Laplace Transform

拉普拉斯变换 Laplace Transform

定义 Definitions

拉普拉斯变换 Laplace Transform

给定一个函数 $f(t)$，其拉普拉斯变换定义为

$$\mathcal{L}\{f(t)\}=F(s)={\int }_0^{\infty }f(t)e^{-st}\mathrm{d}t$$

其中 $s$ 是复数变量。

Given a function $f(t)$, its Laplace transform is defined as

$$\mathcal{L}\{f(t)\}=F(s)={\int }_0^{\infty }f(t)e^{-st}\mathrm{d}t$$

where $s$ is a complex variable.

拉普拉斯逆变换 Inverse Laplace Transform

已知 $F(s)$ 是 $f(t)$ 的拉普拉斯变换，则 $f(t)$ 的拉普拉斯逆变换定义为

$${\mathcal{L}}^{-1}\{F(s)\}=f(t)=\frac{1}{2\pi i}{\int }_{\gamma -i\infty }^{\gamma +i\infty }F(s)e^{st}\mathrm{d}s$$

其中 $\gamma$ 是使得积分路径位于 $F(s)$ 的奇点右侧的实数。

Given $F(s)$ as the Laplace transform of $f(t)$, the inverse Laplace transform is defined as

$${\mathcal{L}}^{-1}\{F(s)\}=f(t)=\frac{1}{2\pi i}{\int }_{\gamma -i\infty }^{\gamma +i\infty }F(s)e^{st}\mathrm{d}s$$

where $\gamma$ is a real number such that the path of integration is to the right of all singularities of $F(s)$.

性质 Properties

线性 Linearity

拉普拉斯变换是线性的，即

$$\mathcal{L}\{af(t)+bg(t)\}=a\mathcal{L}\{f(t)\}+b\mathcal{L}\{g(t)\}$$

The Laplace transform is linear, i.e.,

$$\mathcal{L}\{af(t)+bg(t)\}=a\mathcal{L}\{f(t)\}+b\mathcal{L}\{g(t)\}$$

平移定理 Shift Theorem

如果 $f(t)$ 的拉普拉斯变换是 $F(s)$，则

$$\mathcal{L}\{e^{at}f(t)\}=F(s-a)$$

If the Laplace transform of $f(t)$ is $F(s)$, then

$$\mathcal{L}\{e^{at}f(t)\}=F(s-a)$$

导数 Derivative

如果 $f(t)$ 的拉普拉斯变换是 $F(s)$，则

$$\mathcal{L}\{f^{\prime }(t)\}=sF(s)-f(0)$$

If the Laplace transform of $f(t)$ is $F(s)$, then

$$\mathcal{L}\{f^{\prime }(t)\}=sF(s)-f(0)$$

证明 Proof

拉普拉斯变换的导数性质 Proof of Derivative Property of Laplace Transform

$$\begin{array}{rl}\mathcal{L}\{f^{\prime }(t)\}& ={\int }_0^{\infty }{f}^{\prime }(t)e^{-st}\mathrm{d}t\\ & ={\left[f(t)e^{-st}\right]}_0^{\infty }+s{\int }_0^{\infty }f(t)e^{-st}\mathrm{d}t\\ & =-f(0)+sF(s)\end{array}$$

利用拉普拉斯变换解微分方程 Solving Differential Equations Using Laplace Transform

例子 Example: 热方程 Heat Equation

考虑如下热方程 Consider the heat equation:

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

其中 $\alpha$ 是热扩散系数 where $\alpha$ is the thermal diffusivity.

初始条件 Initial Condition:

$$u(x,0)=f(x)$$

边界条件 Boundary Conditions:

$$u(0,t)=0,u(L,t)=0$$

1. 对时间变量 $t$ 进行拉普拉斯变换 Taking the Laplace transform with respect to $t$:

   $$\mathcal{L}\left\{\frac{\partial u}{\partial t}\right\}=\alpha \mathcal{L}\left\{\frac{{\partial }^{2}u}{\partial x^2}\right\}$$

记 $U(x, s) = \mathcal{L}\{u(x, t)\}$，利用拉普拉斯变换的导数性质得到： Let $U(x, s) = \mathcal{L}\{u(x, t)\}$, using the derivative property of the Laplace transform:

sU(x, s) - u(x, 0) = \alpha \frac{\partial^2 U}{\partial x^2}

代入初始条件 Substituting the initial condition $u(x, 0) = f(x)$:

sU(x, s) - f(x) = \alpha \frac{\partial^2 U}{\partial x^2}

$$2.\ast \ast 求解常微分方程Solvingtheordinarydifferentialequation\ast \ast :$$

\frac{\partial^2 U}{\partial x^2} - \frac{s}{\alpha} U = -\frac{f(x)}{\alpha}

$$这是一个非齐次常系数二阶微分方程Thisisanon-homogeneoussecond-orderdifferentialequationwithconstantcoefficients.其解为Thesolutionis:$$

U(x, s) = A(s) \sinh\left(\sqrt{\frac{s}{\alpha}} x\right) + B(s) \cosh\left(\sqrt{\frac{s}{\alpha}} x\right) + \frac{f(x)}{s}

$$3.\ast \ast 应用边界条件Applyingboundaryconditions\ast \ast :$$

\begin{cases}

U(0, s) = 0 \

U(L, s) = 0

\end{cases}

可以求得 $A(s)$ 和 $B(s)$ We can solve for $A(s)$ and $B(s)$:

\begin{cases}

B(s) = 0 \

A(s) \sinh\left(\sqrt{\frac{s}{\alpha}} L\right) = -\frac{f(x)}{s}

\end{cases}

$$因此Therefore:$$

U(x, s) = -\frac{f(x)}{s} \frac{\sinh\left(\sqrt{\frac{s}{\alpha}} x\right)}{\sinh\left(\sqrt{\frac{s}{\alpha}} L\right)}

$$4.\ast \ast 进行拉普拉斯逆变换TakingtheinverseLaplacetransform\ast \ast :$$

u(x, t) = \mathcal{L}^{-1}\left{ U(x, s) \right}

利用拉普拉斯逆变换，我们可以得到 $u(x, t)$ 的表达式 Using the inverse Laplace transform, we can obtain the expression for $u(x, t)$. 通过拉普拉斯变换和逆变换，我们可以解决一类偏微分方程 This demonstrates the method of solving a class of partial differential equations using Laplace transform and its inverse.