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

Lizarazo, Vladimir; De la cruz, Richard; Lizarazo, Julio

doi:https://doi.org/10.1155/2022/5891265

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Research Article | Open Access

Volume 2022 | Article ID 5891265 | https://doi.org/10.1155/2022/5891265

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

Vladimir Lizarazo,¹Richard De la cruz,²and Julio Lizarazo²

Academic Editor: Daiji Endoh

Received30 Dec 2021

Accepted16 Apr 2022

Published28 May 2022

Abstract

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 . If , the local well-posedness is given for and for if .

1. Introduction

In this paper, we show the local well-posedness for the following Cauchy problem:where is a constant diffusion coefficient, and 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, models the population density, is a positive integer, and 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 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 , 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 , the equation (1) is a generalized Burgers equation (without viscosity) and nonlinear source term. Therefore, from the mathematical viewpoint, the case 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 for if , and for if . 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 . 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 , 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: for the real numbers; for the Schwartz’s space usual; denotes the Fourier transform of ; the inverse Fourier transform will be denoted by ; by , , the set of all such that . is called the Sobolev space and it is a Hilbert space with respect to the inner product ; for the space of all continuous functions on an interval into the Banach space ; if is compact, is seen as a Banach space with the sup norm; for the space of all weakly continuous functions on an interval into Banach space ; for the space of all weakly differentiable functions on an interval into Banach space . We also denote by , , the semigroup in generated by the operator where , i.e., is a -semigroup of contractions in , . Moreover, is the unique solution to the linear problem associated with (1), i.e., is the unique solution to the following problem.

Proposition 1. Let , , , and . Then, there exists a constant , depending only on , such that

In particular, for all .

When there is no risk of confusion, we will use the notations for , for , and .

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

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 for . In addition, the uniqueness of the solution and continuous dependence are established.

Proposition 2. Let be fixed. Then, is a continuous map from into and satisfies the estimates as follows:for all , where is a continuous function, nondecreasing with respect to each of their arguments. In particular,

Proof. Observe that . Then, as is a Banach algebra for , we have the following:whereThe 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 be fixed, , , and is defined by (2). Then, there exists and a unique function satisfying the following integral equation:

Sketch of proof. Let be fixed, but arbitrary. Consider the following:which is a complete metric space with distance . Define on the space the following map:We have the following:(1)If then (2)We can choose sufficiently small such that (3)There exists such that is a contraction on So, has a unique fixed point in which satisfies the integral equation (40) where .

Proposition 4. The problem (1) is equivalent to the integral equation (40). More precisely, if and is a solution of (1), then satisfies the integral equation (40). Conversely, if and is a solution of (40) then and satisfies (1).

proof. Assume that is a solution of (1). Then, , . So, satisfies the integral equation (40). Conversely, assume that is a solution of (40). For , let . Then, for arbitrary,However,and the right hand side of (57) is a integrable function of in . Thus, using the dominated convergence theorem, we have as follows:Now, from the mean value theorem for integrals, there exists a value on the interval such thatand therefore, .
After, in . where is the right derivative. In similar way, we can conclude that the left derivative is in . So, and . As is the solution of the linear problem (3), we conclude that and satisfies (1).

Lemma 1. Suppose , , , , , is nonnegative and is locally integrable on . Ifi.e., in , thenwhere , with and for .

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

Proposition 5. Let and be the corresponding solutions of equation (9). If , thenwhere , and (here is given by previous lemma).

proof. Let and as in the statement of the proposition. Let . From (9) we have as follows:By Propositions 1 and 2, we obtain the following:where . Let . Observe that is finite. In fact, where . Therefore, and from Lemma 1 we have thatAs for all , with , we have that and is bounded for . From (21), we obtain as follows:So, and , with , is convergent.

Proposition 6. Let . Then, the map is continuous in the following sense: if in and , where , are the corresponding solutions (of the problem (1) with ). Let . Then, there exists a positive integer such that for all and

proof. As is a continuous function a , then there exists such that for all . Let . Therefore, is defined on for all . It follows that for all and satisfies where . Therefore, for all and . Now, similar to the proof of the previous proposition, we have as follows:Let . Thus, is finite (where 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 . The problem (1) is locally well-posed in .

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 using a priori estimate and the parabolic regularization method, the so-called vanishing viscosity method (for more details see [12]).

Lemma 2. Let , and be real valued positive continuous functions defined on . Let and be positive continuous functions for , with strictly increasing and nondecreasing. Define and Then, the inequalityimplies the inequalitywhere , , and .

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

Proposition 7. Let be fixed. Then, satisfies the estimatefor all , where .

proof. We define . As thus and are Banach algebras. Moreover, we have that and . Thus, using the Cauchy–Schwartz inequality, we have as follows:

Lemma 3. (T. Kato). Let and be fixed and are real valued functions. Then, there exists a constant such that

In particular, .

proof. See Lemma A.5. in [13].

Theorem 2. Let be fixed. For , consider the initial value problem (1) with initial data and let be the corresponding solution of (1) for some . Then, there exists a , depending on , such that can be extended to the interval , and there is a function such that and .

proof. Using the inner product in and Lemma 3 we have thatThen, for all , where is the maximally extended solution of the following problem.For , from the problem (32) we obtain as follows:and integrating from 0 to we have as follows:From Lemma 2 with , , and we have the following bound:for , where . Observe that , since the function is strictly decreasing for , and there exists an unique such that . Therefore, we can choose such that . Moreover, the function is increasing on , and therefore we have thatfor all .
For the case , from (32) we have thatand from Lemma 2 we have , for where . So, we can choose such that , and therefore, we conclude thatfor all .
As, for all , and since and do not depend on , the usual extension method shows that we must have for all , where is any positive number satisfying .

Theorem 3. Let be fixed. If , then there exists a and a function such that , and satisfies (1) with , in the weak sense, i.e.,for all and .

Moreover, for all , where is as in Theorem 2.

proof. Let be as in Theorem 2. Now, we will split the proof into four steps:

Step 1. First we will show that is a net which converges to a function in the norm, uniformly over .
Let . Then,Let , where 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 .
Finally, we have(where is defined in the proof of Proposition 7). From Theorem 2, we obtain as follows:for all .
Therefore, from the above bounds, we have as follows:Applying Gronwall’s inequality to the last relation, we show that there is a constant satisfying for all , and since is complete there exists the limit in uniformly with respect to , i.e.and so .

Step 2. Now we show that . Let . Since in , as , then there exists a subsequence such thatWe obtain by Fatou’s Lemma as follows:

Step 3. We must show that in for all as .
First of all, we will show that is a weak Cauchy net in , uniformly with respect to . In fact, given and , choosing such that , thenand therefore, we have .
Thus, we have that for all , i.e.,for all . Moreover, since the convergence is uniform for all , we can conclude that .

Step 4. Finally, we show that .
Let . Then,for all . Since in and in , we have in and in uniformly on . Observe that if , in and in then in . After, we haveuniformly on . Thereby, taking the limit as in (51), we obtain as follows:

Corollary 1. Let be as in the preceding theorem, then .

proof. Since is weakly continuous in and the Sobolev space is separable, then applying the Bochner–Pettis theorem, it is a strongly measurable function in . Therefore,exists as a Bochner integral. So, from (53) we conclude thatand therefore, .

Theorem 4. Let and be fixed, , , and two weak sense solutions to (1) with such that , . Then,where is as in the Proposition 7 and

proof. Let . Since , we have and , and also . Using the fact that is real valued we havewhere is fixed, is such that , and is the duality bracket. As is bounded andexists in the norm of , from (58) and (59) we have as follows:From Proposition 7 and (60) we have as follows:where is given by (57). Applying Gronwall’s inequality to (61), and we have proved the theorem.

Theorem 5. Let with . Then, there exists a and a unique such that

proof. From the previous results, there exists a unique solution of (62) in the class described in Theorem 4. Now, we will show that .
Let be such that . Then, we have , for all and for all . Additionally,for all . As we have , then taking supremum over in (63) we have , i.e., the limit of exists as and . Since weakly in as , it follows that in the norm of . Let be fixed. Then, there exists , with , and a unique satisfying , with . We have noticed that the uniqueness of solutions implies that , for . Since is continuous from the right at , then is continuous from the right at . Now, let be fixed. Observe that the following problemhas a unique solution with , because the equation in problem (64) is similar to the equation in problem (1) with and it is easy to show similar results to those obtained for problem (1) with , specially the uniqueness results.
In particular, for the problem (64) there are results analogous to Theorems 3 and 4. Therefore, since is continuous from the right at , then is continuous from the left at . So, . Moreover, we have . From (55) we also conclude that and that it is the unique strong solution of (1) with .

Theorem 6. Let , and be the corresponding solution of the problem (1) for , defined in the interval which is independent of . Then, can be extended, if necessary, to the interval , with viewed as an element of .

proof. Applying (30) with to obtainNow, 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 , we have as follows:and applying the Gronwall’s inequality to obtainObserve that on the right-hand side of (70) is well-defined for and therefore we can extend (if necessary) to as a solution in . Thus, we conclude that . So, for . From (70) also we have thatObserve that the last inequality is independent of and since weakly converges and uniformly to in , then we have .
Following Lemma 5 in [15] we have the next lemma.

Lemma 4. Let . For and , we define

Then, , and there exists a constant such that

Moreover, uniformly on compact subsets of .

proof. Notice that Then, using Lebesque’s dominated convergence theorem, we obtain . Now, to prove the uniformity on compact subsets, it is enough to show that in implies uniformly for , since sequential compactness is equivalent to compactness in metric spaces. Thus, observe thatLet be given and choose such that if , then . Thus, for small enough that we havefor . Now, if then we haveHence (75) holds for all .
On the other hand, we haveFinally, using the mean value theorem, , and then we haveThe proof is complete.

Proposition 8. Let , , (for ) be as in the preceding lemma. If is solution of the problem (62) with , for all , then there are constants and such thatfor sufficiently small and .

proof. Let be such that is well-define in for all . 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 thatFinally, we estimate (80c). As , there is such that . From the Cauchy–Schwartz inequality, we obtainNow, we will estimate each term on the right-hand side of the last inequality. First, observe thatwhere is defined in the proof of Proposition 7.
We also estimate . From Lemma 4 and the inequality (71), we havefor all .
From Lemma 4, we havewhere . To estimate the term , observe thatwhere , and from Gronwall inequality we have as follows:From Lemma 4,for . Therefore, we haveand the term (80c) is bounded forOf 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 be a compact subset in . Suppose that , and are defined as in the preceding result. Then, converges uniformly to , for all , as .

Theorem 7. The map is continuous in the following sense: let , such that in and are the corresponding solutions of the problem (62) with initial condition . Let . Then, there exists a positive integer such that for all and

proof. Consider and let be a sequence in such that converges to . Suppose that , , , are the corresponding solutions of (62) with initial values , , , , respectively. As , from Corollary 2 we have that converges uniformly to and converges uniformly to , as . Thus, given , for sufficiently small, we havefor all . Now, we will show that converges uniformly to zero, as . 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 , we haveTherefore, for sufficiently large, we have , for all .
Finally, the results obtained above can be summarized as follows:

Theorem 8. Let . For , the problem (1) is locally well-posed in .

Data Availability

No data were used to support this study.

Conflicts of Interest

The authors declare that they have no conflicts of interest.

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Copyright © 2022 Vladimir Lizarazo et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

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