First Order System Continued

June 14, 2019

Rewrite the dependent variables $T_1,\: T_2$ into $x,\: y$ , the system of equations from last lecture is

$\begin{align} x'&=-2x+2y\\ y'&=2x-5y \end{align}\tag{1}$

And the final solution we got

$\begin{align} x&=c_1e^{-6t}+c_2e^{-t}\\ y&=-2c_1e^{-6t}+\frac{1}{2}c_2e^{-t} \end{align}\tag{2}$

Let’s rewrite the system (1) into matrices

$\begin{bmatrix} x\\ y \end{bmatrix}'=\begin{bmatrix} -2 & 2\\ 2 & -5 \end{bmatrix}\begin{bmatrix} x\\ y \end{bmatrix}\tag{3}$

Here $x',y'$ are taken w.r.t to independent variable $t$ . And put the (2) into matrices again, but we won’t exploit the fact that $c_1,\: c_2$ can be columns:

$\begin{bmatrix} x\\ y \end{bmatrix} =c_1\begin{bmatrix} 1 \\ -2 \end{bmatrix}e^{-6t}+c_2\begin{bmatrix} 1 \\ \frac{1}{2} \end{bmatrix}e^{-t}\tag{4}$

We will utilize the matrices to solve differential equations. And similar to before, trial solutions: $e^{rt}$ , but we will do it in a way that resembles the solution in (4).

$\begin{align} \begin{bmatrix} x\\ y \end{bmatrix}&=\begin{bmatrix} a_1\\ a_2 \end{bmatrix}e^{\lambda t}\\ \begin{bmatrix} x\\ y \end{bmatrix}'&=\lambda\begin{bmatrix} a_1\\ a_2 \end{bmatrix}e^{\lambda t} \end{align}\tag{5}$

Now we’ve got these two equations and we substitute them into (3):

$\begin{align} \lambda\begin{bmatrix} a_1\\ a_2 \end{bmatrix}e^{\lambda t}&=\begin{bmatrix} -2 & 2\\ 2 & -5 \end{bmatrix}\begin{bmatrix} a_1\\ a_2 \end{bmatrix}e^{\lambda t}\\ \\ \lambda\begin{bmatrix} a_1\\ a_2 \end{bmatrix}&=\begin{bmatrix} -2 & 2\\ 2 & -5 \end{bmatrix}\begin{bmatrix} a_1\\ a_2 \end{bmatrix}\\ \\ \mathbf 0&=\begin{bmatrix} -2-\lambda & 2\\ 2 & -5-\lambda \end{bmatrix}\begin{bmatrix} a_1\\ a_2 \end{bmatrix} \end{align}$

This can look quite similar to you if you remember eigenvalues, if you don’t, no problem. The only way to get non-trivial solutions, that is $a_1\neq 0,\:a_2\neq0$ , is the matrix on the left is singular. That is, the determinant is zero

$\begin{align} \begin{vmatrix} -2-\lambda & 2\\ 2 & -5-\lambda \end{vmatrix}&=0\\ (-2-\lambda)(-5-\lambda)-4&=0\\ \lambda^2+7\lambda+6&=0\\ \lambda_1=-6,\lambda_2=-1 \end{align}$

The equation $\lambda^2+7\lambda+6=0$ is also called the characteristic equation. Then after finding the $\lambda$ , for corresponding $\lambda_1$ and $\lambda_2$ , we need to find the corresponding eigenvectors (basis) $\mathbf x_1=\begin{bmatrix} a_1\\ a_2 \end{bmatrix}$ and another $\mathbf x_2$ with different $a_1,a_2$ (they have nothing to do with the dependent variables). So

$\begin{bmatrix} 4 & 2\\ 2&1 \end{bmatrix}\begin{bmatrix} a_1\\ a_2 \end{bmatrix}=\mathbf0\\ \Rightarrow\\ \lambda_1=-6\sim\mathbf x_1=\begin{bmatrix} 1\\ -2 \end{bmatrix}$

And:

$\begin{bmatrix} -1 & 2\\ 2&-4 \end{bmatrix}\begin{bmatrix} a_1\\ a_2 \end{bmatrix}=\mathbf0\\ \Rightarrow\\ \lambda_2=-1\sim\mathbf x_2=\begin{bmatrix} 2\\ 1 \end{bmatrix}$

And we have a similar solution as (4):

$\begin{bmatrix} x\\ y \end{bmatrix} =c_1\begin{bmatrix} 1 \\ -2 \end{bmatrix}e^{-6t}+\tilde c_2\begin{bmatrix} 2 \\ 1 \end{bmatrix}e^{-t}\tag{6}$ $\tilde c_2=2c_2$ , but they are just constants, you got it.

In short, from a second order DE, with coefficient matrix as:

$A(t)=\begin{bmatrix} a & b\\ c & d \end{bmatrix}$

We can construct the characteristic equation

$\lambda^2-(a+d)\lambda+ad-bc=0\\ \lambda^2-\mathrm{tr}(A)+\det A=0$

To write (3) into a more concise form

$\mathbf x'=A\mathbf x\tag{7}$

And the trial solutions (4), not to confuse the $\mathbf x$ , let the eigenvectors be $\mathbf v$ ,

$\mathbf x=\mathbf ve^{\lambda t}\\ \mathbf x'=\lambda\mathbf ve^{\lambda t}\tag{8}$

Substitute in we get

$\lambda\mathbf ve^{\lambda t}=A\mathbf ve^{\lambda t}\\ \lambda\mathbf v=A\mathbf v$