Notes on Boussinesq Equation

28 3. NONLINEAR PROBLEM. LOCAL THEORY
large class
X = C ( [0, T ] : L 2 (R) ) ∩ L 4( [0, T ] : L ∞ (R) ) .
In fact, suppose ũ ∈ X satisfying the initial data, then it is easy to see that for T ′ < T
sufficiently small ũ ∈ X a T ′ . Therefore u = ũ in R × [0, T ′ ]. Reapplying this argument we
obtain the desired result.
To prove that for any T 0 < T the map from Y ↦→ C ( [0, T 0 ] : L 2 (R) ) ∩ L 4( [0, T 0 ] :
L ∞ (R) ) is continuous, let us take u and v solutions of (3.1) with data (f 0 , h ′ 0 ) and (f 1, h ′ 1 ),
respectively then
u(t) − v(t) = V 1 (t)(f 0 − f 1 ) + V 2 (t)(h ′ 0 − h ′ 1)
∫ T 0
−
0
V 2 (t − τ)∂ 2 x(
|u| α u − |v| α v ) (τ) dτ.
Using the same argument as in (3.4) and (3.8) we have
sup ‖(u − v)(t)‖ 2 ≤ c ( )
‖f 0 − f 1 ‖ 2 + ‖h 0 − h 1 ‖ −1,2
[0,T ]
+ c T (4−α)/4
0 (‖u‖ α L + 4 T L∞ ‖v‖α x L ) sup ‖(u − v)(t)‖ 4 2
T L∞ x
≤ c ( ‖f 0 − f 1 ‖ 2 + ‖h 0 − h 1 ‖ −1,2
)
[0,T ]
+ 2c T (4−α)/4
0 (4cδ) α sup ‖(u − v)(t)‖ 2 .
[0,T ]
On the other hand, the arguments used (3.6) and (3.9) imply that
1/4
‖u − v‖ L 4
T0 L ∞ ≤ c (1 + T
x 0 ) ( )
‖f 0 − f 1 ‖ 2 + ‖h 0 − h 1 ‖ −1,2
+ T (4−α)/4
0 (‖u‖ α L + 4 T L∞ ‖v‖α x L ) sup ‖(u − v)(t)‖ 4 2
T L∞ x
[0,T ]
≤ c (1 + T 1/4
0 ) ( )
‖f 0 − f 1 ‖ 2 + ‖h 0 − h 1 ‖ −1,2
+ 2T (4−α)/4
0 (4cδ) α sup ‖(u − v)(t)‖ 2 .
[0,T 0 ]
Now using (3.7) we obtain
sup { }
sup ‖(u − v)(t)‖ 2 , ‖u − v‖ L 4
T0 L ∞ x
[0,T 0 ]
≤ c α (T 0 ) ( )
‖f 0 − f 1 ‖ 2 + ‖h 0 − h 1 ‖ −1,2
(3.10)
(3.11)
(3.12)
which yields the result.
□
Corollary 3.3. If 0 < α < 4 then for all f ∈ L 2 (R) and g = h ′ ∈ Ḣ−1 (R) there
exist T = T (δ, α) > 0 and a unique solution u of the integral equation (3.1) in [0, T ] with
and
u ∈ C([0, T ] : L 2 (R)) ∩ L 4 ([0, T ] : L ∞ (R))
D 1/2
x u ∈ L ∞ (R : L 2 ([0, T ])).