Research article Special Issues

Linking state-and-transition simulation and timber supply models for forest biomass production scenarios

  • Received: 30 January 2015 Accepted: 18 March 2015 Published: 24 March 2015
  • We linked state-and-transition simulation models (STSMs) with an economics-based timber supply model to examine landscape dynamics in North Carolina through 2050 for three scenarios of forest biomass production. Forest biomass could be an important source of renewable energy in the future, but there is currently much uncertainty about how biomass production would impact landscapes. In the southeastern US, if forests become important sources of biomass for bioenergy, we expect increased land-use change and forest management. STSMs are ideal for simulating these landscape changes, but the amounts of change will depend on drivers such as timber prices and demand for forest land, which are best captured with forest economic models. We first developed state-and-transition model pathways in the ST-Sim software platform for 49 vegetation and land-use types that incorporated each expected type of landscape change. Next, for the three biomass production scenarios, the SubRegional Timber Supply Model (SRTS) was used to determine the annual areas of thinning and harvest in five broad forest types, as well as annual areas converted among those forest types, agricultural, and urban lands. The SRTS output was used to define area targets for STSMs in ST-Sim under two scenarios of biomass production and one baseline, business-as-usual scenario. We show that ST-Sim output matched SRTS targets in most cases. Landscape dynamics results indicate that, compared with the baseline scenario, forest biomass production leads to more forest and, specifically, more intensively managed forest on the landscape by 2050. Thus, the STSMs, informed by forest economics models, provide important information about potential landscape effects of bioenergy production.

    Citation: Jennifer K. Costanza, Robert C. Abt, Alexa J. McKerrow, Jaime A. Collazo. Linking state-and-transition simulation and timber supply models for forest biomass production scenarios[J]. AIMS Environmental Science, 2015, 2(2): 180-202. doi: 10.3934/environsci.2015.2.180

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  • We linked state-and-transition simulation models (STSMs) with an economics-based timber supply model to examine landscape dynamics in North Carolina through 2050 for three scenarios of forest biomass production. Forest biomass could be an important source of renewable energy in the future, but there is currently much uncertainty about how biomass production would impact landscapes. In the southeastern US, if forests become important sources of biomass for bioenergy, we expect increased land-use change and forest management. STSMs are ideal for simulating these landscape changes, but the amounts of change will depend on drivers such as timber prices and demand for forest land, which are best captured with forest economic models. We first developed state-and-transition model pathways in the ST-Sim software platform for 49 vegetation and land-use types that incorporated each expected type of landscape change. Next, for the three biomass production scenarios, the SubRegional Timber Supply Model (SRTS) was used to determine the annual areas of thinning and harvest in five broad forest types, as well as annual areas converted among those forest types, agricultural, and urban lands. The SRTS output was used to define area targets for STSMs in ST-Sim under two scenarios of biomass production and one baseline, business-as-usual scenario. We show that ST-Sim output matched SRTS targets in most cases. Landscape dynamics results indicate that, compared with the baseline scenario, forest biomass production leads to more forest and, specifically, more intensively managed forest on the landscape by 2050. Thus, the STSMs, informed by forest economics models, provide important information about potential landscape effects of bioenergy production.


    The convection-diffusion equation(CDE) governs the transmission of particles and energy caused by convection and diffusion. The CDE is given by

    $ vt+ωvz=ϑ2vz2,czd,t>0,
    $
    (1.1)

    where $ \omega $ denotes coefficient of viscosity and $ \vartheta $ is the phase velocity respectively and both are positive. The Eq (1.1) is subject to the IC,

    $ v(z,0)=ϕ(z),czd
    $
    (1.2)

    and the BCs,

    $ {v(c,t)=f0(t)v(d,t)=f1(t)t>0,
    $
    (1.3)

    where, $ \phi $, $ f_0 $ and $ f_1 $ are given smooth functions.

    Numerous computational procedures have been developed in the literature for the CDE. Chawla et al. [1] developed extended one step time-integration schemes for the CDE. Daig et al. [2] discussed least-squares finite element method for the advection-diffusion equation(ADE). Mittal and Jain [3] revisited the cubic B-splines collocation procedure for the numerical treatment of the CDE. The characteristics method with cubic interpolation for ADE was presented by Tsai et al. [4]. Sari et al. [5] worked on high order finite difference schemes for solving the ADE. Taylor-Galerkin B-spline finite element method for the one-dimensional ADE was developed by Kadalbajoo and Arora [6]. Kara and Zhang [7] presented ADI method for unsteady CDE. Feng and Tian [8] found numerical solution of CDE using the alternating group explicit methods with exponential-type. Dehghan [9] used weighted finite difference techniques for the CDE. Further, Dehghan [10] developed a technique for the numerical solution of the three-dimensional advection-diffusion equation.

    A second-order space and time nodal method for CDE was conducted by Rizwan [11]. Mohebbi and Dehghan [12] presented a high order compact solution of the one-dimensional heat equation and ADE. Karahan [13,14] worked on unconditional stable explicit and implicit finite difference technique for ADE using spreadsheets. Salkuyeh [15] used finite difference approximation to solve CDE. Cao et. al [16] developed a fourth-order compact finite difference scheme for solving the CDEs. The generalized trapezoidal formula is used by Chawla and Al-Zanaidi [17] to solve CDE. Restrictive Taylor approximation has been used by Ismail et al. [18] to solve CDE. A boundary element method for anisotropic-diffusion convection-reaction equation in quadratically graded media of incompressible flow was studied by Salam et al. [19]. Azis [20] obtained standard-BEM solutions o two types of anisotropic-diffusion convection reaction equations with variable coefficients. The principle inspiration of this investigation is that the introduced scheme offers the solution as a piecewise adequately smooth continuous function enabling us to discover a numerical solution at any point in the solution domain. Moreover, it is simple to implement and has incredibly diminished computational expense. The refinement of the scheme with new approximations has elevated the accuracy of the scheme. The methodology used by von Neumann is utilized to affirm that the introduced scheme is unconditionally stable. The scheme is implemented to various test problems and the results are contrasted with the ones revealed in [25,26,27]. For further related studies, the interested reader is referred to [28,29] and references therein.

    The remaining part of the paper is organized in the following sequence. The numerical scheme is presented in section 2 which is based on the cubic B-spline collocation method. Section 3 deals with the Scheme's stability and convergence analysis. The comparison of our numerical results with the ones presented in [25,26,27] is also presented in this section. The study's finding is summed up in section 4.

    Derivation of the scheme

    Define $ \Delta t = \frac{T}{N} $ to be the time and $ h = \frac{d-c}{M} $ the space step sizes for positive integers $ M $ and $ N $. Let $ t_n = n \Delta t, \; n = 0, 1, 2, ..., N $, and $ z_j = jh, \; j = 0, 1, 2... M $. The solution domain $ c\leq z\leq d $ is evenly divided by knots $ z_{j} $ into $ M $ subintervals $ [z_{j}, z_{j+1}] $ of uniform length, where $ c = z_{0} < z_{1} < ... < z_{n-1} < z_{M} = d $. The scheme for solving (1.1) assumes approximate solution $ V(z, t) $ to the exact solution $ v(z, t) $ to be [23]

    $ V(z,t)=M+1j=1Dj(t)B3j(z),
    $
    (2.1)

    where $ D_j(t) $ are unknowns to be calculated and $ B^3_j (z) $ [23] are cubic B-spline basis functions given by

    $ B3j(z)=16h3{(zzj)3,z[zj,zj+1],h3+3h2(zzj+1)+3h(zzj+1)23(zzj+1)3,z[zj+1,zj+2],h3+3h2(zj+3z)+3h(zj+3z)23(zj+3z)3,z[zj+2,zj+3],(zj+4z)3,z[zj+3,zj+4],0,otherwise,
    $
    (2.2)

    Here, just $ B^3_{j-1}(z), B^3_{j}(z) $ and $ B^3_{j+1}(z) $ are last on account of local support of the cubic B-splines so that the approximation $ v_j^n $ at the grid point $ (z_j, t_n) $ at $ n^{th} $ time level is given as

    $ V(zj,tn)=Vnj=k=j+1k=j1Dnj(t)B3j(z).
    $
    (2.3)

    The time dependent unknowns $ D_j^n(t) $ are determined using the given initial and boundary conditions and the collocation conditions on $ B^3_j (z) $. Consequently, the approximations $ v_j^n $ and its required derivatives are found to be

    $ {vnj=α1Dnj1+α2Dnj+α1Dnj+1,(vnj)z=α3Dnj1+α4Dnj+α3Dnj+1,
    $
    (2.4)

    where $ {\alpha_1 = \frac{1}{6}, } \; {\alpha_2 = \frac{4}{6}, } \; {\alpha_3 = \frac{1}{2h}, } \; {\alpha_4 = 0.} $

    The new approximation [24] for $ (v_j^n)_{zz} $ is given as

    $ {(vn0)zz=112h2(14Dn133Dn0+28Dn114Dn2+6Dn3Dn4),(vnj)zz=112h2(Dnj2+8Dnj118Dnj+8Dnj+1+Dnj+2),j=1,2,...,M1(vnM)zz=112h2(DnM4+6DnM314DnM2+28DnM133DnM+14DnM+1),
    $
    (2.5)

    The problem (2.1) subject to the weighted $ \theta $-scheme takes the form

    $ (vnj)t=θhn+1j+(1θ)hnj,
    $
    (2.6)

    where $ h_j^{n} = \vartheta (v_j^n)_{zz}-\omega (v_j^n)_{z} $ and $ n = 0, 1, 2, 3, ... $. Now utilizing the formula, $ (v_j^n)_t = \frac{v_j^{n+1}-v_j^n}{k} $ in (2.6) and streamlining the terms, we obtain

    $ v_j^{n+1}+k \omega \theta (v_j^{n+1})_{z}-k \vartheta \theta (v_j^{n+1})_{zz} = v_j^{n}-k \omega (1-\theta) (v_j^n)_{z}+k \vartheta (1-\theta) (v_j^n)_{zz}, $ (2.7)

    Observe that $ \theta = 0 $, $ \theta = \frac{1}{2} $ and $ \theta = 1 $ in the system (2.7) correspond to an explicit, Crank-Nicolson and a fully implicit schemes respectively. We use the Crank-Nicolson approach so that (2.7) is evolved as

    $ vn+1j+12kω(vn+1j)z12kϑ(vn+1j)zz=vnj12kω(vnj)z+12kϑ(vnj)zz.
    $
    (2.8)

    Subtituting (2.4) and (2.5) in (2.8) at the knot $ z_0 $ returns

    $ (α1kωα327kϑ12h2)Dn+11+(α2+kωα42+11kϑ8h2)Dn+10+(α1+kωα327kϑ6h2)Dn+11+(7kϑ12h2)Dn+12(kϑ4h2)Dn+13+(kϑ24h2)Dn+14=(α1+kωα32+7kϑ12h2)Dn1+(α2kωα4211kϑ8h2)Dn0+(α1kωα32+7kϑ6h2)Dn1(7kϑ12h2)Dn2+(kϑ4h2)Dn3(kϑ24h2)Dn4.
    $
    (2.9)

    Substituting (2.4) and (2.5) in (2.8) produces

    $ (kϑ24h2)Dn+1j2+(α1kωα32kϑ3h2)Dn+1j1+(α2+kωα42+3kϑ4h2)Dn+1j+(α1+kωα32kϑ3h2)Dn+1j+1(kϑ24h2)Dn+1j+2=(kϑ24h2)Dnj2+(α1+kωα32+kϑ3h2)Dnj1+(α2kωα423kϑ4h2)Dnj+(α1kωα32+kϑ3h2)Dnj+1+(kϑ24h2)Dnj+2,j=1,2,3,...,M1.
    $
    (2.10)

    Subtituting (2.4) and (2.5) in (2.8) at the knot $ z_M $ yields

    $ (kϑ24h2)Dn+1M4(kϑ4h2)Dn+1M3+(7kϑ12h2)Dn+1M2+(α1kωα327kϑ6h2)Dn+1M1+(α2+kωα42+11kϑ8h2)Dn+1M+(α1+kωα327kϑ12h2)Dn+1M+1=(kϑ24h2)DnM4+(kϑ4h2)DnM3(7kϑ12h2)DnM2+(α1+kωα32+7kϑ6h2)DnM1+(α2kωα4211kϑ8h2)DnM+(α1kωα32+7kϑ12h2)DnM+1.
    $
    (2.11)

    From (2.9), (2.10) and (2.11), we acquire a system of $ (M+1) $ equations in $ (M+3) $ unknowns. To get a consistent system, two additional equations are obtained using the given boundary conditions. Consequently a system of dimension $ (M +3)\times(M +3) $ is obtained which can be numerically solved using any numerical scheme based on Gaussian elimination.

    Initial State: To begin iterative process, the initial vector $ D^0 $ is required which can be obtained using the initial condition and the derivatives of initial condition as follows:

    $ {(v0j)z=ϕ(zj),j=0,(v0j)=ϕ(zj),j=0,1,2,...,M,(v0j)z=ϕ(zj),j=M.
    $
    (2.12)

    The system (2.12) produces an $ (M+3)\times(M+3) $ matrix system of the form

    $ HC0=b,
    $
    (2.13)

    where,

    $ H = [α3α4α30...00α1α2α10...000α1α2α10...00...0α1α2α10...0α3α4α3]
    , $

    $ D^0 = [D_{-1}^0, D_{0}^0, D_{1}^0, ..., D_{M+1}^0]^T $ and $ b = [\phi'(z_0), \phi(z_0), ..., \phi(z_M), \phi'(z_M)]^T $.

    The von Neumann stability technique is applied in this section to explore the stability of the given scheme. Consider the Fourier mode, $ D_j^n = \sigma^n e^{i \beta h j} $, where $ \beta $ is the mode number, $ h $ is the step size and $ i = \sqrt{-1} $. Plug in the Fourier mode into equation (2.8) to obtain

    $ γ1σn+1eiβ(j4)h+γ2σn+1eiβ(j3)h+γ3σn+1eiβ(j2)h+γ4σn+1eiβ(j1)hγ1σn+1eiβ(j)h=γ1σneiβ(j4)h+γ5σneiβ(j3)h+γ6σneiβ(j2)h+γ7σneiβ(j1)h+γ1σneiβ(j)h,
    $
    (3.1)

    where,

    $ {\gamma_1 = \frac{k \vartheta}{24h^{2}}, } $

    $ {\gamma_2 = \alpha_1- \frac{k \omega \alpha_3}{2} -\frac{k\vartheta}{3h^{2}}, } $

    $ {\gamma_3 = \alpha_2+ \frac{k \omega\alpha_4}{2} +\frac{3k \vartheta}{4h^{2}}, } $

    $ {\gamma_4 = \alpha_1+ \frac{k \omega \alpha_3}{2} -\frac{k \vartheta}{3h^{2}}, } $

    $ {\gamma_5 = \alpha_1+ \frac{k \omega \alpha_3}{2} +\frac{k\vartheta}{3h^{2}}, } $

    $ {\gamma_6 = \alpha_2- \frac{k \omega\alpha_4}{2} -\frac{3k \vartheta}{4h^{2}}, } $

    $ {\gamma_7 = \alpha_1- \frac{k \omega \alpha_3}{2} +\frac{k \vartheta}{3h^{2}}.} $

    Dividing equation (3.1) by $ \sigma^n e^{i \beta j h} $ and rearranging, we obtain

    $ σ=γ1e2iβh+γ5eiβh+γ6+γ7eiβh+γ1e2iβhγ1e2iβh+γ2eiβh+γ3+γ4eiβhγ1e2iβh.
    $
    (3.2)

    Using $ \cos(\beta h) = \frac{e^{i \beta h}+e^{-i \beta h}}{2} $ and $ \sin(\beta h) = \frac{e^{i \beta h}-e^{-i \beta h}}{2i} $ in equation (3.2) and simplifying, we obtain

    $ σ=2γ1cos(2βh)+γ6+2E1cos(βh)i2F1sin(βh)2γ1cos(2βh)+γ3+2E2cos(βh)+i2F2sin(βh),
    $
    (3.3)

    where, $ {E_1 = \alpha_1 + \frac{k \vartheta}{3h^{2}}, } \; {F_1 = \frac{ k \omega \alpha_3}{2}, } \; {E_2 = \alpha_1-\frac{k \vartheta}{3h^{2}}, } \; {F_2 = \frac{ k \omega \alpha_3}{2}.} $

    Notice that $ \beta \in [-\pi, \pi] $. Without loss of generality, we can assume that $ \beta = 0 $, so that Eq (3.3) reduces to

    $ σ=2γ1+γ6+2E12γ1+γ3+2E2,=kϑ12h2+α23kϑ4h2+2α1+2kϑ3h2kϑ12h2+α2+3kϑ4h2+2α12kϑ3h2,=2α1+α22α1+α2=1,
    $

    which proves unconditional stability.

    In this section, we present the convergence analysis of the proposed scheme. For this purpose, we need to recall the following Theorem [21,22]:

    Theorem 1. Let $ v(z)\in C^4[c, d] $ and $ c = z_{0} < z_{1} < ... < z_{n-1} < z_{M} = d $ be the partition of $ [c, d] $ and $ V^*(z) $ be the unique B-spline function that interpolates $ v $. Then there exist constants $ \lambda_i $ independent of $ h $, such that

    $ \|(v-V^*)\|_\infty \leq \lambda_i h^{4-i},\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, i = 0,1,2,3. $

    First, we assume the computed B-spline approximation to (2.1) as

    $ V^*(z,t) = \sum\limits_{j = -3}^{M-1} D^*_j(t) B_j^3 (z). $

    To estimate the error, $ \|v(z, t)-V(z, t)\|_\infty $ we must estimate the errors $ \|v(z, t)-V^*(z, t)\|_\infty $ and $ \|V^*(z, t)-V(z, t)\|_\infty $ separately. For this purpose, we rewrite the equation (2.8) as:

    $ v+12kω(v)z12kϑ(v)zz=r(z),
    $
    (3.4)

    where, $ v^* = v_j^{n+1}, (v^*)_{z} = (v_j^{n+1})_{z}, (v^*)_{zz} = (v_j^{n+1})_{zz} $ and $ r(z) = v_j^{n}-\frac{1}{2}k \omega (v_j^n)_{z}+\frac{1}{2}k \vartheta (v_j^n)_{zz} $. Equation (3.4) can be written in matrix form as:

    $ AD=R,
    $
    (3.5)

    where, $ R = ND^n+h $ and

    $ A = [α1α2α10............0q1q2q3q4q5q60...0γ1γ2γ3γ4γ10......00......0γ1γ2γ3γ4γ10...0q6q5q4q10q2q110......0α1α2α1]
    $

    ,

    $ N = [0000............0q7q8q9q4q5q60...0γ1γ5γ6γ7γ10......00......0γ1γ5γ6γ7γ10...0q6q5q4q12q8q13000......0]
    , $

    $ h = [f_0(t_{n+1}), 0, ...0, f_1(t_{n+1})]^T $, $ D^n = [D_{-1}^n, D_{0}^n, D_{1}^n, ..., D_{M+1}^n]^T $,

    $ {q_1 = \alpha_1- \frac{k \omega \alpha_3}{2} - \frac{7k\vartheta}{12h^{2}}, } \; {q_2 = \alpha_2+ \frac{k \omega \alpha_4}{2} +\frac{11k\vartheta}{8h^{2}}, } \; {q_3 = \alpha_1+ \frac{k \omega \alpha_3}{2} - \frac{7k\vartheta}{6h^{2}}, } \; {q_4 = \frac{7k\vartheta}{12h^{2}}, } $

    $ {q_5 = \frac{k\vartheta}{4h^{2}}, } \; {q_6 = \frac{k\vartheta}{24h^{2}}, } \; {q_7 = \alpha_1+ \frac{k \omega \alpha_3}{2} +\frac{7k\vartheta}{12h^{2}}, } \; {q_8 = \alpha_2- \frac{k \omega \alpha_4}{2} -\frac{11k\vartheta}{8h^{2}}, } $

    $ {q_9 = \alpha_1- \frac{k \omega \alpha_3}{2} +\frac{7k\vartheta}{6h^{2}}, } \; {q_{10} = \alpha_1- \frac{k \omega \alpha_3}{2} - \frac{7k \vartheta}{6h^{2}}, } \; {q_{11} = \alpha_1+ \frac{k \omega \alpha_3}{2} - \frac{7k \vartheta}{12h^{2}}, } $

    $ {q_{12} = \alpha_1+ \frac{k \omega \alpha_3}{2} + \frac{7k \vartheta}{6h^{2}}, } \; {q_{13} = \alpha_1- \frac{k \omega \alpha_3}{2} +\frac{7 k \vartheta}{12h^{2}}.} $

    If we replace $ v^{*} $ by $ V^{*} $ in (3.4), then the resulting equation in matrix form becomes

    $ AD=R.
    $
    (3.6)

    Subtracting (3.6) from (3.5), we obtain

    $ A(DD)=(RR).
    $
    (3.7)

    Now using (3.4), we have

    $ |r(zj)r(zj)|=|(v(zj)v(zj))+kω2(vz(zj)vz(zj))kϑz(vzz(zj)vzz(zj))||(v(zj)v(zj))|+|kω2(vz(zj)vz(zj))|+|kϑ2(vzz(zj)vzz(zj))|.
    $
    (3.8)

    From (3.8) and Theorem (1), we have

    $ RRλ0h4+kω2λ1h3+kϑ2λ2h2=(λ0h2+kω2λ1h+kϑ2λ2)h2=M1h2,
    $
    (3.9)

    where $ M_1 = \lambda_0 h^2+\frac{k\omega}{2}\lambda_1 h+\frac{k\vartheta}{2} \lambda_2. $ It is obvious that the matrix $ A $ is diagonally dominant and thus nonsingular, so that

    $ (DD)=A1(RR).
    $
    (3.10)

    Now using (3.9), we obtain

    $ DDA1RRA1(M1h2).
    $
    (3.11)

    Let $ a_{j, i} $ denote the entries of $ A $ and $ \eta_j, 0 \leq j \leq M+2 $ is the summation of $ jth $ row of the matrix $ A $, then we have

    $ η0=M+2i=0a0,i=2α1+α2,
    $
    $ η1=M+2i=0a1,i=2α1+α2+kωα42,
    $
    $ ηj=M+2i=0aj,i=2α1+α2+kωα42,2jM
    $
    $ ηM+1=M+2i=0aM+1,i=2α1+α2+kωα42,
    $
    $ ηM+2=M+2i=0aM+2,i=2α1+α2.
    $

    From the theory of matrices we have,

    $ M+2j=0a1k,jηj=1,k=0,1,...,M+2,
    $
    (3.12)

    where $ a_{k, j}^{-1} $ are the elements of $ A^{-1} $. Therefore

    $ A1=M+2j=0|a1k,j|1minηk=1ξl1|ξl|,0k,lM+2.
    $
    (3.13)

    Substituting (3.13) into (3.11) we see that

    $ DDM1h2|ξl|=M2h2,
    $
    (3.14)

    where $ M_2 = \frac{M_1}{|\xi_l|} $ is some finite constant.

    Theorem 2. The cubic B-splines $ \{B_{-1}, B_{0}, ....B_{M+1}\} $ defined in relation (2.2) satisfy the inequality

    $ \sum \limits_{j = -3}^{M-1}|B^3_j (z)|\leq \frac{5}{3},\,\,\,\,\, 0 \leq z \leq 1. $

    Proof. Consider,

    $ |M1j=3B3j(z)|M1j=3|B3j(z)|=|B3j1(z)|+|B3j(z)|+|B3j+1(z)|=16+46+16=1.
    $

    Now for $ z\in[z_{j+1}, z_{j+2}] $, we have

    $ {|B^3_{j-2} (z)| \leq \frac{4}{6}} $

    $ {|B^3_{j-1} (z)| \leq\frac{1}{6}, } $

    $ {|B^3_{j} (z)| \leq \frac{4}{6}, } $

    $ {|B^3_{j+1} (z)| \leq\frac{1}{6}.} $

    Then, we have

    $ M1j=3|B3j(z)|=|B3j2(z)|+|B3j1(z)|+|B3j(z)|+|B3j+1(z)|53
    $

    as required.

    Now, consider

    $ V(z)V(z)=M1j=3(DjDj)B3j(z).
    $
    (3.15)

    Using (3.14) and Theorem 2, we obtain

    $ V(z)V(z)=M1j=3(DjDj)B3j(z)|M1j=3B3j(s)|(DjDj)53M2h2.
    $
    (3.16)

    Theorem 3. Let $ v(z) $ be the exact solution and let $ V(z) $ be the cubic collocation approximation to $ v(z) $ then the provided scheme has second order convergence in space and

    $ \|v(z)-V(z)\| \leq \mu h^2, \,\,\,\,\, \mathit{\text{where}}\,\,\, \mu = \lambda_0h^2+ \frac{5}{3}M_2h^2. $

    Proof. From Theorem 1, we have

    $ v(z)V(z)λ0h4.
    $
    (3.17)

    From (3.16) and (3.17), we obtain

    $ v(z)V(z)v(z)V(z)+V(z)V(z)λ0h4+53M2h2=μh2.
    $
    (3.18)

    where $ \mu = \lambda_0h^2+ \frac{5}{3} M_2. $

    In this section, some numerical calculations are performed to test the accuracy of the offered scheme. In all examples, we use the following error norms

    $ L=maxj|Vnum(zj,t)vexact(zj,t)|.
    $
    (3.19)
    $ L2=hj|Vnum(zj,t)vexact(zj,t)|2.
    $
    (3.20)

    The numerical order of convergence $ p $ is obtained by using the following formula:

    $ p=Log(L(n)/L(2n))Log(L(2n)/L(n)),
    $
    (3.21)

    where $ L_\infty(n) $ and $ L_\infty(2n) $ are the errors at number of partition $ n $ and $ 2n $ respectively.

    Example 1. Consider the CDE,

    $ vt+0.1vz=0.012vz2,0z1,t>0,
    $
    (3.22)

    with IC,

    $ v(z,0)=exp(5z)sin(πz)
    $
    (3.23)

    and the BCs,

    $ v(0,t)=0,v(1,t)=0.
    $
    (3.24)

    The analytic solution of the given problem is $ v(z, t) = \exp(5z-(0.25+0.01\pi^2)t)\sin(\pi z) $. To acquire the numerical results, the offered scheme is applied to Example 1. The absolute errors are compared with those obtained in [25] at various time stages in Table 1. In Table 2, absolute errors and error norms are presented at time stages $ t = 5, 10,100 $. Figure 1 displays the comparison that exists between exact and numerical solutions at various time stages. Figure 2 depicts the 2D and 3D error profiles at $ t = 1 $. A 3D comparison between the exact and numerical solutions is presented to exhibit the exactness of the scheme in Figure 3.

    Table 1.  Absolute errors are when $ k = 0.001 $ and $ h = 0.005 $ for Example 1.
    $ t $ $ z=0.1 $ $ z=0.3 $ $ z=0.5 $ $ z=0.7 $ $ z=0.9 $
    PM CNM[25] PM CNM[25] PM CNM[25] PM CNM[25] PM CNM[25]
    0.2 4.42 $ \times10^{-10} $ 1.88 $ \times10^{-5} $ 4.14 $ \times10^{-9} $ 3.61 $ \times10^{-5} $ 1.50 $ \times10^{-8} $ 1.39 $ \times10^{-5} $ 3.54 $ \times10^{-8} $ 2.05 $ \times10^{-4} $ 4.44 $ \times10^{-8} $ 9.67 $ \times10^{-4} $
    0.4 8.55 $ \times10^{-10} $ 3.23 $ \times10^{-5} $ 7.73 $ \times10^{-9} $ 6.76 $ \times10^{-5} $ 2.80 $ \times10^{-8} $ 2.61 $ \times10^{-5} $ 6.60 $ \times10^{-8} $ 3.84 $ \times10^{-4} $ 8.12 $ \times10^{-8} $ 1.64 $ \times10^{-3} $
    0.6 1.23 $ \times10^{-9} $ 4.15 $ \times10^{-5} $ 1.08 $ \times10^{-8} $ 9.45 $ \times10^{-5} $ 3.92 $ \times10^{-8} $ 3.66 $ \times10^{-5} $ 9.24 $ \times10^{-8} $ 5.37 $ \times10^{-4} $ 1.12 $ \times10^{-7} $ 2.10 $ \times10^{-3} $
    0.8 1.57 $ \times10^{-9} $ 4.81 $ \times10^{-5} $ 1.34 $ \times10^{-8} $ 1.17 $ \times10^{-4} $ 4.87 $ \times10^{-8} $ 4.56 $ \times10^{-5} $ 1.15 $ \times10^{-7} $ 6.64 $ \times10^{-4} $ 1.37 $ \times10^{-7} $ 2.42 $ \times10^{-3} $
    1.0 1.86 $ \times10^{-9} $ 5.26 $ \times10^{-5} $ 1.57 $ \times10^{-8} $ 1.35 $ \times10^{-4} $ 5.68 $ \times10^{-8} $ 5.31 $ \times10^{-5} $ 1.34 $ \times10^{-7} $ 7.68 $ \times10^{-4} $ 1.57 $ \times10^{-7} $ 2.63 $ \times10^{-3} $
    10 9.67 $ \times10^{-10} $ 7.17 $ \times10^{-6} $ 7.10 $ \times10^{-9} $ 2.87 $ \times10^{-5} $ 2.46 $ \times10^{-8} $ 2.31 $ \times10^{-5} $ 5.58 $ \times10^{-8} $ 1.11 $ \times10^{-4} $ 5.96 $ \times10^{-8} $ 2.86 $ \times10^{-4} $
    20 6.10 $ \times10^{-11} $ 2.57 $ \times10^{-7} $ 4.41 $ \times10^{-10} $ 1.13 $ \times10^{-6} $ 1.51 $ \times10^{-9} $ 1.41 $ \times10^{-6} $ 3.36 $ \times10^{-9} $ 2.10 $ \times10^{-6} $ 3.55 $ \times10^{-9} $ 7.60 $ \times10^{-6} $

     | Show Table
    DownLoad: CSV
    Table 2.  Absolute errors when $ M = 40 $ and $ k = 0.01 $ for Example 1.
    $ z $ $ t=5 $ $ t=10 $ $ t=100 $
    0.1 8.17 $ \times10^{-7} $ 3.13 $ \times10^{-7} $ 8.20 $ \times10^{-20} $
    0.3 6.55 $ \times10^{-6} $ 2.38 $ \times10^{-6} $ 5.87 $ \times10^{-19} $
    0.5 2.42 $ \times10^{-5} $ 8.46 $ \times10^{-6} $ 1.99 $ \times10^{-18} $
    0.7 5.79 $ \times10^{-5} $ 1.96 $ \times10^{-5} $ 4.39 $ \times10^{-18} $
    0.9 6.57 $ \times10^{-5} $ 2.15 $ \times10^{-5} $ 1.15 $ \times10^{-17} $

     | Show Table
    DownLoad: CSV
    Figure 1.  The approximate (stars, circles, triangles) and exact (solid lines) solutions at various time stages when $ M = 200, k = 0.001 $ for Example 1.
    Figure 2.  2D and 3D error profiles when $ t = 1, M = 100, k = 0.01 $ for Example 1.
    Figure 3.  The exact and approximate solutions when $ t = 1, M = 100, k = 0.01 $ for Example 1.

    The approximate solution when $ t = 1, k = 0.01 $ and $ M = 20 $ for Example 1 is given by

    $ V(z,1)={1.73472×1018+2.21674z+11.0117z2+26.9257z3,z[0,120]0.000773044+2.26312z+10.0841z2+33.1101z3,z[120,110]0.00726607+2.45791z+8.13619z2+39.6031z3,z[110,320]804.1213145.31z+4185.85z21843.97z3,z[1720,910]1242.214605.6z+5808.4z22444.91z3,z[910,1920]1863.6565.99z+7871.97z23168.97z3,z[1920,1].
    $

    Example 2. Consider the CDE,

    $ vt+0.22vz=0.52vz2,0z1,t>0,
    $
    (3.25)

    with IC,

    $ v(z,0)=exp(0.22z)sin(πz)
    $
    (3.26)

    and the BCs,

    $ v(0,t)=0,v(1,t)=0.
    $
    (3.27)

    The analytic solution is $ v(z, t) = \exp(0.22z-(0.0242+0.5\pi^2)t)\sin(\pi z) $. The numerical results are obtained by utilizing the proposed scheme. In Table 3, the comparative analysis of absolute errors is given with that of [25]. Absolute errors and errors norms at time levels $ t = 5, 10,100 $ are computed in Table 4. Figure 4 shows a very close comparison between the exact and numerical solutions at various stages of time. Figure 5 plots 2D and 3D absolute errors at $ t = 1 $. In Figure 6, a tremendous 3D contrast between the exact and approximate solutions is depicted.

    Table 3.  Absolute errors when $ k = 0.002 $ and $ h = 0.002 $ for Example 2.
    $ z $ $ t=0.8 $ $ t=1.0 $
    Present scheme CNM[25] Present scheme CNM[25]
    0.1 1.94 $ \times10^{-7} $ 2.12 $ \times10^{-7} $ 9.01 $ \times10^{-8} $ 1.79 $ \times10^{-7} $
    0.3 5.32 $ \times10^{-7} $ 5.85 $ \times10^{-7} $ 2.47 $ \times10^{-7} $ 4.92 $ \times10^{-7} $
    0.5 6.87 $ \times10^{-7} $ 7.62 $ \times10^{-7} $ 3.19 $ \times10^{-7} $ 6.40 $ \times10^{-7} $
    0.7 5.81 $ \times10^{-7} $ 6.49 $ \times10^{-7} $ 2.69 $ \times10^{-7} $ 5.45 $ \times10^{-7} $
    0.9 2.32 $ \times10^{-7} $ 2.61 $ \times10^{-7} $ 1.07 $ \times10^{-7} $ 2.19 $ \times10^{-7} $

     | Show Table
    DownLoad: CSV
    Table 4.  Absolute errors when $ M = 20 $ and $ k = 0.01 $ for Example 2.
    $ z $ $ t=5 $ $ t=10 $ $ t=100 $
    0.1 2.75 $ \times10^{-14} $ 1.41 $ \times10^{-23} $ 1.36 $ \times10^{-216} $
    0.3 7.52 $ \times10^{-14} $ 3.04 $ \times10^{-25} $ 1.32 $ \times10^{-23} $
    0.5 9.71 $ \times10^{-14} $ 1.81 $ \times10^{-24} $ 3.31 $ \times10^{-24} $
    0.7 8.21 $ \times10^{-14} $ 6.98 $ \times10^{-22} $ 4.24 $ \times10^{-22} $
    0.9 3.28 $ \times10^{-14} $ 2.17 $ \times10^{-19} $ 2.17 $ \times10^{-19} $

     | Show Table
    DownLoad: CSV
    Figure 4.  The approsimate (stars, circles, triangles) and exact (solid lines) solutions at various time stages when $ M = 500, k = 0.002 $ for Example 2.
    Figure 5.  2D and 3D error profiles when $ t = 1, M = 100, k = 0.01 $ for Example 2.
    Figure 6.  The exact and approximate solutions when $ t = 1, M = 100, k = 0.01 $ for Example 2.

    The numerical solution when $ t = 1, k = 0.01 $ and $ M = 20 $ for Example 2 is given by

    $ V(z,1)={2.71051×1020+0.0220426z+0.0044697z20.0330726z3,z[0,120]6.20887×107+0.0220053z+0.0052147z20.0380397z3,z[120,110]7.22875×107+0.0220457z+0.0048116z20.0366959z3,z[110,320]0.0164802+0.0968639z0.116698z2+0.0363097z3,z[1720,910]0.0199915+0.108568z0.129703z2+0.0411262z3,z[910,1920]0.0196901+0.107616z0.128701z2+0.0407746z3,z[1920,1].
    $

    Example 3. Consider the CDE,

    $ vt+0.1vz=0.22vz2,0z1,t>0,
    $
    (3.28)

    with IC,

    $ v(z,0)=exp(0.25z)sin(πz)
    $
    (3.29)

    and the BCs,

    $ v(0,t)=0,v(1,t)=0.
    $
    (3.30)

    The analytic solution is $ v(z, t) = \exp(0.25z-(0.0125+0.2\pi^2)t)\sin(\pi z) $. By utilizing the proposed scheme the numerical results are acquired. An excellent comparison between absolute errors computed by our scheme and the scheme of [25] is discussed in Table 5. In Table 6, absolute errors and error norms are computed at different time stages. A close comparison between the exact and numerical solutions at different time stages is depicted in Figure 7. Figure 8 plots 2D and 3D absolute errors at $ t = 1 $. Figure 9 deals with the 3D comparison that occurs between the exact and numerical solutions. The numerical solution when $ t = 1, k = 0.01 $ and $ M = 20 $ for Example 3 is given by

    $ V(z,1)={4.33681×1019+0.430962z+0.107784z20.708327z3,z[0,120]2.81208×106+0.430793z+0.111159z20.730824z3,z[120,110]2.06532×106+0.430816z+0.110935z20.730077z3,z[110,320]0.337164+1.95012z2.33266z2+0.719619z3,z[1720,910]0.398612+2.15495z2.56025z2+0.803911z3,z[910,1920]0.451717+2.32264z2.73678z2+0.865849z3,z[1920,1].
    $
    Table 5.  Absolute errors when $ k = 0.005 $ at $ h = 0.01 $ for Example 3.
    $ t $ $ z=0.1 $ $ z=0.3 $ $ z=0.5 $ $ z=0.7 $ $ z=0.9 $
    PM CNM[25] PM CNM[25] PM CNM[25] PM CNM[25] PM CNM[25]
    0.2 6.95 $ \times10^{-7} $ 3.72 $ \times10^{-6} $ 1.91 $ \times10^{-6} $ 1.02 $ \times10^{-5} $ 2.49 $ \times10^{-6} $ 1.32 $ \times10^{-5} $ 2.12 $ \times10^{-6} $ 1.12 $ \times10^{-5} $ 8.50 $ \times10^{-7} $ 4.48 $ \times10^{-6} $
    0.4 9.35 $ \times10^{-7} $ 2.80 $ \times10^{-6} $ 2.57 $ \times10^{-6} $ 7.68 $ \times10^{-6} $ 3.34 $ \times10^{-6} $ 9.93 $ \times10^{-6} $ 2.84 $ \times10^{-6} $ 8.41 $ \times10^{-6} $ 1.14 $ \times10^{-6} $ 3.36 $ \times10^{-6} $
    0.6 9.43 $ \times10^{-7} $ 1.57 $ \times10^{-6} $ 2.59 $ \times10^{-6} $ 4.29 $ \times10^{-6} $ 3.37 $ \times10^{-6} $ 5.55 $ \times10^{-6} $ 2.87 $ \times10^{-6} $ 4.69 $ \times10^{-6} $ 1.15 $ \times10^{-6} $ 1.88 $ \times10^{-6} $
    0.8 8.45 $ \times10^{-7} $ 7.77 $ \times10^{-7} $ 2.33 $ \times10^{-6} $ 2.13 $ \times10^{-6} $ 3.02 $ \times10^{-6} $ 2.75 $ \times10^{-6} $ 2.57 $ \times10^{-6} $ 2.33 $ \times10^{-6} $ 1.03 $ \times10^{-6} $ 9.29 $ \times10^{-7} $
    1.0 7.10 $ \times10^{-7} $ 3.61 $ \times10^{-7} $ 1.95 $ \times10^{-6} $ 9.88 $ \times10^{-7} $ 2.54 $ \times10^{-6} $ 1.28 $ \times10^{-6} $ 2.16 $ \times10^{-6} $ 1.08 $ \times10^{-6} $ 8.67 $ \times10^{-7} $ 4.31 $ \times10^{-7} $

     | Show Table
    DownLoad: CSV
    Table 6.  Absolute errors when $ M = 20 $ and $ k = 0.01 $ for Example 3.
    $ z $ $ t=5 $ $ t=10 $ $ t=100 $
    0.1 5.21 $ \times10^{-9} $ 5.07 $ \times10^{-13} $ 3.18 $ \times10^{-22} $
    0.3 1.44 $ \times10^{-8} $ 1.40 $ \times10^{-12} $ 4.69 $ \times10^{-87} $
    0.5 1.87 $ \times10^{-8} $ 1.82 $ \times10^{-12} $ 8.27 $ \times10^{-24} $
    0.7 1.59 $ \times10^{-8} $ 1.55 $ \times10^{-12} $ 8.47 $ \times10^{-22} $
    0.9 6.39 $ \times10^{-9} $ 6.21 $ \times10^{-13} $ 2.17 $ \times10^{-19} $

     | Show Table
    DownLoad: CSV
    Figure 7.  The approximate (stars, circles, triangles) and exact (solid lines) solutions at various time stages when $ M = 100, k = 0.005 $ for Example 3.
    Figure 8.  2D and 3D error profiles when $ t = 1, M = 100, k = 0.01 $ for Example 3.
    Figure 9.  The exact and approximate solutions when $ t = 1, M = 100, k = 0.01 $ for Example 3.

    Example 4. Consider the CDE,

    $ vt+0.8vz=0.12vz2,0z1,t>0,
    $
    (3.31)

    with IC,

    $ v(z,0)=exp((z2)280)
    $
    (3.32)

    and the BCs,

    $ v(0,t)=2020+texp((20.8t)20.4(20+t)),
    $
    (3.33)
    $ v(1,t)=2020+texp((10.8t)20.4(20+t)).
    $
    (3.34)

    The analytic solution of the given problem is $ v(z, t) = \sqrt{\frac{20}{20+t}}\exp(-\frac{(z-2-0.8t)^2}{0.4(20+t)}) $. In Table 7, a comparison between absolute errors calculated by our scheme and the scheme of [26] is presented. Absolute errors and errors norms at time levels $ t = 5, 10,100 $ are presented in Table 8. Figure 10 illustrates the behavior of numerical solutions at various time stages. Figure 11 depicts the 2D and 3D graphs of absolute errors. In Figure 12, an excellent 3D contrast between the exact and numerical solutions shows the enormous accuracy of the scheme.

    Table 7.  Absolute errors and convergence orders when $ t = 1.0 $ and $ k = 0.001 $ for Example 4.
    $ h $ Present scheme CuTBS[26]
    $ L_2 $ $ L_\infty $ $ p $ $ L_2 $ $ L_\infty $
    $ 1/4 $ 7.10 $ \times10^{-7} $ 1.20 $ \times10^{-6} $ 1.20 $ \times10^{-4} $ 1.09 $ \times10^{-4} $
    $ 1/8 $ 5.33 $ \times10^{-8} $ 8.50 $ \times10^{-8} $ 2.21360 2.98 $ \times10^{-5} $ 2.35 $ \times10^{-5} $
    $ 1/16 $ 5.44 $ \times10^{-9} $ 8.49 $ \times10^{-9} $ 2.02852 7.56 $ \times10^{-6} $ 5.76 $ \times10^{-6} $
    $ 1/32 $ 2.34 $ \times10^{-9} $ 3.61 $ \times10^{-9} $ 2.01005 1.91 $ \times10^{-6} $ 1.43 $ \times10^{-6} $
    $ 1/64 $ 2.41 $ \times10^{-9} $ 3.30 $ \times10^{-9} $ 2.01012 4.77 $ \times10^{-7} $ 3.55 $ \times10^{-7} $
    $ 1/128 $ 2.13 $ \times10^{-9} $ 3.29 $ \times10^{-9} $ 2.03705 1.17 $ \times10^{-7} $ 8.65 $ \times10^{-8} $

     | Show Table
    DownLoad: CSV
    Table 8.  Absolute errors when $ M = 20 $ and $ k = 0.01 $ for Example 4.
    $ z $ $ t=5 $ $ t=10 $ $ t=100 $
    0.1 1.29 $ \times10^{-8} $ 3.75 $ \times10^{-10} $ 4.14 $ \times10^{-25} $
    0.3 4.20 $ \times10^{-8} $ 1.43 $ \times10^{-9} $ 3.10 $ \times10^{-25} $
    0.5 7.29 $ \times10^{-8} $ 2.87 $ \times10^{-9} $ 1.94 $ \times10^{-25} $
    0.7 9.66 $ \times10^{-8} $ 4.30 $ \times10^{-9} $ 6.62 $ \times10^{-24} $
    0.9 7.31 $ \times10^{-8} $ 3.51 $ \times10^{-9} $ 1.26 $ \times10^{-60} $

     | Show Table
    DownLoad: CSV
    Figure 10.  The approximate (stars, circles, triangles) and exact (solid lines) solutions at various time stages when $ M = 100, k = 0.001 $ for Example 4.
    Figure 11.  2D and 3D error profiles when $ t = 1, M = 100, k = 0.01 $ for Example 4.
    Figure 12.  The exact and approximate solutions when $ t = 1, M = 100, k = 0.01 $ for Example 4.

    The approximate solution when $ t = 1, k = 0.01 $ and $ M = 20 $ for Example 4 is given by

    $ V(z,1)={0.383764+0.255843z+0.0396075z20.0119477z3,z[0,120]0.383764+0.255836z+0.039733z20.0127849z3,z[120,110]0.383765+0.255811z+0.0399901z20.0136416z3,z[110,320]0.385493+0.247569z+0.0547152z20.0241989z3,z[1720,910]0.385817+0.24649z+0.0559143z20.024643z3,z[910,1920]0.386119+0.245535z+0.0569187z20.0249954z3,z[1920,1].
    $

    Example 5. Consider the CDE,

    $ vt+0.1vz=0.022vz2,0z1,t>0,
    $
    (3.35)

    with IC,

    $ v(z,0)=exp(1.17712434446770z)
    $
    (3.36)

    and the BCs,

    $ v(0,t)=exp(0.09t),
    $
    (3.37)
    $ v(1,t)=exp(1.177124344467700.09t).
    $
    (3.38)

    The analytic solution of the given problem is $ v(z, t) = \exp(1.17712434446770z-0.09t) $. The numerical outcomes are acquired by using the introduced scheme. The absolute errors and errors norms are computed in Tables 9 by applying the presented scheme on Example 5 and are compared with those obtained in [27]. In Table 10, absolute errors and error norms are presented at various time stages. Figure 13 plots the behavior of exact and numerical solutions at various time stages. Figure 14 depicts the 2D and 3D absolute error profiles at $ t = 1 $. In Figure 15, a 3D contrast between the exact and numerical solutions is presented and all the graphs are in good agreement.

    Table 9.  Absolute errors and error norms when $ h = 0.01 $ and $ k = 0.001 $ for Example 5.
    $ z $ t=1 t=2
    CuBQI[27] Present method CuBQI[27] Present method
    $ 0.1 $ 2.1506 $ \times10^{-6} $ 2.5219 $ \times10^{-11} $ 2.8217 $ \times10^{-6} $ 3.3391 $ \times10^{-11} $
    $ 0.5 $ 7.0601 $ \times10^{-6} $ 7.6365 $ \times10^{-11} $ 1.2276 $ \times10^{-5} $ 1.3300 $ \times10^{-10} $
    $ 0.9 $ 7.6594 $ \times10^{-6} $ 7.8200 $ \times10^{-11} $ 1.1643 $ \times10^{-5} $ 1.1782 $ \times10^{-10} $
    $ L_2 $ 6.4790 $ \times10^{-7} $ 6.9230 $ \times10^{-11} $ 1.0719 $ \times10^{-6} $ 1.1447 $ \times10^{-10} $
    $ L_\infty $ 9.1107 $ \times10^{-6} $ 9.7331 $ \times10^{-11} $ 1.5204 $ \times10^{-5} $ 1.6236 $ \times10^{-10} $

     | Show Table
    DownLoad: CSV
    Table 10.  Absolute errors when $ M = 40 $ and $ k = 0.01 $ for Example 5.
    $ z $ $ t=5 $ $ t=10 $ $ t=100 $
    0.1 4.46 $ \times10^{-9} $ 3.42 $ \times10^{-9} $ 1.12 $ \times10^{-12} $
    0.3 1.41 $ \times10^{-8} $ 1.14 $ \times10^{-8} $ 3.81 $ \times10^{-12} $
    0.5 2.34 $ \times10^{-8} $ 1.99 $ \times10^{-8} $ 6.73 $ \times10^{-12} $
    0.7 2.89 $ \times10^{-8} $ 2.50 $ \times10^{-8} $ 8.51 $ \times10^{-12} $
    0.9 1.94 $ \times10^{-8} $ 1.65 $ \times10^{-8} $ 5.59 $ \times10^{-12} $

     | Show Table
    DownLoad: CSV
    Figure 13.  The approximate (stars, circles, triangles) and exact (solid lines) solutions at various time stages when $ M = 100, k = 0.001 $ for Example 5.
    Figure 14.  2D and 3D error profiles when $ t = 1, M = 100, k = 0.01 $ for Example 5.
    Figure 15.  The exact and approximate solutions when $ t = 1, M = 100, k = 0.01 $ for Example 5.

    The approximate solution when $ t = 1, k = 0.01 $ and $ M = 20 $ for Example 5 is given by

    $ V(z,1)={0.913931+1.07581z+0.632996z2+0.255851z3,z[0,120]0.913929+1.07593z+0.630674z2+0.271329z3,z[120,110]0.913913+1.07642z+0.625737z2+0.287787z3,z[110,320]0.815184+1.50737z0.052669z2+0.695802z3,z[1720,910]0.78445+1.60981z0.166497z2+0.73796z3,z[910,1920]0.746014+1.73119z0.294263z2+0.78279z3,z[1920,1].
    $

    This research uses a new approximation for second-order derivatives in the cubic B-spline collocation method to obtain the numerical solution of the CDE. The smooth piecewise cubic B-splines have been used to approximate derivatives in space whereas a usual finite difference has been used to discretize the time derivative. Special consideration is paid to the stability and convergence analysis of the scheme to ensure the errors do not amplify. The approximate solutions and error norms are contrasted with those reported previously in the literature. From this analysis, we can conclude that the estimated solutions are in perfect accord with the actual solutions. The scheme can be applied to a wide range of problems in science and engineering.

    The authors are thankful to the worthy reviewers and editors for the useful and valuable suggestions for the improvement of this paper which led to a better presentation.

    The authors have no conflict of interest.

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