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NANOSYSTEMS: PHYSICS, CHEMISTRY, MATHEMATICS, 2015, 6 (6), P. 757-761
Linearized KdV equation on a metric graph
Z. A. Sobirov1'2, M.I. Akhmedov2, O.V. Karpova3'4, B. Jabbarova5
1Faculty of Mechanics and Mathematics, National University of Uzbekistan, Vuzgorodok,
100047 Tashkent, Uzbekistan 2Applied Mathematics Department of Tashkent Financial Institute, 100000 Tashkent, Uzbekistan 3Faculty of Physics, National University of Uzbekistan, Vuzgorodok, 100047 Tashkent,
Uzbekistan
4Turin Polytechnic University in Tashkent, Uzbekistan 5Urganch State University, Urganch, Uzbekistan sobirovzar@gmail.com, ola_july@mail.ru
PACS 02.30.Ik, 05.45.Yv DOI 10.17586/2220-8054-2015-6-6-757-761
We address a linearized KdV equation on metric star graphs with one incoming finite bond and two outgoing semi-infinite bonds. Using the theory of potentials, we reduce the problem to systems of linear integral equations and show that they are uniquely solvable under conditions of the uniqueness theorem. Keywords: KdV, IBVP, PDE on metric graphs, exact solution, third order differential equations. Received: 1 November 2015
Introduction
The Korteweg - de Vries (KdV) equation is of importance for many problems of physics and related fields. In particular, soliton solutions of KdV equation have found applications in fluid mechanics [1-11]. A pioneering study of KdV equation dates back to Scott Russell, who was able to model the propagation of solitary wave on the water surface in 1834. The linearized KdV provides an asymptotic description of linear, weakly dispersive long waves, such as, e.g., shallow water waves. Earlier, it was proven that via the normal form transforms the solution of the KdV equation can be reduced to the solution of the linear KdV equation [12]. Namely, Belashov and Vladimirov [12] numerically investigated evolution of a single disturbance u(0,x) = u0 exp(-x2/l2) and showed that in the limit l ^ 0, u012 = const, the solution of the KdV equation is qualitatively similar to that of the linearized KdV equation. The boundary value problems for KdV equation on half lines are considered in [2,5,7].
In this paper, we address the linearized KdV equation on a star graph r with one bounded bond and two semi-infinite bonds connected at one point, called the vertex. The bonds are denoted by Bj, j = 1, 2, 3, the coordinate x1 on B1 is defined from —1 to 0, and coordinates x2 and x3 on the bonds B2 and B3 are defined from 0 to such that on each bond the vertex corresponds to 0. On each bond we consider the linear equation:
d d3 ,
uj(xj,t) = fj(x, t), t> 0, xj e Bj, j = 1, 2, 3. (1)
dt Sx3
Below, we will also use the notation x instead of xj (j = 1, 2, 3). We treat a boundary value problem and using the method of potentials, reduce it to a system of integral equations. The solvability of the obtained system of integral equations is proven.
1. Formulation of the problems
To solve the linear KdV equation on an interval, one needs to impose three boundary conditions (BC): two on the left end of thex-interval and one on the right end, (see, e.g., [5-6] and references therein). For the above star graph, we need to impose 5 BCs at the vertex point, which should provide also connection between the bonds and 2 BCs at the left side of B\. In detail, we require:
ui(-1; t) = <Mt), M-l; t) = (t), (2)
«i(0,t) = a2u2(0,t) = a3u3(0,t), (3)
uix(0; t) = b2U2x(0; t) = b3u3x(0; t), (4)
uixx(0; t) = a-1U2xx(0; t) + a-1U3xx(0; t), (5)
for 0 < t <T, T = const.
Furthermore, we assume that the functions fj(x,t),j = 1, 2, 3, are smooth enough and bounded. The initial conditions are given by:
Uj(x, 0) = 0, x e Bj, (j = 1, 2, 3). (6)
It should be noted that the above vertex conditions are not the only possible ones. The main motivation for our choice is caused by the fact that they guarantee uniqueness of the solution and, if the solutions decay (to zero) at infinity, the norm (energy) conservation.
2. Existence and uniqueness of solutions
Lemma 1. Let br + A ^ 1. Then the problem (1)-(6) has at most one solution.
2 b3
Proof of Lemma 1. Using the equation (1) one can easily get:
b
— I u2j(x,t)dx = (2ujUjxx - + 2 I fj(x,t)uj(x,t)dx
a
for appropriate values of constants a and bon each bond. We put 0o(t) = 0. Then, the above equalities and vertex conditions (2)-(5) yield:
dt (e-£i Nf) ^ e-£t (¿j llf f + 0j(t)) , t
l|u||j e-£(t-T^^1j llf (-,T )|j + $(t)) dT, (7)
o
0 _
where (u,v) = f u1v1dx1 + f ujvjdxj + f u3v3dx3, ||u|| = \J(u, u) are L2 scalar product -10 0 and norm defined on graph, e is an arbitrary positive number. Uniqueness of the solution follows from (7).
Theorem 1. Let aj + a3 + aja3 + ^ + ^ = 0, b2 + b2 ^ 1,0o(t) G Cj[0,T], 0i(t) G
C1[0,T]. Then the problem (1) - (6) has a unique solution in C1 ([0,T], C3(r)). Proof of Theorem 1.
To prove the theorem, we use the following functions are called fundamental solutions of the equation ut — uxxx = 0 (see [1, 3, 5, 12, 16]):
f ((Ï--73) , t>n
U(x,t; i,rj)={ (t-n)17^ V(t-n)17V
0 t ^ n,
b
1
V (x,t; £,n) = i (t-n1)173 K (*V1/3) ,
0 t ^ n,
where f (x) = 31/3 Ai (—3173), <^(x) = 3173Bi (—373) for x ^ 0, <^(x) = 0 for x < 0 and Ai(x) and Bi(x) are the Airy functions. The functions f (x) and <^(x) are integrable and
0 +C» +
j f(x)dx = I, / f (x)dx = 2f, / ^(x)dx = 0. 0 0
Below, we also use fractional integrals [9]: ^ _ 1 rt
'(0,i) + '
J(W(t) := yo (t - T)a-1f (t)dT, 0 < a < 1,
D(o,t)/ (t):=^-I (t - t )-a/ (t )dT, 0 <a< 1.
and the inverse of this operator, i.e. the Riemann-Liouville fractional derivatives [8, 9] defined by:
1 d rt r(1 — a) dt J0 We look for solution in the form:
t t t
«i(x,t) = y U(x, t; 0, + y U(x,t; — 1,n)a(n)dn + J V(x,t; — (n)dn + Fi(x,t),
000 t t «2(x,t) = y U(x,t;0,n)^2(n)d^ + f V(x,t;0,n)^2(n)dn + F2(x,t), (8)
00 t t
ua(x,t) = y U(x,t;0,n)^a(n)dn + J V(x, t; 0, + Fj(x,t), (9)
00
where Ffc(x,t) = ± // U(x,t; (£,n)d£dn, k = 1, 2, 3. 0 B
Satisfying the conditions (2) - (4), we have:
/(0)a(t) + p(0)0(t) + D2/3 J ^i(n)/(--^-r)dt = -LD(2^3)[00(t) — Fi(0,t)], (10)
7 (t — n) /3 r( 3)
o
/'(0)a(t) + ^ (0)0 (t) + / ^i(n)/'(--^p)dt - Fix(0,t)], (11)
J (+-rn\ /3 i( 3)
0 (t - n)^ r(3)
/(0)^i(t) - 02/(0)^(t) - fl2^(0)^2(t) + / Kia(n)dn + /K^(n)dn
2, (12) D(o;t )[F2(
i8jDçO3»)[F2(0,t) - Fi(0,t)], /(0)^i(t) - 03/(0)^(t) - a3^(0)^3(t) + / Kia(n)dn + /K^(n)dn
ri)D2/3)[F3(0,t) - Fi(0,t)].
0 0 (13)
We take derivatives from (10) - (11) to obtain:
t o
1 (' 1 2 d
f (0)a'(t) + v(0)P'(t) + rrr)J = ^TyD*t) Jf fo>(t) - F1(0,t)], (14)
1 ^/3 d
f'(0)a'(t) + <ff(0)P'(t) + ^1)7 Km(n)dn = r(iyD^ Jf №i(t) - Fix(0,t)]. (15)
From conditions (4) and (5), it follows:
f'(0)pi(t) + b2f'(0)^2(t) - &2^(0)^2(t) + p(12y / K5a(n)dn + riT I K6P(V)dV
r( 3)
= f(i)D(0/3)[62F2x(0,t) - Fix(0,t)],
(16)
f'(0)^(t) + 63f'(0)^3(t) - bs^'(0)^3(t) + rmJ K5«(n)dn + rk J K6^(v)dv 1,
V3y 0 3 0
r(f)D(o/3)[63Fax(0,t) - Fix(0,t)],
-3^i(t) - ¥^2(t) - i?^3(t) + /K7a'(n)dn + '(n)dn
0 0
= F2xx(0,t) + a3F3xx(0,t) - Fixx(0,t),
where the kernels of integral operators defined as:
K
K
K
K
1 = / -2-T"
n (t - t)/3(t - n)/3 t
2 = -2-1"
n (t - T) /3(t - n) /3 t
= i 1
3 2 2
n (t - t)/3(t - n)/3 t
4 = 1 2~
n (t - T) /3(t - n)/3
f'
f'
f''
K
1
K
5 = -1-1"
n (t - t) /3(T - n)/3 t
6 = -1-r~
n (t - t)/3(t - n)/3
f'
(T - n)1/3,
N
(T - n)1/3,
(T - n)1/3,
1
(T - n)1/3
(T - n)1/3,
dT,
dT,
dT,
dT,
dT,
dT,
K7 = J U(y + 1; t - n)dy , K8 0
(t - n)/3,
x
[ V(y +1; t - n)dy.
(17)
(18)
t
t
1
t
x
We obtained the system of integral equations (12) - (18) with respect to unknowns (a'(t), 0'(t), ^1(t), <^2(t), <£3(t), ^2(t), ^3(t)). The matrix of the coefficients M of these unknowns on the off integral part of the system has a determinant:
A 4. A/f 2 , 2 , 2 2, a2 . «A det M = —- «2 + a2 + «2^3 + 77 + 7- .
81a2«3 V 03 62 /
Under conditions of the theorem this determinant is not singular.
According to the asymptotes of Airy functions the kernels of the integral operators are integrable (see [14, 15]). Hence, it follows from the uniqueness theorem and Fredholm alternatives that the system of equations has a unique solution. Thus the solvability of the problem is proved.
Acknowledgments
This work is partly supported by a grant of the Volkswagen Foundation. References
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