Lower bound on the solution of a Schrödinger-type equation - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T16:22:20Z http://mathoverflow.net/feeds/question/94996 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/94996/lower-bound-on-the-solution-of-a-schrodinger-type-equation Lower bound on the solution of a Schrödinger-type equation timur 2012-04-24T04:40:31Z 2012-05-09T17:02:28Z <p>Consider the equation</p> <p>$-\Delta u + Vu=f$,</p> <p>on a closed manifold (or on a bounded domain with homogeneous Neumann condition). Here one can assume whatever integrability or smoothness conditions on $V$ and $f$ one likes. One can show that if $V$ and $f$ are both nonnegative and not identically zero, then the unique solution $u$ is strictly positive. Moreover, $u$ is bounded from below by a constant $C$ times an appropriate norm of $f$, where $C$ does not depend on $f$. My question is, where or how can I find more information on how $C$ depends on $V$?</p> http://mathoverflow.net/questions/94996/lower-bound-on-the-solution-of-a-schrodinger-type-equation/96458#96458 Answer by Michael Renardy for Lower bound on the solution of a Schrödinger-type equation Michael Renardy 2012-05-09T15:01:37Z 2012-05-09T17:02:28Z <p>Here are a couple of ideas:</p> <p>1 This yields a lower bound, but it is not in terms of a norm of f:</p> <p>Let $\lambda=\min(f/V)$. Then $u-\lambda$ satisifies $$-\Delta(u-\lambda)+V(u-\lambda)=f-\lambda V\ge 0.$$ Hence $u\ge\lambda$ by the maximum principle.</p> <p>2 Assume V is bounded, and let $V_M$ be its maximum: Let $v$ be the solution of $-\Delta v+V_Mv=f$. Then $$-\Delta(u-v)+V(u-v)=(V_M-V)v\ge 0,$$ so $u\ge v$. If the geometry of the domain is simple, you may be able to determine the Green's function for $-\Delta+V_M$ explicitly.</p>