# Subdivision Collocation Method for One-Dimensional Bratu’s Problem

Authors: Ghulam Mustafa; Syeda Tehmina Ejaz; Sabila Kouser; Shafqat Ali; Muhammad Aslam
Journal: Journal of Mathematics
 (2021)

---

## Abstract

The purpose of this article is to employ the subdivision collocation method to resolve Bratu’s boundary value problem by using approximating subdivision scheme. The main purpose of this researcher is to explore the application of subdivision schemes in the field of physical sciences. Our approach converts the problem into a set of algebraic equations. Numerical approximations of the solution of the problem and absolute errors are compared with existing methods. The comparison shows that the proposed method gives a more accurate solution than the existing methods.


---

## Body

## 1. Introduction

The general expression of Liouville–Bratu–Gelfand equation [1, 2]:(1)Δχr+αexpχr=0,r∈Ω1,χr=0,r∈∂Ω1,where parameter α>0 and Ω1 is a bounded domain. We consider the Bratu’s boundary value problem in one-dimensional planar coordinates [2–4] of the form(2)χ⋅⋅r+αexpχr=0,0<r<1,α>0,with conditions at the ends of the domain(3)χ0=0,χ1=0.The detailed information of problem (2) is given in [4, 5]. The exact solution of (2) is(4)χr=−2lncosh0.5θr−0.5cosh0.25θ,whereθ=2αcosh0.25θ.The exponential term guarantees nonlinearity and the bifurcation phenomenon that follows up. In particular, one can verify the following for different values ofα, i.e., problem (2) has no solution for α>αc, unique solution for α=αc, and two bifurcated solutions have been obtained for 0<α<αc, where αc is the critical value given as αc=3.51380719. It is the solution of 1=0.252αcsinh0.25θ.In science and engineering, Bratu’s problem is often used to characterize complex physical and chemical models. For example, Bratu’s problem is used in a wide range of applications, including the thermal combustion theory’s fuel ignition model, the model of the thermal reaction mechanism, the Chandrasekhar model of the universe’s expansion, chemical reaction theory, radiative heat transfer, and nanotechnology.Many researchers have developed analytical and numerical methods to solve the Bratu’s problem, including the B-spline method [6], Adomian decomposition method [7, 8], Chebyshev polynomial approximation method [2], homotopy analysis method [9], homotopy perturbation method [10, 11], differential transform method [12], Laplace transformed decomposition method [13], method of weighted residuals [14], and variational iteration method [15, 16]. Moreover, the solutions of the problem have been reported by Jalilian [17] using the nonpolynomial spline method, by Boyd [18] with the one-point pseudospectral collocation method, and by Abbasbandy et al. [19] with the Lie-group shooting method.Our goal is to make use of subdivision schemes for solving Bratu’s problem. Subdivision schemes-based algorithms are not frequently used to find numerical solutions of boundary value problems. The approximate solutions of boundary value problems have been found by subdivision-based algorithms. Initially, these algorithms were constructed by Qu and Agarwal [20, 21]. Their constructed algorithms were based on an interpolatory subdivision algorithm and formulated only for the second-order two-point boundary value problems. After that, Ejaz et al. [22, 23] constructed subdivision schemes-based algorithm for solutions of boundary value problems of third and fourth order. We present a subdivision collocation algorithm for solving Bratu’s problem in this paper.We organize our paper in the following way. In Section2, we present some important properties of 6-point binary approximating subdivision scheme. In Section 3, subdivision collocation algorithm is formulated for the solution of (2). The convergence and error estimation of the proposed algorithm are also discussed in this section. Numerical results based on the proposed algorithm, comparison with other existing methods, and conclusion based on the obtained results are given in Section 4.


## 2. Subdivision Scheme and Derivatives of Its Two-Scale Relation

In this section, we define 6-point binary approximating subdivision scheme (6PBASS) [24] as(5)Z2ik+1=a0Zi−2k+a1Zi−1k+a2Zik+a3Zi+1k+a4Zi+2k,Z2i+1k+1=b0Zi−2k+b1Zi−1k+b2Zik+b3Zi+1k+b4Zi+2k+b5Zi+3k,with  a0=a4=−1/32α, a1=a3=1/8α, a2=1−3/16α, b0=b5=1−αβ, b1=b4=−1/16−3β1−α, b2=b3=9/16+2β1−α, where α and β are tension parameters. The 6PBASS scheme possesses some of the following properties:(i)
The scheme (5) is C2-continuous for α=1/5,  β=13/1000.(ii)
It has support width5,5.(iii)
Its approximation order is fourth.(iv)
Its fundamental solution is(6)νi=1,fori=0,0,fori≠0,and it satisfies the two-scale relation(7)νx=∑kakν2x−k,whereak is the mask of the scheme (5). Since 6PBASS is C2-continuous by [24], so its 2-scale relations νx are also C2-continuous.(v)
For the computation of the first- and second-order derivatives of (7), we adopt similar approach of [22, 23].The first two derivatives of (7) are given in the following equations:(8)ν⋅⋅0=0,ν⋅⋅±1=±308375375103575526,ν⋅±2=∓89875573155363289,ν⋅±3=±60378751035755526,ν⋅±4=±186056155363289,(9)ν⋅0=−203024100865127633,ν⋅⋅±1=125938400065127633,ν⋅⋅±2=−9651000021709211,ν⋅⋅±3=4350100065127633,ν⋅⋅±4=176550465127633.


## 3. Subdivision Collection Algorithm for Bratu’s Problem

In this section, we have constructed a subdivision collocation algorithm for the solution of (2), which is based upon the fundamental solution of the subdivision scheme and its derivatives. Convergence and error estimations results are also presented in this section.


### 3.1. Formulation of Subdivision Collection Algorithm for Bratu’s Problem

The detail of the proposed algorithm is given as follows:Let(10)Wy=∑i=−4N+4wiνy−yih,0≤y≤1,be an approximate solution of (2) and N must be greater than or equal to four, h is the step size and is defined as h=1/N, yi=ih, where i=−4 to N+4, and wi are the unknowns to be determined. From (10), we get(11)D2Wyi=1h2∑i=−4N+4wiν⋅⋅y−yih.By using (10) and (11) in (2), we get(12)1h2∑i=−4N+4wiν⋅⋅yj−yih+αexp∑i=−4N+4wiνyj−yih=0,where j=0,1,…,N, and the conditions given at the ends of the domain (3) become(13)W0=w0=0,W1=wN=0.The matrix representation of equation (12) is(14)AW+αD1=0,where(15)A=ν4⋅⋅ν3⋅⋅ν2⋅⋅ν1⋅⋅ν0⋅⋅ν−1⋅⋅ν−2⋅⋅⋯0000ν4⋅⋅ν3⋅⋅ν2⋅⋅ν1⋅⋅ν0⋅⋅ν−1⋅⋅⋯00000ν4⋅⋅ν3⋅⋅ν2⋅⋅ν1⋅⋅ν0⋅⋅⋯000⋮0000000⋯νN−3⋅⋅νN−4⋅⋅00000000⋯νN−2⋅⋅νN−3⋅⋅νN−4⋅⋅N+1×N+9,(16)W=w−4,w−3,…,wN+3,wN+4T,(17)D1=h2×expWy0,expWy1,…,expWyN−1,expWyNT.Since system (14) is underdetermined because it has fewer equations than unknowns, so it requires eight more equations to get a unique solution. Two conditions are given in (13) at the ends of the domain of (2) and the detail of the remaining six conditions is given in the next section.


### 3.2. Forced Conditions

As we require six more conditions to get a unique solution of (14), so we will construct three conditions at the left and three conditions at the right end of the domain. Since 6PBASS reproduces third degree polynomial with order of approximation four, so the order of new conditions is four and these conditions are known as forced conditions. Let w−3,w−2,w−1 and wN+1,wN+2,wN+3 represent the left end points and right end points. These left and right end points can be computed by using polynomial of degree three which interpolates the data yi,wi, for 0≤i≤3, i.e., left end conditions are obtained from(18)w−i=S−yi,i=1,2,3,where(19)Syi=∑j=144j−1j+1Wyi−j.Since by (10), Wyi=wi for i=1,2,3 and substituting yi by −yi in (19), we have(20)S−yi=∑j=144j−1j+1wj−i.So, the following conditions can be used at the left end and(21)∑j=044j−1jwj−i=0,i=3,2,1.Similarly, at the right end, we have the following conditions:(22)∑j=044j−1jwi−j=0,i=N+3,N+2,N+1.Finally, we get system ofN+9×N+9 nonlinear equations(23)JcW+αD=0,or(24)JcW=−αDw,where(25)Jc=Jc0T,AT,JcNT,where A is defined in (15) and Jc0 and JcN are constrained as follows: the matrix Jc0 is obtained from all the conditions defined at the left end of the domain, i.e., first three rows obtained from (21) and fourth row of Jc0 obtained from (13) at W0=w0=0. Similarly, the matrix JcN is obtained from all the conditions defined at the right end of the domain, i.e., first row comes from (13) at W1=wN=0 and remaining rows come from (22). Hence,(26)Jc0=01−46−4100⋯0001−46−410⋯00001−46−41⋯000001000⋯0,JcN=0⋯000100000⋯1−46−410000⋯01−46−41000⋯001−46−410.The column vectorW is defined in (16) and D is defined as(27)D=0,0,0,W0,D1T,WN,0,0,0T,where D1 is given in (17).


### 3.3. Iterative Algorithm

To find the numerical solutions of nonlinear system of equation (24), we define an iterative algorithm. The iterative algorithm includes the following steps:(i)
Formulation of initial solution: the initial approximate solutionW0 is selected to find the following system:(28)JcW0=−αℝ0,where(29)ℝ0=0,0,0,W0,δ0,δ1,…,δN,WN,0,0,0T,δi=h3Diyi,Li,D0∗,Li=W0+3ihWN−W0,D0∗=WN−W0,wherei∈0,1,…,N. The column vector ℝ0 is the linear approximation of the column vector (27).(ii)
Iterative scheme: the following iterative scheme is used to find the approximate solutionW∗,(30)JcWk+1=Dwk,k=0,1,2,….(iii)
Terminating criteria: the following condition is used to stop the iteration atk−th level, for any ɛ; let ɛ=10−4,(31)Wk+1−Wk∞≤ɛ.


### 3.4. Convergence and Error Estimation

In this section, we present results of convergence and error estimation of the proposed iterative algorithm. The convergence of the iterative algorithm is guaranteed by the following proposition.Proposition 1.
The approximate solutionWk founded by (28) and (30) linearly converges to the approximate solution W∗ of (24) with the supposition that step size and the Lipschitz constants r0,r1 are small, i.e.,(32)Jc−1∞r0h2+r1h2977310000≤1.The proof is similar to [23].The main result of error estimation is given by the following proposition.Theorem 1.
Let exact solutiongy∈C40,1 and wi be obtained by solving (24) with the fourth order boundary treatment at the end points. Then, we have(33)wyj−gyj∞=Oh4−j,j=0,1,2.


## 3.1. Formulation of Subdivision Collection Algorithm for Bratu’s Problem

The detail of the proposed algorithm is given as follows:Let(10)Wy=∑i=−4N+4wiνy−yih,0≤y≤1,be an approximate solution of (2) and N must be greater than or equal to four, h is the step size and is defined as h=1/N, yi=ih, where i=−4 to N+4, and wi are the unknowns to be determined. From (10), we get(11)D2Wyi=1h2∑i=−4N+4wiν⋅⋅y−yih.By using (10) and (11) in (2), we get(12)1h2∑i=−4N+4wiν⋅⋅yj−yih+αexp∑i=−4N+4wiνyj−yih=0,where j=0,1,…,N, and the conditions given at the ends of the domain (3) become(13)W0=w0=0,W1=wN=0.The matrix representation of equation (12) is(14)AW+αD1=0,where(15)A=ν4⋅⋅ν3⋅⋅ν2⋅⋅ν1⋅⋅ν0⋅⋅ν−1⋅⋅ν−2⋅⋅⋯0000ν4⋅⋅ν3⋅⋅ν2⋅⋅ν1⋅⋅ν0⋅⋅ν−1⋅⋅⋯00000ν4⋅⋅ν3⋅⋅ν2⋅⋅ν1⋅⋅ν0⋅⋅⋯000⋮0000000⋯νN−3⋅⋅νN−4⋅⋅00000000⋯νN−2⋅⋅νN−3⋅⋅νN−4⋅⋅N+1×N+9,(16)W=w−4,w−3,…,wN+3,wN+4T,(17)D1=h2×expWy0,expWy1,…,expWyN−1,expWyNT.Since system (14) is underdetermined because it has fewer equations than unknowns, so it requires eight more equations to get a unique solution. Two conditions are given in (13) at the ends of the domain of (2) and the detail of the remaining six conditions is given in the next section.


## 3.2. Forced Conditions

As we require six more conditions to get a unique solution of (14), so we will construct three conditions at the left and three conditions at the right end of the domain. Since 6PBASS reproduces third degree polynomial with order of approximation four, so the order of new conditions is four and these conditions are known as forced conditions. Let w−3,w−2,w−1 and wN+1,wN+2,wN+3 represent the left end points and right end points. These left and right end points can be computed by using polynomial of degree three which interpolates the data yi,wi, for 0≤i≤3, i.e., left end conditions are obtained from(18)w−i=S−yi,i=1,2,3,where(19)Syi=∑j=144j−1j+1Wyi−j.Since by (10), Wyi=wi for i=1,2,3 and substituting yi by −yi in (19), we have(20)S−yi=∑j=144j−1j+1wj−i.So, the following conditions can be used at the left end and(21)∑j=044j−1jwj−i=0,i=3,2,1.Similarly, at the right end, we have the following conditions:(22)∑j=044j−1jwi−j=0,i=N+3,N+2,N+1.Finally, we get system ofN+9×N+9 nonlinear equations(23)JcW+αD=0,or(24)JcW=−αDw,where(25)Jc=Jc0T,AT,JcNT,where A is defined in (15) and Jc0 and JcN are constrained as follows: the matrix Jc0 is obtained from all the conditions defined at the left end of the domain, i.e., first three rows obtained from (21) and fourth row of Jc0 obtained from (13) at W0=w0=0. Similarly, the matrix JcN is obtained from all the conditions defined at the right end of the domain, i.e., first row comes from (13) at W1=wN=0 and remaining rows come from (22). Hence,(26)Jc0=01−46−4100⋯0001−46−410⋯00001−46−41⋯000001000⋯0,JcN=0⋯000100000⋯1−46−410000⋯01−46−41000⋯001−46−410.The column vectorW is defined in (16) and D is defined as(27)D=0,0,0,W0,D1T,WN,0,0,0T,where D1 is given in (17).


## 3.3. Iterative Algorithm

To find the numerical solutions of nonlinear system of equation (24), we define an iterative algorithm. The iterative algorithm includes the following steps:(i)
Formulation of initial solution: the initial approximate solutionW0 is selected to find the following system:(28)JcW0=−αℝ0,where(29)ℝ0=0,0,0,W0,δ0,δ1,…,δN,WN,0,0,0T,δi=h3Diyi,Li,D0∗,Li=W0+3ihWN−W0,D0∗=WN−W0,wherei∈0,1,…,N. The column vector ℝ0 is the linear approximation of the column vector (27).(ii)
Iterative scheme: the following iterative scheme is used to find the approximate solutionW∗,(30)JcWk+1=Dwk,k=0,1,2,….(iii)
Terminating criteria: the following condition is used to stop the iteration atk−th level, for any ɛ; let ɛ=10−4,(31)Wk+1−Wk∞≤ɛ.


## 3.4. Convergence and Error Estimation

In this section, we present results of convergence and error estimation of the proposed iterative algorithm. The convergence of the iterative algorithm is guaranteed by the following proposition.Proposition 1.
The approximate solutionWk founded by (28) and (30) linearly converges to the approximate solution W∗ of (24) with the supposition that step size and the Lipschitz constants r0,r1 are small, i.e.,(32)Jc−1∞r0h2+r1h2977310000≤1.The proof is similar to [23].The main result of error estimation is given by the following proposition.Theorem 1.
Let exact solutiongy∈C40,1 and wi be obtained by solving (24) with the fourth order boundary treatment at the end points. Then, we have(33)wyj−gyj∞=Oh4−j,j=0,1,2.


## 4. Numerical Examples and Comparison

The numerical technique discussed previously is illustrated in this section by applying subdivision collection algorithm to the planar one-dimensional Bratu’s problem (2) for three distinct values of α, which guarantee the existence of two locally unique solutions. We have created comparison tables using α=1,2 and 3.51 to show the consistency of our approach in comparison to the exact solution as well as the solutions of other methods. All calculations have been performed using MATLAB .(i)
The fact regarding the solution of Bratu’s problem forα=1 is obtained after third iteration, as shown in Table 1. Comparison between the numerical results and absolute errors obtained by our subdivision collection algorithm and decomposition method [25] are presented in Tables 2 and 3, respectively. From the tabulated results, we observed that the numerical results obtained by our subdivision collection algorithm are better than the decomposition method [25].(ii)
The fact regarding the solution of Bratu’s problem forα=2 is obtained after fifth iteration, as shown in Table 4. Comparison between the numerical results and absolute errors obtained by our subdivision collection algorithm, decomposition [25], and Laplace method [5] are presented in Tables 5 and 6, respectively. From the tabulated results, we observed that the numerical results obtained by our subdivision collection algorithm are better than [5, 25].(iii)
The fact regarding the solution of Bratu’s problem forα=3.15 is obtained after forty-two iterations, as shown in Table 7. Comparisons between the absolute errors obtained by our subdivision collection algorithm and B-spline [6] method are presented in Table 8. From the tabulated results, we observed that the numerical results obtained by our subdivision collection algorithm give better approximation than [6].Table 1
Numerical results of (2) by subdivision collection algorithm for case α=1.
xExact solutionBy subdivision collection algorithmAbsolute error0.00.00000000000.0000000000000000.000000000000000.10.04984679000.0498359426666641.08473333363529e − 50.20.08918993500.0891570369807903.28980192095263e − 50.30.11760909560.1175603799255584.87156744417711e − 50.40.13479025260.1347317836661515.84689338488098e − 50.50.14053921420.1404773414427476.18727572535005e − 50.60.13479025260.1347317836661515.84689338488931e − 50.70.11760909560.1175603799255584.87156744419098e − 50.80.08918993500.0891570369807903.28980192097483e − 50.90.04984679000.0498359426666631.08473333366513e − 51.00.00000000000.0000000000000000.00000000000085e − 5Table 2
Comparison between the numerical results of (2) for α=1.
xExact solutionBy [25]By subdivision collection algorithm0.00.00000000000.00000000000.0000000000000000.10.04984679000.04716168750.0498359426666640.20.08918993500.08716800000.0891570369807900.30.11760909560.11776143750.1175603799255580.40.13479025260.13699200000.1347317836661510.50.14053921420.14355468750.1404773414427470.60.13479025260.13699200000.1347317836661510.70.11760909560.11776143750.1175603799255580.80.08918993500.08716800000.0891570369807900.90.04984679000.04716168750.0498359426666631.00.00000000000.00000000000.000000000000000Table 3
Comparison between the absolute errors of (2) for case α=1.
xBy [25]By subdivision collection algorithm0.00.0000000000000000000.000000000000000.12.685102500000001e − 31.08473333363529e − 50.22.021935000000003e − 33.28980192095263e − 50.31.523418999999915e − 44.87156744417711e − 50.42.201747400000009e − 35.84689338488098e − 50.53.015473299999988e − 36.18727572535005e − 50.62.201747400000009e − 35.84689338488931e − 50.71.523418999999915e − 44.87156744419098e − 50.82.021935000000003e − 33.28980192097483e − 50.92.685102500000001e − 31.08473333366513e − 51.00.0000000000000000000.00000000000085e − 5Table 4
Numerical results of (2) for case α=2.
xExact solutionBy subdivision collection algorithmAbsolute error0.00.00000000000.0000000000000000.000000000000000.10.11441074400.1145887729998541.78028999853849e − 40.20.20641911560.2067172981795222.98182579522244e − 40.30.27387931160.2742802059628764.00894362875881e − 40.40.31508936460.3155592433607574.69878760757103e − 40.50.32895242140.3294463009014184.93879501417605e − 40.60.31508936460.3155592433607574.69878760757159e − 40.70.27387931160.2742802059628764.00894362875992e − 40.80.20641911560.2067172981795222.98182579522410e − 40.90.11441074400.1145887729998541.78028999854057e − 41.00.00000000000.0000000000000000.00000000000017e − 4Table 5
Comparison between numerical results of (2) for case α=2.
xExact solutionBy subdivision collection algorithmBy [25]By [5]0.00.00000000000.0000000000000000.00000000000.00000000000.10.11441074400.1142484492728340.09919350000.11228171410.20.20641911560.2060650664116530.19174400000.20220941620.30.27387931160.2733754076426150.26799150000.26769250580.40.31508936460.3144894105247460.31833600000.30708745060.50.32895242140.3283190285579750.33593750000.31935322940.60.31508936460.3144894105247460.31833600000.30415984030.70.27387931160.2733754076426150.26799150000.26194589090.80.20641911560.2060650664116530.19174400000.19404130720.90.11441074400.1142484492728350.09919350000.10353737851.00.00000000000.0000000000000000.00000000000.0000000000Table 6
Comparison between absolute errors of (2) for case α=2.
xBy subdivision collection algorithmBy [25]By [5]0.00.00000000000200e − 40.0000000000000000000.0000000000000000000.11.78028999853849e − 41.521724399999999e − 22.129029899999996e − 30.22.98182579522244e − 41.467511560000001e − 24.209699400000017e − 30.34.00894362875881e − 45.887811600000015e − 36.186805800000028e − 30.44.69878760757103e − 43.246635400000031e − 38.001913999999999e − 30.54.93879501417605e − 46.985078600000028e − 39.599191999999979e − 30.64.69878760757159e − 43.246635400000031e − 31.092952429999999e − 20.74.00894362875992e − 45.887811600000015e − 31.193342070000003e − 20.82.98182579522410e − 41.467511560000001e − 21.237780840000000e − 20.91.78028999854057e − 41.521724399999999e − 21.087336650000000e − 21.00.00000000000017e − 40.0000000000000000000.000000000000000000Table 7
Numerical results of (2) for case α=3.51.
xExact solutionBy subdivision collection algorithmAbsolute error0.00.00000000000.0000000000000020.0000000000000020.10.39580569900.3958385084259700.0000328094259700.20.73909741000.7396538677103300.0005564577103300.31.00875826001.0096428351471830.0008845751471830.41.18253666001.1835951913715330.0010585313715330.51.24274269001.2438554616307910.0011127716307910.61.18253666001.1835951913715320.0010585313715320.71.00875826001.0096428351471820.0008845751471820.80.73909741000.7396538677103290.0005564577103290.90.39580569900.3958385084259680.0000328094259681.00.00000000000.0000000000000000.000000000000000Table 8
Comparison between absolute errors of (2) for case α=3.51.
xBy [6]By subdivision collection algorithm0.00.000000000000000.000000000000020.13.84172369550e − 23.280942597000e − 50.27.48135367780e − 25.564577103300e − 40.31.05827422823e − 18.845751471830e − 40.41.27116880861e − 11.058531371533e − 30.51.34752877607e − 11.112771630791e − 30.61.27116880861e − 11.058531371532e − 30.71.05827422823e − 18.845751471820e − 40.87.48135367780e − 25.564577103290e − 40.93.84172369550e − 23.280942596800e − 51.00.000000000000000.000000000000


## 5. Concluding Remarks

In this paper, we have established a subdivision collocation algorithm for the solution of one-dimensional nonlinear Bratu’s problem. The numerical results obtained by subdivision collection algorithm showed that the algorithm is suitable for the approximate solution of (2). We have concluded that the numerical results converge to the exact solution for the small step size. We have also presented a comparison of absolute errors of the solution obtained from subdivision collection algorithm with decomposition method [25], Laplace method, [5] and B-spline method [6] for different values of α. We conclude that our algorithm gives smaller absolute errors as compared with the other existing methods [5, 6, 25].



---

*Source: 3497017-2021-08-03.xml*