Green’s Function for Second Order Elliptic Equations in Non-divergence Form

Abstract

We construct the Green function for second order elliptic equations in non-divergence form when the mean oscillations of the coefficients satisfy the Dini condition and the domain has C^1,1 boundary. We also obtain pointwise bounds for the Green functions and its derivatives.

Appendix

The following lemma is well known to experts and essentially due to Campanato. We provide its proof for the reader’s convenience.

Lemma A.1

Let${\Omega } \in \mathbb {R}^{n}$bea domain satisfying the following condition: there exists a constantA₀ ∈ (0, 1] such thatfor everyx ∈Ω and 0 < r < diam Ω, wehave

$$\lvert{{\Omega}(x, r)}\rvert \ge A_{0} \lvert{B(x,r)}\rvert,\quad \text{where } {\Omega}(x,r):= {\Omega} \cap B(x, r). $$

Suppose that a function$u \in L_{\text {loc}}^{1}(\overline {\Omega })$is of Dini mean oscillation in Ω, then there exists a uniformly continuous functionu^∗on Ω such thatu^∗ = ua.e. in Ω.

Proof

In the proof we shall denote

By taking the average over ${\Omega }(x, \frac 12 r)$ to the triangle inequality

$$\lvert{\bar u_{x,r} - \bar u_{x,\frac12 r}}\rvert \le \lvert{u- \bar u_{x,r}}\rvert + \lvert{u-\bar u_{x, \frac12 r}}\rvert $$

and using $\lvert {{\Omega }(x, r)}\rvert / \lvert {{\Omega }(x,\frac 12 r)}\rvert \le 2^{n} /A_{0}$, we get

$$\lvert{\bar u_{x,r} - \bar u_{x,r/2}}\rvert \le (2^{n}/A_{0}) \omega(r) + \omega\left( \tfrac12 r\right) \le (2^{n}/A_{0}) \left( \omega(r)+ \omega\left( \tfrac12 r\right) \right). $$

By telescoping, we get

$$\begin{array}{@{}rcl@{}} \lvert{\bar u_{x,r}- \bar u_{x, 2^{-k}r}}\rvert &\le& \sum\limits_{j = 0}^{k-1} \lvert{\bar u_{x,2^{-j}r}-\bar u_{x, 2^{-(j + 1)}r}}\rvert\\ & \le& (2^{n}/A_{0})\left( \omega(r)+ 2\omega\left( \tfrac12 r\right)+ {\cdots} + 2 \omega\left( \tfrac{1}{2^{k-1}} r\right) + \omega\left( \tfrac{1}{2^{k}} r\right)\right) \\ & \le& (2^{n + 1}/A_{0}) \sum\limits_{j = 0}^{\infty} \omega\left( \tfrac{1}{2^{j}} r\right) \lesssim {{\int}_{0}^{r}} \frac{\omega(t)}{t} dt, \end{array} $$

(A.2)

where in the last step we used the fact that $\omega (t) \simeq \omega \left (\frac {1}{2^{j}}r\right )$ when $t \in \left (\frac {1}{2^{j + 1}}r, \frac {1}{2^{j}}r\right ]$; see [4]. Note that the last inequality also implies that

$$ \omega(r) \lesssim {{\int}_{0}^{r}} \frac{\omega(t)}{t} dt. $$

(A.3)

Now, we define the function u^∗ on Ω by setting $u^{\ast }(x)=\lim _{r\to 0} \bar u_{x,r}$. By the Lebesgue differentiation theorem, we have u = u^∗ a.e. By letting k →∞ in (A.2), we obtain

$$ \lvert{u^{*}(x)-\bar u_{x,r}}\rvert \lesssim {{\int}_{0}^{r}} \frac{\omega(t)}{t} dt\quad\text{for a.e. \(x \in {\Omega}\).} $$

(A.4)

For any x, y in Ω, let r = |x − y|, $z=\frac 12 (x+y)$, and use Eq. A.4 to get

$$\lvert{u^{\ast}(x) - u^{\ast}(y)}\rvert \!\le\! \lvert{u^{\ast}(x) - \bar u_{x,r}}\rvert+ \lvert{u^{\ast}(y) - \bar u_{y,r}}\rvert + \lvert{\bar u_{x,r}-\bar u_{y,r}}\rvert \!\lesssim\! {\int}_{0}^{\lvert{x-y}\rvert} \frac{\omega(t)}{t} dt +\lvert{\bar u_{x,r}-\bar u_{y,r}}\rvert. $$

By taking the average over ${\Omega }(z,\frac 12 r)$ to the triangle inequality

$$\lvert{\bar u_{x,r} - \bar u_{y,r}}\rvert \le \lvert{u- \bar u_{x,r}}\rvert + \lvert{u-\bar u_{y, r}}\rvert $$

and noting that ${\Omega }(z, \frac 12 r) \subset {\Omega }(x,r)\cap {\Omega }(y,r)$, we get

$$\lvert{\bar u_{x,r} - \bar u_{y,r}}\rvert \le (2^{n + 1}/A_{0}) \omega(\lvert{x-y}\rvert). $$

Combining together and using Eq. A.3, we conclude that

$$\lvert{u^{\ast}(x)-u^{\ast}(y)}\rvert \lesssim {\int}_{0}^{\lvert{x-y}\rvert} \frac{\omega(t)}{t} dt + \omega(\lvert{x-y}\rvert) \lesssim {\int}_{0}^{\lvert{x-y}\rvert} \frac{\omega(t)}{t} dt. $$

Therefore, we see that u^∗ is uniformly continuous with it the modulus of continuity dominated by the function $\displaystyle \rho (r):={\int }_0^r \frac {\omega (t)}{t},dt$. □