Electronic Journal of Differential Equations, Vol. 2009(2009), No. 08, pp. 1–9.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu (login: ftp)
THE CAUCHY PROBLEM FOR A SHORT-WAVE EQUATION
SILVIO MARQUES A. GAMA, GUEORGUI SMIRNOV
Abstract. We prove existence and uniqueness of solutions for the Cauchy
problem of the simplest nonlinear short-wave equation, utx = u − 3u2 , with
periodic boundary condition.
1. Introduction
In this paper we consider the Cauchy problem for the short-wave equation
utx = u − 3u2 ,
(1.1)
with the boundary condition (L > 0)
u(0, t) = u(L, t),
t ≥ 0,
(1.2)
∀x ∈ R.
(1.3)
and the L-periodic initial condition
u(x, 0) = φ(x),
Here, u(x, t) represents a small amplitude depending on one-dimensional (fast)
space variable x and (slow) time t.
Nonlinear evolution of long waves in dispersive media with small amplitude in
shallow water is a well known subject. It has been described by many mathematical models such as the Boussinesq equation [3, 8], the KdV equation [5], or the
Benjamin-Bona-Mahony-Peregrine equation (BBMP) [1, 7]. In contrast, for shortwaves, commonly called ripples, only a few results exist [6, 4, 2]. When we speak of
long or short-waves, we are referring to an underlying spacescale, X, to which all
space variables have been compared. Thus, for instance, for the surface-wave motion of a fluid, the unperturbed depth serves as a natural parameter. The shortness
of the waves is referred to this underlying parameter.
The short-wave equation (1.1) is derived in [6] via multiple-scale perturbation
theory from BBMP and governs the leading order term of the asymptotic dynamics
of short-waves sustained by BBMP. A first study of equation (1.1) was done in [4].
We sketch here its derivation. Start from BBMP
UT + UX − UXXT = 3(U 2 )X ,
(1.4)
which is the model equation for the unperturbed equation to which we will find
the short-wave limit. Here, U (X, T ) represents a small amplitude depending on
2000 Mathematics Subject Classification. 34A12, 34A34, 35Q35, 35Q53.
Key words and phrases. Cauchy problem; Benjamin-Bona-Mahony-Perigrine equation;
short-waves.
c
2009
Texas State University - San Marcos.
Submitted July 18, 2008. Published January 6, 2009.
1
2
S. M. A. GAMA, G. SMIRNOV
EJDE-2009/08
one-dimensional space variable X and time T . Its linear dispersion relation, ω(k),
is real (this means that we are not dealing with dissipative effects) and is given by
ω(k) =
k
,
1 + k2
(1.5)
having zero limit when k → ∞. The phase and group velocity are all bounded in
the short-wave limit k → ∞. This property allows BBMP to sustain short-waves.
In fact, let us consider a short-wave with characteristic length ` = ε ∼ k −1 , with
k 1. Define the scaled (fast) space variable x = ε−1 X (ε 1). The characteristic
time associated with short-waves is given by looking at the dispersive relation of
the linear part for the time variable. In our case, ω(ε−1 ) = ε − ε3 + ε5 − . . . . In
this way, we obtain the scaled (slow) time variable t = εT . We are lead thus to the
scaled variables x = ε−1 X and t = εT , which transforms the X and T derivatives
into ∂X = ε−1 ∂x and ∂T = ε∂t . Assume now the expansion U = u0 + εu1 + . . . .
Passing to the x and t variables and integrating in x, we have the lowest order in
(1.4) in the form
u0tx = u0 − 3(u0 )2 .
(1.6)
For simplicity, writing u0 as u, we obtain (1.1).
In the next section, under certain conditions, we prove the existence and uniqueness of solutions for (1.1)-(1.3).
2. Main result
Let u = u(x, t) be a classical solution to the Cauchy problem, that is, a twice
continuously differentiable function satisfying (1.1)-(1.3). Integrating the left-hand
side of (1.1) in x, from 0 to L, and using (1.2), we get
Z
d
d L
ux (x, t)dx =
(u(L, t) − u(0, t)) = 0.
dt 0
dt
Therefore, from (1.1), we have
Z
Z L
d L
0=
ux (x, t)dx =
u(x, t) − 3u2 (x, t) dx.
dt 0
0
(2.1)
Thus, it is natural to consider only initial conditions satisfying (2.1).
Note also that the L2 -norm of ux (·, t) is a constant. Indeed, multiplying both
sides of (1.1) by ux and integrating in x, from 0 to L, we obtain
Z
d L u2x (x, t)
1 d
|ux (·, t)|22 =
dx
2 dt
dt 0
2
Z L
=
u(x, t) − 3u2 (x, t) ux (x, t)dx
0
(2.2)
Z L
∂ u2 (x, t)
3
=
− u (x, t) dx
2
0 ∂x
u2 (L, t)
u2 (0, t)
=
− u3 (L, t) −
− u3 (0, t) = 0.
2
2
This observation is of importance in the proof of a global existence.
EJDE-2009/08
SHORT-WAVE EQUATION
3
We will seek for solutions to problem (1.1)-(1.3) in a generalized sense. Namely,
consider a formal Fourier series
∞
X
u(x, t) =
un (t)ei2πnx/L , u−n = un ,
(2.3)
n=−∞
with coefficients depending on t. Assume that
x ∈ R,
u(x, 0) = φ(x),
where φ is an L-periodic function. It is assumed that u−n = un or, equivalently,
u(x, t) ∈ R. Formally substituting Fourier series (2.3) in the differential equation
we obtain a system of ordinary differential equations
X
dun (t)
iL un (t) − 3
uα (t)uβ (t) , n 6= 0.
(2.4)
=−
dt
2πn
α+β=n, n∈Z
(Denote un (t) simply by un .) Note that, for n = 0, we do not obain a differential
equation for u0 , but a constraint relating u0 to all the others Fourier modes. Since
u0 is the real function u average value over the domain of periodicity, we obtain
the equation
X
u0 − 3u20 = 3
|un |2 .
(2.5)
n∈Z, n6=0
This equation admits real solutions
s
1
u0 =
1 ± 1 − 36
6
X
|un |2 ,
(2.6)
n∈Z, n6=0
P
only if n∈Z, n6=0 |un |2 ≤ 1/36. For definiteness assume from now on that the sign
in formula (2.6) is plus, for example. The other choice is essentially the same, the
major difference being the fact that it results in waves travelling in the opposite
direction [4].
Rewrite (2.4), in the integral form
Z t
X
iL
un (t) = φn −
un (s) − 3
uα (s)uβ (s) ds, n 6= 0,
(2.7)
2πn 0
α+β=n, n∈Z
Denote by H the space of complex sequences v = {vn }n∈Z with the norm
1/2
X
|v| = |v0 |2 +
n2 |vn |2
.
n∈Z, n6=0
The space of L-periodic functions u with Fourier coefficients {un }∞
n=−∞ ∈ H, we
shall also denote by H. Let
∞
X
φ(x) =
φn ei2πnx/L ∈ H,
n=−∞
with φ−n = φn . We say that a function u ∈ C [0, ∞), H ,
t → u(t) =
∞
X
un (t)ei2πnx/L ,
u−n = un ,
n=−∞
is a solution to problem (1.1)-(1.3), if u̇ ∈ L∞ [0, ∞), H , and the Fourier coefficients un satisfy (2.6), (2.7), and un (0) = φn , for all n.
4
S. M. A. GAMA, G. SMIRNOV
EJDE-2009/08
Now we are in a position to formulate the main result of this paper.
Theorem 2.1. If φ ∈ H satisfies
X
n2 |φn |2 < 1/72
L
Z
φ(x) − 3φ2 (x) dx = 0,
and
0
n∈Z, n6=0
then problem (1.1)-(1.3) has one and only one solution. For all t ≥ 0, Fourier
series (2.3) converges uniformly in x. Its sum is differentiable in x for almost
all x ∈ [0, L]. The derivative satisfies the conditions ux (·, t) ∈ L2 ([0, L], R) and
ux (x, ·) ∈ C([0, ∞[, R). Moreover, ux is differentiable in t and (1.1) holds for
almost all x ∈ [0, L].
Remark. The uniform convergence of Fourier series (2.3) implies that u(·, t) is a
continuous L-periodic function.
The proof of Theorem 2.1 is divided in several steps. First note that the condition
Z L
(φ(x) − 3φ2 (x))dx = 0,
0
implies
X
φ0 = 3|φ0 |2 + 3
|φn |2 .
n∈Z, n6=0
From this, we get
φ0 =
1
1±
6
s
X
1 − 36
|φn |2 .
(2.8)
n∈Z, n6=0
Since
X
|φn |2 ≤
n∈Z, n6=0
X
n2 |φn |2 < 1/72,
n∈Z, n6=0
it follows that φ0 is well defined. Let v(·) ∈ L∞ ([0, T ], H). The norm in this space
we shall denote by kvk. Define an operator f : L∞ ([0, T ], H) → L∞ ([0, T ], H) as
follows:
Z t
∞
X
iL
vk (s)vn−k (s) ds, n 6= 0,
(2.9)
fn (v(·))(t) = φn −
vn (s) − 3
2πn 0
k=−∞
s
X
1
f0 (v(·))(t) =
1 + 1 − 36
|fn (v(·))(t)|2 .
(2.10)
6
n∈Z, n6=0
Let M > 0. Denote by Φ ∈ L∞ ([0, T ], H) the constant function Φ(t) ≡ φ and
consider a complete metric space
VT M = {v(·) ∈ L∞ ([0, T ], H) : kv − Φk ≤ M }
with the metric induced by L∞ ([0, T ], H). We need the following auxiliary results.
P
Proposition 2.2. If n6=0 n2 |φn |2 < 1/72 and T is sufficiently small, then f is
well defined and is a contractive map from VT M into VT M .
Proof. Since
fn (v)(t) − fn (w)(t) = −
iL
2πn
Z th
0
(vn (s) − wn (s)) + 3
∞
X
((vk (s) − wk (s))vn−k (s)
k=−∞
i
+ wk (s)(vn−k (s) − wn−k (s)) ds,
n 6= 0,
EJDE-2009/08
we have
X
SHORT-WAVE EQUATION
5
n2 |fn (v)(t) − fn (w)(t)|2
n∈Z, n6=0
X
hZ th
n∈Z, n6=0
0
≤ (const)
+
∞
X
|vn (s) − wn (s)|
i i2
(|vk (s) − wk (s)||vn−k (s)| + |wk (s)||vn−k (s) − wn−k (s)| ds
k=−∞
Z th
X
≤ (const)t
+
∞
X
|vn (s) − wn (s)|
0
n∈Z, n6=0
i2
|vk (s) − wk (s)|(|vn−k (s)| + |wn−k (s)|) ds
k=−∞
Z th
X
≤ (const)t
n∈Z, n6=0
+
∞
X
|vn (s) − wn (s)|2
0
2 i
|vk (s) − wk (s)|(|vn−k (s)| + |wn−k (s)|)
ds
k=−∞
Z th
X
≤ (const)t
|vn (s) − wn (s)|2 + |v0 (s) − w0 (s)|2 (|vn (s)|2 + |wn (s)|2 )
0
n∈Z, n6=0
∞
i
X 1 X
k 2 |vk (s) − wk (s)|2 (|vn−k (s)|2 + |wn−k (s)|2 ) ds
+
2
k
k=−∞
k6=0
Z th X
X
≤ (const)t
|vn (s) − wn (s)|2 + |v0 (s) − w0 (s)|2
(|vn (s)|2
0
+ |wn (s)|2 )
n∈Z, n6=0
∞
X
n∈Z, n6=0
k 2 |vk (s) − wk (s)|2
k=−∞
≤ (const)t
Z th
0
1+
X
i
(|vn (s)|2 + |wn (s)|2 ) ds
n∈Z, n6=0
i
(|vn (s)|2 + |wn (s)|2 ) dskv − wk2
X
n∈Z, n6=0
2
≤ (const)T (1 + kvk2 + kwk2 )kv − wk2 .
We have thus proved the inequality
X
n2 |fn (v)(t) − fn (w)(t)|2 ≤ (const)T 2 (1 + kvk2 + kwk2 )kv − wk2 . (2.11)
n∈Z, n6=0
We also have
|f0 (v)(t) − f0 (w)(t)|2
s
X
1 =
1 − 36
36
|fn
(v)(t)|2
s
− 1 − 36
n∈Z, n6=0
≤ q
1 − 36
(const)
P
2
|fn (w)(t)|2 n∈Z, n6=0
2
n∈Z, n6=0 (|fn (v)(t)|
+ |fn (w)(t)|2 )
2
P
1 − 36 n∈Z, n6=0 |fn (w)(t)|2 P
n∈Z, n6=0
X
|fn (v)(t)|2 +
q
(2.12)
6
S. M. A. GAMA, G. SMIRNOV
X
×
EJDE-2009/08
|fn (v)(t) − fn (w)(t)|2 .
n∈Z, n6=0
The inclusion v ∈ VT M implies kvk2 ≤ (kΦk + kΦ − vk)2 ≤ (kΦk + M )2 . Since
Φ = f (0), from (2.11) we get
X
n2 |fn (v)(t) − φn |2 ≤ (const)T 2 (1 + (kΦk + M )2 )2 .
n∈Z, n6=0
Therefore
X
X
|fn (v)(t)|2 ≤ 2
n∈Z, n6=0
n2 |φn |2 + 2
n∈Z, n6=0
X
≤2
X
n2 |fn (v)(t) − φn |2
n∈Z, n6=0
2
2
n |φn | + (const)T 2 (1 + (kΦk + M )2 )2
n∈Z, n6=0
1
,
36
whenever T > 0 is small enough. Thus the map f is well defined (see (2.9) and
(2.10)). From (2.11) and (2.12) we obtain
X
|f0 (v)(t) − f0 (w)(t)|2 ≤ (const)
|fn (v)(t) − fn (w)(t)|2
≤σ<
n∈Z, n6=0
X
≤ (const)
n2 |fn (v)(t) − fn (w)(t)|2
n∈Z, n6=0
≤ (const)T 2 (1 + kvk2 + kwk2 )kv − wk2 .
Invoking again (2.11), we get
kf (v) − f (w)k2 ≤ (const)T 2 (1 + kvk2 + kwk2 )kv − wk2
≤ (const)T 2 (1 + (kΦk + M )2 )kv − wk2 .
(2.13)
In particular, we have
kf (v) − Φk2 ≤ (const)T 2 (1 + (kφk + M )2 )2 ≤ M 2 ,
for small T > 0. Thus we see that f : VT M → VT M and from (2.13) it follows that
f is a contraction, whenever T > 0 is small enough.
Proposition 2.3. Let u ∈ L∞ ([0, T ], H) be a solution to the equation u = f (u).
Assume that
X
n2 |un (t)|2 ≤ δ < 1/36.
n∈Z, n6=0
Then u ∈ C([0, T ], H) and u̇ ∈ L∞ ([0, T ], H).
Proof. Similarly to inequality (2.12) we have
|u(t2 ) − u(t1 )|2 = |u0 (t2 ) − u0 (t1 )|2 +
X
n2 |un (t2 ) − un (t1 )|2
n∈Z, n6=0
X
≤ (const)
n2 |un (t2 ) − un (t1 )|2 .
n∈Z, n6=0
From (2.9) we see that the right side of the inequality is less than or equal to
∞
2 X
X Z t2 (const)|t2 − t1 |
|u
(s)
+
3
|uk (s)||un−k (s)| ds
n
n∈Z, n6=0
t1
k=−∞
EJDE-2009/08
SHORT-WAVE EQUATION
+
X
k∈Z, k6=0
1
k2
X
1 + |u0 (s)|2
t1
n∈Z, n6=0
t2
Z
X
≤ (const)|t2 − t1 |
7
k 2 |uk |2
k∈Z, k6=0
X
|un (s)|2 ds
n∈Z, n6=0
2
≤ (const)|t2 − t1 | .
This proves the continuity of u(t). Since
2
P
9 n∈Z, n6=0 (u̇n u−n + un u̇−n )
P
≤ (const)
|u̇0 |2 =
1 − 36 n∈Z, n6=0 |un |2
X
X
|u̇n |2
n∈Z, n6=0
|un |2
n∈Z, n6=0
and
X
X
n2 |u̇n |2 =
n∈Z, n6=0
n∈Z, n6=0
L 2 un − 3
2π
X
2
uk un−k ,
n∈Z, n6=0
we have
X
|u̇0 |2 +
n2 |u̇n |2
n∈Z, n6=0
X
≤ (const)
un − 3
n∈Z, n6=0
X
≤ (const)
2
uk un−k X
n∈Z, n6=0
1 X
2
2
2
k
|u
|
|u
|
k
n−k
k2
k∈Z, k6=0
k∈Z, k6=0
X
X
2
2
k |uk |
|un |2 ≤ (const).
|un |2 + |u0 |2 |un |2 +
n∈Z, n6=0
≤ (const) 1 + |u0 |2 +
X
k∈Z, k6=0
1
k2
X
k∈Z, k6=0
n∈Z, n6=0
Thus u̇ ∈ L∞ ([0, T ], H).
Note that we also proved that the function u ∈ C([0, T ], H) is Lipschitzian. Now
show that generalized solutions also satisfy property (2.2).
Proposition 2.4. Assume that u ∈ L∞ ([0, T ], H) satisfies (2.6). Then
X
n2 |un (t)|2 = (const).
n∈Z, n6=0
Proof. Indeed, we have
∞
d X 2πn 2
|un |2
dt n=−∞ L
=
∞
X
n=−∞
=
2πn 2
(u̇n u−n + un u̇−n )
L
∞
∞
∞
i
X
X
2πi X h n un u−n − 3
uk u−n−k − u−n un − 3
uk un−k
L n=−∞
k=−∞
=−
6πi
S,
L
k=−∞
8
S. M. A. GAMA, G. SMIRNOV
where
S=
∞
X
n un
n=−∞
Observe that
∞
X
S=
∞
X
h
uk u−n−k − u−n
k=−∞
n,k=−∞
∞
X
i
uk un−k .
k=−∞
∞
X
nun uk u−n−k −
EJDE-2009/08
nu−n uk un−k = 2
n,k=−∞
∞
X
nun uk u−n−k .
n,k=−∞
On the other hand, introducing a new summation index m = n − k, we can rewrite
S in the form
∞
∞
∞
X
X
X
S=
nun uk u−n−k −
(m+k)u−m−k uk um = −
ku−m−k uk um .
n,k=−∞
m,k=−∞
m,k=−∞
Combining this with the previous equality, we get S = −S/2. Thus S = 0.
Proof of Theorem 2.1. From Proposition 2.2 we see that the problem under consideration has one and only one solution u ∈ L∞ ([0, T ], H), whenever T > 0 is
small enough. By Proposition 2.3 u ∈ C([0, T ], H) and u̇ ∈ L∞ ([0, T ], H). Finally,
Proposition 2.4 implies the existence of the solution for all t ≥ 0.
Show that, u(x, t), the sum of Fourier series (2.3) satisfies (1.1). From the
inequality
s
X
X
1 X
n2 |un (t)|2 = (const)
|un (t)| ≤
n2
n∈Z, n6=0
n∈Z, n6=0
n∈Z, n6=0
we see that Fourier series (2.3) converges uniformly in x for all t ≥ 0. The inequality
∞ X
∞
∞
∞
X
X
X
uk (t)un−k (t) ≤
|uk (t)|
|un (t)|
n=−∞ k=−∞
k=−∞
n=−∞
implies that the series
∞ X
∞
X
n=−∞
uk (t)un−k (t) ei2πnx/L
k=−∞
converges for all t ≥ 0. Multiplying (2.7) by ei2πnx/L and adding the obtained
equalities, we get
∞
∞
X
X
X Z t
2π
2π
i nun (t)ei2πnx/L =
i nφn ei2πnx/L +
un (s)
L
L
n=−∞
n=−∞
n∈Z, n6=0 0
X
−3
uα (s)uβ (s) ei2πnx/L ds
α+β=n, n∈Z
From the Lebesgue dominated convergence theorem and the above estimates we
have
Z t X X
ux (x, t) = φx (x) +
un (s) − 3
uα (s)uβ (s) ei2πnx/L ds.
0 n∈Z, n6=0
α+β=n, n∈Z
Combining this with (2.5), we obtain
Z
ux (x, t) = φx (x) +
0
t
(u(x, s) − 3u2 (x, s))ds.
EJDE-2009/08
SHORT-WAVE EQUATION
This completes the proof.
9
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Centro de Matemática da Universidade do Porto and Departamento de Matemática
Aplicada. Faculdade de Ciências da Universidade do Porto. Rua do Campo Alegre, 687,
4169-007 Porto, Portugal
E-mail address, S. M. A. Gama: smgama@fc.up.pt
E-mail address, G. Smirnov: gsmirnov@fc.up.pt