 Solution

ID: past-exam-of-the-elasticity-and-plasticity-course-of-the-university-of-oxford/2008/4/c/solution

The small-slope excess length is

\int (w^{'})^{2} /2

, so the first term is the work needed against uniform tension. The second is the potential energy of the downward distributed load. Their sum is total potential energy subject to the obstacle constraint.

Expand the difference exactly:

U [v] - U [w] = T \int w^{'} (v^{'} - w^{'}) d x + \int p (v - w) d x + \frac{T}{2} \int (v^{'} - w^{'})^{2} d x .

The first two integrals together are nonnegative by part (b), and the square is nonnegative. Thus

U [w] \leq U [v] for every admissible v .

For

T > 0

this also proves uniqueness: equality for two minimizers forces identical derivatives, and the fixed endpoints remove their additive constant. The convex admissible set and strictly convex energy make the variational inequality a global minimization condition.

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