# An inverse problem of fourth-order partial differential equation with nonlocal integral condition – Advances in Continuous and Discrete Models

#### ByM. J. Huntul and Muhammad Abbas

Sep 1, 2022

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)

□