Tag Archives: newtonian potential

Regularity of Newtonian Potential

Suppose {\Gamma(\cdot)} is the fundamental solution of laplace equation.

\displaystyle \Gamma(x-y)=\begin{cases}\frac{1}{n(2-n)\omega_n}|x-y|^{2-n}, n>2\\\frac{1}{2\pi}\log|x-y|, n=2.\end{cases}

We know it has property

\displaystyle |D_{y_i}\Gamma(x-y)|\leq C\frac{1}{|x-y|^{n-1}}

We can define the Newtonianial potential of {f}

\displaystyle Nf(x)=\int_{\Omega}\Gamma(x-y)f(y)dy\quad x\in\mathbb{R}^n

{Nf} is well defined. We have the following propery of {Nf}

Thm: Suppose {\Omega\subset\mathbb{R}^n} is a bounded domain, {n\geq 2}. Assume {f} is bounded and local integrable in {\Omega}, then {Nf} is {C^1(\mathbb{R}^n)} and

\displaystyle D_{x_i}Nf=\int_{\Omega} D_{x_i}\Gamma(x-y)f(y)dy

Proof: Fix {0<\epsilon<1}, let {B_\epsilon(x)=\{y\in\Omega||x-y|<\epsilon\}} and {B_\epsilon^c(x)=\{y\in\Omega||x-y|\geq\epsilon\}}.

\displaystyle Nf=\int_{B_\epsilon(x)}\Gamma(x-y)f(y)dy+\int_{B^c_\epsilon(x)}\Gamma(x-y)f(y)dy=I+II

Since {D_{x_i}\Gamma(x-y)} is uniformly bounded in {B^c_\epsilon(x)}, by the Lebesgue dominating theorem, {II} is differentiable and

\displaystyle D_{x_i}II=\int_{B^c_\epsilon(x)}D_{x_i}\Gamma(x-y)f(y)dy

Considering {I}, we will prove if {|x-z|<\epsilon/2}

\displaystyle |I(x)-I(z)|\leq \begin{cases}C\epsilon|x-z|\quad\quad\text{ if } n>2\\ C\epsilon|\log \frac{\epsilon}{2}||x-z|\text{ if }n=2\end{cases}

where {C=C(||f||_{L^\infty},n,\Omega)}.

\displaystyle \int_{B_\epsilon(x)}\Gamma(x-y)f(y)dy-\int_{B_\epsilon(z)}\Gamma(z-y)f(y)dy=\int_{B_\epsilon(x)\cap B^c_\epsilon(z)}\Gamma(x-y)f(y)dy

\displaystyle +\int_{B_\epsilon(x)\cap B_\epsilon(z)}[\Gamma(x-y)-\Gamma(z-y)]f(y)dy-\int_{B^c_\epsilon(x)\cap B_\epsilon(z)}\Gamma(z-y)f(y)dy


For any {y\in B_\epsilon(x)\cap B^c_\epsilon(z)}, {|y-x|\geq |y-z|-|z-x|\geq \epsilon/2}. So

\displaystyle |\mathcal{A}|\leq C\frac{1}{\epsilon^{n-2}} |B_\epsilon(x)\cap B^c_\epsilon(z)|\leq C\frac{\epsilon^n-(\epsilon-|x-z|/2)^n}{\epsilon^{n-2}}\leq C\epsilon |x-z|\text{ if } n>2

\displaystyle |\mathcal{A}|\leq C|\log\frac{\epsilon}{2}| |B_\epsilon(x)\cap B^c_\epsilon(z)|\leq C\epsilon|\log\frac{\epsilon}{2}||x-z|\text{ if }n=2

Similarly for {C}, we also have

\displaystyle |\mathcal{C}|\leq\begin{cases}C\epsilon|x-z|\quad \quad \text{ if } n>2\\ C\epsilon|\log\frac{\epsilon}{2}||x-z|\text{ if }n=2\end{cases}

Considering {\mathcal{B}}, we have

\displaystyle |\mathcal{B}|\leq C\int_{B_\epsilon(x)\cap B_\epsilon(z)}\int_0^1|D_{x_i}\Gamma(tx+(1-t)z-y)||x-z|dtdy

\displaystyle =C|x-z|\int_0^1\int_{B_\epsilon(x)\cap B_\epsilon(z)}|D_{x_i}\Gamma(tx+(1-t)z-y)|dydt

\displaystyle \leq C|x-z|\int_0^1\int_{B_\epsilon(x)\cap B_\epsilon(z)} \frac{1}{|tx+(1-t)z-y|^{n-1}}dydt

\displaystyle \leq C|x-z|\int^1_0 C\epsilon dt\leq C\epsilon |x-z|

Combing the fact about {\mathcal{A},\mathcal{B},\mathcal{C}}, we get

\displaystyle |I(x)-I(z)|\leq \begin{cases}C\epsilon|x-z|\quad\quad\text{ if } n>2\\ C\epsilon|\log \frac{\epsilon}{2}||x-z|\text{ if }n=2\end{cases}


\displaystyle \left|\frac{Nf(x)-Nf(z)}{x-z}-\int_{\Omega}D_{x_i}\Gamma(x-y)f(y)dy\right|

\displaystyle \leq \left|\frac{II(x)-II(z)}{x-z}-\int_{\Omega}D_{x_i}\Gamma(x-y)f(y)dy\right|+\left|\frac{I(x)-I(z)}{x-z}\right|

Applying the fact that {II} is differentiable and the fact about {I}

\displaystyle \overline{\lim\limits_{z\rightarrow x}}\left|\frac{Nf(x)-Nf(z)}{x-z}-\int_{\Omega}D_{x_i}\Gamma(x-y)f(y)dy\right|

\displaystyle \leq \left|\int_{B_\epsilon(x)}D_{x_i}\Gamma(x-y)f(y)dy\right|+\begin{cases}C\epsilon\quad\quad\text{ if } n>2\\ C\epsilon|\log \frac{\epsilon}{2}|\text{ if }n=2\end{cases}

\displaystyle \leq C\epsilon+\begin{cases}C\epsilon\quad\quad\text{ if } n>2\\ C\epsilon|\log \frac{\epsilon}{2}|\text{ if }n=2\end{cases}

Let {\epsilon\rightarrow 0}, we get {D_{x_i}Nf} exists and

\displaystyle D_{x_i}Nf=\int_{\Omega}D_{x_i}\Gamma(x-y)f(y)dy

Next we shall prove {D_{x_i}Nf} is continuous for {i=1,\cdots,n}, then {Nf\in C^1(\mathbb{R}^n)}

\displaystyle \left|\int_{\Omega}D_{x_i}\Gamma(x-y)f(y)dy-\int_{\Omega}D_{x_i}\Gamma(z-y)f(y)dy\right|

\displaystyle \leq\left|\int_{B^c_\epsilon(x)}D_{x_i}\Gamma(x-y)f(y)dy-\int_{B^c_{\epsilon}(z)}D_{x_i}\Gamma(z-y)f(y)dy\right|

\displaystyle +\left|\int_{B_\epsilon(x)}D_{x_i}\Gamma(x-y)f(y)dy\right|+\left|\int_{B_\epsilon(z)}D_{x_i}\Gamma(z-y)f(y)dy\right|

By the dominating theorem

\displaystyle \lim\limits_{x\rightarrow z}\left|\int_{B^c_\epsilon(x)}D_{x_i}\Gamma(x-y)f(y)dy-\int_{B^c_{\epsilon}(z)}D_{x_i}\Gamma(z-y)f(y)dy\right|=0

\displaystyle \left|\int_{B_\epsilon(x)}D_{x_i}\Gamma(x-y)f(y)dy\right|\leq C\epsilon

\displaystyle \left|\int_{B_\epsilon(z)}D_{x_i}\Gamma(z-y)f(y)dy\right|\leq C\epsilon

where {C=C(||f||_{L^\infty}, n)}. So {\displaystyle \int_{\Omega}D_{x_i}\Gamma(x-y)f(y)dy=D_{x_i}Nf} is continuous everywhere.

Remark: This revised version also have errors. It is a shame. What do you mean {II} is differentiable? Don’t you notice that the integral domain of {II} also has relation with {x}?

To prove this theorem rigorously and neatly, we need the following lemma from calculus

Lemma: Suppose {u_n{x}} is a sequence of differntiable function on {[a,b]\rightarrow \mathbb{R}}. If {u_n} converges to another function {u} uniformly({u} is finite somewhere) and {u'_n} converges uniformly to {v} in {\Omega} then {u} is differentiable and {u'=v}.

Proof: Let {\xi(r):C^1(\mathbb{R}_+^1)\rightarrow \mathbb{R}^1} and {0\leq\eta'\leq 2}, \xi(r)=0\text{ if } r\leq 1, \xi(r)=1\text{ if }r>2.


{\displaystyle w_\epsilon=\int_{\Omega}\xi\left(\frac{|x-y|}{\epsilon}\right)\Gamma(x-y)f(y)dy}

Let {\xi_\epsilon(r)=\xi_{\epsilon}(r/\epsilon)} Since

\displaystyle \left|D_{x_i}\left(\xi_\epsilon\left(|x-y|\right)\Gamma(x-y)f(y)\right)\right|\leq \frac{C}{\epsilon^{n-1}}f(y)

which is in {L^1(\Omega)}. By the Lebesgue Differentiable theorem, {w_\epsilon} is differntiable and

\displaystyle D_{x_i}w_\epsilon=\int_{\Omega}D_{x_i}\left(\xi_\epsilon\left(|x-y|\right)\Gamma(x-y)\right)f(y)dy

\displaystyle \left|D_{x_i}w_\epsilon-\int_{\Omega}D_{x_i}\Gamma(x-y)f(y)dy\right|\leq \sup|f|\int_{\Omega}\left|D_{x_i}((1-\xi_\epsilon)\Gamma(x-y))\right|dy

\displaystyle \leq \sup|f|\int_{|x-y|\leq 2\epsilon}|D_{x_i}\Gamma|+\frac{2}{\epsilon}|\Gamma|dy

\displaystyle \leq \sup|f|\begin{cases}C\epsilon\quad \quad \text{ if }n>2\\ C\epsilon (1+|\log\epsilon|)\text{ if }n=2\end{cases}

So {D_{x_i}w_\epsilon} converges to {\displaystyle \int_{\Omega}D_{x_i}\Gamma(x-y)f(y)dy} uniformly on any compact subset of {\mathbb{R}^n}. And it is easy to prove {w_\epsilon} converges to {Nf} uniformly on any compact subset of {\mathbb{R}^n}. So by the lemma, we have {Nf} is differentiable and has

\displaystyle D_{x_i}Nf(x)=\int_{\Omega}D_{x_i}\Gamma(x-y)f(y)dy,\quad i=1,2\cdots,n


L^p estimate of lapacian operator

Refering to GT’s book corollary 9.10

\textbf{Thm:} \Omega is a bounded domain in \mathbb{R}^n. If u\in W^{2,p}(\Omega) and vanish on the boundary, then

(1) ||D^2u||_p\leq ||\Delta u||_p , 1<p<\infty.

where C=C(n,p). If p=2

(2) ||D^2u||_2= ||\Delta u||_2

The proof of this theorem is using the following fact, if u\in C^2(\mathbb{R}^n) has compact support, then

\displaystyle u(x)=\int_{\mathbb{R}^n}\Gamma(x-y)\Delta u(y)dy

\Gamma(x) is the fundamental solution of \Delta(Give a second thought why the RHS has compact support). This is exactly the Newtonian potential. By the result of Thm 9.9

||D^2u||_p\leq C||\Delta u||_p

is true for u\in C^2_0(\Omega). Since C^2_0(\Omega) is dense in W^{2,p}(\Omega) with zero boundary, then 9.10 is true.

Counterexample of regularity of newtonian potential

Define the Newtonial potential in a bounded domain \Omega by

\displaystyle N(x)=\int_{\Omega}\Gamma(x-y)f(y)dy

If f is bounded and integrable on \Omega, then N\in C^1(\mathbb{R}^n). And if f is bounded and locally holder continuous in \Omega then N\in C^2(\Omega) and \Delta N=f. But when f is only continuous, N is not necessarily secondly differentiable. Here is a counterexample.

\mathbf{Problem:} Let P be a homogeneous harmonic polynomial of degree 2. Suppose D^\alpha P\neq 0 for some multi-index \alpha=2, for instance P=x_1x_2, D_{12}P\neq 0. Choose a cut-off function \eta\in C^\infty_0(\{x||x|<2\}) with \eta=1 when |x|<1. Denote t_k=2^k, and let c_k\to 0 as k\to \infty and \sum c_k divergent. Define

\displaystyle f(x)=\sum\limits_0^\infty c_k\left(\Delta(\eta P)\right)(t_kx)

Prove f is a continuous function but \Delta u=f has no C^2 solution near the origin.

\mathbf{Proof:} If x=0, then f=0. If x\neq 0, then there exists only one k_0 such that 1\leq |t_{k_0}x|\leq 2. Since P is a harmonic polynomial and \eta has compact support,

f(x)=c_{k_0}\left(\Delta(\eta P)\right)(t_{k_0}x)\leq Const.c_{k_0}

So f\in C^\infty(\mathbb{R}^n\backslash \{0\}). As x\to 0, c_k\to 0, so f(x)\to 0, which means f is continuous at 0.

Let \displaystyle w(x)=\sum\limits_{0}^\infty t^{-2}_kc_k(\eta P)(t_kx), then it is easy to prove w is well defined and for x\neq 0 and the unique k_0 such that 1\leq |t_{k_0}x|\leq 2,

 \displaystyle D^\alpha w(x)=\sum\limits_0^{k_0}c_k+c_{k_0}D^\alpha(\eta P)(t_{k_0}x)\quad (1)

\Delta w(x)=c_{k_0}\left(\Delta(\eta P)\right)(t_{k_0}x)=f(x) \quad (2)

(1) means w\in C^2(\mathbb{R}^n\backslash\{0\}) but w\not\in C^2(\mathbb{R}^n), otherwise D^\alpha w(0)=\sum c_k which does not exist.

Suppose there exists u is a C^2 solution at B_{\epsilon}(0), (2) means that \Delta (u-w)=0 on B_{\epsilon}\backslash\{0\}. Since u-w is bounded, by the removable singularity theorem,  u-w is a harmonic function in B_\epsilon (0), thus an analytic function. However this means w=u-(u-w) is a C^2 function on B_\epsilon (0). Contradiction.

\text{Q.E.D}\hfill \square

\mathbf{Remark:} Gilbarg Trudinger’s book. Chapter 4. Exercise 4.9

\mathbf{Erratum:} u-w is not bounded in general, because w can not be proved bounded directly. But we have w(x)=o(\log r). In fact suppose 2^{-l}\leq|x|< 2^{-l+1}, some l\geq 1, then

\displaystyle w(x)=\sum_{k=0}^{l-1} c_k+c_l\frac{(\eta P)(t_lx)}{t_l^2}=\sum_{k=0}^{l-1} c_k+c_l\eta (t_lx)P(x)

Because c_k\to 0,

\displaystyle \left|\frac{w(x)}{\log |x|}\right|\leq \frac{|\sum_{k=0}^{l-1} c_k+c_l\eta (t_lx)P(x)|}{l\log 2}\to 0\text{ as }l\to \infty

So u-w=o(\log r) as r\to 0. Then one can use the Removable singularity theorem(Bochner), or the conclusion of 3.7 in GT’s book.