[SOLVED] Amath 573 homework 4

1. Consider the Modified Vector Derivative NLS equation
Bt + (∥B∥
2B)x + γ(e1 × B0) (e1 · (Bx × B0)) + e1 × Bxx = 0.
This equation describes the transverse propagation of nonlinear Alfv´en waves in
magnetized plasmas. Here B = (0, u, v), e1 = (1, 0, 0), B0 = (0, B0, 0), and γ
is a constant. The boundary conditions are B → B0, Bx → 0 as |x| → ∞.
By looking for stationary solutions B = B(x − W t), one obtains a system of
ordinary differential equations. Integrating once, one obtains a first-order system
of differential equations for u and v.
a) Show that this system is Hamiltonian with canonical Poisson structure, by
constructing its Hamiltonian H(u, v).b) Find the value of the Hamiltonian such that the boundary conditions are satisfied. Then H(u, v) equated to this constant value defines a curve in the
(u, v)-plane on which the solution lives. In the equation of this curve, let
U = u/B0, V = v/B0, and W0 = W/B2
0
. Now there are only two parameters
in the equation of the curve: W0 and γ.c) With γ = 1/10, plot the curve for W0 = 3, W0 = 2, W0 = 1.1, W0 = 1,
W0 = 0.95, W0 = 0.9. All of these curves have a singular point at (1, 0).
This point is an equilibrium point for the Hamiltonian system, corresponding
to the constant solution which satisfies the boundary condition. The curves
beginning and ending at this equilibrium point correspond to soliton solutions
of the Modified Vector Derivative NLS equation. How many soliton solutions
are there for the different velocity values you considered? Draw a qualitatively
correct picture of the solitons for all these cases.2. Show that the canonical Poisson bracket
{f, g} =
X
N
j=1

∂f
∂qj
∂g
∂pj
−
∂f
∂pj
∂g
∂qj

satisfies the Jacobi identity
{{f, g}, h} + {{g, h}, f} + {{h, f}, g} = 0.3. Show that the Sine-Gordon equation
utt − uxx + sin(u) = 0
is Hamiltonian with canonical Poisson structure and Hamiltonian
H =
Z 
1
2
p
2 +
1
2
q
2
x + 1 − cos(q)

dx,
where q = u, and p = ut
.4. Check explicitly that the conserved quantities F−1 =
R
udx, F0 =
R
1
2
u
2dx, F1 = R
1
6
u
3 −
1
2
u
2
x


dx, F2 =
R
1
24u
4 −
1
2
uu2
x +
3
10u
2
xx

dx are mutually in involution with
respect to the Poisson bracket defined by the Poisson structure given by ∂5. Find the fourth conserved quantity for the KdV equation ut = uux + uxxx, i.e., the
conserved quantity which contains 1
24
R
u
4dx6. Recursion operator For a Bi-Hamiltonian system with two Poisson structures
given by B0, B1, one defines a recursion operator R = B1B
−1
0
, which takes one
element of the hierarchy of equations to the next element. For the KdV equation
with B0 = ∂x and B1 = ∂xxx +
1
3
(u∂x + ∂xu), we get B
−1
0 = ∂
−1
x
, integration with
respect to x. Write down the recursion operator. Apply it to ux (the zero-th KdV
flow) to obtain the first KdV flow. Now apply it to uux+uxxx to get (up to rescaling
of t2) the second KdV equation. What is the third KdV equation?7. Consider the function U(x) = 2∂
2
x
ln
1 + e
kx+α


. Show that for a suitable k, U(x)
is a solution of the first member of the stationary KdV hierarchy (as you’ve already
seen, it is the one-soliton solution):
6uux + uxxx + c0ux = 0.
(Note: it may be easier to define c0 in terms of k, instead of the other way around)
Having accomplished this, let u(x, t1, t2, t3, . . .) = U(x)|α=α(t1,t2,t3,…)
. Determine the
dependence of α on t1, t2 and t3 such that u(x, t1, t2, t3, . . .) is simultaneously a
solution of the first, second and third KdV equations:
ut1 = 6uux + uxxx,
ut2 = 30u
2ux + 20uxuxx + 10uuxxx + u5x,
ut3 = 140u
3ux + 70u
3
x + 280uuxuxx + 70uxxuxxx + 70u
2uxxx + 42uxuxxxx + 14uu5x + u7x
Based on this, write down a guess for the one-soliton solution that solves the entire
KdV hierarchy+1