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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

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