# Hint for the first part write as y ay 0 and multiply

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Hint: For the first part write as y ay = 0 and multiply both sides by e at . Then explain why you get d dt ( e at y ( t ) ) = 0 , y ( t 0 ) = 0 . Now you finish the argument. To show uniqueness in the second part, suppose y = ( a + ib ) y, y (0) = 0 and verify this requires y ( t ) = 0. To do this, note y = ( a ib ) y, y (0) = 0 and that d dt | y ( t ) | 2 = y ( t ) y ( t ) + y ( t ) y ( t ) = ( a + ib ) y ( t ) y ( t ) + ( a ib ) y ( t ) y ( t ) = 2 a | y ( t ) | 2 , | y | 2 ( t 0 ) = 0 Thus from the first part | y ( t ) | 2 = 0 e 2 at = 0 . Finally observe by a simple computation that (16.8) is solved by (16.9). For the last part, write the equation as y ay = f and multiply both sides by e at and then integrate from t 0 to t using the initial condition. 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 (16.10) 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 (16.10). Saylor URL: The Saylor Foundation
16.4. EXERCISES 349 25. Now suppose Φ ( t ) is an n × n matrix of the form Φ ( t ) = ( x 1 ( t ) · · · x n ( t ) ) (16.11) where x k ( t ) = A x k ( t ) . Explain why Φ ( t ) = A Φ ( t ) if and only if Φ ( t ) is given in the form of (16.11). Also explain why if c F n , y ( t ) Φ ( t ) c solves the equation y ( t ) = A y ( t ) . 26. In the above problem, consider the question whether all solutions to x = A x (16.12) are obtained in the form Φ ( t ) c for some choice of c F n . In other words, is the general solution to this equation Φ ( t ) c for c F n ? Prove the following theorem using linear algebra. Theorem 16.4.1 Suppose Φ ( t ) is an n × n matrix which satisfies Φ ( t ) = A Φ ( t ) . Then the general solution to (16.12) is Φ ( t ) c if and only if Φ ( t ) 1 exists for some t. Fur- thermore, if Φ ( t ) = A Φ ( t ) , then either Φ ( t ) 1 exists for all t or Φ ( t ) 1 never exists for any t. (det (Φ ( t )) is called the Wronskian and this theorem is sometimes called the Wronskian alter- native.) Hint: Suppose first the general solution is of the form Φ ( t ) c where c is an arbitrary constant vector in F n . You need to verify Φ ( t ) 1 exists for some t . In fact, show Φ ( t ) 1 exists for every t . Suppose then that Φ ( t 0 ) 1 does not exist. Explain why there exists c F n such that there is no solution x to c = Φ ( t 0 ) x By the existence part of Problem 24 there exists a solution to x = A x , x ( t 0 ) = c but this cannot be in the form Φ ( t ) c . Thus for every t , Φ ( t ) 1 exists. Next suppose for some t 0 , Φ ( t 0 ) 1 exists. Let z = A z and choose c such that z ( t 0 ) = Φ ( t 0 ) c Then both z ( t ) , Φ ( t ) c solve x = A x , x ( t 0 ) = z ( t 0 ) Apply uniqueness to conclude z = Φ ( t ) c . Finally, consider that Φ ( t ) c for