PDE — Homework 3

Problem 1.

证明 $e^{-x^{2}}$ 是 $\mathscr{S}$ 函数.

证明.

(i) $e^{-x^2}\in C^{\infty}$
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\frac{d(e^{-x^2})}{dx}=-2xe^{-x^2}, \frac{d^2(e^{-x^2})}{dx^2}=(4x^2-2)e^{-x^2}

由求导的乘积法则我们可以知道：

\frac{d^n(e^{-x^2})}{dx^n}=P_n(x)e^{-x^2}, \deg(P_n)=n

故 $e^{-x^2}$ 的任意阶导数存在，即 $e^{-x^2}\in C^{\infty}$ .

(ii) 对 $\forall\alpha,\beta$ 有： $\sup_{x\in\mathbb{R}}|x^\alpha D^\beta f(x)|<c(\alpha,\beta)$
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\Rightarrow x^{\alpha}D^\beta e^{-x^2}=x^{\alpha}P_{\beta}(x)e^{-x^2}=P_{\beta+\alpha}(x)e^{-x^2}=\sum_{k=0}^{\alpha+\beta}p_k x^k e^{-x^2}

而对于 $x^k e^{-x^2}$ ，我们可知：

\frac{d(x^k e^{-x^2})}{dx}=(k x^{k-1}-2x^{k+1})e^{-x^2}=0\Rightarrow x=\pm\sqrt{\frac{k}{2}}\Rightarrow |x^k e^{-x^2}|\leq(\frac{k}{2})^k e^{-\frac{k}{2}}=M_k.

则

|x^{\alpha}D^\beta e^{-x^2}|=\left|\sum_{k=0}^{\alpha+\beta}p_k x^k e^{-x^2}\right|\leq\sum_{k=0}^{\alpha+\beta}|p_k x^k e^{-x^2}|\leq\sum_{k=0}^{\alpha+\beta}M_k|p_k|=c(\alpha,\beta)<\infty

故

\sup_{x\in\mathbb{R}}|x^\alpha D^\beta f(x)|<c(\alpha,\beta)

综上我们可知： $e^{-x^2}$ 是 $\mathscr{S}$ 函数.

∎

Problem 2.

设 $P(\xi)$ ， $Q(\xi)$ 皆为常系数多项式，证明以下各个命题等价：

(1) $\varphi(x) \in \mathscr{S}$ ;
(2) 对任意 $P(\xi)$ ， $Q(\xi)$ ， $P(x)Q(D)\varphi \in \mathscr{S}$ ;
(3) 对任意 $P(\xi)$ ， $Q(\xi)$ ， $Q(D)[P(x)\varphi(x)] \in \mathscr{S}$ .

证明.

由于 $P(\xi)$ ， $Q(\xi)$ 皆为常系数多项式，故我们可设 $P(x)=\sum_{k=0}^{m}a_k x^k$ ， $Q(x)=\sum_{k=0}^{n}b_k x^k$ .

(I) (1) $\Rightarrow$ (2)
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|P(x)Q(D)\varphi(x)|=\left|\sum_{k=0}^{m}a_k x^k\psi(x)\right|=\sum_{k=0}^{m}|a_k|\cdot|x^k\psi(x)|

而由于 $\psi(x)\in\mathscr{S}$ ，故 $\exists m_k>0$ ，使得 $|x^k\psi(x)|<m_k$ ，则

|P(x)Q(D)\varphi(x)|<\sum_{k=0}^{m}|a_k|\cdot m_k<\infty

故 $P(x)Q(D)\varphi(x)\in\mathscr{S}$ .

(II) (2) $\Rightarrow$ (3)
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Q(D)[P(x)\varphi] = \sum_{\gamma \leq \beta} \binom{\beta}{\gamma} D^\gamma P(x) \cdot D^{\beta - \gamma} \varphi

其中 $Q(D) = \sum_{|\beta| \leq k} b_\beta D^\beta$ ， $\binom{\beta}{\gamma}$ 为组合系数， $D^\gamma P(x)$ 是 $P(x)$ 的 $\gamma$ 阶导数（仍为多项式）.

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(III) (3) $\Rightarrow$ (1)
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∎

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Problem 1.

证明 $e^{-x^{2}}$ 是 $\mathscr{S}$ 函数.

证明.

(i) $e^{-x^2}\in C^{\infty}$

我们对<span data-mx-katex="e%5E%7B-x%5E2%7D"></span>求导，得到：

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\frac{d(e^{-x^2})}{dx}=-2xe^{-x^2}, \frac{d^2(e^{-x^2})}{dx^2}=(4x^2-2)e^{-x^2}

由求导的乘积法则我们可以知道：

\frac{d^n(e^{-x^2})}{dx^n}=P_n(x)e^{-x^2}, \deg(P_n)=n

故 $e^{-x^2}$ 的任意阶导数存在，即 $e^{-x^2}\in C^{\infty}$ .

(ii) 对 $\forall\alpha,\beta$ 有： $\sup_{x\in\mathbb{R}}|x^\alpha D^\beta f(x)|<c(\alpha,\beta)$
由(i)我们可知<span data-mx-katex="D%5E%5Cbeta%20e%5E%7B-x%5E2%7D%3DP_%7B%5Cbeta%7D(x)e%5E%7B-x%5E2%7D"></span>
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\Rightarrow x^{\alpha}D^\beta e^{-x^2}=x^{\alpha}P_{\beta}(x)e^{-x^2}=P_{\beta+\alpha}(x)e^{-x^2}=\sum_{k=0}^{\alpha+\beta}p_k x^k e^{-x^2}

而对于 $x^k e^{-x^2}$ ，我们可知：

\frac{d(x^k e^{-x^2})}{dx}=(k x^{k-1}-2x^{k+1})e^{-x^2}=0\Rightarrow x=\pm\sqrt{\frac{k}{2}}\Rightarrow |x^k e^{-x^2}|\leq(\frac{k}{2})^k e^{-\frac{k}{2}}=M_k.

则

|x^{\alpha}D^\beta e^{-x^2}|=\left|\sum_{k=0}^{\alpha+\beta}p_k x^k e^{-x^2}\right|\leq\sum_{k=0}^{\alpha+\beta}|p_k x^k e^{-x^2}|\leq\sum_{k=0}^{\alpha+\beta}M_k|p_k|=c(\alpha,\beta)<\infty

故

\sup_{x\in\mathbb{R}}|x^\alpha D^\beta f(x)|<c(\alpha,\beta)

综上我们可知： $e^{-x^2}$ 是 $\mathscr{S}$ 函数.

∎

Problem 2.

设 $P(\xi)$ ， $Q(\xi)$ 皆为常系数多项式，证明以下各个命题等价：

(1) $\varphi(x) \in \mathscr{S}$ ;
(2) 对任意 $P(\xi)$ ， $Q(\xi)$ ， $P(x)Q(D)\varphi \in \mathscr{S}$ ;
(3) 对任意 $P(\xi)$ ， $Q(\xi)$ ， $Q(D)[P(x)\varphi(x)] \in \mathscr{S}$ .

证明.

由于 $P(\xi)$ ， $Q(\xi)$ 皆为常系数多项式，故我们可设 $P(x)=\sum_{k=0}^{m}a_k x^k$ ， $Q(x)=\sum_{k=0}^{n}b_k x^k$ .

(I) (1) $\Rightarrow$ (2)

首先由于<span data-mx-katex="%5Cvarphi(x)%5Cin%5Cmathscr%7BS%7D"></span>故<span data-mx-katex="%5Cpsi(x)%3DQ(D)%5Cvarphi(x)%20%5Cin%20%5Cmathscr%7BS%7D"></span>,进而有

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|P(x)Q(D)\varphi(x)|=\left|\sum_{k=0}^{m}a_k x^k\psi(x)\right|=\sum_{k=0}^{m}|a_k|\cdot|x^k\psi(x)|

而由于 $\psi(x)\in\mathscr{S}$ ，故 $\exists m_k>0$ ，使得 $|x^k\psi(x)|<m_k$ ，则

|P(x)Q(D)\varphi(x)|<\sum_{k=0}^{m}|a_k|\cdot m_k<\infty

故 $P(x)Q(D)\varphi(x)\in\mathscr{S}$ .

(II) (2) $\Rightarrow$ (3)

我们对乘积 <span data-mx-katex="P(x)%5Cvarphi"></span> 求导 <span data-mx-katex="Q(D)"></span>，有：

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Q(D)[P(x)\varphi] = \sum_{\gamma \leq \beta} \binom{\beta}{\gamma} D^\gamma P(x) \cdot D^{\beta - \gamma} \varphi

其中 $Q(D) = \sum_{|\beta| \leq k} b_\beta D^\beta$ ， $\binom{\beta}{\gamma}$ 为组合系数， $D^\gamma P(x)$ 是 $P(x)$ 的 $\gamma$ 阶导数（仍为多项式）.

对任意项 <span data-mx-katex="D%5E%5Cgamma%20P(x)%20%5Ccdot%20D%5E%7B%5Cbeta%20-%20%5Cgamma%7D%20%5Cvarphi"></span>，令 <span data-mx-katex="P_1(x)%20%3D%20D%5E%5Cgamma%20P(x)"></span>为多项式，<span data-mx-katex="Q_1(D)%20%3D%20D%5E%7B%5Cbeta%20-%20%5Cgamma%7D"></span>；则我们由 <span data-mx-katex="(2)"></span> 的条件，可知：<span data-mx-katex="P_1(x)Q_1(D)%5Cvarphi%20%3D%20D%5E%5Cgamma%20P(x)%20%5Ccdot%20D%5E%7B%5Cbeta%20-%20%5Cgamma%7D%20%5Cvarphi%20%5Cin%20%5Cmathscr%7BS%7D"></span>.

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(III) (3) $\Rightarrow$ (1)

取 <span data-mx-katex="P(x)%20%3D%201"></span>，<span data-mx-katex="Q(D)%20%3D%20D%5E%5Cbeta"></span>.我们由 <span data-mx-katex="(3)"></span> 的条件，<span data-mx-katex="Q(D)%5B1%20%5Ccdot%20%5Cvarphi%5D%20%3D%20D%5E%5Cbeta%20%5Cvarphi%20%5Cin%20%5Cmathscr%7BS%7D"></span>，故 <span data-mx-katex="D%5E%5Cbeta%20%5Cvarphi"></span> 存在且连续.由 <span data-mx-katex="%5Cbeta"></span> 的任意性，<span data-mx-katex="%5Cvarphi%20%5Cin%20C%5E%5Cinfty(%5Cmathbb%7BR%7D%5En)"></span>.

取 <span data-mx-katex="Q(D)%20%3D%201"></span>，<span data-mx-katex="P(x)%20%3D%20x%5E%5Calpha"></span>.我们由 <span data-mx-katex="(3)"></span> 的条件，<span data-mx-katex="1%20%5Ccdot%20%5Bx%5E%5Calpha%20%5Cvarphi%5D%20%3D%20x%5E%5Calpha%20%5Cvarphi%20%5Cin%20%5Cmathscr%7BS%7D"></span>，故 <span data-mx-katex="%5Csup_%7Bx%20%5Cin%20%5Cmathbb%7BR%7D%5En%7D%20%5Cleft%7C%20x%5E%5Calpha%20%5Cvarphi(x)%20%5Cright%7C%20%3C%20%2B%5Cinfty"></span>.

结合无穷可微性，对任意多重指标 <span data-mx-katex="%5Cbeta"></span>，需证 <span data-mx-katex="%5Csup_%7Bx%20%5Cin%20%5Cmathbb%7BR%7D%5En%7D%20%5Cleft%7C%20x%5E%5Calpha%20D%5E%5Cbeta%20%5Cvarphi%20%5Cright%7C%20%3C%20%2B%5Cinfty"></span>.取 <span data-mx-katex="P(x)%20%3D%20x%5E%5Calpha"></span>，<span data-mx-katex="Q(D)%20%3D%20D%5E%5Cbeta"></span>，由 <span data-mx-katex="(3)"></span> 的条件，<span data-mx-katex="D%5E%5Cbeta%20%5Bx%5E%5Calpha%20%5Cvarphi%5D%20%5Cin%20%5Cmathscr%7BS%7D"></span>.展开 <span data-mx-katex="D%5E%5Cbeta%20%5Bx%5E%5Calpha%20%5Cvarphi%5D"></span>可知其为有限项的和，每项形如 <span data-mx-katex="D%5E%5Cgamma%20x%5E%5Calpha%20%5Ccdot%20D%5E%7B%5Cbeta%20-%20%5Cgamma%7D%20%5Cvarphi"></span>（<span data-mx-katex="%5Cgamma%20%5Cleq%20%5Cbeta"></span> ），其中 <span data-mx-katex="D%5E%5Cgamma%20x%5E%5Calpha"></span> 是多项式，<span data-mx-katex="D%5E%7B%5Cbeta%20-%20%5Cgamma%7D%20%5Cvarphi"></span> 是 <span data-mx-katex="%5Cvarphi"></span> 的导数.由施瓦茨函数定义，这些项的上确界有限，故 <span data-mx-katex="%5Csup_%7Bx%20%5Cin%20%5Cmathbb%7BR%7D%5En%7D%20%5Cleft%7C%20x%5E%5Calpha%20D%5E%5Cbeta%20%5Cvarphi%20%5Cright%7C%20%3C%20%2B%5Cinfty"></span>.

故<span data-mx-katex="%5Cvarphi%5Cin%5Cmathscr%7BS%7D"></span>.

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∎

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PDE — Homework 3

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PDE — Homework 3

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