CDS 140a Winter 2013 Homework 6

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R. Murray, D. MacMartin Issued: 12 Feb 2013 (Tue)
ACM 101/AM 125b/CDS 140a, Winter 2013 Due: 19 Feb 2013 (Tue)


Note: In the upper left hand corner of the second page of your homework set, please put the number of hours that you spent on this homework set (including reading).

  1. Perko, Section 2.14, problem 1
    (a) Show that the system

    \dot x&=a_{11}x+a_{12}y+Ax^2-2Bxy+Cy^2\\ \dot y&=a_{21}x-a_{11}y+Dx^2-2Axy+By^2


    is a Hamiltonian system with one degree of freedom; i.e., find the Hamiltonian function $H(x,y)$ for this system.
    (b) Given $f\in C^2(E)$, where $E$ is an open, simply connected subset of $\mathbb R^2$, show that the system $\dot{x}=f(x)$ is a Hamiltonian system on $E$ iff $\nabla\cdot f(x)=0$ for all $x\in E$.

  2. Perko, Section 2.14, problem 7. Show that if $x_0$ is a strict local minimum of $V(x)$ then the function $V(x)-V(x_0)$ is a strict Lyapunov function (i.e., $\dot{V}<0$ for $x\neq 0$) for the gradient system $\dot x=-\mathrm{grad}V(x)$.
  3. Perko, Section 2.14, problem 12. Show that the flow defined by a Hamiltonian system with one degree of freedom is area preserving. Hint: Cf. Problem 6 in Section 2.3
  4. Perko, Section 3.3, problem 5. Show that

    \dot x &=y+y(x^2+y^2)\\ \dot y &=x-x(x^2+y^2)


    is a Hamiltonian system with $4H(x,y)=(x^2+y^2)^2-2(x^2-y^2)$. Show that $dH/dt=0$ along solution curves of this system and therefore that solution curves of this system are given by




    Show that the origin is a saddle for this system and that $(\pm 1,0)$ are centers for this system. (Note the symmetry with respect to the $x$-axis.) Sketch the two homoclinic orbits corresponding to $C=0$ and sketch the phase portrait for this system. (You need not comment on the compound separatrix cycle.)

  5. A planar pendulum (in the $x$-$z$ plane) of mass $m$ and length $\ell$ hangs from a support point that moves according to $x=a\cos (\omega t)$. Find the Lagrangian, the Hamiltonian, and write the first-order equations of motion for the pendulum.