Notes on Boussinesq Equation

12 2. LINEAR PROBLEM
Lemma 2.1. There exists C > 0 such that for any x, t ∈ R we have
|I(x, t)| ≤ C(1 + β)|t| −1/2 , x, t ∈ R. (2.5)
Remark 2.2. The result for a general class of functions φ was proved by Kenig, Ponce
and Vega in [26] (see Lemma 2.7 in [26]).
To prove Lemma 2.1 we will use of the following result.
Lemma 2.3 (Van der Corput). Let ψ ∈ C0 ∞(R) and ϕ ∈ C2 (R) satisfy that ϕ ′′ (ξ) >
λ > 0 on the support of ψ. Then
∫
∣ e iϕ(ξ) ψ(ξ) dξ∣ ≤ 10λ −1/2 {‖ψ‖ L ∞ + ‖ψ ′ ‖ L 1}. (2.6)
Proof. See [44].
□
Proof of Lemma 2.1.
Case 1. We first consider the following situation:
0 < m ≤ |φ ′′ (ξ)| ≤ M and Ω bounded and
∫
I(x, t) = e i(tφ(ξ)+xξ) |φ ′′ (ξ)| 1/2+iβ dξ.
Lemma 2.3 implies then that
|I(x, t)| ≤ c (m|t|) −1/2{ ∫
M +
Ω
≤ c φ (1 + |β|)|t| −1/2 .
Case 2: 1 ≤ |φ ′′ (ξ)| ≤ 2, Ω = {ξ ∈ R : |ξ| > 1/ɛ}.
∫
∣
|ξ|>1/ɛ
e itφ(ξ)+ixξ |φ ′′ (ξ)| 1/2+iβ dξ∣
Ω
| 1 }
2 + iβ||φ′′ (ξ)| −1/2 |φ ′′′ (ξ)| dξ
1
∣
≤ ∣
i(tφ ′ (ξ) + x) eitφ(ξ)+ixξ |φ ′′ (ξ)| 1/2+iβ ∣∣|ξ|>1/ɛ
∫
+ ∣ e itφ(ξ)+ixξ d ( |φ ′′ (ξ)| 1/2+iβ )
|ξ|>1/ɛ dξ i(tφ ′ dξ∣
(ξ) + x)
∫
≤ c |φ ′′ (ξ)| −1/2 |φ ′′′ (ξ)|
|ξ|>1/ɛ
|tφ ′ (ξ) + x| + t|φ′′ (ξ)| 3/2
|tφ ′ (ξ) + x| 2 dξ
∫
{
≤ c |ξ| −6 + |ξ| −2} dξ < ∞.
|ξ|>1/ɛ
(2.7)
(2.8)
Here we have used that |φ ′ (ξ)| = ∞ as |ξ| → ∞. This shows that the integral
is convergent. Therefore we can write the set Ω as the union of bounded intervals
and apply the argument in Case 1.
Case 3: 0 ≤ |φ ′′ (ξ)| ≤ 2, Ω = {ξ ∈ R : |ξ| ≤ ɛ}.