УСЛОВИЕ ЭКСТРЕМАЛЬНОСТИ ПОВЕРХНОСТИ ВРАЩЕНИЯ ДЛЯ ФУНКЦИОНАЛА ТИПА ПЛОЩАДИ

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  • 517.957 + 514.752

    .. , ..

    , - . - - .

    R3, M C2 ,S = . , R3 C2- (x) (x). , [3]

    J1(M) =

    S

    (|x|)dS, J2(M) =

    (|x|)dx,

    x = (x1, x2, x3), dS . - M ,

    J(M) = J1(M) J2(M).

    - J1(M) , J2(M) = const. -, [1] - , .

    1.

    M , x3 -, = (t), P t(. 1). P t . - , t (a, b), 0 2, = (t) .

    . 1. . 11. 20072008

    .

    .

    ,..,

    20072008

    39

  • r(t, ) = {(t) cos , (t) sin , t},rt = { cos , sin , 1},r = { sin , cos , 0}.

    E = (rt, rt) = 2 + 1, F = (rt, r) = 0, G = (r, r) =

    2.

    dS2 = (2 + 1)dt2 + 2d2.

    . 1

    J1(M) =

    S

    (|x|)dS =2

    0

    d

    b

    a

    EG F 2dt = 2b

    a

    2 + 1dt,

    = ((t)).

    J2(M) =

    (|x|)dx =b

    a

    dt

    2

    0

    d

    0

    d = 2

    b

    a

    h((t))dt,

    40 .. , .. .

  • = ((t))

    h() =

    0

    (y)ydy.

    - . M J2(M), (t),

    (a) = A, (b) = B,

    , J1(M). , - J(M) = J1(M) J2(M). (t) , ,

    J [(t)] =

    b

    a

    (

    2 + 1 + h())dt. (1)

    t, F = F (, ), [2]

    d

    dt(F F) = 0.

    d

    dt

    (

    2 + 1 h() 2

    2 + 1

    )= 0. (2)

    = u. -

    (u)u

    u2 + 1 h(u) (u)uu2

    u2 + 1

    =

    (u)uu2 + 1

    h(u) = .

    = const.

    u2 =

    ((u)u

    + h(u)

    )2 1. (3)

    dt = du((u)u

    +h(u)

    )2 1

    .

    . 1. . 11. 20072008 41

  • t =

    du((u)u

    +h(u)

    )2 1

    + C. (4)

    , C.

    1. M R3 ,

    t =

    du((u)u

    +h(u)

    )2 1

    + C

    (2) (1).

    M .

    2.

    1. () = 1, () = 0.

    J [(t)] =

    b

    a

    2 + 1dt.

    (4)

    t =

    du(u

    )2 1

    + C = archu

    + C.

    = u = ch

    (t C

    )

    , .

    2. () = u32 , () = 0.

    J [(t)] =

    b

    a

    1

    2 + 1dt.

    t =

    du1

    2u 1

    + C.

    42 .. , .. .

  • 1/2 = k, u = k, = sin2 2

    t = k

    1 d + C =k

    2

    (1 cos )d + C = k

    2( sin ) + C.

    = u = k sin2

    2=

    k

    2(1 cos )

    .

    3.

    - , -. , () = () = 1. , - , , . , .

    2. (t) (2) t0 (a, b) . t = t0 , .

    . , , t0 = 0 , , t > 0 (t) , t < 0 .

    F (u) =1(

    (u)u+h(u)

    )2 1

    .

    (3)

    u2 = F2(u).

    t1 = , t2 = , > 0. u1 = u(t1),u2 = u(t2), u0 = u(0).

    u1

    u0

    F (z)dz = t1 = ,

    u0

    u2

    F (z)dz = t2 = .

    . 1. . 11. 20072008 43

  • ,

    u2

    u1

    F (z)dz = 0,

    F (u) : u1 = u2. - (2).

    Summary

    EXTREMALITY CONDITION OF SURFACE OF REVOLUTIONFOR AREA-TYPE FUNCTIONAL

    V.A. Klyachin, T.V. Tkacheva

    Present article is devoted to investigation of extremal rotation surfaces for squaretype functional. The solutions of differentional Euler-Lagrange equation are obtained.Also, the symmetry property of these surface is proved and demonstrated examplesfunctionals and its corresponding solutions are constructed.

    1. . . . .:, 1989. 312 .

    2. .., .., .. : . .: , 1986.

    3. .. // . . . . 2006. . 70, 4. C. 7790.

    4. .. // .. 260. 1981. 2. . 293295.

    44 .. , .. .