MCQuestion about numerical methods
 

Can someone help me with this multiple choice question?

Spatial discretization methods M1 (order1) and M2 (order 2) are applied on the same grid with uniform spacing h. Which is the correct statement?

1) the truncation error depends on the grid spacing h;

2) the truncation error of method M2 is always higher than M1 ;

3) the truncation error of method M1 is always higher than M2 ;

4) the truncation error does not depend on the grid spacing h, but on the order of the methods

thanks in advance

Please, do not post here homework.

I apologise, I was discussing about this question with a colleague of mine and I thought that someone could have helped us. We're not just looking for the right ans, but we're interested in the explanation. However, sorry for posted It

Study the topic, this is well explained in a lot of textbook. You can propose a question on fundamental CFD topics but here you have not tried to give at least your opinion to discuss.

Sure, I'll give you my point of view. Truncation error is usually proportional to a power of the grid spacing \delta{h}^{n_th}. Thus, It is function of the grid spacing and function of nth order approximation of the method used. If on the same grid, I discretize with a 2nd order scheme, I expect a lower truncation error than the one I have with a 1st order scheme.

However, in the MCQ I reported, just one answer is correct... maybe there is some special case for which 1st order leads to a lower truncation error but I'm not sure

The truncation error depends on the order of the methods.

In your case, since both grids have uniform spacing, the 2nd order FVM code will be more accurate.

I understand why you may be confused in this matter, as books mostly say 2nd order methods are better, and give a mathematical reasoning to why that is, but they do not give a physical intuition to a CFD engineer on why higher order methods are better.

In short: With higher order methods, you can get accurate solutions, with coarse meshes.

Or its corollary: With lower order methods, you need extremely dense meshes to get accurate solutions.

This is mathematically verified by plotting the error graph for 1st order, 2nd order, and 3rd order solvers.

As can be seen, the 3rd order solver's error plot has a higher steep, meaning, it is more accurate than 2nd order solvers on the same grid.

For more understanding you would need to study more about this.

The local truncation error is a sum of infinite terms, derivatives multiplied by the step sizes.
For a fixed grid size you know only how the lte has some error magnitudine but you cannot say that for that grid size the error is lower for the higher order discretization. The errror decreases asimptotically with h^p.
Have a look to the book of Ferziger, Peric, Street as example of such cases.