1.1. Newtonian Potentials

By | January 16, 2021

The goal of this note is to study solvability of second-order elliptic and parabolic equations in Hölder spaces. The prototype of elliptic equation is the Poisson equation

    \[   -\triangle u =f. \]

To understand the property of solution u, one of the easiest ways is to use Newtonian potential. In Section 1.1, we derive Newtonian potential and study some basic properties on this object. Also, we study a maximum principle for Poisson equation in Section 1.2. Such property plays a crucial role when we study second-order elliptic and parabolic equations.  Next, we introduce Hölder spaces which illustrate some smoothness of solutions in Section 1.3. After introducing Hölder spaces, in Section 1.4, we estimate a Hessian of Newtonian potential and a Hessian of solution for Poisson equation in Hölder space. This will lead us to think our main object of this note, so called Schauder estimate

To study properties of solution of the Poisson equation, we first seek a solution satisfying

    \[   -\triangle u =0.\]

A natural candidate for such solution is to assume that

    \[  u(x)=\phi(|x|), \]

i.e. radial solution. Write r=|x|. In spherical coordinate, the Laplacian is written as

    \[ \triangle u = \frac{\partial^2}{\partial r^2} u +\frac{n-1}{r}\frac{\partial u}{\partial r} + \frac{1}{r^2} \triangle_{\mathbb{S}^{n-1}}u,  \]

where \triangle_{\mathbb{S}^{n-1}} is the spherical Laplacian (or Laplace-Beltrami operator on unit sphere). Since we seek a radial solution, this implies that \phi should satisfy

    \[    \phi''(r)+\frac{n-1}{r} \phi'(r)=0.\]

Suppose that \phi'(r)\neq 0. Then

    \[ \log(|\phi'|)'=\frac{\phi''}{\phi'}=\frac{1-n}{r}  \]

and hence \phi'(r)=\frac{a}{r^{n-1}} for some constant a. Hence if r>0, then

    \[  \phi(r)=\begin{cases}  b\log r +c &\quad (n=2),\\  \frac{b}{r^{n-2}}+c &\quad (n\geq 3),  \end{cases}\]

where b and c are constants. This motivates us to introduce following one.

Definition 1.1. The function 

    \[  \Gamma(x)=\begin{cases}    \frac{1}{2\pi} \log |x| &\quad (n=2),\\    \frac{1}{(2-n)n \omega_n} \frac{1}{|x|^{n=2}}&\quad (n\geq 3),\end{cases}\]


defined for x\in \mathbb{R}^nx\neq 0, is the fundamental solution of Laplacian. Here \omega_n denotes the volume of the unit ball in \mathbb{R}^n

We wil use the following computation:

    \[ D_{i} \frac{1}{|y|^{n-2}} = -\frac{(n-2)y_i}{|y|^n}  \]

and 

    \begin{align*}    D_{j} \frac{y_i}{|y|^{n-2}}&=\frac{\delta_{ij}|y|^2-ny_i y_j}{|y|^{n+2}}.\end{align*}

From this, we see that \Gamma is harmonic in \mathbb{R}^n\setminus \{0\}
For f and g, we define the convolution of f and g by

    \[  (f*g)(x) = \int_{\mathbb{R}^n}  f(x-y)g(y) \myd{y}. \]

Theorem 1.1. Let f\in C_c^\infty(\mathbb{R}^n) and define 

    \[   u(x)=(\Gamma *f)(x).\]


Then 

    \[  u \in C^\infty(\mathbb{R}^n)\quad \text{and}\quad -\triangle u=f\quad \text{in } \mathbb{R}^n. \]

 

Proof. For simplicity, we prove the theorem when n\geq 3. The case n=2 can be similarly proved. We first show that u is well-defined.

Since f\in C_c^\infty(\mathbb{R}^n), there exists R>0 such that \supp f \subset B_R. A change of variable into polar coordinate gives

    \begin{align*}|u(x)|&\leq c_n \int_{\mathbb{R}^n} \frac{1}{|y|^{n-2}}|f(x-y)|\myd{y}\\&\leq c_n \norm{f}{\Leb{\infty}(\mathbb{R}^n)} \int_{B_{R+|x|}} \frac{1}{|y|^{n-2}}\myd{y}\\&=c_n \norm{f}{\Leb{\infty}(\mathbb{R}^n)} \int_0^{R+|x|} \rho \myd{\rho} \\&=c \norm{f}{\Leb{\infty}(\mathbb{R}^n)} (R+|x|)^2<\infty.\end{align*}

Hence u is well-defined.
Fix 1\leq j\leq n. For 0<|h|<R, we have 

    \[   \supp f(x+he_j-\cdot) \subset B_{2R+|x|}\]

since \supp f \subset B_R. So 

    \begin{align*}  \frac{u(x+he_j)-u(x)}{h}&=\int_{\mathbb{R}^n} \Gamma(y)\left[\frac{f(x+he_j-y)-f(x-y)}{h} \right] \myd{y}\\  &=\int_{B_{2R+|x|}} \Gamma(y)\left[\frac{f(x+he_j-y)-f(x-y)}{h} \right] \myd{y}.\end{align*}

By mean value property, it follows that 

    \begin{align*}  &\left|\int_{B_{2R+|x|}} \Gamma(y)\left[\frac{f(x+he_j-y)-f(x-y)}{h} \right] \myd{y} \right|\\  &\leq \int_{B_{2R+|x|}} \Gamma(y) \norm{D_j f}{\Leb{\infty}(\mathbb{R}^n)}\myd{y}\\  &\leq C(R,|x|) \norm{D_j f}{\Leb{\infty}(\mathbb{R}^n)}<\infty.\end{align*}

Since 

    \[   \lim_{h\rightarrow 0}\frac{f(x+he_j-y)-f(x-y)}{h}=D_j f(x-y)\quad \text{pointwise on }  B_{2R+|x|} \]

it follows from dominated convergence theorem that 

    \[    D_ju(x)=\int_{\mathbb{R}^{n}} \Gamma(y)D_j f(x-y)\myd{y}.   \]

By induction, one can show that u\in C^\infty(\mathbb{R}^n). It remains to show that -\triangle u=f
Fix \varepsilon>0. Then

    \begin{align*}D_{ij}u(x)&=\int_{\mathbb{R}^n} \Gamma(x-y) D_{ij}f(y)\myd{y}\\&=\int_{B_\varepsilon} \Gamma(y) D_{ij}f(x-y)\myd{y}+\int_{\mathbb{R}^n\setminus B_\varepsilon}  \Gamma(y) D_{ij}f(x-y)\myd{y}\\&=I_1+I_2.\end{align*}

Then

    \[   I_1 \leq C  \norm{D_{ij}f}{\Leb{\infty}(\mathbb{R}^n)}\int_{B_\varepsilon} \frac{1}{|y|^{n-2}} \myd{y} = C\norm{D_{ij}f}{\Leb{\infty}(\mathbb{R}^n)}\varepsilon^2. \]

Since f has compact support, integration by part gives

    \begin{align*}  I_2&=-\int_{\mathbb{R}^n\setminus B_\varepsilon} D_j\Gamma(y) D_i f(x-y)\myd{y}-\int_{\partial B_\varepsilon} \Gamma(y) D_i f(x-y) \left(\frac{y_j}{|y|}\right)\myd{\sigma(y)}\\  &=I_{21}+I_{22}.\end{align*}

A change of variable gives that

    \begin{align*}  I_{22}&=-\int_{\partial B_\varepsilon} \Gamma(y) D_i f(x-y) \left(\frac{y_j}{|y|}\right)\myd{y}\\  &\leq \frac{C\norm{\nabla f}{\Leb{\infty}(\mathbb{R}^n)}}{\varepsilon^{n-2}} \varepsilon^{n-1}\\  &=C \varepsilon.\end{align*}

Integration by part again gives

    \begin{align*} I_{21} &=\int_{\mathbb{R}^n\setminus B_\varepsilon} D_{ji}\Gamma(y) f(x-y)\myd{y} +\int_{\partial B_\varepsilon} D_j \Gamma(y) f(x-y) \left(\frac{y_i}{|y|} \right)\myd{\sigma(y)}.\end{align*}

From this calculation, we have

(1)   \begin{align*} \triangle u(x) &=\int_{B_\varepsilon} \Gamma(y)\triangle f(x-y)\myd{y}\\&\relphantom{=}+\int_{\mathbb{R}^n\setminus B_\varepsilon} \triangle \Gamma(y)f(x-y)\myd{y}+\int_{\partial B_\varepsilon} \left(\nabla \Gamma(y)\cdot \frac{y}{|y|}\right) f(x-y)\myd{\sigma(y)}\nonumber. \end{align*}

Since \Gamma is harmonic in \mathbb{R}^n\setminus B_\varepsilon, it follows that

    \[  \int_{\mathbb{R}^n\setminus B_\varepsilon} \triangle \Gamma(y)f(x-y)\myd{y}=0. \]

Since

    \[   \nabla \Gamma(y)\cdot \frac{y}{|y|} = \frac{1}{(2-n)n\omega_n} \nabla \left(\frac{1}{|y|^{n-2}}\right) \cdot \frac{y}{|y|} =-\frac{1}{n\omega_n} \frac{1}{|y|^{n-1}} , \]

it follows that

    \[  \int_{\partial B_\varepsilon} \left(\nabla \Gamma(y)\cdot \frac{y}{|y|}\right) f(x-y)\myd{\sigma(y)}=-\frac{1}{n\omega_n \varepsilon^{n-1}}\int_{\partial B_\varepsilon} f(x-y)\myd{\sigma(y)}. \]

  Since f is continuous and \sigma(\partial B_\varepsilon)=n\omega_n \varepsilon^{n-1}, it follows that   

    \[  \lim_{\varepsilon \rightarrow 0+} \int_{\partial B_\varepsilon} \left(\nabla \Gamma(y)\cdot \frac{y}{|y|}\right) f(x-y)\myd{\sigma(y)}=-f(x).\]

  Hence letting \varepsilon \rightarrow 0+ in (1), we conclude that   

    \[  \triangle u(x)=-f(x) \]

  This completes the proof. 

Definition 1.2. Let f\in C_c^\infty(\mathbb{R}^n). The function \mathcal{N}f defined by  

    \[   \mathcal{N}f(x)=(\Gamma *f)(x) \]

  is called the Newtonian potential of f.

Remark. Here we obtain another calculation related to some fundamental studies on Poisson equation. Following the above argument, we have

    \begin{align*}D_{ij}\mathcal{N}f(x)&=\int_{\mathbb{R}^n} \Gamma(y)D_{ij}f(x-y)\myd{y}\\&=-\int_{\mathbb{R}^n} D_j \Gamma (y) D_i f(x-y)\myd{y}.\end{align*}

For simplicity, we set x=0. Choose a radial function \zeta \in C^\infty_c(\mathbb{R}^n) such that \zeta=1 near at the origin. Then

    \begin{align*}  D_{ij}\mathcal{N}f(0)=\int_{\mathbb{R}^n} D_i \Gamma(y)D_j\left(f(-y)-f(0)\zeta(y)\right)\myd{y}+f(0)C_{ij},\end{align*}

where

    \[  C_{ij}=\int_{\mathbb{R}^n} D_i \Gamma(y) D_j \zeta(y)\myd{y}.  \]

We write

    \[   K_{ij}(y)=-D_{ij}\Gamma(y) =\frac{1}{n\omega_n} \left(\frac{n y_i y_j-\delta_{ij}|y|^2}{|y|^{n+2}} \right).   \]

Note that

    \[   |K_{ij}(y)|\leq \frac{C}{|y|^n}\quad \text{and}\quad |f(y)-f(0)\zeta(y)|=|f(y)-f(0)+f(0)(1-\zeta(y))|\leq C|y|.  \]

Since f(y)-f(0)\zeta(y) has compact support, integration by part gives

    \[  D_{ij}\mathcal{N}f(0)=\lim_{r\rightarrow 0+}\int_{|y|>r} K_{ij}(y) \left(f(-y)-f(0)\zeta(y)\right)\myd{y}+f(0)C_{ij}. \]

For any r>0, we have

    \[    \int_{|y|> r} K_{ij}(y)\zeta(y)\myd{y}=0. \]

A change of variable gives

    \begin{align*}\int_{|y|> r} K_{ij}(y)\zeta(y)\myd{y}&=\int_r^\infty \int_{\mathbb{S}^{n-1}} K_{ij}(\rho\omega)\myd{\sigma(\omega)} \zeta(\rho) \rho^{n-1}\myd{\rho}.\end{align*}

Hence it suffices to show that

    \[   \int_{\mathbb{S}^{n-1}} K_{ij}(\omega)\myd{\sigma(\omega)}=0.\]

If i\neq j, the integral is zero since K_{ij} is antisymmetric. If i=j, then a change of variable and harmonicity of \Gamma give

    \[     n\int_{\mathbb{S}^{n-1}}K_{ii}(\omega)\myd{\sigma(\omega)}=\int_{\mathbb{S}^{n-1}} K_{11}(\omega)\myd{\sigma(\omega)}+\cdots+\int_{\mathbb{S}^{n-1}} K_{nn}(\omega)\myd{\sigma(\omega)} =0 \]

for 1\leq i\leq n. Integration by part gives

    \begin{align*}  C_{ij}&=\int_{B_\varepsilon} D_i \Gamma(y) D_j \zeta (y)\myd{y}+\int_{\mathbb{R}^n\setminus B_\varepsilon} D_i \Gamma(y)D_j \zeta(y)\myd{y}\\  &=-\int_{\mathbb{R}^n\setminus B_\varepsilon}  D_{ij} \Gamma(y)  \zeta (y) \myd{y}-\int_{\partial B_\varepsilon} D_i \Gamma(y)\zeta(y)\frac{y_j}{|y|}\myd{\sigma(y)}\\  &=-\int_{\mathbb{R}^n\setminus B_\varepsilon}  D_{ij} \Gamma(y)  \zeta (y) \myd{y}-\frac{1}{n\omega_n}\int_{\partial B_\varepsilon} \frac{y_i y_j}{|y|^{n+1}} \zeta(y)\myd{\sigma(y)}\\  &=-\int_{\mathbb{R}^n\setminus B_\varepsilon}  D_{ij} \Gamma(y)  \zeta (y) \myd{y}-\frac{1}{n\omega_n \varepsilon^{n+1}} \int_{\partial B_\varepsilon} y_i y_j \zeta(y)\myd{\sigma(y)}.\end{align*}

If i\neq j, the quantity C_{ij} is zero due to antisymmetric. If i=j, then a change of coordinate and harmonicity of \Gamma give

    \[n C_{11}=C_{11}+C_{22}+\cdots + C_{nn}=-\frac{1}{n\omega_n \varepsilon^{n-1}}\int_{\partial B_\varepsilon} \zeta(y)\myd{\sigma(y)}.  \]

Letting \varepsilon \rightarrow 0+, we conclude that

    \[ C_{ii}=- \frac{1}{n}\quad \text{for all } 1\leq i\leq n.  \]

This shows that

(2)   \begin{equation*}    D_{ij} \mathcal{N}f(x)=\lim_{r\rightarrow 0+} \int_{|y|>r} K_{ij}(y)f(x-y)\myd{y} -\frac{1}{n}\delta_{ij}f(x) \end{equation*}

for f\in C_{c}^\infty(\mathbb{R}^n).
Now define

    \[   \mathcal{K}_{ij}f(x)=\lim_{r\rightarrow 0+}\int_{|y|>r} K_{ij}(y)f(x-y)\myd{y}.  \]

Note that the kernel K_{ij} satisfies

    \[  |K_{ij}(y)|\leq \frac{C}{|y|^n}\quad \text{and}\quad  \int_{\mathbb{S}^{n-1}} K_{ij}\myd{\sigma}=0. \]

We call such kernel as Calderón-Zygmund singular kernel. Then  it follows from the well-known theory on singular integral operator that for 1<p<\infty, there exists a constant C=C(n,p)>0 such that

    \[   \norm{\mathcal{K}_{ij}f}{\Leb{p}(\mathbb{R}^n)}\leq C \norm{f}{\Leb{p}(\mathbb{R}^n)}\]

for all f\in C_c^\infty(\mathbb{R}^n). Hence by (2), we conclude that

    \[  \norm{D_{ij}\mathcal{N}f}{\Leb{p}(\mathbb{R}^n)}\leq C \norm{f}{\Leb{p}(\mathbb{R}^n)}. \]

We call such estimate as Calderón-Zygmund type estimate. This estimate plays a crucial roles when we study the Sobolev space theory for partial differential equations. Due to our purpose of this note, we will not go to any futher. We end this section by introducing several textbooks. When one has some interests on general theory on singular integrals, see Stein (1970). Modern theory of this theory can be found in Stein (1993). An application for Calderón-Zygmund estimate can be found in Gilbarg-Trudinger (1998) and Krylov (2008)

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