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Review

Pore-forming virulence factors of Staphylococcus aureus destabilize epithelial barriers-effects of alpha-toxin in the early phases of airway infection

  • Staphylococcus aureus (S. aureus) is a human commensal and an opportunistic pathogen that may affect the gastrointestinal tract, the heart, bones, skin or the respiratory tract. S. aureus is frequently involved in hospital- or community-acquired lung infections. The pathogenic potential is associated with its ability to secrete highly effective virulence factors. Among these, the pore-forming toxins Panton-Valentine leukocidin (PVL) and hemolysin A (Hla) are the important virulence factors determining the prognosis of pneumonia cases. This review focuses on the structure and the functions of S. aureus hemolysin A and its sub-lethal effects on airway epithelial cells. The hypothesis is developed that Hla may not just be a tissue-destructive agent providing the bacteria with host-derived nutrients, but may also play complex roles in the very early stages of interactions of bacteria with healthy airways, possibly paving the way for establishing acute infections.

    Citation: Jan-Peter Hildebrandt. Pore-forming virulence factors of Staphylococcus aureus destabilize epithelial barriers-effects of alpha-toxin in the early phases of airway infection[J]. AIMS Microbiology, 2015, 1(1): 11-36. doi: 10.3934/microbiol.2015.1.11

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  • Staphylococcus aureus (S. aureus) is a human commensal and an opportunistic pathogen that may affect the gastrointestinal tract, the heart, bones, skin or the respiratory tract. S. aureus is frequently involved in hospital- or community-acquired lung infections. The pathogenic potential is associated with its ability to secrete highly effective virulence factors. Among these, the pore-forming toxins Panton-Valentine leukocidin (PVL) and hemolysin A (Hla) are the important virulence factors determining the prognosis of pneumonia cases. This review focuses on the structure and the functions of S. aureus hemolysin A and its sub-lethal effects on airway epithelial cells. The hypothesis is developed that Hla may not just be a tissue-destructive agent providing the bacteria with host-derived nutrients, but may also play complex roles in the very early stages of interactions of bacteria with healthy airways, possibly paving the way for establishing acute infections.


    In this paper, a nonlinear compact finite difference scheme is studied for the following the pseudo-parabolic Burgers' equation [1]

    $ ut=μuxx+γuux+ε2uxxt,0<x<L,0<tT, $ (1.1)

    subject to the periodic boundary condition

    $ u(x,t)=u(x+L,t),0xL,0<tT, $ (1.2)

    and the initial data

    $ u(x,0)=φ(x),0xL, $ (1.3)

    where $ \mu > 0 $ is the coefficient of kinematic viscosity, $ \gamma $ and $ \varepsilon > 0 $ are two parameters, $ \varphi(x) $ is an $ L $-periodic function. Parameter $ L $ denotes the spatial period. Setting $ \varepsilon = 0 $, Eq (1.1) reduces to a viscous Burgers' equation [2]. Equation (1.1) is derived by the degenerate pseudo-parabolic equation [3]

    $ ut=(uα+uβux+ε2uκ(uγut)x)x, $ (1.4)

    where $ \alpha $, $ \beta $, $ \kappa $, $ \gamma $ are nonnegative constants. The derivative term $ \{u^{{\kappa}}(u^\gamma u_t)_x\}_x $ represents a dynamic capillary pressure relation instead of a usual static one [4]. Equation (1.4) is a model of one-dimensional unsaturated groundwater flow.

    Here, $ u $ denotes the water saturation. We refer to [5] for a detailed explanation of the model. Equation (1.1) is also viewed as a simplified edition of the Benjamin-Bona-Mahony-Burgers (BBM-Burgers) equation, or a viscous regularization of the original BBM model for the long wave propagation [6]. The problem (1.1)–(1.3) has the following conservation laws

    $ Q(t)=L0u(x,t)dx=Q(0),t>0, $ (1.5)
    $ E(t)=L0[u2(x,t)+ε2u2x(x,t)]dx+2μt0L0 u2x(x,s)dxds=E(0),t>0. $ (1.6)

    Based on (1.6), by a simple calculation, the exact solution satisfies

    $ max{u,εux,εu}c0, $

    where $ c_0 = \big(1+\frac{\sqrt{L}}{2}\big)\sqrt{E(0)} $.

    Numerical and theoretical research for solving (1.1)–(1.3) have been extensively carried out. For instance, Koroche [7] employed the the upwind approach and Lax-Friedrichs to obtain the solution of In-thick Burgers' equation. Rashid et al. [8] employed the Chebyshev-Legendre pseudo-spectral method for solving coupled viscous Burgers' equations, and the leapfrog scheme was used in time direction. Qiu al. [9] constructed the fifth-order weighted essentially non-oscillatory schemes based on Hermite polynomials for solving one dimensional non-linear hyperbolic conservation law systems and presented numerical experiments for the two dimensional Burgers' equation. Lara et al. [10] proposed accelerate high order discontinuous Galerkin methods using Neural Networks. The methodology and bounds are examined for a variety of meshes, polynomial orders, and viscosity values for the 1D viscous Burgers' equation. Pavani et al. [11] used the natural transform decomposition method to obtain the analytical solution of the time fractional BBM-Burger equation. Li et al. [12] established and proved the existence of global weak solutions for a generalized BBM-Burgers equation. Wang et al. [13] introduced a linearized second-order energy-stable fully discrete scheme and a super convergence analysis for the nonlinear BBM-Burgers equation by the finite element method. Mohebbi et al. [14] investigated the solitary wave solution of nonlinear BBM-Burgers equation by a high order linear finite difference scheme.

    Zhang et al. [15] developed a linearized fourth-order conservative compact scheme for the BBMB-Burgers' equation. Shi et al. [16] investigated a time two-grid algorithm to get the numerical solution of nonlinear generalized viscous Burgers' equation. Li et al. [17] used the backward Euler method and a semi-discrete approach to approximate the Burgers-type equation. Mao et al. [18] derived a fourth-order compact difference schemes for Rosenau equation by the double reduction order method and the bilinear compact operator. It offers an effective method for solving nonlinear equations. Cuesta et al. [19] analyzed the boundary value problem and long-time behavior of the pseudo-parabolic Burgers' equation. Wang et al. [20] proposed fourth-order three-point compact operator for the nonlinear convection term. They adopted the classical viscous Burgers' equation as an example and established the conservative fourth-order implicit compact difference scheme based on the reduction order method. The compact difference scheme enables higher accuracy in solving equations with fewer grid points. Therefore, using the compact operators to construct high-order schemes has received increasing attention and application [21,22,23,24,25,26,27,28,29].

    Numerical solutions for the pseudo-parabolic equations have garnered widespread attention. For instance, Benabbes et al. [30] provided the theoretical analysis of an inverse problem governed by a time-fractional pseudo-parabolic equation. Moreover, Ilhan et al. [31] constructed a family of travelling wave solutions for obtaining hyperbolic function solutions. Di et al. [32] established the well-posedness of the regularized solution and gave the error estimate for the nonlinear fractional pseudo-parabolic equation. Nghia et al. [33] considered the pseudo-parabolic equation with Caputo-Fabrizio fractional derivative and gave the formula of mild solution. Abreu et al. [34] derived the error estimates for the nonlinear pseudo-parabolic equations basedon Jacobi polynomials. Jayachandran et al. [35] adopted the Faedo-Galerkin method to the pseudo-parabolic partial differential equation with logarithmic nonlinearity, and they analyzed the global existence and blowup of solutions.

    To the best of our knowledge, the study of high-order difference schemes for Eq (1.1) is scarce. The main challenge is the treatment of the nonlinear term $ uu_x $, as well as the error estimation of the numerical scheme. Inspired by the researchers in [15] and [20], we construct an implicit compact difference scheme based on the three–point fourth-order compact operator for the pseudo-parabolic Burgers' equation. The main contribution of this paper is summarized as follows:

    ● A fourth-order compact difference scheme is derived for the pseudo-parabolic Burgers' equation.

    ● The pointwise error estimate ($ L^\infty $-estimate) of a fourth-order compact difference scheme is proved by the energy method [36,37] for the pseudo-parabolic Burgers' equation.

    ● Numerical stability, unique solvability, and conservation are obtained for the high-order difference scheme of the pseudo-parabolic Burgers' equation.

    In particular, our numerical scheme for the special cases reduces to several other ones in this existing paper (see e.g., [38,39]).

    The remainder of the paper is organized as follows. In Section 2, we introduce the necessary notations and present some useful lemmas. A compact difference scheme is derived in Section 3 using the reduction order method and the recent proposed compact difference operator. In Section 4, we establish the key results of the paper, including the conservation invariants, boundedness, uniqueness of the solution, stability, and convergence of the scheme. In Section 4.4, we present several numerical experiments to validate the theoretical findings, followed by a conclusion in Section 5.

    Throughout the paper, we assume that the exact solution $ u(x, t) $ satisfies $ u(x, t)\in \mathcal{C}^{6, 3}\big([0, L]\times [0, T]\big) $.

    In this section, we introduce some essential notations and lemmas. We begin by dividing the domain $ [0, L]\times[0, T] $. For two given positive integers, $ M $ and $ N $, let $ h = L/M, \, \tau = T/N $. Additionally, denote $ x_i = ih, \, 0\leq i\leq M, \, t_k = k\tau, \, 0\leq k\leq N $; $ \mathcal{V}_{h} = \{\, v\, |\, v = \{{v}_{i}\}, \, v_{i+M} = v_i\, \} $. For any grid function $ u $, $ v\in \mathcal{V}_{h} $, we introduce

    $ vk+12i=12(vki+vk+1i),δtvk+12i=1τ(vk+1ivki),δxvki+12=1h(vki+1vki),Δxvki=12h(vki+1vki1),δ2xvki=1h(δxvki+12δxvki12),ψ(u,v)i=13[uiΔxvi+Δx(uv)i]. $

    Moreover, we introduce the discrete inner products and norms (semi-norm)

    $ (u,v)=hMi=1uivi,u,v=hMi=1(δxui+12)(δxvi+12),u=(u,u),|u|1=u,u,u=max1iM|ui|. $

    The following lemmas play important roles in the numerical analysis later, and we collect them here.

    Lemma 1. [15,40] For any grid functions $ u $, $ v\in \mathcal{V}_h $, we have

    $ \Vert v\Vert_\infty\leq \frac{\sqrt{L}}{2}\vert v\vert_1, \quad \|v\|\leq \frac{L}{\sqrt{6}}|v|_1, \quad (u, \delta_x^2 v) = -\langle u, v\rangle, \quad\big(\psi( u, v) , v\big) = 0. $

    Lemma 2. [40] For any grid function $ v\in \mathcal{V}_h $ and arbitrary $ \xi > 0 $, we have

    $ \vert v\vert_1\leq \frac{2}{h} \Vert v \Vert, \quad \Vert v\Vert_\infty^2\leq \xi\vert v\vert_1^2+\Big(\frac{1}{\xi}+\frac{1}{L}\Big)\Vert v\Vert^2. $

    Lemma 3. [20] Let $ g(x)\in C^5[x_{i-1}, x_{i+1}] $ and $ G(x) = g''(x) $, we have

    $ g( x_i) g'( x_i) = \psi(g, g) _i-\frac{h^2}{2}\psi(G, g)_i+\mathcal{O}(h^4). $

    Lemma 4. [15,18] For any grid functions $ u $, $ v\in \mathcal{V}_h $ and $ S\in \mathcal{V}_h $ satisfying

    $ vk+12i=δ2xuk+12ih212δ2xvk+12i+Sk+12i,1iM,0kN1, $ (2.1)

    we have the following results:

    $ ({\rm{I}}) $

    $ (vk+12,uk+12)=|uk+12|21h212vk+122+h4144|vk+12|21+h212(Sk+12,vk+12)+(Sk+12,uk+12), $ (2.2)
    $ (vk+12,uk+12)|uk+12|21h218vk+122+h212(Sk+12,vk+12)+(Sk+12,uk+12), $ (2.3)
    $ (δtvk+12,uk+12)=12τ(|uk+1|21|uk|21)h224τ(vk+12vk2)+h4288τ(|vk+1|21|vk|21)+(δtSk+12,uk+12)+h212(δtvk+12,Sk+12). $ (2.4)

    $ ({\rm{II}}) $

    $ |uk+12|21uk+12(vk+12+Sk+12),h212vk+12245uk+12+h25Sk+12, $ (2.5)
    $ vk+12218h2|uk+12|21+92Sk+122. $ (2.6)

    Proof. The result in (2.2)–(2.3) has been described in [15], and (2.5) has been proven in [18], we only need to only prove (2.4) and (2.6). Using the definition of the operator, we have

    $ (δtvk+12,uk+12)=(δt(δ2xuk+12h212δ2xvk+12+Sk+12),uk+12)=12τ(|uk+1|21|uk|21)h212(δtvk+12,vk+12+h212δ2xvk+12Sk+12)+(δtSk+12,uk+12)=12τ(|uk+1|21|uk|21)h224τ(vk+12vk2)+h4288τ(|vk+1|21|vk|21)+(δtSk+12,uk+12)+h212(δtvk+12,Sk+12). $

    Taking the inner product of (2.1) with $ v^{k+\frac{1}{2}} $, we have

    $ vk+122=(δ2xuk+12,vk+12)h212(δ2xvk+12,vk+12)+(Sk+12,vk+12)δ2xuk+12vk+12+h212|vk+12|21+Sk+12vk+1216vk+122+32δ2xuk+122+13vk+122+16vk+122+32Sk+12223vk+122+6h2|uk+12|21+32Sk+122. $

    Therefore, the result (2.6) is obtained.

    Remark 1. [18] Denote $ {\mathbf 1} = (1, 1, \cdots, 1)^T\in \mathcal{V}_h $. If $ S = 0 $ in (2.1), then we further have

    $ (ψ(u,u),1)=0,(ψ(v,u),1)=0. $

    Let $ v = u_{xx} $, then the problem (1.1) is equivalent to

    $ \left\{ {ut=μv+γuux+ε2vt,0<x<L,0<tT,(3.1)v=uxx,0<x<L,0<tT,(3.2)u(x,0)=φ(x),0xL,(3.3)u(x,t)=u(x+L,t),0xL,0<tT.(3.4)} \right. $

    According to (3.2) and (3.4), it is easy to know that

    $ v(x,t)=v(x+L,t),0xL,0<tT. $ (3.5)

    Define the grid functions $ U = \big\lbrace \, U_i^k\, |\, 1\leq i\leq M, \, 0\leq k\leq N\, \big\rbrace $ with $ U_i^k = u(x_i, t_k) $, $ V = \lbrace\, V_i^k\, |\, 1\leq i\leq M, \, 0\leq k\leq N \, \rbrace $ with $ V_i^k = v(x_i, t_k) $. Considering (3.1) at the point $ (x_i, t_{k+\frac{1}{2}}) $ and (3.2) at the point $ (x_i, t_{k}) $, respectively, we have

    $ \left\{ {ut(xi,tk+12)=μv(xi,tk+12)+γu(xi,tk+12)ux(xi,tk+12)+ε2vt(xi,tk+12),1iM,0kN1,v(xi,tk)=uxx(xi,tk),1iM,0kN.} \right. $

    Using the Taylor expansion and Lemma 3, we have

    $ \left\{ {δtUk+12i=μVk+12i+γ(ψ(Uk+12,Uk+12)ih22ψ(Vk+12,Uk+12)i)+ε2δtVk+12i+Pk+12i,1iM,0kN1,Vki=δ2xUkih212δ2xVki+Qki,1iM,0kN.} \right. $ (3.6)

    Noticing the initial-boundary value conditions (3.3)–(3.5), we have

    $ {U0i=φ(xi),1iM;(3.7)Uki=Uki+M,Vki=Vki+M,1iM,1kN.(3.8) $

    There is a positive constant $ c_1 $ such that the local truncation errors satisfy

    $ {|Pk+12i|c1(τ2+h4),1iM,0kN1,|Qki|c1h4,1iM,0kN,|δtQk+12i|c1(τ2+h4),1iM,0kN1. $

    Omitting the local truncation error terms in (3.6) and combining them with (3.7) and (3.8), the difference scheme for (3.1)–(3.5) as follows

    $ \left\{ {δtuk+12i=μvk+12i+γ(ψ(uk+12,uk+12)ih22ψ(vk+12,uk+12)i)+ε2δtvk+12i,1iM,0kN1,(3.9)vki=δ2xukih212δ2xvki,1iM,0kN,(3.10)u0i=φ(xi),1iM,(3.11)uki=uki+M,vki=vki+M,1iM,1kN.(3.12)} \right. $

    Remark 2. As we see from the difference equations (3.9) and (3.10), only three points for each of them are utilized to generate fourth-order accuracy for the nonlinear pseudo-parabolic Burgers' equation without using additional boundary message. This is the reason we call this scheme the compact difference scheme. In addition, a fast iterative algorithm can be constructed, as shown in the numerical part in Section 4.4.

    Theorem 1. Let $ \{\, u_i^k, \, v_i^k \, |\, 1\leq i\leq M, \, 0\leq k\leq N \} $ be the solution of (3.9)–(3.12). Denote

    $ Q^k = (u^k, {\mathbf 1}). $

    Then, we have

    $ Qk=Q0,0kN. $

    Proof. Taking an inner product of (3.9) with $ {\mathbf 1} $, we have

    $ (δtuk+12,1)=μ(vk+12,1)+γ(ψ(uk+12,uk+12)h22ψ(vk+12,uk+12),1)+ε2(δtvk+12,1),0kN1. $

    By using Remark 1 in Lemma 4, the equality above deduces to

    $ (u^{k+1}, {\mathbf 1})-(u^{k}, {\mathbf 1}) = 0, $

    namely

    $ Q^{k+1} = Q^{k}, \quad 0\leq k\leq N-1 . $

    Theorem 2. Let $ \{\, u_i^k, \, v_i^k\, |\, 1\leq i\leq M, \, 0\leq k\leq N\, \} $ be the solution of (3.9)–(3.12). Then it holds that

    $ E^{k} = E^0, \quad 1\leq k\leq N, $

    where

    $ Ek=uk2+ε2|uk|21+ε2h212vk2ε2h4144|vk|21+2τμ(k1l=0|ul+12|21+h212k1l=0vl+122h4144k1l=0|vl+12|21). $

    Proof. Taking the inner product of (3.9) with $ u^{k+\frac{1}{2}} $, and applying Lemma 1, we have

    $ (δtuk+12,uk+12)=μ(vk+12,uk+12)+ε2(δtvk+12,uk+12). $

    With the help of (2.2) and (2.4) in Lemma 4, the equality above deduces to

    $ 12τ(uk+12uk2)=μ(|uk+12|21h212vk+122+h4144|vk+12|21)ε22τ((|uk+1|21|uk|21)+h212(vk+12vk2)h4144(|vk+1|21|vk|21)). $

    Replacing the superscript $ k $ with $ l $ and summing over $ l $ from $ 0 $ to $ k-1 $, we have

    $ (uk2+ε2|uk|21+ε2h212vk2ε2h4144|vk|21)(u02+ε2|u0|21+ε2h212v02ε2h4144|v0|21)+2τμ(k1l=0|ul+12|21+h212k1l=0vl+122h4144k1l=0|vl+12|21)=0, $

    which implies that

    $ E^k = E^0, \quad 1\leq k\leq N. $

    Remark 3. Combining Lemma 1 with Theorem 2, it is easy to know that there is a positive constant $ c_2 $ such that

    $ ukc2,ε|uk|1c2,εukc2,1kN. $ (4.1)

    Next, we recall the Browder theorem and consider the unique solvability of (3.9)–(3.12).

    Lemma 5 (Browder theorem[41]). Let $ (H, (\cdot, \cdot)) $ be a finite dimensional inner product space, $ \Vert \cdot\Vert $ be the associated norm, and $ \Pi:H\to H $ be a continuous operator. Assume

    $ \exists\, \alpha > 0, \quad\forall z\in H, \quad\Vert z\Vert = \alpha, \quad \Re\big(\Pi(z), z \big)\geq 0. $

    Then there exists a $ z^*\in H $ satisfying $ \Vert z^*\Vert\leq \alpha $ such that $ \Pi(z^*) = 0. $

    Theorem 3. The difference scheme (3.9)–(3.12) has a solution at least.

    Proof. Denote

    $ u^k = \left(u_1, u_2, \cdots, u_M \right), \quad v^k = \left(v_1, v_2, \cdots, v_M\right), \quad 0\leq k\leq N. $

    It is easy to know that $ u^0 $ has been determined by (3.11). From (3.10) and (3.11), we can get $ v^0 $ by computing a system of linear equations as its coefficient matrix is strictly diagonally dominant. Suppose that $ \lbrace\, u^k, \, v^k \, \rbrace $ has been determined, then we may regard $ \lbrace \, u^{k+\frac{1}{2}}, \, v^{k+\frac{1}{2}} \, \rbrace $ as unknowns. Obviously,

    $ u_i^{k+1} = 2 u_i^{k+\frac{1}{2}}-u_i^k, \quad v_i^{k+1} = 2 v_i^{k+\frac{1}{2}}-v_i^k, \quad 1\leq i\leq M, \quad 0\leq k\leq N-1. $

    Denote

    $ X_i = u_i^{k+\frac{1}{2}}, \quad Y_i = v_i^{k+\frac{1}{2}}, \quad 1\leq i\leq M, \quad 0\leq k\leq N-1. $

    Then the difference scheme (3.9)–(3.10) can be rewritten as

    $ \left\{ {2τ(Xiuki)μYiγ(ψ(X,X)ih22ψ(Y,X)i)2τε2(Yivki)=0,1iM,0kN,(4.2)Yi=δ2xXih212δ2xYi,1iM.(4.3)} \right. $

    Define an operator $ \Pi $ on $ \mathcal{V}_h $:

    $ \Pi( X_i) = \frac{2}{\tau}(X_i-u_i^k ) -\mu Y_i-\gamma \Big(\psi( X, X)_i -\frac{h^2}{2}\psi( Y, X)_i \Big)- \frac{2}{\tau}\varepsilon^2(Y_i-v_i^k ), \quad 1\leq i\leq M, \quad 0\leq k\leq N. $

    Taking an inner product of $ \Pi(X) $ with $ X $, we have

    $ (Π(X),X)=2τ(X2(uk,X))μ(Y,X)2τε2((Y,X)(vk,X)). $ (4.4)

    In combination of the technique from (2.2) in Lemma 4 and the Cauchy-Schwartz inequality, we have

    $ (δxY,δxX)=(δx(δ2xXh212δ2xY),δxX)=δ2xX2+h212(δ2xY,δ2xX)=δ2xX2+h212(δ2xY,Y+h212δ2xY)=δ2xX2h212δxY2+h4144δ2xY2 $

    and

    $ (δxuk,δxX)δxukδxX14δxuk2+δxX2=14δxuk2+|X|21. $

    Correspondingly,

    $ (δxuk,δxX)14δxuk2|X|21. $

    Then

    $ (Y,X)+(vk,X)=(δ2xXh212δ2xY,X)+(δ2xukh212δ2xvk,X)=|X|21(δxuk,δxX)h212((δxY,δxX)+(δ2xvk,X))14δxuk2+h212(δ2xX2+h212δxY2h4144δ2xY2(δ2xvk,X))14δxuk2+h212(h212δxY2h4144δ2xY214vk2)14δxuk2h248vk2, $

    and

    $ \Vert X\Vert^2-(u^k, X)\geq \Vert X\Vert^2-\frac{1}{2}(\Vert u^k\Vert^2+ \Vert X\Vert^2) \geq \frac{1}{2}( \Vert X\Vert^2-\Vert u^k\Vert^2). $

    Substituting the equality above into (4.4) and according to (2.3) in Lemma 4, we have

    $ (Π(X),X)1τ(X2uk2)+2ε2τ(14δxuk2h248vk2)1τ(X2uk2ε22δxuk2ε2h224vk2). $

    Thus, when $ \Vert X\Vert = \alpha^k $, where $ \alpha^k = \sqrt{\Vert u^k\Vert^2+\frac{\varepsilon^2}{2} \Vert \delta_xu^k\Vert^2+\frac{\varepsilon^2h^2}{24}\Vert v^k\Vert^2} $, then $ (\Pi(X), X)\geq 0 $. By Lemma 5, there exists a $ X^*\in \mathcal{V}_h $ satisfying $ \Vert X^*\Vert\leq\alpha^k $ such that $ \Pi(X^*) = 0 $. Consequently, the difference scheme (3.9)–(3.12) exists at least a solution $ u^{k+1} = 2X^*-u^k $. Observing, when $ (X_1^*, X_2^*, \cdots, X_M^*) $ is known, $ (Y_1^*, Y_2^*, \cdots, Y_M^*) $ can be determined by (4.3) uniquely. Thus, we know $ v_i^{k+1} = 2Y_i^*-v_i^k $, $ 1\leq i\leq M $ exists.

    Now we are going to verify the uniqueness of the solution of the difference scheme. We have the following result.

    Theorem 4. When $ \gamma = 0 $, the solution of the difference scheme (3.9)–(3.12) is uniquely solvable for any temporal step-size; When $ \gamma\neq 0 $ and $ \tau \leq {\min}\Big\{\, \frac{4L}{c_2|\gamma|(L+1)}, \, \frac{2\varepsilon^2}{3c_2|\gamma|(2L+1)}\, \Big\} $, the solution of the difference scheme (3.9)–(3.12) is uniquely solvable.

    Proof. According to Theorem 3, we just need to prove that (4.2)–(4.3) has a unique solution. Suppose that both $ \lbrace u^{(1)}, v^{(1)}\rbrace\in \mathcal{V}_h $ and $ \lbrace u^{(2)}, v^{(2)}\rbrace\in \mathcal{V}_h $ are the solutions of (4.2)–(4.3), respectively. Let

    $ u_i = u_i^{(1)}-u_i^{(2)}, \quad v_i = v_i^{(1)}-v_i^{(2)}, \quad 1\leq i\leq M. $

    Then we have

    $ \left\{ {2τuiμviγ(ψ(u(1),u(1))iψ(u(2),u(2))i)+γh22(ψ(v(1),u(1))iψ(v(2),u(2))i)2ε2τvi=0,1iM,(4.5)vi=δ2xuih212δ2xvi,1iM.(4.6)} \right. $

    Taking an inner product of (4.5) with $ u $, we have

    $ 2τu2μ(v,u)γ(ψ(u(1),u(1))ψ(u(2),u(2)),u)+γh22(ψ(v(1),u(1))ψ(v(2),u(2)),u)2ε2τ(v,u)=0. $

    With the application of Lemma 2 and (2.3) in Lemma 4, it follow from the equality above that

    $ 2τu2+(μ+2ε2τ)(|u|21+h218v2)γ(ψ(u(1),u(1))ψ(u(2),u(2)),u)γh22(ψ(v(1),u(1))ψ(v(2),u(2)),u). $ (4.7)

    By the definition of $ \psi(\cdot, \cdot) $ and (4.1), we have

    $ h22(ψ(v(1),u(1))ψ(v(2),u(2)),u)=h22(ψ(v(1),u(1))ψ(v(1)v,u(1)u),u)=h22(ψ(v,u(1)),u)=h36Mi=1[viΔxu(1)i+Δx(vu(1))i]ui=h36Mi=1[u(1)iΔx(uv)i+(vu(1))iΔxui]=h36Mi=1[u(1)i12h(ui+1vi+1ui1vi1)+(vu(1))iΔxui]=h36Mi=1[u(1)i12h(vi+1(ui+1ui)+ui(vi+1vi1)+vi1(uiui1))+(vu(1))iΔxui]=h36Mi=1[u(1)i(uiΔxvi+12vi+1δxui+12+12vi1δxui12)+(vu(1))iΔxui]c2h26(|v|1u+2v|u|1). $

    Using the Cauchy-Schwarz inequality, Lemmas 1 and 2, we have

    $ h22(ψ(v(1),u(1))ψ(v(2),u(2)),u)c26(h44|v|21+u2)+c23(h4v2+14|u|21)c224(L+2)|u|21+c2h26(1+2L)v2. $ (4.8)

    Similarly, we have

    $ (ψ(u(1),u(1))ψ(u(2),u(2)),u)=(ψ(u(1),u(1))ψ(u(1)u,u(1)u),u)=(ψ(u,u(1)),u)=h3Mi=1[uiΔxu(1)i+Δx(uu(1))i]ui=h3Mi=1[u(1)iΔx(uu)i+(uu(1))iΔxui]=h3Mi=1[u(1)i(uiΔxui+12ui+1δxui+12+12ui1δxui12)+(uu(1))iΔxui]c2|u|1uc22(u2+|u|21)c22(1+1L)u2+c2|u|21. $ (4.9)

    Substituting (4.8) and (4.9) into (4.7), we can obtain

    $ 2τu2+(μ+2ε2τ)|u|21+h218(μ+2ε2τ)v2c2|γ|2L(L+1)u2+c2|γ|24(L+26)|u|21+c2h2|γ|6(2L+1)v2. $

    When $ \tau \leq {\min}\{\, \frac{4L}{c_2|\gamma| (L+1)}, \, \frac{2\varepsilon^2}{3c_2 |\gamma| (2L+1)}\, \} $, we have $ u_i = 0, \, 1\leq i \leq M $.

    Let $ h_0 > 0 $ and denote

    $ c3=max(x,t)[0,L]×[0,T]{|u(x,t)|,|ux(x,t)|},c4=3+c3|γ|+3c23γ2h204μ+3c23γ2μ,c5=c21LT(1+μ2+ε4+3μh2016+ε2h2012)+1312ε2h20c21L,c6=32c5e32c4T, $

    and error functions

    $ eki=Ukiuki,fki=Vkivki,1iM,1kN, $

    we have the following convergence results.

    Theorem 5. Let $ \lbrace \, u(x, t), \, v(x, t)\, \rbrace $ be the solution of (3.1)–(3.5) and $ \lbrace \, u_i^k, \, v_i^k\, \vert\, 0\leq i\leq M, \, 0\leq k\leq N \, \rbrace $ be the solution of the difference scheme (3.9)–(3.12). When $ c_4\tau\leq \frac{1}{3} $ and $ h\leq h_0 $, we have

    $ \|e^{k}\|\leq c_6(\tau^2+h^4), \quad\varepsilon|e^{k}|_1\leq c_6(\tau^2+h^4) , \quad\varepsilon\|e^{k}\|_\infty\leq \frac{c_6\sqrt{L}}{2}(\tau^2+h^4), \quad 0\leq k\leq N. $

    Proof. Subtracting (3.9)–(3.12) from (3.6)–(3.8), we can get an error system

    $ \left\{ {δtek+12i=μfk+12i+γ(ψ(Uk+12,Uk+12)iψ(uk+12,uk+12)i)γh22(ψ(Vk+12,Uk+12)iψ(vk+12,uk+12)i)+ε2δtfk+12i+Pk+12i,1iM,0kN1,(4.10)fki=δ2xekih212δ2xfki+Qki,1iM,0kN,(4.11)e0i=0,1iM,(4.12)eki=eki+M,fki=fki+M,1iM,1kN.(4.13)} \right. $

    Taking an inner product of (4.10) with $ e^{k+\frac{1}{2}} $, we have

    $ (δtek+12,ek+12)=μ(fk+12,ek+12)+γ(ψ(Uk+12,Uk+12)ψ(uk+12,uk+12),ek+12)γh22(ψ(Vk+12,Uk+12)ψ(vk+12,uk+12),ek+12)+ε2(δtfk+12,ek+12)+(Pk+12,ek+12). $ (4.14)

    Applying (2.3) in Lemma 4, we have

    $ (fk+12,ek+12)|ek+12|21h218||fk+12||2+h212(Qk+12,fk+12)+(Qk+12,ek+12). $ (4.15)

    Similar to the derivation in (4.8) and (4.9), we have

    $ (ψ(Uk+12,Uk+12)ψ(uk+12,uk+12),ek+12)=(ψ(ek+12,Uk+12),ek+12)=h3Mi=1[ek+12iΔxUk+12i+Δx(ek+12Uk+12)i]ek+12i=h3Mi=1[(ek+12i)2ΔxUk+12iek+12iUk+12iΔxek+12i]=h3Mi=1(ek+12i)2ΔxUk+12i+h6Mi=1Uk+12i+1Uk+12ihek+12iek+12i+1c32ek+122 $ (4.16)

    and

    $ (ψ(Vk+12,Uk+12)ψ(vk+12,uk+12),ek+12)=(ψ(fk+12,Uk+12),ek+12)=h3Mi=1[fk+12iΔxUk+12i+Δx(fk+12Uk+12)i]ek+12i=h3Mi=1fk+12iek+12iΔxUk+12i+h3Mi=1fk+12iUk+12iΔxek+12i13c3||fk+12||||ek+12||+13c3||fk+12||||Δxek+12||. $ (4.17)

    Substituting (4.15)–(4.17) into (4.14), and using (2.3)–(2.4) in Lemma 4, we obtain

    $ 12τ(ek+12ek2)μ(|ek+12|21h218||fk+12||2+h212(Qk+12,fk+12)+(Qk+12,ek+12))+12c3|γ|ek+122+c3h2|γ|6(||fk+12||||ek+12||+||fk+12||||Δxek+12||)ε22τ(|ek+1|21|ek|21+h212(fk+12fk2)h4144(|fk+1|21|fk|21)2τ(δtQk+12,ek+12))+ε2h212(δtfk+12,Qk+12)+(Pk+12,ek+12). $

    Then, we have

    $ ek+12ek2+ε2(|ek+1|21|ek|21+h212(fk+12fk2)h4144(|fk+1|21|fk|21))2μτ|ek+12|21μτh29||fk+12||2+μτh26(Qk+12,fk+12)+2μτ(Qk+12,ek+12)+c3τ|γ|ek+122+c3τh2|γ|3||fk+12||||ek+12||+c3τh2|γ|3||fk+12||||Δxek+12||+2τε2(δtQk+12,ek+12)+2τ(Pk+12,ek+12)+τε2h26(δtfk+12,Qk+12). $

    Using Cauchy-Schwartz inequality, we can rearrange the inequality above into the following form

    $ ek+12ek2+ε2(|ek+1|21|ek|21)+ε2h212[(fk+12h212|fk+1|21)(fk2h212|fk|21)]2μτ|ek+12|21μτh29||fk+12||2+μτh26Qk+12fk+12+2μτQk+12ek+12+c3τ|γ|ek+122+c3|γ|τh23||fk+12||||ek+12||+c3|γ|τh23||fk+12||||Δxek+12||+2τε2δtQk+12ek+12+2τPk+12ek+12+τε2h26(δtfk+12,Qk+12)2μτ|ek+12|21μτh29||fk+12||2+3μτh216Qk+122+μτh227fk+122+μ2τQk+122+τek+122+c3|γ|τek+122+μτh227||fk+12||2+3c23γ2τh24μ||ek+12||2+μτh227||fk+12||2+3c23γ2τμek+122+τε4δtQk+122+τek+122+τPk+122+τek+122+τε2h26(δtfk+12,Qk+12)τ(32+c3|γ|2+3c23γ2h28μ+3c23γ22μ)(ek+12+ek2)+τ(μ2+3μh216)Qk+122+τε4δtQk+122+τPk+122+τε2h26(δtfk+12,Qk+12). $

    Replacing the superscript $ k $ with $ l $ and summing over $ l $ from 0 to $ k $, we get

    $ \begin{align} &\|e^{k+1}\|^2+\varepsilon^2|e^{k+1}|_{1}^{2} +\frac{\varepsilon^2h^{2}}{12}\Big(\|f^{k+1}\|^{2}-\frac{h^{2}}{12}|f^{k+1}|_{1}^{2}\Big) -\frac{\varepsilon^2h^{2}}{12}\Big(\|f^{0}\|^{2}-\frac{h^{2}}{12}|f^{0}|_{1}^{2}\Big)\\ \leq& \tau\Big(\frac{3}{2}+\frac{ c_3|\gamma|}{2}+\frac{3c_3^2\gamma^2h^2}{8\mu} +\frac{3c_3^2\gamma^2}{2\mu}\Big) \sum\limits_{l = 0}^{k}(\|e^{l+1}\|^2+\|e^{l}\|^2) +\tau(k+1)\Big(\mu^2+\frac{3\mu h^2}{16}\Big)Lc_1^2h^8\\ & +\tau\varepsilon^4(k+1)Lc_1^2(\tau^2+h^4)^2+\tau(k+1)Lc_1^2(\tau^2+h^4)^2 +\frac{\tau\varepsilon ^2h^2}{6}\sum\limits_{l = 0}^{k}(\delta_tf^{l+\frac{1}{2}}, Q^{l+\frac{1}{2}}) , \quad 0\leq k\leq N-1. \end{align} $ (4.18)

    For the last item on the right-hand side of (4.18), we have

    $ \begin{align} \sum\limits_{l = 0}^{k}(\delta_{t}f^{l+\frac{1}{2}}, Q^{l+\frac{1}{2}}) & = \frac{1}{\tau}\Big[\sum\limits_{l = 0}^{k}(f^{l+1}, Q^{l+\frac{1}{2}}) -\sum\limits_{l = 0}^{k}(f^{l}, Q^{l+\frac{1}{2}})\Big]\\ & = \frac{1}{\tau}\big[(f^{k+1}, Q^{k+\frac{1}{2}}) -(f^0 , Q^{\frac{1}{2}})\big]-\sum\limits_{l = 1}^k(f^l, \delta_tQ^l)\\ &\leq\frac{1}{\tau}\Big(\frac{1}{6}\|f^{k+1}\|^2 +\frac{3}{2}\|Q^{k+\frac{1}{2}}\|^2\Big) +\frac{1}{2\tau}(\|f^0\|^2+\|Q^{\frac{1}{2}}\|^2) +\frac{1}{2}\sum\limits_{l = 1}^k(\|f^l\|^2+\|\delta_tQ^l\|^2). \end{align} $ (4.19)

    Substituting (4.19) into (4.18) and using Lemma 2, we get

    $ \begin{align} &\|e^{k+1}\|^2+\varepsilon^2|e^{k+1}|_{1}^{2} +\frac{\varepsilon^2h^{2}}{18}\|f^{k+1}\|^{2}\\ \leq& \frac{\varepsilon^2h^{2}}{12}\Big(\|f^{0}\|^{2}-\frac{h^{2}}{12}|f^{0}|_{1}^{2}\Big) +2\tau\Big(\frac{3}{2}+\frac{ c_3|\gamma|}{2}+\frac{3c_3^2\gamma^2h^2}{8\mu} +\frac{3c_3^2\gamma^2}{2\mu}\Big) \sum\limits_{l = 0}^{k}\|e^{l+1}\|^2\\ &+\tau\Big(\mu^2+\frac{3\mu h^2}{16}\Big)(k+1)Lc_1^2h^8 +\tau\varepsilon^4(k+1)Lc_1^2(\tau^2+h^4)^2+\tau(k+1)Lc_1^2(\tau^2+h^4)^2\\ &+\frac{\tau\varepsilon ^2h^2}{6}\Big[\frac{1}{2}\sum\limits_{l = 1}^{k}\|f^l\|^2 +\frac{1}{2}kLc_1^2(\tau^2+h^4)^2 +\frac{1}{\tau}\Big(\frac{1}{6}\|f^{k+1}\|^2+\frac{3}{2}Lc_1^2h^8\Big) +\frac{1}{2\tau}(\|f^0\|^2+Lc_1^2h^8)\Big]. \end{align} $

    We can rearrange the inequality above into the following form

    $ \begin{align} &\|e^{k+1}\|^2+\varepsilon^2|e^{k+1}|_{1}^{2} +\frac{\varepsilon^2h^{2}}{36}\|f^{k+1}\|^{2}\\ \leq&\tau\Big(3+c_3|\gamma|+\frac{3c_3^2\gamma^2h^2}{4\mu} +\frac{3c_3^2\gamma^2}{\mu}\Big) \Big[\sum\limits_{l = 0}^{k}\Big(\|e^{l}\|^2+\varepsilon^2|e^{l}|_{1}^{2} +\frac{\varepsilon^2h^{2}}{36}\|f^{l}\|^{2}\Big)+\|e^{k+1}\|^2\Big]\\ &+\frac{\varepsilon^2h^{2}}{12}\|f^{0}\|^{2}+\tau\Big(\mu^2+\frac{3\mu h^2}{16}\Big)(k+1)Lc_1^2h^8 +\tau\varepsilon^4(k+1)Lc_1^2(\tau^2+h^4)^2+\tau(k+1)Lc_1^2(\tau^2+h^4)^2\\ &+\frac{\tau\varepsilon^2h^2}{12}kLc_1^2(\tau^2+h^4)^2 +\frac{\varepsilon^2h^2}{4}Lc_1^2h^8+\frac{\varepsilon^2h^{2}}{12}\|f^{0}\|^{2} +\frac{\varepsilon^2h^{2}}{12}Lc_1^2h^8. \end{align} $ (4.20)

    Denote

    $ F^k = \|e^k\|^2+\varepsilon^2|e^{k}|_{1}^{2} +\frac{\varepsilon^2h^{2}}{36}\|f^{k}\|^{2}, \quad 1\leq k\leq N. $

    In combination of (2.6) in Lemma 4 with (4.12), we have

    $ \begin{align} \frac{\varepsilon^2h^{2}}{12}\|f^{0}\|^{2} \leq \frac{3\varepsilon^2h^{2}}{8}\|Q^{0}\|^{2} \leq \frac{3}{8}c_1^2\varepsilon^2h^{10}L. \end{align} $ (4.21)

    Substituting (4.21) into (4.20), when $ h\leq h_0 $ and $ c_4\tau\leq\frac{1}{3} $, (4.20) can be rewritten as

    $ F^{k+1}\leq \tau c_4\sum\limits_{l = 0}^kF^{l}+\tau c_4F^{k+1}+c_5(\tau^2+h^4)^2, $

    which implies that

    $ F^{k+1}\leq \frac{3}{2}c_4\tau\sum\limits_{l = 0}^k F^{l}+\frac{3}{2}c_5(\tau^2+h^4)^2. $

    According to the Gronwall inequality, we have

    $ \begin{align} F^{k+1}\leq \frac{3}{2}c_5 {\rm e}^{\frac{3}{2}c_4T}(\tau^2+h^4)^2 = c_6^2(\tau^2+h^4)^2, \quad 0\leq k\leq N-1. \end{align} $

    Thus, it holds that

    $ \|e^{k}\|\leq c_6(\tau^2+h^4), \quad\varepsilon|e^{k}|_1\leq c_6(\tau^2+h^4) , \quad\varepsilon\|e^{k}\|_\infty\leq \frac{c_6\sqrt{L}}{2}(\tau^2+h^4), \quad 0\leq k\leq N. $

    Below, we consider the stability of the difference scheme (3.9)–(3.12). Suppose that $ \lbrace \, \tilde{u}_i^k, \, \tilde{v}_i^k\, | \, 1\leq i\leq M, \, 0\leq k\leq N \, \rbrace $ is the solution of

    $ \left\{ {\begin{array}{l} \delta_t \tilde{u}_i^{k+\frac{1}{2}} = \mu \tilde{v}_i^{k+\frac{1}{2}}+\gamma\Big(\psi(\tilde{u}^{k+\frac{1}{2}}, \tilde{u}^{k+\frac{1}{2}})_i -\frac{h^2}{2}\psi(\tilde{v}^{k+\frac{1}{2}}, \tilde{u}^{k+\frac{1}{2}})_i\Big) +\varepsilon^2\delta_t\tilde{v}_i^{k+\frac{1}{2}}, \notag\\ \qquad \qquad 1\leq i\leq M, \quad0\leq k\leq N-1, \notag\\ \tilde{v}_i^k = \delta_x^2\tilde{u}_i^k-\frac{h^2}{12}\delta_x^2\tilde{v}_i^k, \quad1\leq i\leq M, \quad0\leq k\leq N, \\ \tilde{u}_i^0 = \varphi(x_i)+r(x_i), \quad 1\leq i\leq M, \notag\\ \tilde{u}_i^k = \tilde{u}_{i+M}^k, \quad \tilde{v}_i^k = \tilde{v}_{i+M}^k , \quad 1\leq i\leq M, \quad 1\leq k\leq N.\notag \end{array}} \right. $ (4.22)

    Denote

    $ \xi_i^k = \tilde{u}_i^k-u_i^k, \quad\eta_i^k = \tilde{v}_i^k-v_i^k, \quad 1\leq i\leq M, \quad0\leq k\leq N. $

    Subtracting (3.9)–(3.12) from (4.22), we obtain the perturbation equation as follows

    $ \left\{ {\begin{array}{l} \delta_t \xi_i^{k+\frac{1}{2}} = \mu \eta_i^{k+\frac{1}{2}}+\gamma\big[\psi(\tilde{u}^{k+\frac{1}{2}}, \tilde{u}^{k+\frac{1}{2}})_i -\psi(u^{k+\frac{1}{2}}, u^{k+\frac{1}{2}})_i\big] -\frac{\gamma h^2}{2}\big[\psi(\tilde{v}^{k+\frac{1}{2}}, \tilde{u}^{k+\frac{1}{2}})_i-\psi(v^{k+\frac{1}{2}}, u^{k+\frac{1}{2}})_i \big] & (4.23)\\ \;\;\;\;\;\;\;\;+\varepsilon^2\delta_t\eta_i^{k+\frac{1}{2}}, \quad 1\leq i\leq M, \quad0\leq k\leq N-1, \\ \eta_i^k = \delta_x^2\xi_i^k-\frac{h^2}{12}\delta_x^2\eta_i^k, \quad 1\leq i\leq M, \quad0\leq k\leq N, &(4.24)\\ \xi_i^0 = r(x_i), \quad 1\leq i\leq M, &(4.25) \\ \xi_i^k = \xi_{i+M}^k, \quad\eta_i^k = \eta_{i+M}^k , \quad 1\leq i\leq M, \quad1\leq k\leq N. &(4.26) \end{array}} \right. $

    Theorem 6. Let $ \lbrace\, \xi_i^k, \, \eta_i^k\, \vert\, 1\leq i\leq M, \, 0\leq k\leq N\, \rbrace $ be the solution of (4.23)–(4.26). When $ c_4\tau\leq\frac{1}{3} $ and $ h\leq h_0 $, we have

    $ \|\xi^{k}\|\leq c_7|r|_1, \quad\varepsilon|\xi^{k}|_1\leq c_7|r|_1, \quad\varepsilon\|\xi^{k}\|_\infty\leq \frac{c_7\sqrt{L}}{2}|r|_1, $

    where $ c_7 = \frac{1}{2}\sqrt{{\rm e}^{\frac{3}{2}c_4T}\Big(L^2+15\varepsilon^2\Big)}. $

    Proof. Taking an inner product of (4.23) with $ \xi^{k+\frac{1}{2}} $, we have

    $ \begin{align*} (\delta_t\xi^{k+\frac{1}{2}}, \xi^{k+\frac{1}{2}}) = &\, \mu(\eta^{k+\frac{1}{2}}, \xi^{k+\frac{1}{2}}) +\gamma\big(\psi(\tilde{u}^{k+\frac{1}{2}}, \tilde{u}^{k+\frac{1}{2}}) -\psi(u^{k+\frac{1}{2}}, u^{k+\frac{1}{2}}), \xi^{k+\frac{1}{2}}\big)\notag\\ &-\frac{\gamma h^2}{2}\big(\psi(\tilde{v}^{k+\frac{1}{2}}, \tilde{u}^{k+\frac{1}{2}}) -\psi(v^{k+\frac{1}{2}}, u^{k+\frac{1}{2}}), \xi^{k+\frac{1}{2}}\big) +\varepsilon^2(\delta_t\eta^{k+\frac{1}{2}}, \xi^{k+\frac{1}{2}}). \end{align*} $

    Similar to the analysis technique in Theorem 5, we obtain

    $ \begin{align*} &\|\xi^{k+1}\|^2-\|\xi^k\|^2+\varepsilon^2(|\xi^{k+1}|_{1}^{2}-|\xi^{k}|_{1}^{2}) +\frac{\varepsilon^2h^{2}}{12}\Big[\Big(\|\eta^{k+1}\|^{2}-\frac{h^{2}}{12}|\eta^{k+1}|_{1}^{2}\Big) -\Big(\|\eta^{k}\|^{2}-\frac{h^{2}}{12}|\eta^{k}|_{1}^{2}\Big)\Big]\\ \leq& -\frac{\mu\tau h^2}{9}||\eta^{k+\frac{1}{2}}||^2 +c_3\tau|\gamma|\cdot\|\xi^{k+\frac{1}{2} }\|^2 +\frac{ c_3\tau h^2|\gamma|}{3}||\eta^{k+\frac{1}{2}}||\cdot||\xi^{k+\frac{1}{2}}|| +\frac{ c_3\tau h^2|\gamma|}{3}||\eta^{k+\frac{1}{2}}||\cdot||\Delta_x \xi^{k+\frac{1}{2}}||\nonumber\\ \leq& c_3\tau|\gamma|\cdot||\xi^{k+\frac{1}{2}}||^2 +\frac{c_3^2\tau h^2\gamma^2}{2\mu}||\xi^{k+\frac{1}{2}}||^2 +\frac{2c_3^2\tau \gamma^2}{\mu}\|\xi^{k+\frac{1}{2}}\|^2\nonumber\\ \leq& \tau\Big(\frac{c_3|\gamma|}{2} +\frac{c_3^2\gamma^2 h^2}{4\mu}+\frac{c_3^2\gamma^2}{\mu}\Big) (\|\xi^{k+1}\|^2+\|\xi^k\|^2). \end{align*} $

    Replacing the superscript $ k $ with $ l $ and summing over $ l $ from 0 to $ k $, we get

    $ \begin{align*} &\|\xi^{k+1}\|^2+\varepsilon^2|\xi^{k+1}|_{1}^{2} +\frac{\varepsilon^2h^{2}}{12}\Big(\|\eta^{k+1}\|^{2}-\frac{h^{2}}{12}|\eta^{k+1}|_{1}^{2}\Big) -\|\xi^0\|^2-\varepsilon^2|\xi^{0}|_{1}^{2} -\frac{\varepsilon^2h^{2}}{12}\Big(\|\eta^{0}\|^{2}-\frac{h^{2}}{12}|\eta^{0}|_{1}^{2}\Big)\\ \leq& 2\tau\Big(\frac{c_3|\gamma|}{2} +\frac{c_3^2\gamma^2 h^2}{4\mu}+\frac{c_3^2\gamma^2}{\mu}\Big) \Big(\sum\limits_{l = 0}^k\|\xi^{l}\|^2+\|\xi^{k+1}\|^2\Big). \end{align*} $

    Using Lemma 2 and when $ h\leq h_0 $, we can rearrange the inequality above into the following form

    $ \begin{align} &\|\xi^{k+1}\|^2+\varepsilon^2|\xi^{k+1}|_{1}^{2} +\frac{\varepsilon^2h^{2}}{18}\|\eta^{k+1}\|^{2} \\ \leq& \tau\Big(c_3|\gamma|+\frac{c_3^2\gamma^2 h^2}{2\mu}+\frac{2c_3^2\gamma^2}{\mu}\Big) \Big(\sum\limits_{l = 0}^k\|\xi^{l}\|^2+\|\xi^{k+1}\|^2\Big) +\|\xi^0\|^2+\varepsilon^2|\xi^{0}|_{1}^{2} +\frac{\varepsilon^2h^{2}}{12}\|\eta^{0}\|^2\\ \leq& \tau c_4 \Big(\sum\limits_{l = 0}^k\|\xi^{l}\|^2+\|\xi^{k+1}\|^2\Big) +\|\xi^0\|^2+\varepsilon^2|\xi^{0}|_{1}^{2} +\frac{\varepsilon^2h^{2}}{12}\|\eta^{0}\|^2. \end{align} $ (4.27)

    Denote

    $ \tilde{F^k} = \|\xi^k\|^2+\varepsilon^2|\xi^{k}|_{1}^{2} +\frac{\varepsilon^2h^{2}}{18}\|\eta^{k}\|^{2}, \quad 0\leq k\leq N. $

    Combining (2.6) in Lemma 4 with (4.25), we have

    $ \begin{align} \frac{\varepsilon^2h^{2}}{12}\|\eta^{0}\|^{2} \leq \frac{3\varepsilon^2}{2}|\xi^{0}|_1^{2} = \frac{3\varepsilon^2}{2}|r|_1^{2}. \end{align} $ (4.28)

    Substituting (4.28) into (4.27), (4.27) can be rewritten as

    $ \begin{align} \|\xi^{k+1}\|^2+\varepsilon^2|\xi^{k+1}|_{1}^{2} +\frac{\varepsilon^2h^{2}}{18}\|\eta^{k+1}\|^{2} \leq \tau c_4 \Big(\sum\limits_{l = 0}^k\|\xi^{l}\|^2+\|\xi^{k+1}\|^2\Big) +\frac{L^2}{6}|r|_1^2+\varepsilon^2|r|_{1}^{2} +\frac{3\varepsilon^2}{2}|r|_1^{2}. \end{align} $

    Then, we have

    $ (1-c_4\tau)\tilde{F}^{k+1}\leq \tau c_4 \sum\limits_{l = 0}^k\tilde{F}^{l} +\Big(\frac{5\varepsilon^2}{2}+\frac{L^2}{6} \Big)|r|_1^{2}, \quad 0\leq k\leq N-1. $

    According to the Gronwall inequality, when $ c_4\tau\leq\frac{1}{3} $, we have

    $ \tilde{F}^{k}\leq c_7^2|r|_1^2, \quad 0\leq k\leq N, $

    where $ c_7 = \frac{1}{2}\sqrt{{\rm e}^{\frac{3}{2}c_4T}\Big(L^2+15\varepsilon^2\Big)}. $

    Therefore, it holds that

    $ \|\xi^{k}\|\leq c_7|r|_1, \quad\varepsilon|\xi^{k}|_1\leq c_7|r|_1, \quad\varepsilon\|\xi^{k}\|_\infty\leq \frac{c_7\sqrt{L}}{2}|r|_1. $

    In this section, we perform numerical experiments to verify the effectiveness of the difference scheme and the accuracy of the theoretical results. Before conducting the experiments, we first introduce an algorithm for solving the nonlinear compact scheme. Denote

    $ \begin{align*} u^k = (u_1^k, u_2^k, \cdots, u_M^k)^T, \quad\nu^k = (v_1^k, v_2^k, \cdots, v_M^k)^T, \quad w = (w_1, w_2, \cdots, w_M)^T, \quad z = (z_1, z_2, \cdots, z_M)^T, \end{align*} $

    where $ 0\leq k\leq N $. The algorithm of the compact difference scheme (3.9)–(3.12) can be described as follows:

    $ \textbf{Step 1} $ Solve $ u^0 $ and $ v^0 $ based on (3.10) and (3.11).

    $ \textbf{Step 2} $ Suppose $ u^k $ is known, the following linear system of equations will be used to approximate the solution of the difference scheme (3.9)–(3.12), for $ 1\leq i\leq M $, we have

    $ \left\{ {\begin{array}{l} \frac{2}{\tau}(w_i^{(l+1)}-u_i^k) = \mu z_i^{(l+1)}+\gamma\Big[\psi(w^{(l)}, w^{(l+1)})_i-\frac{h^2}{2}\psi(z^{(l)}, w^{(l+1)})_i\Big] +\frac{2}{\tau}\varepsilon^2(z_i^{(l+1)}-\nu_i^k), \notag\\ z_i^{(l)} = \delta_x^2w_i^{(l)}-\frac{h^2}{12}\delta_x^2z_i^{(l)}, \notag\\ w_i^{(l)} = w_{i+M}^{(l)}, \quad z_i^{(l)} = z_{i+M}^{(l)}, \notag \end{array}} \right. $

    until

    $ \max\limits_{1\leq i\leq M}|w_i^{(l+1)}-w_i^{(l)}|\leq \epsilon, \quad l = 0, 1, 2, \ldots. $

    Let

    $ u_i^{k+1} = 2w_i^{(l+1)}-u_i^k, \quad v_i^{k+1} = 2z_i^{(l+1)}-v_i^k, \quad0\leq i\leq M. $

    In the following numerical experiments, we set the tolerance error $ \epsilon = 1\times 10^{-12} $ for each iteration unless otherwise specified.

    When the exact solution is known, we define the discrete error in the $ L^\infty $-norm as follows:

    $ {\rm E}_\infty(h, \tau) = \max\limits_{1\leq i\leq M, \, 0\leq k\leq N}|U_i^k-u_i^k|, $

    where $ U_i^k $ and $ u_i^k $ represent the analytical solution and the numerical solution, respectively. Additionally, the convergence orders in space and time are defined as follows:

    $ {\rm Ord}_\infty^h = \log_2\frac{{\rm E}_\infty\left(2h, \tau\right)}{{\rm E}_\infty\left(h, \tau\right)}, \quad {\rm Ord}_\infty^\tau = \log_2\frac{{\rm E}_\infty\left(h, 2\tau\right)}{{\rm E}_\infty\left(h, \tau\right)}. $

    When the exact solution is unknown, we use a posteriori error estimation to verify the convergence orders in space and time. For a sufficiently small $ h $, we denote

    $ {\rm F}_\infty(h, \tau) = \max\limits_{1\leq i\leq M, \, 0\leq k\leq N}|u_i^k(h, \tau)-u_{2i}^k(h/2, \tau)|, \quad {\rm Ord}_\infty^h = \log_2\frac{{\rm F}_\infty(2h, \tau)}{{\rm F}_\infty(h, \tau)}. $

    Similarly, for sufficiently small $ \tau $, we denote

    $ {\rm G}_\infty(h, \tau) = \max\limits_{1\leq i\leq M, \, 0\leq k\leq N}|u_i^k(h, \tau)-u_i^{2k}(h, \tau/2)|, \quad {\rm Ord}_\infty^\tau = \log_2\frac{{\rm G}_\infty(h, 2\tau)}{{\rm G}_\infty(h, \tau)}. $

    Example 1. We first consider the following equation

    $ \left\{ {\begin{array}{l} u_t = u_{xx}+uu_x+u_{xxt}+f\left(x, t\right), \quad 0 < x < 2, \quad 0 < t\leq 1, \notag\\ u\left(x, 0\right) = \sin\left(\pi x\right), \quad 0\leq x\leq 2, \notag\\ u\left(0, t\right) = u\left(2, t\right), \quad 0 < t\leq 1, \notag \end{array}} \right. $

    where

    $ f\left(x, t\right) = {\rm e}^t\sin\left(\pi x\right)+2\pi^2{\rm e}^t\sin\left(\pi x\right)-\pi {\rm e}^{2t}\sin\left(\pi x\right)\cos(\pi x). $

    The initial condition is determined by the exact solution $ u\left(x, t\right) = {\rm e}^t\sin(\pi x) $ with the period $ L = 2 $ and $ T = 1 $.

    The numerical results are reported in Table 1 and Figure 1.

    Table 1.  Convergence orders versus numerical errors for the scheme (3.5)–(3.12) with reduced step-size under $ L = 2 $ and $ T = 1 $.
    $ \tau = 1/1000 $ $ h = 1/50 $
    $ h $ $ {\rm E}_\infty(h, \tau) $ $ {\rm Ord}_\infty^h $ $ \tau $ $ {\rm E}_\infty(h, \tau) $ $ {\rm Ord}_\infty^\tau $
    1/2 6.1769e-02 * 1/4 6.7342e-03 *
    1/4 7.4321e-03 3.0550 1/8 1.6884e-03 1.9959
    1/8 4.8805e-04 3.9287 1/16 4.2259e-04 1.9983
    1/16 3.1790e-05 3.9404 1/32 1.0587e-04 1.9970
    1/32 2.0894e-06 3.9274 1/64 2.6669e-05 1.9890

     | Show Table
    DownLoad: CSV
    Figure 1.  The numerical solution $ u(x, t) $ and the numerical error surface $ |U(x, t)-u(x, t)| $ with $ \tau = 1/1000 $, $ h = 1/50 $, $ L = 2 $ and $ T = 1 $.

    Table 1, we progressively reduce the spatial step-size $ h $ half by half $ (h = 1/2, 1/4, 1/8, 1/16, 1/32) $ while keeping the time step-size $ \tau = 1/1000 $. Conversely, we gradually decrease the time step-size $ \tau $ half by half $ (\tau = 1/4, 1/8, 1/16, 1/32, 1/64) $ while maintaining the spatial step-size $ h = 1/50 $.

    As we can see, the spatial convergence order approaches to the four order approximately, and the temporal convergence order approaches to two orders in the maximum norm, which are consistent with our convergence results. Comparing our numerical results with those in [42] from Table 2, we find our scheme is more efficient and accurate.

    Table 2.  Convergence orders versus numerical errors for the scheme [42] with reduced step-size under $ L = 2 $ and $ T = 1 $.
    $ \tau = 1/1000 $ $ h = 1/200 $
    $ h $ $ {\rm E}_\infty(h, \tau) $ $ {\rm Ord}_\infty^h $ $ \tau $ $ {\rm E}_\infty(h, \tau) $ $ {\rm Ord}_\infty^\tau $
    1/2 5.0747e-01 * 1/4 6.7850e-03 *
    1/4 1.1891e-01 2.0935 1/8 1.7378e-03 1.9651
    1/8 3.0838e-02 1.9471 1/16 4.7159e-04 1.8816
    1/16 7.6531e-03 2.0106 1/32 1.5477e-04 1.6074
    1/32 1.9208e-03 1.9943 1/64 7.5545e-05 1.0347

     | Show Table
    DownLoad: CSV

    By observing the first subgraph in Figure 1, the evolutionary trend surface of the numerical solution $ u(x, t) $ with $ \tau = 1/1000 $, $ h = 1/50 $, $ L = 2 $ and $ T = 1 $ is illustrated. This figure successfully reflects the panorama of the exact solution. In order to verify the accuracy of the difference scheme (3.9)–(3.12), we have drawn the numerical error surface in the second subgraph in Figure 1 with $ \tau = 1/1000 $, $ h = 1/50 $, $ L = 2 $ and $ T = 1 $.

    We observe that the rates of the numerical error in the maximum norm approaches a fixed value, which verifies that the difference scheme (3.9)–(3.12) is convergent. It took us 2.03 seconds to compute the spatial order of accuracy and 0.37 seconds to determine the temporal order of accuracy.

    Example 2. We further consider the problem of the form

    $ \left\{ {\begin{array}{l} u_t = u_{xx}+uu_x+\varepsilon^2u_{xxt}, \quad -25 < x < 25, \quad 0 < t\leq {T}, \notag\\ u\left(x, 0\right) = \frac{1}{2} \mathrm{sech}\left(\frac{x}{4}\right), \quad -25\leq x\leq 25, \notag\\ u\left(-25, t\right) = u\left(25, t\right), \quad 0 < t\leq {T}, \notag \end{array}} \right. $

    where the exact solution is unavailable.

    $ \textbf{Case I} $ $ \varepsilon = 1 $:

    The numerical results are reported in Table 3 and Figure 2. The two discrete conservation laws of the difference scheme (3.9)–(3.12) are reported in Table 4. In the following calculations, we set $ T = 1 $. First, we fix the temporal step-size $ \tau = 1/1000 $ and reduce the spatial step-size $ h $ half by half $ (h = 50/11, 50/22, 50/44, 50/88) $. Second, we fix the spatial step-size $ h = 1/2 $, meanwhile, reduce the temporal-step size $ \tau $ half by half $ (\tau = 1/2, 1/4, 1/8, 1/16) $.

    Table 3.  Convergence orders versus numerical errors for the scheme (3.9)–(3.12) with reduced step sizes under $ L = 50 $ and $ T = 1 $.
    $ \tau = 1/1000 $ $ h = 1/2 $
    $ h $ $ {\rm F}_\infty(h, \tau) $ $ {\rm Ord}_\infty^h $ $ \tau $ $ {\rm G}_\infty(h, \tau) $ $ {\rm Ord}_\infty^\tau $
    50/11 4.5583e-03 * 1/2 2.7427e-05 *
    50/22 5.0140e-04 3.1845 1/4 6.8356e-06 2.0045
    50/44 4.0505e-05 3.6298 1/8 1.7076e-06 2.0011
    50/88 4.6251e-06 3.1305 1/16 4.2681e-07 2.0003

     | Show Table
    DownLoad: CSV
    Figure 2.  The numerical solutions $ u(x, t) $ with $ \tau = 1/1000 $, $ h = 1/2 $, $ L = 50 $ (Left: $ \varepsilon = 1 $ and $ T = 1 $; Right: $ \varepsilon = 0.1 $ and $ T = 10 $).
    Table 4.  The discrete conservation laws of the difference scheme (3.9)–(3.12) with reduced step sizes under $ h = 1/2 $ and $ \tau = 1/1000 $.
    $ t $ $ Q $ $ E $
    0 6.267721589835858 2.041650615050223
    0.125 6.267721589832776 2.041650615048104
    0.250 6.267721589829613 2.041650615045897
    0.375 6.267721589826562 2.041650615043805
    0.500 6.267721589823495 2.041650615041712
    0.625 6.267721589820407 2.041650615039603
    0.750 6.267721589816996 2.041650615037288
    0.875 6.267721589813823 2.041650615035165
    1.000 6.267721589810754 2.041650615033137

     | Show Table
    DownLoad: CSV

    As we can see, the spatial convergence order approaches to four orders, approximately, and the temporal convergence order approaches to two orders in the maximum norm, which is consistent with our convergence results. It took us 6.74 seconds to compute the spatial order of accuracy, and 0.30 seconds to determine the temporal order of accuracy.

    From Table 4, we can see that the discrete conservation laws in Theorems 1 and 2 are also satisfied. In the first graph of Figure 2, we depict the evolutionary trend surface of the numerical solution $ u(x, t) $ with $ \tau = 1/1000 $, $ h = 1/2 $, $ L = 50 $ and $ T = 1 $, and this figure successfully reflects the panorama of the exact solution.

    When simulating a short duration of time $ T = 1 $, the impact of values $ \varepsilon = 1 $ and $ \varepsilon = 0.1 $ on the numerical simulation is relatively small. Therefore, in the following Case Ⅱ, we take $ \varepsilon = 0.1 $ and $ T = 10 $ to observe the impact of $ \varepsilon $ on the numerical simulation situation.

    $ \textbf{Case II} $ $ \varepsilon = 0.1 $:

    The numerical results are reported in Table 5 and Figure 2. The two discrete conservation laws of the difference scheme (3.9)–(3.12) are reported in Table 6.

    Table 5.  Convergence orders versus numerical errors for the scheme (3.9)–(3.12) with reduced step-sizes under $ L = 50 $ and $ T = 10 $.
    $ \tau = 1/1000 $ $ h = 1/2 $
    $ h $ $ {\rm F}_\infty(h, \tau) $ $ {\rm Ord}_\infty^h $ $ \tau $ $ {\rm G}_\infty(h, \tau) $ $ {\rm Ord}_\infty^\tau $
    25/3 5.1657e-02 * 1/2 9.3665e-03 *
    25/6 8.1492e-03 2.6642 1/4 2.6282e-03 1.8334
    25/12 6.5306e-04 3.6414 1/8 5.8168e-04 2.1758
    25/24 4.8822e-05 3.7416 1/16 1.3874e-04 2.0679

     | Show Table
    DownLoad: CSV
    Table 6.  The discrete conservation laws of the difference scheme (3.9)–(3.12) with reduced step-sizes under $ h = 1/2 $ and $ \tau = 1/1000 $.
    $ t $ $ Q $ $ E $
    0 6.267721589835858 2.000401671877802
    1.25 6.267721589837073 2.000401671878745
    2.50 6.267721589838247 2.000401671879621
    3.75 6.267721589839301 2.000401671880389
    5.00 6.267721589840338 2.000401671881143
    6.25 6.267721589840971 2.000401671881638
    7.50 6.267721589841729 2.000401671882164
    8.75 6.267721589842555 2.000401671882761
    10.0 6.267721589843263 2.000401671883252

     | Show Table
    DownLoad: CSV

    First, we fix the temporal step-size $ \tau = 1/1000 $ and reduce the spatial step-size $ h $ half by half $ (h = 25/3, 25/6, 25/12, 25/24) $. Second, we fix the spatial step-size $ h = 1/2 $ and reduce the temporal step size $ \tau $ half by half $ (\tau = 1/2, 1/4, 1/8, 1/16) $.

    As we can see, the spatial convergence order approaches to four orders, approximately, and the temporal convergence order approaches to two orders in the maximum norm, which is consistent with our convergence results. It took us 33.76 seconds to compute the spatial order of accuracy and 0.33 seconds to determine the temporal order of accuracy.

    In the second subgraph of Figure 2, we depict the evolutionary trend surface of the numerical solution $ u(x, t) $ with $ \tau = 1/1000 $, $ h = 1/2 $, $ L = 50 $ and $ T = 10 $. Compared with the first subgraph of Figure 2, smaller $ \varepsilon $ amplifies sharper transitions and wave-like behavior, whereas the larger $ \varepsilon $ makes the solution smoother.

    Example 3. In the last example, we consider the problem

    $ \left\{ {\begin{array}{l} u_t = u_{xx}+uu_x+u_{xxt}, \quad 0 < x < 30, \quad 0 < t\leq 20, \notag\\ u\left(0, t\right) = u\left(30, t\right), \quad 0 < t\leq 20.\notag \end{array}} \right. $

    with the Maxwell initial conditions $ u\left(x, 0\right) = {\rm e}^{-\left(x-7\right)^2}, \quad 0\leq x\leq 30. $

    Figure 3 reflects the behavior of the solutions to the pseudo-parabolic Burgers' equation. During the propagation process, we observe that the pseudo-parabolic Burgers' equation exhibits characteristics of both diffusion and advection. As we can see, the peak gradually spreads out and flattens as time progresses. Additionally, the solution moves to the right, indicating propagation direction.

    Figure 3.  The numerical solution $ u(x, t) $ with $ \tau = 1/500 $, $ h = 3/10 $, $ L = 30 $ and $ T = 20 $.

    The numerical scheme ensures stability, convergence, and the preservation of physical properties, which can be observed from the smooth transitions over time. This phenomenon indicates that the numerical scheme preserves the physical properties, ensuring stability and convergence.

    In Figure 4, we observe that the pseudo-parabolic Burgers' equation exhibits propagation characteristics coupled with gradual damping.

    Figure 4.  The numerical solution $ u(x, t) $ with $ \tau = 1/500 $, $ h = 3/10 $, $ L = 30 $ and $ T = 20 $.

    From Table 7, we can see that the discrete conservation law agrees well with Theorems 1 and 2. The value of $ Q $ remains almost constant throughout the simulation, which is crucial for maintaining the physical integrity of the solution. Similarly, the phenomenon is suitable for the energy $ E $, and these results further verify the correctness and reliability of the high-order compact difference scheme.

    Table 7.  The discrete conservation laws of the difference scheme (3.9)–(3.12) with reduced step sizes under $ h = 3/10 $ and $ \tau = 1/500 $.
    $ t $ $ Q $ $ E $
    0 1.772453850905516 2.505978912117327
    2.50 1.772453850935547 2.505978912141013
    5.00 1.772453850964681 2.505978912152665
    7.50 1.772453850994238 2.505978912161075
    10.0 1.772453851021927 2.505978912167129
    12.5 1.772453851052832 2.505978912173078
    15.0 1.772453851083987 2.505978912178467
    17.5 1.772453851115895 2.505978912183606
    20.0 1.772453851148868 2.505978912188604

     | Show Table
    DownLoad: CSV

    We propose and analyze an implicit compact difference scheme for the pseudo-parabolic Burgers' equation, achieving second-order accuracy in time and fourth-order accuracy in space. Using the energy method, we provide a rigorous numerical analysis of the scheme, proving the existence, uniqueness, uniform boundedness, convergence, and stability of its solution. Finally, the theoretical results are validated through numerical experiments. The experimental results demonstrate that the proposed scheme is highly accurate and effective, aligning with the theoretical predictions. As part of our ongoing research [42,43,44,45,46,47,48], we aim to extend these techniques and approaches to other nonlocal or nonlinear evolution equations [49,50,51,52,53,54,55].

    The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.

    The authors would like to thank the supervisor Qifeng Zhang, who provide this interesting topic and detailed guidance. They are also guilt to Baohui Xie's work when he studied in Zhejiang Sci-Tech Universtiy. The authors are grateful to the editor and the anonymous reviewers for their careful reading and many patient checking of the whole manuscript.

    The work is supported by Institute-level Project of Zhejiang Provincial Architectural Design Institute Co., Ltd. (Research on Digital Monitoring Technology for Central Air Conditioning Systems, Institute document No. 18), and National Social Science Fund of China (Grant No. 23BJY006) and General Natural Science Foundation of Xinjiang Institute of Technology(Grant No. ZY202403).

    The authors declare there is no conflicts of interest.

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