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Mar 1, 2022 at 21:36 comment added Willie Wong For your comment: Suppose $- \Delta (u-v) + m(u-v) + \lambda (u^p - v^p) = 0$. Assuming you are looking at real valued functions: if $p$ is odd then $u\mapsto u^p$ is strictly increasing, and so $(u^p - v^p)(u-v) \geq 0$. Then the same argument tells you that $\|u-v\|_{L^2} = 0$.
Mar 1, 2022 at 21:18 comment added S.Z. I wonder if a similar argument could show that the nonlinear operator on the left side of the equation is injective or one needs some global inverse function theorem for that?
Mar 1, 2022 at 20:53 vote accept S.Z.
Mar 1, 2022 at 19:20 history answered Willie Wong CC BY-SA 4.0