Use of eigenvalues and eigenvectors for solving coupled differential equations
Example:
Consider a system of two first-order coupled linear ODEs for two functions of time, v(t) and w(t):
(
dv
dt = 4v(t) − 5w(t)
dw
dt = 2v(t) − 3w(t)
with initial conditions
(
v(0) = 8
w(0) = 5
.
We can re-write this system in a matrix form:
A d v(t)
4 −5 v(t)
=
2 −3 w(t)
dt w(t)
| {z } | {z }
~ (t)
U
and the initial conditions
~0 = 8
U
5
Then we have
d~
~ (t),
U (t) = AU
dt
a linear coupled system with constant coefficients (A6= A(t)).
v(t)
~
A general solution for this type of system U (t) =
represents a linear combination of its
w(t)
linearly independent particular solutions. In this case, i.e. for two unknown variables, we need two
of them:
~ (t) = c1 U
~ 1 (t) + c2 U
~ 2 (t)
U
.
The particular solutions for the linear coupled equations are sought in the same form as a solution
for a scalar form of the first-order linear ODE:
du
= au
dt
u(0) = u0
u(t) = u0 eat
Indeed, this exponent is a solution:
d
d
u0 eat = a u0 eat = au(t).
u(t) =
dt
dt
Now let us assume particular solutions in the exponential form:
(
v(t) = x1 eλt
w(t) = x2 eλt
10
Then the system will look like this
(
λx1 eλt = 4x1 eλt − 5x2 eλt
λx2 eλt = 2x1 eλt − 3x2 eλt
,
where eλt is a common non-zero term. We can cancel it and obtain a system of two linear equations:
(
λx1 = 4x1 − 5x2
.
λx2 = 2x1 − 3x2
In matrix from we have
A~x = λ~x,
i.e. the definition of eigenvectors and eigenvalues! We have already solved such problems. First,
find eigenvalues:
det (A − λI) = 0
(4 − λ)
−5
= (4 − λ)(−3 − λ) − 2(−5) = 0
2
(−3 − λ)
λ1,2
λ2 − λ − 2 = 0
p
√
1 ± 1 + 4(1)(2)
−b ± b2 − 4ac
1±3
=
=
=
2a
2(1)
2
λ1 = −1
λ2 = 2.
Now let us find eigenvectors:
(A − λI) ~x = 0.
For λ1 = −1, the linearly dependent equations are:
(
(4 + 1)x1 − 5x2 = 0
2x1 + (−3 + 1)x2 = 0
(
5x1 − 5x2 = 0
2x1 − 2x2 = 0
The respective eigenvector
~xλ1
For λ2 = 2, the first equation is
1
=
1
(4 − 2)x1 − 5x2 = 2x1 − 5x2 = 0
and
~xλ2
Now we found two particular solutions:
1
=
0.4
λ1 t
−t 1
~
U1 = e ~xλ1 = e
1
11
~ 2 = eλ2 t ~xλ2 = e2t
U
1
0.4
Let us write the general solution as a linear combination of the particular ones
1
−t 1
2t
~
~
~
U = c1 U1 + c2 U2 = c1 e
+ c2 e
1
0.4
8
and use the initial conditions U (0) =
to find the unknown scaling coefficients c1,2 . All expo5
nential terms will equate to 1, so we get a linear system again:
1
1
8
c1
+ c2
=
1
0.4
5
or
(
c1 + c2 = 8
c1 + 0.4c2 = 5
Subtract the second equation from the first one
0.6c2 = 3 → c2 = 5
Then
c1 + 5 = 8 → c1 = 3.
Finally, we have got
1
−t 1
2t
~
U = 3e
+ 5e
1
0.4
so that the solutions for v(t) and w(t) are
(
v(t) = 3e−t + 5e2t
w(t) = 3e−t + 2e2t .
Note: for a system first-order coupled linear equations, the eigenvalues represent growth (positive
power of the exponent) or decay (negative power) rates.
This analysis can be extended to the second-order linear systems.
12
Decoupling coupled equations using matrix diagonalisation
There is another way to apply eigenvalues and eigenvectors for solving systems of coupled linear
ODEs. Let us consider a following example:
(
dx
dt = 4x + 2y
dy
dt = −x + y
with initial conditions
x(0) = 1, y(0) = 0.
Let use the same notation as before
~ = x
U
y
and write the matrix-vector form of the system
d~
~,
U = AU
dt
where
A=
4 2
.
−1 1
Its eigenvalues and eigenvectors can be easily determined:
1
λ1 = 3 → ~xλ1 =
−0.5
1
λ1 2 = 3 → ~xλ2 =
−1
We can diagonalise matrix A:
A = SΛS −1
where
3 0
1
1
.
, Λ=
S=
0 2
−0.5 −1
Now let us introduce another vector
for which
a(t)
~
V =S=
,
b(t)
~ = SV
~.
U
Back to the matrix equation:
d~
d ~
d~
U=
SU = S V
dt
dt
dt
because S is time-independent. On the other hand,
d~
~
U = AU
dt
so that
S
d~
~
~) → dV
~ = S −1 AS V
V = A(S V
| {z } .
dt
dt
Λ
13
Now we have a new system with a diagonal system matrix:
d~
~,
V = ΛV
dt
which means that the variables a(t) and b(t) are actually decoupled
d a(t)
3 0 a(t)
=
0 2 b(t)
dt b(t)
(
d
dt a(t) = 3a(t)
d
dt b(t) = 2b(t)
We can write separate solutions
a(t) = k1 e3t , b(t) = k2 e2t .
Now let us find x and y from a and b. By definition,
~ = SV
~,
U
so that
(
x(t)
a(t)
1
1
a(t)
=S
=
,
y(t)
b(t)
−0.5 −1 b(t)
x(t) = a + b = k1 e3t + k2 e2t
y(t) = −0.5a − b = −0.5k1 e3t − k2 e2t
.
In order to find the unknown coefficients k1,2 , we need to use the initial conditions at t = 0. The
exponents will be equal to 1 again, so that we get:
(
x(0) = 1 = k1 + k2
y(0) = 0 = −0.5k1 − k2
From the second equation, k1 = −2k2 so that k2 = −1 and k1 = 2.
The general solution is
x(t) = 2e3t − e2t
y(t) = −e3t + e2t .
Again, the eigenvalues here are growth/decay rates.
14
Solving second-order coupled systems of linear ODE
It is possible to perform almost exactly the same procedure to solve a second-order coupled system
of linear ordinary differential equations. However, in this case the eigenvalues will represent characteristic frequencies of the system rather than its growth/decay rates, in contrast to first-order ODEs.
Consider a mechanical system consisting of two masses, m1 and m2 , on a horizontal surface. The
masses are connected by a spring with a spring constant k. Let us find time dependence of their
displacements ∆x1 (t) and ∆x2 (t) from the respective equilibrium positions x1,0 and x2,0 (see Fig.1)
assuming that the friction forces are negligible.
Figure 1: Spring and two masses.
Recalling the Hooke’s Law
F = −k∆x,
we can write the balance of forces equations for both masses accounting for only for the force
exerted by the spring:
(
2
1
= −k∆x1 + k∆x2
m1 d dt∆x
2
.
2
d ∆x2
m2 dt2 = k∆x1 − k∆x2
Re-write the system as a matrix-vector product:
#
"
k
− mk1
d2 ∆x1
∆x1
m1
= 2
,
k
− mk2 ∆x2
dt ∆x2
m2
d2
∆~x.
dt2
Now, by analogy with the scalar second-order linear DE (wave equation), write particular solutions
for both displacements as periodic functions (complex exponents):
A∆~x =
∆x1 (t) = c1 eiωt
∆x2 (t) = c2 eiωt .
The equations will look like
"
− mk1
k
m2
k
m1
− mk2
#
✟
iωt
✟
e
c1
∆x1
✟
2 iωt
= −ω ✟
e
,
c2
∆x2
so that −ω 2 will represent an eigenvalue,
A∆~x = −ω 2 ∆~x.
15
The characteristic frequencies an be found as square roots of the eigenvalues from
det A + ω 2 I = 0,
which in this case would represent a fourth-order polynomial with respect to ω but only a secondorder one with respect to λ = −ω 2 :
k
k
k2
2
2
ω −
= 0,
ω −
−
m1
m2
m1 m2
ω4 − k
1
k2
1
k2
+
−
= 0,
ω2 +
m1 m1
m1 m2
m1 m2
1
1
2
2
+
ω ω −k
= 0.
m1 m2
The eigenvalues are
ω12 = 0,
1
1
2
+
ω2 = k
m1 m2
and the respective characteristic frequencies
ω1 = 0,
s 1
1
+
.
ω2 = k
m1 m2
The eigenvectors are calculated from the system equation:
A + ω 2 I ∆~xω = 0,
by substituting ω12 and ω22 .
For ω12 = 0 we have
"
− mk1
so that
−
k
m2
k
m1
− mk2
∆~xω1
1
m1
+
1
m2
∆x1
=0
∆x2
k
k
∆x1 +
∆x2 = 0
m1
m2
so that ∆x1 = ∆x2 and
For ω22 = k
#
1
=
.
1
, the respective eigenvector will be (check it yourself!)
∆~xω2 =
1
.
−1
Then the general solution for the system will be a linear combination of the particular solutions:
~ = b1 e−iω1 t ∆~xω1 + b2 e−iω2 t ∆~xω2
U
16
r −i k m1 + m1 t
1
1
1
2
,
+ b2 e
= b1
−1
1
where unknown coefficients b1,2 can be found from specifying the initial conditions at t = 0, i.e.
∆x1 (0) and ∆x2 (0). The first term is time-independent (ω = 0) and corresponds to a constant
displacement of the masses from their initial position. In our case,
b1 =
∆x1 (0) + ∆x2 (0)
,
2
b2 =
∆x1 (0) − ∆x2 (0)
.
2
17