MA2327, Problem set #3 (Practice problems with solutions)

Transcription

1 MA2327, Problem set #3 (Practice problems with solutions) 5 2 Compute the matrix exponential e ta in the case that A = Compute the matrix exponential e ta in the case that A = 5 3 Find the unique solution of the initial value problem y (t) 3y (t)+4y(t) = 0, y(0) =, y (0) = 4, y (0) = 3 4 Use the matrix exponential to solve the initial value problem 6 3 y (t) = Ay(t), A =, y(0) = Suppose that Φ(t) is a fundamental matrix for the system y (t) = A(t)y(t) Find the matrix A(t) and determine the unique solution that satisfies y(0) = y 0 when sint cost Φ(t) = e sint, y cost sint 0 = 0 6 Suppose that Φ(t) is a fundamental matrix for the system y (t) = Ay(t) Find the constant matrix A and compute its matrix exponential e ta when sint cost Φ(t) = cost sint Compute the matrix exponential e ta in the case that A = The characteristic polynomial of the given matrix is f(λ) = λ 2 (tra)λ+deta = λ 2 6λ+9 = (λ 3) 2 Thus, λ = 3 is the only eigenvalue and it is easy to check that the only eigenvector is v =

2 Pick any nonzero vector v that is not an eigenvector and let v 2 = (A λi)v, say 2 v =, v 0 2 = (A 3I)v = 2 These vectors form a Jordan basis for A and we also have 2 3 e B = = J = B AB = = e tj 3t = te 3t e 3t As for the exponential of the original matrix A, this is given by 2 +2t 2t e ta = Be tj B = e 3t = e 3t 0 2 t 0 /2 2t 2t 2 Compute the matrix exponential e ta in the case that A = The characteristic polynomial of the given matrix is 5 5 f(λ) = λ 2 (tra)λ+deta = λ 2 6λ+0 = (λ 3) 2 + Thus, the eigenvalues are λ = 3±i and the corresponding eigenvectors turn out to be v =, v 2 i 2 = 2+i This implies that A is diagonalisable and that we also have 3+i B = = J = B AB = 2 i 2+i 3 i e = e tj = 3t e it e 3t e it As for the exponential of the original matrix A, this is given by e e ta = Be tj B = e 3t it 2 i 2+i e it 2+i 2i i 2 = e3t (i+2)e it +(i 2)e it e it e it 2i 5e it 5e it (i 2)e it +(i+2)e it Using the formula e ±it = cost±isint, we may thus conclude that e ta = e3t 2icost+4isint 2isint 2i 0isint 2icost 4isint cost+2sint sint = e 3t 5sint cost 2sint

3 3 Find the unique solution of the initial value problem y (t) 3y (t)+4y(t) = 0, y(0) =, y (0) = 4, y (0) = 3 This is a homogeneous equation and the associated characteristic polynomial is λ 3 3λ 2 +4 = 0 Noting that λ = is a root, it is easy to check that λ 3 3λ 2 +4 = (λ+)(λ 2 4λ+4) = (λ+)(λ 2) 2 In particular, λ = 2 is a double root and λ = is a simple root, so y(t) = c e 2t +c 2 te 2t +c 3 e t for some constants c,c 2,c 3 Next, we turn to the initial conditions and we note that y(t) = c e 2t +c 2 te 2t +c 3 e t, y (t) = 2c e 2t +c 2 e 2t +2c 2 te 2t c 3 e t, y (t) = 4c e 2t +4c 2 e 2t +4c 2 te 2t +c 3 e t This gives rise to a system of three equations in three unknowns, namely = y(0) = c +c 3, 4 = y (0) = 2c +c 2 c 3, 3 = y (0) = 4c +4c 2 +c 3 On the other hand, row reduction of the associated augmented matrix gives Thus, the unique solution of the system is (c,c 2,c 3 ) = (2,, ) and we finally get y(t) = c e 2t +c 2 te 2t +c 3 e t = 2e 2t te 2t e t 4 Use the matrix exponential to solve the initial value problem 6 3 y (t) = Ay(t), A =, y(0) = 2 3 The characteristic polynomial of the given matrix is f(λ) = λ 2 (tra)λ+deta = λ 2 8λ+5 = (λ 5)(λ 3)

4 Since the eigenvalues are distinct, A is diagonalisable, and it is easy to check that 3 v =, v 2 = are eigenvectors corresponding to λ = 5 and λ = 3, respectively This implies that 3 5 e B = = J = B AB = = e tj 5t = 3 e 3t In particular, the matrix exponential of the original matrix A is given by e ta = Be tj B = 3 e 5t 2 e 3t 3 = 3e 5t e 3t 3e 3t 3e 5t 2 e 5t e 3t 3e 3t e 5t Next, we turn to the initial value problem Since Φ(t) = e ta is a fundamental matrix, every solution has the form y(t) = Φ(t)c = e ta c for some vector c and this implies that y(0) = e 0 c = c = y(t) = e ta y(0) = y(t) = 3e 5t e 3t 3e 3t 3e 5t 2 e 5t e 3t 3e 3t e 5t = 3 [ 4e 3t 3e 5t 4e 3t e 5t ] 5 Suppose that Φ(t) is a fundamental matrix for the system y (t) = A(t)y(t) Find the matrix A(t) and determine the unique solution that satisfies y(0) = y 0 when sint cost Φ(t) = e sint, y cost sint 0 = 0 The fundamental matrix is itself a matrix solution of the system, so one has Φ (t) = A(t)Φ(t) = A(t) = Φ (t) Φ(t) When it comes to the derivative Φ (t), an application of the product rule gives sint cost cost sint Φ (t) = e sint cost +e sint cost sint sint cost On the other hand, the inverse of Φ(t) is easily seen to be sint cost Φ(t) = e sint cost sint

5 Once we now combine the last three equations, we find that sint cost sint cost cost sint sint cost A(t) = cost + cost sint cost sint sint cost cost sint Using this fact together with the identity sin 2 t+cos 2 t =, we conclude that 0 0 cost A(t) = cost + = 0 0 cost Next, we turn to the initial value problem Since Φ(t) is a fundamental matrix, we have y(t) = Φ(t)c = y(0) = Φ(0)c = c = Φ(0) y(0) = y(t) = Φ(t)Φ(0) y(0) Since y(0) = e by assumption, the unique solution of the problem is thus sint cost 0 y(t) = e sint cost sint 0 0 sint cost 0 cost = e sint = e sint cost sint 0 0 sint 6 Suppose that Φ(t) is a fundamental matrix for the system y (t) = Ay(t) Find the constant matrix A and compute its matrix exponential e ta when sint cost Φ(t) = cost sint The fundamental matrix is itself a matrix solution of the system, so one has Φ (t) = AΦ(t) = A = Φ (t) Φ(t) Using the given formula for Φ(t), it is thus easy to check that cost sint sint cost 0 A = = sint cost cost sint 0 To compute the matrix exponential e ta, one may now proceed in the usual way to find the eigenvalues and eigenvectors of A However, it is also possible to compute e ta by relating it to the fundamental matrix Φ(t) as follows Consider the initial value problem y (t) = Ay(t), y(0) = y 0

6 Since the unique solution can be expressed in the form y(t) = Φ(t)c, we have y(0) = Φ(0)c = c = Φ(0) y(0) = y(t) = Φ(t)c = Φ(t)Φ(0) y(0) On the other hand, e ta is itself a fundamental matrix, so the solution is also given by y(t) = e ta c = y(0) = c = y(t) = e ta c = e ta y(0) These two expressions must obviously coincide for any y(0) by uniqueness, hence ] [ sint cost 0 e ta = Φ(t)Φ(0) = cost sint 0 sint cost 0 cost sint = = cost sint 0 sint cost