Hi I have gotten a hard time solving this problem.
$$ \begin{array}{c} f(x,y) \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y} = 1 \\\\ f(s,s) = \frac{s}{2} \quad , 0<s<1 \end{array} $$ Looking for steps for this since I am trying to learn
$$f(x,y) \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y} = 1$$ The system of characteristic ODEs is : $$\begin{cases} \frac{dx}{dt}=f\\ \frac{dy}{dt}=1\\ \frac{df}{dt}=1\\ \end{cases}\quad\iff\quad \frac{dx}{f}=\frac{dy}{1}=\frac{df}{1}=dt$$ A first characteristic equation comes from $\frac{dx}{f}=\frac{df}{1}$ $$f^2-2x=c_1$$ A second characteristic equation comes from $\frac{dy}{1}=\frac{df}{1}$ $$f-y=c_2$$ General solution of the PDE on the form of implicit equation $c_1=\Phi(c_2)$ : $$f^2-2x=\Phi(f-y)$$ $\Phi$ is an arbitrary function to be determined according to the specified condiion $f(s,s)=\frac{s}{2}$
$\left(\frac{s}{2}\right)^2-2s=\Phi(\frac{s}{2}-s)$
Let $X=-\frac{s}{2}\quad\implies\quad s=-2X$
$X^2+4X=\Phi(X)$
Now the function $\Phi$ is determined : $$\Phi(X)=X^2+4X$$ We put it into the above general solution where $X=f-y$ : $$f^2-2x=(f-y)^2+4(f-y)$$ $$\boxed{f(x,y)=\frac{y-2}{2}+\frac{x-2}{y-2}}$$ This solution isn't valid for any $x,y$ since the condition $f(s,s)=\frac{s}{2}$ is limited to $0<s<1$
$-\frac12<X<0$
$-\frac12<f-y<0$
$-\frac12<\frac{-y-2}{2}+\frac{x-2}{y-2}<0$
$$\frac12<-\frac{y}{2}+\frac{x-2}{y-2}<1$$ From this one can find the range of $(x,y)$ where the above solution is valid with respect to the specified condition.
