第十章 学习笔记 （The Virbrating Membrane）
这一章研究的是一个薄膜的振动问题。基本假设条件与上一章类似。
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首先是振动幅度很小。
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薄膜的张力 T 认为是恒定的。
类似弦振动问题推导，将其推广到二维平面上，就可以得到膜的振动方程。
T σ ( ∂ 2 u ∂ x 2 + ∂ 2 u ∂ y 2 ) = ∂ 2 u ∂ t 2 Δ u − 1 c 2 ∂ 2 u ∂ t 2 = 0 \frac{T}{\sigma}(\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2}) = \frac{\partial^2 u}{\partial t^2} \\ \Delta u - \frac{1}{c^2} \frac{\partial^2 u}{\partial t^2} = 0 σT(∂x2∂2u+∂y2∂2u)=∂t2∂2uΔu−c21∂t2∂2u=0
如果这个膜是方形的，那么在直角坐标系下很容易用分离变量法求解。这里只介绍圆形的膜。
The circular membrane
膜是圆形的，用极坐标会更方便。也就是 u = ψ ( r , φ , t ) u = \psi(r, \varphi,t) u=ψ(r,φ,t)。 那么遇到的第一个问题就是要先写出 Laplace 方程在极坐标系下的表达式。极坐标和直角坐标之间的关系还是比较简单的。
{ x = r cos ( φ ) y = r sin ( φ ) { r = x 2 + y 2 φ = arctan ( y / x ) \left\{ \begin{aligned} x &= r \cos(\varphi) \\ y &= r \sin(\varphi) \end{aligned} \right. \\ \left\{ \begin{aligned} r &= \sqrt{x^2 + y^2}\\ \varphi &= \arctan(y/x) \end{aligned} \right. {xy=rcos(φ)=rsin(φ){rφ=x2+y2=arctan(y/x)
我们可以先求 r r r 对 x x x 和 y y y 的偏导数。这两个相对简单。
∂ r ∂ x = x r = cos φ ∂ r ∂ y = y r = sin φ \frac{\partial r}{\partial x} = \frac{x}{r} = \cos \varphi \\ \frac{\partial r}{\partial y} = \frac{y}{r} = \sin \varphi \\ ∂x∂r=rx=cosφ∂y∂r=ry=sinφ
然后再求 φ \varphi φ 对 x x x 和 y y y 的偏导数。
∂ φ ∂ x = − y x 2 + y 2 = − r sin φ r 2 = − sin φ r ∂ φ ∂ y = x x 2 + y 2 = r cos φ r 2 = cos φ r \begin{aligned} \frac{\partial \varphi}{\partial x} &= -\frac{y}{x^2+y^2} =-\frac{r \sin \varphi}{r^2} = -\frac{\sin \varphi}{r}\\ \frac{\partial \varphi}{\partial y} &= \frac{x}{x^2+y^2} =\frac{r \cos \varphi}{r^2} = \frac{\cos \varphi}{r}\\ \end{aligned} ∂x∂φ∂y∂φ=−x2+y2y=−r2rsinφ=−rsinφ=x2+y2x=r2rcosφ=rcosφ
当然也可以按照书上的推导方法：
tan ( φ ) = y x ∂ tan φ ∂ x = 1 cos 2 φ ∂ φ ∂ x = − y x 2 = − r sin φ r 2 cos 2 φ = − sin φ r cos 2 φ ∂ tan φ ∂ y = 1 cos 2 φ ∂ φ ∂ y = 1 x = 1 r cos φ \tan(\varphi) = \frac{y}{x} \\ \frac{\partial \tan \varphi}{\partial x} = \frac{1}{\cos^2 \varphi} \frac{\partial \varphi}{\partial x}= - \frac{y}{x^2} = -\frac{r \sin \varphi}{r^2 \cos^2 \varphi} = -\frac{\sin \varphi}{r \cos^2 \varphi}\\ \frac{\partial \tan \varphi}{\partial y} = \frac{1}{\cos^2 \varphi} \frac{\partial \varphi}{\partial y}= \frac{1}{x} = \frac{1}{r \cos \varphi} \\ tan(φ)=xy∂x∂tanφ=cos2φ1∂x∂φ=−x2y=−r2cos2φrsinφ=−rcos2φsinφ∂y∂tanφ=cos2φ1∂y∂φ=x1=rcosφ1
整理一下就得到：
∂ φ ∂ x = − sin φ r ∂ φ ∂ y = cos φ r \frac{\partial\varphi}{\partial x} = -\frac{\sin \varphi}{r} \\ \frac{\partial\varphi}{\partial y} = \frac{\cos \varphi}{r} ∂x∂φ=−rsinφ∂y∂φ=rcosφ
可以看到，两种方法得到的结果是一致的。下面利用求偏导的链式法则。
∂ ∂ x = ∂ ∂ r ∂ r ∂ x + ∂ ∂ φ ∂ φ ∂ x = cos φ ∂ ∂ r − sin φ r ∂ ∂ φ \begin{aligned} \frac{\partial}{\partial x} &= \frac{\partial}{\partial r} \frac{\partial r}{\partial x} + \frac{\partial}{\partial \varphi} \frac{\partial \varphi}{\partial x} \\ &= \cos \varphi \frac{\partial}{\partial r} -\frac{\sin \varphi}{r} \frac{\partial}{\partial \varphi} \\ \end{aligned} ∂x∂=∂r∂∂x∂r+∂φ∂∂x∂φ=cosφ∂r∂−rsinφ∂φ∂
再求一次导可以得到二阶导数的结果。
∂ 2 ∂ x 2 = ∂ ∂ x ( ∂ ∂ r ∂ r ∂ x + ∂ ∂ φ ∂ φ ∂ x ) = ( cos φ ∂ ∂ r − sin φ r ∂ ∂ φ ) ( cos φ ∂ ∂ r − sin φ r ∂ ∂ φ ) = cos 2 φ ∂ 2 ∂ r 2 + ( cos φ sin φ r 2 ∂ ∂ φ − cos φ sin φ r ∂ ∂ r ∂ ∂ φ ) + ( sin 2 φ r 2 ∂ 2 ∂ φ 2 + sin φ cos φ r 2 ∂ ∂ φ ) + ( sin 2 φ r ∂ ∂ r − sin φ cos φ r ∂ ∂ r ∂ ∂ φ ) = cos 2 φ ∂ 2 ∂ r 2 + sin 2 φ r 2 ( ∂ 2 ∂ φ 2 + r ∂ ∂ r ) + 2 cos φ sin φ r 2 ( ∂ ∂ φ − r ∂ ∂ r ∂ ∂ φ ) \begin{aligned} \frac{\partial^2}{\partial x^2} &= \frac{\partial}{\partial x}(\frac{\partial}{\partial r} \frac{\partial r}{\partial x} + \frac{\partial}{\partial \varphi} \frac{\partial \varphi}{\partial x}) \\ &= (\cos \varphi \frac{\partial}{\partial r} -\frac{\sin \varphi}{r} \frac{\partial}{\partial \varphi})(\cos \varphi \frac{\partial}{\partial r} -\frac{\sin \varphi}{r} \frac{\partial}{\partial \varphi})\\ &= \cos^2 \varphi \frac{\partial^2}{\partial r^2}+(\frac{\cos \varphi \sin \varphi}{r^2} \frac{\partial}{\partial \varphi} - \frac{\cos \varphi \sin \varphi}{r} \frac{\partial}{\partial r}\frac{\partial}{\partial \varphi})+ (\frac{\sin^2 \varphi}{r^2} \frac{\partial^2}{\partial \varphi^2} + \frac{\sin \varphi \cos \varphi}{r^2} \frac{\partial }{\partial \varphi})+ (\frac{\sin^2 \varphi}{r} \frac{\partial}{\partial r} - \frac{\sin \varphi \cos \varphi}{r} \frac{\partial}{\partial r}\frac{\partial}{\partial \varphi}) \\ &=\cos^2 \varphi \frac{\partial^2}{\partial r^2} + \frac{\sin^2 \varphi}{r^2}(\frac{\partial^2}{\partial \varphi^2} + r \frac{\partial}{\partial r}) + \frac{2\cos \varphi \sin \varphi}{r^2} (\frac{\partial}{\partial \varphi} - r \frac{\partial}{\partial r}\frac{\partial}{\partial \varphi} ) \end{aligned} ∂x2∂2=∂x∂(∂r∂∂x∂r+∂φ∂∂x∂φ)=(cosφ∂r∂−rsinφ∂φ∂)(cosφ∂r∂−rsinφ∂φ∂)=cos2φ∂r2∂2+(r2cosφsinφ∂φ∂−rcosφsinφ∂r∂∂φ∂)+(r2sin2φ∂φ2∂2+r2sinφcosφ∂φ∂)+(rsin2φ∂r∂−rsinφcosφ∂r∂∂φ∂)=cos2φ∂r2∂2+r2sin2φ(∂φ2∂2+r∂r∂)+r22cosφsinφ(∂φ∂−r∂r∂∂φ∂)
同理，对 y y y 求一次导可以得到：
∂ ∂ y = ∂ ∂ r ∂ r ∂ y + ∂ ∂ φ ∂ φ ∂ y = sin φ ∂ ∂ r + cos φ r ∂ ∂ φ \frac{\partial}{\partial y} = \frac{\partial}{\partial r} \frac{\partial r}{\partial y} + \frac{\partial}{\partial \varphi} \frac{\partial \varphi}{\partial y} =\sin{\varphi} \frac{\partial}{\partial r} + \frac{\cos \varphi}{r}\frac{\partial }{\partial \varphi} \\ ∂y∂=∂r∂∂y∂r+∂φ∂∂y∂φ=sinφ∂r∂+rcosφ∂φ∂
再求一次导得到：
∂ 2 ∂ y 2 = ( sin φ ∂ ∂ r + cos φ r ∂ ∂ φ ) ( sin φ ∂ ∂ r + cos φ r ∂ ∂ φ ) = sin 2 φ ∂ 2 ∂ r 2 + ( − sin φ cos φ r 2 ∂ ∂ φ + sin φ cos φ r ∂ ∂ r ∂ ∂ φ ) + ( cos 2 φ r ∂ ∂ r + sin φ cos φ r ∂ ∂ r ∂ ∂ φ ) + ( − sin φ cos φ r 2 ∂ ∂ φ + cos 2 φ r 2 ∂ 2 ∂ φ 2 ) = sin 2 φ ∂ 2 ∂ r 2 + 2 sin φ cos φ r 2 ( − ∂ ∂ φ + r ∂ ∂ r ∂ ∂ φ ) + cos 2 φ r 2 ( r ∂ ∂ r + ∂ 2 ∂ φ 2 ) \begin{aligned} \frac{\partial^2}{\partial y^2} &= (\sin{\varphi} \frac{\partial}{\partial r} + \frac{\cos \varphi}{r}\frac{\partial }{\partial \varphi})(\sin{\varphi} \frac{\partial}{\partial r} + \frac{\cos \varphi}{r}\frac{\partial }{\partial \varphi}) \\ &= \sin^2 \varphi \frac{\partial^2}{\partial r^2} + (-\frac{\sin \varphi \cos \varphi}{r^2} \frac{\partial}{\partial \varphi} + \frac{\sin \varphi \cos \varphi}{r} \frac{\partial}{\partial r}\frac{\partial}{\partial \varphi}) + (\frac{\cos^2 \varphi}{r} \frac{\partial}{\partial r} + \frac{\sin \varphi \cos \varphi}{r} \frac{\partial}{\partial r}\frac{\partial}{\partial \varphi}) + (-\frac{\sin \varphi \cos \varphi}{r^2}\frac{\partial}{\partial \varphi} + \frac{\cos^2 \varphi}{r^2} \frac{\partial^2}{\partial \varphi^2}) \\ &= \sin^2 \varphi \frac{\partial^2}{\partial r^2} + \frac{2 \sin \varphi \cos \varphi}{r^2}(-\frac{\partial}{\partial \varphi}+r \frac{\partial}{\partial r}\frac{\partial}{\partial \varphi}) + \frac{\cos^2 \varphi}{r^2} (r\frac{\partial}{\partial r} + \frac{\partial^2}{\partial \varphi^2}) \end{aligned} ∂y2∂2=(sinφ∂r∂+rcosφ∂φ∂)(sinφ∂r∂+rcosφ∂φ∂)=sin2φ∂r2∂2+(−r2sinφcosφ∂φ∂+rsinφcosφ∂r∂∂φ∂)+(rcos2φ∂r∂+rsinφcosφ∂r∂∂φ∂)+(−r2sinφcosφ∂φ∂+r2cos2φ∂φ2∂2)=sin2φ∂r2∂2+r22sinφcosφ(−∂φ∂+r∂r∂∂φ∂)+r2cos2φ(r∂r∂+∂φ2∂2)
将这两部分加起来，可以得到：
∂ 2 ∂ x 2 + ∂ 2 ∂ y 2 = cos 2 φ ∂ 2 ∂ r 2 + sin 2 φ r 2 ( ∂ 2 ∂ φ 2 + r ∂ ∂ r ) + 2 cos φ sin φ r 2 ( ∂ ∂ φ − r ∂ ∂ r ∂ ∂ φ ) + sin 2 φ ∂ 2 ∂ r 2 + 2 sin φ cos φ r 2 ( − ∂ ∂ φ + r ∂ ∂ r ∂ ∂ φ ) + cos 2 φ r 2 ( r ∂ ∂ r + ∂ 2 ∂ φ 2 ) = ∂ 2 ∂ r 2 + 1 r 2 ( ∂ 2 ∂ φ 2 + r ∂ ∂ r ) = ∂ 2 ∂ r 2 + 1 r 2 ∂ 2 ∂ φ 2 + 1 r ∂ ∂ r \begin{aligned} \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} &= \cos^2 \varphi \frac{\partial^2}{\partial r^2} + \frac{\sin^2 \varphi}{r^2}(\frac{\partial^2}{\partial \varphi^2} + r \frac{\partial}{\partial r}) + \frac{2\cos \varphi \sin \varphi}{r^2} (\frac{\partial}{\partial \varphi} - r \frac{\partial}{\partial r}\frac{\partial}{\partial \varphi} ) \\ &+ \sin^2 \varphi \frac{\partial^2}{\partial r^2} + \frac{2 \sin \varphi \cos \varphi}{r^2}(-\frac{\partial}{\partial \varphi}+r \frac{\partial}{\partial r}\frac{\partial}{\partial \varphi}) + \frac{\cos^2 \varphi}{r^2} (r\frac{\partial}{\partial r} + \frac{\partial^2}{\partial \varphi^2}) \\ &= \frac{\partial^2}{\partial r^2} + \frac{1}{r^2}(\frac{\partial^2}{\partial \varphi^2} + r \frac{\partial}{\partial r}) \\ &= \frac{\partial^2}{\partial r^2} + \frac{1}{r^2}\frac{\partial^2}{\partial \varphi^2} + \frac{1}{r} \frac{\partial}{\partial r} \end{aligned} ∂x2∂2+∂y2∂2=cos2φ∂r2∂2+r2sin2φ(∂φ2∂2+r∂r∂)+r22cosφsinφ(∂φ∂−r∂r∂∂φ∂)+sin2φ∂r2∂2+r22sinφcosφ(−∂φ∂+r∂r∂∂φ∂)+r2cos2φ(r∂r∂+∂φ2∂2)=∂r2∂2+r21(∂φ2∂2+r∂r∂)=∂r2∂2+r21∂φ2∂2+r1∂r∂
至此，柱坐标系下的 Laplace 算子的表达式就推导出来了。因此，波动方程可以写为：
∂ 2 u ∂ r 2 + 1 r 2 ∂ 2 u ∂ φ 2 + 1 r ∂ u ∂ r = 1 c 2 ∂ 2 u ∂ t 2 \frac{\partial^2 u}{\partial r^2} + \frac{1}{r^2}\frac{\partial^2 u}{\partial \varphi^2} + \frac{1}{r} \frac{\partial u}{\partial r} = \frac{1}{c^2} \frac{\partial ^2 u}{\partial t^2} ∂r2∂2u+r21∂φ2∂2u+r1∂r∂u=c21∂t2∂2u
下面用分离变量法来求解，首先 u ( r , φ , t ) = V ( r , φ ) Z ( t ) u(r, \varphi, t) = V(r, \varphi) Z(t) u(r,φ,t)=V(r,φ)Z(t)。带入 Laplace 方程，简单整理一下。
∂ 2 ( V ( r , φ ) Z ( t ) ) ∂ r 2 + 1 r 2 ∂ 2 ( V ( r , φ ) Z ( t ) ) ∂ φ 2 + 1 r ∂ ( V ( r , φ ) Z ( t ) ) ∂ r = 1 c 2 ∂ 2 ( V ( r , φ ) Z ( t ) ) ∂ t 2 Z ( t ) ∂ 2 V ( r , φ ) ∂ r 2 + Z ( t ) 1 r 2 ∂ 2 V ( r , φ ) ∂ φ 2 + Z ( t ) 1 r ∂ V ( r , φ ) ∂ r = V ( r , φ ) 1 c 2 d 2 Z ( t ) d t 2 1 V ( r , φ ) ( ∂ 2 V ( r , φ ) ∂ r 2 + 1 r 2 ∂ 2 V ( r , φ ) ∂ φ 2 + 1 r ∂ V ( r , φ ) ∂ r ) = 1 Z ( t ) 1 c 2 d 2 Z ( t ) d t 2 \frac{\partial^2 (V(r, \varphi) Z(t))}{\partial r^2}+\frac{1}{r^2} \frac{\partial^2 (V(r, \varphi) Z(t))}{\partial \varphi^2} + \frac{1}{r} \frac{\partial (V(r, \varphi) Z(t))}{\partial r} = \frac{1}{c^2} \frac{\partial^2 (V(r, \varphi) Z(t))}{\partial t^2}\\ Z(t)\frac{\partial^2 V(r, \varphi)}{\partial r^2}+ Z(t)\frac{1}{r^2} \frac{\partial^2 V(r, \varphi) }{\partial \varphi^2} +Z(t) \frac{1}{r} \frac{\partial V(r, \varphi)}{\partial r} = V(r, \varphi) \frac{1}{c^2} \frac{d^2 Z(t)}{d t^2}\\ \frac{1}{V(r, \varphi) } \left(\frac{\partial^2 V(r, \varphi)}{\partial r^2}+ \frac{1}{r^2} \frac{\partial^2 V(r, \varphi)}{\partial \varphi^2} +\frac{1}{r} \frac{\partial V(r, \varphi)}{\partial r} \right) = \frac{1}{Z(t) } \frac{1}{c^2} \frac{d^2 Z(t)}{d t^2}\\ ∂r2∂2(V(r,φ)Z(t))+r21∂φ2∂2(V(r,φ)Z(t))+r1∂r∂(V(r,φ)Z(t))=c21∂t2∂2(V(r,φ)Z(t))Z(t)∂r2∂2V(r,φ)+Z(t)r21∂φ2∂2V(r,φ)+Z(t)r1∂r∂V(r,φ)=V(r,φ)c21dt2d2Z(t)V(r,φ)1(∂r2∂2V(r,φ)+r21∂φ2∂2V(r,φ)+r1∂r∂V(r,φ))=Z(t)1c21dt2d2Z(t)
可以看到等号左边是 r r r 和 φ \varphi φ 的函数，右边是 t t t 的函数。如果两边相等，那么等式两边只能是常数。
1 Z ( t ) 1 c 2 d 2 Z ( t ) d t 2 = − k 2 Z ¨ ( t ) + c 2 k 2 Z ( t ) = 0 Z ¨ ( t ) + ω 2 Z ( t ) = 0 \frac{1}{Z(t) } \frac{1}{c^2} \frac{d^2 Z(t)}{d t^2} = -k^2 \\ \ddot Z(t) + c^2k^2 Z(t) = 0 \\ \ddot Z(t) + \omega^2 Z(t) = 0 \\ Z(t)1c21dt2d2Z(t)=−k2Z¨(t)+c2k2Z(t)=0Z¨(t)+ω2Z(t)=0
上面式子里 ω = c k \omega = c \ k ω=c k。这个就是简单的振动方程，大家都知道解是 Z ( t ) = C cos ( ω t + δ ) Z(t) = C \cos(\omega t + \delta) Z(t)=Ccos(ωt+δ) 。
这里需要解释一下，上面为什么认为常数是个负数 − k 2 -k^2 −k2。如果常数是正数，我们也可以试着解一下。
1 Z ( t ) 1 c 2 d 2 Z ( t ) d t 2 = k 2 Z ¨ ( t ) − c 2 k 2 Z ( t ) = 0 \frac{1}{Z(t) } \frac{1}{c^2} \frac{d^2 Z(t)}{d t^2} = k^2 \\ \ddot Z(t) - c^2k^2 Z(t) = 0 \\ Z(t)1c21dt2d2Z(t)=k2Z¨(t)−c2k2Z(t)=0
这个解大家也都熟悉 Z ( t ) = C exp ( ± c k t ) Z(t) = C \exp(\pm ck\ t) Z(t)=Cexp(±ck t)。从数学上来说，这个解完全正确，但是从物理含义来说，这个解不是振荡解，无法描述膜的振动。所以这个解需要被抛弃。这就是为什么常数要选择负数的原因。
下面再看等式左边，还可以接着分离变量 V ( r , φ ) = R ( r ) Φ ( φ ) V(r,\varphi) = R(r) \Phi (\varphi) V(r,φ)=R(r)Φ(φ)：
1 V ( r , φ ) ( ∂ 2 V ( r , φ ) ∂ r 2 + 1 r 2 ∂ 2 V ( r , φ ) ∂ φ 2 + 1 r ∂ V ( r , φ ) ∂ r ) = − k 2 1 R ( r ) Φ ( φ ) ( ∂ 2 ( R ( r ) Φ ( φ ) ) ∂ r 2 + 1 r 2 ∂ 2 ( R ( r ) Φ ( φ ) ) ∂ φ 2 + 1 r ∂ ( R ( r ) Φ ( φ ) ) ∂ r ) = − k 2 1 R ( r ) Φ ( φ ) ( Φ ( φ ) d 2 R ( r ) d r 2 + 1 r 2 R ( r ) d 2 Φ ( φ ) d φ 2 + Φ ( φ ) 1 r d R ( r ) d r ) = − k 2 1 R ( r ) d 2 R ( r ) d r 2 + 1 Φ ( φ ) 1 r 2 d 2 Φ ( φ ) d φ 2 + 1 R ( r ) 1 r d R ( r ) d r = − k 2 1 R ( r ) ( d 2 R ( r ) d r 2 + 1 r d R ( r ) d r ) + 1 Φ ( φ ) 1 r 2 d 2 Φ ( φ ) d φ 2 + k 2 = 0 1 R ( r ) ( r 2 d 2 R ( r ) d r 2 + r d R ( r ) d r ) + r 2 k 2 + 1 Φ ( φ ) d 2 Φ ( φ ) d φ 2 = 0 \frac{1}{V(r, \varphi) } \left(\frac{\partial^2 V(r, \varphi)}{\partial r^2}+ \frac{1}{r^2} \frac{\partial^2 V(r, \varphi)}{\partial \varphi^2} +\frac{1}{r} \frac{\partial V(r, \varphi)}{\partial r} \right) = -k^2 \\ \frac{1}{R(r)\Phi(\varphi) } \left(\frac{\partial^2 (R(r)\Phi(\varphi))}{\partial r^2}+ \frac{1}{r^2} \frac{\partial^2 (R(r)\Phi(\varphi))}{\partial \varphi^2} +\frac{1}{r} \frac{\partial (R(r)\Phi(\varphi))}{\partial r} \right) = -k^2 \\ \frac{1}{R(r)\Phi(\varphi) } \left(\Phi(\varphi) \frac{d^2 R(r)}{d r^2}+ \frac{1}{r^2} R(r)\frac{d^2 \Phi(\varphi)}{d \varphi^2} +\Phi(\varphi)\frac{1}{r} \frac{d R(r)}{d r} \right) = -k^2 \\ \frac{1}{R(r)} \frac{d^2 R(r)}{d r^2}+ \frac{1}{\Phi(\varphi)}\frac{1}{ r^2} \frac{d^2 \Phi(\varphi)}{d \varphi^2} +\frac{1}{R(r)}\frac{1}{r} \frac{d R(r)}{d r} = -k^2 \\ \frac{1}{R(r)} \left(\frac{d^2 R(r)}{d r^2}+\frac{1}{r} \frac{d R(r)}{d r} \right)+\frac{1}{\Phi(\varphi)}\frac{1}{ r^2} \frac{d^2 \Phi(\varphi)}{d \varphi^2}+k^2 = 0\\ \frac{1}{R(r)} \left(r^2\frac{d^2 R(r)}{d r^2}+r \frac{d R(r)}{d r} \right)+r^2 k^2+\frac{1}{\Phi(\varphi)} \frac{d^2 \Phi(\varphi)}{d \varphi^2} = 0 V(r,φ)1(∂r2∂2V(r,φ)+r21∂φ2∂2V(r,φ)+r1∂r∂V(r,φ))=−k2R(r)Φ(φ)1(∂r2∂2(R(r)Φ(φ))+r21∂φ2∂2(R(r)Φ(φ))+r1∂r∂(R(r)Φ(φ)))=−k2R(r)Φ(φ)1(Φ(φ)dr2d2R(r)+r21R(r)dφ2d2Φ(φ)+Φ(φ)r1drdR(r))=−k2R(r)1dr2d2R(r)+Φ(φ)1r21dφ2d2Φ(φ)+R(r)1r1drdR(r)=−k2R(r)1(dr2d2R(r)+r1drdR(r))+Φ(φ)1r21dφ2d2Φ(φ)+k2=0R(r)1(r2dr2d2R(r)+rdrdR(r))+r2k2+Φ(φ)1dφ2d2Φ(φ)=0
整理得到的这个式子前两项是 r r r 的函数，最后一项是 φ \varphi φ 的函数。所以可以拆分为两个部分。
1 R ( r ) ( r 2 d 2 R ( r ) d r 2 + r d R ( r ) d r ) + r 2 k 2 = σ 1 Φ ( φ ) d 2 Φ ( φ ) d φ 2 = − σ \frac{1}{R(r)} \left(r^2\frac{d^2 R(r)}{d r^2}+r \frac{d R(r)}{d r} \right)+r^2 k^2 = \sigma \\ \frac{1}{\Phi(\varphi)} \frac{d^2 \Phi(\varphi)}{d \varphi^2} = - \sigma R(r)1(r2dr2d2R(r)+rdrdR(r))+r2k2=σΦ(φ)1dφ2d2Φ(φ)=−σ
先搞简单的，上面两个式子第二个比第一个简单。
d 2 Φ ( φ ) d φ 2 + σ Φ ( φ ) = 0 \frac{d^2 \Phi(\varphi)}{d \varphi^2} + \sigma \ \Phi(\varphi) = 0 dφ2d2Φ(φ)+σ Φ(φ)=0
我们知道 Φ ( φ ) \Phi(\varphi) Φ(φ) 是周期函数，周期是 2 π 2 \pi 2π。所以 σ \sigma σ 一定是非负数， Φ ( φ ) = C 1 cos ( σ φ + ψ ) \Phi(\varphi) = C_1 \cos(\sqrt \sigma \varphi + \psi) Φ(φ)=C1cos(σφ+ψ)。为了周期是 2 π 2 \pi 2π， σ = m 2 \sigma = m^2 σ=m2。这里 m m m 也是非负整数。
Φ ( φ ) = C m cos ( m φ + ψ ) \Phi(\varphi) = C_m \cos(m \varphi + \psi) Φ(φ)=Cmcos(mφ+ψ)
还剩下最后一部分，也是最难的一部分。
1 R ( r ) ( r 2 d 2 R ( r ) d r 2 + r d R ( r ) d r ) + r 2 k 2 = m 2 ( r 2 d 2 R ( r ) d r 2 + r d R ( r ) d r ) + R ( r ) ( r 2 k 2 − m 2 ) = 0 ( d 2 R ( r ) d r 2 + 1 r d R ( r ) d r ) + ( k 2 − m 2 r 2 ) R ( r ) = 0 \frac{1}{R(r)} \left(r^2\frac{d^2 R(r)}{d r^2}+r \frac{d R(r)}{d r} \right)+r^2 k^2 = m^2 \\ \left(r^2\frac{d^2 R(r)}{d r^2}+r \frac{d R(r)}{d r} \right)+R(r)(r^2 k^2 - m^2) = 0 \\ \left(\frac{d^2 R(r)}{d r^2}+ \frac{1}{r}\frac{d R(r)}{d r} \right)+(k^2 - \frac{m^2}{r^2}) R(r)= 0 R(r)1(r2dr2d2R(r)+rdrdR(r))+r2k2=m2(r2dr2d2R(r)+rdrdR(r))+R(r)(r2k2−m2)=0(dr2d2R(r)+r1drdR(r))+(k2−r2m2)R(r)=0
下面要把 k k k 消掉，做个变量替换 z = k r , d z = k d r z = k\ r, dz = k\ dr z=k r,dz=k dr。说实话，这个变量替换我是没法一眼看出来的。要是我估计是各种替换都试试，运气好就能凑出结果来。
g ( z ) = R ( z k ) R ( r ) = g ( k r ) d R ( r ) d r = d g ( k r ) d r = k d g ( z ) d z d 2 R ( r ) d r 2 = d 2 g ( k r ) d 2 r = k 2 d 2 g ( z ) d z 2 g(z) = R\left(\frac{z}{k}\right) \\ R(r) = g(kr) \\ \frac{dR(r)}{dr} = \frac{d\ g(kr) }{dr} = k \frac{d g(z)}{dz} \\ \frac{d^2R(r)}{dr^2} = \frac{d^2\ g(kr) }{d^2r} = k^2 \frac{d^2 g(z)}{dz^2} g(z)=R(kz)R(r)=g(kr)drdR(r)=drd g(kr)=kdzdg(z)dr2d2R(r)=d2rd2 g(kr)=k2dz2d2g(z)
将上面推导的关系式带入方程。
( d 2 R ( r ) d r 2 + 1 r d R ( r ) d r ) + ( k 2 − m 2 r 2 ) R ( r ) = 0 ( k 2 d 2 g ( z ) d z 2 + k 2 z d g ( z ) d z ) + ( k 2 − k 2 m 2 z 2 ) g ( z ) = 0 ( d 2 g ( z ) d z 2 + 1 z d g ( z ) d z ) + ( 1 − m 2 z 2 ) g ( z ) = 0 \left(\frac{d^2 R(r)}{d r^2}+ \frac{1}{r}\frac{d R(r)}{d r} \right)+(k^2 - \frac{m^2}{r^2}) R(r)= 0 \\ \left(k^2\frac{d^2 g(z)}{d z^2}+ \frac{k^2}{z}\frac{d g(z)}{d z} \right)+(k^2 - \frac{k^2m^2}{z^2}) g(z)= 0 \\ \left(\frac{d^2 g(z)}{d z^2}+ \frac{1}{z}\frac{d g(z)}{d z} \right)+(1 - \frac{m^2}{z^2}) g(z)= 0 (dr2d2R(r)+r1drdR(r))+(k2−r2m2)R(r)=0(k2dz2d2g(z)+zk2dzdg(z))+(k2−z2k2m2)g(z)=0(dz2d2g(z)+z1dzdg(z))+(1−z2m2)g(z)=0
可以看到确实将 k k k 消掉了。得到的这个方程称为 Bessel 方程， m m m 为方程的阶。因为我们这里 m m m 是整数，所以这个方程是Bessel 方程的一个特殊形式，称为整数阶 Bessel 方程。这个方程的结果为 Bessel 函数。Bessel 函数是用幂级数表示的。
g ( z ) = z μ ( ∑ n = 0 ∞ a n z n ) g(z) = z^\mu \left( \sum_{n=0}^{\infin} a_n z^n \right) g(z)=zμ(n=0∑∞anzn)
我们首先考察在原点附近函数的特性。这时 g ( z ) ≈ z μ g(z) \approx z^\mu g(z)≈zμ。为了防止函数在原点是发散的， μ ≥ 0 \mu \geq 0 μ≥0。
将 g ( z ) = z μ g(z) = z^\mu g(z)=zμ 带入Bessel函数。
μ ( μ − 1 ) z μ − 2 + μ z μ − 2 + ( 1 − m 2 z 2 ) z μ = 0 μ 2 z μ − 2 + z μ − m 2 z μ − 2 = 0 μ 2 − m 2 + z 2 = 0 \mu (\mu -1) z^{\mu-2} + \mu z^{\mu-2} + (1 - \frac{m^2}{z^2}) z^\mu = 0\\ \mu^2 z^{\mu-2} + z^\mu - m^2 z^{\mu-2} = 0 \\ \mu^2 - m^2 + z^2 = 0 μ(μ−1)zμ−2+μzμ−2+(1−z2m2)zμ=0μ2zμ−2+zμ−m2zμ−2=0μ2−m2+z2=0
在 z = 0 z= 0 z=0 处，只有 μ = m \mu = m μ=m 才能使上面的等式成立。所以：
g ( z ) = z m ( ∑ n = 0 ∞ a n z n ) = ∑ n = 0 ∞ a n z n + m d g ( z ) d z = ∑ n = 0 a n ( n + m ) z n + m − 1 d 2 g ( z ) d z 2 = ∑ n = 0 a n ( n + m ) ( n + m − 1 ) z n + m − 2 g(z) = z^m \left( \sum_{n=0}^{\infin} a_n z^n \right) = \sum_{n=0}^{\infin} a_n z^{n+m} \\ \frac{dg(z)}{dz} = \sum_{n=0} a_n (n+m)z^{n+m-1}\\ \frac{d^2g(z)}{dz^2} = \sum_{n=0} a_n (n+m)(n+m-1)z^{n+m-2}\\ g(z)=zm(n=0∑∞anzn)=n=0∑∞anzn+mdzdg(z)=n=0∑an(n+m)zn+m−1dz2d2g(z)=n=0∑an(n+m)(n+m−1)zn+m−2
下面的工作就全是体力活了。
( d 2 g ( z ) d z 2 + 1 z d g ( z ) d z ) + ( 1 − m 2 z 2 ) g ( z ) = 0 ∑ n = 0 ∞ a n ( n + m ) ( n + m − 1 ) z n + m − 2 + 1 z ∑ n = 0 ∞ a n ( n + m ) z n + m − 1 + ( 1 − m 2 z 2 ) ∑ n = 0 ∞ a n z n + m = 0 ∑ n = 0 ∞ a n ( n + m ) ( n + m − 1 ) z n + m − 2 + ∑ n = 0 ∞ a n ( n + m ) z n + m − 2 + ∑ n = 0 ∞ a n z n + m − m 2 ∑ n = 0 ∞ a n z n + m − 2 = 0 ∑ n = 0 ∞ a n ( n 2 + 2 m n ) z n + m − 2 + ∑ n = 0 ∞ a n z n + m = 0 \left(\frac{d^2 g(z)}{d z^2}+ \frac{1}{z}\frac{d g(z)}{d z} \right)+(1 - \frac{m^2}{z^2}) g(z)= 0 \\ \sum_{n=0}^{\infin} a_n (n+m)(n+m-1)z^{n+m-2} + \frac{1}{z} \sum_{n=0}^{\infin} a_n (n+m)z^{n+m-1}+(1 - \frac{m^2}{z^2})\sum_{n=0}^{\infin} a_n z^{n+m} = 0 \\ \sum_{n=0}^{\infin} a_n (n+m)(n+m-1)z^{n+m-2} + \sum_{n=0}^{\infin} a_n (n+m)z^{n+m-2}+ \sum_{n=0}^{\infin} a_n z^{n+m} - m^2\sum_{n=0}^{\infin} a_n z^{n+m-2} = 0 \\ \sum_{n=0}^{\infin} a_n ({n}^{2}+2 m n)z^{n+m-2} \ + \sum_{n=0}^{\infin} a_n z^{n+m} =0 (dz2d2g(z)+z1dzdg(z))+(1−z2m2)g(z)=0n=0∑∞an(n+m)(n+m−1)zn+m−2+z1n=0∑∞an(n+m)zn+m−1+(1−z2m2)n=0∑∞anzn+m=0n=0∑∞an(n+m)(n+m−1)zn+m−2+n=0∑∞an(n+m)zn+m−2+n=0∑∞anzn+m−m2n=0∑∞anzn+m−2=0n=0∑∞an(n2+2mn)zn+m−2 +n=0∑∞anzn+m=0
这里还有一个技巧：
∑ n = 0 ∞ a n z n + m = ∑ n = 2 ∞ a n − 2 z n + m − 2 \sum_{n=0}^{\infin} a_n z^{n+m} = \sum_{n=2}^{\infin} a_{n-2} z^{n+m-2} n=0∑∞anzn+m=n=2∑∞an−2zn+m−2
所以上面的式子可以进一步简化：
∑ n = 0 ∞ a n ( n 2 + 2 m n ) z n + m − 2 + ∑ n = 0 ∞ a n z n + m = 0 ∑ n = 0 ∞ a n ( n 2 + 2 m n ) z n + m − 2 + ∑ n = 2 ∞ a n − 2 z n + m − 2 = 0 a 0 × 0 × z m − 2 + a 1 ( 1 + 2 m ) z m − 1 + ∑ n = 2 ∞ ( a n ( n 2 + 2 m n ) + a n − 2 ) z n + m − 2 = 0 \sum_{n=0}^{\infin} a_n ({n}^{2}+2 m n)z^{n+m-2} \ + \sum_{n=0}^{\infin} a_n z^{n+m} =0 \\ \sum_{n=0}^{\infin} a_n ({n}^{2}+2 m n)z^{n+m-2} \ + \sum_{n=2}^{\infin} a_{n-2} z^{n+m-2} =0 \\ a_0\times 0 \times z^{m-2} + a_1 (1+2m)z^{m-1} + \sum_{n=2}^{\infin} (a_n ({n}^{2}+2 m n) + a_{n-2}) z^{n+m-2} =0 n=0∑∞an(n2+2mn)zn+m−2 +n=0∑∞anzn+m=0n=0∑∞an(n2+2mn)zn+m−2 +n=2∑∞an−2zn+m−2=0a0×0×zm−2+a1(1+2m)zm−1+n=2∑∞(an(n2+2mn)+an−2)zn+m−2=0
这是一个关于 z z z 的 多项式，这个多项式恒为 0 等价于这个多项式的每一项的系数都为 0，所以有：
a 1 ( 2 m + 1 ) = 0 a n ( n 2 + 2 m n ) + a n − 2 = 0 a_1 (2m+1) = 0 \\ a_n ({n}^{2}+2 m n) + a_{n-2} = 0 a1(2m+1)=0an(n2+2mn)+an−2=0
所以 a 1 = a 3 = a 5 = ⋯ = 0 a_1 = a_3 = a_5 = \cdots = 0 a1=a3=a5=⋯=0 ， a 0 a_0 a0 为任意的数。
a n = − a n − 2 n 2 + 2 m n = − a n − 2 n ( n + 2 m ) = − a n − 4 ( n ( n + 2 m ) ) ( ( n − 2 ) ( n − 2 + 2 m ) ) a 2 n = − a 2 n − 2 4 n 2 + 4 m n = − a 2 n − 2 ( 2 n ) ( 2 n + 2 m ) = a 2 n − 4 ( ( 2 n ) ( 2 n + 2 m ) ) ( ( 2 n − 2 ) ( 2 n − 2 + 2 m ) ) = ( − 1 ) n a 0 ( ( 2 n ) ( 2 n − 2 ) ⋯ ( 2 ) ) ( ( 2 n + 2 m ) ( 2 n − 2 + 2 m ) ⋯ ( 2 + 2 m ) ) = ( − 1 ) n a 0 2 n n ! 2 n ( n + m ) ! / m ! = ( − 1 ) n a 0 2 2 n n ! ( n + m ) ! / m ! \begin{aligned} a_{n} &= -\frac{a_{n-2}}{n^2 + 2 m n} = -\frac{a_{n-2}}{n(n + 2 m)}\\ &=-\frac{a_{n-4}}{(n(n + 2 m))((n-2)(n-2+2m))} \\ a_{2n} &= -\frac{a_{2n-2}}{4n^2 + 4 m n} = -\frac{a_{2n-2}}{(2n)(2n + 2 m)}\\ &=\frac{a_{2n-4}}{((2n)(2n + 2 m))((2n-2)(2n-2+2m))} \\ & = \frac{(-1)^n a_{0}}{((2n)(2n-2)\cdots(2))((2n + 2 m)(2n-2+2m)\cdots (2+2m)) } \\ & = \frac{(-1)^n a_{0}}{2^n n! 2^n (n+m)! / m!} \\ & = \frac{(-1)^n a_{0}}{2^{2n} n! (n+m)! / m!} \end{aligned} ana2n=−n2+2mnan−2=−n(n+2m)an−2=−(n(n+2m))((n−2)(n−2+2m))an−4=−4n2+4mna2n−2=−(2n)(2n+2m)a2n−2=((2n)(2n+2m))((2n−2)(2n−2+2m))a2n−4=((2n)(2n−2)⋯(2))((2n+2m)(2n−2+2m)⋯(2+2m))(−1)na0=2nn!2n(n+m)!/m!(−1)na0=22nn!(n+m)!/m!(−1)na0
至此，Bessel 函数的解析表达式就出来了。
g m ( z ) = ∑ n = 0 ∞ a n z n + m = ∑ n = 0 ∞ ( − 1 ) n a 0 2 2 n n ! ( n + m ) ! / m ! z 2 n + m = a 0 m ! z m ∑ n = 0 ∞ ( − 1 ) n n ! ( n + m ) ! z 2 n 2 2 n \begin{aligned} g_m(z) &= \sum_{n=0}^{\infin} a_n z^{n+m} \\ &= \sum_{n=0}^{\infin} \frac{(-1)^n a_{0}}{2^{2n} n! (n+m)! / m!} z^{2n+m} \\ &= a_0 m! z^m \sum_{n=0}^{\infin} \frac{(-1)^n}{ n! (n+m)!} \frac{z^{2n}}{2^{2n}} \end{aligned} gm(z)=n=0∑∞anzn+m=n=0∑∞22nn!(n+m)!/m!(−1)na0z2n+m=a0m!zmn=0∑∞n!(n+m)!(−1)n22nz2n
由于 a 0 a_0 a0 是任意常数，我们可以取 a 0 m ! = 2 − m a_0 m! = 2^{-m} a0m!=2−m。这样取上面的表达式还可以再简化一些。
J m ( z ) = a 0 m ! z m ∑ n = 0 ∞ ( − 1 ) n n ! ( n + m ) ! z 2 n 2 2 n = ( z 2 ) m ∑ n = 0 ∞ ( − 1 ) n n ! ( n + m ) ! ( z 2 ) 2 n = ∑ n = 0 ∞ ( − 1 ) n n ! ( n + m ) ! ( z 2 ) 2 n + m \begin{aligned} J_m(z) &= a_0 m! z^m \sum_{n=0}^{\infin} \frac{(-1)^n}{ n! (n+m)!} \frac{z^{2n}}{2^{2n}} \\ &= \left(\frac{z}{2}\right)^m \sum_{n=0}^{\infin} \frac{(-1)^n}{ n! (n+m)!} \left(\frac{z}{2}\right)^{2n} \\ &= \sum_{n=0}^{\infin} \frac{(-1)^n}{ n! (n+m)!} \left(\frac{z}{2}\right)^{2n+m} \end{aligned} Jm(z)=a0m!zmn=0∑∞n!(n+m)!(−1)n22nz2n=(2z)mn=0∑∞n!(n+m)!(−1)n(2z)2n=n=0∑∞n!(n+m)!(−1)n(2z)2n+m
上面式子中我们把 g m ( z ) g_m(z) gm(z) 写成了 J m ( z ) J_m(z) Jm(z) ，因为Bessel 函数通常用 J m ( z ) J_m(z) Jm(z)。下面是用数学软件(Maxima) 画出的m=0,1,2,3 时的函数图像。
上面图上我们可以看出Bessel 函数也是振荡函数。当 m 为偶数时，Bessel 函数是偶函数，m 为奇数时，Bessel 函数为奇函数。
V ( r , φ ) = R ( r ) Φ ( φ ) = C m J m ( k r ) cos ( m φ + ψ ) \begin{aligned} V(r,\varphi) &= R(r) \Phi(\varphi) \\ &= C_m J_m(kr) \cos(m \varphi + \psi) \end{aligned} V(r,φ)=R(r)Φ(φ)=CmJm(kr)cos(mφ+ψ)
当 m = 0 时， V ( r , φ ) V(r,\varphi) V(r,φ) 的函数图像如下图。
当 m = 1 m = 1 m=1 时， V ( r , φ ) V(r,\varphi) V(r,φ) 的函数图像不再是旋转对称的。也就是 Bessel 函数在极坐标中 φ \varphi φ 方向被 cos ( φ ) \cos(\varphi) cos(φ) 函数调制了。
当 m = 2 m = 2 m=2 时， V ( r , φ ) V(r,\varphi) V(r,φ) 的函数图像如下图。
其他的函数图像大家可以自己画出来。简单的说就是随着 m 增大，在 φ \varphi φ 方向上的振荡也越来越多。上面这些图是用 Mathematica 画的。Mathematica 本身不能画极坐标的3D 图，所以需要在直角坐标系中来画这个图。下面是画图时使用的语句。
Plot3D[BesselJ[0, Sqrt[x^2 + y^2]] Cos[0 ArcTan[x, y]], {x, -18.07, 18.07}, {y, -18.07, 18.07}, RegionFunction -> Function[{x, y, z}, x^2 + y^2 <= 18.07*18.07], PlotPoints -> 100, Mesh -> None, ColorFunction -> "DarkRainbow"]Plot3D[BesselJ[1, Sqrt[x^2 + y^2]] Cos[1 ArcTan[x, y]], {x, -19.62, 19.62}, {y, -19.62, 19.62}, RegionFunction -> Function[{x, y, z}, x^2 + y^2 <= 19.62*19.62], PlotPoints -> 100, Mesh -> None, ColorFunction -> "DarkRainbow"]Plot3D[BesselJ[2, Sqrt[x^2 + y^2]] Cos[2 ArcTan[x, y]], {x, -21.12, 21.12}, {y, -21.12, 21.12}, RegionFunction -> Function[{x, y, z}, x^2 + y^2 <= 21.12*21.12], PlotPoints -> 100, Mesh -> None, ColorFunction -> "DarkRainbow"]
u ( r , φ , t ) = V ( r , φ ) Z ( t ) = C m J m ( k r ) cos ( m φ + ψ ) cos ( c k t + δ ) u(r,\varphi, t) = V(r,\varphi) Z(t) \\ =C_m J_m(kr) \cos(m \varphi + \psi) \cos(c\ k \ t + \delta) u(r,φ,t)=V(r,φ)Z(t)=CmJm(kr)cos(mφ+ψ)cos(c k t+δ)
上面式子中的 k k k 并没有确定，需要边界条件来确定 k k k。之后就只有 m m m 一个可变的参数了，不同的 m m m 表示振动的不同模态。至此，这章最重要的知识点就都总结完了。剩余的部分个人认为不是特别重要，所以没有详细写。