To understand and be able to apply Euler’s formula,
\begin{equation*}
e^{i \beta t} = \cos \beta t + i \sin \beta t.
\end{equation*}
To understand that if a \(2 \times 2\) matrix \(A\) has two complex eigenvalues, \(\lambda = \alpha \pm i \beta\text{,}\) then the general solution will involve sines and cosines. Furthermore, the origin will be a spiral sink, a spiral source or a center.
The characteristic polynomial of the system (5.4.1) 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 straight-line solutions (Figure 5.4.1). 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 straight-line solutions
Figure5.4.1.A system with no straight-line solutions
Subsection5.4.1Complex 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,
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, 1
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*}
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
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.
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
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.
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.
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} e^{it} \begin{pmatrix} 1 \\ i \end{pmatrix}\\
\amp = e^{-t/10} (\cos t + i \sin t) \begin{pmatrix} 1 \\ i\end{pmatrix}\\
\amp = e^{-t/10} \begin{pmatrix} \cos t + i \sin t \\ - \sin t + i \cos t \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}
\end{align*}
The real and imaginary parts of this solution are
\begin{equation*}
{\mathbf x}_{\text{Re}}(t) = e^{-t/10} \begin{pmatrix} \cos t \\ - \sin t \end{pmatrix} \quad \text{and} \quad {\mathbf x}_{\text{Im}}(t) = e^{-t/10} \begin{pmatrix} \sin t \\ \cos t \end{pmatrix},
\end{equation*}
respectively. Thus, we have the general solution
\begin{equation*}
\mathbf x(t) = c_1 e^{-t/10} \begin{pmatrix} \cos t \\ - \sin t \end{pmatrix} + c_2 e^{-t/10} \begin{pmatrix} \sin t \\ \cos t \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 t + \sin t \\ \cos t - \sin t \end{pmatrix}.
\end{equation*}
The phase portrait of this solution indicates that we do indeed have a spiral sink (Figure 5.4.5).
a direction field of slope arrows and a solution curve that spirals towards the origin
Find the eigenvalues, \(\lambda\) and \(\overline{\lambda}\) of \(A\text{.}\)
(b)
Find eigenvectors, \(\mathbf v\) and \(\overline{\mathbf v}\) for the eigenvalues \(\lambda\) and \(\overline{\lambda}\text{,}\) respectively.
(c)
Find the complex solution to the system \(d\mathbf x/dt = A \mathbf x\text{.}\)
(d)
Find the real solution to the system \(d\mathbf x/dt = A \mathbf x\text{.}\)
(e)
Is the origin a spiral source or a spiral sink? Sketch a solution curve in the \(xy\)-plane.
Subsection5.4.3Solving 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 characteristic 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.
Example5.4.8.
The system \(\mathbf x' = A \mathbf x\text{,}\) where
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 \(e^{i\theta} = \cos \theta + i \sin \theta\text{,}\)
The characteristic polynomial of the associated matrix is \((3 - \lambda)(1 - \lambda) + 26\text{,}\) which gives eigenvalues \(2 \pm 5i\text{.}\) Choosing eigenvalue \(2 + 5i\) (we could have chosen either one) and subtracting this from the main diagonal of the associated matrix gives \(\begin{pmatrix}1 - 5i \amp -2 \\ 13 \amp -1 - 5i\end{pmatrix}\text{,}\) so we have eigenvector \(\begin{pmatrix}1 + 5i \\ 13\end{pmatrix}\text{.}\) This gives a complex solution (ignoring initial conditions)
Now we are asked to draw a sketch of the solution. Since the real part of the eigenvalue (\(2\)) is positive, we have a spiral source. We also have to determine whether the spiral goes clockwise or counter-clockwise as we spiral out. The matrix associated to the equation is \(A = \begin{pmatrix}3 \amp -2 \\ 13 \amp 1\end{pmatrix}\text{.}\) If we multiply this matrix by \(\begin{pmatrix}1 \\ 0\end{pmatrix}\) we get \(\begin{pmatrix}3 \amp -2 \\ 13 \amp 1\end{pmatrix}\cdot \begin{pmatrix}1 \\ 0\end{pmatrix}\)\(= \begin{pmatrix}3 \\ 13\end{pmatrix}\text{.}\) This means that at the point \((1,0)\) a solution curve will be pointing toward the diretion \(\begin{pmatrix}3 \\ 13\end{pmatrix}\text{,}\) so we are spiraling counter-clockwise away from the origin. So we must draw a curve passing through \(\begin{pmatrix}1 \\ 3\end{pmatrix}\) spiraling counter-clockwise outward.
This is the solution curve for our initial value problem.
Figure5.4.10.Solution curve
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 6.1.
Activity5.4.2.Planar Systems with Complex Eigenvalues.
Consider the system \(d\mathbf x/dt = A \mathbf x\text{,}\) where
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.
Reading Questions5.4.5Reading Questions
1.
When are two complex numbers equal?
2.
What is Euler’s formula?
3.
For a \(2 \times 2\) linear system with complex eigenvalues, what are the three different possibilities for the phase plane of the system?
