Download Solutions to Homework 3: Phase Portraits and Hamiltonian Systems and more Assignments Spanish Language in PDF only on Docsity! Solutions to Homework 3 By H̊akan Nordgren As always, the sketches are all in a separate pdf. Problem 9.8: For the following systems determine whether the system is a gradient system or a Hamiltonian system, and then draw its phase-portrait. 1. x′ = y2 + 2xy y′ = x2 + 2xy. 2. x′ = x2 − 2xy y′ = y2 − 2xy. 3. x′ = x2 − 2xy y′ = y2 − x2. Solution: 1. This system is a gradient system with potential function V1(x, y) = − ( x2y + y2x ) . 2. This system is Hamiltonian, with Hamiltonian H(x, y) = x2y − y2x. 3. This is a gradient system with potential function V2(x, y) = −x 3 3 − y3 3 + x 2y. Problem 9.9.b: Find conditions on a, b, c, and d so that( x y )′ = ( a b c d )( x y ) , is a Hamiltonian system. Solution: For the above to be Hamiltonian, we require that there is a function H : R2 → R such that ∂H ∂y = ax + by −∂H ∂x = cx + dy. Such a function H must have H(x, y) = axy + by 2 2 + c1(x) and H(x, y) = −c x2 2 − dxy + c2(y). Thus we must have a = −d. Problem 9.10: Consider the system x′ = f(x, y) y′ = g(x, y) 1 1. Determine conditions on f and g so that the above system is a gradient system. 2. And determine conditions on f and g so that the above system is a Hamiltonian system. Solution: 1. For the system to be a gradient system there must be a function V : R2 → R such that ∂V ∂x = f(x, y) ∂V ∂y = g(x, y). This means that ∂f∂y = ∂2V ∂y∂x = ∂g ∂x . 2. For the system to be Hamiltonian, there must be a function H : R2 → R such that ∂H ∂y = f(x, y) −∂H ∂x = g(x, y). Thus, ∂f∂x = ∂2H ∂y∂x = − ∂g ∂y . Problem 9.11: Show that the linearisation of a planar Hamiltonian system has eigenvalues of the form ±λ, or ±iλ, where λ is a real number. Solution: The linearised system associated with the planar Hamiltonian system x′ = ∂H ∂y y′ = −∂H ∂x , is ( u v )′ = ( ∂2H ∂y∂x ∂2H ∂y2 −∂ 2H ∂x2 − ∂2H ∂x∂y )( u v ) . The Jacobian matrix is of the form ( a b c −a ) , which has eigenvalues ± √ a2 − bc Problem 9.16: A solution φt(x0) is called recurrent if there is a sequence (tn) such that tn →∞ and φtn(x0) → x0. Show that a gradient system can have no non-constant recurrent solutions. Solution: The reason for this, roughly, is that the potential function is always decreasing along its trajectories. Let us see what a proof of this would look like. Let V be the potential function for the gradient system. We begin by supposing that x0 is not an equilibrium point. If it were then φt(x0) = x0 for all t; that is, φt(x0) would be a constant solution. We may also suppose that φt(x0) does not bump into any equilibrium points along its 2
