Consider the following domain, discretized by 6 nodes.

The solution at node 1 is 20. The solution at node 2 is 10. The solution at node 3 is 5.

Assuming that the distance between nodes is 1, which of the following is the value of the central finite difference approximation of the second derivative of the solution at node 2?

Question

Consider the following domain, discretized by 6 nodes.

The solution at node 1 is 20. The solution at node 2 is 10. The solution at node 3 is 5.

Assuming that the distance between nodes is 1, which of the following is the value of the central finite difference approximation of the second derivative of the solution at node 2?

BlackTom AI · Accepted Answer

To evaluate the central finite difference approximation of the second derivative at node 2, use the standard formula for a uniform grid with spacing h: f''(x2) ≈ [f(x3) − 2 f(x2) + f(x1)] / h^2.

First, identify the given values: f(x1) = 20, f(x2) = 10, f(x3) = 5, with h = 1.

Compute the numerator: f(x3) − 2 f(x2) + f(x1) = 5 − 2*(10) + 20 = 5 − 20 + 20 = 5.

Since h^2 = 1^2 = 1, divide by 1 to obtain f''(x2) ≈ 5.

Now examine the other options to see why they are not correct:
- 2.5: This would result if the combination had been misweighted, for example using a factor of 1/2 somewhere, which is not consistent with the standard second-derivative stencil on a unit grid.
- 10: This would imply either using only f(x3) or double-counting terms, not applying the −2 f(x2) term, which is essential for the central second-derivative formula.
- 15: This would occur if one erroneously added too much weight from the neighboring points, or used an incorrect sign arrangement; the correct balanced combination with h = 1 yields 5, not 15.
- 2: This value would arise if the contributions from f(x1) and f(x3) were severely discounted or if the −2 f(x2) term were replaced with − f(x2), which is not the standard stencil.

Therefore, the only value consistent with the central finite difference second derivative formula on a unit grid and the given node values is 5.

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