Solving the advection equation with artificial damping I'm having trouble with the following system of partial differential equations:
$$
\frac{\partial u_1}{\partial t} + a \frac{\partial u_1}{\partial x}=u_2(x,t)\;, \\
\frac{\partial u_2}{\partial t} +  b u_2(x,t) = ba \frac{\partial u_1}{\partial x}\;, 
$$
where $a$ and $b$ are real positive constants. 
If $b=0$, this reduces to the advection equation, so with the initial condition $u_1(x,0)=f(x)$ the solution is a right-running wave $u_1(x,t)=f(x-at)$. 
This (simplified) system may be used in a finite-difference model with $b>0$ to artificially damp such a wave over time to simulate an infinite medium.  
However, I noticed that with $b>0$, $u_1(x,0)=f(x)$ and $u_2(x,0)=0$ the solution for $u_1$ does not depend on time and becomes $u_1(x,t)\simeq f(x)$. 


*

*Would anyone be able to explain this?

*More specifically, I would like to find the initial condition $u_2(x,0)=g(f(x))$ so that $u_1(x,t)=0$ as $t\rightarrow\infty$.


Thanks!
 A: $$\begin{cases}
\frac{\partial u_1}{\partial t} + a \frac{\partial u_1}{\partial x}=u_2(x,t) \\
\frac{\partial u_2}{\partial t} +  b u_2(x,t) = ba \frac{\partial u_1}{\partial x} 
\end{cases} \quad\to\quad 
b\frac{\partial u_1}{\partial t}+\frac{\partial u_2}{\partial t}=0$$
Integrating with respect to $t$ leads to : $$bu_1(x,t)+u_2(x,t)=\phi(x)$$
where $\: \phi \:$ is any function.
If the initial conditions are 
$\begin{cases}u_1(x,0)=f(x)\\u_2(x,0)=g(x)\end{cases}\quad\to\quad \phi(x)=bf(x)+g(x)$
$$u_1(x,t)=f(x)+\frac{g(x)-u_2(x,t)}{b}$$
CONDITION at $t=\infty$ :
$u_1(x,\infty)=0 \quad\to\quad 0=f(x)+\frac{g(x)-u_2(x,\infty)}{b} \quad\to\quad u_2(x,\infty)=g(x)+bf(x)$
$u_1(x,\infty)=0 \quad\implies\quad \left(\frac{\partial u_1}{\partial t}\right)_{t=\infty}=0\quad$ and $\quad\left(\frac{\partial u_1}{\partial x}\right)_{t=\infty}=0$
$\left(\frac{\partial u_1}{\partial t}\right)_{t=\infty} + a \left(\frac{\partial u_1}{\partial x}\right)_{t=\infty}=u_2(x,\infty)=0$
$u_2(x,\infty)=0=g(x)+bf(x)$
Thus, the initial condition is very simple : $g(x)=u_2(x,0)=-bf(x)$
As a consequence  $\quad u_2(x,t)=-bu_1(x,t)$ 
The first PDE becomes : $\quad \frac{\partial u_1}{\partial t} + a \frac{\partial u_1}{\partial x}=-bu_1(x,t)$
Solving this PDE with the method of characteristics leads to the result:
$$u_1(x,t)=e^{-bt}f(x-at)$$
$$u_2(x,t)=-be^{-bt}f(x-at)$$
according to the initial conditions $u_1(x,0)=f(x)$ and $u_2(x,0)=-bf(x)$
A: For arbitrary differentiable functions $f$ and $g$ you have solutions
$$ \eqalign{u_1(x,t) &= f(x) + \exp(-bx/a) g(x-at)\cr
            u_2(x,t) &= a f'(x) - b \exp(-bx/a) g(x-at)\cr} $$
For arbitrary continuous initial conditions $u_1(x,0)$, $u_2(x,0)$ you can take 
$$g(x) = \exp(bx/a) (-f(x) + u_1(x,0))$$
where $f(x)$ is a solution of the ODE
$$ a f'(x) + b f(x) = u_2(x,0) + b u_1(x,0) $$
namely
$$ f(x) = \frac{\exp(-bx/a)}{a} \left( \int (u_2(x,0)+b u_1(x,0)) \exp(bx/a)\; dx + c\right) $$
