# Maximum principle for an elliptic like operator

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 $$B_1$$?

• 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. – leo monsaingeon Jul 30 '20 at 6:59
• that is a good point... i didn't think of that. thanks – 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.

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