IS Math-2017-02

题目来源：Problem 2 日期：2024-07-28 题目主题：Math-常微分方程-热方程

解题思路

这道题目涉及求解热传导方程，即一维热方程，并满足一定的边界条件和初始条件。首先需要通过分离变量法将偏微分方程化为两个常微分方程，并分别求解这两个常微分方程。然后结合边界条件和初始条件确定解的形式，并通过傅里叶级数展开确定初始条件对应的系数。

Solution

Question 1: Calculate the integral

To calculate the integral

$${\int }_0^1\sin (n\pi x)\sin (m\pi x)\mathrm{d}x,$$

we use the orthogonality of sine functions. Let’s explore this in more detail.

The orthogonality property of sine functions over the interval $[0,1]$ states that:

$${\int }_0^1\sin (n\pi x)\sin (m\pi x)\mathrm{d}x=0\text{if}n\ne m,$$

and

$${\int }_0^1\sin (n\pi x)\sin (m\pi x)\mathrm{d}x=\frac{1}{2}\text{if}n=m.$$

Proof of Orthogonality

To prove this property, we can use trigonometric identities and properties of integrals.

First, consider the product-to-sum identities for sine functions:

$$\sin (n\pi x)\sin (m\pi x)=\frac{1}{2}\left[\cos ((n-m)\pi x)-\cos ((n+m)\pi x)\right].$$

Using this identity, the integral becomes:

$${\int }_0^1\sin (n\pi x)\sin (m\pi x)\mathrm{d}x=\frac{1}{2}{\int }_0^1\left[\cos ((n-m)\pi x)-\cos ((n+m)\pi x)\right]\mathrm{d}x.$$

Now, we evaluate the two integrals separately.

1. Integral of $\cos ((n-m)\pi x)$:

$${\int }_0^1\cos ((n-m)\pi x)\mathrm{d}x.$$

• When $n\ne m$:

$${\int }_0^1\cos ((n-m)\pi x)\mathrm{d}x={\left[\frac{\sin ((n-m)\pi x)}{(n-m)\pi }\right]}_0^1=\frac{\sin ((n-m)\pi )}{(n-m)\pi }-\frac{\sin (0)}{(n-m)\pi }=0.$$

• When $n=m$:

$${\int }_0^1\cos ((n-m)\pi x)\mathrm{d}x={\int }_0^1\cos (0)\mathrm{d}x={\int }_0^{1}1\mathrm{d}x=1.$$

1. Integral of $\cos ((n+m)\pi x)$:

$${\int }_0^1\cos ((n+m)\pi x)\mathrm{d}x.$$

For any positive integers $n$ and $m$:

$${\int }_0^1\cos ((n+m)\pi x)\mathrm{d}x={\left[\frac{\sin ((n+m)\pi x)}{(n+m)\pi }\right]}_0^1=\frac{\sin ((n+m)\pi )}{(n+m)\pi }-\frac{\sin (0)}{(n+m)\pi }=0.$$

Combining these results, we have:

- When $n \neq m$:

 $$

\int_0^1 \sin(n\pi x) \sin(m\pi x) , \mathrm{d}x = \frac{1}{2} \left(0 - 0\right) = 0.

- When $n = m$:

 \int_0^1 \sin(n\pi x) \sin(n\pi x) \, \mathrm{d}x = \frac{1}{2} \left(1 - 0\right) = \frac{1}{2}.

$$

Therefore, the orthogonality property of sine functions is verified:

$${\int }_0^1\sin (n\pi x)\sin (m\pi x)\mathrm{d}x=\{\begin{array}{ll}0& \text{if }n\ne m,\\ \frac{1}{2}& \text{if }n=m.\end{array}$$

This property simplifies many problems involving Fourier series and orthogonal functions, as it allows us to separate coefficients in series expansions cleanly.

Question 2: Separation of variables

Suppose $u(x,t)=\xi (x)\tau (t)$. Substituting this into the PDE:

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

we get:

$$\xi (x)\frac{\mathrm{d}\tau (t)}{\mathrm{d}t}=\tau (t)\frac{{\mathrm{d}}^2\xi (x)}{\mathrm{d}{x}^2}.$$

Dividing both sides by $\xi (x)\tau (t)$, we have:

$$\frac{1}{\tau (t)}\frac{\mathrm{d}\tau (t)}{\mathrm{d}t}=\frac{1}{\xi (x)}\frac{{\mathrm{d}}^2\xi (x)}{\mathrm{d}{x}^2}=-C,$$

where $C$ is a separation constant.

This yields two ordinary differential equations:

1. $\frac{\mathrm{d}\tau (t)}{\mathrm{d}t}=-C\tau (t)$,
2. $\frac{{\mathrm{d}}^2\xi (x)}{\mathrm{d}{x}^2}+C\xi (x)=0$.

Question 3: Solve the ODEs and find $u_n(x,t)$

We will discuss the cases based on the sign of $C$.

Case 1: $C=0$

1. For $\tau (t)$:

   $$\frac{\mathrm{d}\tau (t)}{\mathrm{d}t}=0.$$

$$Thesolutionis:$$

\tau(t) = \text{constant}.

2. For $\xi(x)$:

\frac{\mathrm{d}^2 \xi(x)}{\mathrm{d} x^2} = 0.

$$Thegeneralsolutionis:$$

\xi(x) = Ax + B.$$

Applying the boundary conditions $u(0,t)=0$ and $u(1,t)=0$:

$\xi (0)=0⟹B=0,$

$\xi (1)=0⟹A\cdot 1=0⟹A=0.$

Thus, the solution for $\xi (x)$ is:

$$\xi (x)=0.$$

Therefore, the trivial solution $u(x,t)=0$ is obtained when $C=0$.

Case 2: $C<0$

Let $C=-{\lambda }^2$ where $\lambda$ is a positive constant.

1. For $\tau (t)$:

   $\frac{\mathrm{d}\tau (t)}{\mathrm{d}t}={\lambda }^2\tau (t).$

   The solution is:

   $\tau (t)=A{e}^{{\lambda }^{2}t},$

   where $A$ is a constant.

2. For $\xi (x)$:

   $\frac{{\mathrm{d}}^2\xi (x)}{\mathrm{d}{x}^2}-{\lambda }^2\xi (x)=0.$

   The general solution is:

   $\xi (x)=B{e}^{\lambda x}+C{e}^{-\lambda x}.$

   Applying the boundary conditions $u(0,t)=0$ and $u(1,t)=0$:

   $\xi (0)=0⟹B+C=0⟹C=-B,$

   $\xi (1)=0⟹B{e}^{\lambda }-B{e}^{-\lambda }=0⟹B(e^{\lambda }-e^{-\lambda })=0⟹B=0.$

   Thus, the solution for $\xi (x)$ is:

   $\xi (x)=0.$

Therefore, the trivial solution $u(x,t)=0$ is obtained when $C<0$.

Case 3: $C>0$

Let $C={\lambda }^2$ where $\lambda$ is a positive constant.

1. For $\tau (t)$:

   $\frac{\mathrm{d}\tau (t)}{\mathrm{d}t}=-{\lambda }^2\tau (t).$

   The solution is:

   $\tau (t)=A{e}^{-{\lambda }^{2}t},$

   where $A$ is a constant.

2. For $\xi (x)$:

   $\frac{{\mathrm{d}}^2\xi (x)}{\mathrm{d}{x}^2}+{\lambda }^2\xi (x)=0.$

   The general solution is:

   $\xi (x)=B\sin (\lambda x)+C\cos (\lambda x)$

   Applying the boundary conditions $u(0,t)=0$ and $u(1,t)=0$:

   $\xi (0)=0⟹C=0,$

   $\xi (1)=0⟹B\sin (\lambda )=0⟹\lambda =n\pi ,$

   where $n$ is a positive integer.

Thus, the solution for $\xi (x)$ is:

${\xi }_n(x)=B\sin (n\pi x).$

The solution for ${\tau }_n(t)$ is:

${\tau }_n(t)=A{e}^{-(n\pi {)}^{2}t}$

Therefore, a solution of partial differential equation which satisfies the boundary condition can be given as:

$u_n(x,t)=A_n{e}^{-(n\pi {)}^{2}t}\sin (n\pi x),$

Question 4: Find $c_n$ for the initial condition

To obtain the general solution, we linearly combine the particular solutions for each $n$. Since the heat equation is linear, any linear combination of solutions is also a solution. Therefore, we combine the particular solutions ${\xi }_n(x)$ and ${\tau }_n(t)$ for each $n$ to form the general solution:

$u(x,t)={\sum }_{n=1}^{\infty }{c}_n{e}^{-(n\pi {)}^{2}t}\sin (n\pi x),$

Using the initial condition $u(x,0)=x-x^2$:

$$u(x,0)=\sum _{n=1}^{\infty }{c}_n\sin (n\pi x)=x-x^2.$$

To find $c_n$, we use the orthogonality of sine functions. Multiply both sides by $\sin (m\pi x)$ and integrate from 0 to 1:

$${\int }_0^1(x-x^2)\sin (m\pi x)\mathrm{d}x=\sum _{n=1}^{\infty }{c}_n{\int }_0^1\sin (n\pi x)\sin (m\pi x)\mathrm{d}x.$$

Using the orthogonality property:

$${\int }_0^1(x-x^2)\sin (m\pi x)\mathrm{d}x=c_m\frac{1}{2}.$$

Thus:

$$c_m=2{\int }_0^1(x-x^2)\sin (m\pi x)\mathrm{d}x.$$

To find $c_n$, we need to evaluate the integral:

$$c_n=2{\int }_0^1(x-x^2)\sin (n\pi x)\mathrm{d}x.$$

We split the integral into two parts:

$$c_n=2\left({\int }_0^{1}x\sin (n\pi x)\mathrm{d}x-{\int }_0^1{x}^2\sin (n\pi x)\mathrm{d}x\right)=2(I-J).$$

Part 1: Evaluate $I={\int }_0^{1}x\sin (n\pi x)\mathrm{d}x$

Using integration by parts, let:

$$u=x,\mathrm{d}v=\sin (n\pi x)\mathrm{d}x,$$

then:

$$\mathrm{d}u=\mathrm{d}x,v=-\frac{1}{n\pi }\cos (n\pi x).$$

So, the integral becomes:

$${\int }_0^{1}x\sin (n\pi x)\mathrm{d}x={-\frac{x}{n\pi }\cos (n\pi x)|}_0^1+{\int }_0^1\frac{1}{n\pi }\cos (n\pi x)\mathrm{d}x.$$

Evaluating the first term at the boundaries:

$$-\frac{1}{n\pi }{\left[x\cos (n\pi x)\right]}_0^1=-\frac{1}{n\pi }\left(1\cdot \cos (n\pi )-0\cdot \cos (0)\right)=-\frac{\cos (n\pi )}{n\pi }.$$

For the second term, we integrate:

$${\int }_0^1\frac{1}{n\pi }\cos (n\pi x)\mathrm{d}x=\frac{1}{n\pi }{\frac{\sin (n\pi x)}{n\pi }|}_0^1=\frac{1}{(n\pi {)}^2}\left(\sin (n\pi )-\sin (0)\right)=0.$$

Thus, the integral ${\int }_0^{1}x\sin (n\pi x)\mathrm{d}x$ simplifies to:

$$I={\int }_0^{1}x\sin (n\pi x)\mathrm{d}x=-\frac{\cos (n\pi )}{n\pi }.$$

Part 2: Evaluate $J={\int }_0^1{x}^2\sin (n\pi x)\mathrm{d}x$

Again, using integration by parts, let:

$$u=x^2,\mathrm{d}v=\sin (n\pi x)\mathrm{d}x,$$

then:

$$\mathrm{d}u=2x\mathrm{d}x,v=-\frac{1}{n\pi }\cos (n\pi x).$$

So, the integral becomes:

$${\int }_0^1{x}^2\sin (n\pi x)\mathrm{d}x={-\frac{x^2}{n\pi }\cos (n\pi x)|}_0^1+{\int }_0^1\frac{2x}{n\pi }\cos (n\pi x)\mathrm{d}x.$$

Evaluating the first term at the boundaries:

$$-\frac{1}{n\pi }{\left[x^2\cos (n\pi x)\right]}_0^1=-\frac{1}{n\pi }\left(1\cdot \cos (n\pi )-0\cdot \cos (0)\right)=-\frac{\cos (n\pi )}{n\pi }.$$

For the second term, we use integration by parts again. Let:

$$u=x,\mathrm{d}v=\frac{2}{n\pi }\cos (n\pi x)\mathrm{d}x,$$

then:

$$\mathrm{d}u=\mathrm{d}x,v=\frac{2}{(n\pi {)}^2}\sin (n\pi x).$$

So, the integral becomes:

$${\int }_0^1\frac{2x}{n\pi }\cos (n\pi x)\mathrm{d}x={\frac{2x}{(n\pi {)}^2}\sin (n\pi x)|}_0^1-{\int }_0^1\frac{2}{(n\pi {)}^2}\sin (n\pi x)\mathrm{d}x.$$

Evaluating the first term at the boundaries:

$$\frac{2}{(n\pi {)}^2}{\left[x\sin (n\pi x)\right]}_0^1=\frac{2}{(n\pi {)}^2}\left(1\cdot \sin (n\pi )-0\cdot \sin (0)\right)=0.$$

For the second term, we integrate:

$${\int }_0^1\frac{2}{(n\pi {)}^2}\sin (n\pi x)\mathrm{d}x=\frac{2}{(n\pi {)}^3}{\left[-\cos (n\pi x)\right]}_0^1=\frac{2}{(n\pi {)}^3}\left(-\cos (n\pi )+\cos (0)\right)=\frac{2(1-\cos (n\pi ))}{(n\pi {)}^3}.$$

Thus, the integral ${\int }_0^1{x}^2\sin (n\pi x)\mathrm{d}x$ simplifies to:

$${\int }_0^1{x}^2\sin (n\pi x)\mathrm{d}x=-\frac{\cos (n\pi )}{n\pi }+\frac{2(1-\cos (n\pi ))}{(n\pi {)}^3}.$$

Combining both parts, we get:

$$c_n=2\left(-\frac{\cos (n\pi )}{n\pi }-\left(-\frac{\cos (n\pi )}{n\pi }+\frac{2(1-\cos (n\pi ))}{(n\pi {)}^3}\right)\right)=\frac{4(1-\cos (n\pi ))}{(n\pi {)}^3}=\frac{4(1-(-1{)}^n))}{(n\pi {)}^3}.$$

Finally, the coefficients $c_n$ are:

$$c_n=\frac{8}{(n\pi {)}^3}\text{ for odd }n,\text{ and }{c}_n=0\text{ for even }n.$$

Final solution

The solution to the heat equation with the given initial condition is:

$$u(x,t)=\sum _{\begin{array}{c}n=1\\ n\text{ odd}\end{array}}^{\infty }\frac{8}{(n\pi {)}^3}\exp \left(-(n\pi {)}^{2}t\right)\sin (n\pi x).$$

知识点

常微分方程偏微分方程热方程分离变量法傅里叶级数

重点词汇

1. Heat equation - 热方程
2. Separation of variables - 分离变量法
3. Fourier series - 傅里叶级数
4. Orthogonality - 正交性
5. Boundary conditions - 边界条件
6. Initial conditions - 初始条件

参考资料

1. Kreyszig, E. (2011). Advanced Engineering Mathematics (10th ed.). Wiley. - Chapter 11: Fourier Series and Integrals
2. Strauss, W. A. (2007). Partial Differential Equations: An Introduction (2nd ed.). Wiley. - Chapter 4: The Heat Equation