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LECTURE 27. COMPLEX SOLUTIONS AND THE FUNDAMENTAL MATRIX
Complex eigenvalues. We continue studying
(27.1) �y �^ = A�y,
where A = (aij ) is a constant n × n matrix. In this subsection, further, A is a real matrix. When A has a complex eigenvalue, it yields a complex solution of (27.1). The following principle of equating real parts then allows us to construct real solutions of (27.1) from the complex solution.
Lemma 27.1. If �y(t) = α� (t) + iβ�(t), where α� (t) and β�(t) are real vector-valued functions, is a complex solution of (27.1), then both α� (t) and β�(t) are real solutions of (27.1).
The proof is nearly the same as that for the scalar equation, and it is omitted.
Exercise. If a real matrix A has an eigenvalue λ with an eigenvector �v, then show that A also has an eigenvalue λ¯^ with an eigenvector �v¯.
Example 27.2. We continue studying A = (^) −
- Recall that^ pA(λ) =^
�^ �
λ 1 −
λ
�^ �^ has two
complex eigenvalues 1 ± 2 i.
If λ = 1 + 2i, then A − �
λI = �
i −
2 i has an eigenvector^2
i. The result of the above 1 exercise then ensures that (^) − 2 i is an eigenvector of the eigenvalue λ = 1 − 2 i.
In order to find real solutions of (27.1), we write
e(1+2i)t^21 i^ = et^ −cos 22 sin 2t^ t + iet^ 2 cos 2sin 2t^ t.
cos 2t sin 2t The above lemma then asserts that et −2 sin 2t and et 2 cos 2t are real solutions of (27.1).
Moreover, they are linearly independent. Therefore, the general real solution of (27.1) is
cos 2t sin 2t c 1 et^ −2 sin 2t + c 2 et^ 2 cos 2t.
The fundamental matrix. The linear operator T �y := �y �^ − A�y has a natural extension from vectors to matrices. For example, when n = 2, let
y 11 f 1 y 12 g 1 T (^) y = , T =. 21 f 2 y 22 g 2
Then, (^) � � � � y 11 y 12 f 1 g 1 T (^) y =. 21 y 22 f 2 g 2 In general, if A is an n × n matrix and Y = (y 1 · · · yn) is an n × n matrix, whose j-th column is yj , then T Y = T (y 1 · · · yn) = (T y 1 · · ·T yn).
In this sense, �y �^ = A�y extends to Y �^ = AY.
1
Exercise. Show that T (U + V ) = T U + T V, T (U C) = (T U )C, T (U�c) = (T U )�c, where U, V are n × n matrix-valued functions, C is an n × n matrix, and �c is a column vector.
That means, T is a linear operator defined on the class of matrix-valued functions Y differen tiable on an interval I. The following existence and uniqueness result is standard.
Existence and Uniqueness result.. If A(t) and F (t) are continuous and bounded (matrix-valued functions) on an interval t 0 ∈ I, then for any matrix Y 0 then initial value problem Y �^ = A(t)Y + F (t), Y (t 0 ) = Y 0 has a unique solution on t ∈ I.
Working assumption. A(t), F (t), and f (t) are always continuous and bounded on an interval t ∈ I.
Definition 27.3. A fundamental matrix of T Y = 0 is a solution U (t) for which |U (t 0 )| � = 0 at some point t 0.
We note that the condition |U (t 0 )| =� 0 implies that |U (t)| =� 0 for all t ∈ I. We use this fact to derive solution formulas. As an application of U (t), we obtain solution formulas for the initial value problem
�y �^ = A(t)�y + f�(t), �y(t 0 ) = y� 0. Let U (t) be a fundamental matrix of Y �^ = A(t)Y. In the homogeneous case of f�(t) = 0, let �y(t) = U (t)�c, where �c is an arbitrary column vector. Then, �y �^ = U ��c = (A(t)U )�c = A(t)(U�c) = A(t)�y, that is, y is a solution of the homogeneous system. The initial condition then determines �c and �c = U −^1 (t 0 )�y 0. Next, for a general f�(t), we use the variation of parameters by seting �y(t) = U (t)�v(t), where �v is a vector-valued function. Then, �y �^ = (U�v)�^ = U ��v + U�v �^ = A(t)U�v + U�v �^ = A(t)�y + U�v �.
Hence, U�v �^ = f�(t) and (^) � �y(t) = U (t) U −^1 (t)f�(t) dt.
Liouville’s equation. We prove a theorem of Liouville, which generalizes Abel’s identity for the Wronskian.
Theorem 27.4 (Liouville’s Theorem). If Y �(t) = A(t)Y (t) on an interval t ∈ I, then (27.2) |Y (t)|�^ = trA(t)|Y (t)|.
Proof. First, if |Y (t 0 )| = 0 at a point t 0 ∈ I, then |Y (t)| = 0 for all t ∈ I, and we are done. We therefore assume that |Y (t)| = 0� for all t ∈ I. Let Y (t 0 ) = I at a point t 0. That is, Y (t 0 ) = (y 1 (t 0 ) · · · yn(t 0 )) = (E 1 E 2 · · ·En). Here, Ej are the unit coordinate vectors in Rn, that is, the n-vector Ej has 1 in the j-th position and zero otherwise.
Lecture 27 2 18.034 Spring 2009
