Remark on the well-posedness of weakly dispersive equations

We improve the results about the well-posedness of the regularized fractional dispersive equation (1 + D x α ) u t + u x + uu x = 0 when 0 α ≤ 1. When #x03B1; < 1, the existence and uniqueness of vanishing viscosity solution is proved.


Introduction
Fractional Benjamin-Bona-Mahony (fBBM) equation was introduced by Linares, Pilod and Saut [16] to investigate the role of weak dispersion (0 < α < 1) on the solution of the Burgers equation u t + u x + uu x = 0. They showed the local in time well-posedness in Sobolev spaces using energy estimates. Nevertheless, this method does not provide the uniqueness. Indeed, the difference w = u − v between two solutions satisfies (see inequality (4.31) of [16]) where R |R(−∆) s/2 w|dx ≤ ||u + v|| H s+ α 2 ||w|| 2 H s+ α 2 . But the last term can not be uniformly controlled by the H s+ α 2 −norm if 0 < α < 1. When 0 < α < 1, it seems difficult to obtain the global well-posedness. It has been shown by Bona and Saut [3] that the linearization around 0 has blow-up solution and Klein, Saut numerically observe blow-up when 0 < α < 1 3 [15]. While α = 1, global well-posedness can be obtained thanks to Brezis-Gallouët estimates [5,18]. The paper is organized as follows. In Section 2, we improve the regularity of the global well-posedness when α = 1. In Section 3, we deal with the uniqueness of vanishing viscosity solution for 0 < α < 1.

The regularized Benjamin-Ono equation
When α = 1, the equation can be rewritten as the Benjamin-Ono equation under the form where H is the Hilbert defined by its Fourier symbol H(u)(ξ) = −i sgn(ξ)û(ξ).
Moreover, for all t ∈ R u(t) The proof is done in two steps: first, a compactness argument is used to obtain a weak solution, then the uniqueness of the weak solution provides the strong continuity of the weak solution.
Let (u n,0 ) n∈N be a sequence of H 1 (R) such that u n,0 → u 0 in H 1 2 (R). We denote by u n (t) ∈ C(R; H 1 (R)) the solution of the initial value problem associated with the initial datum u n,0 . Then it is proved that the energy [18] and the sequence (u n ) n∈N is bounded in C(R; H 1 2 (R)). On the other hand, multiplying (1) by ∂ t u n and integrating over space gives and Young's inequality provides, for ε > 0. Thus, and the sequence (u n ) n∈N is uniformly bounded and equicontinuous. We deduce according to the Rellich theorem for all is a weak solution of the equation (1) and Suppose now that the weak solution of the Cauchy problem From inequality (2) and from the uniqueness of the weak solution, we obtain and the solution u belongs to C([0, T ]; H 1 2 (R)). Note that we also obtain the continuity with respect to the initial data since we proved that for u n,0 → u 0 in H 1 2 (R), then the respective solution (u n ) verifies u n → u in C([0, T ]; H 1 2 (R)). It remains to prove the uniqueness of the weak solution. We are inspired by the method introduced by Yudovich [21]. We need a Trudinger-type estimates proved by Gérard and Grellier [9] in the torus T.
Proof. The proof is similar to [9] except that u is split as ; L 2 (R)) be two weak solutions of (1) starting from the same initial datum. Consider the function g defined as Then, for w : We note that the operator P (D x ) is bounded in L p , for 1 < p < +∞ according to the Mikhlin-Hörmander theorem [17,19]. The Sobolev and the Trudinger inequalities imply and from inequality (2) Similarly, we have We can write thanks to the fractional Leibniz rule [14].

Lemma 1.3. We have for
Taking p > 2 large enough so that Finally, g(t) ≡ 0, and u = v.

The weak dispersive equation
Let us come back to the initial value problem, for 0 < α < 1, The existence is proved in [16] using energy estimates.
We briefly remind the proof in order to highlight the loss of uniqueness (see [16] for details). Applying the operator J s to (3), multiplying by J s u and integrating by part over space, we obtain where, thanks to Leibniz's rule, The Hölder inequality provides and we conclude with the Gronwall lemma.
Let us try to prove the uniqueness. Let u and v be two solutions of (3) and denote w = u − v. We have Using the fractional Leibniz rule and integrating by parts, it gets Here, for all 0 < < s and 1 p + 1 q = 1 2 , Even if the three first terms are bounded by ||u + v|| H s+ α 2 ||w|| 2 H s+ α 2 , the last one can not be uniformly controlled by the H s+ α 2 −norm when 0 < α < 1. To avoid this difficulty, we propose the following regularization [11] Lemma 2.2. Let 0 < α < 1 and r ≥ 0. We have Proof. The idea of the proof is introduced in [4,13]. By duality, it is enough to show that for all function w ∈ S(R), Let us denote ξ = (1 + ξ 2 ) Since for x ∈ R, θ ∈ R, we have e −x ≤ 1 x θ , we deduce Proof. Thanks to the Duhamel formula, u ε is solution of (4) if and only if u ε is the fixed point of Φ ε defined as 1+|ξ| α t * u(x). Let B T be the closed ball We prove that Φ ε (B T ) ⊆ B T and Φ ε is a contraction mapping on B T . Let u ε ∈ B T , we have

Lemma 2.2 gives
and Φ ε is a contraction mapping on B T as soon as T < To obtain the continuity with respect to the initial data, for u ε 0 and v ε 0 in H r (R) with u ε 0 H r ≤ M , v ε 0 H r ≤ M , we consider u ε , v ε the respective solution. Let 0 ≤ t ≤ T = 1 Cε,sM , we find similarly or in other words, since 1 − C ε,r T M > 0, Lemma 2.4. Let 0 < α < 1, r ≥ 2 − α 2 and u ε 0 ∈ H r (R), then T ε can be chosen independent of ε.
Proof. Let r = s + α 2 . Applying J s to (4), multiplying by J s u ε and integrating by parts over space, it comes d dt is weakly continuous and we obtain with similar computations and the Gronwall lemma allows to conclude.