📂Mathematical Physics

Solving the Laplace Equation Independent of the Jet Axis in Cylindrical Coordinates Using the Method of Separation of Variables

Theorem

The general solution of the Laplace’s equation with cylindrical symmetry in the cylindrical coordinate system is as follows.

$$V(s,\phi) = A_{0} \ln s +B_{0} +\sum \limits _{k=1} ^\infty ( A_{k} s^k ++ B_{k} s^{-k} )( C_{k}\cos k\phi + D_{k}\sin k\phi)$$

Proof

Step 0

When boundary conditions are more easily represented in cylindrical coordinates, one has to solve the Laplace’s equation for cylindrical coordinates. The Laplace’s equation in cylindrical coordinates is as follows. (Reference1, Reference2)

$$\nabla ^2 V = \dfrac{1}{s} \dfrac{\partial }{ \partial s} \left( s\dfrac{\partial V}{\partial s} \right) + \dfrac{1}{s^2} \dfrac{\partial ^2 V}{\partial \phi^2 } +\dfrac{\partial ^2 V }{\partial z^2 }=0$$

Let’s say that the potential $V$ is a function independent of $z$. In other words, it’s assumed that the value of $V$ does not change when only $z$ varies while other values remain the same. Then, the change in $V$ due to $z$ is $0$, which means the third term disappears.

$$\dfrac{1}{s} \dfrac{\partial }{ \partial s} \left( s\dfrac{\partial V}{\partial s} \right) + \dfrac{1}{s^2} \dfrac{\partial ^2 V}{\partial \phi^2 } =0 \tag{1}$$

Suppose that the potential $V(s,\phi)$ is a function possible to separate variables. It means that $V$ is composed of the product of a function $S(s)$ solely of $s$ and a function $\Phi (\phi)$ solely of $\phi$. If we insert $V(s,\phi)=S(s) \Phi (\phi)$ into $(1)$, it becomes the following.

$$\dfrac{1}{s} \dfrac{d }{d s} \left( s \dfrac{d S}{d s} \right) \Phi + \dfrac{1}{s^2} \dfrac{d ^2 \Phi }{d \phi ^2 } S=0$$

Multiplying both sides by $\dfrac{s^2}{S \Phi}$ yields

$$\dfrac{s}{S} \dfrac{d }{d s} \left( s \dfrac{d S}{d s} \right) + \dfrac{1}{\Phi} \dfrac{d ^2 \Phi }{d \phi ^2 } =0$$

For this equation to hold, both the first and the second terms must be constants. This is because the first term is affected when $s$ changes, but the second term is unaffected by $s$. However, their sum must result in $0$, leading to the entire first term being a constant. Similarly, the second term is also a constant.

$$\dfrac{s}{S} \dfrac{d }{d s} \left( s \dfrac{dS}{ds} \right) = C_{1}$$

$$\dfrac{1}{\Phi} \dfrac{ d^2 \Phi} {d \phi ^2} =C_2$$

$(1)$’s complex partial differential equation is transformed into two simple ordinary differential equations. Solving each differential equation to find $S(s)$ and $\Phi (\phi)$ and multiplying them gives us the desired $V(s,\phi)$. Since $C_{1}+C_2=0$, the two constants are of the same magnitude but opposite in sign. Here, $C_2$ must be a negative constant. This is because the differential equation’s solution turns out to be in the form of $\Phi (\phi)=Ae^{k\phi}+Be^{-k\phi}$ when $C_2$ is positive (Solution Reference). Being a cylindrical coordinate system, $\Phi (\phi) = \Phi (\phi+2\pi)$ must be satisfied, but the aforementioned equation does not. Therefore, $C_{1}$ is a positive constant, and $C_2$ is a negative constant.

$$\dfrac{s}{S} \dfrac{d }{d s} \left( s \dfrac{dS}{ds} \right) =k^2$$

$$\dfrac{1}{\Phi} \dfrac{ d^2 \Phi} {d \phi ^2} =-k^2$$

Step 1

$$\dfrac{1}{\Phi} \dfrac{ d^2 \Phi} {d \phi ^2} =-k^2$$

The solution to the above differential equation is well-known to be $\Phi (\phi)=e^{\pm ik \phi}$. (Reference) Hence, the general solution is

$$\Phi (\phi) = A e^{ik\phi} + Be^{-ik\phi}$$

Using Euler’s formula $e^{i\theta}=\cos \theta + i\sin \theta$, it can be represented as follows.

$$\Phi ( \phi) = A\cos k\phi + B \sin k\phi$$

Here, $A$ and $B$ are complex constants, different from the constants mentioned before. Let’s now check if $\Phi (\phi)=\Phi (\phi+2\pi)$ is satisfied.

$$\begin{align*} \Phi (\phi+2\pi) &= A\cos k(\phi +2\pi) + B \sin k(\phi+2\pi) \\ &= A\cos (k\phi +2k\pi) + B \sin (k\phi+2k\pi) \end{align*}$$

If $k= 0, \pm 1, \pm 2, \cdots$ then

$$\begin{align*} \Phi (\phi) &= A\cos (k\phi) + B \sin (k\phi) \\ &= \Phi ( \phi) \end{align*}$$

Hence, the solution is

$$\Phi ( \phi) = A\cos k\phi + B \sin k\phi\quad (k=0,1,2\cdots)$$

In the case of negative values, it overlaps with the case of positive values, so only the cases of positive values need to be documented. However, the case of $k=0$ is actually not included in the solution. This is because, when solving the differential equation with $k=0$,

$$\dfrac{ d^2 \Phi} {d \phi^2 } =0$$

$$\implies \dfrac{ d \Phi}{d \phi}=C$$

$$\implies \Phi=C\phi +D$$

we get $\Phi$, which lacks the periodicity mentioned earlier. Thus, the final solution obtained is

$$\Phi ( \phi) = A\cos k\phi + B \sin k\phi,\ \ (k=1,2\cdots)$$

Step 2

$$\dfrac{s}{S} \dfrac{d }{d s} \left( s \dfrac{dS}{ds} \right) =k^2$$

Simplifying the above differential equation yields

$$s\left( \dfrac{d S}{ds} + s \dfrac{d^2 S}{d s^2} \right) =k^2 S$$

$$\implies s^2 \dfrac{d^2 S}{d s^2 } + s\dfrac{d S}{ds} -k^2 S=0$$

This is Euler’s equation, which can be solved in the manner described here, but in this document, it’s solved more simply. Utilizing that the solution to the differential equation comes in the form of $s^n$ and substituting $S=s^n$,

$$s^2(n)(n-1)s^{n-2} +sns^{n-1}-k^2s^n=0$$

$$\implies n(n-1)+n-k^2=0$$

$$\implies n^2-k^2=0$$

$$\therefore n=\pm k$$ The two solutions of the differential equation are $s^k$ and $s^{-k}$. The general solution is a linear combination of the two solutions, and adding the condition of $k$ from Step 1,

$$S(s) = Cs^k+Ds^{-k},\quad k=1,2,\cdots$$ However, it’s important that $S(s)$ exists for the case of $k=0$. Solving the differential equation when $k=0$ yields

$$\dfrac{s}{S} \dfrac{d }{d s} \left( s \dfrac{dS}{ds} \right) =0$$

$$\implies \dfrac{ d}{ds} \left( s \dfrac{ dS }{ds} \right) =0$$

$$\implies s\dfrac{dS}{ds}=C$$

$$\implies dS=\dfrac{C}{s} ds$$

$$\implies S(s)=C\ln s +D$$

Thus, the final expression of the general solution is

$$S(s)=A_{0} \ln s +B_{0} +\sum \limits _{k=1} ^\infty ( A_{k} s^k + B_{k} s^{-k})$$

Step 3

Combining the results of Step 1 and Step 2, the potential is as follows.

$$V(s,\phi) = A_{0} \ln s +B_{0} +\sum \limits _{k=1} ^\infty ( A_{k} s^k ++ B_{k} s^{-k} )( C_{k}\cos k\phi + D_{k}\sin k\phi)$$