Notes on Boussinesq Equation

14 2. LINEAR PROBLEM
If ξ ∈ Ω 3 then |φ ′ (ξ) − x/t| ≤ c|φ ′ (ξ)| ≥ c|ξ| 2 and |ξ| > |t| −1/3 . Integration by parts
yields the results. In fact,
∫
∣ e i(tφ(ξ)+xξ) |φ ′′ (ξ)| 1/2+iβ dξ∣
Ω 3
≤ c ∫
|t|
Ω 3
{
≤ c
∫
(1 + |β|)
|t|
( 1 2 + (ξ)| −1/2 |φ ′′′ (ξ)|
|β|)|φ′′ |φ ′ (ξ) − x/t|
|ξ|>|t| −1/3
≤ c(1 + |β|)|t| −1/2 .
|φ ′′′ (ξ)|
|ξ| 3/2
This completes the proof of the lemma.
dξ + c
|t|
+ |φ′′ (ξ)| 3/2
|φ ′ (ξ) − x/t| 2 }
dξ
∫
|ξ|>|t| −1/3
|ξ| −5/2 dξ
(2.13)
□
For φ define
W γ (t)f(x) =
∫ ∞
−∞
Theorem 2.4. Let φ and W γ be defined as before, we have
where γ ∈ [0, 1] with γ = 1 p ′ − 1 p , p = 2
1−γ and p′ = 2
1+γ .
Proof. Consider the operator
W γ+iβ (t)f(x) =
e i(tφ(ξ)+xξ) |φ ′′ (ξ)| γ/2 ˆf(ξ) dξ. (2.14)
‖W γ (t)f‖ p ≤ c |t| −γ/2 ‖f‖ p ′ (2.15)
∫ ∞
−∞
Observe that W 1+iβ (t)f(x) = I(·, t) ∗ f(x) and
e i(tφ(ξ)+xξ) |φ ′′ (ξ)| γ/2+iβ ˆf(ξ) dξ.
T 0+iβ (t)f(x) = (e itφ(ξ) |φ ′′ (ξ)| iβ ̂f(ξ)) ∨ .
Therefore the Young inequality and Lemma 2.1 imply that
‖W 1+iβ (t)f‖ L ∞ ≤ c|t| −1/2 ‖f‖ L 1.
On the other hand, Plancharel’s theorem (A.9) gives
‖W iβ (t)f‖ L 2 = ‖f‖ L 2.
Therefore the Stein interpolation theorem (see Appendix Theorem A.14) yields the result.
□
Theorem 2.5. If W γ (t) is defined as in (2.14) and γ ∈ [0, 1] then
( ∫ ∞
−∞
‖W γ/2 (t)f‖ q p dt) 1/q
≤ c ‖f‖2 , (2.16)