Newtonian Potential

— 1. Definition of Newtonian Potential —

Definition 1 Let ${\Omega}$ be a bounded open set, we define the Newtonian potential of ${f}$ is the function ${N_f}$ on ${{\mathbb R}^n}$ by

$\displaystyle N_f(x) = \int_\Omega \Gamma(x-y) f(y)\, dy,$

where ${\Gamma}$ is the fundamental solution of Laplace’s equation.

We need to check if ${N_f }$ is well-defined. If ${f \in L^\infty(\Omega)}$, then one can easily see ${N_f \in L^\infty ({\mathbb R}^n)}$. We can further relax ${f}$ to be in ${L^p(\Omega)}$, where ${p > \frac{n}{2}}$. Since the Holder conjugate ${p' < \frac{n}{n-2}}$, simply by Holder’s inequality, for any ${x \in R^n}$

$\displaystyle |N_f(x)| = \left| \int_\Omega \Gamma(x-y) f(y) \, dy \right| \le C\| f \|_p \left( \int_\Omega \frac{1}{|x-y|^{(n-2)p'}} \, dy \right)^{\frac{1}{p'}} < \infty.$

Therefore ${N_f \in L^\infty ({\mathbb R}^n)}$ is well-defined in the whole space. If we only require ${N_f}$ to be defined in ${\Omega}$, we only need ${f}$ to be integrable plus a little bit local boundedness near the singularity ${x}$, for example, local Holder continuity or so. In this case, we can see ${N_f}$ is also well defined outside ${\Omega}$, but not defined on ${\partial \Omega}$. So this case will become less interesting when we try to solve Dirichlet problem through Newtonian potential, since the function on boundary needs to be defined.

— 2. Regularity of Newtonian Potential —

Theorem 2 (Representation for first derivative) Let ${f}$ be bounded and integrable in ${\Omega}$. Then ${N_f \in C^1({\mathbb R}^n)}$ and for any ${x \in {\mathbb R}^n}$,

$\displaystyle D_i N_f(x) = \int_{\Omega} D_i \Gamma(x-y) f(y)\, dy, \quad i = 1,2, \cdots, n.$

Proof: Define a function

$\displaystyle v(x):= \int_\Omega D_i \Gamma(x-y) f(y)\, dy.$

Since ${|D_i \Gamma(x)| \le C|x|^{1-n}}$, we know ${v}$ is well-defined. We want to show ${v = D_i N_f}$.We fix a function ${\eta}$ in ${C^1({\mathbb R})}$ satisfying ${0 \le \eta \le 1, 0 \le \eta' \le 2, \eta(t) = 0}$ for ${t \le 1}$, ${\eta(t) =1}$ for ${t \ge 2}$. And define for ${\varepsilon >0}$ small,

$\displaystyle N_f^\varepsilon(x) = \int_{\Omega}\Gamma(x-y) \eta_\varepsilon f(y) \, dy,$

where ${\eta_\varepsilon = \eta (|x-y| / \varepsilon)}$. Clearly ${N_f^\varepsilon \in C^1({\mathbb R}^n)}$ since we make the singularity vanished. And we have

$\displaystyle |v(x)- D_i N_f^\varepsilon(x)| \le \int_{|x-y| \le 2\varepsilon} D_i [(1-\eta_\varepsilon) \Gamma(x-y)] |f(y)| \, dy\le C\|f\|_{L^\infty} \cdot \varepsilon.$

Therefore ${D_i N_f^\varepsilon \rightarrow v}$ uniformly in compact subsets of ${{\mathbb R}^n}$ as ${\varepsilon \rightarrow 0}$. One can also easily see that ${N_f^\varepsilon \rightarrow N_f}$ uniformly as ${\varepsilon \rightarrow 0}$. Therefore, ${N_f \in C^1}$ and

$\displaystyle D_i N_f(x) =v(x)= \int_{\Omega} D_i \Gamma(x-y) f(y)\, dy.$

$\Box$

Theorem 3 (Representation for second derivative) Let ${f \in C_{loc}^\alpha(\Omega)}$, with ${\alpha \le 1}$, and also bounded. Then ${N_f \in C^2(\Omega)}$, ${\Delta N_f = f}$ in ${\Omega}$, and for any ${x \in \Omega}$,

$\displaystyle D_{x_i x_j} N_f(x) = \int_{\Omega_0} D_{x_i x_j} \Gamma(x-y) (f(y)-f(x)) \, dy - f(x) \int_{\partial \Omega_0} D_{x_i} \Gamma \nu_j(y) \, dS_y,$

where ${\Omega_0}$ is any domain containing ${\Omega}$ for which the divergence theorem holds and ${f}$ is extended to be 0 outside ${\Omega}$.

Proof: The proof follows from the same strategy as the previous theorem. We define ${u(x)}$ to be the RHS of the equation stated in theorem. Since ${f}$ is locally Holder continuous, the integrand is integrable near the singularity, hence ${u}$ is well-defined. Let ${\eta_\varepsilon}$ be the same as defined in last theorem, then define

$\displaystyle v_\varepsilon = \int_{\Omega} D_{x_i} \Gamma(x-y) \eta_\varepsilon f(y) \, dy.$

We will have

$\displaystyle D_{x_j} v_\varepsilon (x) = \int_\Omega D_{x_j} (D_{x_i} \Gamma \eta_\varepsilon) f(y) \, dy$

$\displaystyle = \int_\Omega D_{x_j}(D_{x_i} \Gamma \eta_\varepsilon) (f(y) - f(x)) \, dy + f(x)\int_{\Omega_0} D_{x_j}(D_{x_i} \Gamma \eta_\varepsilon) \, dy$

$\displaystyle = \int_\Omega D_{x_j}(D_{x_i} \Gamma \eta_\varepsilon) (f(y) - f(x)) \, dy + f(x)\int_{\Omega_0} D_{x_j x_i} \Gamma \eta_\varepsilon + D_{x_i} \Gamma D_{x_j} \eta_\varepsilon \, dy$

$\displaystyle = \int_\Omega D_{x_j}(D_{x_i} \Gamma \eta_\varepsilon) (f(y) - f(x)) \, dy - f(x)\int_{\Omega_0} D_{y_j x_i} \Gamma \eta_\varepsilon + D_{x_i} \Gamma D_{y_j} \eta_\varepsilon \, dy$

$\displaystyle = \int_\Omega D_{x_j}(D_{x_i} \Gamma \eta_\varepsilon) (f(y) - f(x)) \, dy - f(x)\int_{\partial \Omega_0} D_{x_i} \Gamma \nu_j(y) \, dS_y,$

provided ${2\varepsilon < dist(x, \partial \Omega)}$. Hence, by subtraction

$\displaystyle |u(x) - D_{x_j} v_\varepsilon (x)| = \left| \int_{|x-y| \le 2 \varepsilon} D_{x_j}[(1-\eta_\varepsilon)D_{x_i}\Gamma] (f(y) -f(x)) \, dy \right|$

$\displaystyle \le \left(\frac{n}{\alpha} + 4 \right) [f]_{\alpha;x} (2\varepsilon)^\alpha.$

Therefore ${D_{x_j} v_\varepsilon}$ converges to ${u}$ uniformly on any compact subset of ${\Omega}$ as ${\varepsilon \rightarrow 0}$. One can also easily see that ${v_\varepsilon}$ converges uniformly to ${D_i N_f}$ in ${\Omega}$. Therefore ${N_f \in C^2(\Omega)}$ and ${u = D_{x_i x_j} N_f}$.Finally, setting ${\Omega_0 = B_R(x)}$, we have for sufficiently large ${R}$,

$\displaystyle \Delta w = \int_{B_R(x)} D_{x_i x_i} \Gamma (f(y)-f(x)) \, dy - f(x) \int_{\partial B_R(x)} D_{x_i} \Gamma \nu_i(y) \, dS_y$

$\displaystyle = \int_{B_R(x)} D_{x_i x_i} \Gamma (f(y)-f(x)) \, dy + f(x) \int_{\partial B_R(x)} D_{y_i} \Gamma \nu_i(y) \, dS_y$

$\displaystyle = \int_{B_R(x)} D_{x_i x_i} \Gamma (f(y)-f(x)) \, dy + \frac{f(x)}{n \omega_n R^{n-1}}\int_{\partial B_R(x)} \nu_i(y) \nu_i(y) \, d S_y$

$\displaystyle = \int_{B_R(x)} D_{x_i x_i} \Gamma (f(y)-f(x)) \, dy + f(x).$

But we can estimate

$\displaystyle \left| \int_{B_R(x)} D_{x_i x_i} \Gamma (f(y)-f(x)) \, dy \right| = \left| \int_{|x-y| < \delta} D_{x_i x_i} \Gamma (f(y)-f(x)) \, dy \right|$

$\displaystyle \le \int_{|x-y| < \delta} \frac{1}{|x-y|^{n-\alpha}} \, dy \le \delta^\alpha,$

for any ${\delta > 0}$. Therefore ${\int_{B_R(x)} D_{x_i x_i} \Gamma (f(y)-f(x)) \, dy = 0}$ and hence ${\Delta w =f}$. $\Box$

Remark 1 If the local Holder continuity of ${f}$ is replaced by Dini continuity, we still have the same result. And this theorem suggests that we can use Newtonian potential to get a solution of Poisson equation.

Theorem 4 (interior Schauder estimate) If ${f \in C^\alpha(\bar{\Omega})}$, then ${N_f \in C^{2,\alpha}(\bar{\Omega'})}$ for any ${\Omega' \subset \subset \Omega}$. Furthermore,

$\displaystyle \|N_f\|_{C^{2,\alpha} (\bar{\Omega'})} \le C(n, \Omega ,\Omega') \|f\|_{C^\alpha (\bar{\Omega})}.$

Proof: Without loss of generality we can assume ${\Omega = B_2}$ and ${\Omega' = B_1}$. Take ${x,z \in B_1}$, from the representation we have, we know

$\displaystyle D_{x_i x_j} N_f(x) - D_{x_i x_j} N_f(z)$

$\displaystyle = \int_{B_2} D_{x_i x_j} \Gamma(x-y) (f(y)-f(x)) - D_{x_i x_j} \Gamma(z-y) (f(y)-f(z)) \, dy$

$\displaystyle - f(x) \int_{\partial B_2} D_{x_i} \Gamma(x-y) \nu_j(y) \, dS_y + f(z) \int_{\partial B_2} D_{x_i} \Gamma(z-y) \nu_j(y) \, dS_y.$

If we set ${\delta = |x-z|}$, ${\xi = \frac{x+z}{2}.}$ We can split the RHS into 6 parts, namely,

$\displaystyle D_{x_i x_j} N_f(x) - D_{x_i x_j} N_f(z) = I_1 + I_2 + I_3 + I_4 + I_5 + I_6,$

where

$\displaystyle I_1 = \int_{B_\delta (\xi)} D_{x_i x_j} \Gamma(x-y) (f(y)-f(x)) \, dy;$

$\displaystyle I_2 = \int_{B_2 - B_\delta(\xi)} D_{x_i x_j} \Gamma(x-y) (f(z)-f(x)) \, dy;$

$\displaystyle I_3 = -\int_{B_\delta (\xi)} D_{x_i x_j} \Gamma(z-y) (f(y)-f(z)) \, dy;$

$\displaystyle I_4 = \int_{B_2 - B_\delta(\xi)} [D_{x_i x_j} \Gamma(x-y) - D_{x_i x_j} \Gamma(z-y)] (f(y)-f(z)) \, dy;$

$\displaystyle I_5 = [f(z) - f(x)] \int_{\partial B_2} D_{x_i} \Gamma(z-y) \nu_j(y) \, dS_y;$

$\displaystyle I_6 = f(x) \int_{\partial B_2} [D_{x_i} \Gamma(z-y) - D_{x_i} \Gamma(x-y)] \nu_j(y) \, dS_y.$

We will estimate all these 6 terms. Since ${B_\delta(\xi) \subset B_{2 \delta}(x)}$, we have

$\displaystyle |I_1| \le C \int_{B_{2 \delta}(x)} \frac{1}{|x-y|^n} |f(y) - f(x)| \, dy \le C \delta^\alpha.$

And ${I_3}$ should have a same estimate by this argument. To estimate ${I_2}$, notice that if ${y \in \partial B_2}$, ${|x-y|>1}$, and if ${y \in \partial B_\delta(\xi)}$, ${|x-y|>\frac{\delta}{2}}$. And we will have by divergent theorem

$\displaystyle |I_2| \le |f(x) - f(z)| \left( \int_{\partial B_2} |D_{x_i} \Gamma(x-y) \nu_j(y)| \, dS(y) +\int_{\partial B_\delta(\xi)} |D_{x_i} \Gamma(x-y) \nu_j(y)| \, dS(y) \right)$

$\displaystyle \le \left( \int_{\partial B_2} 1 \, dS(y) + \int_{\partial B_\delta(\xi)} \left(\frac{\delta}{2} \right)^{1-n} \, dS(y) \right) \delta^\alpha \le C \delta^\alpha.$

To estimate ${I_4}$, note that ${D_{x_i x_j} \Gamma(x-y)}$ is smooth when ${y \in B_2 - B_\delta(\xi)}$, by mean value theorem, we know ${|D_{x_i x_j} \Gamma(x-y) - D_{x_i x_j} \Gamma(z-y)| \le |D D_{x_i x_j} \Gamma(\hat{x}-y)| |x-z|}$ for some ${\hat{x}}$ between ${x}$ and ${z}$. And for ${y \in B_2 - B_\delta(\xi)}$, we always have ${|\hat{x} - y| \ge \frac{|y-z|}{3}}$. Then we will have

$\displaystyle |I_4| \le \int_{B_2 - B_\delta(\xi)} |D D_{x_i x_j} \Gamma(\hat{x}-y)| |x-z| |f(y)-f(z)| \, dy$

$\displaystyle \le C \delta \int_{B_2 - B_0.5\delta(z)} \frac{1}{|z-y|^{n+1-\alpha}} \, dy \le C \delta \int_{\frac{1}{2}\delta}^\infty \frac{1}{r^{2-\alpha}} \, dr \le C \delta^\alpha.$

Since ${|z - y| < 1}$ on ${y \in \partial B_2}$, we have trivial estimate for ${I_5}$:

$\displaystyle |I_5| \le C|f(x) - f(z)| \le C \delta^\alpha.$

To estimate ${I_6}$, we again have mean value theorem as in estimate of ${I_4}$, and since ${DD_{x_i} \Gamma(\hat{x}-y)}$ and ${f}$ are bounded, we have

$\displaystyle |I_6| \le C|x-z| \le C \delta^\alpha.$

Combining these 6 estimates we can conclude ${N_f in C^\alpha(\bar{B_1})}$. From the previous theorems we can bounded the ${L^\infty}$ norm of ${N_f}$ and its gradient by ${\|f\|_L^{\infty}}$. Therefore we have the inequality

$\displaystyle \|N_f\|_{C^{2, \alpha}(\bar{B_1})} \le C(n) \|f\|_{C^\alpha(\bar{B_2})}.$

$\Box$

This tells us if ${\Delta u = f}$ and ${f \in C^\alpha}$, then ${u \in C^{2,\alpha}}$. And the following example tells us if ${f}$ only continuous, then ${u}$ is not necessarily ${C^2}$ (${\alpha}$ cannot be 0).

Remark 2 If ${f}$ is only continuous, ${u}$ is ${C^1}$, but not necessarily ${C^2}$.

An example: For ${\alpha}$ with ${|\alpha| =2}$, let ${P}$ be a homogeneous harmonic polynomial of degree 2 with ${D^\alpha P \neq 0}$. Choose ${\eta \in C_0^\infty (\{ x: |x| < 2\})}$ with ${\eta = 1}$ when ${t_k = 2^k}$. Define ${f(x) = \sum_0^\infty \frac{1}{k} \Delta(\eta P)(t_k x).}$ Notice that ${\Delta(\eta P) (t_k x) =0}$, when ${|t_k x|<1}$ or ${|t_k x| > 2}$. Therefore

$\displaystyle f(x) = \begin{cases} \frac{1}{k} \Delta (\eta P)(2^k x), & \frac{1}{2^k} \le |x| < \frac{1}{2^{k-1}};\\ 0, & x =0. \end{cases}$

And since ${f}$ vanishes on the boundary of each annulus ${\{ \frac{1}{2^k} \le |x| < \frac{1}{2^{k-1}} \}}$ and ${\frac{1}{k} \rightarrow 0}$ as ${k \rightarrow \infty}$, we can easily see that ${f}$ is continuous.

\noindent Now for ${x}$ such that ${\frac{1}{2^l} \le |x| < \frac{1}{2^{l-1}}}$ define ${v(x): = \sum_{k=0}^l \frac{1}{2^{2k}} \frac{1}{k} (\eta P)(2^k x)}$. Then by a similar argument we can see ${v}$ is well-defined and smooth away from ${0}$. For any ${x \neq 0}$, we can find an ${l}$ such that ${\frac{1}{2^l} \le |x| < \frac{1}{2^{l-1}}}$, and Notice that

$\displaystyle D^\alpha v(x) = \sum_{k=0}^l \frac{1}{k} D^\alpha (\eta P)(2^k x) = \sum_{k=0}^{l-1} \frac{1}{k} D^\alpha P + \frac{1}{l} D^\alpha (\eta P).$

Since ${D^\alpha P}$ is not ${0}$, ${l \rightarrow \infty}$ as ${|x| \rightarrow 0}$, we can see ${v(x)}$ is not ${C^2}$ around ${x = 0}$. Noticing that

$\displaystyle v(x) = \sum_{k=0}^{l-1} \frac{1}{k} P(x) + \frac{1}{l} \eta(2^l x) P(x),$

since ${P}$ is homogeneous. We can see as ${|x| \rightarrow 0}$, ${\frac{1}{l} \eta(2^l x) P(x) \rightarrow 0.}$ Also ${|\sum_{k=0}^{l-1} \frac{1}{k} P(x)| \le l |P(x)| \le C x^2 \log |x| \rightarrow 0}$ as ${|x| \rightarrow 0}$, therefore ${v(x)}$ stays bounded when ${|x| \rightarrow 0}$. Now suppose ${u}$ is a ${C^2}$ solution of ${\Delta u = f}$. Then ${\Delta (u-v) = 0}$ for ${x \neq 0}$. Since ${u-v}$ is bounded around 0, ${0}$ is a removable singularity for ${u - v}$, which makes it a harmonic function. But this implies ${v \in C^2}$, which leads to a contradiction. So there is no ${C^2}$ solutions in any neighborhood of the origin to ${\Delta u = f}$.

— 3. Solution to Dirichlet problem —

Theorem 5 Let ${\Omega}$ be a bounded domain and suppose each point of ${\partial \Omega}$ is regular (in the sense that Perron’s method will give a harmonic function prescribed continuous boundary value, e.g. ${\partial \Omega}$ is ${C^1}$.) Then if ${f}$ is bounded and locally Holder continuous, the classical Dirichlet problem : ${\Delta u = f}$ in ${\Omega}$, ${u = g}$ on ${\partial \Omega}$ is uniquely solvable for any continuous boundary values ${g}$.

Proof: From Theorem 2 and 3, we know that ${N_f \in C^1({\mathbb R}^n) \cap C^2(\Omega)}$, ${\Delta N_f = f}$ in ${\Omega}$. Consider ${v = u - N_f}$, then the Dirichlet problem is equivalent to ${\Delta v = 0}$ in ${\Omega}$, ${v = g - N_f}$ on ${\partial \Omega}$, which has a unique solution by Perron’s method. $\Box$

The regularity of ${f}$ can be relaxed as following theorem.

Theorem 6 Let ${\Omega}$ be as described above, ${f \in L^p(\Omega)}$ for some ${p > \frac{n}{2}}$ and is local Holder continuous then the classical Dirichlet problem can by uniquely solved.

Proof: From section 1 we know ${N_f}$ is well-defined on ${{\mathbb R}^n}$. And we need to check ${N_f \in C(\bar{\Omega})}$. We fix a function ${\eta}$ as in Theorem 2. And define for ${\varepsilon >0}$ small,

$\displaystyle N_f^\varepsilon(x) = \int_{\Omega}\Gamma(x-y) \eta_\varepsilon f(y) \, dy,$

where ${\eta_\varepsilon = \eta (|x-y| / \varepsilon)}$. Clearly ${N_f^\varepsilon \in C^1({\mathbb R}^n)}$. Extending ${f}$ to be 0 outside ${\Omega}$, then

$\displaystyle |N_f(x)- N_f^\varepsilon(x)| \le \int_{|x-y| \le 2\varepsilon} (1-\eta_\varepsilon) \Gamma(x-y) |f(y)| \, dy \le C\|f\|_{p, B_{2\varepsilon} (x)}.$

Hence ${N_f^\varepsilon(x) \rightarrow N_f(x)}$ uniformly as ${\varepsilon \rightarrow 0}$, and thus ${N_f(x) \in C({\mathbb R}^n)}$. Then follow the proof of Theorem 2 one can see ${w \in C^1(\Omega)}$. Note that we might not have ${w \in C^1({\mathbb R}^n)}$ anymore since we lose boundedness of ${f}$ near the boundary of ${\Omega}$. Then we can follow the exact same proof of Theorem 3 and Theorem 8 to get the result. $\Box$

Remark 3 In order to solve Dirichlet problem, we can’t assume ${f \in L^p(\Omega)}$ for some ${p \le \frac{n}{2}}$, otherwise ${N_f}$ might not be well-defined on boundary.