Consider the equation $$ -\Delta f+mf+\lambda f^p=0$$ on $\mathbb{R}^d$, where $d>2$,$m>0$, $p>1$ is integer, and $\lambda \in \mathbb{R}$. Are there any known results regarding the non-existence of non-zero solutions $f$ of this equation in $\mathcal{S}(\mathbb{R}^d)$ (the Schwartz space of rapidly decreasing smooth functions), at least for some values of $d$, $p$ and $\lambda$?
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$\begingroup$ Well... there's the case when $p$ is odd and $\lambda \geq 0$.... (but I guess that's too trivial for you) $\endgroup$– Willie WongCommented Mar 1, 2022 at 17:58
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$\begingroup$ @WillieWong, actually I'm interested in this case. Can you provide a reference for it? $\endgroup$– S.Z.Commented Mar 1, 2022 at 19:06
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$\begingroup$ Incidentally, the question as you posed is a bit strange: by virtue of Agmon estimates, frequently when you have a solution to the equation of the type you stated, the solution would have exponential (and hence rapid) decay. So if you are primarily interested in decay rates, to rule out rapidly decaying solutions is more or less the same as looking for conditions to make the equation not have any non-trivial solutions. But your question doesn't seem to be focused on regularity issues. $\endgroup$– Willie WongCommented Mar 1, 2022 at 19:33
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$\begingroup$ If you are interested: the standard reference for Agmon estimates is Agmon, S., Lectures on Exponential Decay of Solution of Second-Order Elliptic Equation, Princeton University Press, 1982. $\endgroup$– Willie WongCommented Mar 1, 2022 at 19:35
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$\begingroup$ @WillieWong, thanks a lot. Actually I'm interested in non-existence of classical solutions of polynomial growth at infinity. I asked the question for rapidly decaying functions because I thought that's simpler. So these Agmon estimates, together with your answer, seems to be exactly what I need. $\endgroup$– S.Z.Commented Mar 1, 2022 at 20:52
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(To substantiate my comment above)
If $\lambda \geq 0$ and $p$ is odd, assume $f\in \mathcal{S}(\mathbb{R}^d)$ is a solution, then you can multiply the equation by $f$ and integrate by parts (everything converges appropriately) to find
$$ \int |\nabla f|^2 + m f^2 + \lambda f^{p+1} = 0 $$
Noting that $p+1$ is even and so $f^{p+1}\geq 0$, you find this implies $\|f\|_{L^2} = 0$.
(Alternatively, you can also use the maximum principle: write the equation as $-\triangle f + Vf = 0$ where $V = m + \lambda f^{p-1}$. When $p$ is odd this function $V$ is manifestly positive, and by maximum principle any solution cannot have a positive local max nor a negative local min. Combined with the Schwartz assumption this forces $f\equiv 0$.)
answered Mar 1, 2022 at 19:20
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$\begingroup$ 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? $\endgroup$– S.Z.Commented Mar 1, 2022 at 21:18
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$\begingroup$ 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 $\|u-v\|_{L^2} = 0$. $\endgroup$ Commented Mar 1, 2022 at 21:36