Combining the Shooting Method with an Operational Matrix Method to Solve Two Point Boundary Value Problems

Abstract

An operational matrix method is a well known method to solve an initial value problem (IVP). A recent contribution to the operational matrix method is frame operational matrix (FOM) method. The shooting method is a standard method which converts a boundary value problem (BVP) into an IVP. The BVP is then solved by solving the IVP. In this paper, we combine the shooting method with that of the FOM method to solve two point BVPs. In short, this method is described as the following. First, we convert the BVP into an IVP by shooting method. The IVP is then solved by the FOM method. The approximate solution thus obtained is checked whether it satisfy both the boundary condition. If not, then we take better approximations until they satisfy both the boundary condition. To get a better approximation, we use the Newton’s iteration method.

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Acknowledgements

We would like to express our great appreciation to the reviewers for their valuable and constructive suggestions which improved the quality of the article.

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Correspondence to Kshama Sagar Sahu.

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KSS and MKJ formulated this research to solve boundary value problem. The major part of this research is carried out by KSS.

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Appendix

Appendix

Frame Matrix for \(J=0\left( M=3\right) \)

$$\begin{aligned} F_0= & {} \left( \begin{array}{ccc} \frac{2}{3} &{} 2 &{} \frac{2}{3} \\ \\ -\frac{1121}{1189} &{} 0 &{} \frac{1121}{1189} \\ \\ -\frac{2}{3} &{} 2 &{} -\frac{2}{3} \end{array}\right) .\\ F_1= & {} \left( \begin{array}{ccc} \frac{1}{18} &{} \frac{1}{2} &{} \frac{17}{18} \\ \\ -\frac{419}{5333} &{} -\frac{1189}{3363} &{} -\frac{419}{5333} \\ \\ -\frac{1}{18} &{} 0 &{} \frac{1}{18} \end{array}\right) . \end{aligned}$$

Operational Matrix for \(J=0\left( M=3\right) \)

$$\begin{aligned} P_1= & {} \left( \begin{array}{ccc} \frac{1}{2} &{} \frac{1121}{2378} &{} -\frac{1}{4} \\ \\ -\frac{1138}{7725} &{} 0 &{} -\frac{1155}{39202} \\ \\ 0 &{} \frac{1155}{19601} &{} 0 \end{array}\right) . \end{aligned}$$

Frame Matrix for \(J=1\left( M=7\right) \)

$$\begin{aligned} F_0= & {} \left( \begin{array}{ccccccc} \frac{2}{7} &{} \frac{6}{7} &{} \frac{10}{7} &{} 2 &{} \frac{10}{7}&{} \frac{6}{7} &{} \frac{2}{7}\\ \\ -\frac{398}{985} &{} -\frac{1194}{985} &{} -\frac{796}{985} &{} 0 &{} \frac{796}{985} &{} \frac{1194}{985} &{} \frac{398}{598} \\ \\ -\frac{796}{985} &{} -\frac{398}{985} &{} \frac{1194}{985} &{} 0 &{} 0 &{} 0 &{} 0 \\ \\ 0 &{} 0 &{} 0 &{} 0 &{} -\frac{1194}{985} &{} \frac{398}{985} &{} \frac{796}{985} \\ \\ -\frac{2}{7} &{} -\frac{6}{7} &{} \frac{2}{7} &{} 2 &{} \frac{2}{7} &{} -\frac{6}{7} &{} -\frac{2}{7} \\ \\ -\frac{4}{7} &{} \frac{8}{7} &{} -\frac{4}{7} &{} 0 &{} 0 &{} 0 &{} 0 \\ \\ 0 &{} 0 &{} 0 &{} 0 &{} -\frac{4}{7} &{} \frac{8}{7} &{} -\frac{4}{7} \end{array}\right) .\\ F_1= & {} \left( \begin{array}{ccccccc} \frac{1}{98} &{} \frac{9}{98} &{} \frac{25}{98} &{} \frac{1}{2} &{} \frac{73}{98} &{} \frac{89}{98} &{} \frac{97}{98} \\ \\ -\frac{226}{15661}&{} -\frac{253}{1948} &{}-\frac{447}{1511} &{} -\frac{1189}{3363}&{} -\frac{447}{1511}&{} -\frac{253}{1948}&{}-\frac{226}{15661} \\ \\ -\frac{253}{4383} &{} -\frac{371}{1094} &{} -\frac{393}{1757} &{} 0 &{} 0 &{} 0 &{} 0 \\ \\ 0 &{} 0 &{} 0 &{} 0 &{} -\frac{393}{1757} &{} -\frac{371}{1094} &{} -\frac{253}{4383} \\ \\ -\frac{1}{98} &{} -\frac{9}{98} &{} -\frac{8}{49} &{} 0 &{} \frac{8}{49}&{} \frac{9}{98} &{}\frac{1}{98} \\ \\ -\frac{2}{49} &{} -\frac{11}{98}&{} \frac{15}{98}&{} 0 &{} 0 &{} 0 &{} 0 \\ \\ 0 &{} 0 &{} 0 &{} 0 &{} -\frac{15}{98} &{} \frac{11}{98}&{} \frac{2}{49} \end{array}\right) . \end{aligned}$$

Operational Matrix for \(J=1\left( M=7\right) \)

$$\begin{aligned} P_1= & {} \left( \begin{array}{ccccccc} \frac{1}{2} &{} \frac{1121}{2378} &{} \frac{66}{19601} &{} \frac{65}{152} &{} -\frac{1}{4} &{} \frac{2}{105} &{} -\frac{44}{105}\\ \\ -\frac{580}{3361} &{} 0 &{} -\frac{17}{420} &{} \frac{17}{420} &{} -\frac{165}{39202} &{} -\frac{33}{19601} &{} -\frac{33}{19601}\\ \\ -\frac{336}{3713} &{} \frac{43}{336} &{} -\frac{13}{420} &{} 0 &{} \frac{336}{3713} &{} -\frac{83}{2293} &{} 0 \\ \\ -\frac{336}{3713} &{} -\frac{43}{336} &{} 0 &{} \frac{13}{420} &{} \frac{336}{3713} &{} 0 &{} -\frac{83}{2293}\\ \\ 0 &{} \frac{529}{4834} &{} -\frac{130}{2413} &{} -\frac{130}{2413} &{} 0 &{} \frac{1}{60} &{} -\frac{1}{60}\\ \\ 0 &{} 0 &{} \frac{195}{2032} &{} 0 &{} 0 &{} -\frac{9}{140} &{} 0\\ \\ 0 &{} 0 &{} 0 &{} \frac{195}{2032} &{} 0 &{} 0 &{} \frac{9}{40} \end{array} \right) . \end{aligned}$$

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Sahu, K.S., Jena, M.K. Combining the Shooting Method with an Operational Matrix Method to Solve Two Point Boundary Value Problems. Int. J. Appl. Comput. Math 7, 29 (2021). https://doi.org/10.1007/s40819-021-00967-x

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Keywords

  • Boundary value problems
  • Initial value problems
  • Frame
  • Operational matrix
  • Shooting method

Mathematics Subject Classification

  • 65L05
  • 65L10