ABSTRACT
Proof. Assume Hj(r) = {u eHj(TD) / u = 0 q.e in Tc] and H ( r c ) = {v = v\rc/v 6Hj ( r D ) , v = g q.e in T}. We have (y — g)\r € Hj( r ) , using the weak maximum principle (y — g)+\r € Hj( r ) and IT |Vr(» - 9)+\2 ~ <*f(y - g)+dr < Jr \VT(y - g)\2 - af(y - g)dT As the problem gives u € H o ( r ) n H 2 ( T D ) such that — Aru = af in T has a unique solution, we get (y — g)+ = y — g in T. Similarly we have (y — g)~ = y — g in Fc. L e m m a 3 We have y < g in Yc and so y > g in T.
