Solve PDE with condition

Question

\begin{equation} \begin{cases} \sqrt{x}u_x - \sqrt{y}u_y = u^2 \\ u_{x=y} = \phi (y) \end{cases} \end{equation} I want to change u(x,y) to v$(\tau, s )$ : \begin{equation} \begin{cases} \frac{dx}{ds} = \sqrt{x} , x(s=0) = \tau \implies x = (\frac{s+c}{2})^2\tau^2 \\ \frac{dy}{ds} = -\sqrt{y} , y(s=0) = \tau \implies y = (\frac{-s+c}{2})^2\tau^2 \\ \tau = \frac{\sqrt{x} + \sqrt{y}}{c} \\ s^2 = 2\frac{x+y}{\tau^2} - c^2 = 2\frac{c^2(x+y)}{(\sqrt{x} + \sqrt{y})^2} - c^2 = (c\frac{\sqrt{x}-\sqrt{y}}{\sqrt{x} + \sqrt{y}})^2 \\ s = c\frac{|\sqrt{x}-\sqrt{y}|}{\sqrt{x} + \sqrt{y}} \end{cases} \end{equation} So we change the condition to : \begin{equation} \begin{cases} \frac{dv}{ds} = v^2 \\ v_{s=0} = \phi(\tau^2) \\ \end{cases} \end{equation} We find the solution : \begin{equation} \begin{cases} v = \frac{-1}{s+c} \\ v_{s=0} = \phi(\tau^2) \\ \implies v = \frac{-\phi(\tau^2)}{s\phi(\tau^2)+c} \\ Check : \frac{dv}{ds} = \frac{\phi^2(\tau^2)}{(s\phi(\tau^2)+c)^2} = v^2 \\ \end{cases} \end{equation} Change $v(\tau, s)$ to $u(x,y)$ : \begin{equation} u(x,y) = \frac{-\phi( \frac{(\sqrt{x} + \sqrt{y})^2}{c^2})}{c\frac{|\sqrt{x}-\sqrt{y}|}{\sqrt{x} + \sqrt{y}}\phi( \frac{(\sqrt{x} + \sqrt{y})^2}{c^2}) + c} = \frac{-\phi( \frac{(\sqrt{x} + \sqrt{y})^2}{c^2})}{\frac{|\sqrt{x}-\sqrt{y}|}{\sqrt{x} + \sqrt{y}}\phi( \frac{(\sqrt{x} + \sqrt{y})^2}{c^2}) + 1} \\ \end{equation} Choose simple case c = 1 \begin{equation} \begin{cases} \frac{du}{dx} = -\frac{(\sqrt{x} + \sqrt{y})sgn(\sqrt{x}-\sqrt{y}) - |\sqrt{x} - \sqrt{y}|}{2\sqrt{x}(\sqrt{x} + \sqrt{y})^2(\frac{|\sqrt{x}-\sqrt{y}|}{\sqrt{x} + \sqrt{y}}\phi( \frac{(\sqrt{x} + \sqrt{y})^2}{c^2}) + 1)^2}\phi^2(\frac{(\sqrt{x}+\sqrt{y})^2}{c^2}) \\ \frac{du}{dy} = -\frac{-(\sqrt{x} + \sqrt{y})sgn(\sqrt{x}-\sqrt{y}) - |\sqrt{x} - \sqrt{y}|}{2\sqrt{y}(\sqrt{x} + \sqrt{y})^2(\frac{|\sqrt{x}-\sqrt{y}|}{\sqrt{x} + \sqrt{y}}\phi( \frac{(\sqrt{x} + \sqrt{y})^2}{c^2}) + 1)^2}\phi^2(\frac{(\sqrt{x}+\sqrt{y})^2}{c^2}) \\ \sqrt{x}\frac{du}{dx} - \sqrt{y}\frac{du}{dy} = -\frac{sgn(\sqrt{x}-\sqrt{y}) }{(\sqrt{x} + \sqrt{y})(\frac{|\sqrt{x}-\sqrt{y}|}{\sqrt{x} + \sqrt{y}}\phi( \frac{(\sqrt{x} + \sqrt{y})^2}{c^2}) + 1)^2}\phi^2(\frac{(\sqrt{x}+\sqrt{y})^2}{c^2}) \neq u^2 \end{cases} \end{equation} Where did i do wrong ?

Answer 1

I'll present a solution similar to yours, then you can compare them to find your mistakes.

The equations for the characteristic curves are given by $$ \frac{dx}{ds}=\sqrt{x},\qquad \frac{dy}{ds}=-\sqrt{y}, \qquad \frac{du}{ds}=u^2. \tag{1} $$ Solving them, we find \begin{align} \sqrt{x}&=\sqrt{x_0}+\frac{s}{2}, \tag{2a} \\ \sqrt{y}&=\sqrt{y_0}-\frac{s}{2}, \tag{2b} \\ u&=\frac{u_0}{1-u_0s}. \tag{2c} \end{align} The condition $u(y,y)=\phi(y)$ is equivalent to $$ x(0)=x_0=y_0, \qquad y(0)=y_0, \qquad u(0)=u_0=\phi(y_0), \tag{3} $$ so Eqs. $(2)$ can be rewritten as \begin{align} \sqrt{x}&=\sqrt{y_0}+\frac{s}{2}, \tag{4a} \\ \sqrt{y}&=\sqrt{y_0}-\frac{s}{2}, \tag{4b} \\ u&=\frac{\phi(y_0)}{1-\phi(y_0)s}. \tag{4c} \end{align} Now we solve $(4\text{a})$ and $(4\text{b})$ for $s$ and $y_0$ in order to express them as functions of $x$ and $y$. The result is $$ y_0=\left(\frac{\sqrt{x}+\sqrt{y}}{2}\right)^2, \qquad s=\sqrt{x}-\sqrt{y}. \tag{5} $$ Plugging $(5)$ into $(4\text{c})$ we finally obtain $$ u(x,y)=\frac{\phi\!\left(\left(\frac{\sqrt{x}+\sqrt{y}}{2}\right)^2\right)} {1-(\sqrt{x}-\sqrt{y})\,\phi\!\left(\left(\frac{\sqrt{x}+\sqrt{y}}{2}\right)^2\right)}. \tag{6} $$