<s>[INST] Solving a linear system with an initial condition So there is a linear system
$
\frac{du}{dt} = - \left( \begin{array}{cc}
4 & -3  \\
6 & -5  \end{array} \right) \frac{du}{dx}
$
with an initial condition 
$
u(x, 0)=  \left( \begin{array}{c}
-tanh(\lambda x)  \\
0  \end{array} \right) 
$
Now to start this off I have manipulated the first equation to get this
$
\left( \begin{array}{ccc}
1 & \frac{-3}{4} & -\frac{1}{4} \frac{du_{1}}{dt}\\
0 & \frac{1}{12} & \frac{1}{6} \frac{du_{2}}{dt}-\frac{1}{4} \frac{du_{1}}{dt}\end{array} \right) \frac{du}{dx}
$
which can be rewritten as
$
\left( \begin{array}{ccc}
1 & \frac{-3}{4} & -\frac{1}{4} \frac{du_{1}}{dt}\\
0 & \frac{1}{12} & \frac{1}{6} \frac{du_{2}}{dt}-\frac{1}{4} \frac{du_{1}}{dt}\end{array} \right) \frac{du}{dx} =>\left( \begin{array}{cc}
1 & \frac{-3}{4}\\
0 & \frac{1}{12}\end{array} \right) \frac{du}{dx} = 
\left( \begin{array}{c}
-\frac{1}{4} \frac{du_{1}}{dt} \\
\frac{1}{6} \frac{du_{2}}{dt}-\frac{1}{4} \frac{du_{1}}{dt}  \end{array} \right) 
$
but I'm not sure how to deal with the derivatives. Would I use what I already have but just plug in the initial condition to $u_{1}$ and $u_{2}$ or am I using the wrong method?
 [/INST] If one writes $u=(u_1,u_2)$ and one diagonalizes the matrix $A$, one gets the simpler differential system $\frac{\partial v_1}{\partial t}=-\frac{\partial v_1}{\partial x}$ and $\frac{\partial v_2}{\partial t}=2\frac{\partial v_2}{\partial x}$, with $v_1=u_1+u_2$ and $v_2=u_1+2u_2$. 
For any $a$, the differential equation $\frac{\partial v}{\partial t}=a\frac{\partial v}{\partial x}$ means that $v$ is a function of $x+at$, hence that $v(x,t)=v(x+at,0)$. Here, $v_1(x,t)=v_1(x-t,0)$ and $v_2(x,t)=v_2(x+2t,0)$. 
Finally, $u_1=2v_1-v_2$ and  $u_2=v_2-v_1$ hence
$$
u_1(x,t)=2u_1(x-t,0)+2u_2(x-t,0)-u_1(x+2t,0)-2u_2(x+2t,0),
$$
and
$$
u_2(x,t)=-u_1(x-t,0)-u_2(x-t,0)+u_1(x+2t,0)+2u_2(x+2t,0).
$$
In the special case when $u_1(x,0)=u_0(x)$ and $u_2(x,0)=0$, one gets
$$
u_1(x,t)=2u_0(x-t)-u_0(x+2t),\quad
u_2(x,t)=u_0(x+2t)-u_0(x-t).
$$
</s>