MIT OpenCourseWarehttp://ocw.mit.edu 18.034 Honors Differential Equations Spring 2009 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.� � � � � � � � � � � � � � � � � � � � � � � � � � � � � � 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 = −14 11 . Recall that pA(λ) = ���1 −−4 λ 1 − 1 λ ��� has two complex eigenvalues 1 ± 2i. If λ = 1 + 2i, then A −� λI = � −−24 i −12i has an eigenvector 21 i . The result of the above 1exercise then ensures that −2i is an eigenvector of the eigenvalue λ = 1 − 2i. In order to find real solutions of (27.1), we write e(1+2i)t 1= e t cos 2t + iet sin 2t.2i −2 sin 2t 2 cos 2t cos 2t sin 2tThe 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 c1e t + c2e t . −2 sin 2t 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 y11 f1 y12 g1T = , T = . y21 f2 y22 g2 Then, � � � � y11 y12 f1 g1T = . y21 y22 f2 g2 In general, if A is an n × n matrix and Y = (y1 ) is an n × n matrix, whose j-th column is yj , · · · ynthen T Y = T (y1 yn) = (T y1 T yn).· · · · · · In this sense, �y � = A�y extends to Y � = AY . 1Exercise. Show that T (U + V ) = T U + T V, T (UC) = (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 t0 ∈ I, then for any matrix Y0 then initial value problem Y � = A(t)Y + F (t), Y (t0) = Y0 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 (t0)| �= 0 at some point t0. We note that the condition |U (t0)| =� 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(t0) = 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(t0)�y0. 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 (t0)| = 0 at a point t0 ∈ 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 (t0) = I at a point t0. That is, Y (t0) = (y1(t0) yn(t0)) = (E1 E2 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 2009We use the derivative formula for the determinant d |Y (t)|� = det(y1(t) . . . yn(t))dt = det(y1�(t) y2(t) yn(t)) + det(y1(t) y2�(t) yn(t)) + + det(y1(t) yn�(t)).· · · · · · · · · · · · This formula is based on the Laplace expansion formula for determinant, and we do not prove it here. Since yj� (t0) = A(t0)yj (t0) = A(t0)Ej = Aj (t0), where Aj (t) is the jth column of A(t), evaluating the above determinant formula at t = t0 we obtain |Y (t0)|� = det(A1(t0) E2 · · · En) + det(E1 A2(t0) · · · En) + · · · + det(E1 E2 · · · An(t0)) = a11(t0) + a22(t0) + + ann(t0) = trA(t0).· · · Thus, (27.2) holds at t0. In general, let C = Y (t0)−1. Then U (t) = Y (t)C satisfies U� = A(t)U, U (t0) = I. Therefore, by the argument above |U(t0)|� = trA(t0)|U(t0)| = trA(t0). Since d d dt(|Y (t)C|) = dt(|Y (t)||C|) = |Y (t)|�|C|, at t = t0, it follows that trA(t0) = |Y (t0)|�|Y (t0)|−1. Since t0 is arbitrary, the proof is complete. � Lecture 27 3 18.034 Spring
View Full Document
Unlocking...