Find by the method of characteristic, the integral surface of $$pq=xy$$ which passes through the curve $$z=x,y=0$$
By strip condition, there is a unique initial strip $$x_{0}=s,y_{0}=0,z_{0}=s,p_{0}=1,q_{0}=0$$
And the characteristic equations are
$$\frac{dx}{dt}=f_p,\\ \frac{dy}{dt}=f_q,
\\ \frac{dz}{dt}=pf_p+qf_q,
\\ \frac{dp}{dt}=-f_x-pf_z,
\\ \frac{dq}{dt}=-f_y-qf_z$$
where $f$ is given non linear pde.
Please help me to eliminate s and t find z in x & y.
Best Answer
$$\frac{\partial z}{\partial x}\frac{\partial z}{\partial y}=xy\quad;\quad z(x,0)=x$$
Preliminary comment :
The solution is very easy by separation of variables. We look for particular solution on the form $z(x,y)=f(x)g(y)$ $$(g\frac{df}{dx})(f\frac{dg}{dy})=xy\quad\implies\quad \begin{cases} fdf=xdx \quad f=\pm\sqrt{x^2+c_1}\\ gdg=ydy\quad g=\pm\sqrt{y^2+c_2} \end{cases}$$
$$z=\pm\sqrt{(x^2+c_1)(y^2+c_2)}$$ With condition $z(x,0)=x=\pm\sqrt{(x^2+c_1)c_2} \implies c_1=0$ ; $c_2=1$ ; sign of $x$. $$\boxed{z(x,y)=x\sqrt{y^2+1}}$$
Solving with method of characteristics : $$\frac{dx}{q}=\frac{dy}{p}=\frac{dz}{2pq}=\frac{dp}{y}=\frac{dq}{x}=dt $$ $$x_{0}=s,y_{0}=0,z_{0}=s,p_{0}=1,q_{0}=0$$ Solving the system leads to : $$\begin{cases} x=Ae^{t}+Be^{-t} \\ y=Ce^{t}+De^{-t} \\ p=Ce^{2t}-De^{2t}\\ q=Ae^{2t}-Be^{2t}\\ z=ACe^{2t}-BDe^{2t}+E \end{cases}$$ $pq-xy=0 \implies AB+BC=0$
With boundary condition : $p_0=1$ and $q_0=0$
$A+B=s\quad;\quad C+D=0\quad;\quad AC-BD+E=s\quad;\quad C-D=1\quad;\quad A-B=0\quad;\quad AD+BC=0$.
$A=B=E=\frac{s}{2}$ and $C=-D=\frac12$ $\quad\implies\quad$ $\begin{cases} x=s \cosh(t) \\ y= \sinh(t) \\ z=\frac{s}{2}(\cosh(2t)+1)=s \cosh^2(t) \end{cases}$
$x^2-s^2y^2=s^2\quad\implies\quad s^2=\frac{x^2}{y^2+1}$
$z=\frac{x^2}{s}=x^2\left(\frac{+\sqrt{y^2+1}}{x}\right)$ . The sign $+$ is determined according to $z(x,0)=x$ . $$z=x\sqrt{y^2+1}$$