Solution to Partial Differential Equations: Methods and Applications (Robert McOwen) Section 4.5

1. Let $K(x)=-(4\pi|x|)^{-1}$, $f\in C(\mathbb R^3)$ with $f(x)=O(|x|^{-2-\epsilon})$ as $|x|\to\infty$, and $\displaystyle u(x)=\int_{\mathbb R^3}\!K(x-y)f(y)\,\mathrm dy$.
   1. Show $u\in C^1(\mathbb R^3)$ and $\nabla u(x)=O(|x|^{-1-\epsilon})$ as $|x|\to\infty$.
   2. If additionally $f\in C^1(\mathbb R^3)$, show $u\in C^2(\mathbb R^3)$.
Solution
1. Since $f(x)=O(|x|^{-2-\epsilon})$ as $|x|\to\infty$, there exist positive constants $C$ and $R$ such that $|f(y)|\leq C|y|^{-2-\epsilon}$ for $|y|\geq R$, which implies $\displaystyle M=\max_{y\in\mathbb R^3}|f(y)|$ is finite. For any $x\in\mathbb R^3$, we set $R'=\max\{R,2|x|\}$ and pick $\rho>0$ such that $B_\rho(x)\subseteq B_{R'}(0)$. Then $u$ can be represented as

   $\displaystyle u(x)=\int_{B_{R'}(0)-B_\rho(x)}\!K(x-y)f(y)\,\mathrm dy+\int_{B_\rho(x)}\!K(x-y)f(y)\,\mathrm dy+\int_{\mathbb R^3-B_{R'}(0)}\!K(x-y)f(y)\,\mathrm dy$.

   The first integral is proper so it is well-defined. The second integral can be estimated by

   $\begin{aligned}\left|\int_{B_\rho(x)}\!K(x-y)f(y)\,\mathrm dy\right|&\leq \frac{M}{4\pi}\int_{B_\rho(x)}\!\frac{\mathrm dy}{|x-y|}=\frac M{4\pi}\int_0^\pi\!\int_0^{2\pi}\!\int_0^\epsilon\!\frac{\rho^2\sin\phi}{\rho}\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\phi\\&=\frac{M}{4\pi}\cdot2\cdot2\pi\cdot\frac{\epsilon^2}2=\frac{M\epsilon^2}2.\end{aligned}$

   The third integral can be estimated by

   $\begin{aligned}\left|\int_{\mathbb R^3-B_{R'}(0)}\!K(x-y)f(y)\,\mathrm dy\right|&\leq\frac{C}{4\pi}\int_{\mathbb R^3-B_{R'}(0)}\!\frac{|y|^{-2-\epsilon}}{|x-y|^2}\,\mathrm dy\leq\frac C{2\pi}\int_{\mathbb R^3-B_{R'}(0)}\!|y|^{-3-\epsilon}\,\mathrm dy\\&=\frac C{2\pi}\int_0^\pi\!\int_0^{2\pi}\!\int_{R'}^\infty\!\rho^{-3-\epsilon}\cdot\rho^2\sin\phi\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\phi\\&=\left.\frac{C}{2\pi}\cdot2\cdot2\pi\cdot\frac{\rho^{-\epsilon}}{-\epsilon}\right|_{R'}^\infty=\frac{2C(R')^{-\epsilon}}{\epsilon}.\end{aligned}$

   Here we have used the inequality $|x-y|\geq|y|/2$ for $y\in\mathbb R^3-B_{R'}(0)$. Therefore, the Newtonian potential $u$ of $f$ is well-defined for all $x\in\mathbb R^3$.
   For any $i=1,\dots,n$, let

   $\displaystyle u_{(i)}(x)=\int_{\mathbb R^3}\!\frac{\partial K}{\partial x_i}(x-y)f(y)\,\mathrm dy=\frac1{4\pi}\int_{\mathbb R^3}\frac{x_i-y_i}{|x-y|^3}f(y)\,\mathrm dy,$

   which is well-defined because $\displaystyle\frac{\partial K}{\partial x_i}$ is $O(|x-y|^{-2}$ as $|x-y|\to0$ and $\displaystyle\frac{\partial K}{\partial x_i}(x-y)f(y)$ is $O(|y|^{-4-\epsilon})$ as $|y|\to\infty$ (as for the above estimate of $u$). To show $\displaystyle u_{(i)}=\frac{\partial u}{\partial x_i}$, we first approximate $u$ by a smooth function as follows. Let $\eta\in C^\infty(\mathbb R)$ satisfy $\eta(t)=0$ for $t<1$, $\eta(t)=1$ for $t>2$ and $0\leq\eta(t)\leq1$ and $0\leq\eta'(t)\leq2$ for all $t$. Then we define

   $\begin{aligned}u_\delta(x)&=\int_{\mathbb R^3}\!K_\delta(x-y)f(y)\,\mathrm dy:=\int_{\mathbb R^3}\!K(x-y)\eta\left(\frac{|x-y|}\delta\right)f(y)\,\mathrm dy\\&=-\frac1{4\pi}\int_{\mathbb R^3}\!\frac{\eta(|x-y|/\delta)}{|x-y|}f(y)\,\mathrm dy\quad\text{for}~x\in\mathbb R^3,\end{aligned}$

   where $\displaystyle K_\delta(z)=-\frac{\eta(|z|/\delta)}{|z|}$ for $\delta>0$ and $z\in\mathbb R^3$. Note that $K_\delta$ is smooth so $u_\delta$ is also smooth and satisfies

   $\begin{aligned}\frac{\partial u_\delta}{\partial x_i}(x)=\frac1{4\pi}\int_{\mathbb R^3}\!\left[\frac{x_i-y_i}{|x-y|^3}\eta\left(\frac{|x-y|}{\delta}\right)+\frac{x_i-y_i}{\delta|x-y|}\eta'\left(\frac{|x-y|}{\delta}\right)\right]f(y)\,\mathrm dy\end{aligned}$

   It is straightforward to check that $u_\delta\to u$ uniformly on any compact subsets of $\mathbb R^3$, which means $u$ is continuous in $\mathbb R^3$. Now we prove $\displaystyle\frac{\partial u_\delta}{\partial x_i}\to u_{(i)}$ as $\delta\to0$. Since $\eta(t)=1$ for $t>2$, we observe that

   $\begin{aligned}u_{(i)}(x)-\frac{\partial u_\delta}{\partial x_i}(x)&=\frac1{4\pi}\int_{|x-y|<2\delta}\!\left\{\left[1-\eta\left(\frac{|x-y|}{\delta}\right)\right]\frac{x_i-y_i}{|x-y|^3}-\frac{x_i-y_i}{\delta|x-y|}\eta'\left(\frac{|x-y|}{\delta}\right)\right\}f(y)\,\mathrm dy\\&=\frac1{4\pi}\int_{|z|<2\delta}\left\{\left[1-\eta\left(\frac{|z|}\delta\right)\right]\frac{z_i}{|z|^3}-\frac{z_i}{\delta|z|}\eta'\left(\frac{|z|}\delta\right)\right\}f(x-z)\,\mathrm dz.\end{aligned}$

   It is easy to estimate the following integrals

   $\begin{aligned} &\left|\frac1{4\pi}\int_{|z|<2\delta}\!\frac{z_i}{\delta|z|}\eta'\left(\frac{|z|}\delta\right)f(x-z)\,\mathrm dz\right|\leq\frac{2M}{4\pi\delta}\int_{|z|<2\delta}\mathrm dz=\frac{2M}{4\pi\delta}\cdot\frac{4\pi(2\delta)^3}3=\frac{16M\delta^2}3,\\&\begin{aligned}\left|\frac1{4\pi}\int_{|z|<2\delta}\left[1-\eta\left(\frac{|z|}{\delta}\right)\right]\frac{z_i}{|z|^3}f(x-z)\,\mathrm dz\right|&\leq\frac{M}{2\pi}\int_{|z|<2\delta}\!\frac{dz}{|z|^2}=\frac{M}{2\pi}\int_0^\pi\!\int_0^{2\pi}\int_0^\delta\!\frac{\rho^2\sin\phi}{\rho^2}\,\mathrm d\rho\,\mathrm d\theta\,\mathrm d\phi\\&=\frac{M}{2\pi}\cdot2\cdot2\pi\cdot\delta=2M\delta.\end{aligned}\end{aligned}$

   Thus, we see that

   $\displaystyle\left|u_{(i)}(x)-\frac{\partial u_\delta}{\partial x_i}(x)\right|\leq \left(\frac{16\delta}3+2\right)M\delta\quad\text{for}~x\in\mathbb R^3,$

   which implies $u_{(i)}$ is the uniform limit of the continuous functions $\displaystyle\frac{\partial u_\delta}{\partial x_i}$ as $\delta$ tends to zero. Recall that Theorem 7.12 of "Introduction to Analysis (Fourth Edition)" states that "Let $(a,b)$ be a bounded interval and suppose that $f_n$ is a sequence of functions which converges at some $x_0\in(a,b)$. If each $f_n$ is differentiable on $(a,b)$ and $f_n'$ converges uniformly on $(a,b)$ as $n\to\infty$, then $f_n$ converges uniformly on $(a,b)$ and $\displaystyle\lim_{n\to\infty}f_n'(x)=\left(\lim_{n\to\infty}f_n(x)\right)'$". Therefore we show that $u\in C^1(\mathbb R^3)$ and

   $\displaystyle\frac{\partial u}{\partial x_i}(x)=u_{(i)}(x)=\frac1{4\pi}\int_{\mathbb R^3}\!\frac{x_i-y_i}{|x-y|^3}f(y)\,\mathrm dy$.

   Write $x=|x|\omega$ for $\omega\in\mathbb S^2$ and use the change of variable $y=|x|z$, we find

   $\begin{aligned}|\nabla u(x)|&\leq\frac1{4\pi}\int_{\mathbb R^3}\!\frac{|f(y)|}{|x-y|^2}\,\mathrm dy=\frac1{4\pi}\int_{\mathbb R^3}\!\frac{f(|x|z)}{|x-|x|z|^2}|x|^3\,\mathrm dz\\&=\frac{|x|}{4\pi}\int_{\mathbb R^3}\!\frac{f(|x|z)}{|\omega-z|^2}\,\mathrm dz=\frac{|x|}{4\pi}\left(\int_{B_{R/|x|}(0)}\!\frac{f(|x|z)}{|\omega-z|^2}\,\mathrm dz+\int_{\mathbb R^3-B_{R/|x|}(0)}\!\frac{f(|x|z)}{|\omega-z|^2}\,\mathrm dz\right).\end{aligned}$

   When $|x|\geq2R$, the first integral can be estimated by

   $\displaystyle\int_{B_{R/|x|}(0)}\!\frac{f(|x|z)}{|\omega-z|^2}\,\mathrm dz\leq4M\int_{B_{R/|x|}(0)}\mathrm dz=4M\cdot\frac{4\pi(R/|x|)^3}3=\frac{16M\pi R^3}{3|x|^3}.$

   On the other hand, the second integral is estimated by

   $\begin{aligned}\int_{\mathbb R^3-B_{R/|x|}(0)}\!\frac{f(|x|z)}{|\omega-z|^2}\,\mathrm dz&\leq C\int_{\mathbb R^3-B_{R/|x|}(0)}\frac{|x|^{-2-\epsilon}|z|^{-2-\epsilon}}{|\omega-z|^2}\,\mathrm dz\\&=C|x|^{-2-\epsilon}\int_{\mathbb R^3-B_{R/|x|}(0)}\!\frac{|z|^{-2-\epsilon}}{|\omega-z|^2}\,\mathrm dz\leq\tilde C|x|^{-2-\epsilon}.\end{aligned}$

   Here we have used the fact that improer integral is convergent because $\displaystyle\frac{|z|^{-2-\epsilon}}{|\omega-z|^2}\bigg/\frac1{|z|^{4+\epsilon}}\to1$ as $|z|\to\infty$ and is $O(|\omega-z|^{-2})$ as $z\to\omega$. Therefore, we obtain

   $\displaystyle|\nabla u(x)|\leq\frac{|x|}{4\pi}\left(\frac{16M\pi R^3}{3|x|^3}+\tilde C|x|^{-2-\epsilon}\right)\leq\hat C|x|^{-1-\epsilon}.$

   
   The original statement is that "Show that $u\in C^1(\mathbb R^n)$ and $\nabla u(x)=O(|x|^{-2})$", which might be typos so I correct it directly. As my best knowledge, $\nabla u= O(|x|^{-1-\epsilon})$ is more reasonable because $u=O(|x|^{-\epsilon})$ as $|x|\to\infty$, which is shown in two papers: "On Helmholtz's theorem and the completeness of the Papkovich-Neuber stress functions for infinite domains" (Author: M. E. Gurtin) and "A Constructive Proof of Helmholtz’s Theorem" (Author: Bernardo De La Calle Ysern, José C. Sabina De Lis). If you have a better proof to improve this decaying rate, please tell me! Thank you.
2. Note that

   $\displaystyle\frac{\partial u}{\partial x_i}=\frac1{4\pi}\int_{\mathbb R^3}\!\frac{x_i-y_i}{|x-y|}f(y)\,\mathrm dy=\frac1{4\pi}\int_{\mathbb R^3}\!\frac{y_i}{|y|}f(x-y)\,\mathrm dy$.

   Then it is clear that

   $\displaystyle\frac{\partial^2u}{\partial x_j\partial x_i}=\frac1{4\pi}\int_{\mathbb R^3}\!\frac{y_i}{|y|^3}\frac{\partial f}{\partial x_j}(x-y)\,\mathrm dy,$

   which is continuous in $\mathbb R^3$. Thus, $u\in C^2(\mathbb R^3)$.