Transport equation with variable coefficients using characteristics I want to solve the following pde:
$$x\partial_xu(x,y,z)+y\partial_y(x,y,z)+\partial_zu(x,y,z)=0,(x,y,z)\in \mathbb R^3$$
$$u(x,y,0)=u_0(x,y),(x,y)\in\mathbb R^2$$
using characteristics.
Until now I only know how to solve the (in)homogenous Transport equation with constant coefficients. In that case I considered the path
$\gamma(s)=u(x+sb,t+s)$, where $b\in \mathbb R^n$ is the coefficient-vector.
I know that any solution must be constant on this path, so using the initial value I knew the values of the solution.
How can I do something similar for the above pde?
 A: The equation of the characteristics is
$$
\frac{dx}{x}=\frac{dy}{y}=\frac{dz}{1}=\frac{du}{0}.
$$
This means that $u$ is constant on the curves solution of the first three equations. The solution is
$$
x=C_1\,e^{z},\quad y=C_2\,e^z,\quad z=z.
$$
The solution of the equation is then
$$
u(x,y,z)=u_0(x\,e^{-z},y\,e^{-z}).
$$
A: $$xu_x+yu_y+u_z=0$$
With the method of characteristics :
$$\frac{dx}{x}=\frac{dy}{y}=\frac{dz}{1}=\frac{du}{0}$$
First characteristic equation , from $\frac{dx}{x}=\frac{dz}{1} \quad\to\quad x\:e^{-z}=c_1$
Second characteristic equation , from $\frac{dy}{y}=\frac{dz}{1} \quad\to\quad y\:e^{-z}=c_2$
By necessity $du=0 \quad\to\quad u=c_3$ as a third trivial caracteristic.
The general solution on implicit form is : $\Phi\left(x\:e^{-z}\:,\:y\:e^{-z}\:,\:u\right)=0$
Or on the explicit form :
$$u=F\left(x\:e^{-z}\:,\:y\:e^{-z}\right)$$
where $F$ is any differentiable function of two variables.
Boundary condition :
$u(x,y,0)=F\left(xe^0\:,\:ye^0\right)=u_0(x,y)\quad\to\quad F\equiv u_0$
$$u(x,y,z)=u_0\left(x\:e^{-z}\:,\:y\:e^{-z}\right)$$
