Unformatted text preview: Inhomogeneous Linear Systems
General Properties
tions has the form An inhomogeneous system of linear diﬀerential equa dY
= A(t) Y + G(t),
dt
where A(t) is a known n × n matrix function of t and G(t) is a known n vector function of t. The familiar rules relating solutions of homogeneous and inhomogeneous equations apply. Thus if Y (t, C ) is the general solution of
the homogeneous system, the general solution of the inhomogeneous system
takes the form Y (t, C ) + Yp (t), where Yp (t) is any vector solution of the
inhomogeneous system.
That being the case, the next order of business must be to develop some
techniques for obtaining particular solutions of inhomogeneous linear systems. The ﬁrst of these is the
Method of Variation of Parameters We have seen that, for any fundamental matrix solution Y(t), the general solution of the linear homogeneous
system Y (t) = A(t) Y (t) may be taken to be Y(t) C where C is an arbitrary
nvector. Following our development of the variation of parameters formula
in the ﬁrst and second order scalar cases, it would now be reasonable to think
we might ﬁnd a particular solution of the inhomogeneous system shown above
in the form Yp (t) = Y(t) C (t) where now C (t) is an nvector function of t.
Substituting this solution form into the inhomogeneous system we obtain
Y (t) C (t) + Y(t) C (t) = A(t) Y(t) C (t) + G(t).
Since Y (t) = A(t) Y(t), this gives
Y(t) C (t) = G(t) ⇒ C (t) = Y(t)−1 G(t).
Then, integrating, we have
ˆ
C (t ) = C + t Y(τ )G(τ ) dτ ⇒
1 t ˆ
Yp (t) = Y(t) C + Y(t) Y(τ )−1 G(τ ) dτ ˆ
for an arbitrary constant nvector C . We can obtain a particular solution for
ˆ
any choice of C and any choice of the lower limit of the integral. With the
ˆ
choice C = 0 we obtain the general solution of the inhomogeneous system
in the form
t Y (t, C ) = Y(t) C + Y(t) Y(τ )−1 G(τ ) dτ for an arbitrary nvector C .
If we want to solve an initial value problem Y (t) = Y0 we take the lower
limit of the integral to be t0. Then
Y0 = Y (t0, C ) = Y(t0 ) C + 0 ⇒ C = Y(t0 )−1 Y0
and we have the solution of the initial value problem in the form
Y (t) = Y(t) Y(t0 )−1 Y0 + Y(t) t Y(τ )−1 G(τ ) dτ t0 t = Y(t, t0) Y0 + Y(t, τ ) G(τ ) dτ,
t0 where we recall that Y(τ, τ ) = Y(t, t) = In . This is the variation of
parameters formula for a linear inhomogeneous system with possibly time
varying system matrix A(t). To check, we verify the initial condition
t0 Y (t 0 ) = Y (t 0 , t 0 ) + t0 Y(t0 , τ ) G(τ ) dτ = I Y0 + 0 = Y0 and satisfaction of the diﬀerential system:
t Y (t) = Y (t) Y0 + Y(t, t)G(t) + t0 d
Y(t, τ ) G(τ ) dτ = A(t) Y(t) Y(t) Y0
dt t + I G(t) + A(t) Y(t, τ ) G(τ ) dτ = A(t) Y (t) + G(t).
t0 Example 1
We recall the inhomogeneous system from Example 1 of the
introductory linear systems section of the notes:
d
dt y1
y2 = 1/t 1
0 2/t
2 y1
y2 + t
.
t2 For the corresponding homogeneous system we obtained the fundamental
matrix solution
Y (t ) = t t3/2 − t/2
.
0
t2 This gives, for the inverse matrix at t = τ and Y(t, τ ) = Y(t)Y(τ )−1 ,
Y(τ )−1 = 1/(2τ 2 ) − 1/2
, Y(t, τ ) =
1/τ 2 1/τ
0 t/τ
0 t3/(2τ 2 ) − t/2
.
t2/τ 2 Then the solution of the inhomogeneous system shown above with initial
condition y 1(2) = 1, y 2(2) = −1 is given, using the variation of parameters
formula, by
Y (t ) =
= t/2 t3/8 − t/2
0
t2 /4 t − t3/8
+
−t 2/4 t
2 1
−1 t/τ
0 t + t + t3/2 − t τ 2/2
t2 2 dτ = t3 /(2τ 2 ) − t/2
t2/τ 2 τ
τ2 dτ t4 /3 − 7t3 /8 + t2 − 2 t/3
.
t3 − 9 t2 /4 For a linear inhomogeneous constant coeﬃcient system and initial condition
dY
= A Y + G(t), Y (t0 ) = Y0 ,
dt
the variation of parameters formula shown previously takes the form
Y (t) = eA(t−t0) Y0 + t eA(t−τ ) G(τ ) dτ. t0 It can be used to compute solutions, quite usefully if G(t) has some nonstandard functional form, but it is mostly used as an abstract representation
formula. More widely used is the
Method of Undetermined Coeﬃcients The method of undetermined coeﬃcients, for an inhomogeneous system with constant n × n matrix A,
as just shown, works much the same way for linear inhomogeneous systems
as it does for scalar linear constant coeﬃcient inhomogeneous diﬀerential
equations, except that the ”undetermined coeﬃcients” are now vectors of
dimension n.
3 We consider the inhomogeneous system Example 1 d
dt y1
y2 11
−1 2 = y1
y2 t
.
t2 + Since the ﬁnite family associated with t2 consists of that function together
with t and 1 we try a solution
y 1 (t )
y 2 (t ) = t2 a1
a2 b1
b2 +t c1
.
c2 +1 Substituting this solution form into the system we obtain
2t
11
−1 2 a1
a2 t2 a1
a2 + b1
b2 +t b1
b2 +1 =
c1
c2 1
0 +t + t2 0
.
1 Equating the vector coeﬃcients of t2, t and 1, respectively, we obtain the
equations
11
a1
0
0
=
;
+
−1 2
a2
1
0
11
−1 2 b1
b2 1
0 + 11
−1 2 c1
c2 −2 a1
a2
b1
b2 = 1
3 = 1
3 2 −1
11 2 −1
11 = 0
.
0 11
−1 2 We can see easily that the inverse of
we have = b1
b2 − a1
a2 1
−1 2
3 0
−1
− is
= 1
3 1
0 1
3 0
;
0 2 −1
. Using that
11
1
;
−1
= 1
3 0
;
−1 1 2 −1
1
0
1
1
=
.
3
−1
311
9 −1
Using these vectors in the original solution form, we see that we have the
particular solution
c1
c2 y 1 (t )
y 2 (t ) = t2
3 = 1
−1 + t
3 0
−1 +
4 1
9 1
−1 = t2/3 + 1/9
.
−t2/3 − t/3 − 1/9 This method works with inhomogeneous vector functions G(t) which are
polynomials in t as long as the system matrix A is nonsingular; i.e., has
determinant diﬀerent from zero and is therefore invertible.
The vector version of the method of undetermined coeﬃcients has a particularly neat form when the inhomogeneous term is a constant vector times
an exponential. Thus, if G(t) = ert B , where B is a given constant vector,
the general linear inhomogeneous constant coeﬃcient system becomes
dY
= A Y + ert B.
dt
The method of undetermined coeﬃcients in this case corresponds to trying
for a particular vector solution of the form, for some constant vector C ,
Yp (t) = ert C.
Substituting this solution form into the system we have
r ert C = A ert C + ert B ⇒ (r I − A) C = B.
This system has the unique solution
C = (r I − A)−1 B
if and only if det(r I − A) = 0, that is, if and only if r is not an eigenvalue
of A, just as in the scalar inhomogeneous cases where r is not a root of the
characteristic polynomial associated with the homogeneous equation. The
particular solution in the present case is then
Yp (t) = ert (r I − A)
Example 2 −1 B. We consider a system with the same homogeneous part as in the ﬁrst example; namely:
d
dt y1
y2 = 11
−1 2
5 y1
y2 + ert 1
,
2 where r is any real number. We can verify quite immediately that the eigen√
values of the matrix appearing here are −3 ± i 3 /2, so a real r will never be an eigenvalue of the matrix. We try for a particular solution
y 1 (t )
y 2 (t ) = ert c1
c2 c1
c2 = and obtain the equation
r − 1 −1
1
r−2 1
.
2 Solving for c1 and c2 we have
c1 = r2 r
2r − 3
, c2 = 2
− 3r + 3
r − 3r + 3 and thus the particular solution is
y 1 (t )
y 2 (t ) ert
=2
r − 3r + 3 r
.
2r − 3 This is then added to the general solution of the corresponding homogeneous
system to obtain the general solution of the inhomogeneous system.
Linear constant coeﬃcient systems with inhomogeneous terms of the form
G(t) = eρ t cos σ t B or G(t) = eρ t sin σ t B can be treated in two equivalent
ways. The ﬁrst is to try for a particular solution in the form
Yp (t) = eρt cos σt C + eρt sin σt D,
Where C and D are vectors to be determined. We substitute this form
into the inhomogeneous system and, assuming ρ ± i σ are not eigenvalues
of A, identify C and D. The second way is to realize that eρt cos σt B and
eρt sin σt B are the real and imaginary parts, respectively, of e(ρ+iσ)t B . We
would then replace the original inhomogeneous system by
dZ
= A Z + e(ρ+iσ)t B
dt
and, proceeding as before, obtain the complex solution
Z (t) = e(ρ+iσ)t (r I − A)
6 −1 B. The desired particular solution Yp (t) is then the real part of Z (t) in the
cos σt case and the imaginary part of this solution in the sin σt case. The
arithmetic is much the same either way but it is a little easier to organize
the work following the complex solution route.
Example 3 Consider the inhomogeneous system
d
dt y1
y2 11
−1 2 = y1
y2 + sin ωt 1
.
1 We replace this system by the complex inhomogeneous system
d
dt z1
z2 = 11
−1 2 z1
z2 1
.
1 + eiωt Using the formula developed earlier a solution is then Z (t) = eiωt
where
c1
c2
= (i ω )2 = 1
− 3ω i + 3 =
= iω − 1
−1
1
iω − 2 1
(3 − ω 2 )2 + 9 ω 2 −1 1
1 iω − 2
1
−1
iω − 1 3 − ω 2 + 3ω i
(3 − ω 2 )2 + 9 ω 2 c1
,
c2 1
1 iω − 1
iω − 2 − (3 − ω 2 )2 − 3 ω 2 − 3 ω i + (3 − ω 2)ω i
.
−2 (3 − ω 2 ) − 3 ω 2 − 6 ω i + (3 − ω 2 )ω i Multiplying this vector by eiωt and taking the imaginary part of the result
we obtain a particular solution of the original system in the form
Yp (t) = y 1 (t )
y 2 (t )
− = cos ωt
(3 − ω 2)2 + 9 ω 2 − 3 ω + (3 − ω 2 )ω
−6 ω + (3 − ω 2 )ω (3 − ω 2 ) + 3 ω 2
2(3 − ω 2 ) + 3 ω 2 sin ωt
2
(3 − ω 2 ) + 9 ω 2 . One would again add this to the general solution of the corresponding homogeneous system to obtain the general solution Y (t, C ) of the inhomogeneous
system. 7 ...
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This note was uploaded on 01/23/2012 for the course MATH 4254 taught by Professor Robinson during the Spring '10 term at Virginia Tech.
 Spring '10
 ROBINSON
 Linear Systems

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