I am trying to prove some monotonicity of a solution of a given pde; after considering a quantity like $ \phi(x) = x \cdot \nabla v(x)$ ($v$ is the solution of a given pde) I arrive at something along the lines of

$$-\Delta \phi(x)+ \phi(x) + 2 \int_0^1 \frac{ \phi(tx)}{t} dt = f(x) \ge 0 \qquad B_1$$ with $ \phi=0$ on $ \partial B_1$ where $B_1$ is the unit ball in $ R^N$. Also assume some smoothness in $v$ and hence this integrand is somewhat behaved. So my question is after making various assumptions on smoothness and.... are we able to prove a maximum principle? ie. can we show $ \phi \ge 0$ in $B_1$?

  • $\begingroup$ A first observation is that this nonlocality is kind of singular, not at all the standard form of a convolutional kernel. A second observation is that the answer is liekely no: indeed you already have $\phi(0)=0$, so the strong maximum principle cannot hold. Of course one cannot rule out a very special situation where the weak MP holds but not the strong one. But I doubt it. $\endgroup$ – leo monsaingeon Jul 30 '20 at 6:59
  • $\begingroup$ that is a good point... i didn't think of that. thanks $\endgroup$ – Math604 Jul 30 '20 at 19:53

(For an actual answer, see the edit below.)

Let $\phi$ be smooth near zero and non-negative. Suppose that the Taylor expansion of $\phi$ at zero is non-trivial, and let $P(x)$ be the leading term. Then $P(x)$ is a non-negative homogeneous polynomial of degree $2 k \geqslant 2$. Then $-\Delta P$ is a homogeneous polynomial of degree $2k - 2$ which is not everywhere positive (it is negative along the line $x_0 \mathbf R$ where $x_0$ is a minimum of $P$ over the unit sphere), while both $\phi$ and the integral term are homogeneous of degree $2k$. Since $-\Delta P(x)$ is the leading term in the expansion of $f(x)$, it follows that $f(x)$ cannot be positive in a neighbourhood of the origin. Therefore, if $f \geqslant 0$, then $\phi$ is necessarily not everywhere positive!

I am not sure what happens if $P(x)$ has zero Taylor expansion near zero, though. I bet the answer is similar, but I fail to see an straightforward argument.

Edit: The following seems to be a complete solution, although I did not check it carefully.

Suppose that $\phi$ is smooth in the unit ball, non-negative, and satisfies the integro-differential equation in question. Consider the symmetrisation $\phi^\star$ of $\phi$: $$ \phi^\star(x) = \frac{1}{|x|^{N-1}} \int_{\partial B} \phi(|x| u) du = \int_{SO(N)} \phi(O x) dO $$ (both integrals with respect to normalised measures). Then $\phi^\star$ is a rotationally invariant solution of the equation in question, with $f$ replaced by its symmetrisation $f^\star$. We will prove that $\phi^\star$ is identically zero, so that $\phi$ is identically zero, too.

So the problem becomes one-dimensional: if $\phi^\star(x) = \psi(|x|)$ and $f^\star(x) = g(|x|)$, then we have $\psi \ge 0$ on $(0, 1)$, $\psi(0) = \psi'(0) = \psi(1) = 0$ and $$ -\psi''(r) - \frac{N-1}{r} \psi'(r) + \psi(r) + 2 \int_0^1 \frac{\psi(r t)}{t} \, dt = g(r) \geqslant 0 . $$ Observe that $\psi''(r) + \tfrac{N-1}{r} \psi'(r) = r^{1-N} (r^{N-1} \psi'(r))'$. Thus, $$ (r^{N-1} \psi'(r))' \leqslant r^{N-1} \psi(r) + 2 \int_0^r r^{N-1} \, \frac{\psi(s)}{s} \, ds . $$ Integrating both sides, we get $$ r^{N-1} \psi'(r) \leqslant \int_0^r s^{N-1} \psi(s) ds + 2 \int_0^r \frac{r^N - s^N}{N} \, \frac{\psi(s)}{s} \, ds , $$ and integrating both sides again (assuming that $N \ne 2$, which requires as slightly different argument), $$ \begin{aligned} \psi(r) & \leqslant \int_0^r \frac{s^2 (1 - (s / r)^{N - 2})}{N - 2} \, \frac{\psi(s)}{s} \, ds \\ & + 2 \int_0^r \frac{r^{N + 1} - s^{N + 1} - (N + 1) s^N (r - s)}{N (N + 1)} \, \frac{\psi(s)}{s} \, ds . \end{aligned} $$ In particular, $$ \psi(r) \leqslant C r^2 \int_0^r \frac{\psi(s)}{s} \, ds $$ for some constant $C$. Gronwall's inequality applied to $\psi(r) / r$ implies that $\psi(r) \le 0$, and hence $\psi(r) = 0$, as claimed.

  • $\begingroup$ it will take me some time to digest your answer; thank you very much. $\endgroup$ – Math604 Jul 30 '20 at 19:54
  • $\begingroup$ You're welcome. I may have messed something up, please do ask me if you find anything unclear. $\endgroup$ – Mateusz Kwaśnicki Jul 30 '20 at 20:00

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