10.5: Repeated Eigenvalues with One Eigenvector
Example: Find the general solution of \(\dot{x}_{1}=x_{1}-x_{2}, \dot{x}_{2}=x_{1}+3 x_{2}\) .
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.
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 .