Consider the following system $$\begin{pmatrix} dx/dt \\ dy/dt \end{pmatrix} = \begin{pmatrix} 2 \amp 1 \\ -1 \amp 4 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}.\label{equation-linear05-repeated-eigenvalues}\tag{3.10}$$ The characteristic polynomial of the system (3.10) is $\lambda^2 - 6\lambda + 9$ and $\lambda^2 - 6 \lambda + 9 = (\lambda - 3)^2\text{.}$ This polynomial has a single root $\lambda = 3$ with eigenvector $\mathbf v = (1, -1)\text{.}$ This there is a single straightline solution for this system (Figure 3.33). The strategy that we used in find the general solution to a system with distinct real eigenvalues will clearly have to be modified if we are to find a general solution to a system with a single eigenvalue.

The remaining case the we must consider is when the characteristic equation of a matrix $A$ has repeated roots. The simplest such case is \begin{equation*} \begin{pmatrix} dx/dt \\ dy/dt \end{pmatrix} = \begin{pmatrix} \lambda & 0 \\ 0 & \lambda \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = A \begin{pmatrix} x \\ y \end{pmatrix}. \end{equation*} The eigenvalues of $A$ are both $\lambda\text{.}$ Since $A{\mathbf v} = \lambda {\mathbf v}\text{,}$ any nonzero vector in ${\mathbb R}^2$ is an eigenvector for $\lambda\text{.}$ Thus, solutions to this system are of the form \begin{equation*} {\mathbf x}(t) = \alpha e^{\lambda t} {\mathbf v}. \end{equation*} Each solution to our system lies on a straight line through the origin and either tends to the origin if $\lambda \lt 0$ or away from zero if $\lambda \gt 0\text{.}$

A more interesting case occurs if \begin{equation*} A = \begin{pmatrix} \lambda & 1 \\ 0 & \lambda \end{pmatrix}. \end{equation*} Again, both eigenvalues are $\lambda\text{;}$ however, there is only one linearly independent eigenvector, which we can take to be $(1, 0)\text{.}$ Therefore, we have a single straight-line solution \begin{equation*} {\mathbf x}_1(t) = \alpha e^{\lambda t}\begin{pmatrix} 1 \\ 0 \end{pmatrix}. \end{equation*}

To find other solutions, we will rewrite the system as \begin{align*} x' & = \lambda x + y\\ y' & = \lambda y. \end{align*} This is a partially coupled system. If $y \neq 0\text{,}$ the solution of the second equation is \begin{equation*} y(t) = \beta e^{\lambda t}. \end{equation*} Therefore, the first equation becomes \begin{equation*} x' = \lambda x + \beta e^{\lambda t}, \end{equation*} which is a first-order linear differential equation with solution \begin{equation*} x(t) = \alpha e^{\lambda t} + \beta t e^{\lambda t}. \end{equation*} Consequently, a solution to our system is \begin{equation*} \alpha e^{\lambda t} \begin{pmatrix} 1 \\ 0 \end{pmatrix} + \beta e^{\lambda t} \begin{pmatrix} t \\ 1 \end{pmatrix}. \end{equation*}

##### Example3.34

Consider the linear system \begin{align*} x' \amp = -x + y\\ y' \amp = -y\\ x(0) \amp = 1\\ y(0) \amp = 3. \end{align*} The matrix that corresponds to this system is \begin{equation*} A = \begin{pmatrix} -1 & 1 \\ 0 & -1 \end{pmatrix} \end{equation*} has a single eigenvalue, $\lambda = -1\text{.}$ An eigenvector for $\lambda$ is $\mathbf v = (1, 0)\text{.}$ Thus, the general solution to our system is \begin{align*} x(t) \amp = c_1 e^{-t} + c_2 t e^{-t}\\ y(t) \amp = c_2 e^{-t}. \end{align*} Applying the initial conditions $x(0) = 1$ and $y(0) = 3\text{,}$ the solution to our initial value problem is \begin{align*} x(t) \amp = e^{-t} + 3te^{-t}\\ y(t) \amp = 3e^{-t}. \end{align*} Notice that we have only one straightline solution (Figure 3.35).

# Subsection3.5.2Solving Systems with Repeated Eigenvalues¶ permalink

If the characteristic equation has only a single repeated root, there is a single eigenvalue. If this is the situation, then we actually have two separate cases to examine, depending one whether or not we can find two linearly independent eigenvectors.

##### Example3.36

Suppose we have the system $\mathbf x' = A \mathbf x\text{,}$ where \begin{equation*} A = \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix}. \end{equation*} The single eigenvalue is $\lambda = 2\text{,}$ but there are two linearly independent eigenvectors, $\mathbf v_1 = (1,0)$ and $\mathbf v_2 = (0,1)\text{.}$ In this case our solution is \begin{equation*} \mathbf x(t) = c_1 e^{2t} \begin{pmatrix} 1 \\ 0 \end{pmatrix} + c_2 e^{2t} \begin{pmatrix} 0 \\ 1 \end{pmatrix}. \end{equation*} This is not two surprising since the system \begin{align*} x' & = 2x\\ y' & = 2y \end{align*} is uncoupled and each equation can be solved separately.

##### Example3.37

Now let us consider the example $\mathbf x' = A \mathbf x\text{,}$ where \begin{equation*} A = \begin{pmatrix} 5 & 1 \\ -4 & 1 \end{pmatrix}. \end{equation*} Since the characteristic polynomial of $A$ is $\lambda^2 - 6 \lambda + 9 = (\lambda - 3)^2\text{,}$ we have only a single eigenvalue $\lambda = 3$ with eigenvector $\mathbf v_1 = (1, -2)\text{.}$ This gives us one solution to our system, $\mathbf x_1(t) = e^{3t}\mathbf v_1\text{;}$ however, we still need a second solution.

Since all other eigenvectors of $A$ are a multiple of $\mathbf v\text{,}$ we cannot find a second linearly independent eigenvector and we need to obtain the second solution in a different manner. We must find a vector ${\mathbf v}_2$ such that $(A - \lambda I){\mathbf v}_2 = {\mathbf v}_1\text{.}$ To do this we can start with any nonzero vector ${\mathbf w}$ that is not a multiple of ${\mathbf v}_1\text{,}$ say ${\mathbf w} = (1, 0)\text{.}$ We then compute \begin{equation*} (A - \lambda I) {\mathbf w} = (A - 3I) {\mathbf w} = \begin{pmatrix} 2 & 1 \\ -4 & -2 \end{pmatrix} \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 2 \\ -4 \end{pmatrix} = 2 {\mathbf v}_1. \end{equation*} Thus, we can take ${\mathbf v}_2 = (1/2)\mathbf w = (1/2, 0)\text{,}$ and our second solution is \begin{equation*} {\mathbf x}_2 = e^{\lambda t} ({\mathbf v}_2 + t {\mathbf v}_1) = e^{3t} \begin{pmatrix} 1/2 + t \\ -2t \end{pmatrix} \end{equation*} Thus, our general solution is \begin{equation*} {\mathbf x} = c_1 {\mathbf x}_1 + c_2 {\mathbf x}_2 = c_1 e^{3t} \begin{pmatrix} 1 \\ -2 \end{pmatrix} + c_2 e^{3t} \begin{pmatrix} 1/2 + t \\ -2t \end{pmatrix}. \end{equation*}

If the eigenvalue is positive, we will have a nodal source. If it is negative, we will have a nodal sink. Notice that we have only given a recipe for finding a solution to $\mathbf x' = A \mathbf x\text{,}$ where $A$ has a repeated eigenvalue and any tow eigen vectors are linearly dependent. We will justify our procedure in the next section (Section 3.6).

• If \begin{equation*} A = \begin{pmatrix} \lambda & 1 \\ 0 & \lambda \end{pmatrix}, \end{equation*} then $A$ has one repeated real eigenvalue. The general solution to the system ${\mathbf x}' = A {\mathbf x}$ is \begin{equation*} {\mathbf x}(t) = \alpha e^{\lambda t} \begin{pmatrix} 1 \\ 0 \end{pmatrix} + \beta e^{\lambda t} \begin{pmatrix} t \\ 1 \end{pmatrix}. \end{equation*} If $\lambda \lt 0\text{,}$ then the solutions tend towards the origin as $t \to \infty\text{.}$ For $\lambda \gt 0\text{,}$ the solutions tend away from the origin.