10.2: Linear Systems of Differential Equations
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( \newcommand{\kernel}{\mathrm{null}\,}\)
A first order system of differential equations that can be written in the form
y′1=a11(t)y1+a12(t)y2+⋯+a1n(t)yn+f1(t)y′2=a21(t)y1+a22(t)y2+⋯+a2n(t)yn+f2(t)⋮y′n=an1(t)y1+an2(t)y2+⋯+ann(t)yn+fn(t)
is called a linear system.
The linear system Equation ??? can be written in matrix form as
[y′1y′2⋮y′n]=[a11(t)12(t)⋯1n(t)a21(t)22(t)⋯2n(t)⋮⋮⋱⋮an1(t)n2(t)⋯nn(t)][y1y2⋮yn]+[f1(t)f2(t)⋮fn(t)],
or more briefly as
y′=A(t)y+f(t),
where
y=[y1y2⋮yn],A(t)=[a11(t)12(t)⋯1n(t)a21(t)22(t)⋯2n(t)⋮⋮⋱⋮an1(t)n2(t)⋯nn(t)],andf(t)=[f1(t)f2(t)⋮fn(t)].
We call A the coefficient matrix of Equation ??? and f the forcing function. We’ll say that A and f are continuous if their entries are continuous. If f=0, then Equation ??? is homogeneous; otherwise, Equation ??? is nonhomogeneous.
An initial value problem for Equation ??? consists of finding a solution of Equation ??? that equals a given constant vector
k=[k1k2⋮kn].
at some initial point t0. We write this initial value problem as
y′=A(t)y+f(t),y(t0)=k.
The next theorem gives sufficient conditions for the existence of solutions of initial value problems for Equation ???. We omit the proof.
Theorem 10.2.1 : Existence
Suppose the coefficient matrix A and the forcing function f are continuous on (a,b), let t0 be in (a,b), and let k be an arbitrary constant n-vector. Then the initial value problem
y′=A(t)y+f(t),y(t0)=k
has a unique solution on (a,b).
Example 10.2.1
- Write the system y′1=2y1+2y2+2e4ty′2=2y1+2y2+2e4t in matrix form and conclude from Theorem 10.2.1 that every initial value problem for Equation ??? has a unique solution on (−∞,∞).
- Verify that y=15[87]e4t+c1[11]e3t+c2[1−1]e−t is a solution of Equation ??? for all values of the constants c1 and c2.
- Find the solution of the initial value problem y′=[1221]y+[21]e4t,y(0)=15[322].
Solution a
The system Equation ??? can be written in matrix form as
y′=[12]y+[21]e4t.
An initial value problem for Equation ??? can be written as
y′=[1221]y+[21]e4t,y(t0)=[k1k2].
Since the coefficient matrix and the forcing function are both continuous on (−∞,∞), Theorem 10.2.1 implies that this problem has a unique solution on (−∞,∞).
Solution b
If y is given by Equation ???, then
Ay+f=15[1221][87]e4t+c1[1221][11]e3t+c2[1221][1−1]e−t+[21]e4t=15[2223]e4t+c1[33]e3t+c2[−11]e−t+[21]e4t=15[3228]e4t+3c1[11]e3t−c2[1−1]e−t=y′.
Solution c
We must choose c1 and c2 in Equation ??? so that
15[87]+c1[11]+c2[1−1]=15[322],
which is equivalent to
[111−1][c1c2]=[−13].
Solving this system yields c1=1, c2=−2, so
y=15[87]e4t+[11]e3t−2[1−1]e−t
is the solution of Equation ???.
Note
The theory of n×n linear systems of differential equations is analogous to the theory of the scalar n-th order equation P0(t)y(n)+P1(t)y(n−1)+⋯+Pn(t)y=F(t) as developed in Sections 9.1. For example by rewriting Equation ??? as an equivalent linear system it can be shown that Theorem 10.2.1 implies Theorem 9.1.1.