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According to Theorem 1.1' in this paper we have the following estimate on classical solutions $u \in C^2(\overline{B_1^+})$ of $-\Delta u = f \text{ in } B_1^+ = B_1 \cap \{x _n \ge 0 \}$ and $u = 0 \text{ on } \partial B_1^+ \cap \{x_n = 0\}$

$$|D^2u(x) - D^2u(y)| \le C\left(r\lVert u \rVert_{L^\infty(B_{1}^+)} + \int_0^{r} \frac{\omega_f(t)}{t}\,dt + r\int_{r}^{1} \frac{\omega_f(t)}{t^2}\,dt\right) \tag{1}$$ $\forall \, x,y \in B_{1/2}^{+}$ with $r = |x-y|$ where, $\omega_f$ denotes the modulus of continuity of $f$ which we assume is Dini continuous.

The proof is supposed to be similar to proof of Theorem 1.1 which is the interior estimate. Following the lines of this proof we can get all analogous 'boundary' harmonic function estimates in $B_{1}^+$ v.i.a. applying the estimates in Theorem 1.1 to the odd extensions of the harmonic functions to the full ball $B_1$. But in the last step (going along the lines of eqn $(1.13)$ in the paper) it seems we need the estimate $$|D^2u_0(x) - D^2u_0(y)| \le C \lVert u \rVert_{L^{\infty}(B_1^+)}|x-y|, \, \forall \, x,y \in B_{1/2}^+ \tag{2}$$ where, $u_0$ satisfies $-\Delta u_0 = f(0)$ in $B_{1}^+$ and $u_0 = 0$ on $\partial B_1^+ \cap \{x_n = 0\}$ where, $C$ is supposed to be independent of $f$.

I can't seem to get the estimate $(2)$ for half-ball.

In case of interior estimate we can consider $v_0 := u_0 - \frac{f(0)}{2n}(1 - |x|^2)$ which is harmonic in $B_1$ and write \begin{align*} |D^2u_0(x) - D^2u_0(y)| = |D^2v_0(x) - D^2v_0(y)| &\le r\lVert D^3v_0\rVert_{L^{\infty}(B_{1/2})} \\ &\le Cr\lVert v_0\rVert_{L^{\infty}(\partial B_{1})} = Cr\lVert u_0\rVert_{L^{\infty}(\partial B_{1})}\end{align*} where, $r = |x-y|$ which proves the interior analogue of $(2)$ using only gradient estimate for harmonic function $v_0$.

But a similar approach doesn't seem to be working for the boundary case (for example, considering $u_0 - \frac{f(0)}{2}x_n^2$ which is harmonic and applying odd extension to this.)

Is there a different way of approaching the estimate $(2)$?

Any help is appreciated. Thanks.

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    $\begingroup$ I think you should subtract the function $v$ which solves $\Delta v=f(0)$ with Dirichlet boundary conditions on $\partial B_1^+$. Please, let me know. $\endgroup$ – Giorgio Metafune Apr 10 '20 at 13:24
  • $\begingroup$ @GiorgioMetafune I was trying that initially and couldn't use the maximum principle directly. I got $v(x) = \frac{f(0)}{2}x_n^2 - \frac{f(0)}{2}\int_{\mathbb{S}^{n-1}} K(x,y) y_n^2\operatorname{sgn}(y_n)\,d\sigma(y)$ (where, $K$ is the Poisson Kernel for unit sphere).. which is deg $2$ polynomial + $f(0) \times$ harmonic and I didn't know how to bound the $f(0)$ factor with $\lVert v \rVert_{L^{\infty}(B_1^+)}$. $\endgroup$ – jeanboi Apr 10 '20 at 17:41
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One approach is to observe that $$\|u_0\|_{L^{\infty}(B_1^+)} \geq \frac{1}{16n}|f(0)|.$$ It suffices by linearity to prove this when $f(0) = 1$. Consider the barrier $$b(x) = \frac{1}{2n}\left(\left|x - \frac{1}{2}e_n\right|^2 - \frac{1}{8}\right).$$ Since $b \geq \frac{1}{16n}$ on $\partial B_1^+$, either $u_0 \geq b$ somewhere on $\partial B_1^+$ and we are done, or $u_0 \leq b$ on $\partial B_1^+$ in which case the comparison principle implies that $$u\left(\frac{e_n}{2}\right) \leq -\frac{1}{16n}$$ and we are again done.

With this estimate in hand, your suggestion to use reflection and interior estimates for the harmonic function $v_0 = u_0 - \frac{f(0)}{2}x_n^2$ works because $$\|v_0\|_{L^{\infty}(B_1^+)} \leq (1+8n)\|u_0\|_{L^{\infty}(B_1^+)}.$$ More precisely, $$\|D^3u_0\|_{L^{\infty}(B_{1/2}^+)} = \|D^3v_0\|_{L^{\infty}(B_{1/2}^+)} \leq C(n)\|v_0\|_{L^{\infty}(B_1^+)} \leq C(n)(1+8n)\|u_0\|_{L^{\infty}(B_1^+)}.$$

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  • $\begingroup$ Thank you very much! :) I couldn't bound $|f(0)|$ with $\lVert u_0 \rVert_{L^{\infty}(B_{1/2}^+)}$. Could you tell me how you guessed the barrier function? $\endgroup$ – jeanboi Apr 10 '20 at 16:02
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    $\begingroup$ @jeanboi: Sure! My thought process was: it should be impossible for $u_0$ to be very "flat" while $f(0) = 1$, since the solution in $B_{1/2}(e_n)$ with zero boundary data is a paraboloid that bends a fair amount. $\endgroup$ – Connor Mooney Apr 10 '20 at 18:03
  • $\begingroup$ Thanks for the insight! :) $\endgroup$ – jeanboi Apr 10 '20 at 18:39
  • $\begingroup$ This is better than what I proposed. $\endgroup$ – Giorgio Metafune Apr 10 '20 at 21:22

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