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47
Cauchy problem for the linearized KdV equation on general metric star graphs
Z.A. Sobirov1,2, M.I. Akhmedov2, H. Uecker3
1Faculty of Mathematics, National University of Uzbekistan, Vuzgorodok, 100047 Tashkent, Uzbekistan 2Applied Mathematics Department of Tashkent Financial Institute, 100000 Tashkent, Uzbekistan 3Institut fUr Mathematik, Universitat Oldenburg, D26111 Oldenburg, Germany
Keywords: KdV, initial value problem, PDE on metric graphs, exact solution, third order differential equation. Received: 2 February 2015
1. Introduction
The Korteweg-de Vries (KdV) equation has attracted much attention in the literature, both in the context of physics and mathematics. This equation was found to permit soliton solutions and allow the modeling of solitary wave propagation on a water surface, a phenomena first discovered by Scott Russell in 1834. The KdV equation is also used, e.g., to model the unidirectional propagation of small amplitude long waves in nonlinear dispersive systems such as ion-acoustic waves in a collisionless plasma, and magnetosonic waves in a magnetized plasma etc [11]. The linearized KdV provides an asymptotic description of linear, undirectional, weakly dispersive long waves, for example, shallow water waves. Earlier, it was proven that via normal form transforms, the solution of the KdV equation can be reduced to the solution for the linear KdVequation [12]. Belashov and Vladimirov [12] numerically investigated the evolution of a single disturbance u(0,x) = u0 exp(-x2/l2) and showed that in the limit l ^ 0, u0l2 = const, the solution of the KdV equation is qualitatively similar to the solution of the linearized KdV equation. Boundary value problems on half lines were considered in [2,5,7].
Here, summarizing and extending the results in [13], we investigate the linearized KdV equation on star graphs r with m + k semi-infinite bonds connected at one point, called the vertex. The bonds are denoted by Bj, j = 1, 2,..., k + m, the coordinate Xj on Bj is defined from to 0 for j = 1, 2,..., k, and from 0 to for j = k + 1,..., k + m such that on each bond, the vertex corresponds to 0. On each bond we consider the linear equation:
PACS 02.30.Em, 02.30.Jr
DOI 10.17586/2220-8054-2015-6-2-198-204
(1)
Below, we will also use x instead of Xj (j = 1, 2,..., k + m). We investigate an initial value problem, and using the method of potentials, construct solution formulas.
2. Formulation of the problems
To solve linear KdV equation on an interval, one needs to impose three boundary conditions (BC): two on the left end of the x-interval and one on the right end, (see, e.g., [5,6] and references therein). For the star graph with m + k semi-infinite bonds, we need to impose k + 2m BCs, which should also provide connection between the bonds. In detail, we require:
ui(0; t) = aUj(0; t), j = 2,k + m, (2)
u+(+0; t) = Bu-(—0; t), (3)
k k+m
J^a-Vxx(-0;t) = ^ a-1 u*xx(+0;t), (4)
i=1 i=k+1
for t > 0, where u-(x; t) = (u1(x, t),..., uk(x, t))T, u+(x; t) = (uk+1(x, t),..., uk+m(x, t))T, subscripts x and double x mean the first and the second order partial derivatives with respect to x, ak are non-zero constants, B is a m x k matrix.
Furthermore, we assume that the fj(x,t) and the initial conditions:
uj(x, 0) = u0j(x), x G Bj, (j = 1, 2, ...,k + m), (5)
are sufficiently smooth enough and bounded, and that u0 j satisfies the vertex conditions (2) -
(4).
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.
Here, we introduce some notation that will be useful in the following. For any vector,
v = (V1,V2, . . . ,Vk+m)T we put V = (V2, . . . , Vk+m)T, V- = (V1, . . . , Vk)T, V- = (V2, . . . , Vk)T, V+ = (Vk+1, . . . ,Vk+m)T.
3. Existence and uniqueness of solutions
Lemma 1. Let Ik — BTB be negatively defined matrix. Then the problem has at most one solution in H3(r).
Proof of Lemma 1. Using the equation (1) one can easily get:
b b d i „ x=b
— u2(x, t)dx = (2uj ujxx — u?J +2 fj (x,t)uj (x, t)dx
d t J x=a i
a a
for appropriate values of constants a and b. From this equality and vertex conditions (2) - (4) we have:
dt ||u(.,t)HÎ < (ôu"(0,t)) {Ik — BTB du-(0,t) + 2 ||u(,t)||r ||f(-,t)| According to condition of Lemma 1, we get:
t
||u(-,t)||r < ||u(-, 0) ||r + j ||f (-,r )|r dr. (6)
0
The inequality (6) proves the lemma.
Notice that equality in (6) (i.e. energy conservation) holds iff BTB = Im. We shall construct solutions and prove existence theorems for data from the Schwartz class of smooth decreasing functions, and for data in Sobolev classes.
Let S(Bk) be the Schwartz space of rapidly decaying functions on the closure of Bk,
dv
k = 1, 2, 3. We say v(x,t) e C1 ([0,T]; S(Bfc)) (T > 0) if v and — in C([0,T]; S(Bfc)). Theorem 1. Assume that Ik — BTB is negatively defined matrix, u0k(x) e S(Bk), fk (x, t) e
C1 ([0, T]; S(Bk)) /or íome T > 0 and that u$ =
d3p ( ) d3p -uofc (x) and
fk (x,t) satis/
dx3p k 5x3p"
vertex conditions (1) - (5) for any nonnegative integer p. Then (1) - (5) has a solution in
C1 ([0, T]; S(Bfc)).
To treat the case of Sobolev data consider function v = (v1(x1), v2(x2),..., vk+m(xk+m))
( ) d3p
defined on the graph. We suppose that vk e S(Bk) and the functions vk = 3p vk (x) satisfy
vertex conditions (2) -(4) for any non-negative integer p. We denote the set of all such functions v by S-(r) (S+(r)), and define W-(r) (or W +(r)) as the closure of the set S-(r) ( or S + (r)) with respect to the norm ||v||3r = k=1 ||vk||H3(Bfc).
Theorem 2. Let Ik — BTB be negatively defined matrix,
u0 = (u01(x1), u02(x2),..., u0 k+m(xk+m)) e W±(r). Then (1) - (5) has a unique solution in L^(0,T,W ±(r)).
First, we construct exact solutions, using the results from the theory of potentials for the linearized KdV equation. For that purpose, we give some preliminaries from [1,3,5].
4. Some preliminaries from potentials theory
The following functions are called fundamental solutions of the equation ut — uxxx = 0 (see [1,3,5,12]):
f
U(x,t; £,n)=< (t - n)1/3 V(t - n)1/37'
x —
U
0,
0
x - C
V(x,t; C,n)= <( (t - n)1/^ V(t - n)1/3
0,
if t > n; if t < n,
if t > n; if t < n,
where f (x)
n / x\ n / x\
3171)' 0(x) = 3171for x > 0, 0(x) = 0 for x < 0
31/3
and Ai(x) and Bi(x) are the Airy functions. The functions f (x) and 0(x) are integrable and
ro s , n ,,„, , , 2n
f-oo f (x)dx = —, f0 ™ f (x)dx = —, f0 ^ 0(x)dx = 0. We summarize some properties of
33 potentials for (1) from [3,5]. For given w, f and 0 let:
b t b u(x,t) = y U(x, t; C, 0M£)d£, v(x,t) = J j U(x,t; £,r)f (£,r)d£dr,
a 0 a
t t w(1)(x,t) = y Uxg(x,n; a,t)0(n)dn, w(2)(x,t) = J Vx?(x,n; a,t)0(n)dn.
n n
1
1
Lemma 2. a) Let u G BV([a, b]). Then u(x, t) satisfies ut — uxxx = 0 for t > 0 and:
lim u(x,t)=|nu(xo), if x0 G (a,b); (x,y)^(xo,o) I 0, if xo G (a,b).
b) Let f G L2((a, b) x (0, T). Then, v(x, t) satisfies ut — uxxx = nf (x, t) in (a, b) x (0, T], T > 0 and initial condition u(x, 0) = 0, x G (a, b).
c) If 0 G H 1(0,T), then :
lim w(1)(x,t) = —0(y), lim w(1)(x,t) = — n0(y), lim w(2)(x,t) = 0.
x^-a+0 3 x^a—0 3 x^-a+0
Now, we are ready to construct exact solutions for the considered problems. We assume that initial data and source terms in each bond are sufficiently smooth and bounded functions.
5. Integral formula for exact solution
We look for solution in the form:
t t u(*, t) = / U(x, t;0, ,)*(,,),*, + / V(*,0, + F(x, (), (7)
00
where
t
Fk(x, t) = 1 j U(x, t; £, 0)uk(£, 0)d£ + 1JJ U(x, t; £, n)fk(C, nRdrç. B 0 Bfc
0(t) = (01 (t),... ,0k+m(t))T, ^(t) = (0,..., 0,^k+1(t),... ,^k+m(t))T are unknown vector functions.
According to vertex conditions (2) - (4), we get:
t t t f (0) t f (0) ^ , t ^(0)
C J fv3 01(T )dT = J ^(T )dT+J (t^ ^)dT — (c F — f)
, (8)
x=0
0 0 0 t t
f'(0) , i ^'(0) .,,+ ,
0+ (t)dT + / , ,2/3^+(T)dT
(t — t)2/3^ w J (t — t)2/3
(9)
B/ fp0—(t)dT — (dF+(0,t) — B ■ dF"(0,t))
3
2(C—)Tf —(t) + (C+)Tf+(t) = 3 (—C"!C+) d2F(0,t), (10)
n v
where C = ( 1, —, —,..., —1
a2 a3 afc+m,
Abel's integral equations (8), (9) can be written in terms of fractional integrals [9]:
t
J(0,t)/(t) := pjO) /(t - r)a-1/(r0 < « < 1
o
and solved using the inverse operators, i.e. the Riemann-Liouville fractional derivatives [8,9] defined by:
D")fM-fïT-O)s/(t-T)-"f(t)dT, 0<a< 1
Using the relation D?o t)Ja
equations:
( o,t) ( o,t)
I from (8) and (9), we obtain the linear algebraic
f (0) (C| - Im+fc_l) 0(t) - (0(m+fc-1)xl|lm+fc-l) <^(0)^(t)
r(l/3)
c| - W-i) D2o/3t)F(0,i), (11)
f'(0) (B| - Im) 0(t) + ^'(0) (0mxfc| - Im) ^(t) = -
1 n1/3 (B| _ I
r(2/3)D(0'i) lB| Im
dF(0,t). (12)
We rewrite the system of equations (10), (11) and (12) in the following matrix form:
M
0(t)
G(t), G(t)
l
C| - Im+fc-^ D /
r(1/3)
1 ni/3 iB\ I
r(2/3)D(0'i) (B|
3
(o,t)F(0,t) d F(0,t)
-(-C-|C+)32F(0,t) n 1
/
where
M
( f(0)C-f (0)C+
-f (0)4-1
0
V
f '(0)B
2C -
mx(k-l)
0(fc-1) xm
-f (0)Im -f '(0)Im C+
°(fc- 1)xm
-0(0)Im -0'(0)Im
0
1xm J
Now we must prove that det(M) = 0.
Suppose that det(M) = 0. Then, the homogenous equation Ma = 0 has a nontrivial, time independent solution a = (a1, a2,..., ak+m) (notice that M is constant matrix). Therefore, putting in (7) 0(t) = 0O = (a1,..., ak+m)T = const, ^(t) = = (0,..., 0, ak+m+1,..., ak+2m) = const, we obtain a solution for the problem with u0(x) = 0, f(x,t) = 0. According to the uniqueness theorem, we have:
00
f
(t - T)1/3' V(t - T)1/3
x
dT +
(t - T)1/3 ^ (t - T)1/3
x
dT = 0,
or
0of + =0,
for any fixed t. The last equality contradicts the condition of linear independence for the Airy functions Ai(x) and Bi(x). This proves the statement det(M) = 0.
1
m
t
t
1
1
Summarizing the above studies, we obtain:
t
u(x,t) = F(x,t) + y U(x,t - T)M-1G(t)dr 0
with
tt/ U(x,t;0,0)Ifc | 0fc xm | 0fc xm
(X' Omxfc | U(x,t;0, 0)Im | V(x,t;0, 0)Im
6. Proof of existence theorems
Proof of Theorem 1. According to the theory of potentials [3,5], the solutions constructed in
the previous sections and their x-derivatives, up to the second order, are continuous functions
in the closure of each bondof the graph.
Now, we consider the functions Vj (xj , t) that are solutions of the considered problem
d3 d3 with initial conditions Vj(x, 0) = 3u0j(x), and with fj replaced by —3fj. According to the
d 3u-
conditions of the theorem, one can easily obtain . j (x, t) = v7 (x,t). From this, we conclude
dx3
that the Unctions Uj(x, t), j = 1, 2, 3 and their x-derivatives of any order are continuous functions in the closure of Bj.
Now, we consider the half lines corresponding to each bond separately. Notice that Uj (x,t) is a solution of the linearized KdV equation on the half line Bj and satisfies compatibility conditions at the point x = 0, t = 0. Applying Theorem 1.1 from [7], we get that these solutions define a C1 map from [0,T] into S(Bj).
Proof of Theorem 2. Above, we proved the estimate:
t
||u(,t)||r < ||u(, 0) ||r + y ||f (-,t )|r dT.
0
Note that for the function v, constructed above, the same estimate holds. Summing up these two estimates, we have:
||u(-,t)||3>r < ||u(-, 0) 113,r + J ||f (-,T )|3,r dT. (13)
0
By construction, S±(r) is dense in W±(r). This, together with the a priori estimate (13) proves the theorem. Thus, we have shown the existence and uniqueness of the solution for the linearized KdV equation on a metric star graph and derived its explicit solution. The above approach can also be extended to cases of graphs with different topologies.
t
Acknowledgement
We gratefully thank the Volkswagen Stiftung for financial support.
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