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# Numerical solutions of Volterra integro-differential equations using General Linear Method

The reviewing process of the paper is handled by Gafurjan Ibragimov, Siti Hasana Sapar and Siti Nur Iqmal Ibrahim

• In this paper, a third order General Linear Method for finding the numerical solution of Volterra integro-differential equation is considered. The order conditions of the proposed method are derived based on techniques of B-series and 'rooted trees'. The integral operator in Volterra integro-differential equation approximated using Simpson's rule and Lagrange interpolation is discussed. To illustrate the efficiency of third order General Linear Method, we compare the method with a third order Runge-Kutta method.

Mathematics Subject Classification: Primary: 58F15, 58F17; Secondary: 53C35.

 Citation: • • Figure 1.  Log maximum error versus number of functions evaluations for Problem 1

Figure 2.  Log maximum error versus number of functions evaluations for Problem 2

Figure 3.  Log maximum error versus number of functions evaluations for Problem 3

Figure 4.  Log maximum error versus number of functions evaluations for Problem 4

Figure 5.  Log maximum error versus number of functions evaluations for Problem 5

Table 1.  Matrix representation of coefficients of GLM.

 $A_{s\times s}$ $U_{s\times r}$ $B_{r\times s}$ $V_{r\times r}$

Table 2.  Matrix coefficients of GLM with $s = 3$, $r = 2$.

 $\left[ {\begin{array}{*{20}{l}} 0&0&0\\ {{a_{21}}}&0&0\\ {{a_{31}}}&{{a_{32}}}&0 \end{array}} \right]$ $\left[ {\begin{array}{*{20}{l}} {{u_{11}}}&{{u_{12}}}\\ {{u_{21}}}&{{u_{22}}}\\ {{u_{31}}}&{{u_{32}}} \end{array}} \right]$ $\left[ {\begin{array}{*{20}{l}} {{b_{11}}}&{{b_{12}}}&{{b_{13}}}\\ {{b_{21}}}&{{b_{22}}}&{{b_{23}}} \end{array}} \right]$ $\left[ {\begin{array}{*{20}{c}} 1&{{v_{12}}}\\ {{v_{21}}}&0 \end{array}} \right]$

Table 3.  Order conditions of GLM up to order three.

 No Order conditions 1 $b_{11}(u_{11}+u_{12})+b_{12}(u_{21}+u_{22})+b_{13}(u_{31}+u_{32})-v_{12}=1$ 2 $b_{21}(u_{11}+u_{12})+b_{22}(u_{21}+u_{22})+b_{23}(u_{31}+u_{32})=0$ 3 $-b_{11}u_{12}+b_{12}(a_{21}(u_{11} + u_{12})-u_{22})+b_{13}\big(a_{31}(u_{11} + u_{12})+a_{32}(u_{21} + u_{22})-u_{32}\big)$ $\qquad+v_{12}\xi_{22}=\frac{1}{2}$ 4 $-b_{21}u_{12}+b_{22}(a_{21}(u_{11} + u_{12})-u_{22})+b_{23}\big(a_{31}(u_{11} + u_{12})+a_{32}(u_{21} + u_{22})-u_{32}\big)=0$ 5 $b_{11}u_{12}^{2}+b_{12}\big(a_{21}(u_{11}+u_{12})-u_{22}\big)^{2}+b_{13}\big(a_{31}(u_{11} + u_{12})+a_{32}(u_{21} + u_{22})-u_{32}\big)^{2}$ $\qquad+v_{12}\xi_{23}=\frac{1}{3}$ 6 $b_{21}u_{12}^{2}+b_{22}\big(a_{21}(u_{11}+u_{12})-u_{22}\big)^{2}+b_{23}\big(a_{31}(u_{11} + u_{12})+a_{32}(u_{21} + u_{22})-u_{32}\big)^{2}=0$ 7 $b_{11}u_{12}\xi_{22}+b_{12}\big(\xi_{22}u_{22} - a_{21}u_{12}\big)+b_{13}\big(-a_{31}u_{12} + a_{32}(a_{21}(u_{11} + u_{12})-u_{22})$ $\qquad +u_{32}\xi_{22}\big)+v_{12}\xi_{24}=\frac{1}{6}$ 8 $b_{21}u_{12}\xi_{22}+b_{22}\big(\xi_{22}u_{22} - a_{21}u_{12}\big)+b_{23}\big(-a_{31}u_{12} + a_{32}(a_{21}(u_{11} + u_{12})-u_{22})$ $\qquad+u_{32}\xi_{22}\big)=0$ 9 $-b_{11}u_{12}^{3}+b_{12}(a_{21}(u_{11} + u_{12})-u_{22})^{3}+b_{13}\big(a_{31}(u_{11} + u_{12}) +a_{32}(u_{21} + u_{22})-u_{32}\big)^{3}$ $\qquad+v_{12}\xi_{25}\frac{1}{4}=\frac{1}{4}$ 10 $-b_{21}u_{12}^{3}+b_{22}(a_{21}(u_{11} + u_{12})-u_{22})^{3}+b_{23}\big(a_{31}(u_{11} + u_{12}) +a_{32}(u_{21} + u_{22})$ $\qquad-u_{32}\big)^{3}=0$ 11 $-b_{11}u_{12}^{2}\xi_{22} + b_{12}(a_{21}(u_{11} + u_{12})-u_{22})(\xi_{22}u_{22}-a_{21}u_{12}) +b_{13}\big(a_{31}(u_{11} + u_{12})+a_{32}$ $\qquad (u_{21}+u_{22})-u_{32}\big)\big(-a_{31}u_{12}+a_{32}(a_{21}(u_{11}+u_{12})-u_{22})+u_{32}\xi_{22}\big)+v_{12}\xi_{26} =\frac{1}{8}$ 12 $-b_{21}u_{12}^{2}\xi_{22} + b_{22}(a_{21}(u_{11} + u_{12})-u_{22})(\xi_{22}u_{22}-a_{21}u_{12}) +b_{23}\big(a_{31}(u_{11} + u_{12})$ $\qquad+a_{32}(u_{21} + u_{22})-u_{32}\big)\big(-a_{31}u_{12}+a_{32}(a_{21}(u_{11}+u_{12})-u_{22})+u_{32}\xi_{22}\big)=0$ 13 $b_{11}u_{12}\xi_{23} + b_{12}(a_{21}u_{12}^{2}+\xi_{23}u_{22}) + b_{13}\big(a_{31}u_{12}^{2} + a_{32}(a_{21}(u_{11} + u_{12})-u_{22})^{2}+u_{32}\xi_{23}\big)$ $\qquad+v_{12}\xi_{27}=\frac{1}{12}$ 14 $b_{21}u_{12}\xi_{23} + b_{22}(a_{21}u_{12}^{2}+\xi_{23}u_{22}) + b_{23}\big(a_{31}u_{12}^{2} + a_{32}(a_{21}(u_{11} + u_{12})-u_{22})^{2}$ $\qquad+u_{32}\xi_{23}\big)=0$ 15 $b_{11}u_{12}\xi_{24} + b_{12}(\xi_{22}a_{21}u_{12}+\xi_{24}u_{22}) + b_{13}\big(a_{31}u_{12}\xi_{22} + a_{32}(\xi_{22}u_{22} -a_{21}u_{12})+u_{32}\xi_{24}\big)$ $\qquad+v_{12}\xi_{28}=\frac{1}{24}$ 16 $b_{21}u_{12}\xi_{24} + b_{22}(\xi_{22}a_{21}u_{12}+\xi_{24}u_{22}) + b_{23}\big(a_{31}u_{12}\xi_{22} + a_{32}(\xi_{22}u_{22} -a_{21}u_{12})$ $\qquad+u_{32}\xi_{24}\big)=0$

Table 4.  Coefficients Set 1 of third order GLM

 $u_{11}=1$ $u_{12}=0$ $a_{21}=\frac{13}{18}$ $u_{21}=\frac{7}{9}$ $u_{22}=\frac{2}{9}$ $a_{31}=\frac{-17}{9}$ $a_{32}=2$ $u_{31}=\frac{17}{9}$ $u_{32}=\frac{-8}{9}$ $b_{11}=\frac{1}{6}$ $b_{12}=\frac{2}{3}$ $b_{13}=\frac{1}{6}$ $v_{11}=1$ $v_{12}=0$ $b_{21}=0$ $b_{22}=0$ $b_{23}=0$ $v_{21}=1$ $v_{22}=0$

Table 5.  Coefficients Set 2 of third order GLM

 $u_{11}=1$ $u_{12}=0$ $a_{21}=\frac{2}{3}$ $u_{21}=\frac{5}{6}$ $u_{22}=\frac{1}{6}$ $a_{31}=\frac{-5}{3}$ $a_{32}=2$ $u_{31}=\frac{5}{3}$ $u_{32}=\frac{-2}{3}$ $b_{11}=\frac{1}{6}$ $b_{12}=\frac{2}{3}$ $b_{13}=\frac{1}{6}$ $v_{11}=1$ $v_{12}=0$ $b_{21}=0$ $b_{22}=0$ $b_{23}=0$ $v_{21}=1$ $v_{22}=0$

Table 6.  Coefficients Set 3 of third order GLM

 $u_{11}=1$ $u_{12}=0$ $a_{21}=\frac{5}{6}$ $u_{21}=\frac{2}{3}$ $u_{22}=\frac{1}{3}$ $a_{31}=\frac{-7}{3}$ $a_{32}=2$ $u_{31}=\frac{7}{3}$ $u_{32}=\frac{-4}{3}$ $b_{11}=\frac{1}{6}$ $b_{12}=\frac{2}{3}$ $b_{13}=\frac{1}{6}$ $v_{11}=1$ $v_{12}=0$ $b_{21}=0$ $b_{22}=0$ $b_{23}=0$ $v_{21}=1$ $v_{22}=0$

Table 7.  Maximum global errors for Problem 1

 GLM, $s=3$ RK, $s=3$ Step size MAXE $h=0.1$ $1.2347\times 10^{-6}$ $4.7137\times 10^{-6}$ $h=0.025$ $9.9859\times 10^{-9}$ $7.1772\times 10^{-8}$ $h=0.01$ $5.6041\times 10^{-10}$ $4.6094\times 10^{-9}$ $h=0.005$ $6.7079\times 10^{-11}$ $5.7715\times 10^{-10}$ $h=0.001$ $5.1845\times 10^{-13}$ $4.6243\times 10^{-12}$

Table 8.  Maximum global errors for Problem 2

 GLM, $s=3$ RK, $s=3$ Step size MAXE $h=0.1$ $2.4606\times 10^{-6}$ $6.9906\times 10^{-6}$ $h=0.025$ $1.6319\times 10^{-8}$ $1.0137\times 10^{-7}$ $h=0.01$ $8.3870\times 10^{-10}$ $6.4622\times 10^{-9}$ $h=0.005$ $9.7077\times 10^{-11}$ $8.0749\times 10^{-10}$ $h=0.001$ $7.2935\times 10^{-13}$ $6.4604\times 10^{-12}$

Table 9.  Maximum global errors for Problem 3

 GLM, $s=3$ RK, $s=3$ Step size MAXE $h=0.1$ $3.9332\times 10^{-6}$ $4.8141\times 10^{-5}$ $h=0.025$ $1.4323\times 10^{-7}$ $1.4400\times 10^{-6}$ $h=0.01$ $1.0939\times 10^{-8}$ $1.0134\times 10^{-7}$ $h=0.005$ $1.4325\times 10^{-9}$ $1.3052\times 10^{-8}$ $h=0.001$ $1.1851\times 10^{-11}$ $1.0688\times 10^{-10}$

Table 10.  Maximum global errors for Problem 4

 GLM, $s=3$ RK, $s=3$ Step size MAXE $h=0.1$ $4.5416\times 10^{-6}$ $1.4256\times 10^{-5}$ $h=0.025$ $1.7061\times 10^{-8}$ $2.5908\times 10^{-7}$ $h=0.01$ $1.4270\times 10^{-9}$ $1.7075\times 10^{-8}$ $h=0.005$ $2.1002\times 10^{-10}$ $2.1553\times 10^{-9}$ $h=0.001$ $1.8836\times 10^{-12}$ $1.7377\times 10^{-11}$

Table 11.  Maximum global errors for Problem 5

 GLM, $s=3$ RK, $s=3$ Step size MAXE $h=0.1$ $6.3429\times 10^{-6}$ $3.3432\times 10^{-5}$ $h=0.025$ $4.4142\times 10^{-8}$ $6.3237\times 10^{-7}$ $h=0.01$ $4.0164\times 10^{-9}$ $4.1259\times 10^{-8}$ $h=0.005$ $5.4245\times 10^{-10}$ $5.1811\times 10^{-9}$ $h=0.001$ $4.5689\times 10^{-12}$ $4.1572\times 10^{-11}$

Table 12.  Total number of function evaluations Problems 1 - 5

 GLM, $s=3$ RK, $s=3$ Step size TFE $h=0.1$ $34$ $34$ $h=0.025$ $124$ $124$ $h=0.01$ $304$ $304$ $h=0.005$ $604$ $604$ $h=0.001$ $3004$ $3004$
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