Cauchy problem for nonlinear first order hyperbolic PDE with source via method of characteristics. (Work and characteristic plots included) Working on the following problem.
$y u_x -u u_y = x \\ u(x,x)=-2x$
I've went after this with the method of characteristics. I'm using $(t,s$) as my parametrization variables. In parameterize space I find the following solution. 
 
So now my issue. I want to take it back to $(x,y)$ coordinates, i.e solve the cauchy problem. . However it seems pretty complicated to solve for s as a function of x,y, and similarly, solve t as a function of x,y. I graphed the characteristics since things seem sort of strange to me.( I set a specific value of s, then let t vary) to plot it on the x,y plane.

Looking at this picture gives me two insights I believe. The first is that the characteristic curves appear to not interest the initial data curve- but is tangent to it at one point, which seems to violate method of characteristics. Secondly, the characteristics are crossing. This makes me think that shocks are occurring in this solution quite a lot. So I think that this problem actually doesn't have a classical analytic solution. Moreover it seems like it may have a weak solution. 
I'm seeing if someone can give me tips on this problem. Since the structure of x,y, and u are such in $(t,s)$ space, it makes me sort of want to believe that a solution does exist since they are all so similar to eachother. However with the characteristic plot, it has me questioning a lot of things. 
Thanks for the tips. 
 A: I think that your approach is correct. Below an equivalent approach with slightly different notations for simplicity.
$$yu_x-uu_y=x$$
$$\frac{dx}{y}=\frac{dy}{-u}=\frac{du}{x}=ds\quad\implies\quad\begin{cases}
\frac{du}{ds}=x\\
\frac{dx}{ds}=y\\
\frac{dy}{ds}=-u
\end{cases}$$
$$\frac{d^3u}{ds^3}=\frac{d^2x}{ds^2}=\frac{dy}{ds}=-u$$
Solving $\frac{d^3u}{ds^3}+u=0$ :
$$u=C_1e^{r_1s}+C_2e^{r_2s}+C_3e^{r_3s}$$
$r_1=-1\quad ;\quad r_2=\frac12(1+i\sqrt{3})\quad ; \quad r_3=\frac12(1-i\sqrt{3})\quad $ are the cubic roots of $-1$.
$C_1$ , $C_2$ , $C_3$ are constants.
$$x=\frac{du}{ds}=r_1C_1e^{r_1s}+r_2C_2e^{r_2s}+r_3C_3e^{r_3s}$$
$$y=\frac{dx}{ds}=r_1^2C_1e^{r_1s}+r_2^2C_2e^{r_2s}+r_3^2C_3e^{r_3s}$$
$$\left(\begin{matrix}
  e^{r_1s}  \\
  e^{r_2s}  \\
  e^{r_3s} 
 \end{matrix}\right)=
\left(\begin{matrix}
  C_1 & C_2 & C_3 \\
  r_1C_1 & r_2C_2 & r_3C_3 \\
  r_1^2C_1 & r_2^2C_2 & r_3^2C_3
 \end{matrix}\right)^{-1}
\left(\begin{matrix}
  u  \\
  x  \\
  y 
 \end{matrix}\right)$$
After carrying out and simplification :
$$\begin{cases}
e^{r_1s}=\frac{1}{3C_1}(u+y-x)\\
e^{r_2s}=\frac{1}{3C_2}(u-r_2y+r_3x)\\
e^{r_3s}=\frac{1}{3C_3}(u-r_3y+r_2x)\\
\end{cases}$$
$$e^{r_1s}e^{r_2s}e^{r_3s}=1=\frac{1}{3C_1}\frac{1}{3C_2}\frac{1}{3C_3}(u+y-x)(u-r_2y+r_3x)(u-r_3y+r_2x)$$
With $C=27C_1C_2C_3$
$$\boxed{u^3+3\,x\,y\,u+y^3-x^3=C}$$
$C$ is an arbitrary constant. This is a solution of the PDE on the form of implicit equation.
$u(x,y)$ can be explicitly expressed in solving the cubic equation for $u$.
Particular solutions can be derived in the case $C=0$
$$u^3+3\,x\,y\,u+y^3-x^3=(u+y-x)(u^2-y\,u+x\,u+x\,y+y^2+x^2)=0$$
$$u(x,y)=x-y\quad\text{is a particular real solution}$$
$$u(x,y)=\frac{-1+i\sqrt{3}}{2}x+\frac{1+i\sqrt{3}}{2}y\quad\text{is a particular complex solution}$$
$$u(x,y)=\frac{-1-i\sqrt{3}}{2}x+\frac{1-i\sqrt{3}}{2}y\quad\text{is a particular complex solution}$$
None of these particular solutions satisfy the condition $u(x,x)=-2x$. Even the more general solution $u^3+3\,x\,y\,u+y^3-x^3=C$ doesn't satisfy the condition.
If the problem is an academic exercise, solving should not be very arduous. There could be a typo in the condition which might be $u(-x,x)=-2x$ or $u(x,-x)=2x$. Or a typo in the PDE which might be $yu_x+uu_y=x$.
