Local Well-Posedness to the Cauchy Problem for an Equation of the Nagumo Type.

In this paper, we show the local well-posedness for the Cauchy problem for the equation of the Nagumo type in this equation (1) in the Sobolev spaces H s (ℝ). If D > 0, the local well-posedness is given for s > 1/2 and for s > 3/2 if D=0.

1. Introduction

In this paper, we show the local well-posedness for the following Cauchy problem:where D > 0 is a constant diffusion coefficient, α ∈ (0, 1/2) and ϵ > 0 is a small positive quantity. In [1], the equation (1) was used to model chemotaxis (see equation (55) in [1]). Organisms which use chemotaxis to locate food sources include amoebae of the cellular slime mold Dictyostelium discoideum, and the motile bacterium Escherichia coli [1]. Therefore, u=u(x, t) models the population density, n is a positive integer, and α is a parameter which determines the minimal required density for a population to be able to survive (for normalized population density, i.e., such that u=1 is the maximum sustainable population). Balasuriya and Gottwald [1] studied the wave speed of travelling waves for the equation (1). Also, they have the numerical evidence for the wave speed of travelling waves for the equation (1). Other results related to the equation (1) can be found in [2].

When ϵ=0, the equation (1) is called a Nagumo equation or bistable equation [3-7] in which case the model describes an active pulse transmission line simulating a nerve axon.

Also, we can see the equation (1) as a generalized viscous Burgers equation with a source term. Dix [8] proved local well-posedness of the viscous Burgers equation with a source term using a contraction mapping argument. Moreover, for the classical Burgers equation (without viscosity) is well known that classical solutions cannot exits for all time, but weak global solutions can be established [9]. In addition, the uniqueness of the weak solution depends on some entropy condition. Observe that when D=0, the equation (1) is a generalized Burgers equation (without viscosity) and nonlinear source term. Therefore, from the mathematical viewpoint, the case D=0 is very interesting to study the existence and uniqueness of classical solution.

In this paper, we show the local well-posedness for the Cauchy problem to the equation of the Nagumo type (1) in the Sobolev spaces H(ℝ) for s > 1/2 if D > 0, and for s > 3/2 if D=0. Our proof of local well-posedness is based on the results given in [10-12]. We use the Banach fixed point in a suitable complete space to guarantee the existence of local solutions to the problem (1) with D > 0. The Banach fixed point technique has been widely used to show existence and uniqueness of solutions to differential equations in Banach spaces (for instance, see [10-14] for more details). When D=0, we use the parabolic regularization method to show local well-posedness for the Cauchy problem (1) (e.g., [12,15]).

We will use the following notation: ℝ for the real numbers; 𝒮(ℝ) for the Schwartz's space usual; denotes the Fourier transform of f; the inverse Fourier transform will be denoted by ∨; by H(ℝ), s ∈ ℝ, the set of all f ∈ 𝒮′(ℝ) such that . H(ℝ) is called the Sobolev space and it is a Hilbert space with respect to the inner product ; C(I; X) for the space of all continuous functions on an interval I into the Banach space X; if I is compact, C(I; X) is seen as a Banach space with the sup norm; C(I; X) for the space of all weakly continuous functions on an interval I into Banach space X; C1(I; X) for the space of all weakly differentiable functions on an interval I into Banach space X. We also denote by V(t)=e, t ≥ 0, the semigroup in H(ℝ) generated by the operator tQ where Q=(D∂2 − αId), i.e.,{V(t)} is a C0-semigroup of contractions in H(ℝ), s ∈ ℝ. Moreover, u(x, t)=V(t)ψ(x) is the unique solution to the linear problem associated with (1), i.e., u(x, t)=V(t)ψ(x) is the unique solution to the following problem.

Proposition 1 .

Let ψ ∈ H(ℝ), s ∈ ℝ, λ ≥ 0, D > 0 and t > 0. Then, there exists a constant C, depending only on λ, such that

In particular, V(t)ψ ∈ 𝒮(ℝ) for all t > 0.

When there is no risk of confusion, we will use the notations u(t) for u(x, t), ϕ for ϕ(x), and F(u)=ϵuu+u3 − (α+1)u2.

2. Local Well-Posedness of the Problem (1) with D > 0

In this section, we use the Banach fixed point in a suitable complete metric space to show the existence of local solutions for integral equation (9) in Sobolev space H(ℝ) for s > 1/2. In addition, the uniqueness of the solution and continuous dependence are established.

Proposition 2 .

Let s > 1/2 be fixed. Then, F(u) is a continuous map from H(ℝ) into H(ℝ) and satisfies the estimates as follows:for all u, v ∈ H(ℝ), where L(·, ·) is a continuous function, nondecreasing with respect to each of their arguments. In particular,

Proof

Observe that F(u)=(ϵ/n+1)(u)+u3 − (α+1)u2. Then, as H(ℝ) is a Banach algebra for s > 1/2, we have the following:where

The following result is to prove the existence of solutions. The proof is based in standard arguments [10,11]. We only present a sketch of proof.

Proposition 3 .

Let D > 0 be fixed, s > 1/2, ψ ∈ H(ℝ), and V(t) is defined by (2). Then, there exists T=T(‖ψ‖, M) > 0 and a unique function u ∈ C([0, T]; H(ℝ)) satisfying the following integral equation:

Let M, T > 0 be fixed, but arbitrary. Consider the following:which is a complete metric space with distance d(u, v)=sup[0, ‖u(t) − v(t)‖. Define on the space 𝒳(M, T, ψ) the following map:

We have the following:

If g ∈ 𝒳(M, T, ψ) then 𝒜g ∈ 𝒳(M, T, ψ)

We can choose sufficiently small such that

There exists such that 𝒜 is a contraction on

So, 𝒜 has a unique fixed point u in which satisfies the integral equation (40) where .

Proposition 4 .

The problem (1) is equivalent to the integral equation (40). More precisely, if s > 1/2 and u ∈ C([0, T]; H(ℝ))∩C1((0, T]; H(ℝ)) is a solution of (1), then u satisfies the integral equation (40). Conversely, if s > 1/2 and u ∈ C([0, T]; H(ℝ)) is a solution of (40) then u ∈ C1([0, T]; H(ℝ)) and satisfies (1).

Assume that u ∈ C([0, T]; H(ℝ))∩C1((0, T]; H(ℝ)) is a solution of (1). Then, (d/dτ)(V(t − τ)u(τ))=−V(t − τ)F(u(τ)), 0 < τ < t. So, u satisfies the integral equation (40). Conversely, assume that u ∈ C([0, T]; H(ℝ)) is a solution of (40). For t > 0, let η(t) : −∫0V(t − τ)F(u(τ))dτ. Then, for h > 0 arbitrary,

However,and the right hand side of (57) is a integrable function of τ in [0, t]. Thus, using the dominated convergence theorem, we have as follows:

Now, from the mean value theorem for integrals, there exists a value c on the interval (t, t+h) such thatand therefore, lim+(1/h)∫‖V(t+h − τ)F(u(τ)) − F(u(t))‖dτ=0.

After, ∂+η(t)=Qη(t) − F(u(t)) in H(ℝ). where ∂+ is the right derivative. In similar way, we can conclude that the left derivative is ∂−η(t)=Qη(t) − F(u(t)) in H(ℝ). So, η ∈ C1((0, T]; H(ℝ)) and ∂η(t)=Qη(t) − F(u(t)). As V(t)ψ(x) is the solution of the linear problem (3), we conclude that u(t)=V(t)ψ+η(t) ∈ C1((0, T]; H(ℝ)) and satisfies (1).

Lemma 1 .

Suppose β > 0, γ > 0, β+γ > 1, a ≥ 0, b ≥ 0, u is nonnegative and tu(t) is locally integrable on [0, T). Ifi.e., in (0, T), thenwhere ν=β+γ − 1 > 0, E(s)=∑cs with c0=1 and c/c=Γ(mν+γ)/Γ(mν+γ+β) for m ≥ 0.

The proof of this lemma is given in Lemma 7.1.2 in [16].

Proposition 5 .

Let ψ, ϕ ∈ H(ℝ) and u, v ∈ C([0, T]; H(ℝ)) be the corresponding solutions of equation (9). If s > 1/2, thenwhere K=E1/2,1((bΓ(1/2))2T), and L=max{sup[0, ‖u‖, sup[0, ‖v‖} (here E1/2,1 is given by previous lemma).

Let ψ, ϕ, u and v as in the statement of the proposition. Let s > 1/2. From (9) we have as follows:

By Propositions 1 and 2, we obtain the following:where L=max{sup[0, ‖u‖, sup[0, ‖v‖}. Let . Observe that E1/2,1((bΓ(1/2))2T) is finite. In fact, E1/2,1((bΓ(1/2))2T)=∑a where a=c(b2πT). Therefore, (a/a)=(c/c)(b2πT)1/2 and from Lemma 1 we have that

As for all x > 0, with 0 < θ(x) < 1, we have that 1 < e < e and e( is bounded for m ≥ 1. From (21), we obtain as follows:

So, lim|a/a|=0 and E1/2,1((bΓ(1/2))2T)=∑a, with a=c(b2πT), is convergent.

Proposition 6 .

Let s > 1/2. Then, the map ψ ↦ u is continuous in the following sense: if ψ⟶ψ in H(ℝ) and u ∈ C([0, T]; H(ℝ)), where T=T(‖ψ‖, M) > 0, are the corresponding solutions (of the problem (1) with u(0)=ψ). Let T ∈ (0, T). Then, there exists a positive integer N=N(D, ψ, T) such that T ≥ T for all n ≥ N and

As T=T(‖ψ‖, M) > 0 is a continuous function a ‖ψ‖, then there exists N ∈ ℕ such that T ≤ T for all n ≥ N. Let T=min{T, T1, T2,…, T}. Therefore, u is defined on [0, T] for all n. It follows that u ∈ 𝒳(M, T, ψ) for all n and satisfies ‖u(t)‖ ≤ ‖ψ‖+M ≤ γ+M where γ=sup‖ψ‖. Therefore, sup‖u(t)‖ ≤ γ+M for all n and sup‖u(t)‖ ≤ γ+M. Now, similar to the proof of the previous proposition, we have as follows:

Let . Thus, E1/2,1((bΓ(1/2))2T) is finite (where E1/2,1 is given in Lemma 1) and we have as follows:

This finishes the proof.

Finally, from Propositions 3, 5 and 6, we can summarize in the following theorem:

Theorem 1 .

Let s > 1/2. The problem (1) is locally well-posed in H(ℝ).

3. Local Well-Posedness of the Problem (1) with D = 0

In this section, we show the local well-posedness of the problem (1) with D=0 using a priori estimate and the parabolic regularization method, the so-called vanishing viscosity method (for more details see [12]).

Lemma 2 .

Let η(t), a(t) and b(t) be real valued positive continuous functions defined on [0, T]⊆[0, +∞). Let G(r) and H(r) be positive continuous functions for r ≥ 0, with G strictly increasing and H nondecreasing. Define A(t)=sup0≤a(s) and B(t)=sup0≤b(s). Then, the inequalityimplies the inequalitywhere Ω(r)=∫(dζ/H(G−1(ζ))), ϵ > 0, r > 0, and T=sup{τ ∈ [0, T] : Ω(A(τ))+τBτΩ(limG(r))}.

This is a particular case of the theorem given in [17] [pp. 78].

Proposition 7 .

Let s > 3/2 be fixed. Then, F(u) satisfies the estimatefor all u, w ∈ H(ℝ), where L0(x, y)=ϵx∑xy+(ϵn/2)y+x2+xy+y2+(α+1)(x+y).

We define q(u, w)=∑uw. As s > 3/2 thus H(ℝ) and H(ℝ) are Banach algebras. Moreover, we have that H(ℝ)⟶H(ℝ) and H(ℝ)⟶L(ℝ). Thus, using the Cauchy–Schwartz inequality, we have as follows:

Lemma 3 .

(T. Kato). Let r ≥ 1 and s > 3/2 be fixed and h, v are real valued functions. Then, there exists a constant C=C(r, s) such that

In particular, |(v, h∂v)| ≤ C‖∂h‖‖v‖2.

See Lemma A.5. in [13].

Theorem 2 .

Let s > 3/2 be fixed. For D > 0, consider the initial value problem (1) with initial data + and let u ∈ C([0, T]; H(ℝ)) be the corresponding solution of (1) for some T > 0. Then, there exists a T=T(ψ) > 0, depending on ‖ψ‖, such that u can be extended to the interval [0, T(ψ)], and there is a function ρ ∈ C([0, T(ψ)]; [0, +∞)) such that ρ(0)=‖ψ‖2 and ‖u(t)‖2 ≤ ρ(t),  for all t ∈ [0, T(ψ)].

Using the inner product in H(ℝ) and Lemma 3 we have that

Then, ‖u(t)‖2 ≤ ρ(t) for all t ∈ [0, T), where ρ ∈ C([0, T); [0, +∞)) is the maximally extended solution of the following problem.

For n ≥ 2, from the problem (32) we obtain as follows:and integrating from 0 to t we have as follows:

From Lemma 2 with a(t)=‖ψ‖2+2C(α+ϵ+2)t, b(t)=8C, G(r)=r and H(r)=r( we have the following bound:for t ≤ T, where T=sup{t ∈ [0, +∞) : (‖ψ‖2+2C(α+ϵ+2)t)− ≥ 4nCt}. Observe that T > 0, since the function Φ(t)=(‖ψ‖2+2C(α+ϵ+2)t)− − 4nCt is strictly decreasing for t ≥ 0, Φ(0)=(1/‖ψ‖) and there exists an unique t ∈ (0, +∞) such that Φ(t)=0. Therefore, we can choose T(ψ) such that 0 < T(ψ) < T. Moreover, the function W(t)=(Φ(t))−2/ is increasing on [0, T(ψ)], and therefore we have thatfor all t ∈ [0, T(ψ)].

For the case n=1, from (32) we have thatand from Lemma 2 we have ρ(t) ≤ (1/(‖ψ‖2+2C(α+ϵ+2)t)−1 − 8Ct), for 0 ≤ t ≤ T where . So, we can choose T(ψ) such that 0 < T(ψ) < T, and therefore, we conclude thatfor all t ∈ [0, T(ψ)].

As, ‖u(t)‖2 ≤ ρ(t) for all t ∈ [0, T), and since ρ(t) and T do not depend on D, the usual extension method shows that we must have T(D, ψ) ≥ T(ψ) for all D > 0, where T(ψ) is any positive number satisfying 0 < T(ψ) < T.

Theorem 3 .

Let s > 3/2 be fixed. If ψ ∈ H(ℝ), then there exists a T=T(ψ) and a function u0 ∈ C([0, T]; H(ℝ))∩C1([0, T]; H(ℝ)) such that u0(0)=ψ, and u0 satisfies (1) with D=0, in the weak sense, i.e.,for all φ ∈ H(ℝ) and t ∈ [0, T].

Moreover, ‖u0‖2 ≤ ρ(t) for all t ∈ [0, T], where ρ(t) is as in Theorem 2.

Let T=T(ψ) be as in Theorem 2. Now, we will split the proof into four steps:

Step 1 .

First we will show that (u(t)) is a net which converges to a function u0 ∈ C([0, T]; L2(ℝ)) in the L2− norm, uniformly over [0, T].

Let D1, D2 ∈ (0, +∞). Then,

Let , where ρ is the function defined in the proof of Theorem 2. We bound separately each term on the right-hand side of (40) as follows:

In order to bound the first term, we have as follows:

We can bound the second term by the following:

The third term is bounded by |(u − u|u2 − u2|)0| ≤ 2M‖u − u‖02.

Finally, we have(where q is defined in the proof of Proposition 7). From Theorem 2, we obtain as follows:for all t ∈ [0, T].

Therefore, from the above bounds, we have as follows:

Applying Gronwall's inequality to the last relation, we show that there is a constant C > 0 satisfying ‖u(t) − u(t)‖02 ≤ C|D1 − D2| for all t ∈ [0, T], and since L2(ℝ) is complete there exists the limit u0(t)=limu(t) in L2(ℝ) uniformly with respect to t ∈ [0, T], i.e.and so u0 ∈ C([0, T]; L2(ℝ)).

Step 2 .

Now we show that u0 ∈ H(ℝ). Let t ∈ [0, T]. Since u⟶u0 in L2(ℝ), as D⟶0+, then there exists a subsequence {D(} such that

We obtain by Fatou's Lemma as follows:

Step 3 .

We must show that u⇀u0 in H(ℝ) for all t ∈ [0, T] as D⟶0+.

First of all, we will show that (u(t)) is a weak Cauchy net in H(ℝ), uniformly with respect to t ∈ [0, T]. In fact, given φ ∈ H(ℝ) and ϵ > 0, choosing φ ∈ H(ℝ) such that ‖φ − φ‖ ≤ ϵ, thenand therefore, we have lim+sup(u(t) − u(t)|φ|)=0.

Thus, we have that u⇀u0 for all t ∈ [0, T], i.e.,for all φ ∈ H(ℝ). Moreover, since the convergence is uniform for all φ ∈ H(ℝ), we can conclude that u0 ∈ C([0, T]; H(ℝ)).

Step 4 .

Finally, we show that u0 ∈ C1([0, T]; H(ℝ)).

Let φ ∈ H(ℝ). Then,for all t ∈ [0, T]. Since u⟶u0 in L2(ℝ) and u⇀u0 in H(ℝ), we have ∂u⇀∂u0 in H(ℝ) and ∂2u⇀∂2u0 in H(ℝ) uniformly on [0, T]. Observe that if r > 1/2, f⇀f in H(ℝ) and g⇀g in H(ℝ) then fg⇀fg in H(ℝ). After, we haveuniformly on [0, T]. Thereby, taking the limit as D⟶0+ in (51), we obtain as follows:

Corollary 1 .

Let u0 be as in the preceding theorem, then u0 ∈ AC([0, T]; H(ℝ)).

Since t ∈ [0, T(ψ)] ↦ u(u − α)(u − 1)+ϵuu is weakly continuous in H(ℝ) and the Sobolev space is separable, then applying the Bochner–Pettis theorem, it is a strongly measurable function in H(ℝ). Therefore,exists as a Bochner integral. So, from (53) we conclude thatand therefore, u0 ∈ AC([0, T]; H(ℝ)).

Theorem 4 .

Let s > 3/2 and T > 0 be fixed, ψ ∈ H(ℝ), j=1,2, and v ∈ C([0, T]; L2(ℝ))∩C([0, T]; H(ℝ))∩AC([0, T]; H(ℝ)) two weak sense solutions to (1) with D=0 such that v(0)=ψ, j=1,2. Then,where L0 is as in the Proposition 7 and

Let w(t)=v1(t) − v2(t). Since s > 3/2, we have s − 2 > −1/2 > 1 − s and 1 − s > −s, and also H(ℝ)⟶H−(ℝ). Using the fact that w(t) is real valued we havewhere t ∈ [0, T] is fixed, h is such that t+h ∈ [0, T], and 〈·|·|〉 is the H duality bracket. As t ∈ [0, T] ↦ w(t) ∈ H(ℝ) is bounded andexists in the norm of H(ℝ)⟶H−(ℝ), from (58) and (59) we have as follows:

From Proposition 7 and (60) we have as follows:where R is given by (57). Applying Gronwall's inequality to (61), and we have proved the theorem.

Theorem 5 .

Let ψ ∈ H(ℝ) with s > 3/2. Then, there exists a T=T(ψ) > 0 and a unique u0 ∈ C([0, T]; H(ℝ)) such that

From the previous results, there exists a unique solution of (62) in the class described in Theorem 4. Now, we will show that u0 ∈ C([0, T]; H(ℝ)).

Let φ ∈ H(ℝ) be such that ‖φ‖=1. Then, we have |(u0|φ|)| ≤ ‖u0(t)‖ ≤ ρ(t)1/2, for all φ ∈ H(ℝ) and for all t ∈ [0, T]. Additionally,for all φ ∈ H(ℝ). As we have ‖ψ‖=sup‖=1|(ψ|φ|)|, then taking supremum over ‖φ‖=1 in (63) we have liminf+‖u0(t)‖=limsup+‖u0(t)‖=‖ψ‖, i.e., the limit of ‖u0(t)‖ exists as t⟶0+ and lim+‖u0(t)‖=‖ψ‖. Since u(t)⟶ψ weakly in H(ℝ) as t⟶0+, it follows that lim+u0(t)=ψ in the norm of H(ℝ). Let t′ ∈ [0, T) be fixed. Then, there exists , with , and a unique satisfying ∂v(t)+v(t)(v(t) − α)(v(t) − 1)+v(t)∂v=0, with v(0)=u(t′). We have noticed that the uniqueness of solutions implies that v(t)=u0(t+t′), for . Since v is continuous from the right at t=0, then u0 is continuous from the right at t=t′. Now, let t′ ∈ (0, T] be fixed. Observe that the following problemhas a unique solution w(t, x)=u0(t′ − t, −x) with , because the equation in problem (64) is similar to the equation in problem (1) with D=0 and it is easy to show similar results to those obtained for problem (1) with D=0, specially the uniqueness results.

In particular, for the problem (64) there are results analogous to Theorems 3 and 4. Therefore, since w is continuous from the right at t=0, then u0 is continuous from the left at t′. So, u0 ∈ C([0, T]; H(ℝ)). Moreover, we have u0(u0 − α)(u0 − 1)+ϵu0∂u0 ∈ C([0, T]; H(ℝ)). From (55) we also conclude that u0 ∈ C1([0, T]; H(ℝ)) and that it is the unique strong solution of (1) with D=0.

Theorem 6 .

Let s > 3/2, ψ ∈ H(ℝ) and u ∈ C([0, T(ψ)]; H(ℝ))∩C1([0, T(ψ)]; H(ℝ)) be the corresponding solution of the problem (1) for D ≥ 0, defined in the interval [0, T(ψ)] which is independent of D. Then, u can be extended, if necessary, to the interval [0, T], with ψ viewed as an element of H(ℝ).

Applying (30) with r=s+1 to obtain

Now, from inequality (3.12) and Theorem 3.2 in [14], we obtain as follows:

Therefore, using (66), (67) in (65), we have as follows:and integrating from 0 to t, we have as follows:and applying the Gronwall's inequality to obtain

Observe that on the right-hand side of (70) is well-defined for t ∈ [0, T(ψ)] and therefore we can extend (if necessary) u=u(t) to [0, T(ψ)] as a solution in H(ℝ). Thus, we conclude that T(ψ) ≤ T(ψ). So, u ∈ C([0, T(ψ)]; H(ℝ)) for D > 0. From (70) also we have that

Observe that the last inequality is independent of D > 0 and since u weakly converges and uniformly to u0 in H(ℝ), then we have u0 ∈ C([0, T(ψ)]; H(ℝ)).

Following Lemma 5 in [15] we have the next lemma.

Lemma 4 .

Let s > 3/2. For ψ ∈ H(ℝ) and τ > 0, we define

Then, lim+‖ψ − ψ‖=0, and there exists a constant C=C(s) such that

Moreover, lim+‖ψ − ψ‖=0 uniformly on compact subsets of H(ℝ).

Notice that Then, using Lebesque's dominated convergence theorem, we obtain lim+‖ψ − ψ‖=0. Now, to prove the uniformity on compact subsets, it is enough to show that ψ⟶ψ in H(ℝ) implies lim+‖ψ − ψ‖=0 uniformly for n=1,2,…, since sequential compactness is equivalent to compactness in metric spaces. Thus, observe that

Let ϵ > 0 be given and choose N such that if n ≥ N, then ‖ψ − ψ‖ < (1/3)ϵ. Thus, for τ0 > 0 small enough that 0 < τ < τ0 we havefor 1 ≤ n ≤ N. Now, if n ≥ N then we have

Hence (75) holds for all n.

On the other hand, we have

Finally, using the mean value theorem, |e− − e−| ≤ |τ − θ|(1+ξ2), and then we have

The proof is complete.

Proposition 8 .

Let s > 3/2, ψ ∈ H(ℝ), ψ (for τ > 0) be as in the preceding lemma. If u0 is solution of the problem (62) with u0(0)=ψ, for all τ > 0, then there are constants C=C(s, ‖ψ‖, T) > 0 and η=η(s) ∈ (0,1) such thatfor τ sufficiently small and 0 ≤ θ ≤ 2τ.

Let τ0 > 0 be such that u0(t) is well-define in [0, T] for all 0 < τ ≤ τ0. Then,

Now, the right-hand side of the inequality (3) will be estimated.

First, we will estimate (80a). Applying the Cauchy–Schwartz inequality to (80a) we havewhere and .

Now, we will estimate (80b). Observe that

Finally, we estimate (80c). As s > 3/2, there is s0 such that 3/2 < s0+1 < s. From the Cauchy–Schwartz inequality, we obtain

Now, we will estimate each term on the right-hand side of the last inequality. First, observe thatwhere q is defined in the proof of Proposition 7.

We also estimate ‖(u0) − (u0)‖‖u0‖. From Lemma 4 and the inequality (71), we havefor all τ ≤ τ0.

From Lemma 4, we havewhere ϱ=(s0/s). To estimate the term ‖u0 − u0‖0, observe thatwhere , and from Gronwall inequality we have as follows:

From Lemma 4,for 0 ≤ θ ≤ τ. Therefore, we haveand the term (80c) is bounded for

Of the bounds that were found for (80a), (80b), (80c) we conclude thatand using Gronwall inequality, we obtain (79).

The following corollary follows immediately from Proposition 8 and Lemma 4.

Corollary 2 .

Let F be a compact subset in H(ℝ). Suppose that ψ ∈ F, ψ and u0 are defined as in the preceding result. Then, u0 converges uniformly to u0, for all t ∈ [0, T], as τ⟶0+.

Theorem 7 .

The map ψ ↦ u0 is continuous in the following sense: let ψ ∈ H(ℝ), j=1,2,3,… such that ψ⟶ψ in H(ℝ) and u0, ∈ C((0, T); H(ℝ))∩C1((0, T]; H(ℝ)) are the corresponding solutions of the problem (62) with initial condition u0,(0)=ψ. Let T ∈ (0, T). Then, there exists a positive integer N0=N0(s, ψ) such that T ≥ T for all j ≥ N0 and

Consider ψ ∈ H(ℝ) and let {ψ} be a sequence in H(ℝ) such that converges to ψ. Suppose that u0, u0,, u0, u0, are the corresponding solutions of (62) with initial values ψ, ψ, ψ, ψ, respectively. As ψ⟶ψ, from Corollary 2 we have that u0, converges uniformly to u0, and u0 converges uniformly to u0, as τ⟶0+. Thus, given ϵ > 0, for τ sufficiently small, we havefor all t ∈ [0, T]. Now, we will show that ‖u0, − u0‖ converges uniformly to zero, as j⟶∞. In fact, from Lemma 3 and the Cauchy–Schwartz inequality, we have as follows:

Applying Gronwall inequality to the last relation and the fact that ‖ψ − ψ‖ ≤ ‖ψ − ψ‖, we have

Therefore, for j sufficiently large, we have ‖u0, − u0‖ ≤ ϵ, for all t ∈ [0, T].

Finally, the results obtained above can be summarized as follows:

Theorem 8 .

Let s > 3/2. For D=0, the problem (1) is locally well-posed in H(ℝ).