Differential Equations
Grinshpan
Two-Dimensional Homogeneous Linear Systems with Constant
Coefficients. Purely Imaginary Eigenvalues.
Recall the equation mẍ + kx = 0 of a simple harmonic oscillator with frequency ω =
Let y = ẋ denote the velocity. Then ẏ = ẍ = − mk x and we obtain a system of two
first-order linear equations
(
ẋ = y
(1)
ẏ = − mk x.
q
k
.
m
This system is coupled. Our goal is to understand the motion of (x(t), y(t)) in the xy-plane
(phase plane).
At each point (x, y), the system (1) prescribes a velocity vector,
y
ẋ
=
.
ẏ
−ω 2 x
We may start by plotting the velocity vector field:
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The trajectories appear to be closed (this is a consequence of periodicity of motion),
symmetric about the origin, and oriented clockwise. There are no straight-line solutions.
The constant vector ( xy ) = ( 00 ) clearly satisfies the system. This is the only fixed/stationary
point in the phase plane, it corresponds to the static equilibrium of the oscillator.
The law of conservation of energy allows us to identify the trajectories. Indeed, the
potential energy U = 12 kx2 and the kinetic energy K = 21 mẋ2 of the spring-mass system
must add up to a nonnegative constant:
1
2
kx2 +
1
2
my 2 = C.
Hence those closed trajectories are concentric ellipses.
Actually, the equation
ω 2 xẋ + y ẏ = 0
is a direct consequence of (1). Integrating it with respect to t, we arrive at the same
conclusion:
ω 2 x2 + y 2 = C.
For every C > 0, this is an ellipse centered at the origin with semiaxis ratio 1 : ω.
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How to describe a trajectory as a function of t? Since (ωx, y) stays on a circle, we may
write
x = C cos(α(t)),
y = Cω sin(α(t)),
where α(t) is the polar angle. Using the equation ẋ = y, we conclude that α̇ = −ω and so
α = −ωt + δ. It follows that the general solution of (1) can be written in the form
x
cos(ωt − δ)
=C
,
y
−ω sin(ωt − δ)
where C and δ (phase shift) are arbitrary parameters.
A variant of the preceding parameterization may also be of use. Observe that
cos(ωt)
sin(ωt)
(2)
P =
and Q =
−ω sin(ωt)
ω cos(ωt)
are solutions of (1). They initiate respectively at
1
0
P (0) =
and Q(0) =
,
0
ω
and therefore generate all solutions of the system,
x
(3)
= c1 P + c2 Q.
y
Indeed, any given initial condition (x0 , y0 ) is satisfied by taking x0 P +
P and Q form a fundamental pair of solutions.
2
y0
ω
Q. One says that
Complex-variable method
The matrix of the system,
A=
0 1
,
−ω 2 0
has trace tr A = 0 and determinant det A = ω 2 . The characteristic polynomial
p(r) = r 2 + ω 2 of A has purely imaginary roots,
r = ±iω,
the eigenvalues of A. The eigenvectors of A have complex components:
0 1
1
1
0 1
1
1
= iω
,
= −iω
.
−ω 2 0
iω
iω
−ω 2 0
−iω
−iω
The eigenvectors determine a pair of independent “straight-line” (in the complex sense!)
solutions
1
1
−iωt
iωt
.
,
V =e
U =e
−iω
iω
Separating real and imaginary parts we have
cos(ωt)
sin(ωt)
cos(ωt)
sin(ωt)
U=
+i
,
V =
−i
.
−ω sin(ωt)
ω cos(ωt)
−ω sin(ωt)
ω cos(ωt)
Thus U = P + iQ and V = P − iQ are complex linear combinations of real solutions P and
Q (2). The general solution with real components is given , as in (3), by
x
= c1 P + c2 Q,
y
where c1 , c2 are arbitrary real coefficients.
Systems with purely imaginary eigenvalues
The above ideas apply to any system
ẋ = ax + by
ẏ = cx + dy
whose matrix has purely imaginary eigenvalues,
r = ±iω.
This is the case of zero trace a + d = 0 and positive determinant ad − bc > 0.
We have
ẋ
a b
x
=
,
−a2 − bc = ω 2 .
ẏ
c −a
y
The nonconstant trajectories are given by concentric ellipses
(a2 + ω 2 )x2 + 2abxy + b2 y 2 = C.
The frequency of rotation is given by ω. The fixed point (0, 0) is referred to as center. It is
neutral (neither attracting nor repelling) in the sense of stability.
3
EXAMPLE. Consider the system
ẋ = x − 2y
ẏ = 5x − y.
The underlying matrix has zero trace and positive determinant, its eigenvalues are
λ = ±3i. Hence we are dealing with a center. Let’s find the solution in three ways.
Approach 1. Reduce the system to a single second order equation. Solving the first
equation for y, y = − 12 ẋ + 21 x, and substituting the result into the second equation
deduce that − 21 ẍ + 12 ẋ = 5x + 21 ẋ − 12 x or
ẍ + 9x = 0.
Hence x = c1 cos(3t) + c2 sin(3t) and so y = ( 12 c1 − 32 c2 ) cos(3t) + ( 23 c1 + 21 c2 ) sin(3t). We
may write the answer in the vector form as a linear combination of two independent
solutions:
cos(3t)
sin(3t)
x
= c1 1
+ c2
.
y
cos(3t) + 23 sin(3t)
− 32 cos(3t) + 21 sin(3t)
2
Approach 2. Observe that (5x − y)ẋ − (x − 2y)ẏ = 0. Rearranging terms,
5ẋx − (ẋy + xẏ) + 2ẏy = 0,
and integrating the result with respect to t, we obtain an implicit description of trajectories:
5x2 − 2xy + 2y 2 = C.
This family of ellipses may be parameterized as above.
Approach 3. Determine the eigendirections. For r = 3i, the eigenvector condition is
1 −2
z1
z
= 3i 1 .
5 −1
z2
z2
This gives a system of two dependent linear equations,
(1 − 3i)z1 − 2z2 = 0
5z1 − (1 + 3i)z2 = 0,
satisfied by
z1
2
=
.
z2
1 − 3i
2
Similarly,
is an eigenvector corresponding to the eigenvalue r = −3i. We thus
1 + 3i
have a pair of independent complex solutions
2
2
−3it
3it
.
,
e
e
1 + 3i
1 − 3i
Separating real and imaginary parts, we deduce the general solution:
x
2 cos(3t)
2 sin(3t)
= c1
+ c2
.
y
cos(3t) + 3 sin(3t)
−3 cos(3t) + sin(3t)
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