 Solution

ID: past-exam-of-the-approximation-theory-and-finite-element-methods-course-of-the-university-of-oxford/2007/b/6/iii/solution

Write

u_{h} = U_{B} ϕ_{B}

. On

A B D, BCE, B D E

, the gradients are

(0, - 1)

(- 1, 0)

and

(- 1, - 1)

, respectively. Thus

\int_{Ω} ∣\nabla ϕ_{B} ∣^{2} = \frac{1}{2} + \frac{1}{2} + 1 = 2.

The integral of an affine nodal function over a triangle equals its area times the mean of its three vertex values. Each supported triangle therefore contributes

1/6

\int ϕ_{B}

, giving total

1/2

. The reduced system is

[2] [U_{B}] = [1/2]

, and the complete nodal answer is

U_{B} = \frac{1}{4}, U_{A} = U_{C} = U_{D} = U_{E} = U_{F} = 0.

Only one algebraic unknown remains despite the two-dimensional domain.

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