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ISSN 2070-0466, p-Adic Numbers, Ultrametric Analysis and Applications, 2012, Vol. 4, No. 1, pp. 57–63. 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

Abstract—The paper is concerned with the construction of exact solutions for nonlinear pseudod-ifferential 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 simplified system of nonlinear pseudodifferential equations describing tachyon dynam-ics of open-closed strings was proposed in [1–7]:

ψp2 = e−�/4ψ, (1.1)

ϕpψp(p−1)/2 = e−�/2ϕ, (1.2)

where ψ(t, x) and ϕ(t, x), x = (x1, x2, . . . , xd−1) are tachyonic fields for open and closed strings,� = −∂2

tt +Δ is d-dimensional d’Alembert 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 + . . . + x2d−1, x2 = t2 − |x|2.

In the system of equations (1.1)–(1.2) the equation (1.1) describes the dynamics of closed string ψand equation (1.2) describes the dynamics of open string ϕ for the given ψ.

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

ϕp = e−�/2ϕ. (1.3)

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

fn = e−σ0�f, f = f(t, x), (t, x) ∈ Rd, (1.4)

where for equation (1.1) of closed string one has f = ψ, σ0 = 1/4, n = p2 and for equation (1.3) of openstring one has f = ϕ, σ0 = 1/2, n = p.

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

limt→±∞

ψ(t) = 1, (1.5)

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

limt→±∞

ϕ(t) = ±1. (1.6)

∗The text was submitted by the author in English.**E-mail: [email protected]

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) [6–9]: 1) for allproblems the solution is not of constant sign; 2) problem (1.3)–(1.6) with odd p has an odddecreasing solution ϕp(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 ψ0(t) with multiplicity of zeros of function ψp2

0 less than 2p2

p−1 ;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 infinite number of derivatives has been done in manyphysical and mathematical papers (see [1–21] and bibliography within); many of these used computersimulation. In string field theory the interaction is nonlocal [5] which is completely different from localclassical field theory. These equations are of interest not only to p-adic mathematical physics but also tocosmology [18–21].

2. ONE-DIMENSIONAL CASE, d = 1

One-dimensional equation (1.4):

fn = eσ0∂2ttf ≡ (4πσ0)

−1/2

∫ ∞

−∞exp

[−(t− τ)2

4σ0

]f(τ)d τ , t ∈ R, (2.1)

is related to the heat conduction equation [15]:

U ′σ = σ0U

2tt, 0 < σ � 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 σ and t in (2.1) play the role of t and x in the classical heat conduction equation).Function U(σ, t) is called the interpolating function between f and its power fn. It is represented

by the Poisson formula for heat conduction equation (2.2)

U(σ, t) =1√4πσ

∫ ∞

−∞f(τ) exp

[−(t− τ)2

4σ

]d τ, 0 < σ � 1, t ∈ R. (2.4)

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

e−a∂2tte−bt2 = (1− 4ab)−1/2e−

b1−4ab

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 arereal numbers satisfying, 4ABσ0 < 1. Substituting into (2.1) and using (2.5) we get

AnenBt2 = A(1− 4Bσ0)−1/2 exp

( B

1− 4Bσ0t2),

which implies

An−1 = (1− 4Bσ0)−1/2, nB =

B

1− 4Bσ0.

Therefore

A = n1

2(n−1) , B =n− 1

4nσ0,

and finally

f(t) = n1

2(n−1) exp(n− 1

4nσ0t2). (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

ψ(t) = p1

p2−1 exp(p2 − 1

p2t2)

(2.7)

and for the open string

ϕ(t) = p1

2(p−1) exp(p− 1

2pt2). (2.8)

Now let us find the exact solution ϕ of (1.2) given known ψ which has the form (2.7). Assumingϕ(t) = AeBt2 , and 4σ0B < 1 and substituting into (1.2), we get

e1/2∂2ttAeBt2 = ApepBt2p

p2(p+1) exp

[(p2 − 1)(p − 1)

2pt2]

= A(1 − 2B)−1/2 exp( B

1− 2Bt2),

which implies the equations

pB +(p2 − 1)(p − 1)

2p=

B

1− 2B, Ap−1p

p2(p+1) = (1− 2B)−1/2, (2.9)

with the solutions

A = pp+2

2(p2−1) , B =p2 − 1

2p2(2.10)

and, finally:

ψ(t) = pp+2

2(p2−1) 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, . . . , xd−1)]

fn = eσ0Δ ≡ (4πσ0)− d−1

2

∫Rd−1

exp[−|x− y|2

4σ0

]f(y)dy. (3.1)

Its solution is the product of one-dimensional solutions

f(xi) = n1

2(n−1) exp[−n− 1

4nσ0x2i

], i = 1, 2, . . . , d− 1,

that is

f(x) = nd−1

2(n−1) exp(−n− 1

4nσ0|x|2

). (3.2)

The solutions for (d− 1)-dimensional open (f = ϕ, σ0 = 1/2, n = p) and closed (f = ψ, σ0 = 1/4, n =p2) strings follow from formula (3.2):

ϕ0(x) = pd−1p−1 exp

(−p− 1

p|x|2

), (3.3)

ψ0(x) = pd−1p2−1 exp

(−p2 − 1

p2|x|2

). (3.4)


60 VLADIMIROV

Consider now d-dimensional equation (1.3) of open string. Let ϕ0(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

ϕ(t, x) = ϕ0(t)ϕ1(x). (3.5)

It satisfies the boundary conditions

ϕ(±∞, x) = ±ϕ1(x), x ∈ Rd−1. (3.6)

Analogously the following function is the exact solution of the d-dimensional equation of closed string

ψ(t, x) = ψ0(t)ψ1(x), (t, x) ∈ Rd, (3.7)

where ψ0(t) is continuous (one-dimensional) solution of the boundary problem (1.1)–(1.5) for closedstring (if the solution exists). This solution satisfies the boundary conditions

ψ(±∞, x) = ψ1(x), x ∈ Rd−1. (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(n−1) exp[n− 1

4nσ0(t2 − |x|2)

], (t, x) ∈ R

d. (3.9)

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

ϕ(t, x) = pd

2(p−1) exp[p− 1

2p(t2 − |x|2)

], (3.10)

ψ(t) = pd

p2−1 exp[p2 − 1

p2(t2 − |x|2)

]. (3.11)

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

ψppdp

2(p+1) exp[−(p2 − 1)(p − 1)

2px2

]= e−�/2ϕ,ϕ = ϕ(t, x), (t, x) ∈ R

d. (3.12)

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

get

ApepBx2p

dp2(p+1) exp

[(p2 − 1)(p − 1)

2px2

]= A(1−2B)−d/2

exp( B

1− 2Bx2

),

which imply the equations for A and B

A(p−1)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(p2−1) , B =p2 − 1

2p2.

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

ϕ(t, x) = pd(p+2)

2(p2−1) exp(p2 − 1

2p2x2

). (3.14)

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



4. RELATION TO HYPERBOLIC EQUATIONS

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

σ−10

∂U

∂σ=

∂2U

∂t2−ΔU, U = U(σ, t, x), 0 < σ < 1, (t, x) ∈ R

d, (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(σ, 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, ξ), satisfying the estimate

f(t, ξ) � Cεe1−ε4σ0

t2(1 + |ξ|1−d−ε), (t, ξ) ∈ R

d, (4.3)

for all ε > 0. (By (4.3) function f(t, x) is bounded for x ∈ Rd−1 for all t ∈ R.) In this case the following

theorem holds.Theorem. If the condition (4.3) holds, the equation (1.4) is equivalent to the nonlinear

integral equation [15]

(4πσ0)−(d−1)/2

∫Rd−1

exp[−|x− y|2

4σ0

]fn(t, y)dy

= (4πσ0)−1/2

∫ ∞

−∞exp

[−(t− τ)2

4σ0

]f(τ , x)d τ . (4.4)

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

s−10

∂U

∂σ=

∂2U

∂t2+ |ξ|2U . (4.5)

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

U(σ, t, ξ) = eσ0σ|ξ|2v(σ, t, ξ). (4.6)

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

v′σ = σ0v′′tt

for every ξ ∈ Rd−1 and, thus,

v(σ, t, ξ) = (4πσ0σ)−1/2

∫ ∞

−∞exp

[−(t− τ)2

4σ0σ

]v(0, τ , ξ)d τ . (4.7)

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

v(1, t, ξ) = e−σ0|ξ|2U(1, t, ξ) = exp(−σ0|ξ|2)fn(t, ξ)

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

exp(−σ0|ξ|2)fn(t, ξ) = (4πσ0)−1/2

∫ ∞

−∞exp

[−(t− τ)2

4σ0

]f(τ , ξ)d τ .

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).


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

(4πσ0)− d−1

2 f−11 (x)

∫Rd−1

exp[−(x− y)2

4σ0

]fn1 (y)dy

= (4πσ0)−1/2f−n

0 (t)

∫ ∞

−∞exp

[−(t− τ)2

4σ0

]f0(τ)d τ . (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 λ. We get the following equations:

λf1(x) = (4πσ0)−(d−1)/2

∫Rd−1

exp[−(x− y)2

4σ0

]fn1 (y)dy, (5.2)

λfn0 (t) = (4πσ0)

−1/2

∫ ∞

−∞exp

[−(t− τ)2

4σ0

]f0(τ )d τ . (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 find these solutions in the form f1(x) = Ae−Bx2, −4Bnσ0 < 1. Substituting this

in (5.2), we get

An−1 = (1− 4Bnσ0)1/2λ, 4Bnσ0 = 1− (1− 4Bnσ0)

−1,

which implies

A = (nλ2)1

2(n−1 , B =n− 1

4nσ0.

We found the one-dimensional solution

f1(x) = (nλ2)1

2(n−1) exp(−n− 1

4nσ0x2

), x ∈ R, (5.4)

and the (d− 1)-dimensional solution

f1(x) = (nλ2)d−1

2(n−1) exp(−n− 1

4nσ0|x|2

), x ∈ R

d−1. (5.5)

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

f0(t) = (nλ−2)1

2(n−1) exp(n− 1

4nσ0t2), 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(n−1)λd−2n−1 exp

[n− 1

4nσ0(t2 − |x|2)

], (t, x) ∈ R

d. (5.7)

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

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-ofi-m-2011.



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