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.
Solution method 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.
