What is the effect on the local truncation error when step size is reduce by half for the midpoint method?

Question

BlackTom AI · Accepted Answer

Midpoint method is a second-order Runge–Kutta method (order p = 2). For a method of order p, the local truncation error (LTE) scales like h^{p+1}. Therefore for the midpoint method LTE ∝ h^{3}.

1) Relevant knowledge points:
- Order p of a method gives global error O(h^{p}) and local truncation error O(h^{p+1}).
- Midpoint method is order p = 2, so LTE = C·h^{3} for some constant C depending on the problem.

2) Step-by-step reasoning:
- Let the original step size be h and the original LTE be LTE_old = C·h^{3}.
- When the step size is halved to h_new = h/2, plug into the LTE scaling.

3) Calculations (all intermediate steps):
- LTE_new = C·(h_new)^{3} = C·(h/2)^{3}
- = C·h^{3} · (1/2)^{3}
- = C·h^{3} · 1/8
- = (1/8)·(C·h^{3}) = (1/8)·LTE_old

4) Why the correct answer is correct:
- Because LTE scales with the cube of h for the midpoint method, halving h multiplies the LTE by (1/2)^{3} = 1/8. So the local truncation error is reduced to one-eighth of its previous value.

5) Why the other options are incorrect:
- (a) Error reduces to 1/2: would be true only if LTE scaled linearly with h (O(h)), which corresponds to a first-order local behavior, not the midpoint method.
- (b) Error reduces to 1/4: this factor 1/4 = (1/2)^{2} is what the global error would do for a second-order method (global O(h^{2})), not the local truncation error. That confuses local vs global error.
- (d) No changes: incorrect because error depends on h and changes when h changes.

6) Conclusion:
- Halving the step size for the midpoint method reduces the local truncation error by a factor of 1/8. This matches the provided correct answer: c. Error reduces to 1/8 error.

What is the effect on the local truncation error when step size is reduce by half for the midpoint method?单项选择题

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