Value range of solutions to the chordal Loewner equation with restriction on the driving function Текст научной статьи по специальности «Математика»

92 Probl. Anal. Issues Anal. Vol. 8 (26), No 2, 2019, pp. 92-104

DOI: 10.15393/j3.art.2019.6270

UDC 517.54

A. V. Zherdev

VALUE RANGE OF SOLUTIONS TO THE CHORDAL LOEWNER EQUATION WITH RESTRICTION ON THE DRIVING FUNCTION

Abstract. We consider a value range {g(i,T)} of solutions to the chordal Loewner equation with the restriction |A(t)| ^ c on the driving function. We use reachable set methods and the Pontryagin maximum principle.

Key words: Value range, Loewner equation, Hamilton function, Pontryagin maximum principle

2010 Mathematical Subject Classification: 30C55

1. Introduction. Problems of finding value ranges {f(z0)} are typical for the geometric function theory. Here functions f are taken from some class of analytic functions and z0 is a fixed point in the domain of functions from that class.

A number of problems of this kind have been solved for classes of analytic functions defined in the unit disk D = {z : |z| < 1}. Rogosinski [9] gave a description of the value range {f (z0)} for the class of all analytic functions mapping the unit disk D into itself, f (0) = 0, f '(0) ^ 0. Grunsky [2] described the value range {log(f(z0)/z0) : f G S},z0 G D within the class S of univalent analytic functions f in D, f(0) = 0,f'(0) = 1. Goryainov and Gutlyanski [1] extended this result by describing the set {log(f(z0)/z0) : f G Sm} for the subclass Sm = {f G S : |f | ^ M} of bounded functions.

Roth and Schleissinger [10] described the value range {f (z0)} for all analytic univalent functions f : D ^ D, f (0) = 0,f'(0) > 0, that is, they obtained an analogue of Rogosinski's result for univalent functions. In the same article, they found a description of the set {g(z0)} within the class of all univalent analytic functions g : H ^ H, mapping the upper half-plane H = {z : Imz > 0} into itself and normalized g(z) = z + cz-1 + O(|z|-2),

©Petrozavodsk State University, 2019

z —y to. Value ranges for some classes of analytic univalent functions defined in D were described in [4,5].

Denote by H(T), T > 0 the class of all analytic univalent functions g : H\K — H, normalized near infinity as

2T

g(z) = z + — + O(|z|-2).

Here K C H is a so-called hull, which means that K = H if K and H\K is simply connected. Solutions of the chordal Loewner differential equation

dg(z,t) 2

, = —-r-rr-, g(z, 0) = z, t ^ 0, (1)

dt g(z,t) - A(t) yv ' ; ' ' v y

where A(t) is a real-valued continuous function, form a dense subclass of H(T). We call A(t) the driving function of the chordal Loewner equation (1). Thus, the problem of finding the value range {g(z0) : g E H(T)}, z0 E H, is equivalent to describing the set {g(z0,T)} of attainability of the equation (1). Without loss of generality we can put z0 = i. The set

D(T) = {g(i,T) : g solution (1)}

has been described by Prokhorov and Samsonova in [8] using the Pontrya-gin maximum principle. They proved the following theorems.

Theorem 1. [8] The domain D(T), 0 < T ^ 1, is bounded by two curves Zi and l2 connecting the points i and i\j 1 — 4T. The curve /i in the complex (x,y)-plane is parameterized by the equations

C2(y,T)(4T — 1) + (1 — siny)2 = 1 — siny _n< X(T ) 2C0(y, T) cos y ,y(T ) Co(y,T), 2 <y< 2 ,

where C0(y,T) is the unique root of the equation

2 cos2 y log(1 — sin y) + (1 — sin y)2 = 2 cos2 y log C + C2(1 — 4T).

The curve Z2 is symmetric to Zi with respect to the imaginary axis.

Theorem 2. [8] The domain D(T), T > 4, is bounded by two curves Zi = Zii U Zi2 and Z2, which is symmetric to Zi with respect to the imaginary axis; Zi and Z2 have the common endpoint i. The curve Zii in the complex (x, y)-plane is parameterized by the equations

) = C2(y, T )(4T — 1) + (1 — sin y)2 y = 1 — sin y ( ) 2C0(y,T)cos y , y( ) C0(y,T),

where y0(T) < y < 2. The curve l12 is parameterized by the equations

(T) = C0(y,T)(4T - 1) + (1 - sin y)2 (T) = 1 - sin y X( ) 2C00(y,T) cos y , y( ) C00(y,T),

y0(T) < y < 2. Here C0(y,T) > 0 and C00(y,T) > 0 are the minimal and maximal roots of the equation

2 cos2 y log(1 - sin y) + (1 - sin y)2 = 2 cos2 y log C + C2(1 - 4T), respectively, y0(T) G (-^, §) is the unique solution of the equation

1 - sin y 1 - sin y log , . + -, - ■ + 1 = - log(4T - 1).

1 + sin y 1 + sin y

Continuing this research, we consider a problem of describing the value range

DC(T) = {g(i,T) : g solution (1), |A(t)| ^ c};

that is, we added the restriction |A(t)| ^ c on the driving function, which is piecewise continuous on R. We use the Pontryagin maximum principle as the main tool of the research. See [6, 7] for reachable set methods developed for the radial Loewner differential equation.

2. Preliminary Statements. Due to a well-known property of the Loewner equation (1) (see, for example, [3]) and symmetry of the restriction | A(t) | ^ c, the domain Dc(T) is symmetric with respect to the imaginary axis. Therefore, we can consider only the right half (x ^ 0) of the domain.

Putting z = i in the Loewner differential equation (1) and splitting the resulted equation into the real and imaginary parts, we obtain the system of ordinary differential equations

dx 2(x - A)

dt (x - A)2 + y2;

dy 2y

dt (x - A)2 + y2

x(0) = 0,

y(0) = 1.

Following the Pontryagin maximum principle formalism, we introduce an adjoint vector ^(t) = (^1(t), ^2(t)) = 0 and the Hamilton function

H(x,y, *i, *2.A)=2(x." AA; + 2y*'2. (3)

( x - A) 2 + y2

The adjoint vector satisfies the system d^i <9H 2

[((x - A)2 - - 2(x-A)y^],

dt dx ((x—A)2+y2)2

[(2(x-A)y*i + ((x-A)2-y2)^].

d^2 dH 2 2 2

dt dy ((x- A)2+y2)2'

The domain Dc(T) is a set of attainability for the phase system (2) at t = T. The boundary point A = xA(T) + iyA(T) of Dc(T) is delivered by the solution (xA(t), yA(t)) of the Hamiltonian system (2)-(4) with the driving function A^(t) satisfying the Pontryagin maximum principle

max H(xA(t),y^(t),^A(t), ^A(t), A) =

Ae[-c,c]

= H (xA(t),yA(t), ^f(t), *£(t),AA(t)) at continuity points of AA(t). Note that

lim H(x,y, , A) = lim H(x,y, ^2,A) = 0

for any fixed values of x,y, Therefore, the maximum of H is at-

tained at zeros of the derivative of H with respect to A

dH(x,y, tti, ^2, A) = 2((x - A)2 - y2)^i - 2(x - A)y^2 dA (x - A)2 + y2 .

It is not difficult to show that H has only one local maximum on R for any fixed values of x, y, at

A0 = x-----. (5)

+ ^2 - ^2

Therefore, H attains its maximum on the interval [-c, c] either at A0 if A0 G [-c, c], or at one of the endpoints of the interval, otherwise.

We formulate the following lemma providing differential equations for the phase trajectory (x(t),y(t)) in the case when A0 G [-c, c].

Lemma 1. Let A(t) maximize the Hamilton function (3) on R for t G [t1,t2] C [0, T] ; that is,

max H(x(t), y(t), ^(t), ^(t), A) = H(x(t), y(t), ^(t), ^(t),A(t)), agr

where (x(t),y(t)) is a solution of the phase system (2) and (^1(t), ^2(t)) is a solution of the adjoint system (4). Then

/ X «iy

(a) r-*—-— = const = t G L^i, ^aj,

^«2 + «2 - «2

(b) A(t) = x(t) - p, t G [ti, t?j,

(c) the phase trajectory (x(t),y(t)) satisfies the following differential equations

2y (6)

dt p2 + y2 dx p

d • (7) dyy

Proof. Since A(t) maximizes H on R, it satisfies (5) for t G [t1,t2j. Substituting (5) into (2)-(4), we obtain

dx «1 . .

dt ^y^f+^f'

dy _ ^/«1 + «2 - «2

dt ^v^F + ^I '

d«i = 0 d« = ^«2 + «2 - «2 (10)

dt U, dt y2 ' ( 0)

H(x,y, «1, ^2,A0)= - «2'

y

In view of (10), we have

«?(t) = const = c?,t G [t?, t2j'

Due to a well-known property of the Hamilton function, we have

/«2 + «2 _ «

H(x,y, «1, «2, A) = —-1-2-2 = const = C2,t G [ti, 12j' (11)

y

We put p = —. Thus, we have proved statements (a) and (b). Using (11), c2

we can rewrite (9) as (6). Dividing (8) by (9), we obtain the differential equation (7). □

Note that equations of the Hamiltonian system (2)-(4) are invariant under change of the sign of «1(t), x(t) and A(t) to the opposite. Thus,

-o'.s o.'s

(a) T=0.245 (b) T=0.3

Figure 1: Value ranges D(T).

flipping the sign of ^i(t) (and, due to statement (a) of Lemma 1, equally of p) has the effect of reflecting the phase trajectory (x(t),y(t)) in the imaginary axis. Therefore, we can restrict ourselves to the case of p ^ 0. We will see that this choice will lead us to the right half of the boundary of Dc(T).

In the case of no restrictions on the driving function A(t), we have c = to and the condition A0 E [—c, c] always holds. This allows us to deduce from Lemma 1 a description of the boundary of D(T) in the Cartesian coordinates (X, Y).

Theorem 3. The boundary of the domain D(T),T > 0 is given by the equation

2X2 = log Y(1 - 4T - Y2). (12)

Proof. Since conditions of Lemma 1 are satisfied on the whole interval [0,T], we can integrate equations (6) and (7) over this interval with the conditions x(0) = 0, y(0) = 1, x(T) = X, y(T) = Y. We obtain

2p2 log Y + Y2 = 1 - 4T, X = -p log Y. (13)

Finally, multiplying the first of these equations by log Y and using the second, we obtain (12). □

It is easy to see that there are two essentially different cases. In the case of T ^ 4, the set D(T) is a bounded domain with its boundary crossing the imaginary axis at y = \f1- 4T, y =1. This case corresponds to Theorem 1. If T > 4, the set D(T) is unbounded and its boundary

includes the real axis; this case corresponds to Theorem 2. Starting at this point, we only consider the case of T ^ 1.

Note that the boundary point (0, V1 - 4T) of D(T) is delivered by the driving function A(t) = 0. Therefore, it also belongs to the boundary of Dc(T). It is a reasonable assumption that all points of some arc on dD(T) near (0, V1 - 4T) are delivered by driving functions with ranges within the interval [-c, cj, and since this arc belongs to dDc(T). A precise statement is given by the following lemma.

Lemma 2. A segment of the boundary dDc (T) is given by (12), Y G [1 - 4T, Y0], Y0 is the unique solution of one of the equations

1 - p-4

2c2 log Y + Y2 = 1 - 4T, c2 ^ T---—, (14)

2c2logY + Y2 = 1 - 4T, c2 ^ T - i-^4' (15)

(1 + log Y)2 4

Note that if c2 = T - 1 4 4, both equations (14), (15) have the same root Y0 = e-2.

Proof. Consider a point on the boundary dD(T). Let A0(t) denote the driving function delivering this point. By Lemma 1, we have A0(t) = = x(t) - p. Since p > 0, we see from (8) that x(t) and, hence, A0(t) are increasing functions.

A boundary point of D(T) belongs to the boundary of Dc(T) if it is delivered by a driving function with the range within [—c, c]. Since A0(t) is increasing, this condition is equivalent to inequalities

Ao(0) ^ -c, Ao(T) ^ c. (16)

Note that A0(0) = -p, A0(T) = X - p. Equations (13) allow us to express X and p through Y. Substituting into (16) and squaring the result, we obtain

1 - 4T - y2 2 1 - 4T - y2 2 2

-;-— ^ c2, ---— (1 + logy)2 ^ c2'

2 log y 2 log y

We need to find the greatest value Y0 of Y satisfying both conditions. Define the following functions of Y for Y G [V1 - 4T, 1]

1 4T Y2 1 4T Y2

> = -iiogr-, = (1 + log Y)2-

It is easy to see that /i(Y) ^ /2(Y) for Y e [e 2,1], in particular, it is always true if \J 1 — 4T ^ e-2 or, which is the same, T — ^ 0.

Therefore, in this case Y0 is the solution of /1(Y) = c2, which is equivalent to (14); it remains to prove the case V1 — 4T < e-2.

We have /1(Y) ^ /2(Y), and the equality sign holds only at Y = e-2. Hence, we need to check if /2 attains the value c2 within the interval [Vl — 4T, e-2]. The derivative

/2(Y) = 2+Og2 (—2Y(1 + log Y) log Y + ^(1 — 4T — Y2))

vanishes at Y = e-1 and at roots of the equation

2 log Y + 1 = 1 — 4T 1 log Y — 1 Y2 '

The left-hand side of the equation is an increasing function of Y on |V1 — 4T, e-2] and takes the value — | at Y = e-2, while the right-hand side is decreasing on |V1 — 4T, e-2] and takes the value 1-4r — 1 > — 1 at Y = e-2. Therefore, the derivative /2 does not vanish on the interval [V1 — 4T, e-2]. Since /2(VT—4T) > 0, /2 increases on [V1 — 4T,e-2]. Therefore, /2 attains its maximum at Y = e-2. Hence, Y0 is the solution of /2 (Y) = c2 if the inequality /2 (e-2) > c2 holds. Note that the last inequality gives c2 > T — 1-— to complete the proof. □

If A(t) = ±c, the phase system (2) can be integrated directly. We need the following properties of its solutions stated by the remark below.

Remark 1. If trajectory (x(t),y(t)) satisfies

dx 2(x — a) dy 2y

dt (x — a)2 + y2' dt (x — a)2 + y2 ' where a is a real number, then the following quantities are constant:

(x — a)y, (x — a)2 — y2 — 4t.

Proof. The statement can be proved by direct integration of the system. □

3. The main Theorem. Now we are ready to prove the following theorem describing the value range Dc (T) in the case of c2 ^ T — 1-4—.

Theorem 4. Let c2 ^ T — ±-f—, T ^ 4 and let curves l1 — /4 be defined as follows.

1. The curve /1 is a segment of the boundary dD(T) given by (12), Y G [1 — 4T, Y0], Y0 is the unique solution of (14).

2. The curve /2 is given by solutions (X, Y), X + iY = z, ^ G [0,1] of the equation

z2 + 1 — 2c(2^ — 1)(z — i) + — 1) ln Z + — 1) = 4T. (17)

i + c(2^ — 1)

3. The curve /3 is given by solutions (X, Y) of the system

2p2 log — + Y2 — p2 = 1 — 4T — c2,

c

X = —c + p( 1 — log Y;^ ,

:is)

where p G [c, p0] and

p0 = (V^+ c2 — 1)2 + 4c2 +(4T + c2 — 1}). (19)

The curve /4 is symmetric to /3 with respect to the imaginary axis. If the following equation

c2 / 4c \

— 4pc ^ — exM--— p2 = 1 — 4T — c2 (20)

p2 p

has two solutions p1 < p2 in the interval (c,p0), we also define curves

l5 — l10.

4. The curve /5 is given by solutions (X, Y ) of the system (18), p G [c, p1]. The curve is symmetric to /5 with respect to the imaginary axis.

5. The curve /7 is given by solutions (X, Y) of the system

4cp + (X — c)2 — Y2 — 4T = c2 — 1,

— p log --)— = 2c, ( )

c

where p G [p1 ,p2]. The curve /g is symmetric to /7 with respect to the imaginary axis.

6. The curve /9 is given by solutions (X, Y) of (18), p G [p2,p0]. The curve /10 is symmetric to /9 with respect to the imaginary axis.

k 0.35i b/A

— ¿i

-0.2 0.2

(a) T=0.245, c=1

(b) T=0.245, c=0.1

Figure 2: The boundaries of the value ranges Dc(T)

The following two cases are possible:

(1) Dc(T) is bounded by curves h,l2,k — l10, if (20) has two solutions Pi < p2 in the interval (c,p0).

(2) Dc(T) is bounded by curves l1 —14, if (20) has less than two solutions in the interval (c,p0).

Proof. The curve l1 is already given by Lemma 2. It can be seen from Lemma 1 that, at t = 0, the Hamilton function H is maximized at A0 = —p. Thus, if p > c, H attains its maximum on [—c, c] at A = —c. Therefore, we have the following driving function:

A (t) =

- c, 0 ^ t ^ ti,

x(t) - p, ti <t ^ T.

(22)

Denote x1 = x(t1),y1 = y(t1). Applying Remark 1 to the interval [0,t1], we obtain

(x1 + c)y1 = c, (x1 + c)2 — y\ — 4t1 = c2 — 1.

c

Since A(t) is continuous, (22) gives x1 = p — c. Thus, y1 = - and we can

p

also find t1:

2

4ti = p2 ^^ - c2 + 1. p2

(23)

Integration of (6) and (7) over the interval [t1,T] yields the system (18). The equation (23) shows that t1 increases as a function of p. Therefore, we can rewrite the condition t1 G [0,T] as p G [c,p0], where c and p0 are

0.25Î

¿10 h

//h y 5

h

-0.15 0.15

Figure 3: The boundary of the value range Dc(T), T = 0.247, c = 0.05.

the roots of (23) for ti = 0 and ti = T, respectively. Note that if p = c, equations (18) turn into (13).

We have to satisfy the condition A(t) E [-c, c]. Since A(t) is equal to —c on [0, t1] and increases on [t1, T], we only have to ensure that A(T) ^ c. According to Lemma 2 for p = c, we have A(T) < c. Assume that at some point p E (c,po], A(T) > c. Then, due to continuity of A(T) as a function of p, there is a point p1 E (c,p0), such that A(T) = c. Using (22) and (18), we can rewrite it as

—p log — = 2c.

c

With (18) it gives the equation (20) for p1. Thus, if (20) has no roots in (c,p0), the case (2) takes place. Note, however, the existing of a single root of (20) in (c,p0) does not guarantee violation of the condition A(T) ^ c, and, thus, the case (2) is still possible.

Now consider the driving function

- c, 0 ^ t ^ t1,

A(t)=<( x(t) - p, t1 <t ^ t2, (24)

c, t2 < t ^ T.

Denote x2 = x(t2), y2 = y(t2). Applying Remark 1 to the phase system for the interval [t2,T], we can write

(X - c)Y = (x2 - c)y2, (X - C)2 - Y2 - 4T = (x - c)2 - - 4t2. (25)

The condition A(t2) = c gives x2 = c + p. Integrating (6) and (7) over the

interval [t1,t2], we obtain

o 2 1 y2p . 2 2 i A i 2 , U 1 y2p \ 2p2 log--+ - p2 = 1 - 4t2 - c2, x2 = -c + p - log-J ,

that with (25) leads us to (21). These equations describe the boundary segment governed by the driving functions of the type (24). From (25) and (21), we can deduce the equation for t2

2 2 (X - c)2Y2 , ,

4t2 = 1 - c2 + 4pc + p2 - ^-/-, (26)

p2

and, therefore, we have

n , c2 - (X - c)2Y2

4(t2 - t1) = 4pc +--.

p2

The second equation in (21) implies that c2 > (X - c)2Y2; therefore, the inequality t1 < t2 always holds. Since t1 increases and takes the value t1 = T at p = p0, there is a point p2 E [p1,p0], such that at this point t2 = T. Substituting t2 = T into (26) and using the second equation in (21), we again obtain the equation (20) for p2. Thus, we see that existing of two roots of (20) p1 < p2 in the interval [c,p0] is a necessary condition for the case (1).

It is not difficult to see that the segment of the boundary corresponding to p E [p2,p0] is delivered by the driving functions of the type (22) and, consequently, is described by the system (18).

For the remaining part of the boundary dDc (T) the Hamilton function is maximized outside of the interval [-c, c] and, thus, we have |A(t)| = c. Therefore, we can use the generalized Loewner equation (see [6,7])

dg(z,t) 2 , (1 ) 2

—-+ (1 -

dt g(z,t) - c g(z,t) + c

g(z, 0) = z, ^ E [0,1].

Putting z(t) = g(i,t) and integrating the equation over [0,T] we obtain the equation (17) for the curve /2, parameterized by ^ E [0,1]. □

Acknowledgment. This work was supported by the Russian Science Foundation, project 17-11-01229.

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Received May 3, 2019.

In revised form, May 14, 2019.

Accepted May 18, 2019.

Published online June 10, 2019.

Saratov State University

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Petrozavodsk State University

33 Lenin pr., Petrozavodsk, 185910, Russia

E-mail: [email protected]