I came across a multiple select question:
Q. If $u(x, y)$ satisfy the PDE $$u_x u_y=x y,~~~~~~~~~~~~~~~~~~(*)$$
$ u=x$, along $y=0$. Then, $u(x, y)$ can be of the form
(a) $x^2\left(y^2+1\right)$
(b) $x^2 y^2$
(c) $x^2\left(y^2-1\right)$
(d) $x\sqrt{y^2+1}$
Despite of the easy selection (d), let me try to solve it by method of characteristics. Consider the characteristic system
$$\frac{dx}{dt}=q(t)=u_y(t), $$ $$\frac{dy}{dt}=p(t)=u_x(t),$$ $$\frac{du}{dt}=2p(t)q(t),$$ $$\frac{dp}{dt}=y(t),$$ $$\frac{dq}{dt}=x(t),$$ to get, $$x(t)=Ae^t+Be^{-t},$$ $$ y(t)=Ce^t+De^{-t},$$ $$u(t)=(ACe^{2t}-BDe^{-2t})+(AD+BC)t+E,$$ $p(t)=Ce^t-De^{-t},$ and $q(t)=Ae^t-Be^{-t}., $ where $A,B,C,D,E,F$ and $G$ are arbitrary constants.
Now, apply the initial curve (described by $x(s,0)=s,~y(s,0)=0,~u(s,0)=x(s,0)=s$) to obtain,
$$A+B=s,$$ $$C+D=0,$$ $$AC-BD+E=s,$$ and I hope, taking $p(s,0)=u_x(s,0)=1$ and $q(s,0)=0$ (from $(*)$) is the way to say, $$C-D=1,~\text{and},$$ $$A-B=0.$$ Therefore, we have $A=B=E=s/2,~C=-D=1/2$. But, I am unable to proceed now.
Could you please point out any errors in the deduction so far and suggest the next step to reach a possibly unique solution (I am hesitant about uniqueness in non-linear Cauchy problems as well!) that is provided in option (d)? I would be grateful for your any sort of help.