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 $u_1(x,t)\simeq f(x)$.

  1. Would anyone be able to explain this?
  2. 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 $t\rightarrow\infty$.


  • $\begingroup$ What I remember is that letting $v = \begin{pmatrix}u_1 \\ u_2\end{pmatrix}$ it is of the form $\partial_t v = A v+B \partial_x v$ for some matrices $A,B$. If you can diagonalize them simultaneously then it reduces to $\partial_t w = c w+d \partial_x w$ that you can solve with the Fourier series / transform $\endgroup$ – reuns May 25 '17 at 1:49

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 $$ f(x) = \frac{\exp(-bx/a)}{a} \left( \int (u_2(x,0)+b u_1(x,0)) \exp(bx/a)\; dx + c\right) $$


$$\begin{cases} \frac{\partial u_1}{\partial t} + a \frac{\partial u_1}{\partial x}=u_2(x,t) \\ \frac{\partial u_2}{\partial t} + b u_2(x,t) = ba \frac{\partial u_1}{\partial x} \end{cases} \quad\to\quad b\frac{\partial u_1}{\partial t}+\frac{\partial u_2}{\partial t}=0$$ 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 $\begin{cases}u_1(x,0)=f(x)\\u_2(x,0)=g(x)\end{cases}\quad\to\quad \phi(x)=bf(x)+g(x)$


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 $\quad\left(\frac{\partial u_1}{\partial x}\right)_{t=\infty}=0$

$\left(\frac{\partial u_1}{\partial t}\right)_{t=\infty} + a \left(\frac{\partial u_1}{\partial x}\right)_{t=\infty}=u_2(x,\infty)=0$


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 $u_2(x,0)=-bf(x)$


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