Case II Complex Eigenvalues MAT 275 Recall that We will use this identity when solving systems of differential equations with constant coefficients in which the eigenvalues are complex Example ID: 726147
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Slide1
Systems of Ordinary Differential EquationsCase II: Complex Eigenvalues
MAT 275Slide2
Recall that
. We will use this identity when solving systems of differential equations with constant coefficients in which the eigenvalues are complex.
Example: Solve .Solution: Find the eigenvalues first. Starting with , we get, which has roots .We’ll get the eigenvectors next.
Slide3
The eigenvector for
is such that This simplifies to . The square matrix is singular (you verify). Multiplying the top row with gives
. If we let
, then . Thus, the eigenvector for is .The eigenvector for is found in a similar way, and is .The solution in complex form is .
Slide4
Now we need to rewrite
in real form
.Look at the first term: . Recall that
. Making replacements, we have
Expand the second row by multiplication, and simplify:
Doing the same with the second term,
, gives a scalar multiple of the first term. Once terms are combined and constants renamed, we end up with the same result. Thus, it is sufficient to perform this process just once, as we have done above.
Slide5
From the last slide, we have
.
Now, “stack” the terms into two columns, one real and one imaginary:
Recall that if
are solutions of a homogeneous ODE, then so are and . We can drop the imaginary coefficient now. The solution of is
(Reminder: we’d get the same solution had we performed the tasks on the second term from the last slide.)
Slide6
The solution of
is
We should check that the two vectors are linearly independent by checking its Wronskian:
Since the
Wronskian
is not zero, these are linearly independent solutions. Slide7
Phase Portraits (Direction Field).
Phase portraits of a system of differential equations that has two complex conjugate roots tend to have a “spiral” shape.
Assuming that the eigenvalues are of the form :If , then the direction curves trend away from the origin asymptotically (as t grows to infinity). The origin is an unstable spiral point.If , then the direction curves trend into from the origin asymptotically. The origin is a stable spiral point.If , then the direction curves form concentric ellipses. The origin is a center. Slide8
The phase portrait for
is:
The eigenvalues are . Since a > 0, the direction lines flow away from the origin. Slide9
The phase portrait for
is:
The eigenvalues are . Since a < 0, the direction lines flow into from the origin.Compare this direction field to theone on the previous slide. Do yousee the outward and inward effectbetween the two? Slide10
The phase portrait for
is:
The eigenvalues are . Since a = 0, the direction lines form ellipses.