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Linear Differential Equations
Consider a differential equation of the form
x˙ = A(t)x + h(t) (1) where A(t) is a continuous real or complex n × n matrix valued func- tion and h(t) is a continuous n − vector−valued function. We assume that A(t), h(t) are defined for all t ∈ (−∞, ∞). Let V = Rn^ or Cn. We assume both A(·), h(t) are V − valued. Given an equation (1), the associated homogeneous equation is the equa- tion
x˙ = A(t)x (2) We have already proved that every initial value problem for (1) has solu- tions defined for all t. Basic Facts:
- If x(t), y(t) are solutions of (1), then z(t) = y(t) − x(t) is a solution of (2).
- If x(t) is a solution of (1) and z(t) is a solution of (2), then y(t) = x(t) + z(t) is a solution of (1).
Thus, if one knows all solutions of (2), and one knows one particular solution of (1), then one can get all solutions of (1). The general solution of (1) is a vector-valued expression
φ(t, c) (3) involving a vector c = (c 1 ,... , cn) of constants so that every solution can be represented as (3) for a unique choice of the vector c. We now study the general properties of equation (2). Proposition. The set of solutions to (2) form an n−dimensional linear subspace of the vector space of C^1 dunctions from R to V. Proof. Let S be the set of solutions. Clearly, S ⊂ C^1 (R, V ). (A) S is a linear subspace: Suppose x(t), y(t) are in S, and a, b are scalars.
Then, z(t) = ax(t) + by(t) satisfies
z˙(t) = a x˙(t) + b y˙(t) = aA(t)x(t) + bA(t)y(t) = A(t)(z(t))
so, z(t) ∈ S. (B) S is n−dimensional: We need to find n solutions x 1 (t),... , xn(t) such that every solutions can be uniquely expressed as
x(t) =
∑^ n
i=
αixi(t) (4)
where αi are scalars. Let xi(t) be the unique solution such that xi(0) = ei where ei is the i − th standard basis vector of V. Let x(t) be an arbitrary solution of (2). Then, there are scalars αi such that x(0) =
∑ i αiei. Consider^ y(t) =^
∑ i αixi(t). Then, both x(·), y(·) are solutions and they agree at t = 0. By uniqueness of solutions, we have x(t) = y(t) for all t. Now, if x(t) =
∑ i αixi(t) =^
∑ i βixi(t) for all^ t, then this is true for^ t^ = 0, so αi = βi ∀i since {ei} is a basis for V. QED Definition. A set {y 1 (t),... , y`} of solutions to (2) is called linearly independent if it is a linearly independent subset of C^1 (R, V ). That is, whenever
∑^ `
i=
αiyi(t) = 0 ∀t
we have αi = 0 ∀i. A maximal linearly independent set of solutions is called a fundamental set of solutions. Thus, a set {y 1 (t),... , yn(t)} is a fundamental set of solutions iff the set is a basis for the subspace S of C^1 (R, V ). Note, that, since any two bases of a vector space have the same number of elements, we have ` = n for any fundamental set of solutions. We now want a criterion for a set {y 1 (t),... , yn(t)} of solutions to be a fundamental set.
Lemma 1. Let A(t) = (aij (t)) by a continuous n × n matrix function of t, and let h > 0. Then,
det(I + A(t)h) = 1 + tr(A(t))h + O(h^2 ). (8)
Here I is the n × n identity matrix and O(h^2 ) denotes any function R(h) such that there is a constant C > 0 such that
lim sup h→ 0
R(h) h^2
< C.
Proof of Lemma 1. We use induction of n. It is trivial for n = 1 since we can use the zero function for O(h^2 ). Assume the Lemma is true for n − 1. Let B = I + A(t)h. Expanding by minors down the first column of B gives
det(I + A(t)h) = (1 + a 11 (t)h)det(B(1 | 1)
∑^ n
j=
(−1)1+j^ haj 1 (t)det(B(j | 1). (9)
Now, letting In− 1 denote the (n − 1) × (n − 1) identity matrix, we have
B(1 | 1) = In− 1 + hA(1 | 1).
So, by induction, we have the first entry on the right side of (9) equals
(1 + a 11 (t)h)(1 + h trA(1 | 1) + O(h^2 )) = 1 + a 11 (t)h + h trA(1 | 1) + O(h^2 ) = 1 + h trA(t) + O(h^2 ).
On the other hand, each entry in the sum in the second term in equation (9) has an h in its first column. So, this whole sum is O(h^2 ). QED. We need a standard result from the calculus of maps from Rn^ to R. Lemma 2. Let ψ : Rn^ → R be a C^1 function defined on an open set D ⊂ Rn, and let u, v be vectors in D such that the closed line segment joining
u to u + v is completely contained in D (i.e. {u + tv : 0 ≤ t ≤ 1 } ⊂ D). Then, there is a constant K > 0 such that
| ψ(u + v) − ψ(u) | ≤ K| v |. (10) Proof. Let ∇ψ(x) denote the gradient of ψ at the point x. Since the map t → | ∇ψ(u + tv) | is continuous on the closed unit interval [0, 1], we have that
K def = sup 0 ≤t≤ 1
| ∇ψ(u + tv) | < ∞.
The map η(t) = ψ(u + tv) is a C^1 real-valued function of the variable t on the closed interval [0, 1], so the Mean-Value theorem gives that there is a 0 < t 0 < 1 such that
ψ(u + v) − ψ(u) = η(1) − η(0) = η′(t 0 )(1 − 0) = η′(t 0 ).
By the Chain Rule for functions of several variables, we have that
η′(t 0 ) = ∇ψ(u + t 0 v) · (v)
where · denotes the usual dot product (inner product) in Rn. By the Cauchy-Schwarz inequality, we have
| ψ(u + v) − ψ(u) | = | η′(t 0 ) | = | ∇ψ(u + t 0 v) · (v) | ≤ | ∇ψ(u + t 0 v) || ·(v) | ≤ K| v |
QED. We now apply this to the determinant function, det, on n × n matrices, thinking of these as elements of Rn 2 . Lemma 3. Let A and B be two n×n matrices. Then, there is a constant K such that
| det(A + B) − det(A) | ≤ K| B | (11)
W ′
W
= trA(t)
or
log W (t) |W W^ ((tt) 0 )=
∫ (^) t
t 0
trA(s)ds
or
W (t) = W (t 0 )exp
(∫ (^) t
t 0
trA(s)ds
)
As usual, we let o(h) denote a function r(h) such that
lim h→ 0
r(h) h
Let Φ(t) be the matrix whose columns are the solutions y 1 (t),... , yn(t), so that we have the matrix equation
Φ′(t) = A(t)Φ(t).
Now, we have
Φ(t + h) = Φ(t) + Φ′(t)h + o(h) = Φ(t) + A(t)Φ(t)h + o(h) = (I + A(t)h)Φ(t) + o(h)
which, in turn, gives
W (t + h)det((I + A(t)h)Φ(t) + o(h))
Applying Lemma 3 with A = (I + A(t)h)Φ(t) and B = o(h), we get
det((I + A(t)h)Φ(t) + o(h)) = det((I + A(t)h)Φ(t)) + O(o(h)) = det((I + A(t)h)Φ(t)) + o(h)
Using the fact that the det(AB) = det(A)det(B) for all A, B and Lemma 1 gives
W (t + h) = W (t)(1 + tr A(t)h + O(h^2 )) + o(h) = W (t) + W (t)(tr A(t)h) + o(h).
So,
W (t + h) − W (t) h
= W (t)tr A(t) +
o(h) h
Letting h → 0 gives
W ′^ = trA(t)W
as required. QED. Definition. The determinant W (t) = W (y 1 (t),... , yn(t)) of the set of solutions {y 1 ,... , yn} is called the Wronskian of this set of solutions. This a set of solutions is a fundamental set if and only if its Wronskian is not zero for some (or any ) t. When, {y 1 ,... , yn} is a fundamental set of solutions, we form the matrix Φ(t) whose columns are the y′ is and call this matrix a fundamental matrix for (2). Thus, the general solution to (2) has the form
x(t) = Φ(t)c
where Φ(t) is any fundamnental matrix for (2) and c is a constant vector. Fact: Suppose Φ(t) is a fundamental matrix for (2) and h(t) is not iden- tically zero. Then, one can always find a particular solution for (1) of the form
xp(t) = Φ(t)v(t) (13) where v(t) is some non-constant vector-valued function of t. Proof. To motivate how we find v(t), first suppose that we can find a solution as in (13). Then,
