Lecture 3: Nonhomogeneous Linear Systems of ODEs

```LECTURE 3
Nonhomogeneous Linear Systems
We now turn our attention to nonhomogeneous linear systems of the form
dx
(1)
= A (t) x (t) + g (t)
dt
where A (t) is a (potentially t-dependent) matrix and g (t) is some prescribed vector function of t. As in
the last lecture, we shall concentrate on 2 × 2 linear systems; as they are simple to compute and yet they
still retain the essential features of the general case.
Just as in the case a single nonhomogenous linear ODE, the general solution of (1) will be of the form
x (t) = xp (t) + xo (t)
where xp (t) is a particular solution of (1) and xo (t) is the general solution of the corresponding homogeneous
equation
dx
= A (t) x (t) .
(2)
dt
(This afterall is a consequence of the linearity of the system, not the number of equations.) And so, just as
in the case of a single ODE, we will need to know the general solution of homogeneous system (2) in order
to solve the nonhomogeneous system (1).
1. Diagonalizable Systems with Constant Coefficients
Let’s begin with the simple 2 × 2 system of the form
dx
= Ax (t) + g (t)
(3)
dt
where A is a constant (i.e. t-independent) diagonalizable matrix. Let (r1 , ξ), and (r2 , η) be the eigenvalue/eigenvector pairs for A. We then have
x(1) (t) = er1 t ξ
,
x(2) (t) = er2 t η
as two fundamental solutions of (2). Recall that if we form a matrix C by using the eigenvectors ξ and η
as, respectively, the first and second columns then
r1 0
C−1 AC =
≡D
0 r2
And so if we define
(4)
y = C−1 x
h (t) = C−t g (t)
,
and multiply both sides of (3) from the left by C−1 we get
dy
d
dx
=
C−1 x =
C−1
dt
dt
dt
C−1 (Ax (t) + g (t)) = C−1 Ax (t) + C−1 g (t) = C−1 ACC−1 x (t) + C−1 g (t)
= Dy (t) + h (t)
1
2. GENERAL DIAGONALIZABLE SYSTEMS
2
Thus, we get
(5)
or
or
dy1
dt
dy2
dt
dy
= Dy (t) + h (t)
dt
r1 0
y1 (t)
h1 (t)
=
+
0 0
y1 (t)
h2 (t)
dy1
− r1 y1 = h1 (t)
dt
dy2
(6b)
− r2 y2 = h2 (t)
dt
Thus, the change of variable (4) allows us to convert the original system into an uncoupled pair of inhomogeneous ODEs.
(6a)
Recall that the general solution of
(7)
0
y + p (t) y = g (t)
=⇒
1
y=
µ (t)
Z
Z
C
µ (t) g (t) +
µ (t)
where
µ (t) = exp
p (t) dt
In the cases at hand
p (r)
−→
g (t)
−→
−r1
−r2
,
,
a constant
a constant
, a constant
h1 (t)
h2 (t)
and accordingly the solutions of (6a) and (6b) are
Z
r1 t
y1 (t) = e
e−r1 t h1 (t) dt + c1 er1 t
Z
y2 (t) = er2 t e−r2 t h2 (t) dt + c2 er2 t
Hence, we can write
R
er1 t R e−r1 t h1 (t) dt + c1 er1 t
y (t) =
er2 t e−r2 t h2 (t) dt + c2 er2 t
as the general solution of the auxiliary, decoupled system (5).
To recover the general solution of the original system, all we have to do is multiply the solution y (t) from
the left by C, as
(8)
x (t) = CC−1 x (t) = C C−1 x (t) = Cy (t)
2. General Diagonalizable Systems
Recall that the general solution of a single linear ODE
y 0 + p (x) y = g (x)
(9)
is given by
(10)
y=
1
µ (x)
Z
µ (x) g (x) dx +
C
µ (x)
Z
,
µ (x) = exp
Let me state this result a little differently. First, note that if g (x) = 0, then
C
y=
µ (x)
and so
−1
µ (x)
p (x) dx
2. GENERAL DIAGONALIZABLE SYSTEMS
3
is interpretable as a fundamental solution for the corresponding homogeneous problem
y 0 + p (x) y = 0.
Let me denote by ψ (x) this fundamental solution:
ψ (x) ≡ µ (x)
−1
Z
= exp − p (x) dx
Then in terms of the fundamental solution ψ (x) of the corresponding homogeneous problem, the general
solution of nonhomogeous equation (9) is
Z
−1
(100 )
y = ψ (x) ψ (x) g (x) dx + Cψ (x)
In fact, the general solution of a nonhomgeneous system
dx
= A (t) x (t) + g (t)
dt
(11)
can also be expressed in terms of the fundamental solutions of the corresponding homogeneous problem
dx
= A (t) x (t)
dt
(12)
Suppose that the homogeneous system (12) has been solved in such a way that we can express its general
solution in terms of a fundamental matrix Ψ (t)1
x0 (t) = Ψ (t) c
=⇒
dx0
= A (t) x0 (t)
dt
I claim that
Z
x (t) ≡ Ψ (t)
(13)
Ψ−1 (t) g (t) dt + Ψ (t) c
is then the general solution of
dx
= A (t) x (t) + g (t)
dt
Before demonstrating this claim, however, note that (13) is a very straightforward generalization of the
solution(100 ) of a single nonhomogeneous linear ODE to a system of nonhomogeneous linear ODEs.
1In other words, suppose we have found n independent solutions ψ (1) (t) , . . . , ψ (n) (t) of an n × n homogeous linear system
dxo
dt
= A (t) xo and have rewritten the right hand side of expression of the general solution as a linear combination of the
fundamental solutions
x0 (t) = c1 ψ (1) (t) + · · · + cn ψ (n) (t)
as a matrix product

|
x0 (t) =  ψ (1) (t)
|

c1
 . 

ψ (n) (t)  
 .. 
|
cn
|
···


≡ Ψ (t) c
2. GENERAL DIAGONALIZABLE SYSTEMS
4
As for the demonstration that (13) is a solution of 11, that’s easy
Z
d
d
−1
x (t) =
Ψ (t) Ψ (t) g (t) dt + Ψ (t) c
dt
dt
Z
Z
d
dΨ
d
−1
−1
Ψ (t) g (t) dt +
=
Ψ (t) g (t) dt + Ψ (t)
Ψ (t) c
dt
dt
dt
Z
= (A (t) Ψ (t)) Ψ−1 (t) g (t) dt + Ψ (t) Ψ−1 (t) g (t) + A (t) Ψ (t) c
Z
= A (t) Ψ (t) Ψ−1 (t) g (t) dt + Ψ (t) c + g (t)
= A (t) x (t) + g (t)
The step
Z
d
Ψ−1 (t) g (t) dt
→ Ψ (t) Ψ−1 (t) g (t)
dt
is just the application of the fundamental theorem of calculus.
Ψ (t)
Example 3.1.
dx1
= 2x1 − x2 + et
dt
dx2
= 3x1 − 2x2 + e−t
dt
This set of differential equations corresponds to an inhomogeneous system x˙ = Ax + g (t) with
t 2 −1
e
A=
,
g (t) =
3 −2
e−t
We shall first find the fundamental matrix for the corresponding homogeneous system x˙ = Ax. This means
finding the eigenvalues r1 , r2 of A, the correponding eigenvectors ξ1 , ξ2 and then forming the matrix


|
|
Ψ (t) =  er1 t ξ1 er2 ξ2 
|
|
each column of which being a fundamental solution.
Now
0 = det (A − λI) = det
2−λ
3
−1
−2 − λ
= λ2 − 4 + 3 = (λ − 1) (λ + 1)
λ=1
=⇒
=⇒
vλ=1
N ullSp (A − λI) = N ullSp
1
=
1
λ = −1
=⇒
=⇒
vλ=−1
N ullSp (A − λI) = N ullSp
1
=
3
1
3
−1
−3
−1
−1
3
3
and so when we can take
r1 = 1 ,
r2 = −1
ξ1 =
,
ξ2 =
1
1
1
3
=⇒
= N ullSp
= N ullSp
1
0
3
0
λ = 1, −1
−1
0
−1
0
2. GENERAL DIAGONALIZABLE SYSTEMS
5
the fundamental matrix for the homogenous x˙ = Ax is
t
e
e−t
Ψ (t) =
et 3e−t
Using the formula
−1
1
d −b
=
−c a
we find
−t
1 3e−t
1
3e
−e−t
−1
=
Ψ (t) = t
−et
et
e (3e−t ) − et e−t
2 −et
We can now plug into the formula
Z
x (t) = Ψ (t) Ψ−1 (t) g (t) dt + Ψ (t) c
a
c
b
d
−e−t
et
for the general solution of x˙ = Ax + g (t). We have
t
t
e
e−t
c1
e c1 + e−t c2
Ψ (t) c =
=
et 3e−t
c2
et c1 + 3e−t c2
and
t −e−t
e
dt
et
e−t
1 −2t
3
2 − 2e
=
dt
− 12 e2t + 21
R 3 1 −2t − 2 e dt
≡ R 2 1 2t
− 2 e + 12 dt
3
1 −2t
2t + 4e
=
− 14 e2t + 12 t
Z
Ψ−1 (t) g (t) =
Z
1
2
Z 3e−t
−et
And so
Z
Ψ (t)
Ψ−1 (t) g (t) dt =
=
et
et
1 t
4e
1 t
4e
e−t
3e−t
1 −2t
3
2t + 4e
1 2t
− 4 e + 12 t
−2t
−2t
6t + e
− 1 + 2te 6t + e−2t − 3 + 6te−2t
Thus, finally we have
Z
x (t) = Ψ (t) Ψ−1 (t) g (t) dt + Ψ (t) c
t
1 t
e c1 + e−t c2
e 6t + e−2t − 1 + 2te−2t 4
+
= 1 t
−2t
et c1 + 3e−t c2
− 3 + 6te−2t
4 e 6t + e
: :
```