I have the following PDE
$$xu_x+yu_y=0$$
for which I get the characteristic function $$y=cx$$ along which the u(x,y) is constant. The general solution is $$u(x,y)=f(\frac{y}{x})$$.
I understand that the characteristic functions is a fan of lines in the x,y plane going through (0,0) point. It is unclear to me how the general solution could be presented.
What is the geometric solution/interpretation to the PDE in the x,y,u dimensions?
$$xu_x+yu_y=0$$ FIRST PART, Solving with the method of characteristics :
From $\quad xu_x+yu_y=0 u \quad$ the set of characteristic equations is : $$\frac{dx}{x}=\frac{dy}{y}=\frac{du}{0}$$ because the coefficient of $u_x$ is $x$ , the coefficient of $u_y$ is $y$ and the coefficient of $u$ is $0$ .
From $\frac{dx}{x}=\frac{dy}{y}$ the equation of the set of characteristic curves is $\ln|y|-\ln|x|=$constant, or : $$\frac{y}{x}=c_1$$ To be finite $\frac{du}{0}$ implies $u=$constant. So, the equation of the set of characteristic curves is : $$u=c_2$$ The general solution of the PDE expressed on implicit form is : $$F\left(\frac{y}{x} \:,\: u\right)=0$$ where $F$ is any differentiable function of two variables.
Solving this implicit equation for the second variable leads to the explicit form : $$u=f\left(\frac{y}{x}\right)$$ where $f$ is any differentiable function.
Note : Since $f$ is any function, the solution includes $u=\phi(\tan^{-1}\left(\frac{y}{x}\right))$ where $\phi$ is any function. This is equivalent to $u=\phi(\theta)$ where $\theta=\tan^{-1}\left(\frac{y}{x}\right)$ and with the function $g\equiv f(\tan^{-1})$.
SECOND PART
Answer to the question of graphical interpretation :
Your graph represents the set of characteristic curves $$y=c_1x$$ drawn with various values of $c_1$
On one characteristic curve corresponding to $c_1$ : $$u=f\left(\frac{y}{x}\right)=f(c_1)$$ since $c_1$ is a constant, $f(c_1)$ is constant, so $u$ is constant. $$u=f(c_1)=c_2$$
Thus $u(x,y)$ is constant all along the characteristic curve considered.
But $u(x,y)$ isn't constant if the point $(x,y)$ goes from one curve $(c_1)$ to another curve $(c'_1)$, because $u(x,y)$ varies from $f(c_1)=c_2$ to $f(c'_1)=c'_2$ which are different.
If you draw in 3D the function of two variables $\frac{y}{x}$ you obtain the representation of a particular solution of the PDE : $u(x,y)=f\left(\frac{y}{x}\right)=\frac{y}{x}$ in the case of $f(X)=X$. Of course, this isn't a representation of the characteristic curves.
First of all, the "general solution" $u(x,y)=f(\frac{y}{x})$ is not general. All functions of this form have symmetry $u(-x,-y) = u(x,y)$, which need not be satisfied by solutions of the PDE: for example, $u(x,y)=y/\sqrt{x^2+y^2}$ is a solution that is not of the above form.
The correct form of general solution is $$u(x,y) = f(\theta)$$ where $\theta$ is the polar angle and $f$ is a differentiable $2\pi$-periodic function.
The characteristic curves of the PDE are not lines; they are half-lines. The geometry of the solution is that it's constant on each such line, so it depends on the polar angle only.
It has been shown here by several people that the solutions $(x,y)\mapsto u(x,y)$ have to be constant on rays emanating from $(0,0)$. It follows that either $u$ is globally constant, or undefined at $(0,0)$. This means that the solutions are of one of the following forms: Either $$ u(x,y)\equiv c \in{\mathbb R}\qquad\bigl((x,y)\in{\mathbb R}^2\bigr)\ ,$$ or there is a differentiable $2\pi$-periodic function $f$ such that $$u(x,y)=f\bigl({\rm arg}(x,y)\bigr)\qquad\bigl((x,y)\in\dot{\mathbb R}^2\bigr)\ .\tag{2}$$ It remains to prove that the functions of type $(2)$ are indeed solutions. To this end we write ${\rm arg}(x,y)=:\phi$ and compute $$u_x x+u_y y=f'(\phi)\phi_x\,x+f'(\phi)\phi_y\,y=f'(\phi)\left({-y\,x\over x^2+y^2}+{x\,y\over x^2+y^2}\right)\equiv0\ .$$ Some of the OP's difficulties stem from the fact that the variables $x$ and $y$ were not treated in an equal-rights way.
Rewrite it as
$$xu_x+yu_y=\left(x\frac{\partial}{\partial x} + y\frac{\partial}{\partial y}\right) u =0$$
Now recall that
$$r^2=x^2+y^2$$
so, taking the full derivative
$$rdr=xdx+ydy$$
Dualising this
$$r\frac{\partial}{\partial r}=x\frac{\partial}{\partial x}+y\frac{\partial}{\partial y}$$
So, our equation is
$$r\frac{\partial}{\partial r} u = 0 \Rightarrow \frac{\partial}{\partial r} u=0$$
That is, $u$ is constant when moving along $r$. Introducing polar coordinates we see that a solution depends only on the angle $\theta$.
