Physics 441 Homework 6 solutions

67. This is a trivial exercise in following directions. Apply the Painleve test $q = q_{0} t^{m}$ , $p = p_{0} t^{n}$

m q_{0} t^{m - 1} = 4 p_{0}^{3} t^{3 n}, n p_{0} t^{n - 1} = - 4 q_{0}^{3} t^{3 m}, n = m = - \frac{1}{2}, p_{0}^{4} = - q_{0}^{4}, {(p_{0} q_{0})}^{2} = - \frac{1}{64}

Find resonances

- \frac{1}{2} q_{0} t^{- \frac{3}{2}} + (ρ - \frac{1}{2}) q_{1} t^{ρ - \frac{3}{2}} = 4 (p_{0}^{3} t^{- \frac{3}{2}} + 3 p_{0}^{2} p_{1} t^{ρ - \frac{3}{2}})

- \frac{1}{2} p_{0} t^{- \frac{3}{2}} + (ρ - \frac{1}{2}) p_{1} t^{ρ - \frac{3}{2}} = - 4 (q_{0}^{3} t^{- \frac{3}{2}} + 3 q_{0}^{2} q_{1} t^{ρ - \frac{3}{2}})

|\begin{matrix} ρ - \frac{1}{2} & - 12 p_{0}^{2} \\ 12 q_{0}^{2} & ρ - \frac{1}{2} \end{matrix}| = 0, {(ρ - \frac{1}{2})}^{2} = \frac{144}{64} = (\frac{3}{2})^{2}, ρ = - 1, 2

We conclude that there is an invariant of weight two (quartic in $p$ and $q$ , whose weights are $\frac{1}{2}$ ). Obviously this is the Hamiltonian. However this suggests constructing new variables of weight $1$ from homogeneous quadratic combinations of $q$ and $p$ , remember that ${q, p}_{p b} = 1$ ;

{q^{2} + p^{2}, q p}_{p b} = 2 (q^{2} - p^{2}), {q^{2} - p^{2}, q p}_{p b} = 2 (q^{2} + p^{2}), {q^{2} + p^{2}, q^{2} - p^{2}}_{p b} = 8 p q

{u_{1}, u_{3}}_{p b} = 2 u_{2}, {u_{2}, u_{3}}_{p b} = 2 u_{1}, {u_{1}, u_{2}}_{p b} = 8 u_{3}

H = \frac{1}{2} (u_{1}^{2} + u_{2}^{2}), {\dot{u}}_{1} = - 8 u_{2} u_{3}, {\dot{u}}_{2} = 8 u_{1} u_{3}, {\dot{u}}_{3} = - 4 u_{1} u_{2}

I_{1}^{2} = u_{1}^{2} + u_{2}^{2}, I_{2}^{2} = u_{1}^{2} - 2 u_{3}^{2}

Let ${k^{'}}^{2} = \frac{I_{2}^{2}}{I_{1}^{2}}$ , then let $u_{1} (t) = I_{1} dn (t)$ with this modulus. We have an elliptic quadrature.

68.
Note that

u_{1} {\dot{u}}_{1} - u_{2} {\dot{u}}_{2} = 0, u_{2}^{2} - u_{1}^{2} = c

This has many solutions (trig, hyperbolic, elliptic),

{\dot{u}}_{2} = - u_{1}^{2} = c - u_{2}^{2}, \int \frac{d u_{2}}{c - u_{2}^{2}} = t, \frac{1}{2 c} ln \frac{c + u_{2}}{c - u_{2}} = t, u_{2} = c coth c t, u_{1} = \frac{c}{sinh c t}

and of course $c$ being a constant of integration, you can set it equal to any value of your choosing, creating a whole family of similarity-invariant solutions.

69. You get $u_{1} = cos u_{3} t, u_{2} = sin u_{3} t$ and $g_{1} = g_{2} = g_{3} = 1$ , so $u_{1}^{'} = a cos \frac{(a u_{3}) t}{a}, u_{2}^{'} = a sin \frac{(a u_{3}) t}{a}$ are clearly solutions too.

70. $I_{1} = \frac{1}{2} (B u_{1}^{2} - A u_{2}^{2})$ $I_{2} = \frac{1}{2} (C u_{1}^{2} - A u_{3}^{2})$

71.

|\begin{matrix} ρ - \frac{1}{2} & - 3 c_{2}^{2} \\ 3 c_{1}^{2} & ρ - \frac{1}{2} \end{matrix}| = 0, - \frac{1}{2} c_{1} = c_{2}^{3}, - \frac{1}{2} c_{2} = - c_{1}^{2}

72.

- \sqrt{2} u^{- 2} + q_{1} (r - 1) u^{- 2 + r} = - \sqrt{- 2} u^{- 2} + p_{1} u^{- 2 + s}, 2 \sqrt{- 2} u^{- 3} + p_{1} (s - 2) u^{s - 3} = 2 \sqrt{2} u^{- 3} + 6 q_{1} u^{- 3 + r}

\frac{1}{r - 2} = \frac{r - 1}{6}, (r - 4) (r + 1) = 0

q = \sqrt{- 2} u^{- 1} + a_{4} u^{3} - \frac{3 \sqrt{- 2} a_{4}^{2}}{36} u^{7} + \dots

p = - \sqrt{2} u^{- 2} + 3 a_{4} u^{2} - \frac{21 \sqrt{- 2} a_{4}^{2}}{36} u^{7} + \dots

73. Equations of motion are

{\dot{q}}_{1} = p_{1}, {\dot{q}}_{2} = p_{2}, ṗ_{1} = - 2 q_{1} q_{2}, ṗ_{2} = q_{2}^{2} - q_{1}^{2}

Let

t \to α^{- 1} t, q_{i} \to q_{i} α^{x}, p_{i} \to p_{i} α^{y}

then

x + 1 = y, y + 1 = 2 x, x = 2, y = 3

so a particular solution is

q_{i} = q_{i, 0} t^{- 2}, p_{i} = p_{i, 0} t^{- 3}, - 2 q_{i, 0} = p_{i, 0}, - 3 p_{1, 0} = - 2 q_{1, 0} q_{2, 0}, - 3 p_{2, 0} = q_{2, 0}^{2} - q_{1, 0}^{2}

Then if we start with

- 2 q_{1, 0} = p_{1, 0}, - 3 p_{1, 0} = 6 q_{1, 0} = - 2 q_{1, 0} q_{2, 0}, \Rightarrow q_{2, 0} = - 3, q_{1, 0} = \sqrt{q_{2, 0}^{2} + 6 q_{2, 0}} = \sqrt{9 + 18} = \sqrt{27}

and we get $p_{1, 0} = - 2 \sqrt{27}$ , and $p_{2, 0} = 6$ . If instead we start with

- 3 p_{2, 0} = - 3 (- 2 q_{2, 0}) = q_{2, 0}^{2} - q_{1, 0}^{2}

an obvious solution is $q_{1, 0} = p_{1.0} = 0$ and $q_{2, 0} = 6$ , $p_{2, 0} = - 12$ .
Now solve the variational system (this step is the same as finding resonances)

q_{i} = q_{i, 0} t^{- 2} + ξ_{i}, p_{i} = p_{i, 0} t^{- 3} + η_{i}

with $ξ, η$ small, the variational system is

{\dot{ξ}}_{i} = η_{i}, {\dot{η}}_{1} = - 2 t^{- 2} (q_{1, 0} ξ_{2} + q_{2, 0} ξ_{1}), {\dot{η}}_{2} = 2 (q_{2, 0} ξ_{2} - q_{1, 0} ξ_{1})

Let

ξ_{i} = ξ_{i, 0} t^{ρ - 2}, η_{i} = η_{i, 0} t^{ρ - 3}

(\begin{matrix} ρ - 2 & - 1 & 0 & 0 \\ 2 q_{2, 0} & ρ - 3 & 2 q_{1, 0} & 0 \\ 0 & 0 & ρ - 2 & - 1 \\ 2 q_{1, 0} & 0 & - 2 q_{2, 0} & ρ - 3 \end{matrix}) (\begin{matrix} ξ_{1, 0} \\ η_{1, 0} \\ ξ_{2, 0} \\ η_{2, 0} \end{matrix}) = 0

insert your values (either set) for $q_{i, 0}$ and find the eigenvalues. It is especially easy for the second set

|\begin{matrix} ρ - 2 & - 1 & 0 & 0 \\ 12 & ρ - 3 & 0 & 0 \\ 0 & 0 & ρ - 2 & - 1 \\ 0 & 0 & - 12 & ρ - 3 \end{matrix}| = 0, ((ρ - 2) (ρ - 3) + 12) ((ρ - 2) (ρ - 3) - 12) = 0

giving

ρ = \frac{5 \pm \sqrt{25 - 72}}{2}, too bad, fails Painleve, ρ = - 1, 6

indicating a weight-six invariant which is something like

I = a p_{1}^{2} + b p_{2}^{2} + c p_{1} p_{2} + d q_{1}^{3} + e q_{2} q_{1}^{2} + f q_{2}^{2} q_{1} + g q_{2}^{3}

Gee-whiz, what could it be? Its the Hamiltonian.

74. First establish the leading singularity exponents;

g_{1} + 1 = g_{2} + g_{3} + g_{4}, g_{i} = \frac{1}{2}

c_{2} = c_{3} = c_{4}, - \frac{1}{2} c_{1} = c_{2} c_{3} c_{4}, - \frac{1}{2} c_{2} = - c_{1} c_{3} c_{4}, c_{1} = \frac{1}{\sqrt{2 i}}, c_{2} = c_{3} = c_{4} = \sqrt{\frac{i}{2}}

They are not integers; the system fails the Painleve test but we still have hope, since we can reduce it to quadrature if we can find a sufficient number of invariants. If only there were some technique for finding their weights, we could narrow the search...
The variational equations are

ẏ_{1} = \frac{i}{2 t} (y_{2} + y_{3} + y_{4}), - ẏ_{2} = \frac{1}{t} (\frac{i}{2} y_{1} + \frac{1}{2} y_{3} + \frac{1}{2} y_{4})

- ẏ_{3} = \frac{1}{t} (\frac{i}{2} y_{1} + \frac{1}{2} y_{2} + \frac{1}{2} y_{4}), - ẏ_{4} = \frac{1}{t} (\frac{i}{2} y_{1} + \frac{1}{2} y_{2} + \frac{1}{2} y_{3})

Let y_{j} = y_{j, 0} t^{ρ - \frac{1}{2}}, (\begin{matrix} \frac{1}{2} - ρ & - i ∕ 2 & - i ∕ 2 & - i ∕ 2 \\ i ∕ 2 & \frac{1}{2} - ρ & 1 ∕ 2 & 1 ∕ 2 \\ i ∕ 2 & 1 ∕ 2 & \frac{1}{2} - ρ & 1 ∕ 2 \\ i ∕ 2 & 1 ∕ 2 & 1 ∕ 2 & \frac{1}{2} - ρ \end{matrix}) (\begin{matrix} y_{10} \\ y_{20} \\ y_{30} \\ y_{40} \end{matrix}) = 0

resulting in Kowalewski exponents \frac{1}{2} - ρ = - \frac{1}{2}, - \frac{1}{2}, - \frac{1}{2}, \frac{3}{2}

so there are three invariants of weight one (quadratic in the $u^{'} s$ ), which are easy to find.
76.

Ṗ_{i} = 2 (Q_{i - 1}^{2} - Q_{i}^{2}), {\dot{Q}}_{i} = Q_{i} (P_{i} - P_{i + 1})