Global well-posedness and exponential decay estimates for semilinear Newell–Whitehead–Segel equation

Javed Hussain; Munawar Ali

doi:10.1515/nleng-2024-0041

Article Open Access

Global well-posedness and exponential decay estimates for semilinear Newell–Whitehead–Segel equation

Javed Hussain and Munawar Ali

Published/Copyright: November 11, 2024

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From the journal Nonlinear Engineering Volume 13 Issue 1

Abstract

This article presents the application of the Faedo–Galerkin compactness method to establish the local well-posedness of the Newell–Whitehead–Segel equation. By analyzing a finite-dimensional approximate problem, the existence and uniqueness of a local solution were demonstrated. A priori estimates were derived, enabling the transition to the limit and the recovery of the original problem’s local solution. The study further proves the uniqueness and continuous dependence of the solution on initial data. Additionally, under certain conditions, it is shown that the energy norm of the solution decays exponentially over time, and the L 2 -norm of the time derivative of the solution approaches zero asymptotically.

Keywords: amplitude equations; Faedo–Galerkin method; well-posedness; exponential decay of solution

1 Introduction

In this article, we are concerned with the following version of the initial boundary value problem consisting of Newell–Whitehead–Segel Equation (NWSE),

(1.1) ∂ t u = k Δ u + α u − β u q + f ( t , x ) , ( t , x ) ∈ R + × Ω u = 0 , on δ = ∂ Ω , u ( 0 ) = u 0 ( x ) , for x ∈ Ω ,

where α , β , and k are the positive real numbers with β > α + 1 2 and q is the positive odd integer, Ω ⊂ R n , a bounded domain with smooth boundary Γ = ∂ Ω , u 0 ∈ H 0 1 ( Ω ) and forcing factor f ∈ L 2 ( [ 0 , T ] ; L 2 ( Ω ) ) .

Across various realms, a plethora of phenomena exhibit stripe patterns, as shown in the study by Jakobsen et al. [1], spanning from the distinctive markings on a zebra’s skin to the intricate ridges of human fingerprints, the neural pathways in the visual cortex, the undulating ripples in the sand, and the ornate stripes adorning seashells. These stripe patterns’ emergence and dynamics find elucidation through a set of evolutionary equations known as amplitude equations (AE). Among these foundational equations, the NSWE, cf. [2], and the Swift–Hohenberg equation (SHE), cf. [3], stand out prominently. Problem 1.1 under investigation here also falls within the realm of AEs, closely aligned with NSWE. The NSWE has undergone extensive scrutiny across various domains; for instance, Rosu and Cornejo-Pérez [4] employed it to delineate the process of Bernard–Rayleigh convection in fluid mixtures near a bifurcation point. Furthermore, diverse incarnations of NSWE-type amplitude equations exhibit universality across physical phenomena, spanning from relativity, as shown in Hagberg et al. [5], to plasma physics, as discussed in the study by Tamang et al. [6], and even extending to astrophysics [7] and biological systems [8]. Bibi and Ahmad [9] computed Lie and discrete symmetry transformation groups of linear and nonlinear Newell–Whitehead–Segel (NWS) equations. Gandhi et al. [10] have carried out a comparative analysis of the solutions of time-fractional NWS, obtained through hotopy analysis and fractional reduced transform. Experimental evidence is given to demonstrate the efficiency of these methods. Rehman and co-authors [11] obtained a solution of NWS based on the extended direct algebraic method (EDAM). Angad [12] developed some lifting schemes using different wavelet filter coefficients for numerical computation of the solution of linear and nonlinear NWS. Inan et al. [13], using explicit exponential finite difference methods, obtained the analytical and numerical solutions for the NWS in the context of biology. Tuan et al. [14] studied fractional rheological models and NWS equations with the non-local and non-singular kernel. Elgazery [15] studied the periodic solution of the wave-type NWS equation. Ayata [16] employed the conformable Laplace decomposition method (CLDM), which is applied to fractional NWS. Latif and co-authors [17] used the semi-analytics iterative method for solving NWS. Wasim et al. [18] computed the numerical solution of NWS via the exponential B-spline collocation method. Chu and co-authors [19] obtained the exact solutions of the nonlinear evolution equations, including NWS, using the first integral method. Prakash et al. [20] employed the fractional variational iteration method for solving time-fractional NWS. Korkmaz [8], using homogeneous balance and Sine-Gordon equation expansion method, computed the explicit solution NWS and Zeldovich equation. Saadeh and co-authors [21] studied fractional NWS based on the residual power series algorithm. Saberi et al. [22] carried out the Lie symmetry analysis of the version of the NWS, which is of fractional order in time and space. Vanessa et al. [23] provided the classification of the reduction operators and exact solutions of variable coefficient NWS. Some of the related equations studied can be found in previous studies [24–26].

The main motivation for this work comes from the fact that there have been attempts made to numerically study the NWSE, but abstract well-posedness has not been done. In this work, we aim to fill this gap and study the well-posedness and some dynamical properties of the solution to NWSE.

2 Notations and preliminaries

Let Ω ⊆ R n be a bounded smooth domain. For, p ∈ [ 0 , ∞ ) , let L p ( Ω ) denote the Banach space of [equivalence classes] Lebesgue measurable R -valued pth power integrable functions on the set Ω . The norm on L p ( Ω ) is given by

∥ u ∥ L p ( Ω ) ≔ ∫ O ∣ u ( x ) ∣ p d x 1 p , u ∈ L p ( Ω ) .

For p = 2 , to simplify the presentation, throughout the study, we will use ∥ ⋅ ∥ as the L 2 ( Ω ) -norm and ⟨ ⋅ , ⋅ ⟩ as L 2 ( Ω ) inner product. For k ∈ N and p ∈ [ 0 , ∞ ) , by W k , p ( Ω ) , we denote the Sobolev space of all u ∈ L p ( Ω ) for which the weak derivative D α u ∈ L p ( Ω ) , ∣ α ∣ ≤ k . In particular, for p = 2 , we denote H k ≔ W k , 2 ( Ω ) and its norm by ∣ ⋅ ∣ H k ( Ω ) . In particular, H 1 ( Ω ) is a Hilbert space with the following inner product:

⟨ u , v ⟩ H 1 ≔ ⟨ u , v ⟩ L 2 ( Ω ) + ⟨ ∇ u , ∇ v ⟩ L 2 ( Ω ) , u , v ∈ H 1 ( Ω ) .

We also denote by H 0 1 , 2 ( Ω ) the closure in H 1 , 2 ( Ω ) of the space C 0 ∞ ( Ω ) equipped with the norm

∣ u ∣ H 0 1 , 2 ( Ω ) 2 ≔ ⟨ ∇ u , ∇ u ⟩ L 2 ( Ω ) , u ∈ H 0 1 ( Ω ) ,

which, in view of the Poincaré inequality, is equivalent to the norm induced by the H 1 ( Ω ) -norm.

Finally, let A be the Laplace operator with Dirichlet boundary conditions, i.e., a linear operator defined by

(2.1) D ( A ) = H 0 1 , 2 ( Ω ) ∩ H 2 , 2 ( Ω ) , A u = − Δ u , u ∈ D ( A ) .

It is well known that, cf. [27] (Theorem 4.1.2, page 79), A is a self-adjoint positive operator in H , D ( A 1 ⁄ 2 ) = H 0 1 , 2 ( Ω ) , and

∣ A 1 ⁄ 2 u ∣ L 2 ( Ω ) 2 = ∣ ∇ u ∣ L 2 ( Ω ) 2 , u ∈ H 0 1 , 2 ( Ω ) .

By ℒ ( X , Y ) , we mean the space of all bounded linear operators from Banach X to the Banach space Y . For any b > a ≥ 0 , and a separable Banach space X , let us denote by L 2 ( a , b ; X ) the space of [equivalence classes] of all Borel measurable functions u : [ a , b ] → X , such that

∣ u ∣ L 2 ( a , b ; X ) ≔ ∫ a b ∣ u ( t ) ∣ X 2 d t 1 ⁄ 2 < ∞

and

‖ u ‖ L ∞ ( a , b ; X ) = ess sup a < t < b ‖ u ( t ) ‖ X , for p = ∞ .

3 Faedo–Galerkin well-posedness of NSWE

Theorem 3.1

Let u 0 ∈ H 0 1 , 2 ( Ω ) ∩ L q + 1 ( Ω ) , α , β , and k be the positive real numbers with β > α + 1 2 and q be the positive integer bigger than 1, and then, there exists a unique solution u to problem (1.1) such that

(3.1) u ∈ L ∞ ( [ 0 , T ] , H 0 1 , 2 ( Ω ) ∩ L q + 1 ( Ω ) ) ∩ L 2 ( [ 0 , T ] , H 2 , 2 ( Ω ) ∩ H 0 1 , 2 ( Ω ) )

and

(3.2) u ′ ∈ L 2 ( [ 0 , T ] , L 2 ( Ω ) ) .

Proof

Step 1. The Faedo–Galerkin approximation: Let us fix a basis w j ∈ H 0 1 ( Ω ) satisfying

− k Δ w j = λ j w j , w j ∣ p = 0 , ∥ w j ∥ L 2 = 1 .

By the elliptic operator theory [28] (Section 1.3.4 pages 12–13), we may infer that w j are also the basis for H 0 1 , 2 ( Ω ) ∩ L q + 1 ( Ω ) . Now, suppose m be an integer such that

u m ( t ) = ∑ i = 1 m g i m ( t ) w i ( x ) ,

such that

(3.3) ⟨ ∂ t u m ( t ) , w i ⟩ − ⟨ k Δ u m ( t ) , w i ⟩ + ⟨ b u m q ( t ) − α u m ( t ) , w i ⟩ = ⟨ f , w i ⟩ , ⟨ u m ( 0 ) , w i ⟩ = ξ i m ,

where i = 1 , … , m ,

lim m → + ∞ ∑ i = 1 m ξ i m w i → u 0 in H 0 1 , 2 ( Ω ) ∩ L q + 1 ( Ω ) ,

keep in view that u 0 ∈ H 0 1 , 2 ( Ω ) ∩ L q + 1 ( Ω ) so the existence of ξ i m is guaranteed. Next,

(3.4) ⟨ ∂ t u m ( t ) , w i ⟩ = ∑ j = 1 m d g j m d t ( t ) w j , w i = d g i m d t ∥ w i ∥ L 2 = d g i m d t .

Now,

(3.5) ⟨ − k Δ ∂ t u m ( t ) , w i ⟩ = − k Δ ∑ j = 1 m g j m w j , w i = ∑ j = 1 m g j m ( − k Δ w j ) , w i = ∑ j = 1 m g j m ⟨ λ j w j , w i ⟩ = λ i w i

and

⟨ β u m q − α u m ( t ) , w i ⟩ = β ∑ j = 1 m g j m w j q − α ∑ j = 1 m g j m w j , w i ≕ G i ( g ) .

and

(3.6) ξ m i = ⟨ u m ( 0 ) , w i ⟩ = ∑ j = 1 m g j m ( 0 ) w j , w i = ∑ j = 1 m g j m ( 0 ) ⟨ w j , w i ⟩ = g i m .

From

(3.7) d g i m ( t ) d t + λ i g i m ( t ) + G i ( g ) = f i ( t ) ; i = 1 , 2 , … , m

(3.8) g i m ( 0 ) = ξ i m = 1 , … , m .

As G i ( g ) is a q th degree polynomial, and f i = ⟨ f , w i ⟩ ∈ L 2 [ 0 , T ] . Thus, by using the standard Picard iterative method for ordinary differential equations (ODEs), it follow that there exist a unique local solution, i.e., there exist t 0 ( ∣ ξ i m ∣ ) > 0 such that g i m solution of ODEs in [ 0 , t 0 ] ; moreover, g i m ∈ C [ 0 , t 0 ] and g i m ′ ∈ L 2 [ 0 , T ] .

Step 2 A priori estimates.

The first estimate. Now, we prove key a priori estimates. Multiplying both sides of equation (3.3) by g i m ( t ) and summing over i , we obtain

(3.9) ⟨ ∂ t u m ( t ) , u m ( t ) ⟩ − k ⟨ Δ u m ( t ) , u m ( t ) ⟩ + β ⟨ u m ( t ) q , u m ( t ) ⟩ − α ⟨ u m ( t ) , u m ( t ) ⟩ = ⟨ f , u m ( t ) ⟩ .

Now,

(3.10) ⟨ ∂ t u m ( t ) , u m ( t ) ⟩ = 1 2 d d t ∥ u m ( t ) ∥ 2 , ⟨ Δ u m ( t ) , u m ( t ) ⟩ = − ⟨ ∇ u m , ∇ u m ( t ) ⟩ (using Green identity) , ⟨ u m q , u m ( t ) ⟩ = ∫ Ω u m q + 1 d x = ∥ u m ( t ) ∥ L q + 1 q + 1 , ⟨ u m ( t ) , u m ( t ) ⟩ = ∥ u m ( t ) ∥ 2 .

Using the aforementioned set of identities in (3.10), and using Young’s inequality, we obtain

1 2 d d t ∥ u m ( t ) ∥ 2 + k ∥ ∇ u m ( t ) ∥ 2 + β ∥ u m ( t ) ∥ L q + 1 q + 1 − α ∥ u m ( t ) ∥ 2 = ⟨ f , u m ( t ) ⟩ ≤ ‖ f ‖ 2 2 + ‖ u m ( t ) ‖ 2 2 .

On integration from 0 to t , we obtain

(3.11) 1 2 ∥ u m ( t ) ∥ 2 + k ∫ 0 t ∥ ∇ u m ( s ) ∥ 2 d s + β ∫ 0 t ∥ u m ( s ) ∥ L q + 1 q + 1 d s ≤ α ∫ 0 t ∥ u m ( s ) ∥ 2 + 1 2 ∥ u m ( 0 ) ∥ 2 + ∫ 0 t ‖ f ( s ) ‖ 2 2 d s + ∫ 0 t ‖ u m ( s ) ‖ 2 2 d s , ≤ α + 1 2 ∫ 0 t ∥ u m ( s ) ∥ 2 + 1 2 ∑ i = 1 m ξ i m 2 + ∫ 0 t ‖ f ( s ) ‖ 2 2 d s .

Before proceeding ahead, let us use Young’s inequality for a = u m 2 , b = 1 , and p = q + 1 2 ,

(3.12) ∥ u m ( t ) ∥ 2 ≤ ∫ Ω u m 2 d x ≤ ∫ Ω ( u m 2 ) q + 1 2 q + 1 2 d x + ∫ Ω q + 1 q − 1 d x ≤ 2 q + 1 ∫ Ω u m q + 1 d x + q + 1 q − 1 μ ( Ω ) ≤ ∥ u m ( t ) ∥ L q + 1 q + 1 + C 1 ,

where C 1 ≔ q + 1 q − 1 μ ( Ω ) < ∞ , and note that as q > 1 so 2 q + 1 < 1 . Hence, using the aforementioned last inequality in (3.11) and the fact that f ∈ L 2 ( [ 0 , T ] ; L 2 ( Ω ) ) , we obtain

(3.13) 1 2 ∥ u m ( t ) ∥ 2 + k ∫ 0 t ∥ ∇ u m ( s ) ∥ 2 d s + β − α + 1 2 ∫ 0 t ∥ u m ( s ) ∥ L q + 1 q + 1 d s ≤ 1 2 ∑ i = 1 m ξ i m 2 + ∫ 0 t ‖ f ( s ) ‖ 2 2 d s + α + 1 2 C 1 ≤ 1 2 ∑ i = 1 m ξ i m 2 + ∫ 0 T ‖ f ( s ) ‖ 2 2 d s + α + 1 2 C 1 ≕ C < ∞ ,

where C is the positive constant depending on ∥ u 0 ∥ , ∫ 0 t ‖ f ( s ) ‖ 2 2 d s and T ; also, we used constraint β > α + 1 2 . It is following from the last inequality that

(3.14) sup t ∈ [ 0 , T ] ∥ u m ( t ) ∥ 2 ≤ C , ∫ 0 t ∥ ∇ u m ( s ) ∥ 2 d s ≤ C , ∫ 0 t ∥ u m ( s ) ∥ L q + 1 q + 1 d s ≤ C .

Hence from above, it clearly follows that

(3.15) u m is uniformly bounded L ∞ ( [ 0 , T ] ; L 2 ( Ω ) ) , u m is uniformly bounded L 2 ( [ 0 , T ] ; H 0 1 , 2 ( Ω ) ) , u m is uniformly bounded L q + 1 ( [ 0 , T ] ; L q + 1 ( Ω ) ) .

Second estimate. Now let us move toward some more a priori estimates. Multiplying (3.3) by λ i g i m , taking Δ and summing over i from i = 1 to m with − k Δ u m , and integrating w.r.t x , we obtain

(3.16) ⟨ ∂ t u m ( t ) , − k Δ u m ( t ) ⟩ − k ⟨ Δ u m ( t ) , − k Δ u m ( t ) ⟩ + β ⟨ u m q , − k Δ u m ( t ) ⟩ − α ⟨ u m ( t ) , − k Δ u m ( t ) ⟩ = ⟨ f , − k Δ u m ( t ) ⟩ , ⟨ ∂ t u m ( t ) , − k Δ u m ( t ) ⟩ + k 2 ∥ Δ u m ( t ) ∥ 2 − β k ⟨ u m q ( t ) , Δ u m ( t ) ⟩ = ⟨ f + α u m ( t ) , − k Δ u m ( t ) ⟩ .

Using integration by parts,

(3.17) ⟨ ∂ t u m ( t ) , − k Δ u m ( t ) ⟩ = − k ∫ Ω d u m ( t ) d t Δ u m ( t ) d x = k ∫ Ω ∇ d u m ( t ) d t ∇ u m ( t ) d x = k ∫ Ω d ∇ u m ( t ) d t ∇ u m ( t ) d x = k ∫ Ω 1 2 d d t ( ∇ u m ( t ) ) 2 d x = k 2 d d t ∥ ∇ u m ( t ) ∥ 2 .

Next, again the use of integration parts gives

(3.18) − ⟨ u m q ( t ) , Δ u m ( t ) ⟩ = − ∫ Ω u m q ( t ) Δ u m ( t ) d x = ∫ Ω ∇ ( u m q ( t ) ) ∇ u m ( t ) d x = q ∫ Ω u m q − 1 2 ( t ) ∇ u m ( t ) 2 d x , = q u m q − 1 2 ( t ) ∇ u m ( t ) 2 .

Next, using Young’s inequality, a = ∥ f + α u m ∥ , b = ∥ k Δ u m ∥ , p = q = 2 , and Cauchy Schwartz and then elementary inequality ( x + y ) 2 ≤ 2 ( x 2 + y 2 ) , for all x , y , we obtain

(3.19) ⟨ f + α u m ( t ) , − k Δ u m ( t ) ⟩ ≤ ∥ f + α u m ( t ) ∥ ∥ k Δ u m ( t ) ∥ ≤ ∥ f + α u m ( t ) ∥ 2 2 + ∥ k Δ u m ( t ) ∥ 2 2 ≤ ( ∥ f ∥ 2 + α 2 ∥ u m ( t ) ∥ 2 ) + k 2 2 ∥ Δ u m ( t ) ∥ 2 .

Using Eqs (3.17) and (3.18) and inequality (3.19) in equation (3.16), we obtain

k 2 d d t ∥ ∇ u m ( t ) ∥ 2 + k 2 2 ∥ Δ u m ( t ) ∥ 2 + β k q u m q − 1 2 ( t ) ∇ u m ( t ) 2 ≤ ∥ f ( t ) ∥ 2 + α 2 ∥ u m ( t ) ∥ 2 .

Integrating both sides w.r.t. from 0 to t , and using facts that u m is uniformly bounded in L ∞ ( [ 0 , T ] ; L 2 ( Ω ) ) , u m ( 0 ) ∈ H 0 1 , 2 , and f ∈ L 2 ( [ 0 , T ] ; L 2 ( Ω ) ) , we conclude the following:

k 2 ∥ ∇ u m ( t ) ∥ 2 + k 2 2 ∫ 0 t ∥ Δ u m ( s ) ∥ 2 d s + b k q ∫ 0 t u m q − 1 2 ( s ) ∇ u m ( s ) 2 d s ≤ k 2 ∥ ∇ u m ( 0 ) ∥ 2 + ∫ 0 t ∥ f ( s ) ∥ 2 d s + α 2 ∫ 0 t ∥ u m ( s ) ∥ 2 d s ≤ k 2 ∥ ∇ u m ( 0 ) ∥ 2 + ∫ 0 t ∥ f ( s ) ∥ 2 d s + α 2 C ( T , ‖ u 0 ‖ , q , α ) ≕ K ,

where K depends on T and ‖ u 0 ‖ .

Third estimate. Now, multiplying (3.3) by g i m ′ both sides, and summing w.r.t. i and then integrating w.r.t. t from 0 to t , we have

(3.20) ⟨ ∂ t u m ( t ) , ∂ t u m ( t ) ⟩ − k ⟨ Δ u m ( t ) , ∂ t u m ( t ) ⟩ + β ⟨ u m q , ∂ t u m ( t ) ⟩ = ⟨ f ( t ) + α u m ( t ) , ∂ t u m ( t ) ⟩ .

Next, using Young’s inequality a = ∥ f ( t ) + α u m ( t ) ∥ , b = ∥ ∂ t u m ( t ) ∥ , p = q = 2 , and Cauchy Schwartz and then elementary inequality ( x + y ) 2 ≤ 2 ( x 2 + y 2 ) , for all x , y , we obtain

‖ ∂ t u m ( t ) ‖ 2 + k 2 d d t ∥ ∇ u m ( t ) ∥ 2 + β q + 1 d d t ∥ u m ( t ) ∥ L q + 1 q + 1 ≤ ∥ f + α u m ( t ) ∥ 2 2 + ∥ ∂ t u m ( t ) ∥ 2 2 ≤ ∥ f ∥ 2 + α 2 ∥ u m ( t ) ∥ 2 + ∥ ∂ t u m ( t ) ∥ 2 2 1 2 ‖ ∂ t u m ( t ) ‖ 2 + k 2 d d t ∥ ∇ u m ( t ) ∥ 2 + β q + 1 d d t ∥ u m ( t ) ∥ L q + 1 q + 1 ≤ ∥ f ( t ) ∥ 2 + α 2 ∥ u m ( t ) ∥ 2 .

Integrating both sides from 0 to t , and using facts that u m is uniformly bounded in L ∞ ( [ 0 , T ] ; L 2 ( Ω ) ) , u m ( 0 ) ∈ H 0 1 , 2 ∩ L q + 1 ( Ω ) and f ∈ L 2 ( [ 0 , T ] ; L 2 ( Ω ) ) , we conclude the following:

(3.21) 1 2 ∫ 0 t ‖ ∂ t u m ( s ) ‖ 2 d s + k 2 ∥ ∇ u m ( t ) ∥ 2 + β q + 1 ∥ u m ( t ) ∥ L q + 1 q + 1 ≤ k 2 ∥ ∇ u m ( 0 ) ∥ 2 + β q + 1 ∥ u m ( 0 ) ∥ L q + 1 q + 1 + ∫ 0 t ∥ f ( s ) ∥ 2 d s + α 2 ∫ 0 t ∥ u m ( s ) ∥ 2 d s , ≤ k 2 ∥ ∇ u m ( 0 ) ∥ 2 + β q + 1 ∥ u m ( 0 ) ∥ L q + 1 q + 1 + ∫ 0 t ∥ f ( s ) ∥ 2 d s + α 2 T C ≕ L < ∞ ,

where L depends on ∥ u 0 ∥ , ∥ ∇ u m ( 0 ) ∥ , ∥ u m ( 0 ) ∥ L q + 1 , ∫ 0 t ∥ f ( s ) ∥ 2 d s , and T . From inequalities (3) and (3.21), it readily follows from

(3.22) u m is uniformly bounded L ∞ ( [ 0 , T ] ; H 0 1 , 2 ( Ω ) ∩ L q + 1 ( Ω ) ) , u m is uniformly bounded L 2 ( [ 0 , T ] ; H 2 , 2 ( Ω ) ∩ H 0 1 , 2 ( Ω ) ) , ∂ t u m is uniformly bounded L q + 1 ( [ 0 , T ] ; L 2 ( Ω ) ) .

Using Lemmas 3.1.1 and 3.1.2 from [28],

(3.23) u m ⟶ u weakly star in L ∞ ( [ 0 , T ] , H 0 1 , 2 ∩ L q + 1 ( Ω ) ) , u m ⟶ u weakly in L 2 ( [ 0 , T ] , H 2 , 2 ∩ H 0 1 , 2 ( Ω ) ) , ∂ t u m ( t ) ⟶ u weakly in L 2 ( [ 0 , T ] , L 2 ( Ω ) ) .

Step 3. The limiting process.

Taking B 0 = H 0 1 , 2 ( Ω ) ∩ H 2 , 2 ( Ω ) , B = H 0 2 ( Ω ) , and B 1 = L 2 ( Ω ) , by Theorem 3.1.1 from [28], there is a subsequence (again denoted by) u m of u m , such that

(3.24) u m ⟶ u , strongly in L 2 ( [ 0 , T ] , H 0 1 , 2 ( Ω ) ) ,

and hence, there is a subsequence u m almost everywhere that converges to u in Ω × [ 0 , T ] . It turns out that,

(3.25) u m q ⟶ u q , weakly in L q + 1 q [ 0 , T ] , L q + 1 q ( Ω ) .

By Lemma 3.1.7 and Remark 3.1.7 from [28],

(3.26) u m ( 0 ) = ∑ i = 1 m ξ i m w i ⟶ u ( 0 ) , weakly in L 2 ( Ω ) ,

but u m ( 0 ) strongly converges in H 0 1 , 2 so it must also converge weakly to u 0 in L 2 ( Ω ) . By uniqueness of limit,

(3.27) u ( 0 ) = u 0 ( x ) .

Since each term on left in (3.3) is weakly convergent in L q + 1 q ( Ω ) , by passing the limit, the following holds in L q + 1 q ( Ω ) :

(3.28) ⟨ ∂ t u m ( t ) , w i ⟩ − k ⟨ Δ u m ( t ) , w i ⟩ + β ⟨ u m q , w i ⟩ = ⟨ f ( t ) + α u m ( t ) , w i ⟩ .

Since w i are the base functions in L q + 1 q ( Ω ) ,

(3.29) u ′ + k Δ u + β u q − α u = f , holds in L q + 1 q [ 0 , T ] , L q + 1 q ( Ω ) .

For 1 ≠ q ∈ N , L q + 1 q [ 0 , T ] , L q + 1 q ( Ω ) can be compactly embedded into L 2 ( [ 0 , T ] , L 2 ( Ω ) ) so the following also holds

(3.30) u ′ + k Δ u + β u q − α u = f , holds in L 2 ( [ 0 , T ] , L 2 ( Ω ) ) .

And from (3.23), we have

(3.31) u ∈ L ∞ ( [ 0 , T ] , H 0 1 , 2 ( Ω ) ∩ L q + 1 ( Ω ) ) ∩ L 2 ( [ 0 , T ] , H 2 , 2 ( Ω ) ∩ H 0 1 , 2 ( Ω ) ) ,

and

(3.32) u ′ ∈ L 2 ( [ 0 , T ] , L 2 ( Ω ) ) .

Step 4 uniqueness: Let u and v be two solutions of (1.1), and then u and v satisfy the following:

(3.33) ∂ t u = k Δ u + α u − β u q + f ( t , x ) , ∂ t v = k Δ v + α v − β v q + f ( t , x ) .

Let w = u − v ; using (3.33), we have

(3.34) w t = k Δ w + α w − β ( u q − v q ) , w ∣ δ = 0 , w ( 0 ) = u ( 0 ) − v ( 0 ) = 0 ,

Multiplying both sides of (3.34) by w , we obtain

(3.35) 1 2 d d t ∥ w ( t ) ∥ 2 + k ∥ ∇ w ( t ) ∥ 2 + ∫ Ω w 2 ( t ) ∑ i = 1 q u q − 1 − i ( t ) v i ( t ) d x = α ∥ w ( t ) ∥ 2 .

Since the term k ∥ Δ w ( t ) ∥ 2 + ∫ Ω w 2 ( t ) ∑ i = 1 q u q − 1 − i ( t ) v i ( t ) d x is positive, we have

(3.36) 1 2 d d t ∥ w ( t ) ∥ 2 ≤ α ∥ w ( t ) ∥ 2 ,

(3.37) ∥ w ( t ) ∥ 2 ≤ ∥ w ( 0 ) ∥ 2 + 2 α ∫ 0 t ∥ w ( s ) ∥ 2 d s .

Now, using the Gronwall inequality with a ≔ ∥ w ( 0 ) ∥ 2 , b ≔ 2 α , and U ( t ) ≔ ∥ w ( t ) ∥ 2 , we have for 0 ≤ t ≤ T

(3.38) ∥ w ( t ) ∥ 2 ≤ ∥ w ( 0 ) ∥ 2 e 2 a t .

Since w ( 0 ) = 0 , from above, we have w ( t ) = 0 , which implies u = v .

Continuous dependence on initial data: Let u and v be two solutions of (1.1), and then, u and v satisfy the following:

(3.39) u t = − k Δ u + a u − b u q + f ,

(3.40) v t = − k Δ v + a v − b v q + g ,

subject to initial condition u ( 0 ) = u 0 and v ( 0 ) = v 0 , respectively, and f and g are forging the assumptions mentioned in statement of running theorem. Let w = u − v using (3.39) and (3.40), we have

(3.41) w t + k Δ w + b ( u q − v q ) − a w = f − g , w ∣ Γ = 0 , w ( 0 ) = u ( 0 ) − v ( 0 ) = u 0 − v 0 .

We will follow the lines as uniqueness. For 0 ≤ t ≤ T , it follows that

(3.42) ∥ u ( t ) − v ( t ) ∥ 2 ≤ ∥ u ( 0 ) − v ( 0 ) ∥ 2 + 2 a ∫ 0 t ∥ w ( s ) ∥ 2 d s + ∫ 0 t ∥ f ( s ) − g ( s ) ∥ 2 d s .

Now, using the Gronwall inequality, it follows that there exists a positive constant c 0 depending on T and ∫ 0 t ∥ f ( s ) − g ( s ) ∥ 2 d s , such that

(3.43) ∥ u ( t ) − v ( t ) ∥ ≤ c 0 ∥ u ( 0 ) − v ( 0 ) ∥ .

Next, following precisely the way (3.13) was derived, by replacing u m by u − v , u 0 by u 0 − v 0 and f by f − g , we obtain the following:

1 2 ∥ u ( t ) − v ( t ) ∥ 2 + k ∫ 0 t ∥ Δ u ( s ) − Δ v ( s ) ∥ 2 d s + b − a + 1 2 ∫ 0 t ∥ u ( s ) − v ( s ) ∥ L q + 1 q + 1 d s + q + 1 q − 1 μ ( Ω ) T ≤ ∥ u 0 − v 0 ∥ 2 + 1 2 ∫ 0 T ∥ f ( s ) − g ( s ) ∥ 2 d s

so there exist positive constants c 1 and c 2 such that

(3.44) ∫ 0 t ∥ u ( s ) − v ( s ) ∥ L q + 1 q + 1 d s ≤ c 1 ∥ u 0 − v 0 ∥ 2 + 1 2 ∫ 0 T ∥ f ( s ) − g ( s ) ∥ 2 d s .

Also,

(3.45) ∫ 0 t ∥ Δ u ( s ) − Δ v ( s ) ∥ 2 d s ≤ c 2 ∥ u 0 − v 0 ∥ 2 + 1 2 ∫ 0 T ∥ f ( s ) − g ( s ) ∥ 2 d s .

Now, following the step taken to derive (3), and by replacing u m by u − v , u 0 by u 0 − v 0 , and f by f − g , we obtain the following:

(3.46) k 2 ∥ Δ u ( t ) − Δ v ( t ) ∥ 2 + k 2 2 ∫ 0 t ∥ Δ u ( s ) − Δ v ( s ) ∥ 2 d s ≤ k 2 ∥ Δ u ( 0 ) − Δ v ( 0 ) ∥ 2 + ∫ 0 T ∥ f ( s ) − g ( s ) ∥ 2 d s + c 3 ∥ u 0 − v 0 ∥ 2 ,

where c 3 is a positive constant. Hence, there exist positive constants c 4 and c 5 such that

(3.47) ∥ Δ u ( t ) − Δ v ( t ) ∥ 2 ≤ c 4 ∥ Δ u ( 0 ) − Δ v ( 0 ) ∥ 2 + ∫ 0 T ∥ f ( s ) − g ( s ) ∥ 2 d s + ∥ u 0 − v 0 ∥ 2

and

(3.48) ∫ 0 t ∥ Δ u ( s ) − Δ v ( s ) ∥ 2 d s ≤ c 5 ∥ Δ u ( 0 ) − Δ v ( 0 ) ∥ 2 + ∫ 0 T ∥ f ( s ) − g ( s ) ∥ 2 d s + ∥ u 0 − v 0 ∥ 2 .

Now, following the step taken to derive (3.21), and by replacing u m by u − v , u 0 by u 0 − v 0 , and f by f − g , we obtain the following:

1 2 ∫ 0 t ‖ u ′ ( s ) − v ′ ( s ) ‖ 2 d s + k 2 ∥ Δ u ( t ) − Δ v ( t ) ∥ 2 + b q + 1 ∥ u ( t ) − v ( t ) ∥ L q + 1 q + 1 ≤ k 2 ∥ Δ u 0 − Δ v 0 ∥ 2 + b q + 1 ∥ u 0 − v 0 ∥ L q + 1 q + 1 + ∫ 0 T ∥ f ( s ) − g ( s ) ∥ 2 d s + c 6 ∥ u 0 − v 0 ∥ 2 ,

where c 6 is the positive constant. Hence, there exist positive constants c 7 , c 8 , and c 9 such that

(3.49) 1 2 ∫ 0 t ‖ u ′ ( s ) − v ′ ( s ) ‖ 2 d s ≤ c 7 ∥ Δ u 0 − Δ v 0 ∥ 2 + ∥ u 0 − v 0 ∥ L q + 1 q + 1 + ∫ 0 T ∥ f ( s ) − g ( s ) ∥ 2 d s + ∥ u 0 − v 0 ∥ 2

and

(3.50) ∥ Δ u ( t ) − Δ v ( t ) ∥ 2 ≤ c 8 ∥ Δ u 0 − Δ v 0 ∥ 2 + ∥ u 0 − v 0 ∥ L q + 1 q + 1 + ∫ 0 T ∥ f ( s ) − g ( s ) ∥ 2 d s + ∥ u 0 − v 0 ∥ 2 .

Also,

(3.51) ∥ u ( t ) − v ( t ) ∥ L q + 1 q + 1 ≤ c 9 ∥ Δ u 0 − Δ v 0 ∥ 2 + ∥ u 0 − v 0 ∥ L q + 1 q + 1 + ∫ 0 T ∥ f ( s ) − g ( s ) ∥ 2 d s + ∥ u 0 − v 0 ∥ 2 .

Adding (3.43)–(3.51) and taking maximum c ≔ max { c 0 , … , c 9 } , it follows that

∥ u − v ∥ L ∞ ( [ 0 , T ] , H 0 2 ( Ω ) ∩ L q + 1 ( Ω ) ) + ∥ u − v ∥ L 2 ( [ 0 , T ] , H 2 , 2 ( Ω ) ∩ H 0 2 ( Ω ) ) + ∥ u ′ − v ′ ∥ L 2 ( [ 0 , T ] , L 2 ( Ω ) ) ≤ c ( ∥ u 0 − v 0 ∥ H 0 2 ( Ω ) ∩ L q + 1 ( Ω ) + ∥ u 0 − v 0 ∥ H 2 , 2 ( Ω ) ∩ H 0 2 ( Ω ) + ∥ f − g ∥ L 2 ( [ 0 , T ] ; L 2 ( Ω ) ) ) .

Finally, applying the standard bootstrap argument, we obtain the global existence. This completes the proof.□

4 Exponential decay of solution

Theorem 4.1

Assume that we are in the framework of 3.1. Additionally, we assume that f ∈ L 2 ( 0 , T ; L 2 ( R ) ) and there exist positive constants such that C 0 and γ 0 satisfy γ 0 − a 2 c + a + 2 > 0 , such that ∥ f ( t ) ∥ ≤ C 0 e − γ 0 t . If u be solution to problem (1.1), then u satisfies the following uniform exponential decay estimates, and there exist positive constants C (depending on C 0 and u 0 ) and γ (depending on γ 0 ) such that

∥ u ( t ) ∥ H 2 , 2 ≤ C e − γ t ,

and therefore, the solution is global.

Proof

Let us start with equation (3.10), and dropping the positive terms from the left-hand side, and Young’s inequality, we obtain

(4.1) k 2 d d t ∥ u m ( t ) ∥ 2 − α k ∥ u m ( t ) ∥ 2 ≤ k ⟨ f ( t ) , u m ( t ) ⟩ , k 2 d d t ∥ u m ( t ) ∥ 2 ≤ α k ∥ u m ( t ) ∥ 2 + k 2 ε ∥ f ( t ) ∥ 2 + 2 k ε ∥ u m ( t ) ∥ 2 . ≤ k 2 ε ∥ f ( t ) ∥ 2 + k ( α + 2 ε ) ∥ u m ( t ) ∥ 2 .

Next, from the beginning of (3),

k 2 d d t ∥ Δ u m ( t ) ∥ 2 + k 2 ∥ Δ u m ( t ) ∥ 2 + β k q u m q − 1 2 ( t ) Δ u m ( t ) 2 = k ⟨ f ( t ) + α u m ( t ) , Δ u m ( t ) ⟩

k 2 d d t ∥ Δ u m ( t ) ∥ 2 ≤ k 2 ε ∥ f ( t ) + α u m ( t ) ∥ 2 + 2 k ε ∥ Δ u m ( t ) ∥ 2 ≤ k ε ∥ f ( t ) ∥ 2 + α 2 k ε ∥ u m ( t ) ∥ 2 + 2 k ε ∥ Δ u m ( t ) ∥ 2 ≤ k ε ∥ f ( t ) ∥ 2 + α 2 k ε ∥ u m ( t ) ∥ 2 + k ( α + 2 ε ) ∥ Δ u m ( t ) ∥ 2 .

Using Friedrichs’s inequality, it follows that there exists a positive constant c > 0 such that,

(4.2) k 2 d d t ∥ Δ u m ( t ) ∥ 2 ≤ k ε ∥ f ( t ) ∥ 2 + k a + 2 ε + α 2 c ε ∥ Δ u m ( t ) ∥ 2 .

Adding inequalities (4.1) and (4.2),

k 2 d d t ∥ u m ( t ) ∥ H 2 , 2 2 ≤ 3 k 2 ε ∥ f ( t ) ∥ 2 + k α + 2 ε + α 2 c ε ∥ u m ( t ) ∥ H 2 , 2 2 .

On multiplying both sides by 2 k and using hypothesis that ∥ f ( t ) ∥ ≤ C 0 e − γ 0 t , it follows that

(4.3) d d t ∥ u m ( t ) ∥ H 2 , 2 2 ≤ 3 ε C 0 2 e − 2 γ 0 t + 2 α + 2 ε + α 2 c ε ∥ u m ( t ) ∥ H 2 , 2 2 .

From Gronwall’s inequality, we infer that

(4.4) ∥ u m ( t ) ∥ H 2 , 2 2 ≤ 3 ε C 0 2 e − 2 γ 0 t e 2 α + 2 ε + α 2 c ε t = 3 ε C 0 2 e 2 α + 2 ε + α 2 c ε − γ 0 t .

Taking m → ∞ , we infer

(4.5) ‖ u ( t ) ‖ H 2 , 2 2 ≤ liminf m → + ∞ ∥ u m ( t ) ∥ H 4 2 ≤ 3 ε C 0 2 e 2 α + 2 ε + α 2 c ε − γ 0 t .

The last inequality is true for ε > 0 so in particular, and − γ ≔ γ 0 − α 2 c + α + 2 > 0 and setting C 2 = 3 C 0 2 , we obtain the following last inequality:

(4.6)□ ∥ u ( t ) ∥ H 2 , 2 ≤ C e − γ t .

Theorem 4.2

Suppose that we are in the framework of Theorem 3.1. Additionally, assume that ∂ t f ∈ L 2 ( [ 0 , T ] , L 2 ( Ω ) ) ∪ L ∞ ( [ 0 , T ] , L 2 ( Ω ) ) , and u ( t ) is the solution to problem (1.1), then

(4.7) ∥ ∂ t u ( t ) ∥ ⟶ 0 , as t ⟶ ∞ .

Proof

In order to prove the theorem, we will show that the assumptions of Lemma 6.2.1 are satisfied. It is from known Theorem 3.1 that u ∈ L ∞ ( [ 0 . T ] , H 0 1 , 2 ( Ω ) ∩ L q + 1 ( Ω ) ) ∩ L 2 ( [ 0 , T ] , H 2 , 2 ( Ω ) ∩ H 0 2 ( Ω ) ) , and ∂ t u ∈ L 2 ( [ 0 , T ] , L 2 ( Ω ) ) . Let y ( t ) = ∥ ∂ t u ( t ) ∥ 2 ; then taking the product of (1.1) by ∂ t u on both sides and integrating over Ω , we obtain

(4.8) ⟨ ∂ t u ( t ) , ∂ t u ( t ) ⟩ − k ⟨ Δ u ( t ) , ∂ t u ( t ) ⟩ + β ⟨ u q ( t ) , ∂ t u ( t ) ⟩ − α ⟨ u ( t ) , ∂ t u ( t ) ⟩ = ⟨ f , ∂ t u ( t ) ⟩ , ∥ ∂ t u ( t ) ∥ 2 + k 2 d d t ∥ ∇ u ( t ) ∥ 2 − α 2 d d t ∥ u ( t ) ∥ 2 + β q + 1 d d t ∥ u ( t ) ∥ L q + 1 q + 1 ≤ ∥ f ∥ ∥ ∂ t u ( t ) ∥ ∥ ∂ t u ( t ) ∥ 2 + d d t k 2 ∥ ∇ u ( t ) ∥ 2 − α 2 ∥ u ( t ) ∥ 2 + β q + 1 ∥ u ( t ) ∥ L q + 1 q + 1 ≤ ∥ f ( t ) ∥ 2 2 + ∥ ∂ t ( t ) ∥ 2 2 , 1 2 ∥ ∂ t u ( t ) ∥ 2 + d d t k 2 ∥ ∇ u ( t ) ∥ 2 − α 2 ∥ u ( t ) ∥ 2 + β q + 1 ∥ u ( t ) ∥ L q + 1 q + 1 ≤ ∥ f ∥ 2 2 .

Now, integrating (4.8) from 0 to t :

1 2 ∫ 0 t ∥ ∂ t u ( s ) ∥ 2 d s + k 2 ∥ ∇ u ( t ) ∥ 2 − α 2 ∥ u ( t ) ∥ 2 + β q + 1 ∥ u ( t ) ∥ L q + 1 q + 1 ≤ k 2 ∥ ∇ u ( 0 ) ∥ 2 + α 2 ∥ u 0 ∥ 2 + β q + 1 ∥ u ( 0 ) ∥ L q + 1 q + 1 + 1 2 ∫ 0 t ∥ f ( s ) ∥ 2 d s .

Recall that, for bounded domain Ω ⊂ R n , and for p ≤ r , we have compact embedding L r ( Ω ) ↪ L p ( Ω ) . Since u 0 ∈ H 0 1 , 2 ∩ L q + 1 , ∥ ∇ u ( 0 ) ∥ 2 and ∥ u ( 0 ) ∥ q + 1 are uniformly bounded in H 0 1 , 2 ( Ω ) and L q + 1 ( Ω ) , respectively. And by L q + 1 ( Ω ) ↪ L 2 ( Ω ) for all q > 1 , so ∥ u ( 0 ) ∥ is uniformly bounded in L 2 ( Ω ) . Therefore, there exists a positive constant C 1 such that inequality (4) leads to:

(4.9) ∫ 0 t ∥ ∂ t u ( s ) ∥ 2 d s + k 2 ∥ ∇ u ( t ) ∥ 2 − α 2 ∥ u ( t ) ∥ 2 + β q + 1 ∥ u ( t ) ∥ L q + 1 q + 1 ≤ C 1 .

Dropping the non-negative terms, k 2 ∥ ∇ u ( t ) ∥ 2 and β q + 1 ∥ u ( t ) ∥ L q + 1 q + 1 , from the last inequality,

(4.10) ∫ 0 t ∥ ∂ t u ( s ) ∥ 2 d s − α 2 ∥ u ( t ) ∥ 2 ≤ C 1

(4.11) ∫ 0 t ∥ ∂ t u ( s ) ∥ 2 d s ≤ C 1 + α 2 ∥ u ( t ) ∥ 2 ,

for some constant C 1 .

Now, using Young’s inequality with a = u 2 , b = 1 , p = q + 1 2 , we have

(4.12) ∥ u ( t ) ∥ 2 = ∫ Ω u 2 ( t ) d x , ≤ ∫ Ω 2 u q + 1 q + 1 ( t ) + q − 1 q + 1 d x , = 2 q + 1 ∫ Ω u q + 1 ( t ) d x + ∫ Ω q − 1 q + 1 d x , ≤ ∫ Ω u q + 1 ( t ) d x + ∫ Ω q − 1 q + 1 d x , = ∥ u ( t ) ∥ L q + 1 q + 1 + q − 1 q + 1 μ ( Ω ) .

Since u ∈ L ∞ ( [ 0 , T ] , H 0 1 , 2 ( Ω ) ∩ L q + 1 ( Ω ) ) , ∥ u ( t ) ∥ L q + 1 ( Ω ) q + 1 is uniformly bounded and μ ( Ω ) is finite. Therefore, there exists a positive constant C 2 such that, using (4.11) and (4.12), we have

(4.13) ∫ 0 t ∥ ∂ t u ( s ) ∥ 2 d s ≤ C 2 .

Hence, one of the three conditions of Lemma 6.2.1 is satisfied. Now, differentiating equation (1.1) w.r.t. t and taking its L 2 -inner-product with ∂ t u ( t ) both sides, we obtain

(4.14) ⟨ ∂ t 2 u ( t ) , ∂ t u ( t ) ⟩ − k ⟨ Δ u , ∂ t u ( t ) ⟩ + β ⟨ u q , ∂ t u ( t ) ⟩ − α ⟨ u , ∂ t u ( t ) ⟩ = ⟨ ∂ t f , ∂ t u ( t ) ⟩ , 1 2 d d t ∥ ∂ t u ∥ 2 + k ∥ ∇ ∂ t u ∥ 2 + β ∫ Ω q u q − 1 ( t ) u t 2 ( t ) d x ≤ α ∥ ∂ t u ( t ) ∥ 2 + ∥ ∂ t f ∥ ∥ ∂ t u ( t ) ∥ ≤ α + 1 2 ∥ ∂ t u ( t ) ∥ 2 + ∥ ∂ t f ∥ 2 2 .

By Young’s inequality with a = ∥ ∂ t u ( t ) ∥ 2 , b = 1 , p = 2 , q = 2 , and dropping non-negative terms (second and third terms on the left-hand side) so from the last inequality,

(4.15) 1 2 d d t ∥ ∂ t u ( t ) ∥ 2 ≤ α + 1 2 ∥ ∂ t u ( t ) ∥ 4 + α 2 + ∥ ∂ t f ∥ 2 2 d d t ∥ ∂ t u ( t ) ∥ 2 ≤ ( 2 α + 1 ) ∥ ∂ t u ( t ) ∥ 4 + α + ∥ ∂ t f ∥ 2 .

Now, using hypothesis that ∂ t f ∈ L 2 ( [ 0 , T ] , L 2 ( Ω ) ) ∪ L ∞ ( [ 0 , T ] , L 2 ( Ω ) ) , Gronwall lemma and Lemma 6.2.1 from [28], with y ( t ) = ∥ ∂ t u ( t ) ∥ 2 , a 1 = 2 α + 1 , a 2 = α + ∥ ∂ t f ∥ 2 , and h ( t ) ≡ 0 along with (4.13) and (4.15), we obtain

(4.16) lim t → + ∞ ∥ ∂ t u ( t ) ∥ = 0 .

This completes the proof.□

5 Numerical example and graph

Example: Consider a nonlinear NWS equation. By taking k = 1 , q = 2 , α = 2 , β = 3 , f ( t , x ) = 0 , n = 1 , we arrive at the following initial value problem:

∂ t u = u x x + 2 u − 3 u 2 ,

with the constant initial condition,

u ( 0 , x ) = c .

Indeed, u ( x , 0 ) = c ∈ H 0 1 . The corresponding exact solution of above the problem exists globally in time and is u ( x , t ) = − 2 3 c e 2 t − 2 3 + c − c e 2 t . Graphical representation of the exact solution is shown in Figure 1.

Figure 1

Exact solution of the NWS equation.

6 Conclusion

The key aim of this work is to establish the well-posedness of the NWS equation using the Faedo–Galerkin compactness method. Additionally, we have examined the exponential decay of the solution and demonstrated that the L 2 -norm of the derivative of the solution tends to zero as time approaches infinity. We firmly believe that similar kinds of studies could be done for classes of abstract amplitude equations in future by researchers.

Funding information: This work has not received any external funding.
Author contributions: All authors have accepted responsibility for the entire content of this manuscript and consented to its submission to the journal, reviewed all the results and approved the final version of the manuscript. All authors have contributed equally to this manuscript.
Conflict of interest: The authors declare that there is no conflict of interest.
Ethical approval: This article contains no studies with human participants or animals performed by authors.
Informed consent: This article contains no studies with human participants. So informed consent is not applicable here.
Data availability statement: No data were required for this research.

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Received: 2024-07-03

Revised: 2024-09-09

Accepted: 2024-09-12

Published Online: 2024-11-11

This work is licensed under the Creative Commons Attribution 4.0 International License.

Articles in the same Issue

https://doi.org/10.1515/nleng-2024-0041

Keywords for this article

amplitude equations; Faedo–Galerkin method; well-posedness; exponential decay of solution

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