Linear algebra approach to solving linear systems of differential equations

Systems of differential equations: Linear systems of differential equations

Linear algebra approach to solving linear systems of differential equations

We look again at a pair of coupled homogeneous linear first-order differential equations with constant coefficients of the form

$\left\{\begin{aligned} \frac{\dd x}{\dd t} &= a\, x+ b\, y\\[0.25cm] \frac{\dd y}{\dd t} &= c\, x + d\, y\end{aligned}\right.$ and write this in de matrix-vector form

$\frac{\dd}{\dd t}\cv{x\\y}=\matrix{a & b\\ c & d}\cv{x\\y }$ The matrix

$A=\matrix{a & b\\ c & d}$ actually describes the linear system of differential equations.

Initial Value Problem An initial value problem looks like this:

$\frac{\dd}{\dd t}\cv{x\\y}=A\cv{x\\y }, \quad \cv{x(0)\\y(0)}=\cv{x_0\\y_0}$

We now discuss how linear algebra helps us solve this lineair system of differential equations.

Let $\vec{v}$ be an arbitrary two-dimensional vector and $\lambda$ an arbitrary real number, and then try a solution of the form

$\cv{x(t)\\y(t)} = e^{\lambda t}\,\vec{v}$ Substitution in the matrix-vector form of the linear system of differential equations

$\lambda\,e^{\lambda t}\,\vec{v}=e^{\lambda t}\,A\vec{v}$ You should bear in mind that, for the matrix multiplication $e^{\lambda t}$ is a constant, while for taking derivatives the term $e^{\lambda t}$ is the only thing that depends on $t$ . We find that $\vec{v}$ and $\lambda$ must satisfy

$A\vec{v}=\lambda\vec{v}$ In other words $\vec{v}$ should be an eigenvector corresponding with eigenvalue $\lambda$ .

When the matrix $A$ has two different real eigenvalues $\lambda_1$ and $\lambda_2$ with corresponding eigenvectors $\vec{v}$ and $\vec{w}$ , respectively, then it is the general solution of the linear system of differential equations

$\cv{x(t)\\y(t)} =c_1 e^{\lambda_1 t}\,\vec{v} + c_2e^{\lambda_2 t}\,\vec{w}, \quad\text{with }c_1,c_2\in \mathbb{R}$

For each $\cv{x_0\\y_0}\in\mathbb{R}^2$ exactly two numbers $c_1$ and $c_2$ can be found so that $\cv{x(0)\\y(0)}=\cv{x_0\\y_0}$ . This ensures the existence and uniqueness of the solution of the initial value problem.

With each choice of $\cv{x_0\\y_0}$ there is a solution curve $t\mapsto \cv{x(t)\\y(t)}$ in the phase plane. Two such curves are either the same or do not intersect each other.

One solution catches the eye, namely $\cv{x(t)\\y(t)}=\cv{0\\0}$ , which you get by setting $c_1$ and $c_2$ equal to zero. This is the only solution of the system which does not depend on the independent variable $t$ . If we interpret this variable as time, we have a time-independent solution. The track is a curve that consists of only one point. This is a fixed point or equilibrium of the system. This may be an attracting or repelling equilibrium, or anything else like a semi-stable equilibrium. Here we mainly look at what happens when we choose an initial value near an equilibrium $(0,0)$ . The question is whether the solution curve remains near $(0,0)$ in the course of time or move away from this point. In the following sections we will see that the eigenvalues and eigenvectors of the matrix $A$ play a crucial role herein.