 Weak boundary value problem with an inverse-square potential

ID: weak-boundary-value-problem-with-an-inverse-square-potential

H = {v \in H^{1} (0, 1) : v (0) = 0}

, the form

a (u, v) = \int_{0}^{1} u^{'} v^{'} + \int_{0}^{1} \frac{uv}{x ^{2}} - \int_{0}^{1} u^{'} v

is bounded and coercive. The Hardy inequality on an interval controls its singular term, while the one-sided Poincare inequality controls the first-order term. Thus the Lax-Milgram theorem gives a unique weak solution for every functional

v \mapsto \int_{0}^{1} (f / x) (v / x)

with

f / x \in L^{2} (0, 1)

 New to topics? Read the docs here!