Table 2 Example 4.2, 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 |
|---|---|---|---|---|---|
\(2^{-0}\) | 2.4094e−04 | 4.5191e−05 | 5.3745e−06 | 4.6776e−06 | 3.4511e−06 |
\(2^{-2}\) | 5.8848e−04 | 1.7058e−04 | 1.1039e−04 | 6.2476e−05 | 3.3191e−05 |
\(2^{-4}\) | 2.1117e−03 | 7.2963e−04 | 3.0344e−04 | 1.3611e−04 | 6.4377e−05 |
\(2^{-6}\) | 4.6754e−03 | 2.2578e−03 | 8.2105e−04 | 2.8430e−04 | 1.0734e−04 |
\(2^{-8}\) | 4.8645e−03 | 2.9411e−03 | 1.5511e−03 | 6.7433e−04 | 2.3656e−04 |
\(2^{-10}\) | 4.8682e−03 | 2.9448e−03 | 1.6029e−03 | 8.4404e−04 | 4.2558e−04 |
\(2^{-12}\) | 4.8691e−03 | 2.9452e−03 | 1.6032e−03 | 8.4463e−04 | 4.3880e−04 |
\(2^{-14}\) | 4.8693e−03 | 2.9453e−03 | 1.6032e−03 | 8.4466e−04 | 4.3881e−04 |
\(2^{-16}\) | 4.8693e−03 | 2.9454e−03 | 1.6032e−03 | 8.4467e−04 | 4.3881e−04 |
\(2^{-18}\) | 4.8694e−03 | 2.9454e−03 | 1.6032e−03 | 8.4467e−04 | 4.3881e−04 |
\(2^{-20}\) | 4.8694e−03 | 2.9454e−03 | 1.6032e−03 | 8.4467e−04 | 4.3881e−04 |
\(E^{N,M}\) | 4.8694e−03 | 2.9454e−03 | 1.6032e−03 | 8.4467e−04 | 4.3881e−04 |
\(r^{N,M}\) | 0.7253 | 0.8775 | 0.9245 | 0.9448 | – |