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It is known that the kernel of the (non-negative) Laplacian operator on a closed manifold consists of constant functions. I would like to ask if some similar phenomena happens for the modified operator: $$ Lu=\Delta u+ fu,$$ where $f$ is a smooth function.

More specifically: If $f$ equals to minus an eigenvalue of $\Delta$, then $Lu=0$ has non-trivial solutions. Are these the only $f$ with non-trivial solutions? Can we conclude that $Lu=0$ has only zero (or constant solutions) if we assume $f$ non-constant? Otherwise, can you parametrize its kernel (as you parametrize constant functions by their integrals, or by their value in one point)?

Thank you very much.

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    $\begingroup$ Have you tried some simple examples, e.g. on a circle? $\endgroup$
    – Robert Israel
    Commented Mar 24, 2022 at 22:32
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    $\begingroup$ It seems to me there's a trivial answer: take any smooth function $v$ which is never zero and not an eigenfunction of $\Delta$. Set $f = -\Delta v / v$ which is not a constant. Then $\Delta u + fu = 0$ has a non-trivial solution, namely $u=v$. $\endgroup$
    – Nate Eldredge
    Commented Mar 25, 2022 at 13:46
  • $\begingroup$ @NateEldredge well-pointed, thank you. Do you know if every function $f$ can be realized as $f=-\Delta v/v$? (up to summing a constant at least, since $\int -\Delta v/v>0$) $\endgroup$
    – Llohann
    Commented Mar 25, 2022 at 14:19
  • $\begingroup$ Also asked on Mathematics SX: math.stackexchange.com/questions/4410208/… $\endgroup$
    – J W
    Commented Mar 25, 2022 at 16:11

1 Answer 1

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Q: Can we conclude that $Lu=\Delta u+ fu=0$ has only zero (or constant solutions) if we assume $f$ non-constant?

A: No, a counter example in one dimension is the Mathieu equation, which has non-constant $\pi$-periodic or $2\pi$-periodic solutions $u(x)$ when $f(x)=a-2q\cos 2x$ for any given real $q$ at an infinite sequence of values of $a_n(q)$, $n=1,2,3,\ldots$.

More generally, $-L$ is the Hamiltonian of a particle in the potential $-f$, and we can readily adjust the potential so that it has a bound state at zero energy – simply by adding a constant to the potential to shift the bound state up or down.

Do note that the answer to the question would be affirmative for any generic $f$. To have a nonzero $u$ with $Lu=0$ requires fine tuning of the function $f$, for a generic $f$ such a solution will not exist.

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    $\begingroup$ The answer is negative in a more general situation: for every smooth $f$, there are infinitely many eigenvalues $\lambda$, therefore infinitely many functions $f-\lambda$ for which the equation has non-constant solutions. $\endgroup$
    – Alexandre Eremenko
    Commented Mar 24, 2022 at 22:51
  • $\begingroup$ Dear Carlo and Alexandre, thank you very much for your answer and comment. They are very enlightening. Would you know then any condition on f for the kernel to be trivial? I have reasons to believe that that is the case if f is 'big'. $\endgroup$
    – Llohann
    Commented Mar 25, 2022 at 14:10
  • $\begingroup$ By the way, I set a 50pts bounty on the same question in Math StackExchange. In case you are interested: math.stackexchange.com/questions/4410208/… $\endgroup$
    – Llohann
    Commented Mar 25, 2022 at 14:15
  • $\begingroup$ the set of functions $f$ where the kernel is nontrivial is of measure zero; you will need to fine tune parameters to line up a bound state at zero, this will not happen generically. Incidentally, in physics there is much literature on Hamiltonians where such fine tuning is not needed; one then speaks of a "topologically protected zero-mode"; your $L$ is not of this type. $\endgroup$
    – Carlo Beenakker
    Commented Mar 25, 2022 at 14:26

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