33. Examples We consider examples of conformal mappings which involve combinations of elementary functions.

1) Mapping of a strip onto the unit circle Let there be given in the z-plane the strip D: -p/4 < Re z < p/4, which is to be mapped onto the circle |w| < 1 with correspondence of the three boundary points: f(p/4) = 1, f(i ) = i (i denotes the upper point at infinity of the strip). First of all, we rotate by a right angle and double the width of the strip:

next we use the fact that the exponential function

transforms the strip -p/2 < Im z1 <p /2, onto which (2) maps D, into the right half-plane Re z2 > 0 (really, whence changes from 0 to , arg z2 = y1 from -p/2 to p/2). There remains to map this half-plane onto the unit circle so that z2 = i, -i, 0, corresponding to the points z = p /4, -p /4, i become the points w = 1, -1. i. Such a problem is solved with the aid of (2) of 32. (wehave changed the notation):

Substituting (2) and (1) into (3), we obtain the final solution of the problem

(cf. 9.) We will still explain the correspondence of the lines for this mapping. The family of vertical lines Re(z) = const. for Mapping (1) become the family of horizontal lines, which Mapping (2) converts into the family of rays arg z2 = const., i.e., into the circles passing through the points z2 = 0 and z2 = .The bi-linear mapping transforms these lines into the points w = i and w = -i. Consequently, the rays considered become the family of circles passing through the points w = i. Th orhogonal family of segments Im z = const. become the family of circles with w = i as points of symmetry (Fig. 55).

The inverse function

realizes the inverse mapping of a circle into a strip. Replacing here iz by z1 and iw by pw1/2H, we obtain the mapping of the circle |z| < 1 onto the strip of width H: -H/2 < Im z < H/2. It has the form

(where we have changed to z and w).Mapping (5) takes the points z = 1 into the point w = and the point z = i into w=iH/2.. Its derivative

becomes infinite at the points z = 1.

2) Mapping of half-plane with removed cut onto half-plane. Let there be excluded from the half-plane Im z > 0 the segment (a. a + ih). In order to obtain this mapping, we use the fact that the mapping w = z doubles the angles at the origin and, consequently, one may "straighten" the angle between the excluded segment and the x-axis.

Correspondingly, we move the half-plane z by a to the left: z1 = z - a and obtain with the mapping z2 = z1 the plane with the removed ray -h < Re z2< , Im z2 = 0. Next, we again shift the plane z2 by h to the right: z3 = z2 + h. Finally, applying the mapping z4 = z3, we obtain the upper half-plane. Thus, the mapping becomes

Shifting again the z4 -plane by a to the right, in order to bring the point z = a + ih into the point a, we obtain finally

The derivative of Mapping (7)

vanishes at the points B and D (where z = a) and becomes infinite at the point C (where z = a + ih). The lines v = v0 = const. correspond to the lines of fourth order

symmetric with respect to the line x = a, on which their ordinates attain their maxima. For large values of v0, the curves (9) differ little from straight lines (Fig. 56).

For h = 0, Mapping (7) becomes the identity mapping w = z. We will find the principal part of (7) for small h. For this purpose, we transform (7), neglecting powers of h larger than the second. Applying the well -known formula for roots, we find

The approximate formula (10) ceases to be true for z close to a, because then the quantity h/(z - a)ceases to be small.

3) Mapping the circle with removed radial cut onto the unit circle (Fig. 57). Exclude form the circle |z| < 1 the straight segment [(1 - h)ea, eia]. The transformation of the region obtained onto the unit circle may be reduced with the aid of the supplementary bi-linear mapping to (7). However, it is simpler to employ the properties of the Joukowsky function 7. Turning the region in the z-plane by the angle -a and applying the Joukowsky function z = (z/eia + eia/z), we transform this region into the outside of the segment [-1, 1 + 2h1], where h1 = h/4(1 - h)*. The analogous transformation w = (w/eia + eia/w)

converts the circle in the w-plane into the outside of the segment [-1, 1]. It is readily seen that the linear transformation

maps arcs into arcs of the obtained regions in the z - and w- planes; replacing w and z by their expressions, we find the required mapping:

* In fact, images of the points z = (1 - h)eia are for the mapping under consideration are the points z0 = [1 - h + 1/(1 - h)] = 1 + h/1(1 - h).

For h = 0, we have h1 = 0 and w z. We will find the principal part of Mapping (11) for small h. For this purpose, we substitute w = z + w and neglect smaller than second order terms of h1, noting doing this that w and h1are of the same order (so that quantities of order w may be neglected). We obtain

or

whence

Thus, for small h1 and points z not too close to the point eia , our conformal mapping may be represented by the approximate formula

Differentiating (12), we obtain the principal part of the derivative

Note yet a link between the arguments of the points z = eij and w = eiq of the circles, corresponding to each other by Mapping (11). Separating in (12), after substituting z = eij and w = eiq, the real part and setting

we obtain*

whence

* We take the first two terms of the Taylor series for cos q at the point q = 0.

4) Mapping of plane with removed rays onto the strip 0 < v < H (Fig. 58). For the sake of definiteness, we require that the left of the ray becomes the lower shore , the right the upper shore *.

* These conditions determine only two of the real parameters of the mapping (they reduce to giving the corresponding two pairs of boundary points), the third remains arbitrary (cf. (15)),

We map by the bi-linear mapping

our rays into the single ray (0, ) and then onto the real axis by the transformation

The region under consideration then becomes the upper half-plane. In order to obtain the lower correspondence of the points, we subject this half-plane to a bi-linear mapping onto itself so that the images of the points at infinity A and C of the initial domain in the z-plane (i.e., the points 1) become 0 and :

(k is an arbitrary positive constant). There remains still to apply the logarithm, w = (H/p)ln z3, in order to obtain the mapping onto the strip with lower correspondence of the boundaries

where c = (H/p)ln k/a is an arbitrary real constant. There correspond to the two families of lines u = const. and v = const. in the z-plane for the mapping (15) families of ellipses and hyperbolae with foci a.

5) Mapping of the strip 0 < y < 2H with the cut - x a, y = H onto the strip 0 < v < 2H (Fig.59). The function

maps the strip with its cut onto the upper half-plane with the removed segment (0, bi) of the imaginary axis, where a=eap/2H. Function (7) of Example 2):

(in (7), one must set a = 0, h = b) maps this last domain onto the half-plane. Applying the logarithm, we find the required mapping

6) Mapping of the strip 0 < y < 1 with the cut 0 y h, x = a onto the strip 0 < v < 1 (Fig. 60). The function z = ep(z-a) maps the strip with cut onto the upper half-plane with a cut arc of the unit circle. The mapping

takes this arc into the segment of the imaginary axis (0, bi), where b = 1/i tanh phi/2 = tan ph/2.

Using again (7) of Example 2), we obtain the mapping of the given domain onto the upper half-plane:

The points A and E go for this mapping to the points

by the bi-linear mapping

we take them to 0 and and then apply the logarithm

Thus, we obtain the mapping onto the lower strip. However, obviously, there corresponds to the point C the point z5 = 0; in order order to shift it to the point a on the real axis, one must still move this strip to a. Thus, the required mapping has the form

or, finally,

For small h, the expression in curly brackets, which we will.denote by z , becomes

(we replace cos, tan and by their approximate expressions and neglect during multiplication terms of higher order than h). Using elementary formulae for the hyperbolic functions, we findd:

We now obtain the approximate formula for the conformal mapping (17). For this purpose, we replace on the right hand side of (17) arctanh z by two terms of its Taylor expansion with centre at the point z0 = tanh p(z - a)/2:

Substituting on the right hand side the values z and z0, we find finally

7) Mapping of eccentric circular ring onto concentric ring. We consider first the case when each neighbourhood of the ring lies in the outside of the other (Fig. 61). We construct on the common tangent to these circles, as on the diameter, the semi-circle G ; it intersects the line of the centres of the circumference of the ring at two points a and b which are symmetric simultaneously with respect to the two circles C1 and C2, because there pass through a and b the line of the centres and the line G, orthogonal to both circles. By the property of the bi-linear function

the circles C1 and C2, become the to circles C*1 and C*2, relative to which the points w = 0 and w = , corresponding to the points z = a and z = b, are symmetric. Consequently, the point w = 0 is the common centre of C*1 and C*2. For this, the eccentric ring between C1 and C2 for this goes to the concentric circular ring between C*1 and C*2, Fig. 61 likewise shows the correspondence of the lines for this transformation; the net in the z -plane is a part of the net in Fig. 53.

The additional mapping

where q ranges from - to +, gives the ring obtained onto the strip. This fact does not contradict the statement in 28. regarding the impossibility of mapping doubly-connected domains onto simply-connected ones, because the mapping function (20) is multi-valued. And what is more, the function (20) generates a single-sheeted mapping onto the strip domain on its Riemannian surface, lying on the ring, and , obviously, this domain is simply-connected.

The case when one circle of the ring lies inside the other leads to a study with the aid of an additional linear mapping z=1/(z-c), where c is an arbitrary point, lying between the circles.

34. Mapping of sickles. We will call a circular sickle a domain bounded by two arcs of circles in the entire plane (i.e., in particular, also straight lines). The examples which we will consider here have an important role in applications as well as the further development of the theory.

1) Mapping of the outside of an arc onto the outside of a circle. (This is the degenerate case when the two arcs, bounded by a sickle, coincide). We will assume that the ends AB in the x-plane lie at the points a and that the circles in the w-plane passes through those points. Moreover, we will assume that the centre of the arc lies at the point z = hi, the centre of the circle at the point w = hi so that the tangent to the arc at the point a forms with the negative x-axis the angle a = 2arctan h/a, and the tangent to the circle at the point w = a - angle b = p/2 - a/2) with the positive u-axis (Fig. 62).

We map with the aid of the bi-linear function

the outside of the arc AB onto the outside of some ray. Since |dz1/dz|z=a > 0, the angle of inclination of this ray to the negative axis is likewise a. Moreover, we will seek the mapping onto the outside of the obtained circle in the w-plane. We employ for this purpose again the bi-linear mapping

which maps the circle in the half-plane as well as its neighbourhood into some line. Since |dw1/dw|w=a > 0, the inclination of this line to the positive axis is b. Consequently, the mapping

takes our circle along the outside of the ray, forming with the positive axis the angle 2b = p - a. Thus, this ray coincides with the one obtained for Mapping (1); eliminating z1 between (1) and (2), we obtain the required mapping

We find from this equation

For the mapping under consideration, any circle C ', touching the circle C at the point a, is mapped into a closed curve enveloping the arc AB and having at the point B (z = a) a point of return; this curve resembles the profile of an aerofoil (Fig. 63).

The function (3) realizes the conformal mapping of the outside of this curve onto the outside of the circle, bounded by the circle C '.

The method of obtaining classes of aerofoils, proposed by N.E.Joukowski, is based on this observation; it is especially simple for computations (Joukowski profile).

The shape of the Joukowski profile depends on three parameters: a - characterizing its width, h - its curvature and d the distance between the circles C and C ' (Fig. 63).

2) Mapping of half-plane with deformed segment onto half-plane. The function z1 = z/(a - z) maps the given domain (Fig. 64) onto the sector a < arg z < p. Consequently, the function

maps this domain onto the upper half-plane. We give still the normalization w()=, w'( ) = 1; since for this mapping the point z = goes to the point z2 = -1, one adds the linear transformation

where the constant k is found from the second normalization condition. In fact,

whence, expanding the power inside the brackets by the binomial formula, we find for large |z|

Consequently,

Finally, we have

where C is an arbitrary real constant. In order to obtain the principal part of Mapping (4) for small a and a, we use the first terms of the Taylor series. We have

whence

and, multiplying this by

we find

*Here and hereafter multi-term denotes terms of 4-th order smallness.

We will now compute the area s of the removed section. We have: s = ar - (a/2)r cos a , where r = a/2sin a is the radius of the circle. Neglecting small quantities of higher order, we obtain: s = (a/4)(a/sin a - cotan a) aa/6. Thus, we may finally write (5) in the form

Resorting to a parallel displacement, we obtain the somewhat more general result: The function

generates a conformal mapping onto the half-plane v > 0 of the half-plane y > 0 with removed area s, bounded by the segment (b, b + a) and another circle of smaller curvature.

3) Mapping of a circle with removed circular segment onto a circle. Let the removed sickle be close to eia and the area s of this sickle be small (Fig. 65). We execute the two bi-linear mappings

of the unit circles of the z- and w-planes onto the upper half-planes z and w.

The sickle s becomes the sickle

adjacent to the point z = 0. By (6), we then find:

or, reverting to the variables z and w,

(we disregard everywhere terms of smaller order than s). For this mapping, the point z = 0 becomes the point w 0=s eia/8p; executing the additional bi-linear mapping

of the circle onto itself, so that w 0 goes to 0, we obtain

After substituting for w its approximate value (8) and simple transformations (in which we again neglect small order terms above s), we obtain finally

(we write again w instead of w 1). Mapping (9) establishes the following correspondence of the points z = eij and w=eiq:

or ( if one takes imaginary parts and neglects small higher order terms)

We find for the modulus of the derivative of the mapping on the boundary

and for the derivative at the origin

4) Mapping of a strip with removed sickle onto a strip. Let there be removed from a strip 0 < y < 1 the segment s, bounded by the segment (0, a) of the real axis and the arc of a circle of small curvature (Fig. 66). Performing the supplementary mappings z = ep z, w = ep w of the strip onto the upper half-plane and applying (7),

where s* = |dz/dz|z=0s = ps is the area of the segment*, we find the required mapping

(we neglect everywhere small order terms of s). Shifting yet the w-plane by s/2, we find finally

* We assume in (7) that b = 1 and const. = s*/p.

By an additional shift of the z-plane, we obtain the general result: The function

maps conformally the strip 0<y<1 with the removed circular section s, located on the segment (b,b+a) onto the strip 0<v<1. It establishes the following correspondence of the points y = 0, y = 1 and the lines v = 0, v = 1:

5) Formula for the extension for mapping of a sickle onto a strip. Let the sickle D be bounded by the arcs C1 and C2 of circles intersecting at the points a and let -ih1 and -ih2 be the points of intersection of these arcs with the imaginary axis. Using (5) of 1) in 33., it is simple to construct the conformal mapping of the sickle D on to the strip 0 < v < h:

where l = 2arctan hk/a, k = 1, 2*. Differentiating (16), we find

* It is sufficient for obtaining (16) to note that (5) of 33. , in which z is replaced by z/a, yields the mapping of the given sickle onto the horizontal strip the boundary of which passes through the points

(cf. (11) of 9.). The width of this strip is h = (H/p)(l1 - l2), whence one finds H. There remains to shift the strip so that its lower shore coincides with the real axis; we thus arrive at (16).

One can obtain from (17) an approximate formula, which is important for applications. Let z0 be the point C2, n the segment of the normal to the circle C2 at this point which is inside the sickle D, the angle between the tangents to C1 and C2 at the ends of the segment n and k1 , k2 the curvatures of C1 and C2 (Fig. 67). We assume that h and together with it k1 - k2 and n are infinitely small of first order and the curvatures k1 and k2 are bounded. Then, we may assert that

where one may represent h in the form of a homogeneous third degree polynomial in n, , k1 - k2 with bounded coefficients.

In order to derive (18) from (17), one must express in the latter a, l1, l2, z0 in terms of n, , k1, k2 and then expand |f '(z0)| in powers of p, , k1- k2 to second order inclusively. However, the actual realization of this approach leads to most complicated manipulations. Hence we will obtain (18) for the particular case when the circle C2 coincides with the Ox-axis, i.e., when k2 = 0 (Fig. 68). In this case, (17) yields

where l1= 2 arctan ly/a is half of the angle of the arc C1 (Fig. 68). On the other hand, we have l1= arcsin ak1+ ak1/6 + , x = (sin )/k1, n - cos /k1 - cosl1/k1, whence

Moreover, we have exact to fourth order:

and

Dividing the first of these equations by the second, we obtain, including second order quantities ,

Using to the same order of accuracy the relation

we obtain finally

which coincides with (18) for k2 = 0.

We go for the transition to the general formula when k2 0 to the auxiliary z-plane and in it the sickle D*, bounded by the segment of the real axis C2* and another circle C1* with curvature k1*; we denote by n* the segment of the imaginary axis, included in the sickle D* and by the angle, formed by the arc C1* and the perpendicular to the imaginary axis at the point of their intersection (Fig. 69). We map conformally the lower half-plane z onto the outside of the circle C2 with curvature k2:

Then, the sickle D* becomes the sickle D, bounded by the arc C2 and the other circle C1 with a certain curvature k1, the i,imaginary axis becomes the real axis and the segment n* the segment n of the real axis, where

and the circle C1 forms with the perpendicular to the real axis at the point z1 of their intersection the angle (Fig. 69).

We find now the link between k1 and the other parameters. For this purpose, we denote by ds the length element of the arc C1, by a(s) the angle formed by the tangent to C1 with the x-axis (s is measured from the point z1 so that a(0) = + p/2) and by ds* the element of the arc C1*, corresponding to ds. We have k1 = da/ds. However, the infinitesimal increment of the angle a may be given the form

where the second term darg dz/dz = Im d ln dz/dz denotes the increment of arg dz/dz on the arc dz = eiqds* of the circle C1*, joining to the point z1 = -n*i. By (21), the increment equals

and we find for the curvature

because by the same formula (21)

Hence also by the second formula (21), by which 1/(1 - n) = 1 + k1n/2, we find

Let the function w = f(z) realize the conformal mapping of the sickle, bounded by the arcs C1 and C2 onto the strip of width h. By (21), we have at the point z0 = 1/k2 , corresponding to the point z = 0,

The quantity |dw/dz|z = 0 may be calculated with (20) by replacing in it n and k1 by n* and k1*. Using the found values (22) and (23) for n* and k1*, we obtain to terms of second order inclusively

Since, by the conditions that the difference k1 - k2 is small, the same degree of accuracy has k1k2 n = k2 n and we obtain the required formula

2.3 Symmetry principle and mapping of polygons

We will consider now the methods of great importance for the actual construction of conformal mappings. The first of these methods operates on the so-called symmetry principle, which was formulated by Riemann and established by G. Schwarz. This method , as will be shown in 36., allows to construct in certain cases the solution of the problem of finding conformal mappings.

The second method is especially important for applications, since it yields the possibility of writing (it is true, generally speaking, only in the form of an integral ) the function which realizes the mapping of the upper half-plane onto an arbitrary domain, bounded by a polygon.

35. Symmetry Principle It states in one particular case a simple sufficient condition for the existence of the analytic continuation function, realizing a conformal mapping:

Theorem 1 (B. Riemann, G. Schwarz) Let the boundary of the domain D1 contain an arc of the circle C and the function w = f1(z) realize a conformal mapping of this domain onto the domain D1* such that the arc C becomes the segment C* of the boundary D1* , which is also an arc of a circle. Under these conditions, f1(z) permits the analytic continuation f2(z) through the arc C into the domain D2, symmetric with D1 with respect to C, where the function w=f2(z) realizes the conformal mapping of D2 onto the domain D2*, symmetric with D1* relative to C*, and the function

realizes the conformal mapping * of the domain D1 + C + D2 onto the domain D1* + C* + D2*.

For single-sheetedness of this mapping, one must demand that the domains D1* and D2* do not intersect.

For the proof, we perform the bi-linear mappings

transforming C and C* into the segments G and G ' of the real axes of the z- and w-planes; let the domains D1 and D1* thus become the domains D1 and D1*, and the function w = f1(z) the function w = l*f1l-1(z) = j(z), realizing the conformal mapping of D1 and D1* (Fig. 70)*. We denote by D2 the domain, symmetric with respect to D1 relative to G.

* Euler has formulated the symmetry principle for this case in 1777.

We now construct in D2 the function

and show that it is by analytic continuation the function f1(z). To start with, the function j 2(z) is analytic in D2. In fact, we have for any points z and z + Dz of D2:

where and + are points in D1. By the strength of the analyticity of j1(z) in D1, the right hand side has a limit for
0, whence there exists also the derivative

at any point of the domain D2, i.e., j 2(z) is analytic in D2. By its design, there exist the boundary values of j 2(z) on the segment G:

because, by the boundary correspondence principle for conformal mappings (29.) there exists

Relation (3) assumes the form

but, since the value of j 1(x) is real (G*, by assumption, is a segment of the real axis), then on the segment G

Hence, by the principle of continuous continuation (25.), j1(z) is the analytic continuation through G. of j1(z)

It also follows from the construction of the function j 2(z) that it realizes the conformal mapping of D2 onto D2*, symmetric to Dq with respect to G*. However, the function j(z), constructed from j 1(z) and its analytic continuation j 2(z):

realizes the conformal mapping of D1 + G + D2 onto D1* + G* + D2*.

We return now to the old variables z and w with the aid of the substitutions, inverse to (1). By the strength of the properties of the bi-linear transformations, we obtain in the domain D2, symmetric with respect to D1 relative to the arc C, the function f2(z), analytically continued by f1(z) through the arc C and realizing the conformal mapping of D2 onto D2*, symmetric with D1* with respect to the arc C*, and the theorem has been proved.

As an example of the application of the symmetry principle, we will prove the uniqueness theorem of conformal mapping for the given correspondence of three boundary points, about which we have spoken in 29.

Theorem 2 There exists one and only one conformal mapping w = f(z) of the domain D onto the domain D*, which transfers three boundary points zk of D into three boundary points wk of the domain D*. The points zk and wk are arbitrary, but in the same order following the passage around the borders of the domains.

We consider first the case when D and D* are unit circles. By (2) of 32., one may construct the bi-linear mapping of the circle |z| < 1onto the circle |w| < 1 with the given normalization f(zk) = wk. We will prove the uniqueness of this mapping. Let w = g(z), g(zk) = wk be another mapping of the circle |z| < 1 onto the circle |w| < 1. The function g(z) satisfies the conditions of the symmetry principle and, consequently, continues analytically into the domain, symmetric to the circle |z| < 1 with respect to the circle |z| = 1, i.e., in the outside of the circle |z| > 1. Together with its continuation, w = g(z) realizes a single-sheeted mapping of the entire z-plane and, by Theorem 1 of 31., is a bi-linear function. However, then it may be asserted that g(z)f(z), because, by the bi-linear mapping, proved in 32., it is completely determined by stating the correspondence of the three points.

The general case is readily reduced to the one considered. In fact, let z = j(z), zk = j(zk) and w = y(z), wk = y(zk)be any conformal mappings of the domains D and D* onto the unit circles |z| < 1 and |w| < 1 and w = F(z), F(zk) = wk the mapping of the circle |z| < 1 onto the circle |w| < 1 (its existence and uniqueness were proved above). The mapping of the domain D onto the domain D* with given normalization will obviously be

where y-1 is the inverse mapping of y. The existence of the second mapping w = g(z) of D onto D* with the same normalization would reduce to the existence of the second mapping

of the circle |z| < 1 onto the circle |w| < 1 with normalization G(zk) = wk, which contradicts the statement above, and the theorem has been proved.

We will still consider an application of the principle to the problem of the existence of conformal mappings of multi-connected domains. By the basic theorem of 28., any two simply-connected domains may be transformed singe-sheetedly and conformally onto each other. On the other hand, we have seen that it is impossible to also map a simply-connected domain onto a multi-connected one. There arises the question whether it is possible to map one onto another domain of the same order of connectivity. It turns out that also the answer to this question, generally speaking, is negative. In fact, even in the simplest case of concentric rings there holds

Theorem 3 For the existence of a conformal mapping w = f(z) of the ring r1 < |z| < r2 onto the ring r1 < |z| < r2 , it is necessary and sufficient to satisfy the condition

In order to prove the necessity of Condition (5), we note that the function f(z) satisfies the conditions of the symmetry principle and on the basis of this principle continues analytically in the domains

symmetric with the ring r1 < |z| < r2 with respect to to the circles |z| = r1 and |z| = r2, respectively. The ring under consideration r1 /r2 < |z| < r2/r1 is mapped by f(z) (together with its continuation) onto the ring r1 /r2 < |z| < r2/r1. Hence we apply again to the function f(z) the symmetry principle and may continue it into the ring

Executing such a continuation unlimitedly, we find that f(z) achieves a single-sheeted mapping of the domain 0 < |z| < onto the domain 0 < |w| < , where either

depending on whether the circle |z| = r1 corresponds to the circle |w| = r1 or |w| = r2. Hence we conclude that f(z) is a bi-linear function of one of the two forms

where a is a complex constant. In both cases, obviously, (5) is fulfilled. The sufficiency of (5) follows from the fact that, where it is fulfilled, the rings are similar and may be mapped onto each other by simple stretching.

In addition to the proved theorem, we state that any doubly-connected domain may be all the same mapped onto some ring r1 < |w| < r2, where for a given radius r1 the radius r2 is determined uniquely for a given domain. In the same manner, an arbitrary n-ply-connected domain may be mapped onto some domain, obtained by omission from it of n circles. The proof of these results can be found in M.V.Keldysh's paper [7] or R.Courant's book [5].

In conclusion, we present a generalization of the symmetry principle to the case when the boundary of the mapped domains contains analytic arcs. We call an arc C analytic, if it may be given by the parametric equations

in which x(t) and y(t) are analytic functions of the real variable t on an interval (a, b), i.e., functions which can be expanded in the neighbourhood of every point t0 of this interval in series of power of t - t0. For this, it is assumed that nowhere in the interval x'(t) and y'(t) vanish simultaneously (i.e., there are no special points of C). The curve C may also be closed if x(a) = x(b), y(a) = y(b).

Here applies the so-called Principal of Analytic Continuation

Theorem 3 (G. Schwarz) Let the function w = f(x) yield a conformal mapping of the domain D, the boundary of which contains the analytic arc C, onto some domain D*, where there corresponds to the arc C an analytic arc C* of the boundary of the domain C*. Under those conditions, the function w = f(x) may be analytically continued through the arc C.

In fact, let the point z0 be an arbitrary point on the arc C, z = z(t) = x(t) + iy(t) be the equation of this curve and z0 = z(t0) . Since, by assumption, C is an analytic arc, then in some neighbourhood |t - t0| < d the function z = z(t) may be expanded in a series of powers of t - t0; by Abel's Theorem, this will converge also for complex values of t, which we will denote by z, in the circle |z - t0| < d. Consequently, in this circle, the analytic function z = z(z) is defined. Since z'(t0) 0, then, restricting in the case of need to d, one may assume that the mapping z = z(z) is conformal. By its design, z = z(z) maps the diameter of the circle |z - t0| < d onto some segment of the circle C, containing the point z0; we will assume that the upper semi-circle is mapped for this inside the domain D, the lower one into its outside.

The function w = f(z) yields the conformal mapping

(w = w(w) is the inverse function of w = w(w)) of the upper semi-circle |z - t0| < d onto some part of the upper semi-circle |w-t0| < d1 , where the diameter of the first semi-circle becomes part of the diameter of the second circle. By the symmetry principle of Theorem 1, the function j(z) admits analytic continuation into the lower semi-circle and (together with its continuation) yields the conformal mapping of the entire circle |z - t0| < d onto some part of the circle |w - t0| < d1 , containing a segment of the diameter.

The constructed continuation w = j(z) yields the analytic continuation of f(z) through the segment of the curve C. In fact, in a part of the outside of the domain D, the corresponding lower semi-circle |z - t0| < d, defined by the analytic function w=w[j(z(z))] (z(z) is the function, inverse to z(z)) the boundary value of which on the segment of the arc C coincides with the boundary values of f(z). By the principle of continuous continuation, this function is the analytic continuation of the function f(z). Since z0 is an arbitrary point of the curve C, it may be asserted that f(z) is the analytic continuation through the entire arc C, and the theorem has been proved.

In particular, if the entire boundaries of C and C* of the given domains are analytic curves, then f(z) is analytically continued through the entire boundary of the domain D (and, consequently, is analytic in the closed domain ).

The next section presents several examples of the application of the symmetry principle in the execution of conformal mappings.

36. Examples 1) Mapping of the outside of a cross onto the outside of the unit circle (Fig. 71). We construct the auxiliary (dotted) cut FAB along the imaginary axis and consider in the right half of the figure the mapping z1 = z; it maps this half onto the z1-plane with removed piece from A (-) do D(a) along the real axis. Moreover, we apply the mapping

which maps the region obtained onto the right half-plane. The auxiliary cut then goes to the segment of the imaginary axis containing from F(-fi), where f = (a + c), to the point B(gi), where g = (a + b) (Fig. 71).

The function z2 = (z - a) satisfies the conditions of the symmetry principle, which yields the analytic continuation through FAB into the left half-plane and together with its analytic continuation , which we have denoted by again by z2 = (z - a) yields a mapping of the outside of the given cross onto the outside of the segment BF of the imaginary axis of the z2-plane.

There remains to map the last domain onto the outside of the unit circle. For this purpose, we apply the linear mapping

mapping the outside of BF onto the outside of the unit circle, and then the inverse Joukowsky mapping (cf. 7.):

In particular, for b = c = a, we find

whence

(cf. 30., Example 3).

2) Mapping of the outside of the unit circle with the excluded segment 1 |z| 1 + a, arg z = 2kp/n (k = 0, 1, , n-1) onto the outside of the unit circle (Fig. 72). We insert auxiliary cuts from the points B1 and B2 to infinity by continuation of the radii of the circle and construct the conformal mapping of the obtained sector onto the same sector, but in sucha manner that the points B1 and B2 go to infinity by continuation of the radii of the circle and construct the conformal mapping of the segment obtained onto such a sector, but in such a manner that the points B1 and B2 move to A1 and A2 . This may be realized by the following example: With the aid of the transformation z1 = zn/2 , we map the sector onto the upper half-plane with omitted semi-circle and then with the aid of the Joukowsky transformation z2 = (z1 + 1/z1) onto the upper upper half-plane. Then

B1 and B2 go for this to the points

Moreover, we compress the half-plane z3 = z2/(1 + a1) and apply the inverse Joukowsky mapping z4 = z3 + (z3 - 1). As a result, we obtain again the upper half-plane with removed unit semi-circle, but the points B1 and B2 go now to the points 1. There remains to apply the mapping w = z4 2/n, in order to obtain the lower mapping of the sector onto the sector:

The function (5) satisfies the conditions of the symmetry principle; applying this principle, we find that this function continues through the ray B2 C and together with its continuation yields the mapping of the set of the first and second sectors of the z-plane onto the set of the first and second sectors of the w-plane (Fig. 72). The continuation obtained again continues through the ray B2C and the new continuation maps the third sector of the z-plane onto the third sector of the w-plane (so that the point B3 falls onto the circle). Repeating this argument, we find that the function (3) together with its analytic continuations yields the required mapping.

Fig. 72 shows parts of the images of the circles |w| = r for this mapping for n = 8 and a = 0.25; it is seen that for r = 1.8 and larger infusion of excluded segments there is effectively no difference - practically, these images do not differ from circles.

For a = 0, we have also a1 = 0, whence (5) becomes w z. We will now find the principal part of (5) for small a. We find in correspondence with (4): a1 = na/8.. Neglecting small terms of order higher than a, we obtain from (5) the approximate formula for our mapping:

.

or, finally,

(compare with (10) of Example 4 in 30.). Formula (6) applies for points which are not too close to the nth-degree roots of 1.

3) Mapping of upper half-plane with excluded sections 0 y h, x = ka (k = 0, 1, 2, ) of the upper half-plane. We introduce the additional cuts A-1C and A0C from the ends of the cuts to infinity (broken lines in Fig. 73) and map the resulting half-strips CB-1B0C onto such a half-strip, however in such a manner that the points A-1 and A0 become the vertices of this half-strip. For this purpose, we map first our half-strip onto the half-plane: z1 = cos z/a (cf. 9.) and repeat like the last: z2= z1/cosh ph/a (so that the points A-1 and A0 become the points z2 = 1) and use the mapping inverse to the first:
w = (a/p) arccos z2 . Thus, we obtain the required mapping of a half-strip onto a half-strip:

Applying to this function an unlimited number of times the symmetry principle, we find that it yields the required mapping pattern of Fig. 53 on the half-plane.

Fig 73 displays the images of the lines u = const. and v = const. for the mapping under consideration for h = 0.5 and a = 2; it is seen that for v = 2 and a large effectce of the excluded cuts is practically not noticed - in effect, the prototypes under consideration do not differ from straight lines.

4) Mapping of the plane with the removed segment -a x y = hH (k = 0, 1, 2, ) onto the plane with excluded segments of the real axis (Fig. 74, both domains infinitely connected).

We produce additional (broken lines) cuts along the imaginary axis and map one of the two mapping domains, for example the right one, onto the upper half-plane. For this purpose, we turn the z-plane by 90 and use the results of the preceding problem: The function

yields the conformal mapping of the right half of the domain onto the upper half-plane. Then, the point A0 (z = a) becomes the point w = arccos 1 = 0, the point B-1 (z = 0) the point

the point C-1 (z = -iH) the point

the point A-1 becomes -p, B becomes p + b and, in general, the points Ak become the points w = -kp, but he segments BkCk become the segments [-(k + 1)p + b, -kp - b] (k = 0, 1, 2, ). According to the symmetry principle, we may continue the function (8) through the set of cuts BkCk and discover then that this function together with its continuations maps conformally the given domain onto the w-plane with the cuts (kp - b, kp + b) (k = 0, 1, 2, ) and the problem has been solved.

We will show how to obtain a single-valued, but not single-sheeted mapping of a given (infinitely connected) domain onto the outside of the unit circle (i.e., onto a simply-connected domain). First of all, we map the introduced supplementary cuts along the real axis and line y = H ( the broken lines in Fig. 74) and the strip obtained onto the z1-plane with cut real semi-axis: z1=e2p z/H; the given segments A'0A0 and A'1A1 then become the segment (e-2pa/H; e2p a/H) on the upper and lower shores of the cut. Moreover, by the linear transformation z = (1/sinh(2pa/H){z1 - cosh(2pa/H)}, we move these segments by the single and inverse Joukowsky transformations w = z + (z2-1) into the upper and lower halves of the unit circle. Then, there correspond to the rays EA0' and EA1' in thew-plane the segments (-cotanh pa/h, -1)* and the rays A0F and A1F to the rays (1, ) on the upper and lower shores of the cut along the real axis.By the symmetry principle, the obtained function

admits analytic continuation through the segments EA0' and A0F, where this continuation will map the strip, symmetric to the first one, onto the same domain (the function (9) has the imaginary period Hi). Applying such a continuation an unlimited number of times, we find that (9) yields a single-valued mapping of the given domain onto the outside of the unit circle. However, this mapping is not single-sheeted, because the inverse to (9) is infinitely-valued.

* In fact, the point E (z = - ) becomes, respectively, the point z = 0, z = coth 2a (a = pa/H) and w = -cotanh (a = pa/H) and w=coth 2a - (cotanh 2u-1) = - cotanh a (in the formula of the inverse Joukowsky transformation, one must place the minus sign in front of the root ).

If we choose for the analytic continuation function all new and new examples of the sheets of the w-plane, then in the result the continuation obtains an infinitely-sheeted Riemann surface, lying on the outside of the unit circle in the w-plane. At the points E (w = -coth pa/H) and w = , it has logarithmic branch points. This surface is obtained, if one removes from the Riemann surface the functions, inverse to the function (9), the part lying inside the cylinder projected into the unit circle |w|=1. The function (9) yields a single-sheeted mapping of the given domain onto this surface.

5) Mapping of a domain, bounded by second order curves.

a) Parabola. Let the origin of the co-ordinates initially lie at the focus of the parabola y =2p(x + p/2) (Fig. 75). By the function w = z, the outside of this parabola is mapped onto the half-plane Im w > p/2. In fact, setting z=x+iy, w=u+iv, we obtain

whence one sees that the lines v = c become the parabolas y = 4c(x + c); for c = p/2, we obtain the given parabola. Thus, the function

yields the conformal mapping of the outside of the parabola onto the upper half-plane. Inside the parabola, the function (10) has a branch point.

In order to obtain the mapping of the inside of the parabola, we place the cut along the ray BFG (Fig, 75) and note that the upper half of the parabola is mapped with the aid of the function z1 = z onto the half-strip 0 < y1 < p/2, 0 < x1 < . With the aid of the function .

this half strip is mapped onto the upper half-plane, where the cut BFG corresponds to the ray -1 < x < . Applying the symmetry principle, and then the transformation w = i(1 + z2), we obtain the known mapping of the inside of the parabola. onto the upper half-plane

b) Hyperbola In order to find the conformal mapping onto the upper half-plane of the domain, included between the branches of the hyperbola

(Fig. 76), we make the cut BD along the real axis and note that the function

maps the upper half-plane of the given domain onto the sector

(cf. 7.) However, by the symmetry principle, this function yields a conformal mapping of the entire given domain onto all of the sector

Thus, the function

yields the mapping of the domain, included between the branches of the hyperbola, onto the upper half-plane.

In order to obtain the mapping of the inside of the right branch of the hyperbola, we makes a cut along the ray DFG and notes that the function

yields the conformal mapping of the upper half domain onto the sector

The function

maps this sector onto the upper half-plane, where there corresponds to the ray DEG the ray (-1, ) of the real axis. Applying the symmetry principle and then the supplementary mapping w = i(1 + z2), we obtain the required mapping of the inside of the right branches of the hyperbola onto the upper half-plane

c) Ellipse The conformal mapping of the outside of the ellipse

onto the outside of the unit circle yields the function

where c = (a + b) (cf. Fig. 77).

In order to obtain the mapping of the outside of the ellipse, we make a cut along its larger axis and employ the function

We then obtain the mapping of the upper half of the ellipse onto the upper half ring 1 < |z1| < (a + b)/c, Im z1 > 0, where the cut enters the segment AF1, F2C of the real axis and the unit semi-circle. The function z2 = ln z1 maps.this half ring onto the rectangle 0 < Re z2 < d, 0 < Im z2 < p, where d = ln[(a + b)/2]. The symmetry principle cannot yet be applied, because this domain becomes a single segment. The mapping of the rectangle onto the plane cannot be obtained with the aid of a combination of elementary functions - is exists as a so-called elliptic function (cf. 39., Example 1) - whence also the mapping of the inside of the ellipse onto the half-plane is not described in terms of elementary functions.

37. Mapping of polygons. Prior to deriving the formulae for the mapping of the half-plane onto a polygon * , we will explain the problem of the behaviour of the conformal mapping at angular points of a domain. For the sake of simplicity, we assume that the boundary of the domain D in the neighbourhood of the angular point w0 consists of straight segments; we will denote the angle between these segments by ap, assuming that 0 < a 2 (Fig. 78). Let the function w = f(z) yield the conformal mapping of the upper half-plane onto the domain D, where the angular point w0 corresponds to the point z0 of the real axis.

* cf. 44. for another , more constructive derivation of this formula

In order to explain the character of the function f(z) in the neighbourhood of the point z0, we introduce the auxiliary variable w = (w - w0)1/a. The compound function

realizes the conformal mapping of a part of the neighbourhood of the point z0, belonging to the upper half-plane z, onto the neighbourhood of the point w = 0, belonging to one of the half-planes, whereas the segment of the real axis of the z-plane corresponds to segment of a line (Fig. 78). By the symmetry principle, the function w(z) admits analytical continuation in the complete neighbourhood of the point z0 and can, consequently, be represented by the Taylor series

This series does not contain a free term, because c0' = w'(z0) 0, since the function yields a conformal mapping. Turning with the aid of (1) to the function f(z), we find that it can be represented in the neigbourhood of the point z0 by

Since the expression in the curly brackets is non-zero for z = z0, then onr may separate in some neighbourhood of the point z0 a single-valued analytical branch of the function

Expanding this branch in a Taylor series, we obtain the final representation of the function f(z) in the neighbourhood of the point z0

Whence we see that for a > 1 the derivative f '(z0) = 0, for a < 1 f '(z0) = . For the inverse mapping z = j(w), conversely, j '(w) = 0 for a < 1 and j '(w0) = for a > 1.Hence it is seen from (2) that z0 for a 1, 2 is a branch point of the function f(z) .

We note that in a more general case, when the boundary of the domain D in the neighbourhood of the point w0 consists of smooth or even analytic arcs, intersecting in w0 at an angle ap, the above derivation, generally speaking, is untrue. In this case, the mapping function does not have the form (2): In the principal term, its expansion may exhibit a factor which tends to zero and infinity more slowly than any power of z - z0 .

For example, we consider in the upper semi-circle |z| < r, y 0 the function

where the branch of the logarithm is characterized by the condition 0 < arg z p. It is readily seen that it transforms, for sufficiently small r, the segment (0, r) of the x-axis into the segment (0, r ln 1/r) of the u-axis, the segment (-r, 0) into the ray g : u = x ln (1/-x), v = -x/p, and the semi-circle |z| = r, 0 j p into an arc, close to the semi-circle (Fig. 79). By the boundary correspondence principle, the function (3) is single-sheeted for small r and conformally maps the semi-circle onto the domain D, shown in Fig. 79.

The arc g smoothly joins the segment (0, r ln 1/r) at the point w = 0 so that the angle a = 1; nevertheless, the principal term of the expansion (r) is "spoiled" by the factor ln 1/z. An analogous effect is found for the function

We now proceed to the mapping of polygons. Let there be given in the w-plane the closed polygon A1A2 An without points of intersection which does not contain the point at infinity (we will remove this restriction in 38.)

According to Riemann's Theorem (28.), there exists a function w = f(z) which realizes a conformal mapping of the upper z-half-plane onto the inside D of this polygon. For the sake of definiteness, we give the correspondence of three points of the real axis (for example, a1, a2 and a3) and three points of the boundary of D (for example, the vertices A1, A2 and A3); then, by Theorem 2 of 35. , the function f(z) is uniquely determined. To start with, we assume that this function is known, in particular, that the points a4, , an on the x-axis are known, which become the points A4, , An of the polygon, and pose the problem of finding its analytic expression.

Since the function w = f(z) assumes on any segment (ak, ak+1) of the real axis values which lie on a straight segment Ak, Ak+1, then, by the symmetry principle, it is analytically continued through this segment into the lower half-plane. The analytic continuation of this function yields the conformal mapping of the lower half-plane onto the polygon D ', symmetric to the polygon D with respect to the segment Ak, Ak+1. This analytic continuation may again be continued through any segment (ak', ak'+1) into the upper z-half-plane, where a new analytic continuation will realize a conformal mapping of the upper z-half-plane onto the polygon D", symmetric with the polygon D' with respect to the segment Ak' Ak'+1,

We will assume that we have completed all possible analytic continuations of the form just described. As a result, we obtain, generally speaking, an infinitely-valued analytic function w = F(z) for which the initial function f(z) is in the upper half-plane one of the single-valued branches.

Let w = f*(z) and w = f**(z) be two arbitrary branches of the function F(z) in the upper half-plane. According to our design, these branches yield a conformal mapping of the upper half-plane onto two polygons D* and D**, which differ from each other by an even number symmetry respecting the sides . However, since any pair of symmetry with respect to two arbitrary lines reduces to a certain shift and rotation, then everywhere in the upper half-plane f**(z) = eia f*(z) + a, where a and a are constants. The same is also true for any of the branches of the function F(z) in the lower half-plane.

Moreover, the function

is analytic in the upper half-plane, because f '(z), as the derivative of a function yielding a conformal mapping, vanishes nowhere. This function g(z) remains single-valued for all possible analytic continuations of f(z) by virtue of the above remark relating to the branches of the function F(z) (we have

Thus, we may assert that g(z) is a single-valued function, analytic everywhere in the entire z-plane except at the points z = ak, corresponding to the vertices of the polygon. The analyticity of g(z) at infinity follows from the fact that z = becomes some point on a side of the polygon, and not at its vertex. .

In order to explain the character of the function g(z) at the point z = ak, we select any branch f(z) and employ (2). We find:

whence we readily obtain the Laurent expansion of g(z) in the neighbourhood of the point z = ak:

This expansion shows that the point z = ak is for g(z) a first order pole with residue ak - 1.

Thus, the function g(z) has in the entire plane only n special points. Computing from g(z) the sum of the principal parts of its expansion at these points, we find the function

which is regular in the entire plane and, consequently, constant (cf. the Cauchy-Liouville Theorem in the formulation of 24.). Since the function f(z) holds at the point z = , in the neighbourhood of this point

and

Consequently, g(), and that means also G( ), vanishes, whence

Integrating (5) along any path in the upper half-plane and then taking the exponential, we find

By a further integration, we obtain the required expression for f(z) and we have proved

Theorem 2 (Schwarz - Christoffel*, 1867 - 69) If the function w = f(z) yields the conformal mapping of the half-plane Im(z) > 0 onto the inside of a bounded polygon D with angle akp (0 < ak 2, k = 1, 2, , n) at the vertices, where the points ak of the real axis (- < a1 < a2< < an < are known) are the corresponding vertices of this polygon , then f(z) is represented by the integral

where z0, C and C1 are certain constants.

*Erwin Christoffel (1829 - 1900).

The Schwarz-Christoffel integral is obtained under the assumption that the points ak, corresponding to the vertices of the polygon, are known. However, in problems of conformal mapping, there are only given the vertices Ak of the polygon and the points ak remain unknown. According to what has been stated in 29., three of them (for example, a1, a2 and a3) may be given arbitrarily, while the remaining ones as well as the constants C and C ' must be determined from the conditions of the problem*. This circumstance creates a principal difficulty for the practical use of the Schwarz-Christoffel integral.

* The constant z0 may be fixed for good; for example, one may set z0 = 0, whence in the sequel we will not assume it to be an unknown parameter of (7).

Methods for the determination of the constants ak, C and C1 will be stated below in concrete examples. Principally, the possibility of finding them follows in essence from the proof of Theorem 2, given above. In fact, let there be given a polygon D.. On the basis of the theorem, we may assert that there exists a unique conformal mapping w = f(z) of the half-plane Im z > 0 onto the polygon D, mapping the three given points a1, a2 and a3 of the real axis into three vertices of D,.for example, into A1, A2 and A3 . By what has been proved above, the function for this will be given by (7) for a proper choice of the constants a4, an, C and C1. Thus, when three ak are given, the remaining constants are determined, and besides in a unique manner. We note yet that, according to (6), we have for z = x > an arg(x - ak) on the real axis of the z-plane and, consequently, arg f '(z) = arg C, and since the segment (an, a1) (containing z = ) becomes for the mapping w = f (z) the segment AnA1, then arg C equals the angle q which this segment forms with the u-axis (in Fig. 80, a = q - p). The constant C1 is determined by stating the position of one of the vertices. For the determination of the constants ak and C, one may employ the known lengths of the sides of the polygon

although we do not use this method in by far not all applications. In applications, one often employs approximate methods for the determination of the constants ak and C; the reader can find out about this from the book by P. F. Filchakov [10], the work of G,N. Polozhego [12] or the paper by N.P. Stenin in the collection [8].

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