Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Rotation matrix wikipedia , lookup
Perturbation theory wikipedia , lookup
Simplex algorithm wikipedia , lookup
Computational electromagnetics wikipedia , lookup
Non-negative matrix factorization wikipedia , lookup
Linear algebra wikipedia , lookup
Inverse problem wikipedia , lookup
MAT 274 HW 12 Hints c Bin Cheng MAT 274 HW 12 Hints 1. Problem 15 on Page 395. Solution. To find critical point (equilibrium) solution, set dx/dt = dy/dt = 0 and get an algebraic system x−y =0 5x − 3y − 2 = 0 First equation gives x = y. Plug it into the second equation 5y − 3y − 2 = 0. So y = 1 and thus x = 1. We found the only critical point (x0 , y0 ) = (1, 1). To classify (x0 , y0 ) as to type and stability, we calculate the Jacobian matrix. Let f (x, y) = x − y and g(x, y) = 5x − 3y − 2 so that the original system is of the standard form dx = f (x, y) dt dy = g(x, y) dt Then, the Jacobian matrix, by definition, is ! fx fy J(x, y) = gx gy Plug in the definitions of f and g to get J(x, y) = ! 1 −1 . 5 −3 (Remark: J is a constant matrix, which is expected for linear systems. The Jacobian matrix for nonlinear system is expected to depend on x, y.) Now, evaluate J(x0 , y0 ) at the critical point (1,1) and get the same constant matrix ! 1 −1 J(1, 1) = . 5 −3 Then, calculate the eigenvalues of the Jacobian matrix at (1, 1) to determine the type and stability of (1, 1). ! 1−λ −1 det = λ2 + 2λ + 2 = 0 5 −3 − λ P age 1 MAT 274 HW 12 Hints c Bin Cheng =⇒ λ1,2 = −1 ± i The eigenvalues are complex with negative real part. Therefore, it is a spiral point (by complex) and stable (by negative real part), i.e. a spiral sink. 2. Problem 17 on Page 395. Solution. Follow the same steps as above. Set dx/dt = dy/dt = 0 and get an algebraic system. Solve it to get the critical point (2.5,-0.5). Calculate the Jacobian matrix and evaluate it at this critical point ! 1 −5 J(2.5, −0.5) = . 1 −1 The eigenvalues are ±2i. It is purely imaginary, i.e. complex with zero real part. Since the original system is linear, purely imaginary eigenvalues DO lead to a conclusion; that is, (2.5, −0.5) is a center, and it is stable. 3. Problem 12 on Page 410. Solution. A framework for solving this type of problem, i.e., to determine the type and stability of equilibria/critical points for dx = f (x, y) dt dy = g(x, y) dt is to start with finding all equilibria/critical points by setting dx/dt = dy/dt = 0 and solving the resulting algebraic system. Then, deal with each critical point one by one. It is useful to calculate the Jacobian matrix first ! fx fy . J(x, y) = gx gy Note that for nonlinear system, expect J to depend on x, y, not a constant matrix any more! For a given critical point (x0 , y0 ), evaluate the Jacobian matrix J(x0 , y0 ) so that the original nonlinear system has a linear approximation near (x0 , y0 ) dx = fx (x − x0 ) + fy (y − y0 ) dt dy = gx (x − x0 ) + gy (y − y0 ) dt P age 2 c Bin Cheng MAT 274 HW 12 Hints Upon substitution u = x − x0 , v = y − y0 , it becomes du = fx u + fy v dt i.e. dv = gx u + gy v dt ! ! d u u = J(x0 , y0 ) . dt v v Now, calculate the eigenvalues λ1 , λ2 of J(x0 , y0 ). Then, always relate positive real part to unstable, and negative real part to stable. Saddle point is unstable because one positive real part is enough to destroy the stability. • Real eigenvalues. 3 cases. (1) both postive: nodal souce, unstable. (2) both negative: nodal sink, stable. (3) opposite signs: saddle point, unstable. • Complex eigenvalues. 3 cases. (1) postive real part: spiral souce, unstable. (2) negative real part: spiral sink, stable. (3) zero real part: inconclusive for a genuinely nonlinear system; center and stable for a linear system. Now, let’s look at Problem 12 on Page 410. The equilibria are all given in the problems. For Problem 12, we look at (x0 , y0 ) = (5, 0). The Jacobian matrix is ! ! ∂ ∂ 2 − xy), 2 − xy) (5x − x (5x − x fx fy ∂y J(x, y) = = ∂x ∂ . ∂ gx gy (−2y + xy), ∂x ∂y (−2y + xy) that is J(x, y) = 5 − 2x − y, −x y, −2 + x ! Plug in the critical point (5,0) for (x, y) J(5, 0) = ! −5, −5 . 0, 3 In other words, the linear approximation of the original system near (5, 0) should be P age 3 c Bin Cheng MAT 274 HW 12 Hints du = −5u − 5v dt dv = 0u + 3v. dt The characteristic equation " det(λI − J(5, 0)) = det λ 0 0 λ ! − −5, −5 0, 3 !# = (λ + 5)(λ − 3) = 0 Thus, eigenvalues λ1 = −5, λ2 = 3. Real and opposite signs. So, conclusion: the critical point at (5,0) is a saddle point, and it is unstable. P age 4