38. Supplementary remarks We will solve a number of mapping problems, not considered in 37.

We will study several cases which have not been worked out in 37.

1) One of the vertices of a polygon is the image of the point at infinity. In order to reduce this case to one considered already, we perform the linear mapping z = -1/z + an' of the half-plane Im z > 0 onto the half-plane Im z > 0, moving the points a1, a2, , an = to the finite points a'1, a'2, , an' . Applying (7) of 37., we find

* If one of the points ak = 0, one must use z = -1/(z - a) + an' , where a differs from all the other ak.

We reduce the expression in each brackets to the common denominator and extract from each of them the factor an' - ak' . (k= 1, 2, , n - 1) and find

where ak = 1/(an' - ak') are some real constants and C is a complex constant (all omitted factors are included in it). Using elementary geometric assumptions relating to the sum of angles of an n-fold polygon, according to which

we obtain finally

Thus, if one of the vertices of the polygon D coincides with the point at infinity, then the factor belonging to this point does not occur in the Schwarz-Christoffel formula.

This circumstance is employed in practice to simplify the Schwarz-Christoffel integral (cf. 39. et seq.).

2) One or several vertices of the polygon lie at the point at infinity. Let the vertex Ak of the polygon D lie at the point at infinity. We choose on the rays Ak-1Ak and AkAk+1 arbitrarily points Ak' and Ak", join them by straight segments Ak' and Ak" and consider the (n + 1)-polygon D' obtained (Fig. 81). The function, mapping the half-plane onto the -polygon D' , according to above, is given by

where ak' and ak" are angles measured in fractions of p of the vertices Ak' and Ak" , ak' and ak" are the points of the x-axis, corresponding to these vertices.

Let the segment Ak' Ak" be moved to infinity, remaining parallel to itself; then the points ak' and ak" join at the single point ak, corresponding to the vertex Ak and in the limit the factors of (3), involving ak' and ak", become We will denote by akp, taken with a minus sign, the angle of intersection of the rays Ak-1Ak and Ak-1Ak at the finite point Ak*. We then find from the triangle Ak'Ak"Ak* that ak' +ak" - ak = 1, i.e., ak' +ak" - 2 = ak - 1, and (3) ( it assumes the ordinary form

However, this reasoning may also be applied in the case when several vertices of the polygon lie at infinity.

Thus, the formula of Schwarz-Christoffel remains also true for multiply-connected domains, in which one or several vertices lie at the point at infinity, if for this the angle between the two lines with the vertex at infinity is determined as the angle at the finite point of their intersection, taken with the minus sign (cf. 31.).

For our definition of the angle at infinity, there remains in force Relation (1) for the sum of the angles of the polygon. In fact, for an (n + 1)-polygon D' with finite angles, we have, on the basis of (1), that S ' + ak' +ak" = p - 1, where S ' denotes the sum of all the angles D', but the angle for the vertex Ak= (we retain the notion adopted above). Setting ak'+ak"=ak+1, we find (1) also for the polygon D.

3) Mapping of the outside of a polygon. This case differs from the one in 37. in that at some finite* point a of the upper half-plane, corresponding to the point at infinity of the polygon, the function f(z) has a first order pole (presence of an upper order pole would contradict single-sheetedness). It is shown, as in 37., that at this point g(z) will have a first order pole with residue -2. We will have the same also for the point of the lower half-plane, because is a first order pole for the analytic continuation of the function f(z). Thus, we will have for g(z) the expansion

* If the point at infinity of the polygon corresponds to the point at infinity of the z-plane, this point is a boundary point for the polygon and we have the case already treated in 2).

Hence we will have for the function, realizing a conformal mapping of the upper half-plane onto the outside of a polygon

Here, ak are the angles of the polygon measured in fractions of p, ak the points of the real axis corresponding to its vertices, a is the point of the upper half-plane, corresponding to the point at infinity of the polygon, z0, C, C1 are certain constants.

4) Mapping of the inside (outside) of the unit circle onto the inside (outside) of the polygon represented by the function

Here, ak are the internal (external) angles of the polygon, measured in fractions of p, |ak| =1 are the points of the unit circle corresponding to to its vertices, C and C1 are certain constants. For the mapping of the outside of the circle onto the outside of the polygon, it is assumed, in addition, that the points at infinity of the z- and w-planes correspond to each other.

For the mapping of the inside of the unit circle onto the outside of the polygon, we have

where the notation is the same as in (6) and it is assumed that the point at infinity of the polygon corresponds to the centre of the circle.

Formulae (6) and (7) reduce to the preceding one with the aid of a supplementary bi-linear mapping of the z-plane such as this is done at the start of 38. for the derivation of (2).

5) The inverse problem. Now, let there be given an arbitrary set of real numbers ak and ak, satisfying the conditions

and the arbitrary complex numbers C and C1. We construct with their help the Schwarz-Christoffel integral

It turns out that, under these conditions, the Schwarz-Christoffel function determines the function realizing a conformal mapping of the upper half-plane onto a certain polygon with the vertex angles akp.

In fact, the argument of the derivative of (8)

remains constant on any of the segments (ak, ak+1), k = 1, 2, , n - 1 of the real axis *, and the same derivative dw/dz does not vanish inside such a segment. Consequently, the function (8) maps mutually-single-valuedly the segment (ak, ak+1) onto some straight segment (Ak, Ak+1). The same is also true for the segment (an, a1) of the real axis, which contains z = . In fact, firstly, by the condition Sak = n -2, the segments ( an, ) and ( -, a1) are turned by this transformation by the same angle, secondly, the integral (8) goes to the point z = **; consequently, the function w tends to one and the same limit as z .

* We assume that arg (x - ak) is 0 or p depending of whether x > ak or x < ak , whence on every segment (ak, ak+1) all terms of the sum (9) are constant.

** The first statement follows since dw/dz equals arg C on the segment (an,) and on the segment (-, a1) arg C+(Sak-n)p =argC-2p; the second statement that the principal term of the integrand near the point z = has the form

Thus, the function (8) establishes a mutually-single-valued relationship between the real axis and the contour of some polygon A1, A2, An. Note that certain of the vertices of this polygon may lie at infinity - it will be the vertex Ak for which ak 0 (in fact, for approachment to the corresponding points ak , the function w , because the integral (8) diverges, since the order of infinity of the integrand 1). However, also in this case, we apply the boundary correspondence principle and may assert that this function (7) realizes the conformal mapping of the upper z - half-plane onto the inside of the polygon A1, A2, An. The angle for the vertex Ak of this polygon equals akp, because, by (9), arg dw/dz changes for the passage through each point ak in the direction from the left to the right by -p(ak - 1) = p -akp, whence the corresponding segment Ak-1 Ak, rotates by the angle p -akp counter-clockwise. This proves completely the assertion.

The statement remains valid also in the case when there does not apply the condition

In this case,only there appears an additional (n + 1)-the vertex of the polygon corresponding to the point z = . The reader will verify that this vertex will be finite, if

and moved to infinity if

6) Mapping of the outside of a "star". In conclusion, following L.M.Lakhtin, we find the general formula for the mapping of the outside of the circle |z| > 1 onto the outside of the "star:, formed by n straight segments Lk, extending from the origin of co-ordinates (Fig. 82). We denote the angle between Lk and Lk+1 by akp, the point of the circle, corresponding to the vertex of this angle and the end Lk, by ak and bk , respectively (k = 1, 2, , n; Ln+1 = Ln ).

We consider first the mapping of the upper half-plane z onto the given domain. Let ak' and bk' be the points of the real axis of the z-plane, corresponding to the vertices of the angles LkLk+1and the ends of the segments Lk, and a0 the point in the upper half-plane corresponding to w = . Obviously, the mapping function w = f(z) must have in the neighbourhood of the point z=ak' the form

and, in the neighbourhood of z=bk , the form

where jk(z) and yk(z) are regular in the neighbourhoods of the functions mentioned above. Finally, at the points a and , it must have first order poles and must be regular at the remaining points of the z-plane. Hence, as also above, we conclude that the logarithmic derivative of the mapping function must have first order poles with residues ak at the points z = ak', first order poles with residues -1 at the points z = a and z = , and must be regular at the points z = bk' and the remaining points of the z-plane. Thus,

and, consequently, after integration and raising it to the exponential power, we obtain

where C' is a certain constant.

Completing the supplementary mapping

of the upper half-plane onto the outside of the unit circle |z| < 1, we find the required mapping in the form

where C and ak (| ak| = 1) are certain constants*.

* We have

(ak" - constants),

Side by side with Lakhtin;s formula (10), one must employ the usual formula for the mapping of the outside of a circle onto the outside of a polygon, which in this case has the form

Joined consideration of (10) and (11) permits avoidance of tiresome integrations.

As an example, we consider the particular case n = 2, when the polygon is the outside of two segments, intersecting at the origin of co-ordinates by the angle a1 = a. We have a2 = 2 - a and (10) and (11) assume the forms

Equating the derivatives after simple transformations, we find

whence we find for the determination of the constants b1 and b2 the equation

For example, one may set b1 = 1, when (a - 1)(a2 - a1) = 1 - a1a2 and b2 = - .a1a2. Equation (12) contains four real parameters by choosing them, one may choose the correspondence of the points b1 , b2 and the ends L1 , L2.

We present next several examples of applications of the Schwarz-Christoffel formula to the mapping of polygons.

39. Examples 1) Mapping of the upper half-plane Im z > 0 onto the quadrangle A1A2A3A4 (Fig. 83) Consider to start with the mapping of the first quadrant of the z-plane onto the right half of OA1A2B of the given rectangle with the correspondence of points: O 0, A11, B . We denote the prototype of the point A2 by 1/k, where 0 < k < 1. We may consider the required mapping as the continuation of this mapping by the symmetry principle of the positive semi-axis y, whence we may assume that A3 1/k. Thus, the required mapping can be written in the form

(the constant C1 = 0, by the correspondence of the points O 0). For the determination of the constants C and k, we use the correspondence of the points A1 1:

and likewise of the points A2 1/k:

(we have subdivided the integral from 0 to 1/k into the two from 0 to 1 and from 1 to 1/k and employed (10), whence

We will assume as given the constant k (0 < k < 1) and the dimensions of the rectangles K and K' so that the constants C in (1) and (2) are 1:

Then the mapping of the half-plane onto our rectangle will be given by the function

This integral cannot be expressed in terms of elementary functions; it belongs to the class of the so-called elliptic functions, and its inverse (i.e., the function yielding the mapping of a rectangle onto the half-plane) to the elliptic functions of Jacobi. It has the special notation

and is called the elliptic sine. We will learn its details in 102.; we may consider the solution of the problem of mapping of an arbitrary rectangle onto a half-plane to have been solved (and not with fixed dimensions as here).

We will only note here an interesting property of the function sn z *: It turns out to be a meromorphic function with two periods the ratio of which is purely imaginary.

* We change the notation of the variables.

In order to prove this assertion, we denote our rectangle's width by I and its sides by I, II, II, IV, as is shown in Fig. 84. We continue the function w=sn z, initially defined in the rectangle (I), by the symmetry principle through the side I into the rectangles (2). This continuation realizes the mapping (2) onto the lower half-plane. Continuing this mapping through the side II' of the rectangle (2), we find that w = sn z yields the mapping of the rectangle (3) again onto the upper half-plane, etc. (the rectangles, shaded in Fig. 84, are mapped onto the upper half-plane , those not shaded onto the lower half-plane).

Thus, we continue the function w = sn z into the entire z-plane. For this, the function turns out to be single-valued, because, if for passage along any closed contour we meet again, for example, rectangle (I), then the new value of sn z will coincide with the old ones (they map (I) onto the half-plane with the same normalization as the earlier one). Moreover, this function is regular inside the rectangles and everywhere on their boundaries, except the point iK ' and all points , corresponding to it for the continuations (they are indicated in Fig. 63 by crosses), where it has poles, because these points go to w = during the conformal mapping. Thus, sn z is a meromorphic function.

Moreover, in Fig. 84, small black circles denote the arbitrary point z of the rectangle and all points which are obtained from it by an even number of continuations**. At all these points, having the form z + 4nK + 2 n'K;i, where n, n' = 0, 1, 2, , the functions sn assume the same values:

This property also indicates that sn z has the two periods t = 4K and t' = 2K'.

** By white circles are denoted the points at which our function assumes the value

It also follows from the continuation of sn z that this function is odd

2) Strip with a horizontal cut (Fig. 85). This represents a four-angle figure, the three vertices A1, A2, A3 which lie at infinity and the angles a1 = a2 = a3 = 0. We will assume that a4 = 0, a1 = 1, a2 = and we also take z0 = 0 (when we find immediately from the correspondence of the points a4 and A4 that C1 = 0). Consequently, there corresponds to the point A3 a certain point a3 = -a4 of the negative strip. The Schwarz-Christoffel integral assumes the form:

(the factor corresponding to the point A2 vanishes; 1) in 38.). We employ for. the determination of the constants C ' and a the following reasoning: When the point z circles the point a1 = 1 along half the circle c (i.e., when the vector 1-z=re rotates, changing it argument from 0 to -p ), then the corresponding point w must travel with the ray A4A1 on A1A2 and the increase of w must differ little from -ih1:

where O(r) is infinitely small as r 0. This reasoning is justified by the fact that the image of the semi-circle cr for small r differs little from the segment of the straight line, joining the rays A4A1 and A1A2 and perpendicular to it.

On the other hand, for such a small increment Dz, the increase of the second term in the curly brackets of (1) will likewise be small, because this term is continuous at the point z = 1. However, the increase of the first term ln (1 - z) = ln r + ij equals -ip, whence

Comparing the expressions obtained for Dw and going to the limit r 0, we find

Analogously, when the point z moves around the point a3 = - a along the circle z + a = reij (j changes from p to 0). the increment Dw = -C 'aip + O(r) must differ little from -ih2, whence

Finally, the function realizing the conformal mapping of the half-plane Im z > 0 onto the strip with the cut (Fig. 85) has the form

With the supplementary mapping of our strip witha cut onto the upper half-plane Im w > 0 with a cut inclined segment of length 1 (Fig. 86):

where H = h1 + h2 and, using (9), we find the mapping of the half-plane Im z = 0 onto this region:

(we denote w again by w). After simple transformations, we find:

The removed segment forms with the positive axis the angle ph1/H, which we denote by ap (Fig. 86); introducing the parameter a, we obtain finally the mapping of the upper half-plane Im z > 0 onto the upper half-plane Im w > 0 with the removed segment (0, eiap):

Fig. 86 shows the curves corresponding for this mapping to the straight lines Im z = const.

For a = 1/2, we obtain the old result (33., Example 2).

3) Polygon of Fig. 87 represents a quadrangle with two vertices at infinity. Having in mind an application of the symmetry principle, we restrict consideration to its upper half - the triangle A1A2A3 with the angles a1 = 0, a2 = -a, a3= 1 + a (Sak = 1). We call the points of the x-axis, corresponding to the vertices: a1 = 0, a2 = , a3 = -1; taking into consideration the correspondence of the points a3 and A3, we have

We employ for the determination of the constant C the fact that for the travel of the point z along the semi-circle cr: z = reij (j varies from p to 0) the function, determined by the last integral, receives the increment

(on the circle cr, the function (z + 1)a differs little from 1: (z + 1)a = O(r)). On the other hand, during this circuit, the corresponding point w travels with A1A3 to A1A2, consequently, Dw differs little from -hi. Thus, C = h/p, and the function yielding the conformal mapping of the upper half-plane onto the upper strip of the polygon of Fig. 87 has the form

Replacing here z by ez, we obtain the mapping of the strip 0 < y < p onto the upper half of the polygon

Since, according to our choice of corresponding points, there corresponds to the lower edge of the strip the central line of the polygon, then, by the symmetry principle, (12) yields the conformal mapping of the strip p < y < p onto the entire polygon.

For rational a, Integral (12) is expressed in terms of elementary function (it reduces to the integral of the binomial differential). For n = 1, we obtain already the known mapping (cf. Example 5 in 30.):

For a = 1/2, we have

4) We will find the conformal mapping of the strip p < Im z < p onto the plane with two cut rays(Fig. 88 0 < a < 1). Again, we apply the symmetry principle - the upper half of the region in the w-plane represents itself a triangle with two vertices at infinity and the angles a1 = a -1, a2 = -a, a3 = 2. In order to employ the Schwarz-Christoffel formula, we map the strip 0<y<p onto the half-plane z = ez. Taking into consideration the correspondence of the points, shown in Fig. 88, we take a1=0, a2 = -, when the point a3 falls on the negative semi-axis and we find a3 = -a, where a is as yet an undetermined positive number. The Schwarz-Christoffel formula then assumes the form

here C is a positive constant, because the ray A1A2 does not turn for the mapping and, consequently, arg C = 0 (cf. the note at the end of 37.), C1 is real, because substitution of positive values of z into (15) must yield a real w (cf. Fig. 88). In order to give (15) a simpler form, we set 1/a = - a/(a - 1), i.e., a = (1 - a)/a ; thus

The correspondence of the points z = -a and w = ieiap yields

whence, taking into consideration that C and C1 are real, we find

Substituting into (16) z = ez, we find the required mapping

Fig. 88 likewise shows the correspondence of the lines for this mapping.

5) The polygon in Fig. 89 (0 a 3/2 ) represents itself a quadrangle with two vertices at and and the angles a1 = a - 2, a2 = 2, a3 = - a, a4 = -b. The Schwarz-Christoffel integral has the form

(a 0, a 1). We employ for the determination of the constants Cand b:1) that the ray A2A3 becomes the positive semi-axis, whence arg C = 0, i.e., C is a positive constant; 2) the correspondence of the points z = - b and w = ai. Separating in (18), after substituting z = beip, w = al, the real and imaginary parts, we obtain the two equations

]

which also permit to find (although approximately) the unknown constants.

In particular, for a = 1/2, we find b = 3, C = 33a/32 and the mapping function becomes

For a = 1, we have instead of (18)

the constants are given by

6) The polygon in Fig. 90 represents a pentagon. We consider its right half - the quadrangle with the angles a1=1/2, a2=a4=0, a3=3/2; (Sak=2). With a1 = 0, a2 = 1, a3 = a, a4 = , the conformal mapping of the upper z-half-plane onto this quadrangle is yielded by the function

In order to determine the constants a and C, we consider the change of w as z moves along the semi-circles CR and cr with the centre z = 1 and infinitely large and small radius, respectively. There corresponds to the first circuit the move with the ray A1A4 to the ray A3A4 , whence Dw = H + O(1/R); on the other hand, for large |z|, the root in the integral is approximately equal to 1, whence

comparing these expressions, we find C = Hi/p. There corresponds to the second circuit the move of the ray A1A2 to A3A2, whence Dw = ih + O(r) and the Schwarz-Christoffel integral yields

comparison of these two expressions yields

Substituting (z - a)/z = 1/w reduces the integral to the form

when the integral is readily evaluated. Substituting the found value of a and C and integrating, we obtain

There corresponds to the auxiliary cut along the imaginary axis in the w-plane the cut along the negative semi-axis in the z-plane, whence, by the symmetry principle, the function obtained yields the mapping of the z-plane with removed positive semi-axis onto the entire given pentagon.

Setting z = z, we obtain the final mapping of the upper z-half-plane onto all of the pentagon:

7) The angles of the polygon in Fig. 91 are a1 = a3 = a5 = 0, a2 = a4 = 3/2. Let the points corresponding to the vertices be: a1 = -a, a2 = -1, a3 = -b, a4=0, a5 = , when the function mapping the upper half-plane onto this polygon has the form

Integrating along the infinitely large semi-circle with centre at the origin, we find C(-ip) = -ih3, whence C = h3/p . Integrating along the infinitely small semi-circles with centre at z = -a and z = -b, we find *

whence

The last two equations allow to find a and b; Integral (24) may be expressed in terms of elementary functions.

* It is not difficult to verify that one must take here the negative value of the root.

8) In conclusion, we present the example of the mapping of the circle |z| < 1 onto the inside of the polygon - the five-point star of Fig. 92. This domain represents a decagon, five angles of which are a = 5/7 and five b = 1/5.We will use (6) of 38.; in order to find the points of the circles, corresponding to the vertices of the star; we consider a tenth of it - the triangle A1B1O (Fig. 92). This triangle may be mapped onto the sector 0 < arg z < p/5, |z| < 1 so that the point A1 lies at 1 and B1at eip/5. By the symmetry principle, our mapping continues into the entire star, where the points Ak become ak = ei2p(k-1)/5. (the fifth roots of 1) and bk = eikp/5.(the fifth roots of -1), k = 1, 2, , 5. By the strength of the uniqueness, its mapping may be found from (6) of 38. which consequently has the form

(we have employed the obvious identities P(z - ak) = z5 - 1, P(z - bk) = z5 + 1 .

The constant C can be taken to be real, since it determines the dimension of OBk = R of the star; since the point z = -1 becomes an apex of the star and z = 0 its centre, then

after setting t = {(1 + x5)/(1 - x5)}, this integral becomes one expressible in term of the Euler G-function:

(cf. 90.) Thus,

40. Rounding of corners. In many practical problems, one has to consider that the actual angles of polygons under consideration are always rounded off. Wel present here approximate methods for making allowance for such rounding.

1) Rounding of an angle less than p. We will first find a function which realizes mapping of the upper z-half-plane onto the upper z-half-plane from which a small area has been removed, bounded by the segment (-1, 1) and the arc of a curve resting on this segment and touching the the real axis at its ends * (Fig. 93). We consider for this purpose the mapping

of the upper z-half-plane onto the upper z1-half-plane with half the unit circle removed and in the capacity of our curve we choose the half of an ellipse with semi-axes 1 and 1 + h, near the the semi-circle (Fig. 93). There remains now to find the mapping of the upper z1-half-plane with the removed half ellipse onto the z-half-plane. The last problem is solved in an elementary manner. By the similarity mapping z2 = z1/c, where

we move the foci of the ellipse to the points i; we then apply the mapping z2 = (z3 - 1/z3)/2 and obtain in the z3-plane instead of the ellipse the circle with radius

finally, the mapping z = (z3/r + r/z3)/2 yields the upper half-plane. We have now

or, taking the expressions for r and c into account,

* The function of 2) in 34. is not satisfactory, because its arc does not touch the axis.

With the aid of the supplementary linear transformations

we obtain the more general result: The function**

where b1 = b - a, b2 = b + a, yields the mapping of the upper half-plane Im z > 0 onto the upper half-plane Im z > 0 with removal of the small area, bounded by the segment (b - a, b + a) and the arc lying on this segment and touching it at its ends; the quantity h, proportional to he maximum ordinate value of the curve, is assumed to be small of higher order with respect to a (Fig. 94).

** Instead of we write again z and z..

Now, let the function w = f(z) yield the conformal mapping of the upper half-plane Im z > 0 onto some polygon D. where the point b corresponds to the vertex B of the angle of the polygon less than p. Joining the supplementary mapping z = gb(z) with the aid of function (2), we find the conformal mapping

of the upper z-half-plane onto the domain which is obtained from D by rounding the angle B in a sufficiently small neighbourhood of the width of this angle (Fig. 94). By applying the method of this example, one can round off all corners of D less than p.

2) Rounding of corners larger than p. Without limiting the generality, on may assume that the vertex A1 of the multi-polygon D, the corner of which we round, lies at the point w = 0, the side A1A2 goes along the positive semi-axis and a1 = 0, but the prototypes -a1, -a2 , , -an of the remaining vertices of D are negative (this can always be assured by supplementary bi-linear transformations of the plane). Under these assumptions, one may write the function, yielding the conformal mapping of the upper z-half-plane onto the multi-polygon D, with the aid of the Schwarz-Christoffel integral in the form

where

and C is a positive constant (arg C = 0 due to our choice of the segment A1A2).

In order to round the angle at the vertex A1, we consider instead of (4) the function

where b and g are constants subject to definition; we will assume to b be a small positive number (in each case b < an). According to 5) in 38., the function

yields the mapping of the half-plane Im z > 0 onto a polygon with sides, parallel to the sides of D, where the point z = -b corresponds to the vertex B", lying on the negative axis u (the angle for it equals a1p) and the remaining vertices A2",,An" correspond to the points a1, a2, , an (this polygon is shown by broken lines in Fig. 95).

We consider yet the function

which maps the polygon of Im z > 0 onto the polygon with the vertices A1', A2', , An' (it is shown in Fig. 95 by solid lines). For every fixed z, the vector w determined by (5) is obtained by addition of the vectors f1(z) and f2(z). By executing this addition, we verify the fact that when z describes the real axis, the point w will describe the closed path A1,A2,,AnBA1, which apart from BA1 consists of segments, parallel to the corresponding sides of the given polygon (solid lines in Fig. 95.

In order to obtain for this the parametric equation of the segment BA1, we introduce the positive parameter t = -z (0 < t < b). Equation (5) then yields

whence the tangent (tan) of the angle of inclination of the tangent to BA1 with the u-axis will be

We see from this expression that in the case of an angle larger than p (i.e., for 1 < a1 < 2), tan j will be equal to 0 at t = 0, corresponding to A1 and equal tan a1p at the point t = b, corresponding to B. Thus, with the angle larger than p, the arc BA1 really surrounds the vertex A1*.

* For a1 < 1, we have dv/du|t = 0 = tan a1p , dv/du|t = b = 0 and the arc A1B does not round the angle. You may achieve rounding in this case by taking instead of 5) the function

where b > 0; however, such a method is less convenient than the one given at the start of this section.

According to the boundary correspondence principle, the function (5) yields the conformal mapping of the half-plane Im z > 0 onto the domain bounded by the contours A1,A2,,AnBA1. By varying the constants C, b and g , we may obtain that the domain D will differ conveniently little from the given multi-angle domain D.

We will show by a simple example that it does so. Consider the polygon in Fig. 96, a particular case of the triangle of Example 3 of 39. We will assume that the points A1,A2, and A3 correspond to the points 0, 1 and of the real axis; then the Schwarz-Christoffel integral is written in the form

.

where C is a positive constant (on the segment (0,1), w must have positive values). In correspondence with what has been said above, we set instead

This function transfers the segment (0,1) to the positive semi-axis, and we require that for the passage through the point z = 1 it undergo the increment ih; whence, as in 39., we obtain

Moreover, we require the point z = -b correspond to the point B = r - ir so that for small b the arc BA1 will be close to the arc of a circle of radius r. After setting z = -t, this becomes

Separating in it real and imaginary parts and integrating, we arrive at the relations

The three relations (8) and (9) allow to find r, C and g as functions of the parameter b. We have for small b:

The mapping function then assumes the form (with accuracy to small higher order in b)

We have become familiar in this chapter with several problems of the theory of conformal mapping, relating to a range of the problems formulated at the beginning of 28. The reader will find in the following two chapters more examples of such problems. In Chapter III, conformal mappings will be encountered in connection with the solution of several boundary value problems of the theory of plane vector fields, closely related to applications. Chapter IV is devoted to the variational principles in the theory of conformal mappings; we consider there the behaviour of conformal mappings for changes of the boundaries of the mapped domains (Problem 3) as well as certain approximate formulae.

We will not touch the methods of approximate computations of conformal mappings (except the method of grids in 45., which may be used for this purpose). The reader may become familiar with the analytical methods of such computations through Chapter V of the book by L.B.Kantorovich and V.M. Krylov [9]. For many practical purposes, the preferable methods of computations, using physical analogues - the methods of such calculations with application of not complicated specialized devices and wiring patterns are described in the book by P.F. Filchchakov and V.I. Panchishina [1].

The reader will find practical methods in the book by Koppenfels and Schtalman [13].

Literature of Chapter II

[1] V.I.Smirnov, Course of higher Mathematics, Vol. III, Chapter II, Gostechizdat 1954.
[2] A.I.Markychevich, Theory of analytic functions, Gostechizdat 1950
[3] M.A. Lavrentjev, Conformal mapping with applications to certain problems of mechanics, Gostechizdat 1947
[4] K. Caratheodori, Conformal mapping, translated from english, OHNTI, 1934
[5] R. Courant, Dirichlet Principle, conformal mapping and minimal surfaces, translated from English, IL, 1953
[6] G.M.Golyzin, Geometrical theory of functions of a complex variable, Gostechizdat 1952
[7] M.V.Keldysh, Conformal mapping of multi-connected domains onto canonical domains, Uspekh mat. nauk, vyp. VI, 1939,pp. 90 - 119
[8] G.M.Gluzin, L.V.Kantorovich et al., Conformal mapping of simply-connected and multi-connected domains, OHTI, 1937
[9] L.V.Kantorovich, B.I.Keldysh, Approximate methods of higher analysis, Fizmatgiz, 1962
[10] P.F.Filchchakov, Theory of filtration under hydrotechnical constructions, Vol. I, Izd-vo AN USSR, Kiev, 1959
[11] P.F.Filchchakov, V.I.Paichishin, Integrators. Modelling of potential fields by electronic circuits. Izd vo AN USSR, Kiev 1961
[12] G.N.Polozhii, Effective solution of poblems of approximate conformal mapping of singly- and multi-connected domains and determination of Christoffel-Schwarz coefficients witht the iad of electronic analogues. Ukr. Matem. Journal, 7. No. 4, (1956). 423 - 432.
[13] V. Koppenfels, Tz. Schtalman, Practice of conformal mappings, translated from German, IL, 1963.