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## Section3.4Complex Eigenvalues

Consider the following system,

$$\begin{pmatrix} dx/dt \\ dy/dt \end{pmatrix} = \begin{pmatrix} -3 \amp 1\\ -2 \amp -1 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix}\label{equation-linear04-complex-eigenvalues}\tag{3.4.1}$$

The characteristic polynomial of the system (3.4.1) is $\lambda^2 + 4\lambda + 5\text{.}$ The roots of this polynomial are $\lambda_1 = -2 + i$ and $\lambda_2 = -2 -i$ with eigenvectors $\mathbf v_1 = (1, 1 + i)$ and $\mathbf v_2 = (1, 1- i)\text{,}$ respectively. It is clear from the phase portrait of the system that there are no straightline solutions (Figure 3.4.1). However, we would like to have real solutions for a linear system with real coefficients.

### Subsection3.4.1Complex Eigenvalues

Suppose that we have the system

\begin{equation*} \begin{pmatrix} dx/dt \\ dy/dt \end{pmatrix} = \begin{pmatrix} 0 & \beta \\ -\beta & 0 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = A \begin{pmatrix} x \\ y \end{pmatrix}, \end{equation*}

where $\beta \neq 0\text{.}$ The characteristic polynomial of this system is $\det(A - \lambda I) = \lambda^2 + \beta^2\text{,}$ and so we have imaginary eigenvalues $\pm i \beta\text{.}$ To find the eigenvector corresponding to $\lambda = i\beta\text{,}$ we must solve the system

\begin{equation*} \begin{pmatrix} - i \beta & \beta \\ - \beta & - i \beta \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \end{pmatrix}; \end{equation*}

however, this reduces to solving the equation $i \beta x = \beta y\text{.}$ Thus, we can find a complex eigenvector $(1, i)\text{.}$ Consequently,

\begin{equation*} {\mathbf x}(t) = e^{i \beta t} \begin{pmatrix} 1 \\ i\end{pmatrix} \end{equation*}

is a complex solution to the system ${\mathbf x}' = A {\mathbf x}\text{.}$ The problem is that we have a real system of differential equations and would like real solutions. We can remedy the situation if we use Euler's formula,  15 If you are unfamiliar with Euler's formula, try expanding both sides as a power series to check that we do indeed have a correct identity.

\begin{equation*} e^{i \beta t} = \cos \beta t + i \sin \beta t. \end{equation*}

Let us rewrite our solution as

\begin{align*} {\mathbf x}(t) & = e^{i \beta t} \begin{pmatrix} 1 \\ i \end{pmatrix}\\ & = \begin{pmatrix} \cos \beta t + i \sin \beta t\\ i(\cos \beta t + i \sin \beta t) \end{pmatrix}\\ & = \begin{pmatrix} \cos \beta t + i \sin \beta t\\ - \sin \beta t + i \cos \beta t \end{pmatrix}\\ & = \begin{pmatrix} \cos \beta t \\ - \sin \beta t \end{pmatrix} + i \begin{pmatrix} \sin \beta t\\ \cos \beta t \end{pmatrix} \end{align*}

and consider the real and imaginary parts of the solution:

\begin{equation*} {\mathbf x}_{\text{Re}} = \begin{pmatrix} \cos \beta t \\ - \sin \beta t \end{pmatrix} \qquad \text{and} \qquad {\mathbf x}_{\text{Im}} = \begin{pmatrix} \sin \beta t\\ \cos \beta t \end{pmatrix}. \end{equation*}

Since

\begin{align*} {\mathbf x}'_{\text{Re}}(t) +i {\mathbf x}'_{\text{Im}}(t) & = {\mathbf x}'(t)\\ & = A {\mathbf x}(t)\\ & = A( {\mathbf x}_{\text{Re}}(t) +i {\mathbf x}_{\text{Im}}(t)\\ & = A {\mathbf x}_{\text{Re}}(t) +i A {\mathbf x}_{\text{Im}}(t). \end{align*}

we know that ${\mathbf x}'_{\text{Re}}(t) = A{ \mathbf x}_{\text{Re}}(t)$ and ${\mathbf x}'_{\text{Im}}(t) = A {\mathbf x}_{\text{Im}}(t)$ by setting the real and imaginary parts equal. Thus, both ${\mathbf x}_{\text{Re}}(t)$ and ${\mathbf x}_{\text{Im}}(t)$ are solutions to our system. Moreover, since

\begin{equation*} {\mathbf x}_{\text{Re}}(0) = \begin{pmatrix} 1 \\ 0 \end{pmatrix} \mbox{ and } {\mathbf x}_{\text{Im}}(0) = \begin{pmatrix} 0 \\ 1 \end{pmatrix}, \end{equation*}

we know that the appropriate linear combination of ${\mathbf x}_{\text{Re}}(t)$ and ${\mathbf x}_{\text{Im}}(t)$ will provide a solution to any initial value problem.

We claim that

$${\mathbf x}(t) = c_1 {\mathbf x}_{\text{Re}}(t) + c_2 {\mathbf x}_{\text{Im}}(t)\label{equation-linear04-real-imaginary-solution}\tag{3.4.2}$$

is a general solution to our system. That is, we must be able to write every solution of our system as a linear combination of ${\mathbf x}_{\text{Re}}(t)$ and ${\mathbf x}_{\text{Im}}(t)\text{.}$ If

\begin{equation*} {\mathbf y}(t) = \begin{pmatrix} u(t) \\ v(t) \end{pmatrix} \end{equation*}

is another solution to the system ${\mathbf x}' = A {\mathbf x}\text{,}$ then

\begin{equation*} {\mathbf y}'(t) = \begin{pmatrix} u'(t) \\ v'(t) \end{pmatrix} = \begin{pmatrix} 0 & \beta \\ - \beta & 0 \end{pmatrix} \begin{pmatrix} u(t) \\ v(t) \end{pmatrix} = \begin{pmatrix} \beta v(t) \\ - \beta u(t) \end{pmatrix}. \end{equation*}

In other words, $u'(t) = \beta v(t)$ and $v'(t) = - \beta u(t)\text{.}$ Now, define $f$ by

\begin{equation*} f(t) = (u(t) + i v(t)) e^{i\beta t}. \end{equation*}

The derivative of $f$ is

\begin{align*} f'(t) & = (u'(t) + i v'(t)) e^{i\beta t}+ i \beta (u(t) + i v(t)) e^{i\beta t}\\ & =(\beta v(t) -i \beta u(t)) e^{i\beta t}+ (i \beta u(t) + i^2 \beta v(t)) e^{i\beta t}\\ & = 0. \end{align*}

Therefore, $f(t)$ is a complex constant and $f(t) = (u(t) + i v(t)) e^{i\beta t} = a + bi\text{.}$ We can now write $u(t) + iv(t) = (a + ib) e^{- i \beta t}\text{.}$ Thus,

\begin{align*} u(t) + iv(t) & = (a + ib) e^{- i \beta t}\\ & = (a + ib) (\cos \beta t - i \sin \beta t)\\ & = (a \cos \beta t + b \sin \beta t) + i (b \cos \beta t - a \sin \beta t). \end{align*}

Therefore,

\begin{align*} u(t) & = a \cos \beta t + b \sin \beta t\\ v(t) & = b \cos \beta t - a \sin \beta t. \end{align*}

Consequently,

\begin{align*} \begin{pmatrix} u(t) \\ v(t) \end{pmatrix} & = \begin{pmatrix} a \cos \beta t + b \sin \beta t \\ b \cos \beta t - a \sin \beta t \end{pmatrix}\\ & = a \begin{pmatrix} \cos \beta t \\ - \sin \beta t \end{pmatrix} + b \begin{pmatrix} \sin \beta t \\ \cos \beta t \end{pmatrix}\\ & = a {\mathbf x}_{\text{Re}}(t) + b{\mathbf x}_{\text{Im}}(t). \end{align*}

Notice that the solutions

\begin{equation*} {\mathbf x}(t) = c_1 \begin{pmatrix} \cos \beta t \\ - \sin \beta t \end{pmatrix} + c_2 \begin{pmatrix} \sin \beta t\\ \cos \beta t \end{pmatrix} \end{equation*}

are periodic with period $2 \pi / \beta\text{.}$ This type of system is called a center.

###### Example3.4.2

Consider the initial value problem

\begin{align*} \frac{dx}{dt} \amp = 2y\\ \frac{dy}{dt} \amp = -2x\\ x(0) \amp = 1\\ y(0) \amp = 2. \end{align*}

The eigenvalues of this system are $\lambda = \pm 2i\text{.}$ Therefore, the general solution to the system is

\begin{align*} x(t) \amp = c_1 \cos 2t + c_2 \sin 2t\\ y(t) \amp = - c_1 \sin 2t + c_2 \cos 2t. \end{align*}

Using the initial conditions to solve for $c_1$ and $c_2\text{,}$ the solution to our initial value problem is

\begin{align*} x(t) \amp = \cos 2t + 2 \sin 2t\\ y(t) \amp = - \sin 2t + 2 \cos 2t. \end{align*}

The phase portrait is a circle of radius 2 about the origin (Figure 3.4.3).

### Subsection3.4.2Spiral Sinks and Sources

Now let us consider the system ${\mathbf x}' = A {\mathbf x}\text{,}$ where

\begin{equation*} A = \begin{pmatrix} \alpha & \beta \\ - \beta & \alpha \end{pmatrix} \end{equation*}

and $\alpha$ and $\beta$ are nonzero real numbers. The characteristic equation of $A$ is

\begin{equation*} \lambda^2 - 2 \alpha \lambda + (\alpha^2 + \beta^2) = 0, \end{equation*}

so the eigenvalues are $\lambda = \alpha \pm i \beta\text{.}$ We can use the equation

\begin{equation*} (\alpha - (\alpha + i \beta))x + \beta y = 0 \end{equation*}

to show that $(1, i)$ is an eigenvector for $\alpha + i \beta\text{.}$ Therefore, we have a complex solution of the form

\begin{align*} {\mathbf x}(t) & = e^{(\alpha + i \beta)t} \begin{pmatrix} 1 \\ i \end{pmatrix}\\ & = e^{\alpha t} \begin{pmatrix} \cos \beta t \\ - \sin \beta t \end{pmatrix} + i e^{\alpha t} \begin{pmatrix} \sin \beta t \\ \cos \beta t \end{pmatrix}\\ & = {\mathbf x}_{\text{Re}}(t) + i {\mathbf x}_{\text{Im}}(t). \end{align*}

As before, both

\begin{equation*} {\mathbf x}_{\text{Re}}(t) = e^{\alpha t} \begin{pmatrix} \cos \beta t \\ - \sin \beta t \end{pmatrix} \quad \text{and} \quad {\mathbf x}_{\text{Im}}(t) = e^{\alpha t} \begin{pmatrix} \sin \beta t \\ \cos \beta t \end{pmatrix} \end{equation*}

are real solutions to ${\mathbf x}' = A {\mathbf x}\text{.}$ Furthermore, these solutions are linearly independent. Indeed, ${\mathbf x}_{\text{Re}}$ cannot be a multiple of ${\mathbf x}_{\text{Im}}$ for all values of $t\text{.}$ Thus, we have the general solution

\begin{equation*} {\mathbf x}(t) = c_1 e^{\alpha t} \begin{pmatrix} \cos \beta t \\ - \sin \beta t \end{pmatrix} + c_2 e^{\alpha t} \begin{pmatrix} \sin \beta t \\ \cos \beta t \end{pmatrix}. \end{equation*}

The $e^{\alpha t}$ factor tells us that the solutions either spiral into the origin if $\alpha \lt 0$ or spiral out to infinity if $\alpha \gt 0\text{.}$ In this case we say that the equilibrium points are spiral sinks and spiral sources, respectively.

###### Example3.4.4

Consider the initial value problem

\begin{align*} \frac{dx}{dt} \amp = -x/10 + y\\ \frac{dy}{dt} \amp = -x - y/10\\ x(0) \amp = 2\\ y(0) \amp = 2. \end{align*}

The matrix

\begin{equation*} \begin{pmatrix} -1/10 \amp 1 \\ -1 \amp -1/10 \end{pmatrix} \end{equation*}

has eigenvalues $\lambda = -1/10 \pm i\text{.}$ The eigenvalue $\lambda = -1/10 + i$ has an eigenvector $\mathbf v = (1, i)\text{.}$ The complex solution of our system is

\begin{align*} \mathbf x(t) \amp = e^{(-1/10 + i)t} \begin{pmatrix} 1 \\ i\end{pmatrix}\\ \amp = e^{-t/10} \begin{pmatrix} \cos t \\ - \sin t \end{pmatrix} + i e^{-t/10} \begin{pmatrix} \sin t \\ \cos t\end{pmatrix}\\ \amp = e^{-t/10} (\cos t + i \sin t) \begin{pmatrix} \cos t \\ - \sin t \end{pmatrix} + i e^{-t/10} (\cos t + i \sin t) \begin{pmatrix} \sin t \\ \cos t\end{pmatrix}\\ \amp = e^{-t/10} \begin{pmatrix} \cos^2 t + i \cos t \sin t \\ - \cos t \sin t - i \sin^2 t \end{pmatrix} + e^{-t/10} \begin{pmatrix} - \sin^2 t + i \cos t\sin t \\ - \cos t \sin t + i \cos^2 t \end{pmatrix}\\ \amp = e^{-t/10} \begin{pmatrix} (\cos^2 t - \sin^2 t) + 2 i\, \sin t \cos t \\ - 2 \sin t \cos t + i(\cos^2 t - \sin^2 t) \end{pmatrix}\\ \amp = e^{-t/10} \begin{pmatrix} \cos 2t + i\, \sin 2t \\ - \sin 2t + i \cos 2t \end{pmatrix}\\ \amp = e^{-t/10} \begin{pmatrix} \cos 2t \\ - \sin 2t \end{pmatrix} + i e^{-t/10} \begin{pmatrix} \sin 2t \\ \cos 2t \end{pmatrix} \end{align*}

The real and imaginary parts of this solution are

\begin{equation*} {\mathbf x}_{\text{Re}}(t) = e^{-t/10} \begin{pmatrix} \cos 2t \\ - \sin 2t \end{pmatrix} \quad \text{and} \quad {\mathbf x}_{\text{Im}}(t) = e^{-t/10} \begin{pmatrix} \sin 2t \\ \cos 2t \end{pmatrix}, \end{equation*}

respectively. Thus, we have the general solution

\begin{equation*} \mathbf x(t) = c_1 e^{-t/10} \begin{pmatrix} \cos 2t \\ - \sin 2t \end{pmatrix} + c_2 e^{-t/10} \begin{pmatrix} \sin 2t \\ \cos 2t \end{pmatrix}. \end{equation*}

Applying our initial conditions, we can determine that $c_1 = 2$ and $c_2 = 0\text{;}$ hence, the solution to our initial value problem is

\begin{equation*} \mathbf x(t) = 2 e^{-t/10} \begin{pmatrix} \cos 2t \\ - \sin 2t \end{pmatrix}. \end{equation*}

The phase portrait of this solution indicates that we do indeed have a spiral sink (Figure 3.4.5).

###### Example3.4.6

The initial value problem

\begin{align*} \frac{dx}{dt} \amp = x/10 + y\\ \frac{dy}{dt} \amp = -x + y/10\\ x(0) \amp = 0\\ y(0) \amp = 1/2. \end{align*}

The matrix

\begin{equation*} \begin{pmatrix} 1/10 \amp 1 \\ -1 \amp 1/10 \end{pmatrix} \end{equation*}

has an eigenvector $(1, -i)$ with eigenvalue $\lambda = 1/10 - i\text{.}$ Thus, the complex solution is

\begin{equation*} \mathbf x(t) = e^{(1/10 - i)t} \begin{pmatrix} 1 \\ -i \end{pmatrix}. \end{equation*}

Following the procedure that we used in the previous example, the solution to our initial value problem is

\begin{equation*} \mathbf x(t) = \frac{1}{2} e^{t/10} \begin{pmatrix} \sin t \\ \cos t \end{pmatrix}, \end{equation*}

and he phase portrait is a spiral source (Figure 3.4.7).

### Subsection3.4.3Solving Systems with Complex Eigenvalues

Suppose that we have the linear system $\mathbf x' = A \mathbf x\text{,}$ where

\begin{equation*} A = \begin{pmatrix} a \amp b \\ c \amp d \end{pmatrix}. \end{equation*}

The charactistic polynomial of $A$ is

\begin{equation*} p(\lambda) = \lambda^2 - (a + d)\lambda + (ad - bc). \end{equation*}

If $(a + d)^2 - 4(ad - bc) \lt 0\text{,}$ then the eigenvalues of $A$ are complex, and we cannot apply the strategy that we used to determine the general solution in the case of distinct real roots.

###### Example3.4.8

The system $\mathbf x' = A \mathbf x\text{,}$ where

\begin{equation*} A = \begin{pmatrix} -1 \amp -2 \\ 4 \amp 3 \end{pmatrix}. \end{equation*}

The characteristic polynomial of $A$ is $\lambda^2 - 2 \lambda + 5$ and so the eigenvalues are complex conjugates, $\lambda = 1 + 2i$ and $\overline{\lambda} = 1 - 2i\text{.}$ It is easy to show that an eigenvector for $\lambda = 1 + 2 i$ is $\mathbf v = (1, -1 - i)\text{.}$ Recalling that $e^{i\theta} = \cos \theta + i \sin \theta\text{,}$

\begin{align*} \mathbf x(t) & = e^{(1+2i)t} \mathbf v\\ & = e^{(1+2i)t} \begin{pmatrix} 1 \\ -1 - i \end{pmatrix}\\ & = e^{t} e^{2it} \begin{pmatrix} 1 \\ -1 - i \end{pmatrix}\\ & = e^{t} (\cos 2t + i \sin 2t) \begin{pmatrix} 1 \\ -1 - i\end{pmatrix}\\ & = e^{t} \begin{pmatrix} \cos 2t + i \sin 2t \\ (-1 - i)(\cos 2t + i \sin 2t) \end{pmatrix}\\ & = e^{t} \begin{pmatrix} \cos 2t + i \sin 2t \\ (- \cos 2t + \sin 2t) + i(- \cos 2t - \sin 2t) \end{pmatrix}\\ & = e^{t} \begin{pmatrix} \cos 2t \\ - \cos 2t + \sin 2t \end{pmatrix} + i e^{t} \begin{pmatrix} \sin 2t \\ - \cos 2t - \sin 2t \end{pmatrix} \end{align*}

is a complex solution to our system. Taking the real and imaginary parts of this solution, we obtain the general solution to our system

\begin{equation*} {\mathbf x}(t) = c_1 e^{t} \begin{pmatrix} \cos 2t \\ - \cos 2t + \sin 2t \end{pmatrix} + c_2 e^{t} \begin{pmatrix} \sin 2t \\ - \cos 2t - \sin 2t \end{pmatrix}. \end{equation*}

The nature of the equilibrium solution is determined by the factor $e^{\alpha t}$ in the solution. If $\alpha \lt 0\text{,}$ the equilibrium point is a spiral sink. If $\alpha \gt 0\text{,}$ the equilibrium point is a spiral source. If $\alpha = 0\text{,}$ the equilibrium point is a center.

Although we have outlined a procedure to find the general solution of $\mathbf x' = A \mathbf x$ if $A$ has complex eigenvalues, we have not shown that this method will work in all cases. We will do so in Section 3.6.

### Subsection3.4.4Important Lessons

• If
\begin{equation*} A = \begin{pmatrix} \alpha & \beta \\ -\beta & \alpha \end{pmatrix}, \end{equation*}
then $A$ has two complex eigenvalues, $\lambda = \alpha \pm i \beta\text{.}$ The general solution to the system ${\mathbf x}' = A {\mathbf x}$ is
\begin{equation*} {\mathbf x}(t) = c_1 e^{\alpha t} \begin{pmatrix} \cos \beta t \\ - \sin \beta t \end{pmatrix} + c_2 e^{\alpha t} \begin{pmatrix} \sin \beta t \\ \cos \beta t \end{pmatrix}. \end{equation*}
If $\alpha \lt 0\text{,}$ the equilibrium point is a spiral sink. If $\alpha \gt 0\text{,}$ the equilibrium point is a spiral source.

### SubsectionExercises

###### 1

Solve each of the following linear systems for the given initial values. Sketch the phase plane of each system and determine the nature of the equilibrium solution(s). Do this first without the use of technology, then check your answer with pplane or equivalent software.

1. \begin{align*} x' & = 2x + 2y\\ y' & = -4x + 6y\\ x(0) & = 2\\ y(0) & = -3 \end{align*}
2. \begin{align*} x' & = 2x - 6y\\ y' & = 2x + y\\ x(0) & = 2\\ y(0) & = 1 \end{align*}
3. \begin{align*} x' & = 2x + y\\ y' & = -x + 4y\\ x(0) & = 1\\ y(0) & = -1 \end{align*}
4. \begin{align*} x' & = 4x + 2y\\ y' & = -2x - y\\ x(0) & = 1\\ y(0) & = -1 \end{align*}