IS Math-2019-02

题目来源：Problem 2 日期：2024-07-07 题目主题：Math-偏微分方程

具体题目

The real-valued function $u(x,t)$ is defined for $-\infty <x<\infty$ and $t\ge 0$ with independent variables $x$ and $t$. Consider to solve the partial differential equation

$$\frac{{\partial }^{2}u}{\partial t^2}=c^2\frac{{\partial }^{2}u}{\partial x^2},\text{\hspace{1em}(2.1)}$$

under the initial conditions

$$u(x,0)=\exp (-a{x}^2)\text{\hspace{1em}(2.2)}$$ $$\frac{\partial u}{\partial t}(x,0)=0,\text{\hspace{1em}(2.3)}$$

where $a$ and $c$ are positive real numbers. The imaginary unit is represented by $i$. Answer the following questions.

1. Calculate the following formula by using complex integration

$${\int }_{-\infty }^{\infty }\exp \left(-a(x+id{)}^2\right)\mathrm{d}x,$$

where $d$ is a real number. The following equation may be used.

$${\int }_{-\infty }^{\infty }\exp (-x^2)\mathrm{d}x=\sqrt{\pi }$$

1. The Fourier transform of $u(x,t)$ with respect to $x$, $U(k,t)$, is defined as

$$U(k,t)=\frac{1}{\sqrt{2\pi }}{\int }_{-\infty }^{\infty }u(x,t)\exp (-ikx)\mathrm{d}x.$$

You may assume that integration with respect to $x$ and differentiation with respect to $t$ are interchangeable. Also, you may assume that $u(x,t)$ and $\frac{\partial u}{\partial x}(x,t)$ converge to 0 as $x\to \pm \infty$ for an arbitrary $t$.

(i) Express the partial differential equation of $U(k,t)$ when $u(x,t)$ satisfies Eq. (2.1).

(ii) Show that the solution of (i) takes the following form under the initial condition of Eq. (2.3) using a function $F(k)$ of variable $k$.

$$U(k,t)=F(k)\cos (kct).$$

(iii) Furthermore, using the initial condition of Eq. (2.2), determine $U(k,t)$ by finding $F(k)$. The result of question (1) may be used.

1. Find $u(x,t)$ by calculating the inverse Fourier transform of $U(k,t)$ obtained in question (2). The inverse Fourier transform is defined as

$$u(x,t)=\frac{1}{\sqrt{2\pi }}{\int }_{-\infty }^{\infty }U(k,t)\exp (ikx)\mathrm{d}k.$$

正确解答

1. Complex Integration Calculation

To evaluate the integral

$${\int }_{-\infty }^{\infty }\exp \left(-a(x+id{)}^2\right)\mathrm{d}x,$$

We can use the substitution $y=\sqrt{a}(x+id)$, which simplifies the integral. The differential $\mathrm{d}y=\sqrt{a}\mathrm{d}x$, so $\mathrm{d}x=\frac{1}{\sqrt{a}}\mathrm{d}y$:

$${\int }_{-\infty }^{\infty }\exp \left(-a(x+id{)}^2\right)\mathrm{d}x=\frac{1}{\sqrt{a}}{\int }_{-\infty }^{\infty }\exp \left(-y^2\right)\mathrm{d}y.$$

Using the Gaussian integral result:

$${\int }_{-\infty }^{\infty }\exp (-x^2)\mathrm{d}x=\sqrt{\pi }$$

Thus, the integral evaluates to:

$${\int }_{-\infty }^{\infty }\exp \left(-a(x+id{)}^2\right)\mathrm{d}x=\sqrt{\frac{\pi }{a}}.$$

2. Fourier transform

(i) Expressing the partial differential equation of $U(k,t)$

Assume that $u(x,t)$ satisfies the following partial differential equation (PDE):

$$\frac{{\partial }^{2}u}{\partial t^2}=c^2\frac{{\partial }^{2}u}{\partial x^2}.$$

We need to find the corresponding PDE for $U(k,t)$. Taking the Fourier transform of both sides with respect to $x$:

1. Left-hand side (LHS):

$$\mathcal{F}\left(\frac{{\partial }^{2}u}{\partial t^2}\right)=\frac{{\partial }^2}{\partial t^2}\mathcal{F}\{u(x,t)\}=\frac{{\partial }^{2}U(k,t)}{\partial t^2}.$$

1. Right-hand side (RHS):

$$\mathcal{F}\left(c^2\frac{{\partial }^{2}u}{\partial x^2}\right)=c^2\mathcal{F}\left(\frac{{\partial }^{2}u}{\partial x^2}\right).$$

Using integration by parts twice, assuming boundary terms vanish due to given conditions:

$$\mathcal{F}\left(\frac{{\partial }^{2}u}{\partial x^2}\right)=\frac{1}{\sqrt{2\pi }}{\int }_{-\infty }^{\infty }\frac{{\partial }^{2}u}{\partial x^2}\exp (-ikx)\mathrm{d}x.$$

Integration by parts twice:

$${\int }_{-\infty }^{\infty }\frac{{\partial }^{2}u}{\partial x^2}\exp (-ikx)\mathrm{d}x={\left[\frac{\partial u}{\partial x}\exp (-ikx)\right]}_{-\infty }^{\infty }-{\int }_{-\infty }^{\infty }\frac{\partial u}{\partial x}(-ik)\exp (-ikx)\mathrm{d}x.$$

The boundary term vanishes:

$$=-ik{\int }_{-\infty }^{\infty }\frac{\partial u}{\partial x}\exp (-ikx)\mathrm{d}x.$$

Again, integration by parts:

$${\int }_{-\infty }^{\infty }\frac{\partial u}{\partial x}\exp (-ikx)\mathrm{d}x={\left[u\exp (-ikx)\right]}_{-\infty }^{\infty }-{\int }_{-\infty }^{\infty }u(-ik)\exp (-ikx)\mathrm{d}x.$$

The boundary term vanishes:

$$=ik{\int }_{-\infty }^{\infty }u\exp (-ikx)\mathrm{d}x.$$

Combining both parts, we get:

$$=-k^2{\int }_{-\infty }^{\infty }u\exp (-ikx)\mathrm{d}x.$$

Therefore,

$$\mathcal{F}\left(\frac{{\partial }^{2}u}{\partial x^2}\right)=-k^{2}U(k,t).$$

Therefore,

$$\mathcal{F}\left(c^2\frac{{\partial }^{2}u}{\partial x^2}\right)=-c^2{k}^{2}U(k,t).$$

Combining these results, we get:

$$\frac{{\partial }^{2}U(k,t)}{\partial t^2}=-c^2{k}^{2}U(k,t).$$

(ii) Showing the solution of (i) under the initial condition of Eq. (2.3)

We assume the initial conditions:

$$u(x,0)=f(x),\frac{\partial u}{\partial t}(x,0)=0.$$

Taking the Fourier transform with respect to $x$, we have:

$$U(k,0)=\mathcal{F}\{f(x)\}=F(k),{\frac{\partial U(k,t)}{\partial t}|}_{t=0}=0.$$

The PDE from part (i) is:

$$\frac{{\partial }^{2}U(k,t)}{\partial t^2}+c^2{k}^{2}U(k,t)=0.$$

This is a second-order linear differential equation with the general solution:

$$U(k,t)=A(k)\cos (kct)+B(k)\sin (kct).$$

Using the initial conditions:

1. At $t=0$,

$$U(k,0)=A(k)=F(k).$$

1. At $t=0$,

$${\frac{\partial U(k,t)}{\partial t}|}_{t=0}={\frac{\partial }{\partial t}\left(F(k)\cos (kct)+B(k)\sin (kct)\right)|}_{t=0}=F(k)k\sin (kct)+B(k)kc\cos (kct)=0.$$

Since $\sin (0)=0$ and $\cos (0)=1$,

$$B(k)kc=0⟹B(k)=0.$$

Therefore, the solution is:

$$U(k,t)=F(k)\cos (kct).$$

(iii) Determining $U(k,t)$ using initial condition of Eq. (2.2)

Given the initial condition:

$$u(x,0)=\exp (-a{x}^2),$$

we need to find $F(k)$.

1. Take the Fourier transform of $u(x,0)$:

$$F(k)=\frac{1}{\sqrt{2\pi }}{\int }_{-\infty }^{\infty }\exp (-a{x}^2)\exp (-ikx)\mathrm{d}x.$$

1. This integral is a standard Gaussian integral:

$$F(k)=\frac{1}{\sqrt{2\pi }}{\int }_{-\infty }^{\infty }\exp (-a{x}^2-ikx)\mathrm{d}x.$$

Completing the square in the exponent:

$$-a{x}^2-ikx=-a\left(x^2+\frac{ik}{a}x\right)=-a{\left(x+\frac{ik}{2a}\right)}^2+\frac{k^2}{4a}.$$

Therefore,

$$F(k)=\frac{1}{\sqrt{2\pi }}\exp \left(-\frac{k^2}{4a}\right){\int }_{-\infty }^{\infty }\exp \left(-a{\left(x+\frac{ik}{2a}\right)}^2\right)\mathrm{d}x.$$

Using the Gaussian integral result,

$${\int }_{-\infty }^{\infty }\exp \left(-a{\left(x+\frac{ik}{2a}\right)}^2\right)\mathrm{d}x=\sqrt{\frac{\pi }{a}}.$$

Thus,

$$F(k)=\frac{1}{\sqrt{2\pi }}\exp \left(-\frac{k^2}{4a}\right)\sqrt{\frac{\pi }{a}}=\frac{1}{\sqrt{2a}}\exp \left(-\frac{k^2}{4a}\right).$$

Finally, the solution is:

$$U(k,t)=\frac{1}{\sqrt{2a}}\exp \left(-\frac{k^2}{4a}\right)\cos (kct).$$

3. Inverse Fourier Transform

To find $u(x,t)$:

$$u(x,t)=\frac{1}{\sqrt{2\pi }}{\int }_{-\infty }^{\infty }U(k,t)\exp (ikx)\mathrm{d}k.$$

Substitute $U(k,t)=\frac{1}{\sqrt{2a}}\exp \left(-\frac{k^2}{4a}\right)\cos (kct)$:

$$u(x,t)=\frac{1}{\sqrt{2\pi }}{\int }_{-\infty }^{\infty }\frac{1}{\sqrt{2a}}\exp \left(-\frac{k^2}{4a}\right)\cos (kct)\exp (ikx)\mathrm{d}k.$$

Using the identity $\cos (kct)=\frac{1}{2}(\exp (ikct)+\exp (-ikct))$:

$$u(x,t)=\frac{1}{\sqrt{2\pi }}\frac{1}{\sqrt{2a}}{\int }_{-\infty }^{\infty }\exp \left(-\frac{k^2}{4a}\right)\left(\frac{1}{2}(\exp (ik(ct+x))+\exp (ik(x-ct)))\right)\mathrm{d}k.$$

This splits into two integrals:

$$u(x,t)=\frac{1}{2\sqrt{2\pi a}}\left({\int }_{-\infty }^{\infty }\exp \left(-\frac{k^2}{4a}\right)\exp (ik(ct+x))\mathrm{d}k+{\int }_{-\infty }^{\infty }\exp \left(-\frac{k^2}{4a}\right)\exp (ik(x-ct))\mathrm{d}k\right).$$

Each integral is a Fourier transform of a Gaussian function, which yields:

$$u(x,t)=\frac{1}{2}\left(\exp \left(-a(ct+x{)}^2\right)+\exp \left(-a(x-ct{)}^2\right)\right).$$

Therefore, the solution is:

$$u(x,t)=\frac{1}{2}\left(\exp \left(-a(x+ct{)}^2\right)+\exp \left(-a(x-ct{)}^2\right)\right).$$

知识点

偏微分方程傅里叶变换高斯积分波动方程

傅里叶变换求逆

难点解题思路

这道题的难点在于通过傅里叶变换将偏微分方程转化为常微分方程，并利用初始条件求解傅里叶变换后的方程。关键是理解傅里叶变换的性质，特别是微分运算在傅里叶变换下的表现形式。逆变换时，需要将结果拆分并利用高斯积分公式。

解题技巧和信息

1. 高斯积分：

$${\int }_{-\infty }^{\infty }\exp (-a{x}^2)\mathrm{d}x=\sqrt{\frac{\pi }{a}}$$

2. 傅里叶变换：将偏微分方程变换到频域中处理。
3. 逆傅里叶变换：通过求解频域中的表达式，回到时域中。

重点词汇

partial differential equation 偏微分方程

Fourier transform 傅里叶变换

Gaussian integral 高斯积分

integration by parts 分部积分

参考资料

1. “Partial Differential Equations” by Lawrence C. Evans, Chap. 2
2. “Fourier Transform and Its Applications” by Ronald N. Bracewell, Chap. 3
3. Walter A. Strauss, “Partial Differential Equations: An Introduction”, Chapter 7
4. Gerald B. Folland, “Fourier Analysis and Its Applications”, Chapter 2