Question

According to the Adams-Moulton scheme, the derivative of a function T at time-step t is given by _________

(a) \(\frac{3 T(t)+4T(t-\Delta t)-T(t-2\Delta t)}{2\Delta t}\)

(b) \(\frac{3 T(t)-4T(t-\Delta t)-T(t-2\Delta t)}{2\Delta t}\)

(c) \(\frac{3 T(t)+4T(t-\Delta t)+T(t-2\Delta t)}{2\Delta t}\)

(d) \(\frac{3 T(t)-4T(t-\Delta t)+T(t-2\Delta t)}{2\Delta t}\)

I got this question during an interview.

Origin of the question is Transient Flows topic in section Transient Flows of Computational Fluid Dynamics

Answer 1

Correct answer is (d) \(\frac{3 T(t)-4T(t-\Delta t)+T(t-2\Delta t)}{2\Delta t}\)

Easiest explanation: To find the derivative, the Adams-Moulton method uses the previous and the second previous steps. The mathematical expression is

\(\frac{\partial T(t)}{\partial t}=\frac{3 T(t)-4T(t-\Delta t)+T(t-2\Delta t)}{2\Delta t}\).