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Transcript
M392: Linear systems of ordinary differential equations Definitions: is an eigenvalue of an m by m matrix A provided the eigenvector equation Ax = x with x = the m by 1 matrix x1 x2 , , , , xm T has nontrivial solutions x . The trivial solution is x = 0 = (0 0 ,,, 0). The eigenvector equation, as we will see below, can be rewritten as a linear system of m equations in m unknowns. If is an eigenvalue of A, then every solution x of Ax = x is called an eigenvector of A corresponding to . Problem: 2 2 1) Find eigenvalues and eigenvectors of the matrix A = 1 3 2) Use your answer to solve the system of ordinary differential y ' 2 y1 2 y2 equations 1 for functions y ' y 3 y 2 1 2 y1 (0) 2 y1 (t ) subject to initial conditions y2 (t ) y 2 (0) 1 Solution, Part I. Find eigenvalues and eigenvectors: Method: To solve Ax = x, move everything to the left side and rewrite it as a system of linear equations as follows: Ax x 0 Ax Ix 0 ( A I ) x 0 Recall that a system of m equations in m unknowns has exactly one solution if the determinant of the coefficient matrix is nonzero, and either no solutions or many solutions if that determinant is zero. Since the system of equations Ax = x has a nontrivial solution x by definition of eigenvalues, and also has the trivial solution, the same is true of the rewritten system ( A I ) x 0 . It follows that this system has infinitely many solutions. Therefore its coefficient matrix ( A I ) has determinant zero. Calculate as follows. | A I | 0 2 2 0 0 1 3 0 2 2 . 0 1 3 (2 )(3 ) 2(1) 0 2 5 4 0 ( 1)( 4) 0 and so the eigenvalues are 4 and 1. For each eigenvalue , solve the eigenvector equation x ( A I ) x 0 where x 1 . In other words, solve x2 2 x1 0 2 1 3 x2 0 Start with eigenvalue 4 and plug in to get 2 x1 0 2 4 which says that 1 3 4 x2 0 2 x1 2 x2 0 x1 x2 0 Row ops 0 0 1x1 1x2 0 This is a very simple system with solutions x2 arbitrary r x1 r x r 1 Thus the eigenvalue 4 has eigenvectors 1 r . x2 r 1 2 2 r 4r =4x Check: Ax = 1 3 r 4r Warning: check that the coefficient matrix determinant 2 2 | ( A 4I ) | = is really zero. If not, go back and look for 1 1 your mistake. x s In the same way x1 2 x2 0 Solution 2 x1 2s x 2s 2 the eigenvalue 1 has eigenvectors 1 s x s 1 2 y’ = 3y solution is y(t) = c e^(3t) y '1 2 y1 2 y 2 Solution, Part II. To solve the system let y= y ' y 3 y 2 1 2 and write the system as y′ = A y . Suppose A is a 1 by 1 matrix, ( a number) and y = y(t). Then the solution of y′ = A y is y(t ) ce At y1 y2 To solve the general case, Guess y = et c and plug into y′ = Ay : t y1 t c1 e c1 e t y2 c2 e c2 e t c = A e t c Divide by e t to obtain Ac = c Therefore, is an eigenvalue of A with eigenvector c and a solution is y = e t c . Corresponding to the two eigenvalues (with corresponding eigenvectors) for the matrix A we get two solutions to the system of ODE’s: 1 y For 4 with eigenvectors r we have yI = 1 1 y2 re = 4 t re ------------------------------------------------------------y 2 For 1 with eigenvectors s we have yII = 1 1 y2 4t 2se = se 1t 1t We found that the system of differential equations y′ = Ay has two solutions yI and yII . Thus y ' I Ay I and y ' II Ay II y I ' y II ' Ay I Ay II Add these equations: ( y I y II )' A( y I y II ) It follows that yI yII also solves y′ = Ay . Thus we can add the above two solutions to get solutions 4t t y1 (t ) re 2se (*) 4t t y ( t ) re se 2 It’s possible to show that there are no other solutions, and so (*) is called the general solution to the system of ODE’s. y (0) 2 Finally, plug the given initial conditions 1 into the y ( 0 ) 1 2 2 r 2 s general solution to get 1 r s r 4 / 3 Solve for r and s to get s 1 / 3 4 t 2 4t y ( t ) e e 1 3 3 Plug into (*) to get the particular solution 4 1 y 2 (t ) e t e 4 t 3 3 Exercise: Check this solution by plugging into the original system of ODE’s. Sample problems 1. a) Find the eigenvalues and eigenvectors of the matrix 5 2 A = 2 2 b) Use your answer from a) to solve the following system of differential equations for y1 (t ) and y 2 (t ) subject to the initial conditions y1 (0) = 1 and y 2 (0) = 2: y1 5 y1 2 y 2 y 2 2 y1 2 y 2 2. Solve the following simultaneous differential equations for y1 (t ) and y 2 (t ) subject to initial conditions y1 (0) = 2 and y 2 (0) = -1 . You must proceed by first finding the eigenvalues and eigenvectors of an appropriate matrix. No credit for any other method! y1 2 y1 1 y2 y2 2 y1 5 y2