Axially symmetric flow Текст научной статьи по специальности «Математика»

М

УДК 517.9 A. S. Minenko

Donetsk National Technical University, c. Donetsk 83015, Donetsk, Artema st., 131

AXIALLY SYMMETRIC FLOW

А. С. Миненко

Донецкий национальный технический университет, г. Донецк 83015, г. Донецк, ул. Артема, 131

ОСЕСИММЕТРИЧНЫЙ ПОТОК

О. С. МЫенко

Донецький нацюнальний техшчний ушверситет, м. Донецьк 83015, м. Донецьк, вул. Артема, 131

ОСЕСИМЕТРИЧНИЙ ПОТ1К

An axially symmetric flow with a free boundary within a sufficiently long domain is investigated. This problem is different from the ones considered earlier, that on part of the free boundary the nonlinear Bernoulli's law is set in kind of inequality. The problem of the existence of boundary value problem is equivalent to minimum problem of some functional with a variable domain of integration. Analyticity and monotony of free boundary are proved. The approximate solution of problem, built by Ritz method, converges to exact solution in Z2.

Keywords: differential equation, free boundary, functionality, optimization, numerical methods.

Исследуется осесиметричный поток со свободной границей достаточно длинной области. Эта проблема отличается от ранее изученных тем, что на свободной границе нелинейное условие Бернулли задано в виде неравенства. Проблема решения краевой задачи сводится к проблеме минимума функционала с неизвестной областью интегрирования. Доказана аналитичность и монотонность свободной границы. Приближенное решение, построенное методом Ритца, сходится к точному решению в Z2.

Ключевые слова: дифференциальные уравнения, свободная граница, функционал, оптимизация, численные методы.

Дослщжуеться осесиметричний потк з втьною межею досить довгоТ областк Ця проблема в^зняеться вщ рашше вивчених тим, що на вшьнш межi нелшшноТ' умови Бернулли задано у виглядi нерiвностi. Проблема ршення крайовоТ задачi зводиться до проблеми м^муму функцюнала з невщомою областю штеграци. Доведена аналогична i монотоннють вшьно'Т межк Наближене ршення, побудоване методом Ритца, сходиться до точного ршення в Z2 . Ключовi слова: диферен^альы рiвняння, втьна межа, функцюнал, оптимiзацiя, чисельн методи.

1. Problem. Consider an axially symmetric flow (the x-axis is the axis of symmetry) with vorticity co - const > 0. The flow is moving from left to right. Then the problem is as follows. Let G be the domain, where

dG = Tx kjSvjT2 vjB, Fj = (jc = 0, 0 <y<c), T2=(x = a, 0 <y<b),

B = (0 < JC < a,y = 0), S : y = g(x), x e [0,a]; g(x) e C2[0,a], g(0) = c, g(a) = b, c <b, g'(0) - 0, g'(ci) = 0 and g(x) is the monotonic function. Let y be the sufficiently smooth curve, such that point (0, c) is the left end of y, and y lies on T2. Curve y divides the domain G into two subdomains. We designate the lower part of division GbyGr cG. Gy is the flow region. We will study the following nonlinear problem. It is necessary to find the stream function y/{x, y) and the free boundary of y (G, is simply connected domain), where y/{x,y) satisfies the conditions:

Vxx + Wyy ~ y~XVy = (x,y)eGr, (1)

y/(x,y) is the continuous function in Gy, i//(x,y) is the continuously differentiable function in Gy, except, perhaps, the point {a, h) (c <h<b, (a, h) is the right end of y), such that

V(x,y) = 0, (x,y)eB, (2)

V/Ax,y) = °, (3)

y/(x,y) = 1, (x,y)ey, (4)

y) + '//2(a; y) > \'2j'2, v = const > 0, (a; y) g y, (5)

where on part ofy, lying inside G in (5), there is always an equality. We call (y/,y) the classical solution, if (i//,y) satisfies (l)-(5).

The problem (l)-(5) differs from problems [1-5], because in condition of (5) there is an inequality when (x, y)&yr^S.

2. The minimum problem. Let us consider the functional with a variable domain of integration

J(w, 7) = + V2y +v2y2+ 2coyiw -1)]

Xdxdy

y

on set R, where (y/,y) e R : y is a Jordan curve such that points (0, c) and (a, b) are the ends of y. All the points (x, y) e y, except the point (0, c), are above the horizontal y = c, y e G u S ; i//(x, y) is a continuous function in G,, /// = 1 on ;/, y/ = 0 on B, y/(x,y) is a continuously differentiable function in G , and J(y/,y)< oo, i.e. J is the

bounded functional on R.

Lemma 1. Let the pair (y/,y) be the classical solution of (l)-(5). Then this pair is the stationary point of the functional J or R. Any stationary point of the functional J or R, where y is a smooth curve, is solution of (l)-(5') (Gy is the simply connected domain).

G

M

Proof. Let us consider the first variation J(y/,y) on R [6]. We compute

SJ(y/,Gr,Sy/Jz) = -2 jj 1 dy/

dx

1 dy/

+ -

y & ) dy{y dy

1 dy/

- œ

Sy/dxdy +

+ 2 i

r,ur,y dfl

IT 2 2 2 2 Sy/ds + Jl^v y ~y/x~y/y J-ds.

(6)

Here 8y> is the variation of ///, Sz - dx, ¿>z is variation of fx, y); n is the outward normal. Here we will use the act that the first variation satisfies the condition, that SJ > 0 for the stationary point (y/, y), and < 0, if (x, y) e y n S. In case if (a, h) is the right end of y, where c< h b, then we will consider a minimum problem for the functional J on Rh. If (y/,y) e Rh, the points (0, c) and (a, h) are the ends of y. And if (x,y) <Ey, then c < y <h, except the ends of y: t//(x,y) have the properties which were considered for (// e R (see Theorem 1).

3. Symmetrization. Let us suppose that y is an admissible curve (i.e. Gy is the certain domain). First in this section we will show the existence of linear problem (l)-(4) in domain Gy. This procedure is as follows. Consider set U, where y/ e U \y/(x,y) is a

continuous function, in G , i//(x,y) is a continuously differentiable function in G,, (// = 0 on B, y/ = 1 on y and L(yr) < oo (L is bounded on U),

Oy

dxdy

y

Lemma 2. There is the unique minimum y/ e U of the functional L in U. This function y/(x,y)is the classical solution of (l)-(4). If curve y has the finite length, then

J(V,Y)=\— (x.y) | ^o dx + v2 jjydxdy + co -1 )dxdy.

B\~y J Gy Gy

Proof Here we use variational method and Green's formula (see [7]). Since the solution y/(x,y) of (l)-(4) satisfies the condition y/{x,y) = y2a{x,y), where a(x,j)is a sufficiently smooth function, then

YrVy(x,y) J^o =2a(x,Q).

Let us define symmetrization of domain Gy by Steiner [3] concerning to x- and axes as symmetrization Q = n IGy, T\ = <x <a, 0<y<b^ concerning to x-axis and horizontal y=b. We assume y/ = 1 for all Cj) •

Lemma 3. Let y/(x,y) be the solution of (l)-(4) in domain Gy, and y/*(x,y) is the

* *

solution of (l)-(4) in symmetrized domain G with the free boundary y . Then J*(y/*,y*) < J(y/,y), y/*x(x,y) <0, y/*x(x,y)> 0 in G* and y*: x=x(t), y=y(t), 0<t<T, where x(t), y(t) are monotonically increasing functions.

Proof. The idea of the proof is as follows. In case of symmetrization in relation to x-axis for every x0 e [0,a] we substitute i//(x0,y) for monotonically increasing function

^(Xq,^), such that

mes y: i//(x0,y) <p =mes y: ^(.^j') <p , p e (0,1). Then we obtain

+- vvi^ s J]k2+y>i * v-y2]^

G.,

x . .

y GJ y

The proof of this important inequality has been given in [3], [10]. At last

-1 Wxdy = JJ[0// - Wxcly .

G* G„

This shows that J*Q¥,y*) < J(i//, y). Now if we substitute vf(x,>') for solution if/*

of (l)-(4) in G\ then J(y/, y ) < JQ¥, y) (see Lemma 2) and ij/y (x,y) > 0 in G*. Then it

*

is necessary to carry out symmetrization of G in relation to j'-axis.

4. Theorem of existence. There is a minimal sequence (y/n, yn) e R, neN . The domain G„ has the free boundaryyn. x = xn{t\ y = yjt), 0<t<T, where xn(t), yn(t) are monotonically increasing functions (see Lemma 3). On the other hand, according to Lemma 2, \j/„(x,y) is the solution of (l)-(4) in G„ for every «eN. If the coordinate system (x, y) is turned by angle 3^/4 in positive direction, we obtain the system of coordinates (c,?/ ) where Yn- rtn= rh, ^ - and \j]n ^ qn - for every n(=N. Thus, there exists the limit

monotonic curve y. However, y can also have a common segment with T2 or with horizontal

y=c.

Let us suppose that

1-—>0, (7)

3

then according to the maximum principle 0 < y2 (1 - cob3 / 3) / b2 + coy3 / 3 < < y/n (x, y) < 1 in Gn and y/n(x,y)<y2(l-oc/3)/c2 +coy3/3, x,y eno, where no = 0<x<a, 0<y<c . The function y/„(x,y) is subharmonic in G„ and if/n(x,y) = x,y +coy3 /3 (where (Pnx=yl^ny, (Pny=y~1ZnX)> Pni^y) is the harmonic function in G„. It follows that there is function y/(x,y) such that —»V^// in C(G0), !//„ —»!// in C(G0) (here

o + i.e. G0 is a closed set) for every G0 cz Gy, such that G0 ny - 0 . Then

we obtain that y/ satisfies (l)-(3) in limit domain Gf, and J(y/, y) <cc.

Let z0 - x0 + iy0 be an interior point of y. Now let us choose the ray / with the top in z0 such that the angle between / and the x-axis equals 3,t / 4. Let us make the cut along I.

G*

M

Then we use inequality

\Wn (z) -1| < A Re [e'*/4 (z - z„ ) J2 ? z„ e yn, lim z„ = z0

for every z e G„ and sufficiently large number A>0. Let us assume that z is an interior point Gr. Letting n —> oo, and then z —> z0, we conclude that y = 1 on y. Thus (y/, y) satisfies (l)-(4) in Gy and

J(y/, y) = 21cc(x, 0)dx + v21|ydxdy + co ||(^ - \)dxdy,

where y/(x,y) - y2a(x,y) . It was shown in Lemma 2.

Since y/^y-1 | J=0 = 2an (x, 0) = <pm (x, 0)=> y/yy~l \ y=0 = 2a (x, 0) = <px (x, 0) in C[0, a], where y/„(x,y) = y2an(x,y) and (p{x,y) in harmonic Gv, we can prove that

d = lim

2 (x,0)dx + v2 ^ydxdy + co J|(í//h -1 )dxdy

= J(v,r),

where d = inf ./(y/, y) on R. At last, using Schiffer's method of interior variations [3], we can show that ôu/ôn-v almost everywhere on the part of y, lying inside G. Lemma 4. Let us suppose that there is (7) and we also have inequalities:

1

2 +

mc

„3 \

2 a

co mes G + —-c

í

1-

coc

3 \

\

a _

(8)

J 0

Then G, cannot coincide with G, that is G}, c G, and all the points (x, y) e y are above the horizontal y=c, except point (0,c).

Proof. Let us suppose that Gy =G. Then we have

■1 dy/

Í--ds = 2 (a(x,0)dx + co ÍÍJa'Jv .

S y dn B G

Now, using if/(x,y) = y2a(x,y) <y2(I-coc3 /3)/c2 + ay3 /3, a(x,y) < (1 -coc313)1 c2 + coy /3 for (x,y) e no, we obtain inequality

a _

v j^l + g2, dx < co mes G +

2 a

f 3 A

coc

V

3

y

There is a contradiction, so G„ a G.

Let y0 = (0 < x < a,y = c) and yily0= x < a,y = c (hereGro c Gy ). Then y/0(x.y) = y2( 1 - coc3 /3)/c2 + coy3 /3 is the solution of (l)-(4) in Gy0 = (0 < x < a, 0 < y < c). According to the maximum principle y/ = y/0 in Gy0. Thus, J(y/0= Y) = d. Now

b

G

n >'A

B

G

G

n

0

M

we compute the first variation J on pair (y/0,y0), provided that & = 0 in points (0,c) and (a ,c), (Sz,ri) > Oand the value max|&| is small (see (6)):

ôji,Gr,ôy/,sT= J

r 0

2 2 v c -

1-

coc 3

,3 A

+ cdc

'tis <0.

Then we obtain the domain (},, with the free boundary ;/ , so that JQ/7, f) < J(i//, y) - d, where \jj is the solution of (l)-(4) in Gy. Let G* be the symmetrization of Gy domain in relation to axis and let y/* be the solution of (l)-(4) in G*. Then, according to Lemma 3, J(y/*,y*) <J(ip,y)(here y* is the free boundary of G*). We choose curve y' so

thatG^ cG -. Since y/*(x,y)-y/'(x,y) > Oin Grt, using Lemma 3,"we obtain inequality d<J if/',y' <J(y/*,y*) + v2bmes G,\G, <d

for small value mes Gy, \ G . . Thus, we have a contradiction, since ///', y' e R. The other cases are similar. The lemma is proved.

Let z0 = x0 +iy0 be any interior point of y and Kr :|z-z0| <r, so that c G is for small r. Using methods [3], [7], we will prove the existence of the analytic function g(t), t = <^ + if] in domain Gy nKr, witch is continuous in Gy and g(t) = t on y. Let co = co{t) be the conformal mapping of Gy n Kr on the upper half-plane. According to

Schwartz's principle, the functions Oj(t) = g(t) + t and Q)2(t) = g(t) -1 can be analytically continued through the segments of w - plane, which conform;/. Then

O, -O,

t(w) = -

2

is analytically continued there too. Thus, y is an analytic arc.

Now we can prove the theorem.

Theorem 1. Let S: y = g(x),a <x <b,where g(x) e C2[0,a] and g(x) is the monotonically increasing function, such that g(0) = c, g(a) = b,c<b, #'(0) = 0, g'(a) = 0. Suppose that we have (6), (7) too. Then there exists (¡¿,y __ - the unique solution of (l)-(5) where y is the monotonically increasing arc, analytical in the neighborhood of each of its points, lying inside G, y/(x,y) is the continuous functional in Gr, i//(x,y) is the continuously differentiable function in Gy except, perhaps, (a,h), c <h<b ((a,h) which is the end ofy ).

Proof. Let be the limit pair. We consider only the case when

yr\T2- x-a,h<y<b ,c<h<b. Let y{) be the part of y, such that points (0,c) and {a, h) are the ends for y0. Evidently that J(i//,y) = J(i//,y0) = d. Then we consider the

2

c

M

minimum problem of the functional J on Rh and show that dh = J(ip, y0), where dh = inf J on Rh. We have (y/,y0)eRh and dh<J(y/,y0) = d. Now suppose that {y/l,yl) is any admissible pair, i.e. {y/l,yl)eRh. Let us choose (y/2,y2) g R, so that G cG We can show that

d < J(i//2, y2) <./(///,, n)+v2b mes(Gr2 \Gn)

(see Lemma 4). Then d < dh +1:, where s = v2b mes(G \G ) . Letting 0, we conclude that d < dh. Thus, dh = J(y/, y0). Evidently (if/, y0) is the solution of (l)-(5). The uniqueness of solution (i//,y) follows from [4] (here we also use (9)). Thus, the theorem is proved.

5. Approximate solution. In this section we suppose, that solution (i//,y) of (1) - (5) satisfies the conditions: i//(x,y)&Cl(G \y and S have the finite number of common points. Using the Friedrichs transformation [4], we obtain

(zx + ^z)2 + \Wg2z2z2+2wg{cp-l)zz2

g

g

dxdcp

zz..

where A = (0 < jc < a, 0 < cp < 1), cp(x, z) = y/(x, zg(x)), z(x, <p) is the solution of equation (pipe, z)-cp = 0. We will study a minimum problem of the functional ./, (z) with a constant domain of integration on the following set:

Dz ={z : z e C(A), Jcpza e C(A), z(0,1) = 0, z(x, 0) = 0, min_ Jpza > 0 .

x,q> eA

We define z0(x,<p) as the solution of equation z(x, (p) - z = 0, where cp{x,z) = ip{x,zg{x)) d3\d{i//,y) is the solution of(l)-(5). Evidently, that z0<eDz and

z(](x, cp) - 47prj(x, (p\ (x, cp) <e A, where r](x,(p)\s, the smooth function, such that r/(x, 0) ^ 0, 0 < jc < a.

Lemma 5. The function z0(x, cp), (x, (p) e A is a minimum for ./, (z) on 1)_. Proof. Let us define w(x,<p) according to the formula co = In z. Then we compute,

(wx+^z)2 + e

g_x g

-2 w

g

■ + v2g2e2-w2+2wg^-l)e-e2

dxdcp

de

e~2w-

Ws<pSWx "¿HO;

(p\ sx

.M*. g

g

2 [KSW* + (WWSW + SW<pf J > d£dX3d<P

(9)

+

w]

s<p

+ 2v2 j(l - s) \g2(x)e2w<{x l)&v2(x,l)dsdx, 0 0

where Jj(z) = Jx{ew) = J2(w), we - w0 +söw, Sw = w -w0, w0 - lnz0, Sz - z — z0 = z0Sw,

2

1

a

M

0 < s < 1, z is any element of Dz. Now, using the Friedrich formula [4]

J2(w)^J2(wo)+^-J2(W£)\£__o +J(1 (10)

as ' as

we conclude that Jx{z{)) - J2{w0) < J2{w) - Jx{z) for every z e /)_. Thus the lemma is proved.

We will minimize J (z) on Dz, using the polynomials

_ m mk

= = sup k + mk ,

k=0 j=0 0<k<m

where akJ e Er (the Euclidian space),

m m

r = XK+l), Dr=Er0^D;, Er0-.^ak0-1 = 0,

k=0 k=0

!K = % : mm Jtpz > 0 .

' x,ip eA v

We will seek akj as a solution of nonlinear Ritz system equations (see [8], [9], too).

a)-, it,.

= 0 p=0,l,2,...m;

dap o

dap o

:0,q=l,2„. m ;p=0,l,...m; (11)

m _ m mk

i=0 ¿=0

Theorem 2. There exists a minimum akj e Dr of the functional J3 on Dr, where akj is an interior finite point in Dr.

Proof. Let it be sequence akJp go when p —>cc, i.e. akJp are unbounded in Er. Let us define J4{ckjp) = J3{M pckjp), where

m mk a

2 _ „ _ Ukjp

kjp M k=0 j=0 M p

Then we can show, that J3 oo, if p —» go . On the other hand J3 (akj) is unbounded function on 51),. So the theorem is proved.

Remark. Now, using the Lagrangian function, we conclude that a* is a solution of systems (11). Thus, there exists a solution of (11) for every fixed n. 6. Regularity. At first, we define zn(x,q>;a*k.) - zn.

At ■ 1 — ¿a .

kj J n*

Zn

Lemma 6. The sequence z * is the minimizing sequence for J on Dz

M

Proof. Let us define ||z|| according to the formula

|| z||=max| z|+max| |+max| -yfpz |,

where (x,cp)e A. For sufficiently small value s>0 we can choose the polynomial Pn (x,cp) so that || Zq ~4<pPn ll< £, where

m mk k=0 j=0

Then we obtain Jx (zn) -d < Jx(zn) -d = Jx(zn)-Jl(z0)<s , here d = Jl(z0)< Jl (z ) for every zeDz (see Lemma 5) and s > Ois sufficiently small. Thus the lemma is proved.

The function yn{x)~g{x)zn{x,\alg\ 0<x<a is an approximate equation of free boundary y in problem (l)-(5).

Theorem 3. The sequence zn(xXctkJ)^>z0(x,X) strongly in L2(0,a).

*

Proof. According to Lemma 6 sn =Jl (z* )-J1(z0)—>0, when n —> oo .

'n \\ n) ^lV^O-

Taking into consideration (10), we obtain the inequality

/2

0

which can be written, as

1 a

2v2 J(1 -s) \g2{x)e2wAx l)dw2{x,\)dsdx < en, 0 0

where ws =w0+s(wn-w0), w0=lnz0, wn=\nz*n, 0 <s<\. Then we get the

estimation

2s

Sw2 (x,X)dx < " , öz = z*n-z0, (Sz = Sw exp w0).

1 J S~>

Thus 11 z„ (a; 1, dk]) - z0 (x, 1) | |^(0>fl) -> 0, when w oo.

References

1. Friedman A. Axially symmetric cavities in flows / Friedman A. // Comm. Partial Differential Equations. -8 (1983). - № 9. - P. 949-997.

2. Alt H. W. Axially symmetric jet flows / H. W. Alt, A. Friedman, L. A. Caffareli // Arch. Rational Mech. Anal. -1(1983). - № 28. - P. 97-149.

3. Garabedian P. R. Axially symmetric cavitational flow / Garabedian P. R., Lewy H., Schiffer M. // Ann. Math. -56 (1952). - № 3. - P. 560-607.

4. Friedrichs K. 0. Uber ein Minimumproblem Fur Potentialstromungen mit freien Rand / Friedrichs K. 0. // Math. Ann. - 109 (1934). - P. 60-82.

5. Danyljuk I. I. On integralfuctionals with a variable domain of integration / Danyljuk I. I. // Proc. Steklov Inst, of Math. - 118 (1972). - English: AMS (1976). - P. 1-112.

6. Courant R. Methods in Mathematical physics / R. Courant, D. Hilbert. - New-York. - 1 (1953). - P. 1-260.

7. Minenko A. S. On one thermophysical problem with a free boundary // Dokl. Akad. Nauk. Ukraine. Ser. A. -1979. - №. 6. - P. 413-416. (In Russian).

8. Danyljuk I. I. On the Ritz method in one nonlinear problem with a free boundary / I. I. Danyljuk, A. S. Minenko // Dokl. Akad. Nauk. Ukraine. Ser. A. - 1979. - № 4. - P. 291-294. (In Russian).

9. Danyljuk I. I. On one optimization problem with a free boundary / I. I. Danyljuk, A. S. Minenko // Akad. Nauk. Ukraine. Ser. A. - 1979. - № 5. - P. 389-392. (In Russian).

10. Minenko A. S. Investigation of a problem of whirlwind liquid stream with a free boundary. Nonlinear boundary problems / A. S. Minenko // Akad. Nauk. Ukraine. Inst. Applied Math. Mechanics. - 1993. - № 5. -P. 58-72. (In Russian).

RESUME

A. S. Minenko Axially Symmetric Flow

Background: A new class of problems with a free boundary and hydrodynamic origin is investigated, when on part of the free boundary the nonlinear Bernoulli's law is set in kind of inequality. The existence of a classical solution of respective nonlinear problem is proved. The free boundary proves to be an analytic monotonic curve. The proof is underlain by the Schiffer's method of interior variations and Steiner symmetrization.

Materials and methods: This paper is aimed at development of exact and approximate methods of solution of nonlinear problems with a free boundary. The following methods are devised:

- justification of correctness of class of free boundary problems with variational nature;

- construction of approximate solution of such nonlinear problems by the Ritz method, and investigation of their convergence.

Results: The convergence of the Ritz approximation to the exact solution in the integral metric is ascertained. The developed methods can be applied for studying of the whole class of free boundary problems with variational nature.

Conclusion: The paper proves the existence of a classical solution of the new class of free boundary problems of the Bernoulli's type. The approximate solution converging metrically to the exact one in Z2 is constructed.

The article entered release 25.02.2016.