# The Ordinary Differential Equations Project

## Section3.5Repeated Eigenvalues

Consider the following system
\begin{equation} \begin{pmatrix} dx/dt \\ dy/dt \end{pmatrix} = \begin{pmatrix} 2 \amp 1 \\ -1 \amp 4 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}.\tag{3.5.1} \end{equation}
The characteristic polynomial of the system (3.5.1) 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 $$\mathbf v = (1, 1)\text{.}$$ There is a single straight-line solution for this system (Figure 3.5.1). The strategy that we used to 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.

### Subsection3.5.1Repeated Eigenvalues

The remaining case that 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 (Subsection 2.4.1). 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*}

#### Example3.5.2.

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 $$\mathbf v = (1, 0)\text{.}$$ 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 straight-line solution (Figure 3.5.3).

#### Activity3.5.1.Systems with Repeated Eigenvalues.

Consider the system $$d\mathbf x/dt = A \mathbf x\text{,}$$ where
\begin{equation*} A = \begin{pmatrix} 1 \amp -2 \\ 0 \amp 1 \end{pmatrix} \end{equation*}
##### (a)
Find the eigenvalues of $$A\text{.}$$ There should be a single real eigenvalue $$\lambda\text{.}$$
##### (b)
Find the eigenvectors $$\mathbf v$$ for the eigenvalues $$\lambda\text{.}$$
##### (c)
Find the straight-line solution of $$d\mathbf x/dt = A \mathbf x\text{.}$$ Plot the solution in the $$xy$$-plane.
##### (d)
Find the general solution of $$d\mathbf x/dt = A \mathbf x\text{.}$$
##### (e)
Sketch several solution curves for the system $$d\mathbf x/dt = A \mathbf x\text{.}$$ What do you notice about the solution curves, especially with respect to the straight-line solution?

### Subsection3.5.2Solving Systems with Repeated Eigenvalues

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 on whether or not we can find two linearly independent eigenvectors.

#### Example3.5.4.

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 too surprising since the system
\begin{align*} x' & = 2x\\ y' & = 2y \end{align*}
is uncoupled and each equation can be solved separately.

#### Example3.5.5.

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. Furthermore, since this system is not partially coupled, we will need a more general strategy.
First, 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, the origin is a source. If it is negative, we will have a 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 two eigenvectors are linearly dependent. We will justify our procedure in the next section (Subsection 3.6.6).

#### Activity3.5.2.Systems with Repeated Eigenvalues—Finding a Second Solution.

Consider the system $$d\mathbf x/dt = A \mathbf x\text{,}$$ where
\begin{equation*} A = \begin{pmatrix} 4 \amp 3 \\ -3 \amp -2 \end{pmatrix} \end{equation*}
##### (a)
Find the eigenvalues of $$A\text{.}$$ There should be a single real eigenvalue $$\lambda\text{.}$$
##### (b)
Find the eigenvectors $$\mathbf v_1$$ for the eigenvalues $$\lambda\text{.}$$
##### (c)
Find the straight-line solution of $$d\mathbf x/dt = A \mathbf x\text{.}$$ Plot the solution in the $$xy$$-plane.
##### (d)
Find one solution, $$\mathbf x_1\text{,}$$ of $$d\mathbf x/dt = A \mathbf x\text{.}$$
##### (e)
To find a second solution of $$d\mathbf x/dt = A \mathbf x\text{,}$$ choose a vector $$\mathbf w$$ that is not a multiple of $$\mathbf v_1$$ and compute $$(A - \lambda I) {\mathbf w}\text{.}$$ This should give you a vector of the form $$\alpha \mathbf v_1\text{.}$$ Let $$\mathbf v_2 = (1/\alpha) \mathbf w\text{.}$$ The second solution is $${\mathbf x}_2 = e^{\lambda t} ({\mathbf v}_2 + t {\mathbf v}_1)\text{.}$$ What is the general solution?
##### (f)
Sketch several solution curves for the system $$d\mathbf x/dt = A \mathbf x\text{.}$$ What do you notice about the solution curves, especially with respect to the straight-line solution?

### Subsection3.5.3Important Lessons

• 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.
• Suppose that a system $$d\mathbf x/dt = A \mathbf x$$ has a single eigenvalue with an $$\mathbf v_1$$ and that all other eigenvectors are multiples of $$\mathbf v_1\text{.}$$ Then one solution is $$\mathbf x_1 = e^{\lambda t} {\mathbf v}_1\text{.}$$ To find a second linearly independent solution of $$d\mathbf x/dt = A \mathbf x\text{,}$$ choose a vector $$\mathbf w$$ that is not a multiple of $$\mathbf v_1$$ and compute $$(A - \lambda I) {\mathbf w}\text{.}$$ This should give you a vector of the form $$\alpha \mathbf v_1\text{.}$$ Let $$\mathbf v_2 = (1/\alpha) \mathbf w\text{.}$$ The second solution is $${\mathbf x}_2 = e^{\lambda t} ({\mathbf v}_2 + t {\mathbf v}_1)\text{.}$$ The general solution of $$d\mathbf x/dt = A \mathbf x$$ will be
\begin{equation*} \mathbf x(t) = c_1 e^{\lambda t} {\mathbf v}_1+ c_2 e^{\lambda t} ({\mathbf v}_2 + t {\mathbf v}_1) \text{.} \end{equation*}

#### 1.

Given a $$2 \times 2$$ system with repeated eigenvalues, how many straight-line solutions are there?

#### 2.

Given a $$2 \times 2$$ system with repeated eigenvalues, explain why it is necessary to find a second linearly independent solution.

### Exercises3.5.5Exercises

#### Solving Linear Systems with Repeated Eigenvalues.

Find the general solution of each of the linear systems in Exercise Group 3.5.5.1–8.
##### 1.
\begin{align*} x' & = 9x + 4y\\ y' & = -9x - 3y \end{align*}
##### 2.
\begin{align*} x' & = 5x + 4y\\ y' & = -9x - 7y \end{align*}
##### 3.
\begin{align*} x' & = -x + y\\ y' & = -x - 3y \end{align*}
##### 4.
\begin{align*} x' & = 2x + y\\ y' & = -x \end{align*}
##### 5.
\begin{align*} x' & = 8x + 4y\\ y' & = -9x - 4y \end{align*}
##### 6.
\begin{align*} x' & = 3x + 4y\\ y' & = -9x - 9y \end{align*}
##### 7.
\begin{align*} x' & = 11x + 4y\\ y' & = -9x - y \end{align*}
##### 8.
\begin{align*} x' & = 13x/2 + 4y\\ y' & = -9x - 11y/2 \end{align*}

#### Solving Initial Value Problems.

Solve each of the following linear systems for the given initial values in Exercise Group 3.5.5.9–16.
##### 9.
\begin{align*} x' & = 9x + 4y\\ y' & = -9x - 3y\\ x(0) & = 2\\ y(0) & = -3 \end{align*}
##### 10.
\begin{align*} x' & = 5x + 4y\\ y' & = -9x - 7y\\ x(0) & = 2\\ y(0) & = 1 \end{align*}
##### 11.
\begin{align*} x' & = -x + y\\ y' & = -x - 3y\\ x(0) & = 2\\ y(0) & = 2 \end{align*}
##### 12.
\begin{align*} x' & = 2x + y\\ y' & = -x\\ x(0) & = 0\\ y(0) & = -5 \end{align*}
##### 13.
\begin{align*} x' & = 8x + 4y\\ y' & = -9x - 4y\\ x(0) & = 1\\ y(0) & = -1 \end{align*}
##### 14.
\begin{align*} x' & = 3x + 4y\\ y' & = -9x - 9y\\ x(0) & = -1\\ y(0) & = 0 \end{align*}
##### 15.
\begin{align*} x' & = 11x + 4y\\ y' & = -9x - y\\ x(0) & = -3\\ y(0) & = 2 \end{align*}
##### 16.
\begin{align*} x' & = 13x/2 + 4y\\ y' & = -9x - 11y/2\\ x(0) & = -2\\ y(0) & = -2 \end{align*}

#### 17.

Consider the linear system $$d \mathbf x/dt = A \mathbf x\text{,}$$ where
\begin{equation*} A = \begin{pmatrix} 3 \amp 1 \\ -4 \amp -1 \end{pmatrix}. \end{equation*}
Suppose the initial conditions for the solution curve are $$x(0) = -2$$ and $$y(0) = 5\text{.}$$ We can use the following Sage code to plot the phase portrait of this system, including a solution curve and the straight-line solution.
x, y, t = var('x y t') #declare the variables
F = [3*x + y, -4*x - y] #declare the system
# normalize the vector fields so that all of the arrows are the same length
n = sqrt(F^2 + F^2)
# plot the vector field
p = plot_vector_field((F/n, F/n), (x, -20, 20), (y, -20, 20), aspect_ratio = 1)
# solve the system for the initial condition t = 0, x = -2, y = 5
P1 = desolve_system_rk4(F, [x, y], ics=[0, -2, 5], ivar = t, end_points = 5, step = 0.01)
# grab the x and y values
S1 = [ [j, k] for i, j, k in P1]
# plot the solution
# Setting xmin, xmax, ymin, ymax will clip the window
# Try plotting without doing this to see what happens
p += line(S1, thickness = 2, axes_labels=['$x(t)$','$y(t)$'], xmin = -20, xmax = 20, ymin = -20, ymax = 20)
# plot the straight-line solutions
p += line([(-10, 20), (10, -20)], thickness = 2, color = "red")
p

Use Sage to graph the direction field for the system linear systems $$d\mathbf x/dt = A \mathbf x$$ in Exercise Group 3.5.5.9–16. Plot the straight-line solutions and the solution curve for the given initial condition.