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Autonomous Systems (Part III) 1) Describe the undamped hard spring oscillations modeled by the system or, equivalently, Find the critical points of this system, and determine the type and stability of these points. Solution: Rewrite the given second-order DE as the first-order system Solve for the critical points by solving the homogeneous system of algebraic equations Since is never 0 for any real value of , the bottom equation is only zero when , so the only critical point occurs at (0,0). We'll use the linearization at (0,0) to classify the type and stability of this critical point. Eigenvalues for the Linearized System Type of Critical Point of the Amost Linear System Stable improper node Stable node or spiral point Unstable saddle point Unstable node or spiral point Unstable improper node Stable spiral point Unstable spiral point Stable or unstable, center or spiral point We need to compute the Jacobian of the given system: . At (0,0), the Jacobian is just . It is easy to see that eigenvalues are always has the characteristic polynomial . Notice that . That means the and check that the eigenvalues are in the hard spring oscillation case. Maple tells us what the phase portrait of our autonomous DE looks like. > 3 2 y 1 0 1 2 3 x > That means our critical point (0,0) is a stable center. What does this tell us about the motion of our mass? 2) Describe the undamped soft spring oscillations modeled by the system or, equivalently, Find the critical points of this system, and determine the type and stability of these points. Solution: Rewrite the given second-order DE as the first-order system Solve for the critical points by solving the homogeneous system of algebraic equations Hence, we have critical points at (0,0), (2,0), and (-2,0). We'll use the linearization at each critical point to classify the type and stability of each critical point. We need to compute the Jacobian of the given system: . Case 1: Our critical point is (0,0). At (0,0), the Jacobian is just . It is easy to see that eigenvalues are has the characteristic polynomial . That means the . Case 2: At ( ,0), the Jacobian is just . It is easy to see that has the characteristic polynomial . That means the eigenvalues are . Since the eigenvalues are real with opposite signs, we know are both unstable saddle points. To check our work, let's also have Maple plot the phase portrait. > 4 y 2 0 1 2 x 3 4 > What does this tell us about the motion of our mass? 3) Describe the damped hard spring oscillations modeled by the system or, equivalently, Find the critical points of this system, and determine the type and stability of these points. Solution: Rewrite the given second-order DE as the first-order system Solve for the critical points by solving the homogeneous system of algebraic equations Just as in damped hard oscillation case, the only critical point occurs at (0,0). We need to compute the Jacobian of the given system: . At (0,0), the Jacobian is just . It is easy to see that has the characteristic polynomial . That means the eigenvalues are . Since the eigenvalues are complex conjugates with negative real part, we know (0,0) is an stable spiral point. > 3 2 y 1 0 1 2 3 x > 3) Describe the damped soft spring oscillations modeled by the system or, equivalently, Find the critical points of this system, and determine the type and stability of these points. Solution: Rewrite the given second-order DE as the first-order system Solve for the critical points by solving the homogeneous system of algebraic equations Just as in damped soft spring oscillation case, the critical points occurs at (0,0), (2,0), and (-2,0). We need to compute the Jacobian of the given system: . Case 1: Our critical point is (0,0). At (0,0), the Jacobian is just . It is easy to see that has the characteristic polynomial . That means the eigenvalues are . Since the eigenvalues are complex conjugates with negative real part, we know (0,0) is an stable spiral point. Case 2: At ( ,0), the Jacobian is just . It is easy to see that has the characteristic polynomial . That means the eigenvalues are . Since the eigenvalues are real with opposite signs, we know saddle points. To check our work, let's also have Maple plot the phase portrait. > are both unstable 4 y 2 0 1 2 x 3 4 > Notice that the phase portrait is split into 5 regions. What does our phase portrait tell us about our system?