In mathematical physics, a direct problem is a problem of modeling some physical fields, processes, or phenomena, especially using partial differential equations (PDEs). The aim of solving a direct problem is to obtain the main dependent variable function that describes and governs naturally a physical field or process.
First, the QBS collocation technique is used for solving the proposed problem (1)–(3), when β, (q(t)), (zeta (x)), (eta (x)), and (g(x,t)) are given. For using the QBS method, we divide ([0,1]) into equally spaced grid points, (x_{i+1}-x_{i}), (i=0(1)M) apart. We denote (u(x_{i},t_{j})=u_{i}^{j}), (q(t_{j})=q^{j}), and (g(x_{i},t_{j})=g_{i}^{j}), where (x_{i}=iDelta x), (t_{j}=jDelta t), (Delta x=frac{1}{M}) and (Delta t=frac{T}{N}) for (i=0(1)M) and (j=0(1)N). The (Qs_{i}(x)), (i=-2(1)M+2) are defined as [4, 26]:
$$begin{aligned} Qs_{i}(x)=frac{1}{(Delta x)^{5}}textstylebegin{cases} mu _{i-3}^{5}, & [ {x_{i – 3} ,x_{i -2} }), \ mu _{i-3}^{5}-6 mu _{i-2}^{5}, & [ {x_{i – 2} ,x_{i-1} }), \ mu _{i-3}^{5}-6 mu _{i-2}^{5}+15mu _{i-1}^{5}, & [ {x_{i – 1} ,x_{i} }), \ – mu _{i+3}^{5}+6 mu _{i+2}^{5} – 15 mu _{i+1}^{5},& [ {x_{i} ,x_{i+1} }), \ – mu _{i+3}^{5} + 6 mu _{i+2}^{5},& [ {x_{i + 1} ,x_{i+2} }), \ mu _{i+3}^{5},& [ {x_{i+2} ,x_{i+3} }), \ 0,& text{else}, end{cases}displaystyle end{aligned}$$
(6)
where (mu _{i}=x-x_{i}) and the set of QBS functions ({Qs_{-2}, Qs_{-1},dots ,Qs_{M+2} }) form a basis over ([0,1]). The (Qs_{i}), (Qs’_{i}), (Qs”_{i}), (Qs”’_{i}) and (Qs^{(iv)}_{i}) are defined in Table 1.
Theorem 3
Let β be a positive number and (g(x,t)), (zeta (x)), (eta (x)), (f(x)), (kappa (t)), (q(t)) given functions. Suppose the QBS method and finite difference scheme are used for space and time discritization, respectively. Then, numerical solution (u(x,t)) is given in equations (23), (29), and (32).
Proof
Assume the expression for approximate solution (u(x,t)) at ((x,t_{j})) is defined as
$$ u(x,t_{j})=sum_{k=-2}^{M+2} {C_{k}^{j} Qs_{k} (x)}, $$
(7)
where (C_{k}^{j}) are the time-dependent quantities. The variation of the (u_{M}(x,t)), over the element, can be defined as
$$ u(x,t_{j})=sum_{k=i-2}^{i+2} {C_{k}^{j} Qs_{k} (x)}. $$
(8)
Using (8), function u with its first four derivatives can be defined as:
$$begin{aligned} &u_{i}^{j}= C_{i+2}^{j}+26C_{i+1}^{j}+66C_{i}^{j}+26C_{i-1}^{j}+C_{i-2}^{j}, end{aligned}$$
(9)
$$begin{aligned} &(u_{x} )_{i}^{j}= tau _{1} bigl(C_{i+2}^{j}+10 C_{i+1}^{j}-10 C_{i-1}^{j}- C_{i-2}^{j} bigr), end{aligned}$$
(10)
$$begin{aligned} &(u_{xx} )_{i}^{j}= tau _{2} bigl(C_{i+2}^{j}+2 C_{i+1}^{j}-6 C_{i}^{j}+2 C_{i-1}^{j}+ C_{i-2}^{j} bigr), end{aligned}$$
(11)
$$begin{aligned} &(u_{xxx} )_{i}^{j}= tau _{3} bigl(C_{i+2}^{j}-2 C_{i+1}^{j}+2 C_{i-1}^{j}- C_{i-2}^{j} bigr), end{aligned}$$
(12)
$$begin{aligned} &(u_{xxxx} )_{i}^{j}= tau _{4} bigl(C_{i+2}^{j} – 4 C_{i+1}^{j} + 6 C_{i}^{j} – 4 C_{i-1}^{j} + C_{i-2}^{j} bigr), end{aligned}$$
(13)
where
$$begin{aligned} tau _{1}=frac{5}{Delta x},qquad tau _{2}= frac{20}{(Delta x)^{2}},qquad tau _{3}=frac{60}{(Delta x)^{3}}, qquad tau _{4}=frac{120}{(Delta x)^{4}}. end{aligned}$$
Now the discretization of equation (1) yields
$$begin{aligned}& frac{u_{i}^{j + 1} -2 u_{i}^{j} + u_{i}^{j-1}}{(Delta t)^{2}} + frac{1}{2} bigl( ( u_{xxxx} )_{i}^{j + 1} + (u_{xxxx} )_{i}^{j} bigr) + frac{beta}{2} bigl( (u_{xx} )_{i}^{j + 1} + (u_{xx} )_{i}^{j} bigr) \& quad =frac{1}{2} bigl(q_{j + 1} u^{j + 1} + q_{j} u^{j} bigr)+ frac{1}{2} bigl(g_{i}^{j + 1} + g_{i}^{j}bigr),quad i=0(1)M, j=0(1)N. end{aligned}$$
(14)
Simplifying (14) yields
$$ begin{gathered} biggl( {1 – frac{(Delta t)^{2}}{2} q_{j + 1} } biggr)u_{i}^{j + 1} + frac{beta (Delta t)^{2}}{2} ( {u_{xx} } )_{i}^{j + 1} + frac{(Delta t)^{2}}{2} ( {u_{xxxx} } )_{i}^{j + 1} \ quad = biggl(2+frac{(Delta t)^{2}}{2} q_{j} biggr) u_{i}^{j} – frac{beta (Delta t)^{2}}{2} ( {u_{xx} } )_{i}^{j} – frac{(Delta t)^{2}}{2} ( {u_{xxxx} } )_{i}^{j} + frac{(Delta t)^{2}}{2} bigl( {g_{i}^{j + 1} + g_{i}^{j} } bigr), \ qquad i=0(1)M, j=0(1)N, end{gathered} $$
(15)
which can be written as
$$ begin{gathered} bigl( {1 – A^{j + 1} } bigr)u_{i}^{j + 1}+ beta B ( {u_{xx} } )_{i}^{j + 1} + B ( {u_{xxxx} } )_{i}^{j + 1} \ quad = bigl( {2+ A^{j} } bigr)u_{i}^{j} – beta B ( {u_{xx} } )_{i}^{j} – B ( {u_{xxxx} } )_{i}^{j} + frac{(Delta t)^{2}}{2} bigl( {g_{i}^{j + 1} + g_{i}^{j} } bigr), \ qquad i=0(1)M, j=0(1)N, end{gathered} $$
(16)
where
$$begin{aligned} A^{j} = frac{(Delta t)^{2} }{2} q_{j}, qquad B = frac{(Delta t)^{2}}{2}. end{aligned}$$
Now, using u, (u_{xx}), and (u_{xxxx}) from equations (9)–(13), we get
$$ begin{aligned} &bar{A}^{j + 1} C_{i – 2}^{j + 1} + bar{B}^{j + 1} C_{i – 1}^{j + 1} + bar{D}^{j + 1} C_{i}^{j + 1} + bar{B}^{j + 1} C_{i + 1}^{j + 1} + bar{A}^{j + 1} C_{i + 2}^{j + 1} \ &quad = bar{E}^{j} C_{i – 2}^{j} + bar{F}^{j} C_{i – 1}^{j} + bar{G}^{j} C_{i}^{j} + bar{F}^{j} C_{i + 1}^{j} + bar{E}^{j} C_{i + 2}^{j} -C_{i – 2}^{j-1} -26 C_{i – 1}^{j-1} -66 C_{i}^{j-1} \ &qquad {}-26 C_{i + 1}^{j-1} – C_{i + 2}^{j-1}+ frac{(Delta t)^{2}}{2} bigl( {g_{i}^{j + 1} + g_{i}^{j} } bigr), quad i=2,dots ,M-2, j=0(1)N, end{aligned} $$
(17)
where
$$ begin{aligned} &bar{A}^{j}=1 – A^{j} beta B tau _{2} + B tau _{4},qquad bar{B}^{j}= 26 – 26A^{j} + 2 beta B tau _{2} – 4B tau _{4}, \ &bar{D}=66 – 66A^{j} -6 beta B tau _{2} + 6B tau _{4},qquad bar{E}^{j}=2 + A^{j} -beta B tau _{2} – B tau _{4}, \ &bar{F}^{j}= 52 + 26A^{j} – 2 beta B tau _{2} + 4B tau _{4}, qquad bar{G}^{j}=132 + 66A^{j} +6 beta tau _{2} – 6 B tau _{4}. end{aligned} $$
Now, discretizing the boundary conditions (3), we get
$$ begin{aligned} &C_{-1}^{j}=-C_{1}^{j}-3C_{0}^{j}, qquad C_{-2}^{j}=-C_{2}^{j}-12C_{0}^{j}, \ &C_{M+2}^{j}=12C_{M}^{j}-C_{M-2}^{j}, qquad C_{M+1}^{j}=-3C_{M}^{j}-C_{M-1}^{j}, quad j=0(1)N. end{aligned} $$
(18)
For (i=0), using equation (18) in (17), we get
$$ begin{aligned} & bigl(12 bar{A}^{j+1}-3bar{B}^{j+1}+bar{D}^{j+1} bigr)C_{0}^{j + 1} \ &quad = bigl(12bar{E}^{j} -3 bar{F}^{j}+bar{G}^{j} bigr)C_{0}^{j} + frac{(Delta t)^{2}}{2} bigl( {g_{0}^{j+1}+ g_{0}^{j} } bigr), quad j = 0(1)N. end{aligned} $$
(19)
Now, for (i=1), we get
$$ begin{aligned} & bigl( -3bar{A}^{j+1}+bar{B}^{j+1} bigr)C_{0}^{j + 1} + bigl( -bar{A}^{j+1}+bar{D}^{j+1} bigr)C_{1}^{j + 1} + bar{B}^{j+1} C_{2}^{j+1}+bar{A}^{j+1}C_{3}^{j + 1} \ &quad = bigl( -3bar{E}^{j} +bar{F}^{j} bigr)C_{0}^{j} + bigl(-bar{E}^{j}+ bar{G}^{j} bigr)C_{1}^{j} + bar{F}^{j} C_{2}^{j} + bar{E}^{j} C_{3}^{j} -23C_{0}^{j-1}-65C_{1}^{j-1} \ &qquad {}-26 C_{2}^{j-1} – C_{3}^{j-1}+ frac{(Delta t)^{2}}{2} bigl( {g_{1}^{j + 1} + g_{1}^{j} } bigr), quad j = 0(1)N. end{aligned} $$
(20)
Next, for (i=M-1), we get
$$ begin{aligned} &bar{A}^{j+1} C_{M – 3}^{j + 1} + bar{B}^{j+1} C_{M – 2}^{j + 1} + bigl(-bar{A}^{j+1}+bar{D}^{j+1} bigr)C_{M – 1}^{j + 1} + bigl(-3bar{A}^{j+1}+bar{B}^{j+1} bigr) C_{M}^{j + 1} \ &quad = bar{E}^{j} C_{M – 3}^{j} + bar{F}^{j} C_{M – 2}^{j} + bigl(- bar{E}^{j} +bar{G}^{j} bigr)C_{M – 1}^{j} + bigl( -3bar{E}^{j}+ bar{F}^{j} bigr)C_{M}^{j} -C_{M-3}^{j-1} \ &qquad {}-26 C_{M-2}^{j-1}-65C_{M-1}^{j-1} -23C_{M}^{j-1}+ frac{(Delta t)^{2}}{2} bigl( {g_{M – 1}^{j + 1} + g_{M – 1}^{j} } bigr), quad j=0(1)N. end{aligned} $$
(21)
Finally, for (i=M), we get
$$ begin{aligned} & bigl(12 bar{A}^{j+1}-3bar{B}^{j+1}+bar{D}^{j+1} bigr)C_{M}^{j + 1} \ &quad = bigl(12bar{E}^{j} -3 bar{F}^{j}+bar{G}^{j} bigr)C_{M}^{j} + frac{(Delta t)^{2}}{2} bigl( {g_{M}^{j+1}+ g_{M}^{j} } bigr), quad j = 0(1)N. end{aligned} $$
(22)
At (t_{j+1}), (j=1,dots ,N-1), (19), (20), (17), (21), and (22) can be reformulated as
$$ begin{pmatrix} hat{p}^{j+1} & 0 & 0 & 0 & 0 & 0 & ldots & 0 \ hat{q}^{j+1} & hat{r}^{j+1} & bar{B}^{j+1} & bar{A}^{j+1} & 0 & 0 & ldots & 0 \ bar{A}^{j+1} & bar{B}^{j+1} & bar{D}^{j+1} & bar{B}^{j+1} & bar{A}^{j+1} & 0 & cdots & 0 \ 0 & bar{A}^{j+1} & bar{B}^{j+1} & bar{D}^{j+1} & bar{B}^{j+1} & bar{A}^{j+1} & cdots & 0 \ vdots & vdots & ddots & ddots & ddots & ddots & ddots & vdots \ 0 & ldots & 0 & bar{A}^{j+1} & bar{B}^{j+1} & bar{D}^{j+1} & bar{B}^{j+1} & bar{A}^{j+1} \ 0 & ldots & 0 & 0 & bar{A}^{j+1} & bar{B}^{j+1} & hat{r}^{j+1} & hat{q}^{j+1} \ 0 & ldots & 0 & 0 & 0 & 0 & 0 & hat{p}^{j+1} end{pmatrix}begin{pmatrix} {C_{0}^{j + 1} } \ {C_{1}^{j + 1} } \ {C_{2}^{j + 1} } \ vdots \ {C_{M – 2}^{j + 1} } \ {C_{M – 1}^{j + 1} } \ {C_{M}^{j + 1} } end{pmatrix} = begin{pmatrix} {R_{0}^{j} } \ {R_{1}^{j } } \ {R_{2}^{j} } \ vdots \ {R_{M – 2}^{j} } \ {R_{M – 1}^{j } } \ {R_{M}^{j } } end{pmatrix}, $$
(23)
where
$$begin{aligned} &hat{p}^{j+1} =12 bar{A}^{j+1}-3bar{B}^{j+1}+bar{D}^{j+1}, \ &hat{q}^{j+1} =-3bar{A}^{j+1}+bar{B}^{j+1}, qquad hat{r}^{j+1} =- bar{A}^{j+1}+bar{D}^{j+1}, \ &R_{0}^{j}= bigl(12bar{E}^{j} -3 bar{F}^{j}+bar{G}^{j} bigr)C_{0}^{j}+ frac{(Delta t)^{2}}{2} bigl( {g_{0}^{j+1}+ g_{0}^{j} } bigr), quad j = 0,dots ,N, \ &begin{aligned} R_{1}^{j}&= bigl( -3bar{E}^{j} +bar{F}^{j} bigr)C_{0}^{j} + bigl(- bar{E}^{j}+bar{G}^{j} bigr)C_{1}^{j} + bar{F}^{j} C_{2}^{j} + bar{E}^{j} C_{3}^{j} -23C_{0}^{j-1} \ &quad {}-65C_{1}^{j-1} -26 C_{2}^{j-1} – C_{3}^{j-1}+ frac{(Delta t)^{2}}{2} bigl( {g_{1}^{j + 1} + g_{1}^{j} } bigr), quad j = 0(1)N, end{aligned} \ &begin{aligned} R_{i}^{j}&= bar{E}^{j} C_{i – 2}^{j} + bar{F}^{j} C_{i – 1}^{j} + bar{G}^{j} C_{i}^{j} + bar{F}^{j} C_{i + 1}^{j} + bar{E}^{j} C_{i + 2}^{j} \ &quad {}-C_{i – 2}^{j-1} -26 C_{i – 1}^{j-1} -66 C_{i}^{j-1} -26 C_{i + 1}^{j-1} – C_{i + 2}^{j-1} \ &quad {}+frac{(Delta t)^{2}}{2} bigl( {g_{i}^{j + 1} + g_{i}^{j} } bigr), quad i=2,dots ,M-2, j=0(1)N, end{aligned} \ &begin{aligned} R_{M-1}^{j}&=bar{E}^{j} C_{M – 3}^{j} + bar{F}^{j} C_{M – 2}^{j} + bigl(-bar{E}^{j} +bar{G}^{j} bigr)C_{M – 1}^{j} + bigl( -3 bar{E}^{j}+bar{F}^{j} bigr)C_{M}^{j} \ &quad {}-C_{M-3}^{j-1} -26 C_{M-2}^{j-1}-65C_{M-1}^{j-1}-23C_{M}^{j-1} \ &quad {}+ frac{(Delta t)^{2}}{2} bigl( {g_{M – 1}^{j + 1} + g_{M – 1}^{j} } bigr), quad j=0(1)N, end{aligned} \ &R_{M}^{j}= bigl(12bar{E}^{j} -3 bar{F}^{j}+bar{G}^{j} bigr)C_{M}^{j}+ frac{(Delta t)^{2}}{2} bigl( {g_{M}^{j+1}+ g_{M}^{j} } bigr), quad j = 0(1)N. end{aligned}$$
Now for (j=0), using the initial condition (2) in (17), we have
$$ begin{gathered} bigl(1+bar{A}^{1}bigr) C_{i – 2}^{1} + bigl(26+bar{B}^{1}bigr) C_{i – 1}^{1} +bigl(66+ bar{D}^{1}bigr) C_{i}^{1} +bigl(26+ bar{B}^{1}bigr) C_{i + 1}^{1} + bigl(1+ bar{A}^{1}bigr) C_{i + 2}^{1} \ quad = bar{E}^{0} C_{i – 2}^{0} + bar{F}^{0} C_{i – 1}^{0} + bar{G}^{0} C_{i}^{0} + bar{F}^{0} C_{i + 1}^{0} + bar{E}^{0} C_{i + 2}^{0} +2 (Delta t) eta (x_{i}) + frac{(Delta t)^{2}}{2} bigl( {g_{i}^{1} + g_{i}^{0} } bigr), \ qquad i=2,dots ,M-2. end{gathered} $$
(24)
For (i=0), using (18) in (24), we have
$$begin{aligned} bigl(12 bar{A}^{1}-3bar{B}^{1}+bar{D}^{1} bigr)C_{0}^{1} = bigl(12bar{E}^{0} -3 bar{F}^{0}+bar{G}^{0} bigr)C_{0}^{0}+ frac{(Delta t)^{2}}{2} bigl( {g_{0}^{1}+ g_{0}^{0}} bigr). end{aligned}$$
(25)
Now, for (i=1), we have
$$ begin{aligned} & bigl( 23-3bar{A}^{1}+bar{B}^{1} bigr)C_{0}^{1} + bigl(65 -bar{A}^{1}+bar{D}^{1} bigr) C_{1}^{1} + bigl(26+bar{B}^{1}bigr) C_{2}^{1}+bigl(1+ bar{A}^{1} bigr) C_{3}^{1} \ &quad = bigl( -3bar{E}^{0} +bar{F}^{0} bigr)C_{0}^{0} + bigl(-bar{E}^{0}+ bar{G}^{0} bigr)C_{1}^{0} + bar{F}^{0} C_{2}^{0} + bar{E}^{0} C_{3}^{0} \ &qquad {}+2 (Delta t) eta (x_{1})+ frac{(Delta t)^{2}}{2} bigl( {g_{1}^{j + 1} + g_{1}^{j} } bigr). end{aligned} $$
(26)
Next, for (i=M-1), we get
$$ begin{aligned} &bigl(1+bar{A}^{1}bigr) C_{M – 3}^{1} + bigl(26+bar{B}^{1}bigr) C_{M – 2}^{1} + bigl(65-bar{A}^{1}+bar{D}^{1} bigr)C_{M – 1}^{1} + bigl(23-3 bar{A}^{1}+bar{B}^{1} bigr) C_{M}^{1} \ & quad = bar{E}^{0} C_{M – 3}^{0} + bar{F}^{0} C_{M – 2}^{0} + bigl(- bar{E}^{0} +bar{G}^{0} bigr)C_{M – 1}^{0} + bigl( -3bar{E}^{0}+ bar{F}^{0} bigr) C_{M}^{0} \ &qquad {}+2 (Delta t) eta (x_{M-1}) + frac{(Delta t)^{2}}{2} bigl( {g_{M – 1}^{1} + g_{M – 1}^{0} } bigr). end{aligned} $$
(27)
Finally, for (i=M), we have
$$ bigl(12 bar{A}^{1}-3bar{B}^{1}+bar{D}^{1} bigr)C_{M}^{1} = bigl(12bar{E}^{0} -3 bar{F}^{0} + bar{G}^{0} bigr)C_{M}^{0}+ frac{(Delta t)^{2}}{2} bigl( {g_{M}^{1}+ g_{M}^{0} } bigr). $$
(28)
At time step (t_{1}), (24)–(28) can be reformulated as
$$ begin{aligned} &begin{pmatrix} hat{p}^{1} & 0 & 0 & 0 & 0 & 0 & ldots & 0 \ 23+hat{q}^{1} & 65+hat{r}^{1} & 26+bar{B}^{1} & 1+bar{A}^{1} & 0 & 0 & ldots & 0 \ 1+bar{A}^{1} & 26+bar{B}^{1} & 66+bar{D}^{1} & 26+bar{B}^{1} & 1+ bar{A}^{1} & 0 & ldots & 0 \ 0 & 1+bar{A}^{1} & 26+bar{B}^{1} & 66+bar{D}^{1} & 26+bar{B}^{1} & 1+ bar{A}^{1} & ldots & 0 \ vdots & vdots & ddots & ddots & ddots & ddots & ddots & vdots \ 0 & ldots & 0 & 1+bar{A}^{1} & 26+bar{B}^{1} & 66+bar{D}^{1} & 26+ bar{B}^{1} & 1+bar{A}^{1} \ 0 & ldots & 0 & 0 & 1+bar{A}^{1} & 26+bar{B}^{1} & 65+hat{r}^{1} & 23+ hat{q}^{1} \ 0 & ldots & 0 & 0 & 0 & 0 & 0 & hat{p}^{1} end{pmatrix} \ & quad {}times begin{pmatrix} {C_{0}^{1} } \ {C_{1}^{1} } \ {C_{2}^{1} } \ vdots \ {C_{M – 2}^{1} } \ {C_{M – 1}^{1} } \ {C_{M}^{1} } end{pmatrix} = begin{pmatrix} {R_{0}^{0} } \ {R_{1}^{0}} \ {R_{2}^{0}} \ vdots \ {R_{M – 2}^{0} } \ {R_{M – 1}^{0} } \ {R_{M}^{0} } end{pmatrix}, end{aligned} $$
(29)
where
$$begin{aligned} &R_{0}^{0}= bigl(12bar{E}^{0} -3 bar{F}^{0}+bar{G}^{0} bigr)C_{0}^{0}+ frac{(Delta t)^{2}}{2} bigl( {g_{0}^{1}+ g_{0}^{0} } bigr), \ &begin{aligned} R_{1}^{0}&= bigl( -3bar{E}^{0} +bar{F}^{0} bigr)C_{0}^{0} + bigl(-bar{E}^{0}+bar{G}^{0} bigr)C_{1}^{0} + bar{F}^{0} C_{2}^{0} \ &quad {}+ bar{E}^{0} C_{3}^{0} +2 (Delta t) eta (x_{1})+ frac{(k)^{2}}{2} bigl( {g_{1}^{1} + g_{1}^{0}} bigr), end{aligned} \ &begin{aligned} R_{i}^{0}&= bar{E}^{0} C_{i – 2}^{0} + bar{F}^{0} C_{i – 1}^{0} + bar{G}^{0} C_{i}^{0} + bar{F}^{0} C_{i + 1}^{0} + bar{E}^{0} C_{i + 2}^{0} \ &quad {}+2 (Delta t) eta (x_{i})+frac{(Delta t)^{2}}{2} bigl( {g_{i}^{j + 1} + g_{i}^{j} } bigr), quad i=2,3,dots ,M-2, end{aligned} \ &begin{aligned} R_{M-1}^{0}&=bar{E}^{0} C_{M – 3}^{0} + bar{F}^{0} C_{M – 2}^{0} + bigl(-bar{E}^{0} +bar{G}^{0} bigr)C_{M – 1}^{0} + bigl( -3 bar{E}^{0}+bar{F}^{0} bigr)C_{M}^{0} \ & quad {}+2(Delta t) eta (x_{M-1})+ frac{(Delta t)^{2}}{2} bigl( {g_{M – 1}^{1} + g_{M – 1}^{0} } bigr), end{aligned} \ &R_{M}^{0}= bigl(12 bar{E}^{0} -3 bar{F}^{0}+bar{G}^{0} bigr)C_{M}^{0}+ frac{(Delta t)^{2}}{2} bigl( {g_{M}^{1}+ g_{M}^{0} } bigr). end{aligned}$$
Now, we determine the initial vector ((C_{-2}^{j},C_{-1}^{j}, C_{0}^{0}, dots , C_{M+1}^{0}, C_{M+2}^{0})). For removing (C_{-2}^{0}), (C_{-1}^{0}), (C_{M+1}^{0}), and (C_{M+2}^{0}), we use
$$begin{aligned} &u_{x}(0,0)=zeta _{x}(x_{0}),qquad u_{x}(1,0)= zeta _{x}(x_{M}), end{aligned}$$
(30)
$$begin{aligned} &u_{xx}(0,0)=0, qquad u_{xx}(1,0)=0. end{aligned}$$
(31)
Using (u_{x}) and (u_{xx}) from (10), (11) in equations (30) and (31), we get (C^{0}_{-2}), (C^{0}_{-1}), (C^{0}_{M+1}), and (C^{0}_{M+2}), and removing the unknowns (C_{-2}^{0}), (C_{-1}^{0}), (C_{M+1}^{0}), and (C_{M+2}^{0}), we have an ((M+1) times (M+1)) order system as follows:
$$ begin{aligned} &begin{pmatrix} 54 & 60 & 6 & 0 & 0 & 0 & cdots & 0 \ frac{101}{4} & frac{135}{2} & frac{105}{4} & 1 & 0 & 0& cdots & 0 \ 1 & 26 & 66 & 26 & 1 & 0 & cdots & 0 \ 0 & 1 & 26 & 66 & 26 & 1 &cdots & 0 \ vdots & vdots & ddots & ddots & ddots & ddots & ddots & vdots \ 0 & cdots & 0 & 1 & 26 & 66 & 26 & 1 \ 0 & cdots & 0 & 0 & 1 & frac{105}{4} & frac{135}{2} & frac{101}{4} \ 0 & cdots & 0 & 0 & 0 & 6 & 60 & 54 end{pmatrix} begin{pmatrix} C_{0}^{0} \ C_{1}^{0} \ C_{2}^{0} \ C_{2}^{0} \ vdots \ C_{M-2}^{0} \ C_{M-1}^{0} \ C_{M}^{0} end{pmatrix} \ &quad = begin{pmatrix} zeta (x_{0}) +frac{3.5}{tau _{1}} zeta _{x}(x_{0}) \ zeta (x_{1}) +frac{1}{8 tau _{1}} zeta _{x}(x_{0}) \ zeta (x_{2}) \ zeta (x_{3}) \ vdots \ zeta (x_{M-2 }) \ zeta (x_{M-1}) -frac{1}{8 tau _{1}} zeta _{x}(x_{M}) \ zeta (x_{M}) +frac{3}{tau _{1}} zeta _{x}(x_{M}) end{pmatrix}. end{aligned} $$
(32)
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