Math 567: Assignment 3 Solutions
1. Hamiltonian structure for a Boussinesq system: For the following variant of
the Boussinesq system, for horizontal velocity v = v(x, t) and surface height h =
h(x, t), x ∈ R
ht = −[(1 + αh)v]x + βhxxt
vt = −[h + 12 αv 2 ]x + βvxxt
(α > 0, β > 0), find a Hamiltonian formulation, and identify the conserved energy
and momentum (conserved quantity arising from translation invariance).
Inverting 1 − βd2 /dx2 , we may re-write the system as
ht = −(1 − βd2 /dx2 )−1 [(1 + αh)v]x
vt = −(1 − βd2 .dx2 )−1 [h + 12 αv 2 ]x
h
,
and so as ~ut = J∂H(~u), with ~u :=
v
Z
d2 −1 d
1
0 1
J := −(1 − β 2 )
, H(~u) :=
h2 (x) + (1 + αh(x))v 2 (x) dx
dx
dx 1 0
2 R
2
d
−1
The fact that J is skew adjoint follows from the self-adjointness of (1 − β dx
2)
2
d
d
(since dx
2 is self-adjoint) and the matrix, the skew-adjointness of dx (integration by
parts), and the fact that these three operators all commute. The conserved energy is
H(~u), of course. Since H(u) is invariant by translations ~u(x) 7→ u(x~+ s), a group
of transformations which we may write as
d
d2
0 1
s dx
sJM
e
=e
, M := −(1 − β 2 )
1 0
dx
(and not M = M ∗ ) so there is a corresponding conserved ‘momentum’
Z
1
d2
1
1
h
v
(β 2 − 1)
=
h(−βvxx + v) + v(−βhxx + h)dx
− h~u, M~ui = −
v
h
2
2
dx
2 R
Z
=
(hv + βhx vx ) dx
R
where we used an integration by parts to arrive at the final expression.
1
2. Hamiltonian structure for the BBM equation: Find a Hamiltonian formulation, as well as the conserved energy and momentum, for the ‘Benjamin-BonaMahoney’ equation for u(x, t), x ∈ R:
ut + uux − uxxt = 0.
Again inverting 1 −
d2
,
dx2
d2
1
ut = −(1− 2 )−1 [ u2 ]x = J∂H(u),
dx
2
d2
d
1
J := −(1− 2 )−1 , H(u) :=
dx
dx
6
Z
u3 (x)dx.
R
The conserved energy is H(u) (i.e. kuk2L2 ), and since H(u) is invariant by translations
d2
d
e−s dx = eJ(1− dx2 )s
and since M = 1 −
d2
dx2
is self-adjoint, the conserved ‘momentum’ is
Z
Z
1
1
1
2
hu, M ui =
(u − uuxx )dx =
(u2 + u2x )dx
2
2 R
2 R
(i.e. kuk2H 1 ).
3. Scaling and criticality: Determine the scaling-critical Sobolev index s (in terms of
p and n), and determine the ‘energy’ (i.e. the norm connected to the Hamiltonian)
criticality for each of:
(a) (gKdV) ut + up−1 ux + uxxx = 0
(x ∈ R)
2
The scaling here is uλ (x, t) = λ p−1 u(λx, λ3 t) so
2
kuλ (·, t)kḢ s (R) = λ p−1
+s− 12
ku(·, λ3 t)kḢ s (R)
2
so the critical Sobolev index is sc = 21 − p−1
. There is no energy-critical (sc = 1)
1
case, since we always have sc ≤ 2 .
(b) (NLW) utt − ∆u ± up = 0
(x ∈ Rn )
2
Similalrly, the scaling here is uλ (x, t) = λ p−1 u(λx, λt) so
2
kuλ (·, t)kḢ s = λ p−1
so the critical Sobolev index is sc =
4
s = 1 ↔ p = 1 + n−2
(n ≥ 3).
2
n
2
+s− n
2
−
ku(·, λtkḢ s
2
p−1 ,
and the energy-criritcal case is
(c) (Navier-Stokes) ut +(u·∇)u−∆u = ∇p (x ∈ Rn , u ∈ Rn , p ∈ R, and the role of
‘energy’ here is played by the L2 norm) The scaling here is uλ (x, t) = λu(λx, λ2 t)
(and for the pressure pλ (x, t) = λ2 p(λx, λ2 t)) so
n
kuλ (·, t)kḢ s = λ1+s− 2 ku(·, λ2 t)kḢ s
so the critical Sobolev index is sc = n2 − 1, and the energy-criritcal case is
s = 0 ↔ n = 2. Remark: the super-critical nature of the n = 3 NavierStokes equations is one reason its global well-posedness is a famously difficult
open problem (while for n = 2 the global well-posedness is known).
4. Global Existence for the Generalized KdV equation Consider the IVP for
gKdV in the ‘energy space’:
ut + up−1 ux + uxxx = 0,
u(x, 0) = u0 (x) ∈ H 1 (R).
The ‘standard argument’ gives a good (i.e. ‘subcritical’) local existence theory in
H 1 for any 1 < p < ∞ (note all these exponents are H 1 sub-critical). Use the
Gagliardo-Nirenberg-Sobolev inequality
Z
p+1
u
Z
(x)dx ≤ Cp
R
u2x (x)dx
R
p−1 Z
4
2
p+3
4
u (x)dx
R
together with conservation laws to show that the solution is global if
(a) p is L2 (‘mass’)-subcritical
(b) p is mass-critical and the mass is sufficiently small ( and identify this threshold
mass in terms of Cp )
By the local theory, an a priori bound on the H 1 norm of a solution implies it is
global. We have conservation of ‘mass’ and ‘energy’:
Z
Z 1 2
1
2
p+1
u (x)dx ≡ M,
ux (x) −
u (x) dx ≡ E,
p(p + 1)
R
R 2
so using the G-N-S inequality,
kux k2L2 ≤ 2E +
p−1
p+3
2Cp
2
4 ku k 2
kukp+1
≤
2E
+
M
x
p+1
L
L2
p(p + 1)
p(p + 1)
(a) So if p−1
2 < 2 , i.e. p < 5 (‘mass’ subcritical), this inequality implies kux (·, t)kL2 ≤
const. which, together with mass conservation, gives an a priori H 1 bound, which
implies global existence.
3
(b) If p = 5 (’mass’ critical), and also
2C5 2
µ :=
M < 1 ↔ ku0 kL2 = M 1/2 <
30
15
C5
1/4
then the above inequality implies
(1 − µ)kux k2L2 ≤ 2E
and so again an a priori bound on kux kL2 , hence kukH 1 , hence global existence.
5. Morawetz identities Show that if u(x, t) solves
(N LS3+ ) iut + ∆u = |u|2 u
with n = 3 (i.e. x ∈ R3 ), then
∂
|u|4 + 2(|∇u|2 − |(x̂ · ∇)u|2 )
Im (ūx̂ · ∇u) =
+∇·F
∂t
|x|
for some vector field F (here x̂ = x/|x|).
4
Using ∇ · x̂ =
2
|x| ,
∂
2
− Im (ūx̂ · ∇u) = −Im (ūt x̂ · ∇u + ūx̂ · ∇ut ) = −Im ∇ · [ūut x̂] − 2ut x̂ · ∇ū −
ūut
∂t
|x|
1
2
= ∇ · [−Im(ūut x̂)] + 2Im [x̂ · ∇ū +
ū][i∆u − |u| u]
|x|
1
1
2
4
= ∇ · [−Im(ūut x̂)] − 2Re x̂ · ∇ū∆u − x̂ · ∇ū|u| u +
ū∆u −
|u|
|x|
|x|
1
= ∇ · −Im(ūut x̂) − 2Re x̂ · ∇ū∇u +
ū∇u
|x|
1
1
1
− 2Re −(x̂ · ∇)∇ū · ∇u −
|∇u|2 +
|x̂ · ∇u|2 − |u|2 x̂ · ∇|u|2
|x|
|x|
2
1
1
1
|u|4
− |∇u|2 + 2 ūx̂ · ∇u −
|x|
|x|
|x|
1
= ∇ · −Im(ūut x̂) − 2Re x̂ · ∇ū∇u +
ū∇u
|x|
1
2
1
1
1
1
2
2
2
4
2
4
+ 2Re
x̂ · ∇|∇u| +
|∇u| −
|x̂ · ∇u| + x̂ · ∇|u| −
x̂ · ∇|u| +
|u|
2
|x|
|x|
4
2|x|2
|x|
1
1
1
1
2
= ∇ · −Im(ūut x̂) − 2Re x̂ · ∇ū∇u +
ū∇u + x̂[− |∇u|2 − |u|4 +
|u|
]
|x|
2
4
2|x|2
1
1
1
1
1
+ 2Re
|∇u|2 −
|x̂ · ∇u|2 +
|u|4 + 3 |u|2 − 3 |u|2
|x|
|x|
2|x|
|x|
|x|
1
2
(|∇u|2 − |x̂ · ∇u|2 ) +
|u|4 + ∇ · F
=
|x|
|x|
as required.
6. Decay estimate via ‘Pseudoconformal Energy’: The H 1 scattering theory for
the defocusing NLS (N LSp+ ) using the Morawetz identity which we outlined in class
requires p > 1 + 4/n (mass supercritical) and fails for lower powers. But scattering
can be recovered for some sub-critical powers by changing the space/norm. This
exercise gives an example of how a different identity/estimate can be used to obtain
some decay in the mass critical case p = 1 + 4/n. Take the defocusing cubic NLS in
two space dimensions:
iut + ∆u = |u|2 u,
u(x, 0) = u0 (x).
(a) Show that the ‘psuedoconformal energy’
Z 1
2
2
4
Epc (u, t) :=
|(x + 2it∇)u(x, t)| + t |u(x, t)| dx
2
R2
5
is conserved by solutions of the PDE (assume you are working with a smooth
solution which decays, along with its various derivatives, quickly as x → ∞).
(b) Thus derive the following decay estimate for the L4 norm:
Z
Z
1
4
|u(x, t)| dx ≤ 2
|xu0 (x)|2 dx.
2t
2
2
R
R
Remark: this decay estimate can be used to show that a solution with u0 ∈ L2 and
xu0 ∈ L2 (extra spatial decay!) is global and scatters in L2 . It is MUCH harder
to prove this scattering without the extra xu0 ∈ L2 assumption, and indeed this is a
recent breakthrough development of just the last few years!
You can just compute directly, but here is a slicker way to do it. Set v(x, t) :=
(x + 2it∇)u(x, t) (note v is vector-valued), observe that the operator x + 2it∇ commutes with the operator in the linear part of the Schrödinger equation (using notation
[A, B] := AB − BA)
[i∂t + ∆, x + 2it∇] = [i∂t , 2it∇] + [∆, x] = −2∇ + 2∇ = 0
so
0 = (x + 2it∇)(ut + ∆u − |u|2 u) = (i∂t + ∆)(x + 2it∇)u − (x + 2it∇)|u|2 u
= (i∂t + ∆)v − |u|2 v − 4itRe(ū∇u)u.
So
1
∂t |v|2 = Re(v̄ · vt ) = Re(v̄ · (i∆v − i|u|2 v + 4tRe(ū∇u)u)
2
= −Im ∇ · [v̄ · ∇v] − |∇v|2 + 4tRe(ū∇u) · Re(u[xū − 2it∇ū])
and thus
d1
dt 2
Z
2
While
Z
Z
d 2
t
|u|4 dx = 2t
dt
R2
Z
= 2t
Z
= 2t
Z
= 2t
Z
Z
(x · ∇)|u|4
Z
Z
2
Im(ū∇u) · Re(ū · ∇u) − 2t |u|4 .
= −8t
|v| dx = −8t
R2
2
Im(ū∇u) · Re(ū · ∇u) + t
|u|4 + 4t2 Re
4
2
4
2
Z
|u|2 ūut
Z
|u|2 ū(i∆u − i|u|2 u)
|u| + 4t Re
Z
|u| − 4t Im
∇ · (|u|2 ū∇u) − |u|2 |∇u|2 − 2Re(ū∇u) · ū∇u
Z
4
2
|u| + 8t
Im(ū∇u) · Re(ū∇u).
6
d
Combining the last two equations gives dt
Epc (u, t) = 0, as desired. It then follows
that
Z
Z
1
t2 |u|4 ≤ Epc (u, t) = Epc (u0 , 0) ≤
|xu0 |2
2
so
Z
Z
1
4
|u| ≤ 2 |xu0 |2 .
2t
7