MA282 - DE
3.4 complex Eigenvalues
What happens if the eigenvalues of the systems are complex numbers?
Example. Consider
𝑑𝑌
−3 2
= (
)𝑌
−1 −1
𝑑𝑡
The characteristic polynomial is 𝜆2 + 4𝜆 + 5 = 0.
Its eigenvalues are 𝜆 = −2 ± 𝑖
Eigenvectors:
𝜆1 = −2 + 𝑖 ∶ {
1+𝑖
−3𝑥 + 2𝑦 = (−2 + 𝑖)𝑥
𝑦 =
𝑥
− 𝑥 − 𝑦 = (−2 + 𝑖)𝑦
2
2
)
1+𝑖
Choose (
1−𝑖
−3𝑥 + 2𝑦 = (−2 − 𝑖)𝑥
𝜆2 = −2 − 𝑖 ∶ {
𝑦 =
𝑥
− 𝑥 − 𝑦 = (−2 − 𝑖)𝑦
2
2
)
1−𝑖
Choose (
Straight line solutions:
2
𝑌𝑐1 (𝑡) = 𝑒 (−2+𝑖)𝑡 (
),
1+𝑖
2
𝑌𝑐2 (𝑡) = 𝑒 (−2−𝑖)𝑡 (
)
1−𝑖
Question. What does this formula mean?
What good is it given that we are interested in real solutions?
Euler comes to rescue:
𝑒 𝑎+𝑖𝑏 = 𝑒 𝑎 𝑒 𝑖𝑏 = 𝑒 𝑎 (cos 𝑏 + 𝑖 sin 𝑏)
= 𝑒 𝑎 cos 𝑏 + 𝑖𝑒 𝑎 sin 𝑏
for any real numbers 𝑎, 𝑏.
( Using the power series for exp function:
𝑥2 𝑥3
𝑒 =1+𝑥+ + +⋯
2! 3!
𝑥
we can verify 𝑒 𝑖𝑏 = cos 𝑏 + 𝑖 sin 𝑏. i.e., use this series when 𝑥 = 𝑏𝑖. (Appendix C))
1
We can rewrite the solution
𝑒 (−2+𝑖)𝑡 = 𝑒 −2𝑡 (cos 𝑡 + 𝑖 sin 𝑡)
 𝑌𝑐1 (𝑡) = (
2𝑒 −2𝑡 cos 𝑡 + 𝑖 2𝑒 −2𝑡 sin 𝑡
)
(𝑒 −2𝑡 cos 𝑡 − 𝑒 −2𝑡 sin 𝑡) + 𝑖(𝑒 −2𝑡 cos 𝑡 + 𝑒 −2𝑡 sin 𝑡)
which can be broken into
𝑌𝑐1 (𝑡) = (
𝑒
−2𝑡
−2𝑡
sin 𝑡
2𝑒 −2𝑡 cos 𝑡
) + 𝑖 ( −2𝑡 2𝑒
)
−2𝑡
𝑒
cos 𝑡 + 𝑒 −2𝑡 sin 𝑡
cos 𝑡 − 𝑒
sin 𝑡
Question. Why does this help us solve our differential equation?
Theorem Consider
𝑑𝑌
𝑑𝑡
= 𝐴𝑌, where 𝐴 is a matrix with real entries. If 𝑌𝑐 (𝑡) is a complex-
valued solution, then both
Re𝑌𝑐 (𝑡), Im 𝑌𝑐 (𝑡)
are real-valued solutions, and they are linearly independent.
Proof:
Since 𝑌𝑐 (𝑡) is a solution,
𝑑𝑌𝑐
= 𝐴𝑌𝑐
𝑑𝑡
Replace by 𝑅𝑒𝑌𝑐 + 𝑖 𝐼𝑚 𝑌𝑐 ,
𝑑𝑌𝑐 𝑑(𝑅𝑒𝑌𝑐 + 𝑖 𝐼𝑚 𝑌𝑐 ) 𝑑𝑅𝑒𝑌𝑐
𝑑𝐼𝑚𝑌𝑐
=
=
+𝑖
𝑑𝑡
𝑑𝑡
𝑑𝑡
𝑑𝑡
𝐿𝐻𝑆 =
𝑅𝐻𝑆 = 𝐴𝑌𝑐 = 𝐴 (𝑅𝑒𝑌𝑐 + 𝑖 𝐼𝑚𝑌𝑐 ) = 𝐴𝑅𝑒𝑌𝑐 + 𝑖 𝐴 𝐼𝑚𝑌𝑐

Return to our example now. By Theorem,
𝑅𝑒𝑌1 (𝑡) = (
𝑒
−2𝑡
−2𝑡
sin 𝑡
2𝑒 −2𝑡 cos 𝑡
) , 𝐼𝑚𝑌1 (𝑡) = ( −2𝑡 2𝑒
)
−2𝑡
𝑒
cos 𝑡 + 𝑒 −2𝑡 sin 𝑡
cos 𝑡 − 𝑒
sin 𝑡
are solutions to the original system. Moreover they are linearly independent since
1
0
𝑅𝑒𝑌1 (0) = ( ) , 𝐼𝑚𝑌1 (0) = ( ) are linearly independent. Therefore, the general solution is
1
1
𝑌(𝑡) = 𝑘1 (
2𝑒 −2𝑡 sin 𝑡
2𝑒 −2𝑡 cos 𝑡
)
+
𝑘
(
)
2
𝑒 −2𝑡 cos 𝑡 + 𝑒 −2𝑡 sin 𝑡
𝑒 −2𝑡 cos 𝑡 − 𝑒 −2𝑡 sin 𝑡
2
Remark. Used only one complex solution, but 2 complex solutions are nicely related.
(1) 𝜆 = 𝛼 + 𝑖𝛽 is an eigenvalue then 𝜆̅ = 𝛼 − 𝑖𝛽 is also eigenvalue.
(2) 𝑌𝑐1 (𝑡) = 𝑅𝑒𝑌𝑐1 + 𝑖 𝐼𝑚𝑌𝑐1 is a complex solution associated to 𝜆 then 𝑌𝑐2 (𝑡) = ̅̅̅̅̅̅̅̅
𝑌𝑐1 (𝑡) =
𝑅𝑒𝑌𝑐1 − 𝑖 𝐼𝑚𝑌𝑐1
Qualitative Analysis of Systems with Complex Eigenvalues:
Let 𝜆 = 𝛼 + 𝑖𝛽 be an eigenvalue of
𝑑𝑌
𝑑𝑡
= 𝐴𝑌. Then
𝑌𝑐 (𝑡) = 𝑒 (𝛼+𝑖𝛽)𝑡 𝑉𝑐
is a complex solution where 𝑉𝑐 is a complex eigenvector.
Rewrite as
𝑌𝑐 (𝑡) = 𝑒 (𝛼+𝑖𝛽)𝑡 𝑉𝑐 = 𝑒 𝛼𝑡 (cos 𝛽𝑡 + 𝑖 sin 𝛽𝑡)𝑉𝑐
;it is a combination of exp function and trig. functions.
(1) If 𝛼 > 0, then 𝑒 𝛼𝑡 increases as 𝑡 → ∞, and the solutions spiral away from the origin
(called a spiral source)
(2) If 𝛼 < 0, then 𝑒 𝛼𝑡 decreases as 𝑡 → ∞, and the solutions spiral toward the origin (called
a spiral sink)
(3) If 𝛼 = 0, then the solution is periodic. The origin is a center.
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Example1. Consider
𝑑𝑌
0
= (
1
𝑑𝑡
−1
)𝑌
0
The characteristic polynomial is 𝜆2 + 1 = 0. Its eigenvalues are 𝜆 = ±𝑖
𝑖
The eigenvector associated to 𝜆 = 𝑖 is 𝑉𝑐 = ( )
1
The general solution is of the form
cos 𝑡
− sin 𝑡
) + 𝑘2 (
)
sin 𝑡
cos 𝑡
𝑌(𝑡) = 𝑘1 (
Solution Curve:
𝑥 & 𝑦 −graphs:
Example2. Consider
𝑑𝑌
2
= (
4
𝑑𝑡
−2
)𝑌
−2
The characteristic polynomial is 𝜆2 + 4 = 0. Its eigenvalues are 𝜆 = ±2𝑖
1+𝑖
)
2
The eigenvector associated to 𝜆 = 2𝑖 is 𝑉𝑐 = (
The general solution is of the form
cos 2𝑡 − sin 2𝑡
cos 2𝑡 + sin 2𝑡
) + 𝑘2 (
)
2cos 2𝑡
2sin 2𝑡
𝑌(𝑡) = 𝑘1 (
4
Solution Curve:
𝑥 & 𝑦 −graphs:
Example3. Consider
𝑑𝑌
1.9 −2
= (
)𝑌
4 −2.1
𝑑𝑡
The characteristic polynomial is 𝜆2 + 0.2𝜆 + 4.01 = 0. Its eigenvalues are 𝜆 = −0.1 ± 2𝑖
1+𝑖
)
2
The eigenvector associated to 𝜆 = −0.1 + 2𝑖 is 𝑉𝑐 = (
The general solution is of the form
𝑌(𝑡) = 𝑘1 (𝑒
Solution Curve:
−0.1𝑡
cos 2𝑡 − sin 2𝑡 ) + 𝑘 (𝑒 −0.1𝑡 cos 2𝑡 + sin 2𝑡)
2
𝑒
2cos 2𝑡
𝑒 −0.1𝑡 2sin 2𝑡
−0.1𝑡
𝑥 & 𝑦 −graphs:
5
Example Guess typical 𝑥 − and 𝑦 − graphs when 𝜆 = 𝛼 ± 𝛽𝑖 and 𝛼 > 0 ?
Frequency versus Period:
Solutions in Example3 are not periodic in strict sense.
∵ no time T such that
𝑥(𝑡 + 𝑇) = 𝑥(𝑡) and 𝑦(𝑡 + 𝑇) = 𝑦(𝑡), for 𝑡
However, there is a period associated to these solutions, called the natural period of the
solutions. Perhaps, it is best to think about these solutions as oscillating solutions that are
decaying over time and to measure the oscillations in terms of frequency.
Definition
Natural Period := amount of time to cycle once around the origin, denote by T
Natural Frequency:= the number of cycles that solutions make in one unit of time, denote by F
Example
𝑔(𝑡) = cos 𝛽𝑡
 𝑇=
2𝜋
𝛽
or sin 𝛽𝑡
2𝜋
𝛽
, 𝐹 = 1 cycle / ( 𝛽 ) unit of time = 2𝜋 cycles/unit of time
i.e., 𝑇 ⋅ 𝐹 = 1
Definition Sometimes, measure the frequency in radians rather than in cycles.
This measure of frequency is called angular frequency, denote by 𝑓. Then
𝑓 = 2𝜋 𝐹 (radians /unit of time)
Example
F = β/2π cycles/unit of time
𝛽
𝑓 = 2𝜋 ⋅ 2𝜋 = 𝛽 radians/unit of time
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