2$ which implies the Lipschitz bound $$|v(a)-v(b)|\le \frac12|a-b|\tag{2}$$ The Lipschitz bound, being a pointwise inequality, must extend to the boundary values; however, $v(x,0)=x$ does not satisfy (2).

Solution 2:

Let $u=v^2$ ,

Then $u_x=2vv_x$

$u_y=2vv_y$

$\therefore(2vv_x)^2+(2vv_y)^2=v^2$ with $v(x,0)=x$

$4v^2(v_x)^2+4v^2(v_y)^2=v^2$ with $v(x,0)=x$

$v_x^2+v_y^2=\dfrac{1}{4}$ with $v(x,0)=x$

$v_y^2=\dfrac{1}{4}-v_x^2$ with $v(x,0)=x$

$v_y=\pm\sqrt{\dfrac{1}{4}-v_x^2}$ with $v(x,0)=x$

$v_{xy}=\mp\dfrac{v_xv_{xx}}{\sqrt{\dfrac{1}{4}-v_x^2}}$ with $v(x,0)=x$

Let $w=v_x$ ,

Then $w_y=\mp\dfrac{ww_x}{\sqrt{\dfrac{1}{4}-w^2}}$ with $w(x,0)=1$

$w_y\pm\dfrac{ww_x}{\sqrt{\dfrac{1}{4}-w^2}}=0$ with $w(x,0)=1$

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 $w=w_0$

$\dfrac{dx}{dt}=\pm\dfrac{w}{\sqrt{\dfrac{1}{4}-w^2}}=\pm\dfrac{w_0}{\sqrt{\dfrac{1}{4}-w_0^2}}$ , letting $x(0)=f(w_0)$ , we have $x=\pm\dfrac{w_0t}{\sqrt{\dfrac{1}{4}-w_0^2}}+f(w_0)=\pm\dfrac{wy}{\sqrt{\dfrac{1}{4}-w^2}}+f(w)$ , i.e. $w=F\left(x\mp\dfrac{wy}{\sqrt{\dfrac{1}{4}-w^2}}\right)$

$w(x,0)=1$ :

$F(x)=1$

$\therefore w=1$

$v_x=1$

$v(x,y)=x+g(y)$

$v_y=g_y(y)$

$\therefore1^2+(g_y(y))^2=\dfrac{1}{4}$

$(g_y(y))^2=-\dfrac{3}{4}$

$g_y(y)=\pm\dfrac{i\sqrt3}{2}$

$g_y=\pm\dfrac{i\sqrt3y}{2}+C$

$\therefore v(x,y)=x\pm\dfrac{i\sqrt3y}{2}+C$

$v(x,0)=x$ :

$C=0$

$\therefore v(x,y)=x\pm\dfrac{i\sqrt3y}{2}$

Hence $u(x,y)=\left(x\pm\dfrac{i\sqrt3y}{2}\right)^2=x^2\pm i\sqrt3xy-\dfrac{3y^2}{4}$