I need help with a differential equation, the trouble is I don't think it's separable and I have tried and failed to apply the method of characteristics to figure it out. z is also bound between zero and one.
$$ \frac{\partial u}{\partial x}+z\frac{\partial u}{\partial y}-Cz\frac{\partial u}{\partial z}=0 $$
The problem with characteristics is I couldn't figure out how to apply the boundary conditions.
$$ u(0,y,z)=\delta(y)\delta(z-1) $$
Any and all help would be greatly appreciated.
Parametrized by $x$, characteristic curves have equation $$x=x,\quad y = y_0+\frac{z_0}{C}(1-e^{-Cx}),\quad z = z_0 e^{-Cx} \tag{1}$$ where $(0,y_0,z_0)$ is the point of crossing the plane $x=0$. To check this, differentiate (1) with respect to $x$: $$x'=1,\quad y' =z_0 e^{-Cx} = z ,\quad z' = -C z_0 e^{-Cx} = -Cz \tag{2}$$ The PDE says that $u$ is constant along each curve with equation (1). This implies that $u=0$ outside of the characteristic curve that begins at $y_0=0$, $z_0=1$, which is $$x=x,\quad y = \frac{1}{C}(1-e^{-Cx}),\quad z = e^{-Cx} \tag{3}$$ Along the curve (3), $u$ is something quite singular.
I am inclined to suspect that this PDE+BC problem does not capture the phenomenon you were trying to describe.
Assume $C\neq0$ for the key case:
Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$\dfrac{dx}{dt}=1$ , letting $x(0)=0$ , we have $x=t$
$\dfrac{dz}{dt}=-Cz$ , letting $z(0)=z_0$ , we have $z=z_0e^{-Ct}=z_0e^{-Cx}$
$\dfrac{dy}{dt}=z=z_0e^{-Ct}$ , letting $y(0)=y_0$ , we have $y=y_0+\dfrac{z_0}{C}-\dfrac{z_0e^{-Ct}}{C}=y_0+\dfrac{z_0}{C}-\dfrac{z_0e^{-Cx}}{C}$
$\dfrac{du}{dt}=0$ , letting $u(0)=f(y_0,z_0)$ , we have $u(x,y,z)=f(y_0,z_0)=f\left(y+\dfrac{z}{C}-\dfrac{ze^{Cx}}{C},ze^{Cx}\right)$
$u(0,y,z)=\delta(y)\delta(z-1)$ :
$f(y,z)=\delta(y)\delta(z-1)$
$\therefore u(x,y,z)=\delta\left(y+\dfrac{z}{C}-\dfrac{ze^{Cx}}{C}\right)\delta(ze^{Cx}-1)$
