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

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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

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