Variational inequalities and obstacle problems

On the unit interval let the obstacle $\psi\in H^1(0,1)$ have nonpositive endpoint traces, and let the admissible set be functions in $H_0^1$ lying above it almost everywhere. The positive part of the obstacle is an admissible function, so the set is nonempty. In one dimension continuous representatives make the contact geometry meaningful. The set is norm closed and convex, hence weakly closed. Let the load be square integrable and use derivative energy minus the load pairing.

For any feasible competitor, the segment from a minimizer to that competitor stays feasible by convexity. Differentiate only in the forward segment direction. A minimum at the starting point gives a nonnegative right derivative rather than a zero two-sided derivative. This yields the variational inequality. Conversely, expand the quadratic energy around any candidate satisfying that inequality: the linear term is nonnegative and the remaining squared derivative term is nonnegative, proving global minimality and uniqueness.

The direct method applies: the obstacle set is weakly closed, the quadratic energy is coercive by Poincaré and Cauchy–Schwarz, and its derivative norm term is weakly lower semicontinuous. Strict convexity on zero-trace functions gives uniqueness. The solution is initially a Sobolev function. Statements about classical second derivatives or smooth free boundaries require additional regularity assumptions and cannot be concluded from existence alone. Even smooth data can produce a solution whose second derivative jumps between contact and noncontact regions.

For a sufficiently regular solution, define the reaction $\lambda=-u''-f$. Feasible nonnegative variations show this is nonnegative in the distributional sense. Where the solution lies strictly above a continuous obstacle, compactly supported variations of either sign are permitted, forcing the reaction to vanish and restoring $-u''=f$. Under enough regularity to multiply and interpret the terms, the contact condition is $\lambda(u-\psi)=0$. In the general weak setting, formulate reaction and complementarity using dual pairings rather than assuming a pointwise multiplier function.

An explicit candidate needs more than solving the noncontact differential equation. Check its endpoint traces, its position relative to the obstacle, and continuity of its derivative at a proposed contact boundary. A slope jump creates a distributional point reaction whose sign must be included; it cannot be ignored by checking second derivatives only away from the join. A continuously matched slope avoids such hidden delta contributions in the example below and makes the piecewise reaction a genuine bounded function.

Once a candidate has a nonnegative reaction supported where it touches the obstacle, integrate by parts against a feasible difference. On contact, the competitor cannot lie below the candidate, so the reaction pairing is nonnegative; elsewhere the reaction is zero. This proves the full variational inequality and therefore global minimality. The method is a certificate over all feasible competitors, not merely a plot of two curves or an assertion that the contact region looks plausible. Higher-dimensional regularity and free-boundary geometry are deeper questions beyond this one-dimensional construction.