10.5: Repeated Eigenvalues with One Eigenvector

Example: Find the general solution of $\dot{x}_{1}=x_{1}-x_{2}, \dot{x}_{2}=x_{1}+3 x_{2}$.

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The equations in matrix form are

$$\frac{d}{d t}\left(\begin{array}{l} x_{1} \\ x_{2} \end{array}\right)=\left(\begin{array}{rr} 1 & -1 \\ 1 & 3 \end{array}\right)\left(\begin{array}{l} x_{1} \\ x_{2} \end{array}\right) \nonumber$$

The ansatz $\mathrm{x}=\mathrm{v} e^{\lambda t}$ leads to the characteristic equation

$$\begin{aligned} 0 &=\operatorname{det}(\mathbf{A}-\lambda \mathbf{I}) \\ &=\lambda^{2}-4 \lambda+4 \\ &=(\lambda-2)^{2} . \end{aligned} \nonumber$$

Therefore, $\lambda=2$ is a repeated eigenvalue. The associated eigenvector is found from $-v_{1}-v_{2}=0$, or $v_{2}=-v_{1} ;$ and normalizing with $v_{1}=1$, we have

$$\lambda=2, \quad \mathrm{v}=\left(\begin{array}{r} 1 \\ -1 \end{array}\right) \nonumber$$

We have thus found a single solution to the ode, given by

$$\mathrm{x}_{1}(t)=c_{1}\left(\begin{array}{r} 1 \\ -1 \end{array}\right) e^{2 t} \nonumber$$

and we need to find the missing second solution to be able to satisfy the initial conditions. An ansatz of $t$ times the first solution is tempting, but will fail. Here, we will cheat and find the missing second solution by solving the equivalent secondorder, homogeneous, constant-coefficient differential equation.

Figure 10.3: Phase portrait for example with only one eigenvector.

If $A$ has only a single linearly independent eigenvector $v$, then Equation $\ref{10.13}$ can be solved for $w$ (otherwise, it cannot). Using $A, \lambda$ and $v$ of our present example, Equation $\ref{10.13}$ is the system of equations given by

$$\left(\begin{array}{rr} -1 & -1 \\ 1 & 1 \end{array}\right)\left(\begin{array}{l} w_{1} \\ w_{2} \end{array}\right)=\left(\begin{array}{r} 1 \\ -1 \end{array}\right) \nonumber$$

The first and second equation are the same, so that $w_{2}=-\left(w_{1}+1\right)$. Therefore,

$$\begin{aligned} w &=\left(\begin{array}{c} w_{1} \\ -\left(w_{1}+1\right) \end{array}\right) \\ &=w_{1}\left(\begin{array}{r} 1 \\ -1 \end{array}\right)+\left(\begin{array}{r} 0 \\ -1 \end{array}\right) \end{aligned} \nonumber$$

Notice that the first term repeats the first found solution, i.e., a constant times the eigenvector, and the second term is new. We therefore take $w_{1}=0$ and obtain

$$\mathrm{w}=\left(\begin{array}{r} 0 \\ -1 \end{array}\right) \nonumber$$

as before.

The phase portrait for this ode is shown in Fig. 10.3. The dark line is the single eigenvector $\mathbf{v}$ of the matrix $\mathbf{A}$. When there is only a single eigenvector, the origin is called an improper node.