Some important equations in log-polar coordinates Log-polar coordinates
1 important equations in log-polar coordinates
1.1 laplace s equation
1.2 cauchy–riemann equations
1.3 euler s equation
some important equations in log-polar coordinates
laplace s equation
laplace s equation in 2 dimensions given by
∂
2
u
∂
x
2
+
∂
2
u
∂
y
2
=
0
{\displaystyle {\frac {\partial ^{2}u}{\partial x^{2}}}+{\frac {\partial ^{2}u}{\partial y^{2}}}=0}
in cartesian coordinates. writing same equation in polar coordinates gives more complicated equation
r
∂
∂
r
(
r
∂
u
∂
r
)
+
∂
2
u
∂
θ
2
=
0
{\displaystyle r{\frac {\partial }{\partial r}}\left(r{\frac {\partial u}{\partial r}}\right)+{\frac {\partial ^{2}u}{\partial \theta ^{2}}}=0}
or equivalently
(
r
∂
∂
r
)
2
u
+
∂
2
u
∂
θ
2
=
0
{\displaystyle \left(r{\frac {\partial }{\partial r}}\right)^{2}u+{\frac {\partial ^{2}u}{\partial \theta ^{2}}}=0}
however, relation
r
=
e
ρ
{\displaystyle r=e^{\rho }}
follows
r
∂
∂
r
=
∂
∂
ρ
{\displaystyle r{\frac {\partial }{\partial r}}={\frac {\partial }{\partial \rho }}}
laplace s equation in log-polar coordinates,
∂
2
u
∂
ρ
2
+
∂
2
u
∂
θ
2
=
0
{\displaystyle {\frac {\partial ^{2}u}{\partial \rho ^{2}}}+{\frac {\partial ^{2}u}{\partial \theta ^{2}}}=0}
has same simple expression in cartesian coordinates. true coordinate systems transformation cartesian coordinates given conformal mapping. thus, when considering laplace s equation part of plane rotational symmetry, e.g. circular disk, log-polar coordinates natural choice.
cauchy–riemann equations
a similar situation arises when considering analytical functions. analytical function
f
(
x
,
y
)
=
u
(
x
,
y
)
+
i
v
(
x
,
y
)
{\displaystyle f(x,y)=u(x,y)+iv(x,y)}
written in cartesian coordinates satisfies cauchy–riemann equations:
∂
u
∂
x
=
∂
v
∂
y
,
∂
u
∂
y
=
−
∂
v
∂
x
{\displaystyle {\frac {\partial u}{\partial x}}={\frac {\partial v}{\partial y}},\ \ \ \ \ \ {\frac {\partial u}{\partial y}}=-{\frac {\partial v}{\partial x}}}
if function instead expressed in polar form
f
(
r
e
i
θ
)
=
r
e
i
Φ
{\displaystyle f(re^{i\theta })=re^{i\phi }}
, cauchy–riemann equations take more complicated form
r
∂
log
r
∂
r
=
∂
Φ
∂
θ
,
∂
log
r
∂
θ
=
−
r
∂
Φ
∂
r
,
{\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 in case laplace s equation, simple form of cartesian coordinates recovered changing polar log-polar coordinates (let
p
=
log
r
{\displaystyle p=\log r}
):
∂
p
∂
ρ
=
∂
Φ
∂
θ
,
∂
p
∂
θ
=
−
∂
Φ
∂
ρ
{\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 written in 1 single equation as
(
∂
∂
x
+
i
∂
∂
y
)
f
(
x
+
i
y
)
=
0
{\displaystyle \left({\frac {\partial }{\partial x}}+i{\frac {\partial }{\partial y}}\right)f(x+iy)=0}
by expressing
∂
∂
x
{\displaystyle {\frac {\partial }{\partial x}}}
,
∂
∂
y
{\displaystyle {\frac {\partial }{\partial y}}}
in terms of
∂
∂
ρ
{\displaystyle {\frac {\partial }{\partial \rho }}}
,
∂
∂
θ
{\displaystyle {\frac {\partial }{\partial \theta }}}
equation can written in equivalent form
(
∂
∂
ρ
+
i
∂
∂
θ
)
f
(
e
ρ
+
i
θ
)
=
0
{\displaystyle \left({\frac {\partial }{\partial \rho }}+i{\frac {\partial }{\partial \theta }}\right)f(e^{\rho +i\theta })=0}
euler s equation
when 1 wants solve dirichlet problem in domain rotational symmetry, usual thing use method of separation of variables partial differential equations laplace s equation in polar form. means write
u
(
r
,
θ
)
=
r
(
r
)
Θ
(
θ
)
{\displaystyle u(r,\theta )=r(r)\theta (\theta )}
. laplace s equation separated 2 ordinary differential equations
{
Θ
″
(
θ
)
+
ν
2
Θ
(
θ
)
=
0
r
2
r
″
(
r
)
+
r
r
′
(
r
)
−
ν
2
r
(
r
)
=
0
{\displaystyle {\begin{cases}\theta (\theta )+\nu ^{2}\theta (\theta )=0\\r^{2}r (r)+rr (r)-\nu ^{2}r(r)=0\end{cases}}}
where
ν
{\displaystyle \nu }
constant. first of these has constant coefficients , solved. second special case of euler s equation
r
2
r
″
(
r
)
+
c
r
r
′
(
r
)
+
d
r
(
r
)
=
0
{\displaystyle r^{2}r (r)+crr (r)+dr(r)=0}
where
c
,
d
{\displaystyle c,d}
constants. equation solved ansatz
r
(
r
)
=
r
λ
{\displaystyle r(r)=r^{\lambda }}
, through use of log-polar radius, can changed equation constant coefficients:
p
″
(
ρ
)
+
(
c
−
1
)
p
′
(
ρ
)
+
d
p
(
ρ
)
=
0
{\displaystyle p (\rho )+(c-1)p (\rho )+dp(\rho )=0}
when considering laplace s equation,
c
=
1
{\displaystyle c=1}
,
d
=
−
ν
2
{\displaystyle d=-\nu ^{2}}
equation
r
{\displaystyle r}
takes simple form
p
″
(
ρ
)
−
ν
2
p
(
ρ
)
=
0
{\displaystyle p (\rho )-\nu ^{2}p(\rho )=0}
when solving dirichlet problem in cartesian coordinates, these equations
x
{\displaystyle x}
,
y
{\displaystyle y}
. thus, once again natural choice domain rotational symmetry not polar, rather log-polar, coordinates.
Comments
Post a Comment