Complex Eigenvalues

Section 3.4 Complex Eigenvalues

Consider the following system,

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

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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.

a direction field of slope arrows with a solution curve approaching the origin and no straightline solutions — Figure 3.4.1. A system with no straightline solutions

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Subsection 3.4.1 Complex Eigenvalues

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Suppose that we have the system

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

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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

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

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however, this reduces to solving the equation

$i \beta x = \beta y\text{.}$ Thus, we can find a complex eigenvector

$(1, i)\text{.}$ Consequently,

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

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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, ¹⁵If 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.

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

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Let us rewrite our solution as

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

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and consider the real and imaginary parts of the solution:

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

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Since

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

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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

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

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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.

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We claim that

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

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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

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

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is another solution to the system

${\mathbf x}' = A {\mathbf x}\text{,}$ then

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

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In other words,

$u'(t) = \beta v(t)$ and

$v'(t) = - \beta u(t)\text{.}$ Now, define

$f$ by

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

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The derivative of

$f$ is

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

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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,

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

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Therefore,

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

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Consequently,

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

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Notice that the solutions

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

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are periodic with period

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

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Example 3.4.2.

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).

a direction field of slope arrows and a solution curve that is a circle — Figure 3.4.3. Phase portrait for a center

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Subsection 3.4.2 Spiral Sinks and Sources

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Now let us consider the system

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

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

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and

$\alpha$ and

$\beta$ are nonzero real numbers. The characteristic equation of

$A$ is

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

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so the eigenvalues are

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

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

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to show that

$(1, i)$ is an eigenvector for

$\alpha + i \beta\text{.}$ Therefore, we have a complex solution of the form

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

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As before, both

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

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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

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

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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.

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Example 3.4.4.

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 $c_2 = 2\text{;}$ hence, the solution to our initial value problem is

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

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

a direction field of slope arrows and a solution curve that spirals towards the origin — Figure 3.4.5. Phase portrait for a spiral sink

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Example 3.4.6.

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).

a direction field of slope arrows with a solution curve that spirals out from the origin — Figure 3.4.7. Phase portrait for a spiral source

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Subsection 3.4.3 Solving Systems with Complex Eigenvalues

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Suppose that we have the linear system

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

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

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The charactistic polynomial of

$A$ is

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

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$(a + d)^2 - 4(ad - bc) \lt 0\text{,}$ 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.

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Example 3.4.8.

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

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

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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.

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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.

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Subsection 3.4.4 Important 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.

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Reading Questions 3.4.5 Reading Questions

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1.

When are two complex numbers equal?

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2.

What is Euler's formula?

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3.

For a $2 \times 2$ linear system with distinct real eigenvalues, what are the three different possibilities for the phase plane of the system?

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Exercises 3.4.6 Exercises

Solving Linear Systems with Complex Eigenvalues

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Find the general solution of each of the linear systems in Exercise Group 3.4.6.1–4.

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1.

$\begin{align*} x' & = 2x + 2y\\ y' & = -4x + 6y \end{align*}$

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2.

$\begin{align*} x' & = 2x - 6y\\ y' & = 2x + y \end{align*}$

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3.

$\begin{align*} x' & = -x - 4y\\ y' & = 3x - 2y \end{align*}$

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4.

$\begin{align*} x' & = 2x - 5y\\ y' & = x - 2y \end{align*}$

Solving Initial Value Problems

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Solve each of the following linear systems for the given initial values in Exercise Group 3.4.6.5–8.

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5.

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

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6.

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

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7.

$\begin{align*} x' & = -x - 4y\\ y' & = 3x - 2y\\ x(0) & = 20\\ y(0) & = 20 \end{align*}$

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8.

$\begin{align*} x' & = 2x - 5y\\ y' & = x - 2y\\ x(0) & = 5\\ y(0) & = -5 \end{align*}$

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9.

Consider the linear system $d \mathbf x/dt = A \mathbf x\text{,}$ where

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

Suppose the initial conditions for the solution curve are $x(0) = 1$ and $y(0) = 1\text{.}$ We can use the following Sage code to plot the phase portrait of this system, including a solution curve.

xxxxxxxxxx
 
x, y, t = var('x y t') #declare the variables
F = [3*x + 2*y, -3*x - y] #declare the system
# normalize the vector fields so that all of the arrows are the same length
n = sqrt(F[0]^2 + F[1]^2)
# plot the vector field
p = plot_vector_field((F[0]/n, F[1]/n), (x, -30, 30), (y, -30, 30), aspect_ratio = 1)
# solve the system for the initial condition t = 0, x = 1, y = 1
P1 = desolve_system_rk4(F, [x, y], ics=[0, 1, 1], 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 = -30, xmax = 30, ymin = -30, ymax = 30) 
p

Use Sage to graph the direction field for the system linear systems $d\mathbf x/dt = A \mathbf x$ in Exercise Group 3.4.6.5–8. Plot the solution curve for the given initial condition.

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The Ordinary Differential Equations Project

Section 3.4 Complex Eigenvalues

Subsection 3.4.1 Complex Eigenvalues

Example 3.4.2.

Subsection 3.4.2 Spiral Sinks and Sources

Example 3.4.4.

Example 3.4.6.

Subsection 3.4.3 Solving Systems with Complex Eigenvalues

Example 3.4.8.

Subsection 3.4.4 Important Lessons

Reading Questions 3.4.5 Reading Questions

1.

2.

3.

Exercises 3.4.6 Exercises

Solving Linear Systems with Complex Eigenvalues

1.

2.

3.

4.

Solving Initial Value Problems

5.

6.

7.

8.

9.

Subsection 3.4.7 Project