8.2Fourier 变换的应用习题参考解答
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# 应用习题参考解答
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1、 写出高维热传导方程 Cauchy 问题 (2.8) 的解的表达式 (2.9) 的推导过程.
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\begin\{aligned\} \left\\{\begin\{array\}\{llllllllllll\}u\_t-a^2u\_\{xx\}=0,&x\in\mathbb\{R\}^n, t > 0,\\\\ u|\_\{t=0\}=\varphi(x),&x\in\mathbb\{R\}^n.\end\{array\}\right. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
关于 $\displaystyle x$ 求 Fourier 变换得
\begin\{aligned\} \left\\{\begin\{array\}\{llllllllllll\}\frac\{\mathrm\{ d\}\}\{\mathrm\{ d\} t\}\hat u(\xi,t)+a^2|\xi|^2\hat u(\xi,t)=0,\\\\ \hat u(\xi,0)=\hat \varphi(\xi).\end\{array\}\right. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故
\begin\{aligned\} &\hat u(\xi,t)=\hat \varphi(\xi)\mathrm\{e\}^\{-a^2|\xi|^2t\} =\hat \varphi(\xi)\prod\_\{k=1\}^n \mathrm\{e\}^\{-a^2\xi\_k^2t\} =\hat \varphi(\xi)\prod\_\{k=1\}^n \left\[\frac\{1\}\{\sqrt\{2a^2t\}\} \mathrm\{e\}^\{-\frac\{x\_k^2\}\{4a^2t\}\}\right\]^\wedge(\xi\_k)\\\\ =&\hat \varphi(\xi)\left\[\frac\{1\}\{(a^2t)^\frac\{n\}\{2\}\}\mathrm\{e\}^\{-\frac\{|x|^2\}\{4a^2t\}\}\right\]^\wedge(\xi). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
进而
\begin\{aligned\} u(x,t)=&\frac\{1\}\{(2\pi)^\frac\{n\}\{2\}\} \left\[\varphi(\cdot)\star \frac\{1\}\{(2a^2t)^\frac\{n\}\{2\}\} \mathrm\{e\}^\{-\frac\{|\cdot|^2\}\{4a^2t\}\}\right\] (x) =\frac\{1\}\{(4\pi a^2t)^\frac\{n\}\{2\}\}\int\_\{\mathbb\{R\}^n\} \varphi(y)\mathrm\{e\}^\{-\frac\{|x-y|^2\}\{4a^2t\}\}\mathrm\{ d\} y. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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---
2、 用 Fourier 变换求解非齐次热传导方程的 Cauchy 问题:
\begin\{aligned\} \left\\{\begin\{array\}\{llllllllllll\} u\_t-a^2u\_\{xx\}=f(x,t), x\in\mathbb\{R\}, t > 0,\\\\ u|\_\{t=0\}=\varphi(x), x\in\mathbb\{R\}. \end\{array\}\right. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
[纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/(https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/(https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 关于 $\displaystyle x$ 求 Fourier 变换知
\begin\{aligned\} &\frac\{\mathrm\{ d\}\}\{\mathrm\{ d\} t\}\hat u(\xi,t)+a^2\xi^2 \hat u(\xi,t)=\hat f(\xi,t)\hat u(\xi,0)=\hat \varphi(\xi,t)\\\\ \Rightarrow&\frac\{\mathrm\{ d\}\}\{\mathrm\{ d\} t\}\left\[\mathrm\{e\}^\{a^2\xi^2t\}\hat u(\xi,t)\right\] =\mathrm\{e\}^\{a^2\xi^2t\}\hat f(\xi,t)\\\\ \Rightarrow&\mathrm\{e\}^\{a^2\xi^2t\}\hat u(\xi,t)-\hat \varphi(\xi) =\int\_0^t\mathrm\{e\}^\{a^2\xi^2\tau\}\hat f(\xi,\tau)\mathrm\{ d\} \tau\\\\ \Rightarrow&\hat u(\xi,t) =\hat \varphi(\xi,t)\mathrm\{e\}^\{-a^2\xi^2t\} +\int\_0^t \mathrm\{e\}^\{-a^2\xi^2(t-\tau)\}\hat f(\xi,\tau)\mathrm\{ d\} \tau\\\\ \Rightarrow&u(x,t)=\frac\{1\}\{2a\sqrt\{\pi t\}\} \int\_\{\mathbb\{R\}\} \varphi(y)\mathrm\{e\}^\{-\frac\{(x-y)^2\}\{4a^2t\}\}\mathrm\{ d\} y\\\\ &+\int\_0^t \int\_\{\mathbb\{R\}\} \frac\{1\}\{2a\sqrt\{\pi (t-\tau)\}\} \mathrm\{e\}^\{-\frac\{(x-y)^2\}\{4a^2(t-\tau)\}\}f(y,\tau) \mathrm\{ d\} y \mathrm\{ d\} \tau. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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---
3、 用 Fourier 变换求解下列 Cauchy 问题:
\begin\{aligned\} \left\\{\begin\{array\}\{llllllllllll\}u\_t-a^2u\_\{xx\}-bu\_x-cu=0, x\in\mathbb\{R\}, t > 0,\\\\ u|\_\{t=0\}=\varphi(x), x\in\mathbb\{R\},\end\{array\}\right. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
其中 $\displaystyle a,b,c$ 是常数.
[纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/(https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/(https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 关于 $\displaystyle x$ 求 Fourier 变换知
\begin\{aligned\} &\frac\{\mathrm\{ d\}\}\{\mathrm\{ d\} t\}\hat u(\xi,t) +a^2\xi^2\hat u(\xi,t) -\mathrm\{ i\} b\xi\hat u(\xi,t) -c\hat u(\xi,t)=0, \hat u(\xi,0)+\varphi(\xi)\\\\ \Rightarrow&\frac\{\mathrm\{ d\}\}\{\mathrm\{ d\} t\} \left\[\mathrm\{e\}^\{(a^2\xi^2-\mathrm\{ i\} b\xi-c)t\}\hat u(\xi,t)\right\]=0 \Rightarrow \mathrm\{e\}^\{(a^2\xi^2-\mathrm\{ i\} b\xi-c)t\}\hat u(\xi,t)=\varphi(\xi)\\\\ \Rightarrow&\hat u(\xi,t)=\mathrm\{e\}^\{(-a^2\xi^2+\mathrm\{ i\} b\xi+c)t\}\hat\varphi(\xi). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由习题 7-1 题 3 (2) 知
\begin\{aligned\} u(x,t)=&\frac\{1\}\{\sqrt\{2\pi\}\} \int\_\{\mathbb\{R\}\}\frac\{1\}\{a\sqrt\{2t\}\} \mathrm\{e\}^\{\frac\{-(x-y)^2-2bt(x-y)+(4ac-b^2)t^2\}\{4a^2t\}\} \varphi(y)\mathrm\{ d\} y\\\\ =&\frac\{1\}\{2a\sqrt\{\pi t\}\} \int\_\{\mathbb\{R\}\}\mathrm\{e\}^\{\frac\{-(x-y)^2-2bt(x-y)+(4ac-b^2)t^2\}\{4a^2t\}\} \varphi(y)\mathrm\{ d\} y. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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---
4、 用 Fourier 变换求解下列 Cauchy 问题:
(1)、 $\displaystyle \left\\{\begin\{array\}\{llllllllllll\}u\_t-u\_\{xx\}-4tu=0, x\in\mathbb\{R\}, t > 0,\\\\ u|\_\{t=0\}=x\mathrm\{e\}^x, x\in\mathbb\{R\};\end\{array\}\right.$
[纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/(https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/(https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 设 $\displaystyle v=\mathrm\{e\}^\{-2t^2\}u$, 则
\begin\{aligned\} \left\\{\begin\{array\}\{llllllllllll\}v\_t-v\_\{xx\}=0,&x\in\mathbb\{R\}, t > 0,\\\\ v|\_\{t=0\}=x\mathrm\{e\}^x,&x\in\mathbb\{R\}.\end\{array\}\right. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
关于 $\displaystyle x$ 作 Fourier 变换知
\begin\{aligned\} \hat v\_t+|\xi|^2\hat v=0, \hat v(0)=(x\mathrm\{e\}^x)^\wedge=\mathrm\{ i\}\frac\{\mathrm\{ d\}\}\{\mathrm\{ d\} \xi\}(\mathrm\{e\}^x)^\wedge. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
经过计算后知$\displaystyle u=\mathrm\{e\}^\{t+2t^2+x\}(x+2t)$. 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
(2)、 $\displaystyle \left\\{\begin\{array\}\{llllllllllll\} u\_t-u\_\{xx\}=3t^2u,&x\in\mathbb\{R\}, t > 0,\\\\ u|\_\{t=0\}=x^2+1, &x\in\mathbb\{R\}; \end\{array\}\right.$
[纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/(https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/(https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 设 $\displaystyle v=\mathrm\{e\}^\{-t^3\}u$, 则
\begin\{aligned\} \left\\{\begin\{array\}\{llllllllllll\}v\_t-v\_\{xx\}=0,&x\in\mathbb\{R\}, t > 0,\\\\ v|\_\{t=0\}=x^2+1,&x\in\mathbb\{R\}.\end\{array\}\right. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
类似第 1 小题的计算过程知 $\displaystyle u=\mathrm\{e\}^\{t^3\}(1+2t+x^2)$. 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
(3)、 $\displaystyle \left\\{\begin\{array\}\{llllllllllll\}u\_t-u\_\{xx\}-2u\_x-(2t+1)u=0, x\in\mathbb\{R\}, t > 0,\\\\ u|\_\{t=0\}=x, x\in\mathbb\{R\}.\end\{array\}\right.$
[纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/(https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/(https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 设 $\displaystyle v=\mathrm\{e\}^\{-t^2+x\}u$, 则
\begin\{aligned\} \left\\{\begin\{array\}\{llllllllllll\}v\_t-v\_\{xx\}=0,&x\in\mathbb\{R\}, t > 0,\\\\ v|\_\{t=0\}=x^2+1,&x\in\mathbb\{R\}.\end\{array\}\right. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
类似第 1 小题的计算过程知 $\displaystyle u=\mathrm\{e\}^\{t+t^2\}(x+2t)$. 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
---
5、 求解 Cauchy 问题
\begin\{aligned\} \left\\{\begin\{array\}\{llllllllllll\} u\_t-a(t)u\_\{xx\}=0,& x\in\mathbb\{R\}, t > 0,\\\\ u|\_\{t=0\}=\varphi(x), &x\in\mathbb\{R\},\end\{array\}\right. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
其中 $\displaystyle a(t) > 0$, 将解用 Fourier 逆变换表示出来. 特别地, 求出当 $\displaystyle \varphi(x)=\mathrm\{e\}^\{-|x|\}$, $\displaystyle a(t)=\mathrm\{e\}^\{-t\}$ 时上述问题的解.
[纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/(https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/(https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 关于 $\displaystyle x$ 求 Fourier 变换知
\begin\{aligned\} &\hat u\_t+a(t)|\xi|^2\hat u=0, \hat u(0)=\hat \varphi(\xi) \Rightarrow\left\[\mathrm\{e\}^\{|\xi|^2\int\_0^t a(\tau)\mathrm\{ d\} \tau\} \hat u\right\]\_t=0\\\\ \Rightarrow& \hat u=\hat \varphi(\xi)\mathrm\{e\}^\{-|\xi|^2\int\_0^t a(\tau)\mathrm\{ d\} \tau\} \Rightarrow u=\left\[\hat \varphi(\xi)\mathrm\{e\}^\{-|\xi|^2\int\_0^t a(\tau)\mathrm\{ d\} \tau\}\right\]^\vee. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
当 $\displaystyle \varphi(x)=\mathrm\{e\}^\{-|x|\}$, $\displaystyle a(t)=\mathrm\{e\}^\{-t\}$ 时,
\begin\{aligned\} u=&\left\[(\mathrm\{e\}^\{-|x|\})^\wedge \mathrm\{e\}^\{-|\xi|^2(1-\mathrm\{e\}^\{-t\})\}\right\]^\vee =\mathrm\{e\}^\{-|\cdot|\}\star \left\[\mathrm\{e\}^\{-|\xi|^2(1-\mathrm\{e\}^\{-t\})\}\right\]^\vee\\\\ =&\mathrm\{e\}^\{-|\cdot|\}\star \frac\{1\}\{\sqrt\{1-\mathrm\{e\}^\{-t\}\}\} \frac\{1\}\{\sqrt\{2\}\}\mathrm\{e\}^\{-\frac\{|\cdot|^2\}\{4\sqrt\{1-\mathrm\{e\}^\{-t\}\}\}\}\\\\ =&\frac\{1\}\{\sqrt\{2(1-\mathrm\{e\}^\{-t\})\}\}\int\_\{\mathbb\{R\}\} \mathrm\{e\}^\{-|x-y|\} \mathrm\{e\}^\{-\frac\{y^2\}\{4\sqrt\{1-\mathrm\{e\}^\{-t\}\}\}\}\mathrm\{ d\} y. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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---
6、 求解 KdV 方程的线性化方程的 Cauchy 问题:
\begin\{aligned\} \left\\{\begin\{array\}\{llllllllllll\} u\_t=c^2u\_\{xxx\},&x\in\mathbb\{R\}, t > 0,\\\\ u|\_\{t=0\}=\varphi(x),&x\in\mathbb\{R\}, \end\{array\}\right. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
其中 $\displaystyle c > 0$ 为常数.
[纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/(https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/(https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 关于 $\displaystyle x$ 求 Fourier 变换知
\begin\{aligned\} &\hat u\_t+\mathrm\{ i\} c^2\xi^3\hat u=0, \hat u(0)=\hat \varphi(\xi) \Rightarrow u=\left\[\hat \varphi(\xi)\mathrm\{e\}^\{-\mathrm\{ i\} c^2\xi^2t\}\right\]^\vee. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
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---
7、 用 Fourier 变换求解弹性梁的横振动方程的 Cauchy 问题:
\begin\{aligned\} \left\\{\begin\{array\}\{llllllllllll\} u\_\{tt\}=-c^2u\_\{xxxx\}, x\in\mathbb\{R\}, t > 0,\\\\ u(x,0)=\varphi(x), u\_t(x,0)=\psi(x), x\in\mathbb\{R\},\end\{array\}\right. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
其中 $\displaystyle c > 0$ 为常数.
[纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/(https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/(https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 关于 $\displaystyle x$ 求 Fourier 变换知
\begin\{aligned\} \frac\{\mathrm\{ d\}^2\}\{\mathrm\{ d\} t^2\}\hat u(\xi,t) +c^2\xi^4\hat u(\xi,t) =0, \hat u(\xi, 0)=\hat \varphi(\xi), \hat u\_t(\xi,t)=\hat \psi(\xi). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
特征方程为
\begin\{aligned\} \lambda^2+c^2\xi^4=0\Rightarrow \lambda=\pm c\xi^2\mathrm\{ i\} . \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
故
\begin\{aligned\} \hat u(\xi,t)=A(\xi)\cos c\xi^2t +B(\xi)\sin c\xi^2t. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
将初值条件代入得
\begin\{aligned\} A(\xi)=\hat u(\xi,0)=\hat \varphi(\xi), c\xi^2B(\xi)=\hat u\_t(\xi,0)=\hat \psi(\xi). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
于是
\begin\{aligned\} \hat u(\xi,t)=\hat \varphi(\xi,t)\cos c\xi^2t+\frac\{\hat \psi(\xi)\}\{c\xi^2\} \sin c\xi^2t. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
由
\begin\{aligned\} \left(\mathrm\{e\}^\{-Ax^2\}\right)^\wedge(\xi) =\frac\{1\}\{\sqrt\{2A\}\}\mathrm\{e\}^\{-\frac\{\xi^2\}\{4A\}\}(A > 0) \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
知
\begin\{aligned\} (\mathrm\{e\}^\{-\mathrm\{ i\} x^2\})^\wedge (\xi) =\frac\{1\}\{\sqrt\{2\mathrm\{ i\}\}\} \mathrm\{e\}^\{-\frac\{\xi^2\}\{4\mathrm\{ i\}\}\}. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
从而
\begin\{aligned\} (\cos x^2)^\wedge(\xi)& =\mathrm\{ Re\} \left\[\frac\{1\}\{\sqrt\{2\mathrm\{ i\}\}\} \mathrm\{e\}^\{\frac\{\xi^2\}\{4\}\mathrm\{ i\}\}\right\] =\mathrm\{ Re\}\left\[\frac\{1-\mathrm\{ i\}\}\{2\} \left(\cos\frac\{\xi^2\}\{4\}+\mathrm\{ i\} \sin\frac\{\xi^2\}\{4\}\right)\right\]\\\\ &=\frac\{1\}\{2\}\cos\frac\{\xi^2\}\{4\}+\frac\{1\}\{2\}\sin\frac\{\xi^2\}\{4\} =\frac\{1\}\{\sqrt\{2\}\}\cos\left(\frac\{\xi^4\}\{4\}-\frac\{\pi\}\{4\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
于是
\begin\{aligned\} (\cos ctx^2)^\wedge(\xi) &=\left\[\cos \left(\sqrt\{ct\}x\right)^2\right\]^\wedge(\xi) =\frac\{1\}\{\sqrt\{ct\}\}\cdot \frac\{1\}\{\sqrt\{2\}\}\cos\left\[\frac\{\left(\frac\{\xi\}\{\sqrt\{ct\}\}\right)^2\}\{4\}-\frac\{\pi\}\{4\}\right\]\\\\ &=\frac\{1\}\{\sqrt\{2ct\}\} \cos\left(\frac\{\xi^2\}\{4ct\}-\frac\{\pi\}\{4\}\right), \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
\begin\{aligned\} \left\[\cos(ct\xi^2)\right\]^\vee(x) =(\cos ct\xi^2)^\wedge(-x) =\frac\{1\}\{\sqrt\{2ct\}\} \cos\left(\frac\{x^2\}\{4ct\}-\frac\{\pi\}\{4\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
于是
\begin\{aligned\} u\_1(x,t)&\equiv \left\[\hat \varphi(\xi) \cos c\xi^2t\right\]^\vee(x) =\frac\{1\}\{\sqrt\{2\pi\}\}\int\_\{\mathbb\{R\}\}\frac\{1\}\{\sqrt\{2ct\}\} \cos\left(\frac\{(x-y)^2\}\{4ct\}-\frac\{\pi\}\{4\}\right)\varphi(y)\mathrm\{ d\} y. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
但是可惜 $\displaystyle u\_2(x,t)$ 没有这么好的表达式. 因为
\begin\{aligned\} (\sin x^2)^\wedge(\xi)=\cdots=\frac\{1\}\{\sqrt\{2\}\} \sin\left(\frac\{\xi^2\}\{4\}-\frac\{\pi\}\{4\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
设$\displaystyle \left(\frac\{\sin c\xi^2t\}\{\xi^2\}\right)^\vee(x)=g(x,t)$, 则可由 Fourier 变换的性质知
\begin\{aligned\} \frac\{\mathrm\{ d\} ^2\}\{\mathrm\{ d\} x^2\}g(x)=-\frac\{1\}\{\sqrt\{2ct\}\}\sin\left(\frac\{x^2\}\{4ct\}-\frac\{\pi\}\{4\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
求不定积分再求不定积分, 呜呜. 没有初等表达式哦. 跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
---
8、 用 Fourier 变换求解下面的 Cauchy 问题:
\begin\{aligned\} \left\\{\begin\{array\}\{llllllllllll\} tu\_x+u\_t=0, x\in\mathbb\{R\}, t > 0,\\\\ u(x,0)=\varphi(x), x\in\mathbb\{R\}. \end\{array\}\right. \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
[纸质资料](https://mp.weixin.qq.com/s/ycnPCSqWFlThEnq9ZZ6gBQ)/[答疑](https://mp.weixin.qq.com/s/JGYZG5rsshf7Z2Amo2di8A)/(https://mp.weixin.qq.com/s/Pt6\_h5MqtomrUDYiPEwkxg)/(https://mp.weixin.qq.com/s/dWvpeJFKnFr0WYPoidXXMA) / 关于 $\displaystyle x$ 求 Fourier 变换知
\begin\{aligned\} t\cdot \mathrm\{ i\} \xi \hat u(\xi,t)+\frac\{\mathrm\{ d\}\}\{\mathrm\{ d\} t\}\hat u(\xi,t)=0, \quad\hat u(\xi,0)=\hat \varphi(\xi). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
于是
\begin\{aligned\} &\frac\{\mathrm\{ d\}\}\{\mathrm\{ d\} t\}\left\[\mathrm\{e\}^\{\mathrm\{ i\} \xi \frac\{t^2\}\{2\}\} \hat u(\xi,t)\right\]=0 \Rightarrow \hat u(\xi,t)=\hat \varphi(\xi)\mathrm\{e\}^\{-\mathrm\{ i\} \xi\frac\{t^2\}\{2\}\} \Rightarrow u(x,t)=\varphi\left(x-\frac\{t^2\}\{2\}\right). \tiny\boxed\{\begin\{array\}\{c\}\mbox\{跟锦数学微信公众号\}\\\\\mbox\{zhangzujin.cn\}\end\{array\}\}\end\{aligned\}
跟锦数学微信公众号. [在线资料](https://mp.weixin.qq.com/s/F-TU-uzeo3EjxI5LzjUvRw)/[公众号](https://mp.weixin.qq.com/s/pdC49P5WZXTEpRBa0JBfow)/
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