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Re: Conformal Mapping
*To*: mathgroup at smc.vnet.net
*Subject*: [mg128591] Re: Conformal Mapping
*From*: Roland Franzius <roland.franzius at uos.de>
*Date*: Thu, 8 Nov 2012 02:08:26 -0500 (EST)
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*References*: <k6t7ut$1ni$1@smc.vnet.net> <20121105233941.63E226854@smc.vnet.net> <k7ctg3$drt$1@smc.vnet.net>
Am 07.11.2012 07:01, schrieb Andrzej Kozlowski:
> On 6 Nov 2012, at 00:39, Roland Franzius<roland.franzius at uos.de> wrote:
>
>> Am 01.11.2012 08:22, schrieb MaxJ:
>>> Hi folks,
>>>
>>> I need help finding a Mobius transform such that the region:
>>>
>>> |z-i|< sqrt(2)
>>> &&
>>> |z+i|< sqrt(2)
>>>
>>> in z-plane be mapped conformally into a unit circle in w-plane.
>>>
>>> Any help is appreciated very much.
>>
>> The boundaries are two circles with centers at +-i and radius^2 = 2.
>> Consequently the contours are passing through the six points on the
>> squares with vertices
>>
>> +-1, 2i +-1 and +-1 -2i +-1
>>
>> Consequently the lens shaped area in question is bounded by two
>> symmetric quarter circles with center at +-i intersecting at +-1 with an
>> angle of pi/2.
>>
>> We conclude that the identity w=z is a conformal map of the lens into
>> the inner of the unit circle.
>>
>> The construction of a conformal map onto the unit circle, transforming
>
>> to straight lines the pi/2 vertices at +-1, needs square roots centered
>> at the vertices +-1.
>>
>> This task could generate homework for up to one day approximately.
>>
>> --
>>
>> Roland Franzius
>>
>
>
> There is no need for square roots centered at vertices; ordinary squaring will suffice. And it's less then one day's homework; without Mathematica a maximum of 15 minutes I think, with Mathematica a lot less.
Homework and time was meant for students. In this group, to find the
solution needed about one week.
Thats of course perfect, the students will now have learnt about the
solution after that week by their instructor, probably.
> The standard way to deal with a lens is to use a Moebius transform to send one vertex to 0 and the other to infinity. In this case the map =09
>
> f[z_]:=(z+1)/(z-1)
>
> will do it. We get the angle"
>
> g = InverseFunction[f];
>
> RegionPlot[
> Abs[g[x + I y] - I]< Sqrt[2]&&
> Abs[g[x + I y ] + I]< Sqrt[2], {x, -2, 2}, {y, -2, 2},
> Frame -> False, Axes -> True]
>
> Next, we rotate it to the right by -3Pi I/4 and use squaring (not square root). We will get the upper half plane. We can now map the upper-half plane to the unit disk by a Moebius transformation - fro example
>
> kk[z_]:=(z-I)/(z+I)
>
> The final map is:
>
> gg[z_]=Simplify[kk[(f[z]Exp[-3Pi/4 I])^2]]
> (2 z)/(z^2+1)
>
> We can check this graphically:
>
> RegionPlot[Abs[gg[x + I y]]< 1, {x, -2, 2}, {y, -2, 2}]
>
> We see the lens plus some additional region. Both the lens and the additional region are mapped onto the unit disk. In order to construct the inverse map from the unit disk to the lens we need, of course, a square root.
Yes, approved;-)
--
Roland Franzius
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