Research Article

Block-by-Block Method for Solving Nonlinear Volterra-Fredholm Integral Equation

Table 1


𝑁 𝑥 = 2 0 , 𝑁 𝑡 = 8 , case 1 𝑁 𝑥 = 2 0 , 𝑁 𝑡 = 8 , case 2

𝑥 𝑡 Error 𝑥 𝑡 Error

0.0 0.03125 5 . 6 9 6 8 𝑒 9 0.00.03125 1 . 1 8 7 9 1 1 4 4 8 𝑒 1 3
0.0 0.0625 8 . 2 9 5 6 𝑒 8 0.00.0625 1 . 5 2 0 5 2 6 6 5 3 𝑒 1 1
0.0 0.09375 1 . 1 5 3 4 6 8 𝑒 6 0.00.09375 2 . 5 9 7 9 6 2 3 3 6 𝑒 1 0
0.0 0.125 2 . 4 5 6 2 6 𝑒 6 0.00.125 1 . 9 4 6 2 7 4 1 1 6 𝑒 9
0.0 0.15625 1 . 1 5 3 9 6 8 𝑒 6 0.00.15625 9 . 2 8 0 5 5 8 1 8 3 𝑒 9
0.0 0.1875 1 . 6 7 6 5 6 8 𝑒 5 0.00.1875 3 . 3 2 5 3 9 1 7 9 1 𝑒 8
0.0 0.21875 4 . 9 7 4 7 𝑒 5 0.00.21875 9 . 7 8 2 9 6 1 5 7 4 𝑒 8
0.0 0.25 6 . 2 2 5 6 9 7 𝑒 5 0.00.25 2 . 4 9 1 2 3 0 8 6 9 𝑒 7
0.0 0.28125 1 . 4 4 4 7 9 9 6 2 𝑒 4 0.0 0.28125 5 . 6 8 1 7 4 3 6 2 7 𝑒 7
0.0 0.3125 1 . 6 7 6 5 3 9 4 𝑒 4 0.00.3125 1 . 1 8 7 9 1 1 4 4 7 𝑒 6
0.0 0.34375 3 . 3 5 8 2 9 5 𝑒 4 0.00.34375 2 . 3 1 4 9 0 3 3 4 2 𝑒 6
0.0 0.375 3 . 7 0 8 3 2 9 𝑒 4 0.00.375 4 . 2 5 6 5 0 1 4 6 4 𝑒 6
0.0 0.40625 6 . 7 1 8 0 1 5 𝑒 4 0.00.40625 7 . 4 5 3 9 6 8 0 1 3 𝑒 6
0.0 0.4375 7 . 1 8 5 5 6 0 𝑒 3 0.00.4375 1 . 2 5 2 2 1 9 0 4 5 𝑒 5
0.0 0.46875 1 . 2 1 1 2 9 0 1 𝑒 3 0.00.46875 2 . 0 2 9 6 5 7 9 0 8 𝑒 5