MATH 223 Systems of Differential Equations including example with Complex Eigenvalues
First consider the system of DE’s which we motivated in class using water passing through two
tanks flushing out salt.
1
y1 (t) +
y10 (t) = − 10
1
0
y2 (t) = 10 y1 (t) −
1
y (t)
40 2
1
y (t)
10 2
d
y = Ay where A =
dt
"
,
y1 (0) = 60, y2 (0) = 0
−1/10 1/40
1/10 −1/10
#
We may compute
"
−1/10 1/40
1/10 −1/10
#


#"
#
1 1 "
0
1/2 −1/4

 −3/20
=  −2 2 
= M DM −1
0
−1/20
1/2 1/4
I offer several solutions, the first using change of variable (change of basis). The second considers
the matrix exponential (y = eAt y(0)) and the third solution considers write the final solution as a
linear combination of solutions to the DE to satisfy the initial conditions.
Our first idea was to rewrite dtd y = Ay as dtd y = M DM −1 y and then M −1 dtd y = DM −1 y. Then
using the linearity of differentiation, we have dd M −1 y = DM −1 y. We set
z = M −1 y and so y = M z
and obtain the ‘easy’ system of differential equations
d
z = Dz
dt
d
z1 (t) = (−3/20)z1 (t),
dt
d
z2 (t) = (−1/20)z2 (t),
dt
which we solve as
z1 (t) = z1 (0)e(−3/20)t , z2 (t) = z2 (0)e(−1/20)t .
We may compute z1 (0), z2 (0) using y1 (0) = 60 and y2 (0) = 0 and z = M −1 y to obtain z1 (0) = 30
and z2 (0) = 30. Now we use y = M z and obtain
"
y1 (t)
y2 (t)
#


#
"
#
1 1 "
(−3/20)t
30e(−3/20)t + 30e(−1/20)t

 30e
=  −2 2 
=
30e(−1/20)t
−60e(−1/20)t + 60e(−1/20)t
This solution technique of changing variables (changing basis) to make the system of differential
equations easy to solve (diagonalization) follows our usual pattern.
A second solution involves tackling the problem directly in what first appears a bit unlikely:
d
y = Ay,
dt
Recall that
y = eAt y(0)
1
1
eAt = I + At + A2 t2 + A3 t3 + · · ·
2
3!
1
d At
e = 0 + A + A2 t + A3 t2 + · · · = AeAt
dt
2
where the derivative has been done entrywise. We note that eAt at t = 0 is in fact e0 = I (we are
exponentiating the 2 × 2 zero matrix) and hence eAt y(0) at t = 0 is indeed y(0) as desired.
We have techniques for doing this namely:
y = eAt y(0) = M eDt M −1 y(0)
This technique can be used for non diagonalizable matrices A if we can find a similar matrix B (i.e.
A = M BM −1 ) for which eBt is easy to compute.
A third solution technique is commonly used when solving DE’s, we write our solution in vector
form in terms of the eigenvectors
"
#
y1 (t)
y2 (t)
= 30e
(−3/20)t
"
1
−2
#
+ 30e
(−1/20)t
"
1
2
#
This suggests another solution strategy. We seek solutions of the form
eλt v with dtd eλt v = λeλt v = eλt Av by solving
Av = λv and hence solving for eigenvalues and eigenvectors of A. Then (it needs to proven!)
we write an arbitrary solution to the system of DE’s as
y = ae
(−3/20)t
"
#
1
−2
+ be
(−1/20)t
"
1
2
#
and solving for a, b given the values y(0).
We now consider a system of DE’s that has complex eigenvalues. It arises from considering the
Differential Equation
y 00 = −y,
y(0) = 1, y 0(0) = 0
If we set y1 (t) = y and y2 (t) = y 0 then we can set
y=
y1 (t)
y2 (t)
and then we can write the DE in vector form as
d
0 1
y=
y
−1
0
dt
We can compute eigenvalues and eigenvectors in the natural way using C instead of R.
0 1
−1 0
A
=
i
−i
−1 −1
M
i 0
0 −i
D
− 21 i − 12
1
i − 21
2
M −1
We could use either of the three methods from above. We can use our third method above (that
follows from our change of basis idea). Let vi be an eigenvector of eigenvalue λi . Then as solution
to the DE, ignoring initial conditions, is
y = e λi vi
In order to match the initial conditions, we take the appropriate linear combination of these solutions from eigenvector/eigenvalue pairs. In our case we have
y1 (t)
i
−i
= aeit
+ be−it
y2 (t)
−1
−1
We can solve for a, b by setting t = 0, noting e0 = 1, to obtain
y1 (0)
1
i
−i
a
=
=a
+b
=M
y2 (0)
0
−1
−1
b
We then solve for a, b using M −1 to obtain
1
a
− 1 i − 12
= 12
= M −1
i − 12
0
b
2
1
−1i
= 12
i
0
2
We then can compute the solution.
Once, in a previous version of 223, I solved this by substituting
eit = cos(t) + i sin(t),
e−it = cos(−t) + i sin(−t) = cos(t) − i sin(t)
Then I proceeded to solve for a, b which made things much more complicated. Setting t = 0 first
and then solving for a, b makes things easier. This is easier for computations; both methods spit
out an answer. The solution becomes
1
1
cos(t)
i
−i
+ i(cos(t) − i sin(t))
=
y = − i(cos(t) + i sin(t))
− sin(t)
−1
−1
2
2
Thus the solution to our DE as expected is y = cos(t) which has y(0) = 1 and y 0 (0) = 0.
We can make some additional simplifications to save work. Let z = c+di ∈ C. Use the notation
Re(z) = c and Im(z) = d to denote the real and imaginary part of z although I would caution that
Im(z) ∈ R. We note that z + z̄ ∈ R. Since we are going to get a real solution we can deduce that
in the expression
y1 (t)
i
it
−it −i
= a1 e
+ a2 e
y2 (t)
−1
−1
that a¯1 = a2 . We can get two different real solutions from the Real and Imaginary parts of one
solution
e
it
i
i
− sin t
cos t
= (cos t + i sin t)
=
+i
−1
−1
− cos t
− sin t
Re(e
it
Im(e
it
i
i
− sin t
) = Re((cos t + i sin t)
)=
−1
−1
− cos t
i
i
cos t
) = Im((cos t + i sin t)
)=
−1
−1
− sin t
You may verify that the real part comes from the choice a1 = 1/2, a2 = 1/2 and the imaginary
part comes from the choice a1 = −i/2, a2 = i/2. We now solve taking a linear combination of these
two solutions (which are both real although their origin was complex):
y1 (t)
− sin t
cos t
=a
+b
,
y2 (t)
− cos t
− sin t
y1 (0)
1
=
y2 (0)
0
We solve and get a = 0, b = 1 yielding the solution y1 (t) = cos t, y2 (t) = − sin t.