One real eigenvalue

Systems of differential equations: Linear systems of differential equations

One real eigenvalue

We consider the situation that the ( $2\times 2$ ) matrix $A$ corresponding with the system $\frac{\dd}{\dd t}\cv{x\\ y}=A\cv{x\\ y}$ has one real eigenvalue $\lambda$ . Two cases can be distinguished:

The eigenspace is two-dimensional, that is, there are two eigenvectors that are not multiples of each other.
The eigenspace is one-dimensional, that is, eigenvectors are aligned.

The stability of the equilibrium $(0,0)$ is different in each of these two cases. We always give an example.

Let $\lambda$ be an eigenvalue with corresponding eigenvectors $\vec{v}_1$ and $\vec{v}_2$ that are not multiples of each other. In this case $(0,0)$ is a repelling or attracting equilibrium depending on whether the eigenvalue is positive or negative, respectively.

As an example we consider the system $\frac{\dd }{\dd t}\cv{x\\ y}=A\cv{x\\ y},\quad \text{with }A=\matrix{-1 & 0\\ 0 & -1}\text.$ Eigenvalues and eigenvectors of $A$ are $\lambda_1=-1\text{ with }\vec{v}_1=\cv{1\\0}\quad\text{and}\quad\lambda_2=-1\text{ with }\vec{v}_2=\cv{0\\1}\text.$ The solution is $\cv{x\\y}=c_1e^{-t}\vec{v}_1+ c_2e^{-t}\vec{v}_2$ where $c_1$ and $c_2$ are chosen so that $\cv{x(0)\\ y(0)}=\cv{x_0\\y_0}$ . Trajectories are straight lines moving in the direction of the origin, as the following phase portrait illustrates.

Let $\lambda$ be an eigenvalue with corresponding eigenvector $\vec{v}$ . The following method still leads to two types of solutions. As a matter of fact, there also exists a vector $\vec{w}$ such that $(A-\lambda I)\vec{w}=\vec{v}$ . The vectors $\vec{w}$ are called generalised eigenvectors of $A$ . From the properties of $\vec{w}$ follows (check!) that $\cv{x\\ y}=c(\vec{v}t+\vec{w})e^{\lambda t}$ is a solution, too. The general solution is in this case given by $\cv{x\\ y} = c_1e^{\lambda t}\vec{v}+c_2e^{\lambda t}(\vec{v}t+\vec{w})\text.$ In this case $(0,0)$ is a repelling or attracting equilibrium depending on whether the eigenvalue is positive or negative, respectively.

As a first example we consider the system $\frac{\dd}{\dd t}\cv{x\\ y}=A\cv{x\\ y},\quad \text{with }A=\matrix{-1 & 2\\ 0 & -1}\text.$ Eigenvalue and eigenvector of $A$ is $\lambda=-1\text{ with }\vec{v}=\cv{1\\0}\text.$ A generalised eigenvector, i.e., a solution of $(A-\lambda I)\vec{w}=\vec{v}$ is $\vec{w}=\cv{0\\-\frac{1}{2}}\text.$ Thus the general solution is $\cv{x\\y}=c_1e^{-t}\vec{v}+ c_2e^{-t}(t\vec{v}+\vec{w})$ where $c_1$ and $c_2$ are chosen so that $\cv{x(0)\\ y(0)}=\cv{x_0\\y_0}$ . The trajectories are curves which go in the direction of the origin, as the following phase portrait illustrates. The equilibrium is attracting.

As a second example we consider the system $\frac{\dd}{\dd t}\cv{x\\ y}=A\cv{x\\ y},\quad \text{with }A=\matrix{1 & -1\\ 1 & 3}\text.$ Eigenvalue and eigenvector of $A$ is $\lambda=2\text{ with }\vec{v}=\cv{-1\\1}\text.$ A generalised eigenvector, i.e., a solution of $(A-\lambda I)\vec{w}=\vec{v}$ is $\vec{w}=\cv{1\\0}\text.$ Thus the genera solution is $\cv{x\\y}=c_1e^{2t}\vec{v}+ c_2e^{2t}(t\vec{v}+\vec{w})$ where $c_1$ and $c_2$ are chosen so that $\cv{x(0)\\ y(0)}=\cv{x_0\\y_0}$ . The trajectories are curves which go away from the origin, as the following phase portrait illustrates. The equilibrium is repelling.