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

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

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

${=\mathcal{A}+\mathcal{B}+\mathcal{C}}$

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

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

So

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

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

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 ${0\leq\eta'\leq 2}$, $\xi(r)=0\text{ if } r\leq 1$, $\xi(r)=1\text{ if }r>2$.

Define

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

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.

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.