数学物理方法1

第 501 讲¶

圆柱坐标 \((r, \phi, z)\) 与直角坐标 \((x, y, z)\) 的变换关系：
- \(x = r \cos \phi\)
- \(y = r \sin \phi\)
- \(dx = \cos \phi dr - r \sin \phi d\phi\)
- \(dy = \sin \phi dr + r \cos \phi d\phi\)
直角坐标对圆柱坐标的偏导数：
- \(\frac{\partial r}{\partial x} = \cos \phi\)
- \(\frac{\partial \phi}{\partial x} = -\frac{\sin \phi}{r}\)
- \(\frac{\partial r}{\partial y} = \sin \phi\)
- \(\frac{\partial \phi}{\partial y} = \frac{\cos \phi}{r}\)

由直角坐标系的拉普拉斯算子 \(\nabla^2 \equiv \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2}\) 出发，通过坐标变换和链式法则，推导出圆柱坐标系下的拉普拉斯算子表达式：
- \(\nabla^2 \equiv \frac{\partial^2}{\partial r^2} + \frac{1}{r}\frac{\partial}{\partial r} + \frac{1}{r^2}\frac{\partial^2}{\partial \phi^2}\)
若考虑 z 方向，圆柱坐标系下的拉普拉斯算子为：
- \(\nabla^2 \equiv \frac{\partial^2}{\partial r^2} + \frac{1}{r}\frac{\partial}{\partial r} + \frac{1}{r^2}\frac{\partial^2}{\partial \phi^2} + \frac{\partial^2}{\partial z^2}\)

球坐标 \((r, \theta, \phi)\) 与直角坐标 \((x, y, z)\) 的变换关系：
- \(x = r \sin \theta \cos \phi\)
- \(y = r \sin \theta \sin \phi\)
- \(z = r \cos \theta\)
- \(dx = \sin \theta \cos \phi dr + r \cos \theta \cos \phi d\theta - r \sin \theta \sin \phi d\phi\)
- \(dy = \sin \theta \sin \phi dr + r \cos \theta \sin \phi d\theta + r \sin \theta \cos \phi d\phi\)
- \(dz = \cos \theta dr - r \sin \theta d\theta\)
直角坐标对球坐标的偏导数：
- \(\frac{\partial}{\partial x} = \sin \theta \cos \phi \frac{\partial}{\partial r} + \frac{\cos \theta \cos \phi}{r} \frac{\partial}{\partial \theta} - \frac{\sin \phi}{r \sin \theta} \frac{\partial}{\partial \phi}\)
- \(\frac{\partial}{\partial y} = \sin \theta \sin \phi \frac{\partial}{\partial r} + \frac{\cos \theta \sin \phi}{r} \frac{\partial}{\partial \theta} + \frac{\cos \phi}{r \sin \theta} \frac{\partial}{\partial \phi}\)
- \(\frac{\partial}{\partial z} = \cos \theta \frac{\partial}{\partial r} - \frac{\sin \theta}{r} \frac{\partial}{\partial \theta}\)

经过复杂的运算和化简，得到球坐标系下的拉普拉斯算子表达式：
- \(\nabla^2 \equiv \frac{\partial^2}{\partial r^2} + \frac{2}{r}\frac{\partial}{\partial r} + \frac{1}{r^2}\frac{\partial^2}{\partial \theta^2} + \frac{\cos \theta}{r^2 \sin \theta}\frac{\partial}{\partial \theta} + \frac{1}{r^2 \sin^2\theta}\frac{\partial^2}{\partial \phi^2}\)
- 或写成：\(\nabla^2 \equiv \frac{1}{r^2}\frac{\partial}{\partial r}(r^2\frac{\partial}{\partial r}) + \frac{1}{r^2 \sin \theta}\frac{\partial}{\partial \theta}(\sin \theta \frac{\partial}{\partial \theta}) + \frac{1}{r^2 \sin^2\theta}\frac{\partial^2}{\partial \phi^2}\)
球坐标系与圆柱坐标系的关系：
- 圆柱坐标 \((\rho, \phi, z)\) 与球坐标 \((r, \theta, \phi)\) 的关系为：\(\rho = r \sin \theta\)，\(z = r \cos \theta\)
- 对应的拉普拉斯算子可相互转换。

方程形式：\(\frac{1}{r}\frac{\partial}{\partial r}(r\frac{\partial u}{\partial r}) + \frac{1}{r^2}\frac{\partial^2 u}{\partial \phi^2} + \frac{\partial^2 u}{\partial z^2} + k^2u = 0\)
分离变量法求解：
- 设 \(u(r, \phi, z) = v(r, \phi)Z(z)\)，代入方程并分离变量，得到关于 \(Z(z)\) 和 \(v(r, \phi)\) 的常微分方程：
  - \(Z'' + \lambda Z = 0\)
  - \(\frac{1}{r}\frac{\partial}{\partial r}(r\frac{\partial v}{\partial r}) + \frac{1}{r^2}\frac{\partial^2 v}{\partial \phi^2} + (k^2 - \lambda)v = 0\)
- 进一步设 \(v(r, \phi) = R(r)\Phi(\phi)\)，分离变量后得到：
  - \(\Phi'' + \mu\Phi = 0\)
  - \(R'' + (k^2 - \lambda - \frac{\mu}{r^2})R = 0\)

方程形式：\(\frac{1}{r^2}\frac{\partial}{\partial r}(r^2\frac{\partial u}{\partial r}) + \frac{1}{r^2 \sin \theta}\frac{\partial}{\partial \theta}(\sin \theta \frac{\partial u}{\partial \theta}) + \frac{1}{r^2 \sin^2\theta}\frac{\partial^2 u}{\partial \phi^2} + k^2u = 0\)
分离变量法求解：
- 设 \(u(r, \theta, \phi) = R(r)S(\theta, \phi)\)，代入方程并分离变量，得到关于 \(R(r)\) 和 \(S(\theta, \phi)\) 的方程：
  - \(\frac{1}{r^2}\frac{d}{dr}(r^2\frac{dR}{dr}) + (k^2 - \frac{\lambda}{r^2})R = 0\)
  - \(\frac{1}{\sin \theta}\frac{\partial}{\partial \theta}(\sin \theta \frac{\partial S}{\partial \theta}) + \frac{1}{\sin^2\theta}\frac{\partial^2 S}{\partial \phi^2} + \lambda S = 0\)
- 进一步设 \(S(\theta, \phi) = \Theta(\theta)\Phi(\phi)\)，分离变量后得到：
  - \(\Phi'' + \mu\Phi = 0\)
  - \(\frac{1}{\sin \theta}\frac{d}{d\theta}(\sin \theta \frac{d\Theta}{d\theta}) + (\lambda - \frac{\mu}{\sin^2\theta})\Theta = 0\)