Nonlinear Problem

1. We perform local analysis of the equilibria of dx = −2xy, dt dy = (1 + x)(1 − y), dt and piece together the dynamics of the nonlinear problem based on the linear results. This works provided the equilibria are hyperbolic (by the Hartman-Grobman theorem). Equilibria: ẋ = 0 =⇒ x = 0 or y = 0, ẏ = 0 =⇒ x = −1 or y = 1. Thus, there are 2 equilibria at (x1 , y1 ) = (0, 1) and (x2 , y2 ) = (−1, 0). Stability:  Df (x, y) = −2y −2x 1 − y −(1 + x)  At (x1 , y1 ) = (0, 1): the eigenvalue-eigenvector pairs are given by       0 1 . , (λ2 , v 2 ) = −1, (λ1 , v 1 ) = −2, 1 0 This is a STABLE NODE. At (x2 , y2 ) = (−1, 0): the eigenvalue-eigenvector pairs are given by  √  √    √ √ − 2 2 2, , (λ4 , v 4 ) = . (λ3 , v 3 ) = − 2, 1 1 This is a SADDLE. y x Figure 1: Nullclines are the green dashed lines. Stable node at (0, 1). Strong stable manifold of the stable node indicated by blue double arrows. Saddle at (−1, 0). Stable manifold of saddle drawn in blue. Unstable manifold of saddle drawn in red. 2. We perform local analysis of the equilibria of dx = y 2 − x2 , dt dy = x − 1, dt and piece together the dynamics of the nonlinear problem based on the linear results. This works provided the equilibria are hyperbolic (by the Hartman-Grobman theorem). Equilibria: ẋ = 0 =⇒ y = x or y = −x, ẏ = 0 =⇒ x = 1. Thus, there are 2 equilibria at (x1 , y1 ) = (1, −1) and (x2 , y2 ) = (1, 1). Stability:   −2x 2y Df (x, y) = 1 0 At (x1 , y1 ) = (1, −1): the eigenvalues are λ1,2 = −1 ± i. This is a STABLE FOCUS. At (x2 , y2 ) = (1, 1): the eigenvalue-eigenvector pairs are given by √  √      √ √ −1 − 3 −1 + 3 , (λ4 , v 4 ) = −1 + 3, . (λ3 , v 3 ) = −1 − 3, 1 1 This is a SADDLE. y x Figure 2: Nullclines are the green dashed lines. Stable focus at (1, −1). Saddle at (1, 1). Stable manifold of saddle drawn in blue. Unstable manifold of saddle drawn in red. 3. Once again, we invoke Hartman-Grobman to construct the dynamics of the nonlinear problem dx = x2 + y 2 − 2, dt dy = x + y, dt based on the linearization about the (hyperbolic) invariant sets (equilibria in this case). Equilibria: ẋ = 0 =⇒ x2 + y 2 = 2, ẏ = 0 =⇒ y = −x. Thus, there are 2 equilibria: (x1 , y1 ) = (−1, 1) and (x2 , y2 ) = (1, −1). Stability:   2x 2y Df (x, y) = 1 1 At (x1 , y1 ) = (−1, 1): the eigenvalue-eigenvector pairs are given by √  √      √ √ 1 1 −3 − 17 −3 + 17 (−1 − 17), (−1 + 17), , (λ2 , v 2 ) = . (λ1 , v 1 ) = 2 2 2 2 This is a SADDLE. √ At (x2 , y2 ) = (1, −1): the eigenvalues are λ3,4 = 21 (3 ± i 7). This is an UNSTABLE FOCUS. y x Figure 3: Nullclines are the green dashed lines. Saddle at (−1, 1). Stable manifold of saddle drawn in blue. Unstable manifold of saddle drawn in red. Unstable focus at (1, −1). 4. Yet again, we use Hartman-Grobman to construct the dynamics of the nonlinear problem dx = x, dt dy = −y + x2 , dt by using the linearized flow about a hyperbolic equilibrium. Equilibria: ẋ = 0 =⇒ x = 0, ẏ = 0 =⇒ y = x2 . Thus, there is a single equilibrium at the origin. Stability:  Df (x, y) =  1 0 2x −1 At (x, y) = (0, 0): the eigenvalue-eigenvector pairs are given by       1 0 . , (λ2 , v 2 ) = 1, (λ1 , v 1 ) = −1, 0 1 This is a SADDLE. y x Figure 4: Nullclines are the green dashed lines. The origin is a saddle. Stable manifold of saddle drawn in blue. Unstable manifold of saddle drawn in red. We now compute these solution curves exactly. The nonlinear problem is dx = x, dt dy = −y + x2 . dt Note that the x-equation is decoupled so we can solve it immediately to get x = e t x0 , where x0 = x(0) is a constant. Substituting this result into the y-equation gives the first order linear ODE dy + y = x20 e2t . dt The integrating factor is I(t) := et . Multiplying both sides by I(t) leads to d t
e y = x20 e3t . dt Integrating both sides and solving explicitly for y gives the desired solution:     x(t) et x0
= −t . y(t) e y0 + 13 e2t − e−t x20 The solution above is parameterized by the time variable t. To eliminate t, we write et = x . x0 Substituting into the y component of the solution gives y= x0 y0 − 13 x30 1 2 + x . x 3 Note that x0 and y0 are both constant, so H := x0 y0 − 13 x30 is another constant. These solution curves are drawn in the figure below for various H (compare with Figure 4). y x Figure 5: Solution curves for H = 0 (black), H = ± 14 (green), H = ± 21 (red), H = ±1 (blue).