Notes on Boussinesq Equation
Notes on Boussinesq Equation
Notes on Boussinesq Equation
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4.2. PERSISTENCE 41<br />
We begin by stating the following theorem that deals with the persistence properties<br />
in H s for soluti<strong>on</strong>s of the IVP (4.4). In this secti<strong>on</strong> we use g to denote g = h ′ .<br />
Theorem 4.3. Let (f, g) ∈ H s+1 (R) × H s (R), s ≥ 1, with ‖f‖ 1,2 and ‖g‖ 2 small, then<br />
there exists a unique soluti<strong>on</strong> u ∈ C(R + : H 1 (R)) of the IVP (4.4) such that<br />
u ∈ C b (R + : H 1 (R)) ∩ C(R + : H s+1 (R)).<br />
Proof. Applying ∂x<br />
α to the equati<strong>on</strong> in (4.4), multiplying by ∂x α u t , and integrating<br />
with respect to x, and then integrating by parts we have<br />
∫<br />
1 d {<br />
|∂x α u t | 2 + |∂x α u x | 2 + |∂x α u xx | 2} ∫<br />
dx = ∂x α (u 2 ) xx ∂x α u t dx.<br />
2 dt<br />
Now, using the Cauchy-Schwarz and Gagliardo-Nirenberg inequalities we have<br />
{<br />
}<br />
d<br />
‖∂x α u t ‖ 2 2 + ‖∂x α u x ‖ 2 2 + ‖∂x α u xx ‖ 2 2 ≤ c ‖u‖ ∞ ‖∂x α u xx ‖ 2 ‖∂x α u t ‖ 2 .<br />
dt<br />
Gr<strong>on</strong>wall’s inequality gives<br />
(<br />
‖∂<br />
α<br />
x u t ‖ 2 + ‖∂x α u x ‖ 2 + ‖∂x α )<br />
u xx ‖ 2<br />
sup<br />
[0,T ]<br />
≤<br />
} ( ∫ {‖∂ T<br />
x α g‖ 2 + ‖∂x α f x ‖ 2 + ‖∂x α f xx ‖ 2 exp C<br />
0<br />
)<br />
‖u(t)‖ ∞ dt .<br />
As usual the above formal computati<strong>on</strong> can be justified by using the c<strong>on</strong>tinuous dependence<br />
<strong>on</strong> the data.<br />
Now, using Theorem 4.2 the result follows.<br />
□<br />
Theorem 4.4. C<strong>on</strong>sider the IVP (4.4) with f ∈ H 3 (R), g ∈ H 2 (R). If xf xx , xf x and<br />
xg ∈ L 2 (R), then there exists T > 0 such that<br />
xu ∈ C([0, T ] : H 2 (R)).<br />
Moreover, if ‖f‖ 1,2 , ‖g‖ 2 ≪ 1, the result is global, i.e.<br />
Proof. From (4.4) we have<br />
xu ∈ C(R + : H 2 (R)).<br />
(xu) tt = (xu) xx − (xu) xxxx − 2(xu) x u x − 2(xu) xx u − 2u x + 4u xxx + 6uu x<br />
multiplying by (xu) t and integrating with respect to x, we find<br />
1 d<br />
{ ∫ ∫<br />
∫<br />
}<br />
[(xu) t ] 2 dx + [(xu) x ] 2 dx + [(xu) xx ] 2 dx<br />
2 dt<br />
∫<br />
∫<br />
= −2 u x (xu) x (xu) t dx − 2 u(xu) xx (xu) t dx.<br />
∫<br />
∫<br />
∫<br />
+ 4 u xxx (xu) t dx − 2 u x (xu) t dx + 6 uu x (xu) t