2. Linear Systems of Ordinary Differential Equations
Suppose that an n×n matrix function A(t) is continuous for 0 ≤ t ≤ b, b > 0. for any vector c, consider the initial value problem
dx/dt = A(t) x
(1)
x(0) = c
(2)
There exists, for a given c, a value b0 such that (1) and (2) have a unique solution for 0 ≤ t ≤ b0. The proof of this theorem also shows that if A(t) can be extended to -b ≤ t ≤ b, we have a solution for -b1 ≤ t ≤ 0; that is, we can obtain a unique solution to (1) and (2) for -b0 ≤ t ≤ b0, where 0≤ b0 ≤ b.
Lemma 1. If x(t) satisfies (1) and (2) on -b0 ≤ t ≤ b0, then for any a ∈ R, y(t) = a x(t) satisfies the initial value problem
dy/dt = A(t) y
y(0) = a c
Lemma 2. For i = 1, 2, if x(i)(t) satisfies
dx(i)/dt = A(t) x(i)(t) , x(i)(0) = c(i)
on -bi ≤ t ≤ bi, then y(t) = x1(t) + x(2)(t) satisfies
dy/dt = A(t) t(t) , y(0) = c(1) + c(2)
on -b0 ≤ t ≤ b0, where b0 = min (b1, b2).
[The proofs of these lemmas are elementary.]
Since every c can be written as a linear combination c = a1 c(1) + ... + an c(n) where {c(1), ... , c(n)} is a basis of Rn, we see that there exists a neighbourhood of the origin in which the solutions of (1) form a vector space of dimension n.
Proposition. If A(t) is continuous on [0,b] the set of solutions of (1) on [0,b] form a vector space of dimension n.
Proof. By Lemmas 1 and 2 and the remarks following equation (2), we see that for any s ∈ [0,b], there exists a neighbourhood Ns of s on which solutions of (1) exist and form an n-dimensional vector space. Given two points s1, s2 ∈ [0,b] such that Ns1 and Ns2 intersect, the solutions can be patched up (see Appendix) to give a solution on Ns1 ∪ Ns2. By compactness of [0,b], there exists a set of points si ∈ [0,b], i = 1, ... , M such that Ns1 ∪ ... ∪ NsM = [0,b]. Putting together the solutions on each Nsi, we obtain a vector space of solutions on the whole interval.
Linear systems with constant coefficients. Consider the system
dx/dt = A x
(3)
where A is a constant matrix. The infinite series
exp(tA) = ∑k=1,...,∞tkAk/k!
converges for all values of t and A. Furthermore, it is easy to see, by differentiation of power series, that
(d/dt)[exp(tA)] = A exp(tA)
(*)
If t = 0, exp(tA) = I, the identity matrix. Multiplying (*) by any vector c, we then see that
x(t) = exp(tA) c
(4)
satisfies the initial value problem
dx/dt = A x , x(0) = c
(5)
Higher order linear differential equations. The linear differential equation
a0(t) x + a1(t) dx/dt + a2(t) d2x/dt2 + ... + an(t) dnx/dtn = 0
(6)
can be written as
x1 = x , dx1/dt = x2 , dx2/dt = x3 , ... , dxn-1/dt = xn,
b1(t) dx1/dt + ... + bn(t) dxn/dt = x1
where
bi(t) = - ai(t)/a0(t) , i = 1, ... , n
(7)
or
B(t) dx/dt = x
that is
dx/dt = B-1(t) x
(8)
where x is the vector (x, dx/dt, ... , dn-1x/dtn-1) and
/ | b1(t) | b2(t) | ... | bn-1(t) | bn(t) | \ | |||
| | 1 | 0 | ... | 0 | 0 | | | |||
B(t) = | | | 0 | 1 | ... | 0 | 0 | | | ||
| | . | . | ... | . | . | | | |||
\ | 0 | 0 | ... | 1 | 0 | / |
(9)
A solution of (8) exists in any interval where a0(t) ≠ 0. Thus the higher order differential equation (6) has been reduced to a first order system of linear equations (8).
References. This page only develops the most elementary part of the theory of linear equations, which will be needed in the sequel. Further information may be found in the following books: