 Solution

ID: past-exam-of-the-approximation-theory-and-finite-element-methods-course-of-the-university-of-oxford/2008/a/3/c/solution

The midpoint error annihilates affine functions, so here

n = 1

. Apply the functional to

(x - t)_{+}

K (t) = \frac{1}{2} (h - t)^{2} - h (h /2 - t)_{+} .

Thus

K (t) = {t^{2} /2, (h - t)^{2} /2, 0 \leq t \leq h /2, h /2 \leq t \leq h .

It is nonnegative, and symmetry gives

\int_{0}^{h} K = 2 \int_{0}^{h /2} t^{2} /2 d t = h^{3} /24

. Therefore

∣ X [u] ∣ \leq \frac{h ^{3}}{24} ∥ u^{''} ∥_{\infty} .

For a quadratic with constant nonzero second derivative equality holds, so this midpoint quadrature rule error constant is sharp.

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