On some exact solutions in p-adic open-closed string theory

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  • ISSN 2070-0466, p-Adic Numbers, Ultrametric Analysis and Applications, 2012, Vol. 4, No. 1, pp. 5763. c Pleiades Publishing, Ltd., 2012.

    RESEARCH ARTICLES

    On Some Exact Solutionsin p-Adic Open-Closed String Theory

    V. S. Vladimirov**

    Steklov Mathematical Institute, Russian Academy of Sciences,Gubkina 8, Moscow 119991, Russia

    Received September 7, 2011

    AbstractThe paper is concerned with the construction of exact solutions for nonlinear pseudod-ierential equations which describe tachyon dynamics of open-closed p-adic strings. Existence ofcontinuous solutions and their properties are discussed.

    DOI: 10.1134/S2070046612010074

    Key words: strings, tachyons.

    Dedicated to Igor Vasilievich Volovich on the occasion of his 65th birthday

    1. INTRODUCTION

    The following simplied system of nonlinear pseudodierential equations describing tachyon dynam-ics of open-closed strings was proposed in [17]:

    p2= e/4, (1.1)

    pp(p1)/2 = e/2, (1.2)

    where (t, x) and (t, x), x = (x1, x2, . . . , xd1) are tachyonic elds for open and closed strings, = 2tt + is d-dimensional dAlembert operator, p = 2, 3, 5, . . . is a prime number (nonlinearityparameter). In the following we consider p and p2 as integer larger than one. We will use the standardnotations |x|2 = x21 + x22 + . . . + x2d1, x2 = t2 |x|2.

    In the system of equations (1.1)(1.2) the equation (1.1) describes the dynamics of closed string and equation (1.2) describes the dynamics of open string for the given .

    The equation (1.2) takes the form of equation for open string for = 1

    p = e/2. (1.3)

    We can write (1.1) and (1.3) in the following general form

    fn = e0f, f = f(t, x), (t, x) Rd, (1.4)where for equation (1.1) of closed string one has f = , 0 = 1/4, n = p2 and for equation (1.3) of openstring one has f = , 0 = 1/2, n = p.

    In 1-dimensional case d = 1 the following boundary conditions are proposed. For closed stringequation (1.1)

    limt(t) = 1, (1.5)

    and for open string (1.3) equation (where p is odd)

    limt(t) = 1. (1.6)

    The text was submitted by the author in English.**E-mail: vladim@mi.ras.ru

    57

  • 58 VLADIMIROV

    Let us list several theorems on existence and nonexistence of continuous one-dimensional solutions(d = 1) of mentioned above boundary problems for equations (1.1), (1.2), and (1.3) [69]: 1) for allproblems the solution is not of constant sign; 2) problem (1.3)(1.6) with odd p has an odddecreasing solution p(t) with a simple zero at t = 0; 3) problems (1.1)(1.5) with even p has noeven and increasing on t > 0 solution; 4) problem (1.2)(1.6) has solution when p 1(mod 4)and (1.1)(1.5) has even solution 0(t) with multiplicity of zeros of function

    p2

    0 less than2p2

    p1 ;5) there are no t-almost periodic solutions.

    It is natural to study the stated boundary problems in some algebras of generalized functions(distributions). Real continuous solutions are of interest for physics.

    Study of this new class of equations with innite number of derivatives has been done in manyphysical and mathematical papers (see [121] and bibliography within); many of these used computersimulation. In string eld theory the interaction is nonlocal [5] which is completely dierent from localclassical eld theory. These equations are of interest not only to p-adic mathematical physics but also tocosmology [1821].

    2. ONE-DIMENSIONAL CASE, d = 1

    One-dimensional equation (1.4):

    fn = e02ttf (40)1/2

    exp[(t )

    2

    40

    ]f()d , t R, (2.1)

    is related to the heat conduction equation [15]:

    U = 0U2tt, 0 < 1, t R (2.2)

    with the initial values

    U(0, t) = f(t), U(1, t) = fn(t), t R, (2.3)(notice that and t in (2.1) play the role of t and x in the classical heat conduction equation).

    Function U(, t) is called the interpolating function between f and its power fn. It is representedby the Poisson formula for heat conduction equation (2.2)

    U(, t) =14

    f() exp[(t )

    2

    4

    ]d , 0 < 1, t R. (2.4)

    To obtain an exact solution for (2.1) let us use the following well-known identity

    ea2ttebt

    2= (1 4ab)1/2e b14ab t2 (2.5)

    provided that 4ab < 1. We will search solution f for (2.1) of the form f(t) = AeBt2, where A and B are

    real numbers satisfying, 4AB0 < 1. Substituting into (2.1) and using (2.5) we get

    AnenBt2= A(1 4B0)1/2 exp

    ( B1 4B0 t

    2),

    which implies

    An1 = (1 4B0)1/2, nB = B1 4B0 .

    Therefore

    A = n1

    2(n1) , B =n 14n0

    ,

    and nally

    f(t) = n1

    2(n1) exp(n 14n0

    t2). (2.6)

    p-ADIC NUMBERS, ULTRAMETRIC ANALYSIS AND APPLICATIONS Vol. 4 No. 1 2012

  • ON SOME EXACT SOLUTIONS 59

    In particular, from (2.6) we get the exact solutions for the closed string

    (t) = p1

    p21 exp(p2 1

    p2t2)

    (2.7)

    and for the open string

    (t) = p1

    2(p1) exp(p 1

    2pt2). (2.8)

    Now let us nd the exact solution of (1.2) given known which has the form (2.7). Assuming(t) = AeBt

    2, and 40B < 1 and substituting into (1.2), we get

    e1/22ttAeBt

    2= ApepBt

    2p

    p2(p+1) exp

    [(p2 1)(p 1)2p

    t2]

    = A(1 2B)1/2 exp( B1 2Bt

    2),

    which implies the equations

    pB +(p2 1)(p 1)

    2p=

    B

    1 2B, Ap1p

    p2(p+1) = (1 2B)1/2, (2.9)

    with the solutions

    A = pp+2

    2(p21) , B =p2 12p2

    (2.10)

    and, nally:

    (t) = pp+2

    2(p21) exp(p2 1

    2p2t2). (2.11)

    Formula (2.11) was obtained in [5].

    3. MULTIDIMENSIONAL CASE, d > 1

    Dynamics of open-closed strings is described by the system of equations (1.1)(1.2)(1.3) whend > 1.

    Let us consider (d 1)-dimensional equation for open and closed strings [x = (x1, x2, . . . , xd1)]

    fn = e0 (40)d12

    Rd1

    exp[|x y|

    2

    40

    ]f(y)dy. (3.1)

    Its solution is the product of one-dimensional solutions

    f(xi) = n1

    2(n1) exp[n 14n0

    x2i

    ], i = 1, 2, . . . , d 1,

    that is

    f(x) = nd1

    2(n1) exp(n 14n0

    |x|2). (3.2)

    The solutions for (d 1)-dimensional open (f = , 0 = 1/2, n = p) and closed (f = , 0 = 1/4, n =p2) strings follow from formula (3.2):

    0(x) = pd1p1 exp

    (p 1

    p|x|2

    ), (3.3)

    0(x) = pd1p21 exp

    (p

    2 1p2

    |x|2). (3.4)

    p-ADIC NUMBERS, ULTRAMETRIC ANALYSIS AND APPLICATIONS Vol. 4 No. 1 2012

  • 60 VLADIMIROV

    Consider now d-dimensional equation (1.3) of open string. Let 0(t) be a continuous solution ofboundary problem (1.1)(1.6). This solution always exists if p is odd (see Section 1). Using (3.3) weobtain the exact formula for d-dimensional equation of open string

    (t, x) = 0(t)1(x). (3.5)

    It satises the boundary conditions

    (, x) = 1(x), x Rd1. (3.6)Analogously the following function is the exact solution of the d-dimensional equation of closed string

    (t, x) = 0(t)1(x), (t, x) Rd, (3.7)where 0(t) is continuous (one-dimensional) solution of the boundary problem (1.1)(1.5) for closedstring (if the solution exists). This solution satises the boundary conditions

    (, x) = 1(x), x Rd1. (3.8)The formulas (2.6) and (3.2) imply that the exact solution of d-dimensional equation (1.4) is given byformula

    f(t, x) = nd

    2(n1) exp[n 14n0

    (t2 |x|2)], (t, x) Rd. (3.9)

    In particular, this implies the exact solutions for multidimensional open and closed strings correspond-ingly

    (t, x) = pd

    2(p1) exp[p 1

    2p(t2 |x|2)

    ], (3.10)

    (t) = pd

    p21 exp[p2 1

    p2(t2 |x|2)

    ]. (3.11)

    Let us now nd solution of (1.2) using the known . Substituting (3.11) into (1.2) we obtain theequation for

    ppdp

    2(p+1) exp[(p

    2 1)(p 1)2p

    x2]= e/2, = (t, x), (t, x) Rd. (3.12)

    Let us nd solution of (3.12) in the form (t, x) = AeBx2. Substituting into (3.12) and using (2.5) we

    get

    ApepBx2p

    dp2(p+1) exp

    [(p2 1)(p 1)2p

    x2]= A(12B)

    d/2exp

    ( B1 2Bx

    2),

    which imply the equations for A and B

    A(p1)pdp

    2(p+1) = (1 2B)d/2, pB + (p2 1)(p 1)

    2p=

    B

    1 2B . (3.13)

    These equations coincide with equations (2.9) if we substitute A by A1/d. Therefore by (2.10) thesolutions of (3.13) are the numbers

    A = p(p+2)

    2(p21) , B =p2 12p2

    .

    As a result we obtain the Lorenz-invariant solution of equation (1.2) in the form

    (t, x) = pd(p+2)

    2(p21) exp(p2 1

    2p2x2

    ). (3.14)

    Formula (3.14) generalizes (2.11) to d-dimensional case.

    p-ADIC NUMBERS, ULTRAMETRIC ANALYSIS AND APPLICATIONS Vol. 4 No. 1 2012

  • ON SOME EXACT SOLUTIONS 61

    4. RELATION TO HYPERBOLIC EQUATIONS

    Equation (1.4) is related to the nonlinear boundary problem for parabolic equation (see Section 2)[15]

    10U

    =

    2U

    t2U, U = U(, t, x), 0 < < 1, (t, x) Rd, (4.1)

    with the initial values

    U(0, t, x) = f(t, x), U(1, t, x) = fn(t, x). (4.2)

    The function U(, t, x) is called interpolating function between the solution f(t, x) and its powerfn(t, x).

    We will consider equation (1.4) for the class of distributions f(t, x) with the Fourier transform w.r.t.x, f(t, ), satisfying the estimate

    f(t, ) Ce140

    t2(1 + ||1d), (t, ) Rd, (4.3)

    for all > 0. (By (4.3) function f(t, x) is bounded for x Rd1 for all t R.) In this case the followingtheorem holds.

    Theorem. If the condition (4.3) holds, the equation (1.4) is equivalent to the nonlinearintegral equation [15]

    (40)(d1)/2

    Rd1

    exp[|x y|

    2

    40

    ]fn(t, y)dy

    = (40)1/2

    exp[(t )

    2

    40

    ]f( , x)d . (4.4)

    Proof. Transforming (4.1) with Fourier transform w.r.t. x, we get the equation for U(, t, )

    s10U

    =

    2U

    t2+ ||2U . (4.5)

    Let us make the following substitution in the equation (4.5)

    U(, t, ) = e0||2v(, t, ). (4.6)

    As a result we will get the heat conduction equation for v

    v = 0vtt

    for every Rd1 and, thus,

    v(, t, ) = (40)1/2

    exp[(t )

    2

    40

    ]v(0, , )d . (4.7)

    But by (4.6) and (4.2) one has v(0, t, ) = U(0, t, ) = f(t, ),

    v(1, t, ) = e0||2U(1, t, ) = exp(0||2)fn(t, )

    and this implies for (4.7), when = 1,

    exp(0||2)fn(t, ) = (40)1/2

    exp[(t )

    2

    40

    ]f( , )d .

    Using the inverse Fourier transform and the Fubini theorem which can be applied due to inequality (4.3),we get (4.4).

    Moving backwards this logical sequence given (4.3), starting from (4.4), we get (1.4). This provesthe equivalence of (1.4) and (4.4).

    p-ADIC NUMBERS, ULTRAMETRIC ANALYSIS AND APPLICATIONS Vol. 4 No. 1 2012

  • 62 VLADIMIROV

    5. FOURIER METHOD FOR SOLUTION OF (4.4)

    Following the Fourier method, we look for solution f(t, x) of equation (4.4) in the form of a product ofthe two real functions which depend only on t and x correspondingly, f(t, x) = f0(t)f1(x), which implies

    (40) d1

    2 f11 (x)Rd1

    exp[(x y)

    2

    40

    ]fn1 (y)dy

    = (40)1/2fn0 (t)

    exp[(t )

    2

    40

    ]f0()d . (5.1)

    The left hand side of equation (5.1) depends only on x while the right hand side depends only on t. Thusboth parts are constant, we denote this constant . We get the following equations:

    f1(x) = (40)(d1)/2

    Rd1

    exp[(x y)

    2

    40

    ]fn1 (y)dy, (5.2)

    fn0 (t) = (40)1/2

    exp[(t )

    2

    40

    ]f0( )d . (5.3)

    Let us construct nontrivial (trivial are 0,1) exact solutions of (5.2) and (5.3). We will start from one-dimensional and d 1-dimensional solutions f1(x) of (5.2).

    As above we will nd these solutions in the form f1(x) = AeBx2, 4Bn0 < 1. Substituting this

    in (5.2), we get

    An1 = (1 4Bn0)1/2, 4Bn0 = 1 (1 4Bn0)1,which implies

    A = (n2)1

    2(n1 , B =n 14n0

    .

    We found the one-dimensional solution

    f1(x) = (n2)

    12(n1) exp

    (n 14n0

    x2), x R, (5.4)

    and the (d 1)-dimensional solution

    f1(x) = (n2)

    d12(n1) exp

    (n 14n0

    |x|2), x Rd1. (5.5)

    The equation (5.3) can be solved in the same way as equation (2.1). The solution is

    f0(t) = (n2)

    12(n1) exp

    (n 14n0

    t2), t R. (5.6)

    Multiplying (5.5) and (5.6) we will get the exact solution f(t, x) for (4.4) in the form

    f(t, x) = nd

    2(n1)d2n1 exp

    [n 14n0

    (t2 |x|2)], (t, x) Rd. (5.7)

    Notice that when = 1 formula (5.7) takes the form (3.9), and for d = 2 (5.7) does not depend on .

    ACKNOWLEDGMENTS

    The work is partially supported by the grant of Russian President NSh-2928.2012.1 and grants ofRussian Federation for Basic Research 11-01-00828-a and 11-01-12114-o-m-2011.

    p-ADIC NUMBERS, ULTRAMETRIC ANALYSIS AND APPLICATIONS Vol. 4 No. 1 2012

  • ON SOME EXACT SOLUTIONS 63

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    p-ADIC NUMBERS, ULTRAMETRIC ANALYSIS AND APPLICATIONS Vol. 4 No. 1 2012