# 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?

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I added the tag "differential equations", I think it is more than appropriate here than just "linear-algebra". – Patrick Da Silva Mar 27 '12 at 0:26

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 $$u_1(x,t)=2u_0(x-t)-u_0(x+2t),\quad u_2(x,t)=u_0(x+2t)-u_0(x-t).$$