PDE — Homework 10

Problem 1.

用分离变量法求解：

\begin{cases} \frac{\partial^2 u}{\partial t^2} - a^2 \cdot \frac{\partial^2 u}{\partial x^2} = 0, \, t > 0\\ u|_{t=0} = \sin\frac{3\pi x}{l}, \, \frac{\partial u}{\partial t}|_{t=0} = x(l-x)\\ u(0,t) = u(l,t) = 0 \end{cases}

解.

对于一般的问题：

\begin{cases} \frac{\partial^2 u}{\partial t^2} - a^2 \cdot \frac{\partial^2 u}{\partial x^2} = 0 ,\, t>0\\ u|_{t=0} = \varphi(x), \frac{\partial u}{\partial t}|_{t=0} = g(x) \\ u(0,t) = u(l,t) = 0 \end{cases}

使用分离变量法，我们令 $u(x,t) = X(x)T(t)$ ，代入方程得：

\frac{X''(x)}{X(x)} = \frac{T''(t)}{a^2 T(t)} = -\mu

因此 $X(x)$ 满足如下 Sturm-Liouville 问题：

\begin{cases} X''(x) + \mu X(x) = 0 \\ X(0) = X(l) = 0 \end{cases}

我们可得特征值为 $\mu_k = \left(\frac{k\pi}{l}\right)^2, k = 1, 2, \cdots$ ，特征函数为 $X_k(x) = \sin \frac{k\pi}{l} x$ 。

这些特征函数构成完全正交系 $\left\{\sin \frac{k\pi}{l} x\right\}$ ，满足正交性关系：

\int_0^l \sin\frac{k\pi}{l}x \cdot \sin\frac{m\pi}{l}x dx = \frac{l}{2}\delta_{k,m}= \begin{cases} 0, & m \neq k \\ \frac{l}{2}, & m = k \end{cases}

将特征值 $\mu_k$ 代入与 $t$ 相关的方程：

T_k''(t) + a^2 \mu_k T_k(t) = 0

解得：

T_k(t) = \bar{A}_k \sin \frac{a k \pi}{l} t + \bar{B}_k \cos \frac{a k \pi}{l} t

故：

u_k(x,t) = T_k(t) \cdot X_k(x) = \left(\bar{A}_k \sin \frac{a k \pi}{l} t + \bar{B}_k \cos \frac{a k \pi}{l} t\right) \cdot \sin \frac{k \pi}{l} x

进而我们可知：

u(x,t) = \sum_{k=1}^\infty u_k(x,t) = \sum_{k=1}^\infty \left(A_k \cos \frac{a k \pi}{l} t + B_k \sin \frac{a k \pi}{l} t\right) \cdot \sin \frac{k \pi}{l} x

其中 $A_k = \bar{B}_k$ , $B_k = \bar{A}_k$ 为待定系数。

回到我们的问题：

\begin{cases} \frac{\partial^2 u}{\partial t^2} - a^2 \cdot \frac{\partial^2 u}{\partial x^2} = 0, t > 0 \\ u|_{t=0} = \sin\frac{3\pi x}{l}, \frac{\partial u}{\partial t}|_{t=0} = x(l-x) \\ u(0,t) = u(l,t) = 0 \end{cases}

接下来我们来确定系数 $A_k$ 与 $B_k$ ：

确定系数 $A_k$ ： 由 $u|_{t=0} = \sin\frac{3\pi x}{l}$ 我们可得：

u(x,0) = \sum_{k=1}^{\infty} A_k \sin\frac{k\pi x}{l} = \sin\frac{3\pi x}{l}

我们在上式两边乘上 $\sin\frac{m\pi x}{l}$ 并在 $[0,l]$ 上积分：

A_m \cdot \frac{l}{2} = \int_0^l \sin\frac{3\pi x}{l} \cdot \sin\frac{m\pi x}{l} dx

而

\int_0^l \sin\frac{3\pi x}{l} \sin\frac{m\pi x}{l} dx = \begin{cases} 0, & m \neq 3 \\ \frac{l}{2}, & m = 3 \end{cases}

故

A_k = \begin{cases} 1, & k = 3 \\ 0, & k \neq 3 \end{cases}

确定系数 $B_k$ ： 由 $\frac{\partial u}{\partial t}|_{t=0} = x(l-x)$ 我们可得：

\frac{\partial u}{\partial t}(x,0) = \sum_{k=1}^{\infty} B_k \cdot \frac{ak\pi}{l} \sin\frac{k\pi x}{l} = x(l-x)

我们在上式两边乘上 $\sin\frac{m\pi x}{l}$ 并在 $[0,l]$ 上积分：

\frac{am\pi}{l} B_m \cdot \frac{l}{2} = \int_0^l x(l-x) \sin\frac{m\pi x}{l} dx

即：

B_m = \frac{2}{am\pi} \int_0^l x(l-x) \sin\frac{m\pi x}{l} dx

接下来我们来计算积分：

\begin{aligned} \begin{split} I_m &= \int_0^l x(l-x) \sin\frac{m\pi x}{l} dx \\ &= l \int_0^l x \sin\frac{m\pi x}{l} dx - \int_0^l x^2 \sin\frac{m\pi x}{l} dx \end{split} \end{aligned}

其中：

\int_0^l x \sin\frac{m\pi x}{l} dx = \left[ -\frac{lx}{m\pi} \cos\frac{m\pi x}{l} \right]_0^l + \frac{l}{m\pi} \int_0^l \cos\frac{m\pi x}{l} dx

= -\frac{l^2}{m\pi} \cos(m\pi) + \frac{l^2}{m^2\pi^2} \sin\frac{m\pi x}{l} \Big|_0^l = -\frac{l^2}{m\pi} (-1)^m

\begin{aligned} \begin{split} \int_0^l x^2 \sin\frac{m\pi x}{l} dx &= \left[ -\frac{lx^2}{m\pi} \cos\frac{m\pi x}{l} \right]_0^l + \frac{2l}{m\pi} \int_0^l x \cos\frac{m\pi x}{l} dx \\ &= -\frac{l^3}{m\pi} (-1)^m + \frac{2l}{m\pi} \left[ \frac{lx}{m\pi} \sin\frac{m\pi x}{l} \Big|_0^l - \frac{l}{m\pi} \int_0^l \sin\frac{m\pi x}{l} dx \right] \\ &= -\frac{l^3}{m\pi} (-1)^m + \frac{2l}{m\pi} \left[ 0 + \frac{l^2}{m^2\pi^2} \cos\frac{m\pi x}{l} \Big|_0^l \right] \\ &= -\frac{l^3}{m\pi} (-1)^m + \frac{2l^3}{m^3\pi^3} [(-1)^m - 1] \end{split} \end{aligned}

故

\begin{aligned} \begin{split} I_m &= l \left[ -\frac{l^2}{m\pi} (-1)^m \right] - \left[ -\frac{l^3}{m\pi} (-1)^m + \frac{2l^3}{m^3\pi^3} [(-1)^m - 1] \right] \\ &= -\frac{l^3}{m\pi} (-1)^m + \frac{l^3}{m\pi} (-1)^m - \frac{2l^3}{m^3\pi^3} [(-1)^m - 1] \\ &= \frac{2l^3}{m^3\pi^3} [1 - (-1)^m] \end{split} \end{aligned}

代入 $B_m$ 的表达式：

B_m = \frac{2}{am\pi} \cdot \frac{2l^3}{m^3\pi^3} [1 - (-1)^m] = \frac{4l^3}{am^4\pi^4} [1 - (-1)^m]

即：

B_k = \begin{cases} 0, & 2\mid k \\ \frac{8l^3}{ak^4\pi^4}, & 2\nmid k \end{cases}

最后我们将求解出来的系数代入通解可得：

\begin{aligned} \begin{split} u(x,t) &= A_3 \cos\frac{3a\pi}{l}t \cdot \sin\frac{3\pi x}{l} + \sum_{k=1}^{\infty} B_k \sin\frac{ak\pi}{l}t \cdot \sin\frac{k\pi x}{l} \\ &= \cos\frac{3a\pi}{l}t \cdot \sin\frac{3\pi x}{l} + \frac{8l^3}{a\pi^4} \sum_{\substack{k=1 \\ 2 \nmid k}}^{\infty} \frac{1}{k^4} \sin\frac{ak\pi}{l}t \cdot \sin\frac{k\pi x}{l} \end{split} \end{aligned}

即为我们最终所求的解.

∎

PDF

Download the original write-up here.

Problem 1.

用分离变量法求解：

\begin{cases} \frac{\partial^2 u}{\partial t^2} - a^2 \cdot \frac{\partial^2 u}{\partial x^2} = 0, \, t > 0\\ u|_{t=0} = \sin\frac{3\pi x}{l}, \, \frac{\partial u}{\partial t}|_{t=0} = x(l-x)\\ u(0,t) = u(l,t) = 0 \end{cases}

解.

对于一般的问题：

\begin{cases} \frac{\partial^2 u}{\partial t^2} - a^2 \cdot \frac{\partial^2 u}{\partial x^2} = 0 ,\, t>0\\ u|_{t=0} = \varphi(x), \frac{\partial u}{\partial t}|_{t=0} = g(x) \\ u(0,t) = u(l,t) = 0 \end{cases}

使用分离变量法，我们令 $u(x,t) = X(x)T(t)$ ，代入方程得：

\frac{X''(x)}{X(x)} = \frac{T''(t)}{a^2 T(t)} = -\mu

因此 $X(x)$ 满足如下 Sturm-Liouville 问题：

\begin{cases} X''(x) + \mu X(x) = 0 \\ X(0) = X(l) = 0 \end{cases}

我们可得特征值为 $\mu_k = \left(\frac{k\pi}{l}\right)^2, k = 1, 2, \cdots$ ，特征函数为 $X_k(x) = \sin \frac{k\pi}{l} x$ 。

这些特征函数构成完全正交系 $\left\{\sin \frac{k\pi}{l} x\right\}$ ，满足正交性关系：

\int_0^l \sin\frac{k\pi}{l}x \cdot \sin\frac{m\pi}{l}x dx = \frac{l}{2}\delta_{k,m}= \begin{cases} 0, & m \neq k \\ \frac{l}{2}, & m = k \end{cases}

将特征值 $\mu_k$ 代入与 $t$ 相关的方程：

T_k''(t) + a^2 \mu_k T_k(t) = 0

解得：

T_k(t) = \bar{A}_k \sin \frac{a k \pi}{l} t + \bar{B}_k \cos \frac{a k \pi}{l} t

故：

u_k(x,t) = T_k(t) \cdot X_k(x) = \left(\bar{A}_k \sin \frac{a k \pi}{l} t + \bar{B}_k \cos \frac{a k \pi}{l} t\right) \cdot \sin \frac{k \pi}{l} x

进而我们可知：

u(x,t) = \sum_{k=1}^\infty u_k(x,t) = \sum_{k=1}^\infty \left(A_k \cos \frac{a k \pi}{l} t + B_k \sin \frac{a k \pi}{l} t\right) \cdot \sin \frac{k \pi}{l} x

其中 $A_k = \bar{B}_k$ , $B_k = \bar{A}_k$ 为待定系数。

回到我们的问题：

\begin{cases} \frac{\partial^2 u}{\partial t^2} - a^2 \cdot \frac{\partial^2 u}{\partial x^2} = 0, t > 0 \\ u|_{t=0} = \sin\frac{3\pi x}{l}, \frac{\partial u}{\partial t}|_{t=0} = x(l-x) \\ u(0,t) = u(l,t) = 0 \end{cases}

接下来我们来确定系数 $A_k$ 与 $B_k$ ：

确定系数 $A_k$ ： 由 $u|_{t=0} = \sin\frac{3\pi x}{l}$ 我们可得：

u(x,0) = \sum_{k=1}^{\infty} A_k \sin\frac{k\pi x}{l} = \sin\frac{3\pi x}{l}

我们在上式两边乘上 $\sin\frac{m\pi x}{l}$ 并在 $[0,l]$ 上积分：

A_m \cdot \frac{l}{2} = \int_0^l \sin\frac{3\pi x}{l} \cdot \sin\frac{m\pi x}{l} dx

而

\int_0^l \sin\frac{3\pi x}{l} \sin\frac{m\pi x}{l} dx = \begin{cases} 0, & m \neq 3 \\ \frac{l}{2}, & m = 3 \end{cases}

故

A_k = \begin{cases} 1, & k = 3 \\ 0, & k \neq 3 \end{cases}

确定系数 $B_k$ ： 由 $\frac{\partial u}{\partial t}|_{t=0} = x(l-x)$ 我们可得：

\frac{\partial u}{\partial t}(x,0) = \sum_{k=1}^{\infty} B_k \cdot \frac{ak\pi}{l} \sin\frac{k\pi x}{l} = x(l-x)

我们在上式两边乘上 $\sin\frac{m\pi x}{l}$ 并在 $[0,l]$ 上积分：

\frac{am\pi}{l} B_m \cdot \frac{l}{2} = \int_0^l x(l-x) \sin\frac{m\pi x}{l} dx

即：

B_m = \frac{2}{am\pi} \int_0^l x(l-x) \sin\frac{m\pi x}{l} dx

接下来我们来计算积分：

\begin{aligned} \begin{split} I_m &= \int_0^l x(l-x) \sin\frac{m\pi x}{l} dx \\ &= l \int_0^l x \sin\frac{m\pi x}{l} dx - \int_0^l x^2 \sin\frac{m\pi x}{l} dx \end{split} \end{aligned}

其中：

\int_0^l x \sin\frac{m\pi x}{l} dx = \left[ -\frac{lx}{m\pi} \cos\frac{m\pi x}{l} \right]_0^l + \frac{l}{m\pi} \int_0^l \cos\frac{m\pi x}{l} dx

= -\frac{l^2}{m\pi} \cos(m\pi) + \frac{l^2}{m^2\pi^2} \sin\frac{m\pi x}{l} \Big|_0^l = -\frac{l^2}{m\pi} (-1)^m

\begin{aligned} \begin{split} \int_0^l x^2 \sin\frac{m\pi x}{l} dx &= \left[ -\frac{lx^2}{m\pi} \cos\frac{m\pi x}{l} \right]_0^l + \frac{2l}{m\pi} \int_0^l x \cos\frac{m\pi x}{l} dx \\ &= -\frac{l^3}{m\pi} (-1)^m + \frac{2l}{m\pi} \left[ \frac{lx}{m\pi} \sin\frac{m\pi x}{l} \Big|_0^l - \frac{l}{m\pi} \int_0^l \sin\frac{m\pi x}{l} dx \right] \\ &= -\frac{l^3}{m\pi} (-1)^m + \frac{2l}{m\pi} \left[ 0 + \frac{l^2}{m^2\pi^2} \cos\frac{m\pi x}{l} \Big|_0^l \right] \\ &= -\frac{l^3}{m\pi} (-1)^m + \frac{2l^3}{m^3\pi^3} [(-1)^m - 1] \end{split} \end{aligned}

故

\begin{aligned} \begin{split} I_m &= l \left[ -\frac{l^2}{m\pi} (-1)^m \right] - \left[ -\frac{l^3}{m\pi} (-1)^m + \frac{2l^3}{m^3\pi^3} [(-1)^m - 1] \right] \\ &= -\frac{l^3}{m\pi} (-1)^m + \frac{l^3}{m\pi} (-1)^m - \frac{2l^3}{m^3\pi^3} [(-1)^m - 1] \\ &= \frac{2l^3}{m^3\pi^3} [1 - (-1)^m] \end{split} \end{aligned}

代入 $B_m$ 的表达式：

B_m = \frac{2}{am\pi} \cdot \frac{2l^3}{m^3\pi^3} [1 - (-1)^m] = \frac{4l^3}{am^4\pi^4} [1 - (-1)^m]

即：

B_k = \begin{cases} 0, & 2\mid k \\ \frac{8l^3}{ak^4\pi^4}, & 2\nmid k \end{cases}

最后我们将求解出来的系数代入通解可得：

\begin{aligned} \begin{split} u(x,t) &= A_3 \cos\frac{3a\pi}{l}t \cdot \sin\frac{3\pi x}{l} + \sum_{k=1}^{\infty} B_k \sin\frac{ak\pi}{l}t \cdot \sin\frac{k\pi x}{l} \\ &= \cos\frac{3a\pi}{l}t \cdot \sin\frac{3\pi x}{l} + \frac{8l^3}{a\pi^4} \sum_{\substack{k=1 \\ 2 \nmid k}}^{\infty} \frac{1}{k^4} \sin\frac{ak\pi}{l}t \cdot \sin\frac{k\pi x}{l} \end{split} \end{aligned}

即为我们最终所求的解.

∎

PDF

Download the original write-up here.

PDE — Homework 10

Search

PDE — Homework 10

PDF

PDF