solutions-manual.131-140 - Exercises 131 Thus from the...

• Test Prep
• 10

This preview shows page 1 - 3 out of 10 pages.

Exercises 131 Thus from the product rule which you can easily verify from the usual proof of the product rule to be also true for complex valued functions, d dt | z ( t ) | 2 = d dt ( z z ) = z z + z z = ( a + ib ) z z + z ( a - ib ) z = 2 a | z | 2 , | z | ( t 0 ) = 0 Therefore, from the first part, | z | = 0 and so y = y 1 . Note that this implies that ( e i ( a + ib ) t ) = ( a + ib ) e i ( a + ib ) t where e i ( a + ib ) t is given above. Now consider the last part. Sove y = ay + f, y ( t 0 ) = y 0 . y - ay = f, y ( t 0 ) = y 0 Multiply both sides by e a ( t t 0 ) d dt parenleftBig e a ( t t 0 ) y parenrightBig = e a ( t t 0 ) f ( t ) Now integrate from t 0 to t. Then e a ( t t 0 ) y ( t ) - y 0 = integraldisplay t t 0 e a ( s t 0 ) f ( s ) ds Hence y ( t ) = e a ( t t 0 ) y 0 + e a ( t t 0 ) integraldisplay t t 0 e a ( s t 0 ) f ( s ) ds = e a ( t t 0 ) y 0 + e at integraldisplay t t 0 e as f ( s ) ds 24. Now consider A an n × n matrix. By Schur’s theorem there exists unitary Q such that Q 1 AQ = T where T is upper triangular. Now consider the first order initial value problem x = A x , x ( t 0 ) = x 0 . Show there exists a unique solution to this first order system. Hint: Let y = Q 1 x and so the system becomes y = T y , y ( t 0 ) = Q 1 x 0 (2.4) Now letting y = ( y 1 , ··· , y n ) T , the bottom equation becomes y n = t nn y n , y n ( t 0 ) = ( Q 1 x 0 ) n . Then use the solution you get in this to get the solution to the initial value problem which occurs one level up, namely y n 1 = t ( n 1)( n 1) y n 1 + t ( n 1) n y n , y n 1 ( t 0 ) = ( Q 1 x 0 ) n 1 Continue doing this to obtain a unique solution to 2.4. There isn’t much to do. Just see that you understand it. In the above problem, equations of the above form are shown to have a unique solution and there is even a formula given. Thus there is a unique solution to y = T y , y ( t 0 ) = Q 1 x 0 Saylor URL: The Saylor Foundation
132 Exercises Then let y = Q 1 x and plug this in. Thus Q 1 x = T Q 1 x , Q 1 x ( t 0 ) = Q 1 x 0 Then x = QT Q 1 x = A x , x ( t 0 ) = x 0 . The solution is unique because of uniqueness of the solution for y . You can go backwards, starting with x and defining y Q 1 x and then there is only one thing y can be and so x is also unique. 25. Now suppose Φ ( t ) is an n × n matrix of the form Φ ( t ) = ( x 1 ( t ) ··· x n ( t ) ) (2.5) where x k ( t ) = A x k ( t ) . Explain why
26. In the above problem, consider the question whether all solutions to x = A x (2.6) are obtained in the form Φ ( t ) c for some choice of c F n . In other words, is the general
• • • 