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拉普拉斯算子

## 拉普拉斯算子 
函数 $u$ 的梯度为 $\nabla u$ ，如果再求 $\nabla u$ 的散度，便成为 $\nabla \cdot(\nabla u) \equiv \nabla^2 u$ ，它是拉普拉斯算子 $\nabla^2$ 对 $u$ 的运算，简称拉普拉斯（Laplacian）。在偏微分方程中，$\nabla^2 u$ 扮演着十分重要的角色，特别是，我们在数学物理方法中将要讨论如下方程

$$
\begin{gathered}
\nabla^2 u=0 \quad \text { (拉普拉斯方程) } \\
\nabla^2 u+k^2 u=0 \quad \text { (亥姆霍兹方程) } \\
\frac{\partial^2 u}{\partial t^2}=a^2 \nabla^2 u \quad \text { (波动方程) } \\
\frac{\partial u}{\partial t}=a^2 \nabla^2 u \quad \text { (热传导方程) } \\
i \hbar \frac{\partial \psi}{\partial t}=-\frac{\hbar^2}{2 m} \nabla^2 \psi+V(r) \psi \quad \text { (薛定谔方程) }
\end{gathered}
$$

在这些方程中，$\nabla^2$ 可以是三维形式，也可以是二维形式

$$
\nabla^2=\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}
$$

## 极坐标系下的拉普拉斯算子

极坐标系如图所示，它与直角坐标系的关系为
 ![图片](/uploads/2025-04/2ed230.jpg){width=300px}

$$
x=r \cos \theta, \quad y=r \sin \theta
$$

而

$$
r^2=x^2+y^2, \quad \tan \theta=\frac{y}{x}
$$

利用 $r^2=x^2+y^2$ 计算 $r$ 对 $x$ 的微商，得到

$$
2 r \frac{\partial r}{\partial x}=2 x \Rightarrow \frac{\partial r}{\partial x}=\frac{x}{r}
$$

再次对 $x$ 微商，得到

$$
\frac{\partial^2 r}{\partial x^2}=\frac{r-x \frac{\partial r}{\partial x}}{r^2}=\frac{r-x \frac{x}{r}}{r^2}=\frac{r^2-x^2}{r^3}=\frac{y^2}{r^3}
$$

利用 $\tan \theta=\frac{y}{x}$ 计算 $\theta$ 对 $x$ 的微商，得到

$$
\frac{\partial \theta}{\partial x}=\frac{1}{1+\left(\frac{y}{x}\right)^2}\left(-\frac{y}{x^2}\right)=-\frac{y}{r^2}
$$

再次对 $\theta$ 微商得

$$
\frac{\partial^2 \theta}{\partial x^2}=\frac{2 y}{r^3} \frac{\partial r}{\partial x}=\frac{2 x y}{r^4}
$$

现在对 $y$ 微分，按照类似的过程，我们得到

$$
\frac{\partial r}{\partial y}=\frac{y}{r}, \quad \frac{\partial^2 r}{\partial y^2}=\frac{x^2}{r^3}, \quad \frac{\partial \theta}{\partial y}=\frac{x}{r^2}, \quad \frac{\partial^2 \theta}{\partial y^2}=-\frac{2 x y}{r^4}
$$

从以上结果，容易推出下面两个关系式

$$
\begin{gathered}
\frac{\partial^2 \theta}{\partial x^2}+\frac{\partial^2 \theta}{\partial y^2}=0 \\
\frac{\partial r}{\partial x} \frac{\partial \theta}{\partial x}+\frac{\partial r}{\partial y} \frac{\partial \theta}{\partial y}=0
\end{gathered}
$$

现在我们可以着手将拉普拉斯算子变换到极坐标系。利用复合函数的微分法则，我们有

$$
\frac{\partial u}{\partial x}=\frac{\partial u}{\partial r} \frac{\partial r}{\partial x}+\frac{\partial u}{\partial \theta} \frac{\partial \theta}{\partial x}
$$

再次对 $x$ 微商得

$$
\begin{aligned}
\frac{\partial^2 u}{\partial x^2}= & \frac{\partial}{\partial x}\left(\frac{\partial u}{\partial r}\right) \frac{\partial r}{\partial x}+\frac{\partial u}{\partial r} \frac{\partial^2 r}{\partial x^2}+\frac{\partial}{\partial x}\left(\frac{\partial u}{\partial \theta}\right) \frac{\partial \theta}{\partial x}+\frac{\partial u}{\partial \theta} \frac{\partial^2 \theta}{\partial x^2} \\
= & \left(\frac{\partial^2 u}{\partial r^2} \frac{\partial r}{\partial x}+\frac{\partial^2 u}{\partial r \partial \theta} \frac{\partial \theta}{\partial x}\right) \frac{\partial r}{\partial x}+\frac{\partial u}{\partial r} \frac{\partial^2 r}{\partial x^2} \\
& +\left(\frac{\partial^2 u}{\partial r \partial \theta} \frac{\partial r}{\partial x}+\frac{\partial^2 u}{\partial \theta^2} \frac{\partial \theta}{\partial x}\right) \frac{\partial \theta}{\partial x}+\frac{\partial u}{\partial \theta} \frac{\partial^2 \theta}{\partial x^2} \\
= & \frac{\partial^2 u}{\partial r^2}\left(\frac{\partial r}{\partial x}\right)^2+2 \frac{\partial^2 u}{\partial r \partial \theta} \frac{\partial r}{\partial x} \frac{\partial \theta}{\partial x}+\frac{\partial u}{\partial r} \frac{\partial^2 r}{\partial x^2}+\frac{\partial^2 u}{\partial \theta^2}\left(\frac{\partial \theta}{\partial x}\right)^2+\frac{\partial u}{\partial \theta} \frac{\partial^2 \theta}{\partial x^2}
\end{aligned}
$$

对于 $y$ 有同样的结果

$$
\frac{\partial^2 u}{\partial y^2}=\frac{\partial^2 u}{\partial r^2}\left(\frac{\partial r}{\partial y}\right)^2+2 \frac{\partial^2 u}{\partial r \partial \theta} \frac{\partial r}{\partial y} \frac{\partial \theta}{\partial y}+\frac{\partial u}{\partial r} \frac{\partial^2 r}{\partial y^2}+\frac{\partial^2 u}{\partial \theta^2}\left(\frac{\partial \theta}{\partial y}\right)^2+\frac{\partial u}{\partial \theta} \frac{\partial^2 \theta}{\partial y^2}
$$

将式（1．2．26）与式（1．2．27）相加并注意式（1．2．23）和式（1．2．24），我们得到

$$
\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}=\frac{\partial^2 u}{\partial r^2}\left[\left(\frac{\partial r}{\partial x}\right)^2+\left(\frac{\partial r}{\partial y}\right)^2\right]+2 \frac{\partial^2 u}{\partial r \partial \theta}\left[\frac{\partial r}{\partial x} \frac{\partial \theta}{\partial x}+\frac{\partial r}{\partial y} \frac{\partial \theta}{\partial y}\right]
$$

$$
\begin{aligned}
& +\frac{\partial u}{\partial r}\left[\frac{\partial^2 r}{\partial x^2}+\frac{\partial^2 r}{\partial y^2}\right]+\frac{\partial^2 u}{\partial \theta^2}\left[\left(\frac{\partial \theta}{\partial x}\right)^2+\left(\frac{\partial \theta}{\partial y}\right)^2\right]+\frac{\partial u}{\partial \theta}\left[\frac{\partial^2 \theta}{\partial x^2}+\frac{\partial^2 \theta}{\partial y^2}\right] \\
= & \frac{\partial^2 u}{\partial r^2}\left[\left(\frac{\partial r}{\partial x}\right)^2+\left(\frac{\partial r}{\partial y}\right)^2\right]+\frac{\partial u}{\partial r}\left[\frac{\partial^2 r}{\partial x^2}+\frac{\partial^2 r}{\partial y^2}\right] \\
& +\frac{\partial^2 u}{\partial \theta^2}\left[\left(\frac{\partial \theta}{\partial x}\right)^2+\left(\frac{\partial \theta}{\partial y}\right)^2\right]
\end{aligned}  ...(1.2.28)
$$

进一步利用式（1．2．20）～式（1．2．22），则式（1．2．28）变为

$$
\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}=\frac{\partial^2 u}{\partial r^2}\left(\frac{x^2}{r^2}+\frac{y^2}{r^2}\right)+\frac{\partial u}{\partial r}\left(\frac{x^2}{r^3}+\frac{y^2}{r^3}\right)+\frac{\partial^2 u}{\partial \theta^2}\left(\frac{x^2}{r^4}+\frac{y^2}{r^4}\right)
$$

利用 $r^2=x^2+y^2$ ，我们得到极坐标系中的拉普拉斯

$$
\nabla^2 u=\frac{\partial^2 u}{\partial r^2}+\frac{1}{r} \frac{\partial u}{\partial r}+\frac{1}{r^2} \frac{\partial^2 u}{\partial \theta^2}
$$

它还可以写为更加简洁的形式

$$
\boxed{
\nabla^2 u=\frac{1}{r} \frac{\partial}{\partial r}\left(r \frac{\partial u}{\partial r}\right)+\frac{1}{r^2} \frac{\partial^2 u}{\partial \theta^2}   ...(1.2.31)
}
$$

而极坐标系中的拉普拉斯算子为

$$
\boxed{
\nabla^2=\frac{1}{r} \frac{\partial}{\partial r}\left(r \frac{\partial}{\partial r}\right)+\frac{1}{r^2} \frac{\partial^2}{\partial \theta^2} ...(1.2.32)
}
$$

## 柱坐标系中的拉普拉斯算子
柱坐标系如图 1.8 所示，它与直角坐标系的关系为

![图片](/uploads/2025-04/48ec25.jpg)
$$
x=\rho \cos \phi, \quad y=\rho \sin \phi, \quad z=z
$$

这里我们用 $\rho$ 和 $\phi$ 标记 $x-y$ 平面的极坐标。柱坐标系只是在极坐标系的基础上增加了 $z$坐标，因此利用式（1．2．31）容易写出柱坐标系中的拉普拉斯

$$
\nabla^2 u=\frac{\partial^2 u}{\partial \rho^2}+\frac{1}{\rho} \frac{\partial u}{\partial \rho}+\frac{1}{\rho^2} \frac{\partial^2 u}{\partial \phi^2}+\frac{\partial^2 u}{\partial z^2}
$$

而柱坐标系中的拉普拉斯算子为

$$
\boxed{
\nabla^2=\frac{\partial^2}{\partial \rho^2}+\frac{1}{\rho} \frac{\partial}{\partial \rho}+\frac{1}{\rho^2} \frac{\partial^2}{\partial \phi^2}+\frac{\partial^2}{\partial z^2}
}
$$

## 球坐标系中的拉普拉斯算子
球坐标系如图 1.9 所示，它与直角坐标系的关系为

![图片](/uploads/2025-04/006f5e.jpg)
$$
\begin{aligned}
& x=r \cos \phi \sin \theta \\
& y=r \sin \phi \sin \theta \\
& z=r \cos \theta
\end{aligned}
$$

而

$$
r^2=x^2+y^2+z^2
$$

从图 1.8 ，我们有

$$
\rho=r \sin \theta, \quad x=\rho \cos \phi, \quad y=\rho \sin \phi, \quad \rho^2=x^2+y^2
$$
现在我们要用球坐标变量 $r, \theta$ 和 $\phi$ 来表示拉普拉斯 $\nabla^2 u$ 。首先我们写出 $x-y$ 平面的极坐标系（变量为 $\rho$ 和 $\phi$ ）的拉普拉斯，利用式（1．2．30）我们得到

$$
\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}=\frac{\partial^2 u}{\partial \rho^2}+\frac{1}{\rho} \frac{\partial u}{\partial \rho}+\frac{1}{\rho^2} \frac{\partial^2 u}{\partial \phi^2}
$$

注意到

$$
z=r \cos \theta, \quad \rho=r \sin \theta
$$

与极坐标系－直角坐标系的变换关系（1．2．17a）是类似的。这样在 $z-\rho$ 平面的极坐标系的拉普拉斯为

$$
\frac{\partial^2 u}{\partial z^2}+\frac{\partial^2 u}{\partial \rho^2}=\frac{\partial^2 u}{\partial r^2}+\frac{1}{r} \frac{\partial u}{\partial r}+\frac{1}{r^2} \frac{\partial^2 u}{\partial \theta^2}
$$

在式（1．2．39）两边加入 $\frac{\partial^2 u}{\partial z^2}$ ，并利用（1．2．41），即得到三维情况的拉普拉斯

$$
\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}+\frac{\partial^2 u}{\partial z^2}=\frac{\partial^2 u}{\partial r^2}+\frac{1}{r} \frac{\partial u}{\partial r}+\frac{1}{r^2} \frac{\partial^2 u}{\partial \theta^2}+\frac{1}{\rho} \frac{\partial u}{\partial \rho}+\frac{1}{\rho^2} \frac{\partial^2 u}{\partial \phi^2}
$$

现在需要将式（1．2．42）右边的 $\frac{\partial u}{\partial \rho}$ 表示为球坐标的形式。为此利用关系式 $\tan \theta=$ $\rho / z$ 求出

$$
\frac{\partial \theta}{\partial \rho}=\frac{1}{1+(\rho / z)^2} \frac{1}{z}=\frac{z}{z^2+\rho^2}=\frac{z}{r^2}=\frac{\cos \theta}{r}  ...(1.2.43)
$$

对 $\rho=r \sin \theta$ 两边关于 $\rho$ 微分，并利用式（1．2．43）得到

$$
1=\frac{\partial r}{\partial \rho} \sin \theta+r \cos \theta \frac{\partial \theta}{\partial \rho}=\frac{\partial r}{\partial \rho} \sin \theta+\cos ^2 \theta
$$

由此

$$
\frac{\partial r}{\partial \rho}=\frac{1-\cos ^2 \theta}{\sin \theta}=\sin \theta
$$

现在对 $u$（作为 $r, \theta, \phi$ 的函数）关于 $\rho$ 求微商

$$
\frac{\partial u}{\partial \rho}=\frac{\partial u}{\partial r} \frac{\partial r}{\partial \rho}+\frac{\partial u}{\partial \theta} \frac{\partial \theta}{\partial \rho}+\frac{\partial u}{\partial \phi} \frac{\partial \phi}{\partial \rho}
$$

由于 $x-y$ 平面的极坐标系变量 $\rho$ 和 $\phi$ 是相互独立的，因此 $\partial \phi / \partial \rho=0$ ，再利用式 （1．2．45）和式（1．2．43），式（1．2．46）变成

$$
\frac{\partial u}{\partial \rho}=\frac{\partial u}{\partial r} \sin \theta+\frac{\partial u}{\partial \theta} \frac{\cos \theta}{r}
$$

将式（1．2．47）代入式（1．2．42），并利用 $\rho=r \sin \theta$ ，即得球坐标系的拉普拉斯

$$
\nabla^2 u=\frac{\partial^2 u}{\partial r^2}+\frac{2}{r} \frac{\partial u}{\partial r}+\frac{1}{r^2}\left(\frac{\partial^2 u}{\partial \theta^2}+\cot \theta \frac{\partial u}{\partial \theta}\right)+\frac{1}{r^2 \sin ^2 \theta} \frac{\partial^2 u}{\partial \phi^2}
$$

它还可以写成更简洁的形式

$$
\nabla^2 u=\frac{1}{r^2} \frac{\partial}{\partial r}\left(r^2 \frac{\partial u}{\partial r}\right)+\frac{1}{r^2 \sin \theta} \frac{\partial}{\partial \theta}\left(\sin \theta \frac{\partial u}{\partial \theta}\right)+\frac{1}{r^2 \sin ^2 \theta} \frac{\partial^2 u}{\partial \phi^2}
$$

而球坐标系中的拉普拉斯算子为

$$
\boxed{
\nabla^2=\frac{1}{r^2} \frac{\partial}{\partial r}\left(r^2 \frac{\partial}{\partial r}\right)+\frac{1}{r^2 \sin \theta} \frac{\partial}{\partial \theta}\left(\sin \theta \frac{\partial}{\partial \theta}\right)+\frac{1}{r^2 \sin ^2 \theta} \frac{\partial^2}{\partial \phi^2}
}
$$

`例` 在球坐标系中求函数

$$
u(x, y, z)=\ln \left(x^2+y^2+z^2\right), \quad(x, y, z) \neq(0,0,0)
$$

的拉普拉斯。
解 在球坐标系中

$$
u(r, \theta, \phi)=\ln r^2=2 \ln r
$$

因为 $u$ 不含 $\theta, \phi$ 的依赖性，故

$$
\frac{\partial u}{\partial r}=\frac{\partial}{\partial r}(2 \ln r)=\frac{2}{r}
$$

于是拉普拉斯为

$$
\nabla^2 u=\frac{1}{r^2} \frac{\partial}{\partial r}\left(r^2 \frac{\partial u}{\partial r}\right)=\frac{1}{r^2} \frac{\partial}{\partial r}(2 r)=\frac{2}{r^2}
$$

其他版本
【高等数学】理解：旋度

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