Log-polar coordinates

Log-polar coordinates in the plane consist of a pair of real numbers (ρ,θ), where ρ is the logarithm of the distance between a given point and the origin and θ is the angle between a line of reference (the x-axis) and the line through the origin and the point. The angular coordinate is the same as for polar coordinates, while the radial coordinate is transformed according to the rule

${\displaystyle r=e^{\rho }}$.

Where ${\displaystyle r}$ is the distance to the origin. The formulas for transformation from Cartesian coordinates to log-polar coordinates are given by

${\displaystyle {\begin{cases}\rho =\ln \left({\sqrt {x^{2}+y^{2}}}\right),\\\theta =\operatorname {atan2} (y,\,x).\end{cases}}}$

and the formulas for transformation from log-polar to Cartesian coordinates are

${\displaystyle {\begin{cases}x=e^{\rho }\cos \theta ,\\y=e^{\rho }\sin \theta .\end{cases}}}$

By using complex numbers (x, y) = x + iy, the latter transformation can be written as

${\displaystyle x+iy=e^{\rho +i\theta }}$

i.e. the complex exponential function. From this follows that basic equations in harmonic and complex analysis will have the same simple form as in Cartesian coordinates. This is not the case for polar coordinates.

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Laplace's equation in two dimensions is given by

${\displaystyle {\frac {\partial ^{2}u}{\partial x^{2}}}+{\frac {\partial ^{2}u}{\partial y^{2}}}=0}$

in Cartesian coordinates. Writing the same equation in polar coordinates gives the more complicated equation

${\displaystyle r{\frac {\partial }{\partial r}}\left(r{\frac {\partial u}{\partial r}}\right)+{\frac {\partial ^{2}u}{\partial \theta ^{2}}}=0}$

or equivalently

${\displaystyle \left(r{\frac {\partial }{\partial r}}\right)^{2}u+{\frac {\partial ^{2}u}{\partial \theta ^{2}}}=0}$

However, from the relation ${\displaystyle r=e^{\rho }}$ it follows that ${\displaystyle r{\frac {\partial }{\partial r}}={\frac {\partial }{\partial \rho }}}$ so Laplace's equation in log-polar coordinates,

${\displaystyle {\frac {\partial ^{2}u}{\partial \rho ^{2}}}+{\frac {\partial ^{2}u}{\partial \theta ^{2}}}=0}$

has the same simple expression as in Cartesian coordinates. This is true for all coordinate systems where the transformation to Cartesian coordinates is given by a conformal mapping. Thus, when considering Laplace's equation for a part of the plane with rotational symmetry, e.g. a circular disk, log-polar coordinates is the natural choice.

A similar situation arises when considering analytical functions. An analytical function ${\displaystyle f(x,y)=u(x,y)+iv(x,y)}$ written in Cartesian coordinates satisfies the Cauchy–Riemann equations:

${\displaystyle {\frac {\partial u}{\partial x}}={\frac {\partial v}{\partial y}},\ \ \ \ \ \ {\frac {\partial u}{\partial y}}=-{\frac {\partial v}{\partial x}}}$

If the function instead is expressed in polar form ${\displaystyle f(re^{i\theta })=Re^{i\Phi }}$, the Cauchy–Riemann equations take the more complicated form

${\displaystyle r{\frac {\partial \log R}{\partial r}}={\frac {\partial \Phi }{\partial \theta }},\ \ \ \ \ \ {\frac {\partial \log R}{\partial \theta }}=-r{\frac {\partial \Phi }{\partial r}},}$

Just as in the case with Laplace's equation, the simple form of Cartesian coordinates is recovered by changing polar into log-polar coordinates (let ${\displaystyle P=\log R}$):

${\displaystyle {\frac {\partial P}{\partial \rho }}={\frac {\partial \Phi }{\partial \theta }},\ \ \ \ \ \ {\frac {\partial P}{\partial \theta }}=-{\frac {\partial \Phi }{\partial \rho }}}$

The Cauchy–Riemann equations can also be written in one single equation as

${\displaystyle \left({\frac {\partial }{\partial x}}+i{\frac {\partial }{\partial y}}\right)f(x+iy)=0}$

By expressing ${\displaystyle {\frac {\partial }{\partial x}}}$ and ${\displaystyle {\frac {\partial }{\partial y}}}$ in terms of ${\displaystyle {\frac {\partial }{\partial \rho }}}$ and ${\displaystyle {\frac {\partial }{\partial \theta }}}$ this equation can be written in the equivalent form

${\displaystyle \left({\frac {\partial }{\partial \rho }}+i{\frac {\partial }{\partial \theta }}\right)f(e^{\rho +i\theta })=0}$

When one wants to solve the Dirichlet problem in a domain with rotational symmetry, the usual thing to do is to use the method of separation of variables for partial differential equations for Laplace's equation in polar form. This means that you write ${\displaystyle u(r,\theta )=R(r)\Theta (\theta )}$. Laplace's equation is then separated into two ordinary differential equations

${\displaystyle {\begin{cases}\Theta ''(\theta )+\nu ^{2}\Theta (\theta )=0\\r^{2}R''(r)+rR'(r)-\nu ^{2}R(r)=0\end{cases}}}$

where ${\displaystyle \nu }$ is a constant. The first of these has constant coefficients and is easily solved. The second is a special case of Euler's equation

${\displaystyle r^{2}R''(r)+crR'(r)+dR(r)=0}$

where ${\displaystyle c,d}$ are constants. This equation is usually solved by the ansatz ${\displaystyle R(r)=r^{\lambda }}$, but through use of log-polar radius, it can be changed into an equation with constant coefficients:

${\displaystyle P''(\rho )+(c-1)P'(\rho )+dP(\rho )=0}$

When considering Laplace's equation, ${\displaystyle c=1}$ and ${\displaystyle d=-\nu ^{2}}$ so the equation for ${\displaystyle r}$ takes the simple form

${\displaystyle P''(\rho )-\nu ^{2}P(\rho )=0}$

When solving the Dirichlet problem in Cartesian coordinates, these are exactly the equations for ${\displaystyle x}$ and ${\displaystyle y}$. Thus, once again the natural choice for a domain with rotational symmetry is not polar, but rather log-polar, coordinates.