Posted: January 9th, 2022
For a function defined as
(1) the variation operator and integral operator are exchangeable, i.e.
(2) the variation operator and the differential operator are exchangeable, i.e.
(a) Which mesh will yield more accurate results?
(b) Which will be more computationally expensive?
(c) Suggest a way of meshing which will yield relatively accurate results and, at the same time be less computationally expensive than B?
Place an order in 3 easy steps. Takes less than 5 mins.