📂Partial Differential Equations

Mollifiers

Definition¹

Let’s define the function $\eta \in C^{\infty}(\mathbb{R}^{n})$ as follows.

$$ \begin{equation} \eta (x) := \begin{cases} C \exp \left( \dfrac{1}{|x|^2-1} \right) & |x|<1 \\ 0 & |x| \ge 1\end{cases} \label{1} \end{equation} $$

Such $\eta$ is called a mollifier. In particular, when $C>0$ is a constant satisfying $\displaystyle \int_{\mathbb{R}^{n}} \eta dx=1$, $\eta$ is called a standard mollifier.

Let’s define $\eta_{\epsilon}$ for $\epsilon>0$ as follows.

$$ \eta_\epsilon (x) := \dfrac{1}{\epsilon^n}\eta\left( \dfrac{x}{\epsilon} \right) $$

If $\eta$ is a standard mollifier, then $\int_{\mathbb{R^n}} \eta_{\epsilon} dx=1$ is satisfied. This can be easily verified through a change of variables.

Description

A mollifier serves to smooth out functions that are not smooth through convolution. The word mollify means ’to soothe’, which expresses that $\eta$ approximates a function that is not differentiable to be differentiable, hence the term mollify.

Meanwhile, one may consider that the form of a mollifier does not necessarily have to be like $\eqref{def}$.

Generalization

If $\varphi \in C^{\infty}(\mathbb{R}^{n})$ satisfies the following three conditions, $\varphi$ is called a mollifier.

• Has a compact support.
• $\displaystyle \int_{\mathbb{R^n}} \varphi dx =1$
• $\lim \limits_{\epsilon \rightarrow 0}\varphi_\epsilon = \lim \limits_{\epsilon \rightarrow 0} \dfrac{1}{\epsilon^n}\varphi \left( \dfrac{x}{\epsilon} \right) =\delta (x)$ Here, $\delta (x)$ is the Dirac delta function.

The following condition is called a positive mollifier.

• $\varphi (x) \ge 0, \quad \forall x \in \mathbb{R^n}$

The following condition is called a symmetric mollifier.

• For $\mu \in C^{\infty}$, $\varphi (x)=\mu (|x|)$

Meanwhile, it can be proved that $\eta$ defined by $\eqref{1}$ is actually a smooth function as follows.

Proof

First, let’s define $f$ and $g$ as follows.

$$ \begin{align*} f(s) &:= \begin{cases} C \exp\left( \dfrac{1}{s-1} \right) & s<1 \\ 0 & s \ge 1 \end{cases} \\g(x) &:= |x|^2={x_{1}}^2+{x_{2}}^2+\cdots+{x_{n}}^2 , \quad x\in \mathbb{R}^{n} \end{align*} $$

Then, it aims to show $f\in C^\infty$ and $g\in C^\infty$, ultimately proving $\eta=f \circ g \in C^\infty$.

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• Part 1. $f \in C^\infty$

  First, by calculating the derivatives of $f$ at $(-\infty, 1)$, one finds as follows.

  $$ \begin{align*} f^{\prime}(s) =&\ Ce ^{ \frac{1}{s-1}} \frac{-1}{(s-1)^2} \\ f^{\prime \prime}(s) =&\ Ce ^{ \frac{1}{s-1}} \frac{-1}{(s-1)^4} + Ce ^{ \frac{1}{s-1}} \frac{2}{(s-1)^3} \\ f^{\prime \prime \prime}(s) =&\ Ce ^{ \frac{1}{s-1}} \frac{-1}{(s-1)^6} +Ce ^{ \frac{1}{s-1}} \frac{2}{(s-1)^5}+Ce ^{ \frac{1}{s-1}} \frac{-6}{(s-1)^4} \end{align*} $$

  Henceforth, it aims to prove via induction that for all $k \ge 1$, there exists a constant $a^{(k)}_{0},\ a^{(k)}_{1},\ \cdots,\ a^{(k)}_{2k}$ satisfying the formula below.

  $$ f^{(k)}(s) = \begin{cases} e^{\frac{1}{s-1}} \left( a^{(k)}_{0} + \frac{a^{(k)}_{1}}{s-1}+\cdots+\frac{a^{(k)}_{2k}}{(s-1)^{2k}} \right) & s<1 \\ 0 & s \ge 1 \end{cases} $$

  • Part 1-1. $k=1$

    $$ f^{\prime}(s) = \begin{cases} e^{\frac{1}{s-1} }\frac{-C}{(s-1)^2} & s<1 \\ 0 & s >1\end{cases} $$

    Therefore, $a^{(1)}_{0} =a^{(1)}_{1}=0$, $a^{(1)}_2=-C$ is established. And the following holds.

    $$ \lim \limits_{h \rightarrow 0^+} \dfrac{ f(1+h)-f(1) }{h}=\lim \limits_{h \rightarrow 0^+} \dfrac{Ce^{ \frac{1}{h} } }{h}=\lim \limits_{h \rightarrow 0^+} \dfrac{C\frac{1}{h}}{e^{-\frac{1}{h} } }=0 $$

    $$ \lim \limits_{h \rightarrow 0^-} \dfrac{f(1+h)-f(1) }{h} = \lim \limits_{h \rightarrow 0^-} \dfrac{Ce^{ \frac{1}{h} } }{h}=\lim \limits_{h \rightarrow 0^-} \dfrac{C\frac{1}{h}}{e^{-\frac{1}{h} } }=0 $$

    The limit $0$ can be easily attained using L’Hôpital’s rule. Thus results in $f^{\prime}(1)=0$. Hence, it is valid when $k=1$.

  • Part 1-2. Assuming it holds for an arbitrary $k$, it also holds for $k+1$.

    Assume it holds for any given $k$. Then, there exists a constant $a^{(k)}_{0},\ a^{(k)}_{1},\ \cdots,\ a^{(k)}_{2k}$ which satisfies the following.

    $$ f^{(k)}(s) = \begin{cases} e^{\frac{1}{s-1}} \left( a^{(k)}_{0} + \frac{a^{(k)}_{1}}{s-1}+\cdots+\frac{a^{(k)}_{2k}}{(s-1)^{2k}} \right) & s<1 \\ 0 & s \ge 1 \end{cases} $$

    When calculating $f^{(k+1)}(s)$ at $(-\infty, 1)$, one finds as follows.

    $$ \begin{align*} f^{(k+1)}(s) =&\ e^{\frac{1}{s-1}} \dfrac{-1}{(s-1)^2}\left( a^{(k)}_{0} + \frac{a^{(k)}_{1}}{s-1}+\cdots+\frac{a^{(k)}_{2k}}{(s-1)^{2k}} \right) + e^{\frac{1}{s-1}}\left( \frac{-a^{(k)}_{1}}{(s-1)^2}+\cdots+\frac{-2k a^{(k)}_{2k}}{(s-1)^{2k+1}} \right) \\ =&\ e^{\frac{1}{s-1}} \left( a^{(k+1)}_{0} + \frac{a^{(k+1)}_{1}}{s-1}+\cdots+\frac{a^{(k+1)}_{2k+2}}{(s-1)^{2k+2}} \right) \end{align*} $$

    Here, each $a^{(k+1)}_{0},\ \cdots,\ a^{(k+1)}_{2k+2}$ is determined by $a^{(k)}_{0},\ \cdots,\ a^{(k)}_{2k}$. Also, $(1, \infty)$ yields $f^{(k+1)}(s)=0$. And since the below formula is satisfied, it implies $f^{(k+1)}=0$.

    $$ \begin{align*} \lim \limits_{h \rightarrow 0^+} \dfrac{ f(1+h)-f(1)}{h}=&\ \lim \limits_{h \rightarrow 0^+} \dfrac{e^{\frac{1}{s-1}} \left( a^{(k)}_{0} + \frac{a^{(k)}_{1}}{h}+\cdots+\frac{a^{(k)}_{2k}}{(h)^{2k}} \right) }{h} \\ =&\ \lim \limits_{h \rightarrow 0^+} \dfrac{ \left( \frac{a^{(k)}_{0}}{h} + \frac{a^{(k)}_{1}}{h^2}+\cdots+\frac{a^{(k)}_{2k}}{(h^{2k+1})^{2k}} \right) }{e^{-\frac{1}{h}} } =0 \end{align*} $$

    $$ \begin{align*} \lim \limits_{h \rightarrow 0^-} \dfrac{ f(1+h)-f(1)}{h}=&\ \lim \limits_{h \rightarrow 0^-} \dfrac{e^{\frac{1}{s-1}} \left( a^{(k)}_{0} + \frac{a^{(k)}_{1}}{h}+\cdots+\frac{a^{(k)}_{2k}}{(h)^{2k}} \right) }{h} \\ =&\ \lim \limits_{h \rightarrow 0^-} \dfrac{ \left( \frac{a^{(k)}_{0}}{h} + \frac{a^{(k)}_{1}}{h^2}+\cdots+\frac{a^{(k)}_{2k}}{(h^{2k+1})^{2k}} \right) }{e^{-\frac{1}{h}} }=0 \end{align*} $$

    Repeated application of L’Hôpital’s rule ultimately yields $0$. Thus, by mathematical induction, $f \in C^\infty(\mathbb{R})$ is proven.

• Part 2. $g \in C^\infty( \mathbb{R^n})$

  This is trivial by the definition of $g$.

Therefore, by Part 1., Part 2., the following holds.

$$ \eta = f \circ g \in C^\infty ( \mathbb{R^n} ) $$

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