$yu_x+xu_y=u$ with two conditions using method of characteristics

Question

$\begin{cases} yu_x+xu_y=u\\u(x,0) = x^3 \\ u(0,y)=y^3\end{cases}$

$\dot X(\sigma,s)=Y \quad X(\sigma,0)=\sigma \\ \dot Y(\sigma,s)=X \quad Y(0,s)=s \\ \dot U(\sigma,s)=U \quad U(\sigma,0)=\sigma^3\quad and \quad U(0,s)=s^3$

$\frac{dU}{ds}=U\implies U(\sigma,s) = Ce^s$ (1)

$\frac{dX}{ds}=Y\implies$ $\frac{d^2X}{ds^2}=\frac{dY}{ds}=X\implies \ddot X=X\implies$

$X(\sigma,s)=C_1 e^s+C_2 e^{-s}$ (2) and since $\dot X=Y\implies$

$Y(\sigma,s)=C_1 e^s-C_2 e^{-s}$ (3)

applying $X(\sigma,0)=\sigma$ has no issues to find $C_1+C_2=\sigma$, but what with $Y(0,s)=s$ or did I assume wrong initial conditions?

Answer 1

As I interpret your notation (since you're taking derivatives with respect to $s$), $\sigma$ is a parameter that tells you which characteristic curve you are on, and $s$ is a parameter which tells you where along that characteristic curve you are. But that's not at all how you have actually use these parameters when setting up your conditions (which don't make much sense, to be honest).

For those characteristic curves that pass through a point on the $x$-axis, say $(x,y)=(\sigma,0)$ where $u=x^3=\sigma^3$, you should have the conditions $$ X(\sigma,0)=\sigma ,\quad Y(\sigma,0)=0 ,\quad U(\sigma,0)=\sigma^3 , $$ while for characteristic curves through points on the $y$-axis, $(x,y)=(0,\sigma)$ where $u=y^3=\sigma^3$, it's $$ X(\sigma,0)=0 ,\quad Y(\sigma,0)=\sigma ,\quad U(\sigma,0)=\sigma^3 $$ instead.

Answer 2

You didn't assume wrong conditions. The difficulty comes from two not equivalent conditions leading to a piecewise solution. This might be more understandable with a different presentation (equivalent in fact).

$$yu_x+xu_y=u$$ Charpit-Lagrange characteristic ODEs : $$\frac{dx}{y}=\frac{dy}{x}=\frac{du}{u}=ds$$ A first characteristic equation comes from solving $\frac{dx}{y}=\frac{dy}{x}$ : $$x^2-y^2=c_1$$ A second characteristic equation comes from solving $\frac{du}{u}=\frac{dx+dy}{x+y}$ : $$\frac{u}{x+y}=c_2$$ The general solution of the PDE on implicit form $c_2=F(c_1)$ is : $$\frac{u}{x+y}=F(x^2-y^2)$$ $F$ is an arbitrary function (to be determined according to the boundary conditions). $$\boxed{u(x,y)=(x+y)F(x^2-y^2)}$$

First condition considered alone : $$u(x,0)=x^3=(x+0)F(x^2+0) \quad\implies\quad F(x^2)=x^2$$ Now the function $F$ is known $F(X)=X$ for $X\geq 0$

Second condition considered alone : $$u(0,y)=y^3=(0+y)F(0-y^2)\quad\implies\quad F(-y^2)=y^2$$ Now the function $F$ is known $F(X)=-X \quad\text{for}\quad X\leq 0$

With the two conditions we have : $$F(X)=\begin{cases} X\quad\text{if}\quad X\geq 0 \\ -X\quad\text{if}\quad X\leq 0 \end{cases} \quad\implies\quad F(X)=|X|$$

We put this function into the above general solution where $X=x^2-y^2$ : $$\boxed{u(x,y)=(x+y)\:\big|x^2-y^2\big|}$$