Math 567: Assignment 2 (Due: Monday, Feb. 24)
1. Space-time dispersive estimates for Airy: recall that the solution operator for
the Airy equation
ut + uxxx = 0, u(x, 0) = u0 (x),
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u(x, t) = [e−t(d/dx) u0 ](x)
satisfies the decay estimates
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ke−t(d/dx) u0 kL2x = ku0 kL2x ,
ke−t(d/dx) u0 kL∞
≤ Ct−1/3 ku0 kL1 .
x
(a) Use interpolation to establish the family of estimates
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ke−t(d/dx) u0 kLpx ≤ Cp t−σp ku0 kLp0 ,
x
finding the exponent σp in terms of p (recall
1
p
+
1
p0
= 1).
(b) By observing that uλ (x, t) := u(λx, λ3 t) also solves the Airy PDE, find the
admissibility relation that must hold between exponents r and p in order for
there to hold a homogeneous space-time estimate of the form
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ke−t(d/dx) u0 kLrt ([0,∞);Lpx (R) ≤ Cku0 kL2 (R) .
(c) Use the generalized Young inequality, together with part (a), to establish the
inhomogeneous space-time estimates
Z t
3
(s−t)(d/dx)
e
f (·, s)ds
≤ Ckf kLr0 Lp0
Lrt Lpx
0
t
x
for a pair (r, p) (2 ≤ p ≤ ∞) satisfying the admissibility criterion you found in
part (b).
Remark: you are not asked to do so here, but the full suite of space-time estimates,
as for the Schrödinger equation, can be obtained following the same arguments.
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2. Dispersive decay estimate for the wave equation: The basic decay estimate
for solutions of the wave equation
utt = ∆u,
u(x, 0) = 0,
ut (x, 0) = v0 (x)
(with zero initial displacement) for x ∈ Rn with n odd is
|u(x, t)| ≤ Ct−
m−1
2
k(∇)
n−1
2
v0 kL1 (Rn ) .
Use the solution formulas in the cases n = 1 and n = 3 to prove this estimate.
Hint: for n = 3 you’ll need to use some trick – integration by parts or something like
that – to bound a surface integral of v0 by a bulk integral of ∇v0 .
Remark: notice a couple of things here: 1. for n = 1 there is no decay, as the equation
is not at all dispersive, while for n = 3 the weak dispersion coming from the variation
of the group velocity with the direction (but not the magnitude!) of the frequency
vector does result in some decay (albeit slower than for Schrödinger). 2. this decay
estimate is also weaker than that of (say) Schrödinger or Airy in that we must take
also take some derivatives of the initial data: ‘loss of derivative’.
3. Linearized water waves: linearize the water wave system
− h ≤ x1 ≤ η(x0 , t), x0 ∈ R2
η t + ∇ x0 φ · ∇ x0 η = 0
x1 = η(x0 , t), x0 ∈ R2
φt + 12 |∇φ|2 + gη = 0
∆φ = 0
φ x1 = 0
x1 = −h, x0 ∈ R2
and obtain the dispersion relation ω 2 = g|ξ| tanh(h|ξ|) for the result.
Hint: more precisely, suppose φ, η (and their derivatives) are small perturbations of
the flat, still solution φ ≡ 0, η ≡ 0, drop all terms which are nonlinear in these small
quantities (and in the upper boundary condition, evaluate at the unperturbed surface
0
x1 = 0), and seek solutions of the form η(x0 , t) = ei(ξ·x −ωt) for ξ ∈ R2 .
4. Local existence for the Hartree equation: a variant of the NLS arising in Quantum Mechanics (many-body theory) is the Hartree equation, which has a convolutiontype nonlinearity:
iut + ∆u = (V ∗ |u|2 )u,
u(x, 0) = u0 (x),
where V : R3 → R is the potential (describing the inter-particle interaction).
(a) Use the Hölder and/or Young inequalities to show that if V is a bounded function, then the nonlinearity u 7→ (V ∗ |u|2 )u is locally Lipshitz in L2 (R3 ) (and so
it follows from the standard argument given in class that for u0 ∈ L2 (R3 ) we
can construct a local solution u ∈ C([0, T ]; L2 (R3 ))).
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(b) Physically important potentials such as the Coulomb potential V (x) = ±1/|x|
are not bounded. Show that the Coulomb potential lies in the class L∞ (R3 ) +
L3/2 (R3 ) (that is, it can be written as a sum of two terms, one lying in the first
space, the other in the second).
(c) Use the Hölder and/or Young and/or Sobolev inequalities (more precisely for
the latter, kukL6 (R3 ) ≤ ckukH 1 (R3 ) ) to show that if V ∈ L∞ (R3 )+L3/2 (R3 ), then
the Hartree nonlinearity is locally Lipshitz in H 1 (R3 ) (and so for u0 ∈ H 1 (R3 ),
the standard argument gives a local solution u ∈ C([0, T ], H 1 (R3 )) for this class
of potentials, which includes the Coulomb).
5. Persistence of regularity: consider a nonlinear Schrd̈inger equation with pure
power nonlinearity
iut + ∆u = ±up
u(x, 0) = u0 (x),
with p a positive integer. Recall that if u0 ∈ H s (Rn ) where s > n/2 is (for
simplicity) an integer, we constructed a ‘classical solution’ u ∈ C([0, T max ); H s )
with ku(·, t)kH s → ∞ if T max < ∞. Prove that if T max < ∞, then in fact
kukLp−1 L∞ (Rn ×[0,T max )) = ∞.
t
x
Hint: one (but not the only) way to do this is to argue by contradiction, assuming
that T max < ∞ but kukLp−1 L∞ (Rn ×[0,T max )) < ∞, and apply the Strichartz estimates
x
t
(together with Hölder’s inequality) successively on subintervals I of [0, T max ) chosen
so that kukLp−1 L∞ (Rn ×I) is sufficiently small on each.
t
x
Remark: one implication of an estimate like this one, is ‘persistence of regularity’ –
namely that the constructed H s solutions for different s will all have the same lifespan.
More precisely, an H s2 solution is also an H s1 solution if n/2 < s1 < s2 , but the H s2
norm cannot blow up before the H s1 norm does: clearly T max (H s1 ) ≥ T max (H s2 ), but
in fact we cannot have T max (H s1 ) > T max (H s2 ). For if so, the boundedness of kukH s1
on [0, T max (H s2 )] implies (by Sobolev imbedding) kukLp−1 L∞ (Rn ×[0,T max (H s2 )]) < ∞,
x
t
contradicting what was proved in this exercise.
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