PDE — Homework 1

Problem 1.

对 $n > 1$ 时，

H(x)=\begin{cases} 1,&x_i>0,i=1,2,\dots,n\\0,&\textrm{if not} \end{cases}

则：

\frac{\partial^n H(x)}{\partial x_1\dots\partial x_n}=\delta(x)

Proof.

我们用 $h(x)$ 来表示一维的 Heaviside 函数，则

H(x)=\begin{cases} 1,&x_i>0,i=1,2,\dots,n\\0,&\textrm{if not} \end{cases}=h(x_1)h(x_2)\dots h(x_n)

同时，对于一维的Heaviside函数我们有

\frac{d}{dx_i} h(x_i) = \delta(x_i)

故

\frac{\partial^n H(x)}{\partial x_1 \ldots \partial x_n}=\delta(x_1) \delta(x_2) \cdots \delta(x_n) = \delta(x)

事实上，对 $\forall \varphi \in C_0^{\infty}(\mathbb{R}^n)$ ，由 广义函数的微分运算，我们有：

\left\langle \frac{\partial^n H}{\partial x_1 \cdots \partial x_n}, \varphi \right\rangle = (-1)^n \left\langle H, \frac{\partial^n \varphi}{\partial x_1 \cdots \partial x_n} \right\rangle = (-1)^n \int_{[0,\infty)^n} \frac{\partial^n \varphi}{\partial x_1 \cdots \partial x_n} dx = \varphi(0) = \langle \delta, \varphi \rangle.

∎

Problem 2.

设 $a(x) \in C^{\infty}(\Omega), u \in \mathscr{D}'(\Omega)$ ，证明：

\frac{\partial}{\partial x_i}(a u) = a \frac{\partial u}{\partial x_i} + \frac{\partial a}{\partial x_i} u

Proof.

对 $\forall \varphi \in C_0^{\infty}(\Omega)$ ，由 广义函数的微分运算，我们有：

\left\langle \frac{\partial}{\partial x_i}(au),\varphi\right\rangle=-\left\langle au,\frac{\partial\varphi}{\partial x_i}\right\rangle

由广义函数的乘子运算，我们有：

\left\langle au,\frac{\partial\varphi}{\partial x_i}\right\rangle=\left\langle u,a\frac{\partial\varphi}{\partial x_i}\right\rangle

由于 $a,\varphi\in C^{\infty}(\Omega)$ ，故由微分法则可知：

\frac{\partial (a\varphi)}{\partial x_i}=a\frac{\partial \varphi}{\partial x_i}+\varphi\frac{\partial a}{\partial x_i}

故我们有：

\left\langle u,a\frac{\partial\varphi}{\partial x_i}\right\rangle=\left\langle u,\frac{\partial (a\varphi)}{\partial x_i}-\varphi\frac{\partial a}{\partial x_i}\right\rangle=\left\langle u,\frac{\partial (a\varphi)}{\partial x_i}\right\rangle-\left\langle u,\varphi\frac{\partial a}{\partial x_i}\right\rangle=-\left\langle \frac{\partial u}{\partial x_i},a\varphi\right\rangle-\left\langle u,\varphi\frac{\partial a}{\partial x_i}\right\rangle

我们将其带回最初式子有：

\left\langle \frac{\partial}{\partial x_i}(au),\varphi\right\rangle=-\left\langle au,\frac{\partial\varphi}{\partial x_i}\right\rangle=\left\langle \frac{\partial u}{\partial x_i},a\varphi\right\rangle+\left\langle u,\varphi\frac{\partial a}{\partial x_i}\right\rangle=\left\langle a\frac{\partial u}{\partial x_i},\varphi\right\rangle+\left\langle \frac{\partial a}{\partial x_i}u,\varphi\right\rangle=\left\langle a\frac{\partial u}{\partial x_i}+\frac{\partial a}{\partial x_i}u,\varphi\right\rangle

故由广义相等我们可知：

\frac{\partial}{\partial x_i}(a u) = a \frac{\partial u}{\partial x_i} + \frac{\partial a}{\partial x_i} u

∎

Problem 3.

若广义函数 $u$ 适合 $\check{u} = u$ （ $\check{u} = -u$ ），则称 $u$ 为偶（奇）广义函数。证明：

(1) $\delta(x)$ 与常值函数 $c$ 皆为偶广义函数。
(2) 偶广义函数之微商是奇广义函数。
(3) 任一广义函数 $u \in \mathscr{D}'(\Omega)$ 皆可唯一地分解为一个偶广义函数与一个奇广义函数之和如下：

u(x) = \frac{1}{2}(u(x) + \tilde{u}(x)) + \frac{1}{2}(u(x) - \tilde{u}(x)).

(4) $u \in \mathscr{D}'(\Omega)$ 为偶广义函数当且仅当对任一奇的 $\varphi \in \mathscr{D}(\Omega)$ 均有 $\langle u, \varphi \rangle = 0$ 。

Proof.

对 $\forall \varphi \in C_0^{\infty}(\Omega)$ ：

(1) ① 我们有 $\langle \delta,\varphi\rangle=\varphi(0)$ ， $\langle \check\delta,\varphi\rangle=\langle \delta,\check\varphi\rangle=\check\varphi(0)=\varphi(-0)=\varphi(0)=\langle \delta,\varphi\rangle\Rightarrow\delta(x)=\check\delta(x)$ ，即 $\delta(x)$ 为偶广义函数；
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(2) 设 $u$ 为偶广义函数，满足 $u=\check u$ ，则：

\langle\check{(\frac{\partial u}{\partial x_i})},\varphi\rangle=\langle\frac{\partial u}{\partial x_i},\check\varphi\rangle=-\langle u,\frac{\partial\check\varphi}{\partial x_i}\rangle=-\langle u,\frac{\partial(-\varphi)}{\partial x_i}\rangle=\langle u,\frac{\partial\varphi}{\partial x_i}\rangle=-\langle(\frac{\partial u}{\partial x_i}),\varphi\rangle=\langle-(\frac{\partial u}{\partial x_i}),\varphi\rangle

故由广义相等我们可知：

\check{(\frac{\partial u}{\partial x_i})}=-(\frac{\partial u}{\partial x_i})

即为奇广义函数。

(3) 令 $u_1 = \frac{1}{2}(u + \check{u})$ ， $u_2 = \frac{1}{2}(u - \check{u})$ 。其中： $\check{u_1} = \frac{1}{2}(\check{u} + u) = u_1$ 是偶广义函数； $\check{u_2} = \frac{1}{2}(\check{u} - u) = -u_2$ 是奇广义函数。故：

u_1 + u_2 = \frac{1}{2}(u + \check{u}) + \frac{1}{2}(u - \check{u}) = u

下面我们来证明分解的 唯一性：若 $u = u_1' + u_2'$ ，其中 $u_1'$ 是偶广义函数， $u_2'$ 是奇广义函数，则 $\check{u} = u_1' - u_2'$ 。联立 $u = u_1' + u_2'$ 与 $\check{u} = u_1' - u_2'$ ，解得 $u_1' = \frac{1}{2}(u + \check{u})$ ， $u_2' = \frac{1}{2}(u - \check{u})$ ，故分解唯一。

(4) ( $\Rightarrow$ ) 若 $u$ 是偶广义函数，即 $\check{u} = u$ 。对任一奇的 $\varphi \in \mathscr{D}(\Omega)$ ，有 $\check{\varphi} = -\varphi$ 。则 $\langle u, \varphi \rangle = \langle \check{u}, \varphi \rangle = \langle u, \check{\varphi} \rangle = \langle u, -\varphi \rangle = -\langle u, \varphi \rangle$ ，故 $2\langle u, \varphi \rangle = 0$ ，即 $\langle u, \varphi \rangle = 0$ 。 ( $\Leftarrow$ ) 若对任一奇的 $\varphi \in \mathscr{D}(\Omega)$ ， $\langle u, \varphi \rangle = 0$ 。令 $\varphi = \varphi_1 + \varphi_2$ （其中 $\varphi_1$ 是偶函数， $\varphi_2$ 是奇函数），则 $\langle u, \varphi \rangle = \langle u, \varphi_1 \rangle + \langle u, \varphi_2 \rangle = \langle u, \varphi_1 \rangle$ 。同时 $\langle \check{u}, \varphi \rangle = \langle u, \check{\varphi} \rangle = \langle u, \varphi_1 \rangle$ ，故 $\langle u - \check{u}, \varphi \rangle = 0$ 对所有 $\varphi \in \mathscr{D}(\Omega)$ 成立，即 $u = \check{u}$ ， $u$ 是偶广义函数。

∎

Problem 4.

证明：若 $f(x) = H(x) \cos x$ ， $g(x) = H(x) \sin x$ ，则

f'(x) = \delta(x) - g(x), g'(x) = f(x),

因此 $g$ 与 $f$ 分别适合微分方程

u'' + u = \delta \text{与} u'' + u = \delta'.

Proof.

设

f(x) = H(x)\cos x \Rightarrow f'(x) = H'(x)\cos x + H(x)(\cos x)'

已知 $H'(x) = \delta(x)$ ， $(\cos x)' = -\sin x$ ，故 $f'(x) = \delta(x)\cos 0 - H(x)\sin x = \delta(x) - g(x)$ 。

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g(x) = H(x)\sin x

可得 $g'(x) = H'(x)\sin x + H(x)(\sin x)' = \delta(x)\sin 0 + H(x)\cos x = f(x)$ 。

CodeBlock Loading...

∎

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Download the original write-up here.

Problem 1.

对 $n > 1$ 时，

H(x)=\begin{cases} 1,&x_i>0,i=1,2,\dots,n\\0,&\textrm{if not} \end{cases}

则：

\frac{\partial^n H(x)}{\partial x_1\dots\partial x_n}=\delta(x)

Proof.

我们用 $h(x)$ 来表示一维的 Heaviside 函数，则

H(x)=\begin{cases} 1,&x_i>0,i=1,2,\dots,n\\0,&\textrm{if not} \end{cases}=h(x_1)h(x_2)\dots h(x_n)

同时，对于一维的Heaviside函数我们有

\frac{d}{dx_i} h(x_i) = \delta(x_i)

故

\frac{\partial^n H(x)}{\partial x_1 \ldots \partial x_n}=\delta(x_1) \delta(x_2) \cdots \delta(x_n) = \delta(x)

事实上，对 $\forall \varphi \in C_0^{\infty}(\mathbb{R}^n)$ ，由 广义函数的微分运算，我们有：

\left\langle \frac{\partial^n H}{\partial x_1 \cdots \partial x_n}, \varphi \right\rangle = (-1)^n \left\langle H, \frac{\partial^n \varphi}{\partial x_1 \cdots \partial x_n} \right\rangle = (-1)^n \int_{[0,\infty)^n} \frac{\partial^n \varphi}{\partial x_1 \cdots \partial x_n} dx = \varphi(0) = \langle \delta, \varphi \rangle.

∎

Problem 2.

设 $a(x) \in C^{\infty}(\Omega), u \in \mathscr{D}'(\Omega)$ ，证明：

\frac{\partial}{\partial x_i}(a u) = a \frac{\partial u}{\partial x_i} + \frac{\partial a}{\partial x_i} u

Proof.

对 $\forall \varphi \in C_0^{\infty}(\Omega)$ ，由 广义函数的微分运算，我们有：

\left\langle \frac{\partial}{\partial x_i}(au),\varphi\right\rangle=-\left\langle au,\frac{\partial\varphi}{\partial x_i}\right\rangle

由广义函数的乘子运算，我们有：

\left\langle au,\frac{\partial\varphi}{\partial x_i}\right\rangle=\left\langle u,a\frac{\partial\varphi}{\partial x_i}\right\rangle

由于 $a,\varphi\in C^{\infty}(\Omega)$ ，故由微分法则可知：

\frac{\partial (a\varphi)}{\partial x_i}=a\frac{\partial \varphi}{\partial x_i}+\varphi\frac{\partial a}{\partial x_i}

故我们有：

\left\langle u,a\frac{\partial\varphi}{\partial x_i}\right\rangle=\left\langle u,\frac{\partial (a\varphi)}{\partial x_i}-\varphi\frac{\partial a}{\partial x_i}\right\rangle=\left\langle u,\frac{\partial (a\varphi)}{\partial x_i}\right\rangle-\left\langle u,\varphi\frac{\partial a}{\partial x_i}\right\rangle=-\left\langle \frac{\partial u}{\partial x_i},a\varphi\right\rangle-\left\langle u,\varphi\frac{\partial a}{\partial x_i}\right\rangle

我们将其带回最初式子有：

\left\langle \frac{\partial}{\partial x_i}(au),\varphi\right\rangle=-\left\langle au,\frac{\partial\varphi}{\partial x_i}\right\rangle=\left\langle \frac{\partial u}{\partial x_i},a\varphi\right\rangle+\left\langle u,\varphi\frac{\partial a}{\partial x_i}\right\rangle=\left\langle a\frac{\partial u}{\partial x_i},\varphi\right\rangle+\left\langle \frac{\partial a}{\partial x_i}u,\varphi\right\rangle=\left\langle a\frac{\partial u}{\partial x_i}+\frac{\partial a}{\partial x_i}u,\varphi\right\rangle

故由广义相等我们可知：

\frac{\partial}{\partial x_i}(a u) = a \frac{\partial u}{\partial x_i} + \frac{\partial a}{\partial x_i} u

∎

Problem 3.

若广义函数 $u$ 适合 $\check{u} = u$ （ $\check{u} = -u$ ），则称 $u$ 为偶（奇）广义函数。证明：

(1) $\delta(x)$ 与常值函数 $c$ 皆为偶广义函数。
(2) 偶广义函数之微商是奇广义函数。
(3) 任一广义函数 $u \in \mathscr{D}'(\Omega)$ 皆可唯一地分解为一个偶广义函数与一个奇广义函数之和如下：

u(x) = \frac{1}{2}(u(x) + \tilde{u}(x)) + \frac{1}{2}(u(x) - \tilde{u}(x)).

(4) $u \in \mathscr{D}'(\Omega)$ 为偶广义函数当且仅当对任一奇的 $\varphi \in \mathscr{D}(\Omega)$ 均有 $\langle u, \varphi \rangle = 0$ 。

Proof.

对 $\forall \varphi \in C_0^{\infty}(\Omega)$ ：

(1) ① 我们有 $\langle \delta,\varphi\rangle=\varphi(0)$ ， $\langle \check\delta,\varphi\rangle=\langle \delta,\check\varphi\rangle=\check\varphi(0)=\varphi(-0)=\varphi(0)=\langle \delta,\varphi\rangle\Rightarrow\delta(x)=\check\delta(x)$ ，即 $\delta(x)$ 为偶广义函数；
② 我们有 <span data-mx-katex="%5Clangle%20%5Ccheck%20c%2C%5Cvarphi%5Crangle%3D%5Clangle%20c%2C%5Ccheck%5Cvarphi%5Crangle%3Dc%5Cint%5Ccheck%5Cvarphi(x)dx%3Dc%5Cint%5Cvarphi(-x)dx%3Dc%5Cint%5Cvarphi(x)dx%3D%5Clangle%20c%2C%5Cvarphi%5Crangle%5CRightarrow%20c%3D%5Ccheck%20c"></span>，即常值函数 <span data-mx-katex="c"></span> 为偶广义函数。
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(2) 设 $u$ 为偶广义函数，满足 $u=\check u$ ，则：

\langle\check{(\frac{\partial u}{\partial x_i})},\varphi\rangle=\langle\frac{\partial u}{\partial x_i},\check\varphi\rangle=-\langle u,\frac{\partial\check\varphi}{\partial x_i}\rangle=-\langle u,\frac{\partial(-\varphi)}{\partial x_i}\rangle=\langle u,\frac{\partial\varphi}{\partial x_i}\rangle=-\langle(\frac{\partial u}{\partial x_i}),\varphi\rangle=\langle-(\frac{\partial u}{\partial x_i}),\varphi\rangle

故由广义相等我们可知：

\check{(\frac{\partial u}{\partial x_i})}=-(\frac{\partial u}{\partial x_i})

即为奇广义函数。

(3) 令 $u_1 = \frac{1}{2}(u + \check{u})$ ， $u_2 = \frac{1}{2}(u - \check{u})$ 。其中： $\check{u_1} = \frac{1}{2}(\check{u} + u) = u_1$ 是偶广义函数； $\check{u_2} = \frac{1}{2}(\check{u} - u) = -u_2$ 是奇广义函数。故：

u_1 + u_2 = \frac{1}{2}(u + \check{u}) + \frac{1}{2}(u - \check{u}) = u

(4) ( $\Rightarrow$ ) 若 $u$ 是偶广义函数，即 $\check{u} = u$ 。对任一奇的 $\varphi \in \mathscr{D}(\Omega)$ ，有 $\check{\varphi} = -\varphi$ 。则 $\langle u, \varphi \rangle = \langle \check{u}, \varphi \rangle = \langle u, \check{\varphi} \rangle = \langle u, -\varphi \rangle = -\langle u, \varphi \rangle$ ，故 $2\langle u, \varphi \rangle = 0$ ，即 $\langle u, \varphi \rangle = 0$ 。 ( $\Leftarrow$ ) 若对任一奇的 $\varphi \in \mathscr{D}(\Omega)$ ， $\langle u, \varphi \rangle = 0$ 。令 $\varphi = \varphi_1 + \varphi_2$ （其中 $\varphi_1$ 是偶函数， $\varphi_2$ 是奇函数），则 $\langle u, \varphi \rangle = \langle u, \varphi_1 \rangle + \langle u, \varphi_2 \rangle = \langle u, \varphi_1 \rangle$ 。同时 $\langle \check{u}, \varphi \rangle = \langle u, \check{\varphi} \rangle = \langle u, \varphi_1 \rangle$ ，故 $\langle u - \check{u}, \varphi \rangle = 0$ 对所有 $\varphi \in \mathscr{D}(\Omega)$ 成立，即 $u = \check{u}$ ， $u$ 是偶广义函数。

∎

Problem 4.

证明：若 $f(x) = H(x) \cos x$ ， $g(x) = H(x) \sin x$ ，则

f'(x) = \delta(x) - g(x), g'(x) = f(x),

因此 $g$ 与 $f$ 分别适合微分方程

u'' + u = \delta \text{与} u'' + u = \delta'.

Proof.

设

f(x) = H(x)\cos x \Rightarrow f'(x) = H'(x)\cos x + H(x)(\cos x)'

已知 $H'(x) = \delta(x)$ ， $(\cos x)' = -\sin x$ ，故 $f'(x) = \delta(x)\cos 0 - H(x)\sin x = \delta(x) - g(x)$ 。

同理，由

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g(x) = H(x)\sin x

可得 $g'(x) = H'(x)\sin x + H(x)(\sin x)' = \delta(x)\sin 0 + H(x)\cos x = f(x)$ 。

对 <span data-mx-katex="g(x)"></span>，求二阶导数得 <span data-mx-katex="g''(x)%20%3D%20f'(x)%20%3D%20%5Cdelta(x)%20-%20g(x)"></span>，整理得 <span data-mx-katex="g''%20%2B%20g%20%3D%20%5Cdelta"></span>。

对 <span data-mx-katex="f(x)"></span>，求二阶导数得 <span data-mx-katex="f''(x)%20%3D%20(f')'%20%3D%20(%5Cdelta(x)%20-%20g(x))'%20%3D%20%5Cdelta'(x)%20-%20g'(x)%20%3D%20%5Cdelta'(x)%20-%20f(x)"></span>，整理得 <span data-mx-katex="f''%20%2B%20f%20%3D%20%5Cdelta'"></span>。

因此 <span data-mx-katex="g"></span> 与 <span data-mx-katex="f"></span> 分别适合微分方程 <span data-mx-katex="u''%20%2B%20u%20%3D%20%5Cdelta"></span> 与 <span data-mx-katex="u''%20%2B%20u%20%3D%20%5Cdelta'"></span>。

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∎

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PDE — Homework 1

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PDE — Homework 1

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