8Systems of differential equations
IA Differential Equations
8.1 Linear equations
Consider two dependent variables y
1
(t), y
2
(t) related by
˙y
1
= ay
1
+ by
2
+ f
1
(t)
˙y
2
= cy
1
+ dy
2
+ f
2
(t)
We can write this in vector notation by
˙y
1
˙y
2
=
a b
c d
y
1
y
2
+
f
1
f
2
or
˙
Y = MY + F. We can convert this to a higher-order equation by
¨y
1
= a ˙y
1
+ b ˙y
2
+
˙
f
1
= a ˙y
1
+ b(cy
1
+ dy
2
+ f
2
) +
˙
f
1
= a ˙y
1
+ bcy
1
+ d( ˙y
1
− ay
1
− f
1
) + bf
2
+
˙
f
1
so
¨y
1
− (a + d) ˙y
1
+ (ad − bc)y
1
= bf
2
− df
1
+
˙
f
1
and we know how to solve this. However, this actually complicates the equation.
So what we usually do is the other way round: if we have a high-order
equation, we can do this to change it to a system of first-order equations:
If ¨y + a ˙y + by = f , write y
1
= y and y
2
= ˙y. Then let Y =
y
˙y
Our system of equations becomes
˙y
1
= y
2
˙y
2
= f − ay
2
− by
1
or
˙
Y =
0 1
−b −a
y
1
y
2
0
f
.
Now consider the general equation
˙
Y = MY + F
˙
Y − MY = F
We first look for a complementary solution Y
c
= v
e
λt
, where v is a constant
vector. So we get
λv − M v = 0.
We can write this as
Mv = λv.
So λ is the eigenvalue of M and v is the corresponding eigenvector.
We can solve this by solving the characteristic equation
det
(
M − λI
) = 0.
Then for each λ, we find the corresponding v.
Example.
˙
Y =
−4 24
1 −2
Y +
4
1
e
t
The characteristic equation of
M
is
−4 − λ 24
1 −2 − λ
= 0, which gives (
λ
+
8)(λ − 2) = 0 and λ = 2, −8.
When λ = 2, v satisfies
−6 24
1 −4
v
1
v
2
= 0,
and we obtain v
1
=
4
1
.
When λ = −8, we have
4 24
1 6
v
1
v
2
= 0,
and v
2
=
−6
1
. So the complementary solution is
Y = A
4
1
e
2t
+ B
−6
1
e
−8t
To plot the phase-space trajectories, we first consider the cases where Y is
an eigenvector. If Y is an integer multiple of
4
1
, then it will keep moving
outwards in the same direction. Similarly, if it is an integer multiple of
−6
1
,
it will move towards the origin.
y
1
y
2
v
1
= (4, 1)
v
2
= (−6, 1)
We can now add more (curved) lines based on these two:
y
1
y
2
v
1
= (4, 1)
v
2
= (−6, 1)
To find the particular integral, we try Y
p
= ue
t
. Then
u
1
u
2
−
−4 24
1 −2
u
1
u
2
=
4
1
5 −24
−1 3
u
1
u
2
=
4
1
u
1
u
2
= −
1
9
3 24
1 5
4
1
=
−4
−1
So the general solution is
Y = A
4
1
e
2t
+ B
−6
1
e
−8t
−
4
1
e
t
In general, there are three possible cases of
˙
Y
=
M
Y corresponding to three
different possible eigenvalues of M:
(i)
If both
λ
1
, λ
2
are real with opposite sign (
λ
1
λ
2
<
0). wlog assume
λ
1
>
0.
Then there is a saddle as above:
v
1
= (4, 1)
v
2
= (−6, 1)
(ii)
If
λ
1
, λ
2
are real with
λ
1
λ
2
>
0. wlog assume
|λ
1
| ≥ |λ
2
|
. Then the phase
portrait is
v
1
v
2
If both
λ
1
, λ
2
<
0, then the arrows point towards the intersection and we
say there is a stable node. If both are positive, they point outwards and
there is an unstable node.
(iii) If λ
1
, λ
2
are complex conjugates, then we obtain a spiral
If
Re
(
λ
1,2
)
<
0, then it spirals inwards. If
Re
(
λ
1,2
)
>
0, then it spirals
outwards. If
Re
(
λ
1,2
) = 0, then we have ellipses with common centers
instead of spirals. We can determine whether the spiral is positive (as
shown above), or negative (mirror image of the spiral above) by considering
the eigenvectors. An example of this is given below.