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Table 3 \(H=\frac{1}{24},\ h=\tau \)

From: Parallel algorithm for convection–diffusion system based on least-squares procedure

h

m

\(a=1\)

\(a=1\) e–2

\(a=1e{-}4\)

\(\Vert \cdot \Vert _2\)

\(\Vert \cdot \Vert _\infty \)

\(\Vert \cdot \Vert _2\)

\(\Vert \cdot \Vert _\infty \)

\(\Vert \cdot \Vert _2\)

\(\Vert \cdot \Vert _\infty \)

\(\frac{1}{48}\)

\(*\)

\(2.9866\) e–2

\(1.9250\) e–1

\(8.6118e{-}3\)

\(2.6493\) e–2

\(7.4344e{-}3\)

\(1.5421\) e–2

\(\frac{1}{48}\)

1

\( 2.0573\) e–1

\(4.6896\) e–1

\(9.1964e{-}3\)

\( 2.8774\) e–2

\(8.0709e{-}3\)

\(2.1458\) e–2

\(\frac{1}{48}\)

2

\(2.0179 \) e–1

\(4.5896\) e–1

\(8.9982e{-}3 \)

\(2.1092 \) e–2

\(7.6579e{-}3\)

\(1.5340\) e–2

\(\frac{1}{48}\)

3

\(2.0061 \) e–1

\(4.5878\) e–1

\(9.0836e{-}3 \)

\(2.2172 \) e–2

\(7.6797e{-}3\)

\(1.5379\) e–2

\(\frac{1}{48}\)

4

\(1.9834 \) e–1

\(4.5297\) e–1

\(9.1164e{-}3 \)

\(2.2495 \) e–2

\(7.6884e{-}3\)

\(1.5387\) e–2

\(\frac{1}{96}\)

\(*\)

\(1.4800e{-}2\)

\(9.5488e{-}2\)

\( 4.9541e{-}3\)

\(2.5220e{-}2\)

\(3.6348e{-}3\)

\(7.4154e{-}3\)

\(\frac{1}{96}\)

1

\(1.1518\) e–1

\(2.8223\) e–1

\(4.8606e{-}3\)

\(2.3682e{-}2\)

\(3.6679e{-}3\)

\(7.6455e{-}3\)

\(\frac{1}{96}\)

2

\(1.1342\) e–1

\(2.7485\) e–1

\(4.8758e{-}3\)

\(2.3682e{-}2\)

\(3.6564e{-}3\)

\(7.4153e{-}3\)

\(\frac{1}{96}\)

3

\(1.1266\) e–1

\(2.7333\)

\(4.8861e{-}3\)

\(2.3682e{-}2\)

\(3.6596e{-}3\)

\(7.4153e{-}3\)

\(\frac{1}{96}\)

4

\(1.1186\) e–1

\(2.6995\) e–1

\(4.8874e{-}3\)

\(2.3682e{-}2\)

\(3.6598e{-}3\)

\(7.4153e{-}3\)

\(\frac{1}{192}\)

\(*\)

\(7.5407e{-}3\)

\(4.7396e{-}2\)

\(2.9959e{-}3\)

\(2.0611e{-}2\)

\(1.8231e{-}3\)

\(4.2963e{-}3\)

\(\frac{1}{192}\)

1

\(5.5060e{-}2\)

\(1.4802\) e–1

\(2.9586e{-}3 \)

\(2.0194e{-}2\)

\(1.8251e{-}3\)

\(4.2712e{-}3\)

\(\frac{1}{192}\)

2

\(5.4280e{-}2\)

\(1.4441\) e–1

\(2.9635e{-}3\)

\(2.0194e{-}2\)

\(1.8259e{-}3\)

\(4.2712e{-}3 \)

\(\frac{1}{192}\)

3

\(5.3930e{-}2\)

\(1.4334\) e–1

\(2.9641e{-}3\)

\(2.0194e{-}2\)

\(1.8261e{-}3\)

\(4.2712e{-}3 \)

\(\frac{1}{192}\)

4

\(5.3683e{-}2\)

\(1.4217\) e–1

\(2.9641e{-}3\)

\(2.0194e{-}2\)

\(1.8261e{-}3\)

\(4.2712e{-}3\)

  1. * The numerical results by least-squares algorithm