next up previous
Next: Spherical Coordinates - Up: Basics Potential Solutions Previous: Analytical Functions of a

Cylindrical Coordinates, $F(R,\phi )$

For two dimensional problems that are independent of $z$, $\nabla^{2}F = 0$ can be written as
\begin{displaymath}
{1\over R}{\partial \over \partial R}\left (R{\partial F \o...
... + {1\over R^{2}}{\partial^{2}F\over \partial
\phi^{2}} = 0.
\end{displaymath} (5.17)

There are many particular solutions.
  1. Consider
    \begin{displaymath}
F = F(R) = - c \log R,
\end{displaymath} (5.18)

    where $c$ is a constant, so that

    \begin{displaymath}
-\nabla F = {c\over R}{\bf e}_{R}.
\end{displaymath}

    This corresponds to an isolated line charge on the $z$-axis in electrostatics and an isolated line current on the $z$-axis in MHD. For a fluid this gives a line source ($c>0$) and a line sink ($c<0$)

  2. The above solution may be modified to place the line source at a position ${\bf R}_{0} = ( R_{0}, \phi_{0})$ by using

    \begin{displaymath}
F = - c \log \vert{\bf R} - {\bf R_{0}}\vert,
\end{displaymath}

    and ${\bf R} = (R, \phi)$.

  3. A uniform field or flow, in the $x$-direction at large distances, is
    \begin{displaymath}
F = - R \cos \phi = - cx,
\end{displaymath} (5.19)

    and

    \begin{displaymath}
-\nabla F = c{\bf i}.
\end{displaymath}

  4. A line dipole in either electrostatics or MHD of moment $c$ or a line doublet in fluids has a potential
    \begin{displaymath}
F = c{\cos \phi\over R},
\end{displaymath} (5.20)

    and

    \begin{displaymath}
-\nabla F = {c \cos \phi\over R^{2}}{\bf e}_{R} + {c\sin \phi \over
R^{2}}{\bf e}_{\phi}.
\end{displaymath}

These are the building blocks of our potential solutions and may be added together to get the effect of, for example, a line source, a uniform field and a line dipole,

\begin{displaymath}
F = -c_{1}\log R - c_{2}R\cos \phi + c_{3}{\cos \phi \over R}.
\end{displaymath}

These are just particular solutions of a more general class given by separable solutions to (5.17).

Example 5. .3More general solutions can be obtained by using separation of variables. Hence, we set

\begin{displaymath}
F = h(R) g(\phi ).
\end{displaymath}

Substituting into (5.17) we obtain

\begin{displaymath}
{1\over R}{d\over dR}\left (R{dh\over dR}\right )g + {1\over R^{2}}
h {d^{2}g\over d\phi^{2}} = 0.
\end{displaymath}

Rearranging to into functions of $R$ and $\phi$ alone gives

\begin{displaymath}
{R\over h}{d\over dR}\left (R{dh\over dR}\right ) = -{1\over
g}{d^{2}g\over d\phi^{2}} = n^{2}.
\end{displaymath}

The separation constant is $n^{2}$ and it must be the square of an integer. Solving for $g$ we obtain,

\begin{displaymath}
{d^{2}g\over d\phi^{2}} + n^{2}g = 0, \quad \Rightarrow \quad g =
A\cos n\phi + B\sin n\phi,
\end{displaymath}

and so $n$ must be an integer to ensure that $g(0) = g(2\pi )$. The radial variations are given by the solution to
\begin{displaymath}
R^{2}{d^{2}h\over dR^{2}} + R{dh\over dR} - n^{2}h = 0,
\end{displaymath} (5.21)

with solutions

\begin{displaymath}
h = R^{n} \hbox{ and } h= R^{-n}, \quad n\ne 0.
\end{displaymath}

The case with $n=0$ gives (5.18). Therefore, we have four types of potential solutions (also called harmonic functions)
\begin{displaymath}
R^{n}\cos n\phi, \quad R^{-n}\cos n\phi, \quad R^{n}\sin n\phi,
\quad R^{-n}\sin n\phi,
\end{displaymath} (5.22)

and the general solution is a linear combination of all types. Note the following points.
  1. For $n=0$, $g=1$ and $h = \log R$. Hence,

    \begin{displaymath}
n=0,\qquad \Rightarrow \qquad F = \log R.
\end{displaymath}

  2. From above the $n=1$ solution gives the following possible symmetric, solutions for $F$,

    \begin{displaymath}
n=1, \qquad \Rightarrow F = R\cos \phi, \hbox{ or } F = {\cos
\phi \over R}.
\end{displaymath}

  3. The solutions involving $\cos n\phi$ and $\sin n\phi$ are linearly independent for different values of $n$,
    \begin{displaymath}
\int_{0}^{\pi}\cos n\phi \cos m\phi d\phi = {1\over 2}
\i...
...s (n+m)\phi + \cos (n-m)\phi d\phi = 0, \hbox{
if }n \ne m.
\end{displaymath} (5.23)

Example 5. .4Find the incompressible, irrotational flow of an ideal fluid or plasma, uniform at infinity, past a cylinder of radius $a$.

The situation is shown in Figure 5.2.

Figure 5.2: A uniform flow, at large distances, past a cylinder of radius $a$.
\includegraphics [scale=0.7]{dummy.ps}

The boundary conditions are

\begin{displaymath}
v_{x} = V_{0}, \hbox{ as } R\rightarrow \infty,
\end{displaymath}

and on the surface of the cylinder there can be no radial flow

\begin{displaymath}
v_{R} = 0, \hbox{ on } R=a.
\end{displaymath}

To proceed, we note that

\begin{displaymath}
{\bf v} = - \nabla F.
\end{displaymath}

Thus,

\begin{displaymath}
v_{x} = - {\partial F\over \partial x} = V_{0}, \hbox{ as
...
..., \qquad \Rightarrow \qquad F = - V_{0}x =
-V_{0}R\cos \phi.
\end{displaymath}

The potential must satisfy
\begin{displaymath}
\nabla^{2}F = 0,
\end{displaymath} (5.24)

and the most general solution to (5.24), that is even in $\phi$, is
\begin{displaymath}
F = a_{0}\log R + \sum_{n=1}^{\infty} a_{n}R^{n}\cos n\phi +
c_{n}R^{-n}\cos n\phi.
\end{displaymath} (5.25)

The condition as $R\rightarrow \infty$ shows that all coefficients of the positive powers of $R$ must be zero, except for $n=1$. Hence,

\begin{displaymath}
\hbox{As }R\rightarrow \infty, \quad F \rightarrow -V_{0}R\...
...i +
a_{0}\log R + \sum_{n=1}^{\infty} c_{n}R^{-n}\cos n\phi.
\end{displaymath}

The coefficients $c_{n}$ are determined by the condition on the surface of the cylinder. Here $v_{R}(R=a, \phi) = -\partial
F/\partial R = 0$.

\begin{displaymath}
{\partial F\over \partial R} = - V_{0}\cos \phi + {a_{0}\over R} -
\sum_{n=1}^{\infty}n c_{n}R^{-(n+1)}\cos n\phi.
\end{displaymath}

On $R=a$ we must have

\begin{displaymath}
\left (-V_{0} -{c_{1}\over a^{2}}\right ) \cos \phi + {a_{0...
... a} - \sum_{n=2}^{\infty}{n c_{n}\over a^{n+1}}\cos n\phi = 0.
\end{displaymath}

Using (5.23) each term is orthogonal to each other and so the coefficient of each term in $\cos n\phi$ must be zero. This implies that

\begin{displaymath}
c_{1} = -V_{0}a^{2}, \quad a_{0} = 0, \quad c_{n} = 0, \hbox{ for
}n \geq 2.
\end{displaymath}

From this we find that the solution is
\begin{displaymath}
F = -V_{0}R \cos \phi - V_{0}{a^{2}\over R}\cos \phi.
\end{displaymath} (5.26)

Another way of thinking of this solution is to say that the solution giving a uniform flow at large distances is

\begin{displaymath}
F = -V_{0}R \cos \phi.
\end{displaymath}

To this we must add on a term that is going to make $V_{R}= 0$ on $R=a$ and this must have the same form in $\phi$. Hence the extra term to add on is of the form

\begin{displaymath}
c_{1}{\cos \phi \over R}.
\end{displaymath}

Then we can choose $c_{1}$ so that we do indeed have $v_{R}=0$ on $R=a$.

In this last example we can calculate the pressure from Bernoulli's equation (2.20). Hence, in the absence of gravity,

\begin{displaymath}
p + {1\over 2}\rho v^{2} = p_{0} + {1\over 2}\rho V_{0}^{2} =
\hbox{ constant}.
\end{displaymath}

On the surface of the cylinder $v_{R}=0$ and

\begin{displaymath}
v_{\phi} = -{1\over R}{\partial F\over \partial \phi} = V_{...
...
\phi + V_{0}{a^{2}\over R^{2}}\sin \phi, \hbox{ on }R = a,
\end{displaymath}

Therefore,

\begin{displaymath}
p = p_{0} +{1\over 2}\rho V_{0}^{2} - {1\over 2}\rho
4V_{0}^{2}\sin^{2}\phi.
\end{displaymath}

Thus, the pressure on the surface of the cylinder varies as we move round the cylinder.

Example 5. .5Consider a cylinder rotating with a constant angular speed, $\omega$. In this case the potential must generate a velocity component have $v_{\phi} = a \omega$ on $R=a$. The potential is

\begin{displaymath}
F = - c \phi, \quad \Rightarrow \quad -\nabla F = -{1\over ...
...\over \partial \phi}{\bf e}_{\phi} = {c\over R}{\bf e}_{\phi}.
\end{displaymath}

Hence,

\begin{displaymath}
{c\over a} = a\omega, \quad \Rightarrow \quad c = a^{2}\omega.
\end{displaymath}


next up previous
Next: Spherical Coordinates - Up: Basics Potential Solutions Previous: Analytical Functions of a
Prof. Alan Hood
2000-11-06