ADVANCED DYNAMICS — PHY 4241/5227

HOME AND CLASS WORK – SET 7

Solution for assignment 10c: Double pendulum eigenvectors and explicit orbits.

\frac{g}{l} ( \begin{matrix}−2 (2 ± \sqrt{2}) + 2 & −(2 ± \sqrt{2}) \\ −(2 ± \sqrt{2}) & −(2 ± \sqrt{2}) + 1 \end{matrix} ) ( \begin{matrix}ϕ_{0} \\ ψ_{0} \end{matrix} ) = \frac{g}{l} ( \begin{matrix}a_{±} & b_{±} \\ b_{±} & c_{±} \end{matrix} ) ( \begin{matrix}ϕ_{0} \\ ψ_{0} \end{matrix} ) = 0

with a_{+} = −2−2\sqrt{2}, b_{+} = −2−\sqrt{2}, c_{+} = −1−\sqrt{2} and a_{−} = −2+2\sqrt{2}, b_{−} = −2+\sqrt{2}, c_{−} = −1 + \sqrt{2}. With the ansatz given in the problem we find:

( \begin{matrix}a_{±} & b_{±} \\ b_{±} & c_{±} \end{matrix} ) ( \begin{matrix}1 \\ ψ_{0±} \end{matrix} ) = 0 ⇒ ψ_{0±} = −\frac{a_{±}}{b_{±}} = −\frac{b_{±}}{c_{±}} , ψ_{0+} = −\sqrt{2} and ψ_{0−} = \sqrt{2}

and, therefore, the eigenvectors

Φ_{0+} = ( \begin{matrix}1 \\ −\sqrt{2} \end{matrix} ) and Φ_{0−} = ( \begin{matrix}1 \\ √ & \overline{2} \end{matrix} ) .

They are not orthogonal, because the previously solved eigenvalue problem is of the form

V Φ_{0} = ω^{2} T Φ_{0} with V = ( \begin{matrix}2g/l & 0 \\ 0 & g/l \end{matrix} ) and T = ( \begin{matrix}2 & 1 \\ 1 & 1 \end{matrix} ) .

Multiplying with T^{−1} from the left does not help, because the matrix T^{−1}V will not be symmetric. Instead a generalization of the usual orthogonality definition works, see M&T (12.55) or Goldstein Mechanics. The general solution for the double pendulum problem is given by the real part of our ansatz, which can be written

Φ(t) = (A_{+} cos ω_{+}t + B_{+} sin ω_{+}t) ( \begin{matrix}1 \\ −\sqrt{2} \end{matrix} ) + (A_{−} cos ω_{−}t + B_{−} sin ω_{−}t) ( \begin{matrix}1 \\ √ & \overline{2} \end{matrix} ) .

The four constants are determined by the four initial value, e.g., ϕ_{0}, ϕ̇_{0}, ψ_{0}, ψ̇_{0} at time t = 0:

ϕ_{0} = A_{+} + A_{−} ,  ψ_{0} = \sqrt{2} (−A_{+} + A_{−}) ,
ϕ̇_{0} = ω_{+}B_{+} + ω_{−}B_{−} , ψ̇_{0} = \sqrt{2} (−ω_{+}B_{+} + ω_{−}B_{−}) ,

which gives

A_{+} = \frac{ϕ_{0}}{2} − \frac{ψ_{0}}{2\sqrt{2}} ,  A_{−} = \frac{ϕ_{0}}{2} + \frac{ψ_{0}}{2\sqrt{2}} ,
B_{+} = \frac{ϕ̇_{0}}{2ω_{+}} − \frac{ψ̇_{0}}{2\sqrt{2} ω_{+}} , B_{−} = \frac{ϕ̇_{0}}{2ω_{−}} + \frac{ψ̇_{0}}{2\sqrt{2} ω_{−}} ,

and the time dependence of the angles is

ϕ(t) = +A_{+} cos ω_{+}t + B_{+} sin ω_{+}t + A_{−} cos ω_{−}t + B_{−} sin ω_{−}t ,
ψ(t) = (−A_{+} cos ω_{+}t − B_{+} sin ω_{+}t + A_{−} cos ω_{−}t + B_{−} sin ω_{−}t) \sqrt{2} .

For the initial conditions

ϕ_{0} = 0 , ϕ̇_{0} = 1 , ψ_{0} = 0 , ψ̇_{0} = −1

at time t = 0 see the figure for a plot up to t = 50 \sqrt{l/g} (next page).