Solving a first order nonlinear PDE I have to solve this equation :
$$
\frac{\partial u}{\partial x} \cdot \frac{\partial u}{\partial y} = xy
$$
with initial condition $u(x,0) = x$.
I know it is easy with separation of variables, but I need to do it with the method of characteristics and I can't figure it out.
 A: Let $\begin{cases}p=x^2\\q=y^2\end{cases}$ ,
Then $\dfrac{\partial u}{\partial x}=\dfrac{\partial u}{\partial p}\dfrac{\partial p}{\partial x}+\dfrac{\partial u}{\partial q}\dfrac{\partial q}{\partial x}=2x\dfrac{\partial u}{\partial p}$
$\dfrac{\partial u}{\partial y}=\dfrac{\partial u}{\partial p}\dfrac{\partial p}{\partial y}+\dfrac{\partial u}{\partial q}\dfrac{\partial q}{\partial y}=2y\dfrac{\partial u}{\partial q}$
$\therefore2xu_p2yu_q=xy$ with $u(p,0)=\pm\sqrt p$
$u_pu_q=\dfrac{1}{4}$ with $u(p,0)=\pm\sqrt p$
$u_q-\dfrac{1}{4u_p}=0$ with $u(p,0)=\pm\sqrt p$
$u_{pq}+\dfrac{u_{pp}}{4u_p^2}=0$ with $u(p,0)=\pm\sqrt p$
Let $v=u_p$ ,
Then $v_q+\dfrac{v_p}{4v^2}=0$ with $v(p,0)=\pm\dfrac{1}{2\sqrt p}$
Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$\dfrac{dq}{dt}=1$ , letting $q(0)=0$ , we have $q=t$
$\dfrac{dv}{dt}=0$ , letting $v(0)=v_0$ , we have $v=v_0$
$\dfrac{dp}{dt}=\dfrac{1}{4v^2}=\dfrac{1}{4v_0^2}$ , letting $p(0)=f(v_0)$ , we have $p=\dfrac{t}{4v_0^2}+f(v_0)=\dfrac{q}{4v^2}+f(v)$ , i.e. $v=F\left(p-\dfrac{q}{4v^2}\right)$
$v(p,0)=\pm\dfrac{1}{2\sqrt p}$ :
$F(p)=\pm\dfrac{1}{2\sqrt p}$
$\therefore v=\pm\dfrac{1}{2\sqrt{p-\dfrac{q}{4v^2}}}$
$v^2=\dfrac{1}{4\left(p-\dfrac{q}{4v^2}\right)}$
$v^2=\dfrac{v^2}{4pv^2-q}$
$4pv^4-qv^2=v^2$
$4pv^4-(q+1)v^2=0$
$(4pv^2-(q+1))v^2=0$
$4pv^2-(q+1)=0$ or $v^2=0$ (reject)
$v^2=\dfrac{q+1}{4p}$
$v=\pm\dfrac{\sqrt{q+1}}{2\sqrt p}$
$u_p=\pm\dfrac{\sqrt{q+1}}{2\sqrt p}$
$u(p,q)=\pm\sqrt{p(q+1)}+g(q)$
$u_q=\pm\dfrac{\sqrt p}{2\sqrt{q+1}}+g_q(q)$
$\therefore\left(\pm\dfrac{\sqrt{q+1}}{2\sqrt p}\right)\left(\pm\dfrac{\sqrt p}{2\sqrt{q+1}}+g_q(q)\right)=\dfrac{1}{4}$
$\dfrac{1}{4}\pm\dfrac{\sqrt{q+1}}{2\sqrt p}g_q(q)=\dfrac{1}{4}$
$g_q(q)=0$
$g(q)=C$
$\therefore u(p,q)=\pm\sqrt{p(q+1)}+C$
$u(p,0)=\pm\sqrt p$ :
$C=0$
$\therefore u(p,q)=\pm\sqrt{p(q+1)}$
$u(x,y)=x\sqrt{y^2+1}$
