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Table 1 Example 4.1, maximum absolute errors for the case \(\tau = 0.5\varepsilon\)

From: Fitted computational method for solving singularly perturbed small time lag problem

\(\varepsilon \downarrow\)

N = M =16

32

64

128

256

\(2^{-0}\)

1.3260e−03

7.4390e−04

3.9431e−04

2.0293e−04

1.0294e−04

\(2^{-2}\)

1.3097e−02

7.2784e−03

3.8340e−03

1.9676e−03

9.9661e−04

\(2^{-4}\)

2.7648e−02

1.5726e−02

8.3824e−03

4.3310e−03

2.2011e−03

\(2^{-6}\)

3.7674e−02

1.9780e−02

1.0209e−02

5.3769e−03

2.7671e−03

\(2^{-8}\)

3.7154e−02

2.2029e−02

1.2040e−02

5.6488e−03

2.8289e−03

\(2^{-10}\)

3.7206e−02

2.1942e−02

1.1903e−02

6.2248e−03

3.1986e−03

\(2^{-12}\)

3.7221e−02

2.1951e−02

1.1906e−02

6.1986e−03

3.1627e−03

\(2^{-14}\)

3.7224e−02

2.1953e−02

1.1907e−02

6.1992e−03

3.1627e−03

\(2^{-16}\)

3.7225e−02

2.1953e−02

1.1907e−02

6.1993e−03

3.1627e−03

\(2^{-18}\)

3.7225e−02

2.1953e−02

1.1908e−02

6.1994e−03

3.1627e−03

\(2^{-20}\)

3.7225e−02

2.1953e−02

1.1908e−02

6.1994e−03

3.1627e−03

Proposed scheme

     

\(E^{N,M}\)

3.7225e−02

2.2029e−02

1.2040e−02

6.2248e−03

3.1986e−03

\(r^{N,M}\)

0.77417

0.87156

0.95174

0.96059

–

Method in [14]

     

\(E^{N,M}\)

8.3951e−02

4.9224e−02

2.6666e−02

1.3880e−02

7.0816e−03

\(r^{N,M}\)

0.77019

0.88436

0.94199

0.97086

–