Hindawi Publishing Corporation
International Journal of Mathematics and Mathematical Sciences
Volume 2007, Article ID 37274, 20 pages
doi:10.1155/2007/37274
Research Article
Comparison of KP and BBM-KP Models
Gideon P. Daspan and Michael M. Tom
Received 26 January 2007; Accepted 17 May 2007
Recommended by Nils Ackermann
It is shown that the solutions of the pure initial-value problem for the KP and regularized
KP equations are the same, within the order of accuracy attributable to either, on the
time scale 0 ≤ t ≤ −3/2 , during which nonlinear and dispersive effects may accumulate to
make an order-one relative difference to the wave profiles.
Copyright © 2007 G. P. Daspan and M. M. Tom. This is an open access article distributed
under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
1. Introduction
Bona et al. [1] compared the solutions of the Cauchy problem of the Korteweg-de Vries
(KdV)
Pt + Px + PPx + Pxxx = 0
(1.1)
and the Benjamin-Bona-Mahony (BBM) equation
Qt + Qx + QQx − Qxxt = 0.
(1.2)
In (1.1) and (1.2), P and Q are functions of two real variables x and t.
Such equations have been derived as models for nonlinear dispersive waves in many
different physical contexts, and in most cases where they arise, x is proportional to distance measured in the direction of wave propagation, while t is proportional to elapsed
time. Interest is often focused on the pure initial-value problem for (1.1) and (1.2) in
which P and Q are specified for all real x at some beginning value of t, say t = 0, and then
the evolution equation is solved for t ≥ 0 subject to the restriction that the solution respects the given initial condition. The thrust of their theory was that for suitably restricted
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initial conditions, the solutions P and Q emanating therefrom are nearly identical at least
for values of t in an interval [0,T], where T is quite large.
This paper is concerned with mathematical models representing the unidirectional
propagation of weakly nonlinear dispersive long waves with weak transverse effects. Interest will be directed toward two particular models. One is the Kadomtsev-Petviashvili
(KP) equations
ηt + ηx + ηηx + ηxxx
x
+ αη y y = 0
(1.3)
and the other is a regularized version, the Benjamin-Bona-Mahony (BBM-KP) equation
ξt + ξx + ξξx − ξxxt
x
+ αξ y y = 0,
(1.4)
where α = ±1. If α = +1 in (1.3), the equation is known as the KP II equation, while for
α = −1, it is the KP I equation. The Kadomtsev-Petviashvili equations are two-dimensional extensions of the Korteweg-de-Vries equation. They occur naturally in many physical contexts as “universal” models for the propagation of weakly nonlinear dispersive
long waves which are essentially one directional, with weak transverse effects. Observe
that the linearized dispersion relation for (1.3) is
w(k,l) = k 1 − k2 +
αl2
,
k2
(1.5)
while that of (1.4) is
k2 + αl2
,
w(k,l)
= k 1 + k2
(1.6)
within the scaling assumption that k2 = O(δ), and l = O(δ). As δ → 0, w(k,l) may be
to the same order of δ (since k2 l2 is of higher order).
approximated by w(k,l)
The Cauchy problem for these equations have been studied by a number of authors.
Bourgain [2], using Picard iteration, has proved that the pure initial-value problem
for the KP II equation is locally well-posed, and hence in light of the conservation laws
for the equation, globally well-posed for data in L2 (R). The same method has been used
to extend local well-posedness to some Sobolev spaces of negative indices.
A compactness method that uses only the divergence form of the nonlinearity and the
skew-adjointness of the linear dispersion operator was employed by Iório and Nunes in
[3] to establish local well-posedness for data in H s (R2 ), for s > 2, for the KP I equation.
The Iório-Nunes approach applies equally well to KP II-type equation. Molinet et al. [4]
using the local well-posedness of Iório and Nunes obtained a version of the classical energy method coupled with some of the known conserved quantities and delicate estimates
of Strichartz type for the KP I equation to show global well-posedness for KP I equation
in the space
Z = ϕ ∈ L2 R2 : ϕL2 + ϕxxx L2 + ϕ y L2
+ ϕxy L2 + ∂−x 1 ϕ y L2 + ∂−x 2 ϕ y y L2 < ∞ .
(1.7)
G. P. Daspan and M. M. Tom 3
Kenig [5] improved on this local well-posedness result given by the classical energy estimate by showing local well-posedness in the space
Ys = ϕ ∈ L2 R2 : ϕL2 + J s ϕL2 + ∂−x 1 ϕ y L2 < ∞
(1.8)
s f (k,l) = (1 + k 2 )s/2 f(k,l).
for s > 3/2, where J
In [6], global well-posedness is established in
Z0 = ϕ ∈ L2 R2 : ϕL2 + ∂−x 1 ϕ y L2 + ϕxx L2 + ∂−x 2 ϕ y y L2 < ∞
(1.9)
for the KP I equation. It is worth mentioning the result of Colliander et al. [7], dealing
with the well-posedness of KP I equation when the initial data has low regularity. Bona et
al. [8] have shown that (1.4) can be solved by Picard iteration yielding to local and global
well-posedness results for the associated Cauchy problem. In particular, it is shown that
the pure initial-value problem for (1.4), regardless of the sign of α, is globally well-posed
in
W1 = ϕ ∈ L2 R2 : ϕL2 + ϕx L2 + ϕxx L2 + ∂−x 1 ϕ y L2 + ϕ y L2 < ∞ .
(1.10)
Saut and Tzvetkov [9] improved this global well-posedness to the space
Y = ϕ ∈ L2 R2 : ϕx ∈ L2 R2 .
(1.11)
We remark that provided g satisfies an appropriate constraint, (1.3) and (1.4) are equivalent to the integrated forms
ηt + ηx + ηηx + ηxxx + α∂−x 1 η y y = 0,
ξt + ξx + ξξx − ξxxt + α∂−x 1 ξ y y = 0.
(1.12)
To illustrate the kind of results we have in mind, we briefly outline below what Saut and
Tzvetkov generally discussed concerning the relationship between the two models in [9].
Since both KP and BBM-KP equations model weakly nonlinear dispersive long waves
which are valid to order 2 , they can be considered as order perturbations of the linear
transport equation, and hence can be written as
ηt + ηx + η ηx + ηxxx ± ∂−x 1 η y y = 0,
ξt + ξx + ξ ξx − ξxxt ± ∂−x 1 ξ y y = 0,
(1.13)
(1.14)
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with initial data η0 , ξ0 , respectively, which are of order one. The neglected terms in
the right-hand sides of (1.13) and (1.14) are of order 2 . After performing the change of
variables
M = η,
x1 = −1/2 x,
N = ξ,
y1 = −1 y,
τ = −1/2 t,
(1.15)
one can rewrite (1.12) and (1.13) as
Mτ + Mx1 + MMx1 + Mx1 x1 x1 ± ∂−x11 M y1 y1 = 0,
Nτ + Nx1 + NNx1 − Nx1 x1 τ ± ∂−x11 N y1 y1 = 0,
(1.16)
with initial data, respectively,
M x1 , y1 ,0 = η0 1/2 x1 , y1 ,
(1.17)
N x1 , y1 ,0 = ξ0 1/2 x1 , y1 .
The dispersive and nonlinear terms in (1.13) and (1.14) may have a significant influence
on the structure of the waves on the time scale t1 = −1 (corresponding to τ1 = −3/2 ),
while the neglected order 2 terms might affect the solution at order one on time scale
t2 = −2 (τ2 = −5/2 ). It is therefore of interest to compare the solutions of (1.13) and
(1.14) on time scales between t1 , and t2 . Such analysis was performed for the KdV and
BBM models in [1].
To give an idea of the results one can expect that we consider the Cauchy problem for
the linear versions of (1.13) and (1.14)
ηt + ηx + + ηxxx ± ∂−x 1 η y y = 0,
ξt + ξx + − ξxxt ± ∂−x 1 ξ y y = 0
(1.18)
together with initial condition
η (x, y,0) = ξ (x, y,0) = f (x, y),
(1.19)
where f is of order one. Taking the Fourier transforms in both x and y variables, we have
η t + ikη + (ik)3 η ±
−l2
ik
η = 0,
il2
ξ t + ikξ + k ξ t ± ξ = 0,
k
2
(1.20)
G. P. Daspan and M. M. Tom 5
with η (k,l,0) = ξ (k,l,0) = f(k,l). Solving for η and ξ , there obtains
η = e−it[k−k
3 ± (l 2 /k)]
f(k,l,t),
(1.21)
2
2
ξ = e(−it/(1+k ))(k±(l /k)) f(k,l,t).
It is readily inferred that provided k5 f(k,l), l2 k f(k,l) ∈ L1 (R2 ), then
η (·, ·,t) − ξ (·, ·,t) ≤ η
(k,l,t) − ξ (k,l,t)L1 (R2 ) ≤ 2 tC f ≤ C
∞
(1.22)
since η and ξ are of order one, the estimate (1.22) proves that up to time scale t1 =
−(1+δ) , 0 < δ < 1, η and ξ are 1−δ close to each other.
2. Notations
We will employ the following notations. We will let | · | p , · s denote the norms in
L p (R2 ) and the classical Sobolev spaces H s (R2 ), respectively, where
f 2s =
2
s 1 + k2 + l2 f(k,l) dk dl
R2
(2.1)
Thus, the norm in L2 (R2 ) will simply be denoted
and connotes Fourier transformation.
by · 0 . By H ∞ , we denote s≥0 H s . The elements of H ∞ are infinitely differentiable
functions, all of whose derivatives lie in L2 . If X is an arbitrary Banach space and T > 0, the
space C(0,T,X) is the collection of continuous functions u : [0,T] → X. This collection is
a Banach space with the norm sup0≤t≤T u(t)X , where · X denotes the norm in X.
Define the space H−s 1 (R2 ) = {η ∈ S (R2 ) : ηH−s 1 (R2 ) < ∞} equipped with the norm
ηH−1 (R2 ) =
s
R2
1 + |k|−1
2
s 1 + k2 + l2 η(k,l) dk dl
2 1/2
.
(2.2)
3. Summary of existence theory
As earlier mentioned, the pure initial-value problems for these model evolution equations have been studied. For our analysis, we will use the result of [6] for the Cauchy
problem for KP I equation, and the result for the well-posedness of BBM-KP I equation
is contained in [8].
We will first consider the initial-value problems for the KP I and BBM-KP I equations
ηt + ηx + ηηx + ηxxx
ξt + ξx + ξξx − ξxxt
x
− η y y = 0,
(3.1)
x
− ξ y y = 0,
(3.2)
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International Journal of Mathematics and Mathematical Sciences
with initial condition
η(x, y,0) = ξ(x, y,0) = g(x, y).
(3.3)
The theoretical results relating to the initial-value problems (3.1) and (3.2) are presented
in the following theorems without proof.
Theorem 3.1. Let s ≥ 2. Then for any g ∈ H−s 1 (R2 ), there exist a positive T0 = T0 (| g |L∞ )
(limρ→0 T0 (ρ) = ∞) and a unique η of the integrated KP I equation (3.1) with initial data
g on the time interval [0,T0 ] satisfying η ∈ C([0,T0 ];H−s 1 (R2 )), ηt ∈ C([0,T0 ];H s−3 (R2 )).
Furthermore, the Map g → η is continuous from H−s 1 (R2 ) to C([0,T0 ];H−s 1 (R2 )).
Theorem 3.2. Let g ∈ H−s 1 with s > 3/2. Then, there exist T0 > 0 such that the BBM-KP I
equation (3.2) has a unique solution ξ ∈ C([0,T0 ];H−s 1 (R2 )), ∂−x 1 ξ y ∈ C([0,T0 ];H−s−11 (R2 )),
with ξt ∈ C([0,T0 ];H s−2 (R2 )). Moreover, the map g → ξ is continuous from H−s 1 (R2 ) to
C([0,T0 ] : H−s 1 (R2 )).
4. Main result
In this section, we compare the solutions of the initial-value problems
ηt + ηx + ηηx + ηxxx − ∂−x 1 η y y = 0,
(4.1)
−1
ξt + ξx + ξξx − ξxxt − ∂x ξ y y = 0
both with initial condition
η(x, y,0) = ξ(x, y,0) = g 1/2 x, y .
(4.2)
Our main result in this paper is the following.
2
Theorem 4.1. Let g ∈ H−k+5
1 (R ), where k ≥ 0 and let η and ξ be the unique solutions
guaranteed by Theorems 3.1 and 3.2 for the initial-value problems (4.1). Then there exist
positive constants C and T which depend only on k and g such that the solutions η and ξ of
the initial-value problems
ηt + ηx + ηηx + ηxxx − ∂−x 1 η y y = 0,
ξt + ξx + ξξx − ξxxt − ∂−x 1 ξ y y = 0,
(4.3)
η(x, y,0) = ξ(x, y,0) = g 1/2 x, y ,
G. P. Daspan and M. M. Tom 7
satisfy the inequalities
j
∂x η(·, ·,t) − ∂xj ξ(·, ·,t) ≤ C j/2+5/4 ,
0
(4.4)
j
∂ y η(·, ·,t) − ∂ yj ξ(·, ·,t) ≤ C j+5/4
0
for 0 < ≤ 1, and 0 ≤ t ≤ −3/2 min{T,T0 }, where 0 ≤ j ≤ k.
Before we prove Theorem 4.1, we introduce two new dependent variables
u(x, y,t) = −1 η −1/2 x + −3/2 t, −1 y, −3/2 t ,
(4.5)
v(x, y,t) = −1 ξ −1/2 x + −3/2 t, −1 y, −3/2 t .
A brief calculation shows that u and v satisfy, respectively, the initial-value problems
ut + uux + uxxx − ∂−x 1 u y y = 0,
vt + vvx + vxxx − vxxt − ∂−x 1 v y y = 0,
(4.6)
u(x, y,0) = v(x, y,0) = g(x, y).
By virtue of Theorems 3.1 and 3.2, the existence and uniqueness of u and v are assured.
Theorem 4.2. Let g ∈ H−s 1 (R2 ), where s ≥ 2, then the initial-value problem for both equa2
tions (4.6) have solutions in C([0,T0 ];H−s 1 (R2 )) for some T0 > 0. Moreover, if g ∈ H−k+5
1 (R ),
then there exist positive constants C and T depending only on k and g such that the difference
u − v satisfies
j
∂x u − ∂xj v ≤ C t,
0
j
∂ y u − ∂ yj v ≤ C t
0
(4.7)
for all and t for which 0 ≤ ≤ t and 0 ≤ t ≤ min[T;T0 ] where 0 ≤ j ≤ k.
We will make use of the following anisotropic Sobolev inequalities
1/2 1/4
f y fxx ,
| f |∞ ≤ 2 f 1/4
0
0
0
fx ≤ fxx 1/4 f y 1/2 fxxx 1/4 ,
0
0
0
∞
the proofs of which can be found in [10].
(4.8)
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International Journal of Mathematics and Mathematical Sciences
Proof of Theorem 4.2. Let w = u − v. Then, w is seen to satisfy
wt + wwx + wxxx − wxxt − ∂−x 1 w y y = −uxxt − (wu)x ,
(4.9)
w(x, y,0) = 0.
(4.10)
j
j
We now venture into the task of estimating ∂x w0 + ∂ y w0 , for j = 0,1,2,3,4,....
In light of Theorem 3.1, we note that
uk ≤ cg H−k 1 (R2 ) = Ak .
(4.11)
This fact justifies the following computations.
j
j
We first estimate ∂x w0 . Apply the operator ∂x to both sides of the differential equaj
tion (4.10), multiply the result by ∂x w, and integrate the result over R2 and over [0,t].
After a few integrations by parts, and taking into account the fact that w(x, y,0) ≡ 0, it
follows that
j+1 2 j 2
∂x w + ∂x w dx d y
R2
=2
t
R2
0
j
( j+2)
∂x w ∂x
( j+1)
uτ − ∂x
1
wu + w2
2
dx d y dτ,
(4.12)
j
which holds for j = 0,1,2,.... Similarly, apply the operator ∂ y on (4.10), multiply the
j
result by ∂ y w, and integrate the result over R2 and over [0,t]. After a few integrations by
parts, and taking into account the fact that w(x, y,0) ≡ 0, it follows also that
R2
j
2
j
∂ y w + ∂ y wx
2 dx d y = 2
t
0
1
j
j
j
∂ y w ∂ y uxxτ − ∂ y wu + w2
2
2
R
dx d y dτ
x
(4.13)
for j = 0,1,2,3,4,.... The relations (4.12) and (4.13) will be used repeatedly.
First, for j = 0, (4.12) or (4.13) may be used, since they will both given the same estimate. Making use of (4.12), there appears after two integrations by parts that
R2
w2 + wx2 dx d y =
t
0
R2
2 wuxxτ − ux w2 dx d y dτ,
(4.14)
from this, the following inequality is derived
w 20 ≤
t
0
2w0 uxxτ 0 + ux ∞ w20 dτ.
(4.15)
G. P. Daspan and M. M. Tom 9
By a variant of Gronwall’s lemma, it follows that
C2 C1 t
w 0 ≤ e − 1 ≤ tC2 eC1 = M0 t,
C1
(4.16)
where C1 and C2 are bounds for (1/2)ux ∞ and uxxt 0 , respectively, and t is restricted
to the range [0,1].
Using the equation satisfied by u, the following estimates can be derived:
sup uxxt 0 = sup uxy y − 3ux uxx − uuxxx − uxxxxx 0
t ≥0
t ≥0
1/2 1/4 u y uxx uxxx ≤ sup uxy y 0 + 2u1/4
0
0
0
0
(4.17)
t ≥0
5/4 1/2 1/4 + 6uxx 0 u y 0 uxxxx 0 + uxxxxx 0 .
Hence, C2 may be defined by supt≥0 uxxt 0 ≤ C2 . For C1 , we estimate it from the anisotropic Sobolev inequality
ux ≤ 2uxx 1/4 u y 1/2 uxxxx 1/4 ≤ C1 .
0
0
0
∞
(4.18)
We therefore infer by an application of Gronwall’s lemma that
w 0 ≤ C2 C1 t
e − 1 ≤ tC2 eC1 = M0 t ≤ B
C1
(4.19)
for 0 ≤ t ≤ 1.
For j = 1, integrate (4.12) by parts to get the following relation:
R2
2
wx2 + wxx
dx d y =
t
0
R2
2wx uxxxτ − wx3 − 3wx2 ux − wwx uxx dx d y dτ.
(4.20)
Similarly, for j = 1, integrate (4.13) by parts to get
R2
2
w2y + wxy
dx d y =
t
R2
0
2w y uxxyτ − 2wx w y u y +w2y ux +w y wuxy − w2y wx dx d y dτ.
(4.21)
Adding these two equations above, we obtain
R2
2
2
wx2 + w2y + wxx
+ wxy
dx d y
=
t
0
R2
2wx uxxxτ − wx3 − 3wx2 ux − wx wuxx + 2w y uxxyτ
− 2wx w y u y + w 2y ux + w y wuxy − w 2y wx dx d y dτ.
(4.22)
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International Journal of Mathematics and Mathematical Sciences
The integrand on the right-hand side of (4.22) may be bounded above by
2
2wx 0 uxxxτ 0 + 2w y 0 uxxyτ 0 + 4ux ∞ + vx ∞ wx 0
2 + 2ux ∞ + vx ∞ w y 0 + wx 0 w0 uxx ∞ + w y 0 w0 uxy ∞
(4.23)
+ 2wx 0 w y 0 u y ∞ .
Also from the anisotropic Sobolev inequalities, we infer that
u y ≤ 2u y 1/4 u y y 1/2 uxxy 1/4 ≤ C,
∞
0
0
0
ux ≤ 2uxx 1/4 u y 1/2 uxxx 1/4 ≤ C,
∞
0
0
0
(4.24)
uxx ≤ 2uxxx 1/4 uxy 1/2 uxxxx 1/4 ≤ C,
∞
0
0
0
uxy ≤ 2uxxy 1/4 u y y 1/2 uxxxy 1/4 ≤ C.
∞
0
0
0
Now using the equation satisfied by u, we may derive the following estimates, valid for
0 ≤ τ ≤ 1:
sup uxxxτ 0
t ≥0
1/2 1/4 u y uxx uxxxx ≤ sup uxxy y 0 + 2u1/4
0
0
0
0
t ≥0
1/4 1/2 1/4 2 + 8uxx 0 u y 0 uxxxx 0 uxxx 0 + 3uxx 0 + uxxxxxx 0 ≤ C,
sup uxxyτ 0
t ≥0
3/2 1/4 1/4
≤ sup uxxxy 0 + 4uxy 0 uxxx 0 uxxxx 0
t ≥0
1/4 1/2 1/4 1/4 1/2 1/4 + 6uxx 0 u y 0 uxxx 0 uxxy 0 + 2u y 0 u y y 0 u yxx 0 uxxx 0
1/4 1/2 1/4 + 2ux 0 u y 0 uxx 0 uxxxy 0 + uxxxxxy 0 + u y y y 0 ≤ C.
(4.25)
Putting all these estimates together, the right-hand side of (4.20) may be bounded above
by
2 2 2C4 wx 0 + w y 0 + C3 wx 0 + w y 0 ,
(4.26)
G. P. Daspan and M. M. Tom 11
where C3 and C4 are order-one quantities. The following is then seen to hold from (4.19)
2 2
wx + w y ≤
0
0
t
0
2 2 2C4 wx 0 + w y 0 + C3 wx 0 + w y 0 dτ.
(4.27)
Let A1 (t) = [wx 20 + w y 20 ]1/2 . Observe that
2 2 2
2 2 wx + w y = wx 0 + w y 0 + 2wx 0 w y 0 ≤ 2 wx 0 + w y 0 ,
0
0
(4.28)
so that
√
wx + w y ≤ 2A1 (t).
0
0
(4.29)
Therefore,
d wx 2 + w y 2 ≤ 2C4 wx + w y + C3 wx 2 + w y 2 ,
0
0
0
0
0
0
dt
(4.30)
which is equivalent to
√
d 2
A1 (t) ≤ 2C4 2A1 (t) + C3 A21 (t)
dt
(4.31)
√
d
1
A1 (t) ≤ C4 2 + C3 A1 (t),
dt
2
(4.32)
or
and hence by a variant of Gronwall’s inequality, we obtain
√
A1 (t) ≤ 2 2
√
C4 (1/2)C3 t
e
− 1 ≤ 2 2tC4 e(1/2)C3 = tM1 .
C3
(4.33)
We thus infer that
wx ≤ tM1 ,
0
w y ≤ tM1
0
(4.34)
for 0 ≤ t ≤ 1. Since C3 and C4 are order-one quantities, M1 is also an order-one quantity.
For j = 2, integrate (4.12) and (4.13) by parts, combine the two operations to get the
following:
R2
2
2
2
wxx
+ wxxx
+ w2y y + wxy
y dx d y
=
t
0
R2
2wxx uxxxxτ + 2w y y u y yxxτ + 6wx wxx uxx − 4w y w y y uxy
2
+ 2wwxx uxxx − ww y y uxy y − 4wx wxx
− 2w 2y y ux dx d y dτ.
(4.35)
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International Journal of Mathematics and Mathematical Sciences
The right side of (4.35) may be bounded above by
2 wxx 0 uxxxxτ 0 + w y y 0 uxxy yτ 0 + 6wx 0 wxx 0 uxx ∞
+ 4w y 0 w y y 0 uxy ∞ + 2w0 wxx 0 uxxx ∞ + w0 w y y 0 uxy y ∞ (4.36)
2 2 + 4wxx 0 wx ∞ + 2w y y 0 ux ∞ .
From the anisotropic Sobolev inequalities, we get the following estimates:
uxy y ≤ 2uxxy y 1/4 u y y y 1/2 uxxxy y 1/4 ≤ C,
∞
0
0
0
(4.37)
uxxx ≤ 2uxxxx 1/4 uxxy 1/2 uxxxxx 1/4 ≤ C.
∞
0
0
0
Using the differential equation satisfied by u, we also obtain the following estimates, valid
for 0 ≤ τ ≤ 1:
sup uxxxxτ 0
τ ≥0
5/4 1/2 1/4
≤ sup uxxxxx 0 + uxxx 0 uxy 0 uxxxx 0
τ ≥0
1/4 1/2 1/4 + 4uxx 0 u y 0 uxxx 0 uxxxx 0
(4.38)
1/4 1/2 1/4 + 2ux 0 u y 0 uxx 0 uxxxxx 0
+ uxxxxxxx 0 + uxxy y 0 ≤ C .
Similarly,
sup uxxy yτ 0
τ ≥0
1/4 1/2 1/4
≤ sup uxxxy y 0 + 8uxy y 0 uxxx 0 uxy 0 uxxxx 0
τ ≥0
5/4 1/2 1/4
1/4 1/2 1/4 + 14uxxy 0 u y y 0 uxxxy 0 + 6uxx 0 u y 0 uxxx 0 uxxy y 0
1/4 1/2 1/4 1/4 1/2 1/4 + 2u y y 0 u y y y 0 uxxy y 0 uxxx 0 +4u y 0 u y y 0 uxxy 0 uxxxy 0
1/2 1/4 u y uxx uxxxy y + uxxxxxy y + u y y y y ≤ C .
+ 2u1/4
0
0
0
0
0
0
(4.39)
Putting all these estimates together, the right of (4.35) is bounded by
2 2 2C6 wxx 0 + w y y 0 + C5 wxx 0 + w y y 0 ,
(4.40)
G. P. Daspan and M. M. Tom 13
1 ≤
where from previous estimates we already obtained w0 ≤ tM0 ≤ C0 , wx 0 ≤ t M
C1 , w y 0 ≤ tM1 ≤ C1 for 0 ≤ t ≤ 1. Consequently, we have that
wxx 2 + w y y 2 ≤
0
0
t
0
2 2 2C6 wxx 0 + w y y 0 + C5 wxx 0 + w y y 0 dτ,
(4.41)
where C5 , and C6 are order-one constants. Let
2 2 1/2
A2 (t) = wxx 0 + w y y 0 .
(4.42)
Note that
wxx + w y y 2 = wxx 2 + w y y 2 + 2wxx w y y ≤ 2 wxx 2 + w y y 2 ,
0
0
0
0
0
0
0
0
(4.43)
so that
√
wxx + w y y ≤ 2A2 (t),
0
0
(4.44)
and (4.44) is equivalent to
d wxx 2 + w y y 2 ≤ 2C6 wxx + w y y + C5 wxx 2 + w y y 2 ,
0
0
0
0
0
0
dt
(4.45)
from which we have
√
d 2
A2 (t) ≤ C6 2A2 (t) + C5 A22 (t)
dt
(4.46)
√
d
1
A2 (t) ≤ 2C6 2 + C5 A2 (t).
dt
2
(4.47)
or
Apply Gronwall’s inequality to obtain
√
A2 (t) ≤ 2 2
√
C6 (1/2)C5 t
e
− 1 ≤ 2 2tC6 e(1/2)C5 = tM2 .
C5
(4.48)
From (4.44), we infer that
wxx ≤ tM2 ,
0
w y y ≤ tM2 .
0
(4.49)
14
International Journal of Mathematics and Mathematical Sciences
j
j
For the general case j, the procedure for obtaining a bound on ∂x w0 + ∂ y w0 is similar
to that followed above in the cases j = 0, j = 1, and j = 2. Suppose inductively that for
j < k, where k > 1, there have been established bounds of the form
j j ∂x w + ∂ y w 0
0
≤
C2 j C2 j −1 t
e
− 1 ≤ tC2 j eC2 j −1 = tM j
C2 j −1
(4.50)
for t ∈ [0,1], where C2 j −1 and C2 j are both order-one quantities. The goal now is to
establish the proof for j = k. Before we proceed with the proof, we add (4.12) and (4.13)
together to get the relation
2
2
2
∂kx w + ∂(k+1)
w + ∂ky w + ∂ky wx
x
R2
=
t
0
2 dx d y
1
∂kx w 2∂(k+2)
uτ − 2∂(k+1)
wu + w2
x
x
2
R2
t
+
0
1
∂ky w 2∂ky uxxτ − 2∂ky wu + w2
2
2
R
(4.51)
dx d y dτ.
x
Using Leibniz rule, (4.51) may be written as
R2
2
2
2
∂kx w + ∂(k+1)
w + ∂ky w + ∂ky wx
x
= 2
t
−2
k+1
R2 j =0
0
t
0
R2
t
0
dx d y
∂kx w∂(k+2)
uτ dx d y dτ
x
t
+ 2
−2
R2
0
2 1 (k+1− j) j
(k+1− j)
j
w∂x w + ∂x
w∂x u ∂kx w dx d y dτ
∂x
2
∂ky w∂ky uxxτ dx d y dτ
k
R2
αj
(k− j)
β j ∂y
j
(k− j)
w∂ y wx + ∂ y
j
(k− j)
wx ∂ y u + ∂ y
j
w∂ y ux ∂ky w dx d y dτ.
j =0
(4.52)
G. P. Daspan and M. M. Tom 15
Here, the α j and β j are the constants that appear in Leibniz rule. Separating the top-order
derivatives and directly the rest estimating, we have that
R2
≤ 2
2
2
2
∂kx w + ∂(k+1)
w + ∂ky w + ∂ky wx
x
t
0
k (k+2) ∂ w ∂
uτ 0 dτ − 2
x
x
0
t
+
R2
0
+ 2
−2
k+1
t
0
0
0
w∂kx w∂(k+1)
w + u∂kx w∂(k+1)
w dx d y dτ
x
x
w∂x wdx d y dτ + 2
j
R2
t
0
k+1
R2
j
j
(k+1− j)
α j ∂kx w∂x u∂x
wdx d y dτ
j =1
R2 j =1
k
R2
u∂ky w∂ky wx + ux ∂ky w∂ky w + wx ∂ky w∂ky w dx d y dτ
k
t
+2
R2
k k
∂ w ∂ uxxτ dx d y dτ
y
y
0
0
t
+2
(k+1− j)
0
dx d y
j =1
t
0
α j ∂kx w∂x
t
2 (k− j)
β j ∂ky w∂ y
wx ∂ y udx d y dτ + 2
k− j
j
t
0
k
R 2 j =1
(k− j)
β j ∂ky w∂ y ux ∂ y
wdx d y dτ
β j ∂ky w∂ y w∂ y wx dxd ydτ.
j
j =1
(4.53)
j
The induction hypothesis (4.50) assures us that on the time interval [0, t], ∂x w0 ,
j
j
j
|∂ix w |∞ , ∂ y w 0 , ∂ y wx 0 , ∂ y wx ∞ , and |∂iy w |∞ are bounded by order-one constants
if 0 ≤ j ≤ k − 1, and 0 ≤ i ≤ k − 2. Using the elementary inequalities, the following
estimates may be obtained:
j ∂x u ≤ 2∂xj u1/4 ∂xj u y 1/2 ∂xj+2 u1/4 ≤ C,
0
0
0
∞
(4.54)
j ∂ y u ≤ 2∂ yj u1/4 ∂ yj+1 u1/2 ∂ yj uxx 1/4 ≤ C,
0
0
0
∞
and from the differential equation satisfied by u, we get the following estimates:
k+3 ∂ u + 3
sup ∂k+2
x uτ 0 ≤ sup
x
0
τ ≥0
τ ≥0
+
k
j =0
k
j =0
k+1− j
α j ∂ x
u 0 ∂ x u 0
j+2
k− j j+3 k+1
β j ∂x u0 ∂x u0 + ∂k+5
x u 0 + ∂x u y y 0
≤ C.
(4.55)
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International Journal of Mathematics and Mathematical Sciences
Similarly,
k
k
k− j j
sup ∂ y uxxτ 0 ≤ sup ∂ky uxxx 0 + 3 γ j ∂ y ux 0 ∂ y uxx 0
τ ≥0
τ ≥0
j =0
k
+
j =0
k− j j
δ j ∂ y u0 ∂ y uxxx 0 + ∂ky uxxxxx 0 + ∂k+2
y ux 0
≤C
(4.56)
for 0 ≤ j ≤ k + 1, independent of τ ≥ 0, and the α j , β j , γ j , and δ j are the constants that
appear in Leibniz rule. Because of Theorems 3.1 and 3.2, |wx |∞ , |w y |∞ , and |w|∞ are
similarly bounded. The second and the sixth integral on the right-hand side of (4.53) are
bounded, respectively, as follows:
t
−2
2
∂kx w dx d y dτ
−
wx + ux
0 R2
t 1
0
2
R2
ux + wx
∂ky w
2
≤C
t
0
k 2
∂ w dτ,
dx d y dτ ≤ C
x
t
0
0
k 2
∂ w dτ,
y
(4.57)
0
where, provided 0 ≤ t ≤ 1, C may be inferred to be an order-one quantity. The other
integrals on the right side of (4.53) may be estimated as follows:
t
0
k+1
R 2 j =1
(k+1− j)
j j
(k+1− j) α j ∂kx w∂x
w∂x w + 2∂kx w∂x u∂x
w dx d y dτ
t
+2
k
R 2 j =1
0
t
+2
R2
0
≤C
t
0
k
(k− j)
j j
(k− j) β j ∂ky w∂ y wx ∂ y u + ∂ky w∂ y ux ∂ y w dx d y dτ
k− j
j =1
k 2
wx ∂ w dτ + C x
0
∞
t
+C
+ C
+ C
+ C
0
β j ∂ky w∂ y w∂ y wx dx d y dτ
j
t
k k+1
2
∂ w Mk+1− j M j τ dτ
x
0
0
j =2
t
k k+1
k 2
2
∂ w ux ∂ w dτ + C Nk+1− j N j τ dτ
x
x
0
∞
0
0
t
0
t
0
t
0
j =2
k
k k −1 2
∂ w ux ∂ w dτ +
P
P
τ
dτ
k− j j
y
y
0
∞
0
j =2
k
k k −1 2
∂ w ∂ w uxy +
Qk− j Q j τ dτ
y
y
0
0
∞
j =2
k
k k −1 2
∂ w ∂ w wxy +
Rk− j R j τ dτ,
y
y
0
0
∞
j =2
(4.58)
G. P. Daspan and M. M. Tom 17
valid at least for 0 ≤ t ≤ 1. The constants appearing in these inequalities are order one.
Hence, for 0 ≤ t ≤ 1,
k 2 k 2
∂ w + ∂ w ≤
x
0
y
0
t
2 2 C2k ∂kx w 0 + ∂ky w 0 + C2k−1 ∂kx w 0 + ∂ky w 0 dτ,
0
(4.59)
where C2k−1 and C2 k are order-one quantities. As previously seen, if
2 2 1/2
Ak (t) = ∂kx w0 + ∂ky w0 ,
(4.60)
then an application of Gronwall’s inequality will lead to
√
Ak (t) ≤ 2
√
C2k C2k−1 t
e
− 1 ≤ 2tC2k eC2k−1 = tMk .
C2k−1
(4.61)
We therefore infer that
k ∂ w ≤ tMk ,
x
0
(4.62)
k ∂ w ≤ tMk .
y
0
The result (4.50) for j = k now follows and the inductive step is completed.
It is worth noting that the constants C2k−1 and C2k depend only on g( j) 0 for 0 ≤ j ≤
k + 5. This concludes the proof of Theorem 4.2.
Proof of Theorem 4.1. From (4.5), we have
u(x, y,t) = −1 η −1/2 x + −3/2 t, −1 y, −3/2 t ,
(4.63)
v(x, y,t) = −1 ξ −1/2 x + −3/2 t, −1 y, −3/2 t .
Inverting the above relations, we get
η(x, y,t) = u 1/2 x − 3/2 t, y, 3/2 t ,
(4.64)
ξ(x, y,t) = v 1/2 x − 3/2 t, y, 3/2 t .
It follows that
j
j
j
j
∂x η(x, y,t) = j/2+1 ∂x u 1/2 x − 3/2 t, y, 3/2 t ,
(4.65)
∂x ξ(x, y,t) = j/2+1 ∂x v 1/2 x − 3/2 t, y, 3/2 t .
Thus,
j
∂x η(·, ·,t) − ξ(·, ·,t) 2 = j+2
0
R2
j
∂x u 1/2 x − 3/2 t, y, 3/2 t
2
j − ∂x v 1/2 x − 3/2 t, y, 3/2 t dx d y.
(4.66)
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International Journal of Mathematics and Mathematical Sciences
Now, change variables by letting z1 = 2 x − 3/2 t, z2 = y, and dx d y = |J |dz1 dz2 =
−3/2 dz1 dz2 . The relation above then becomes
j
∂x η(·, ·,t) − ξ(·, ·,t) 2 = j+1/2
0
j
∂x u 1/2 x − 3/2 t, y, 3/2 t
R2
2
j − ∂x v 1/2 x − 3/2 t, y, 3/2 t dz1 dz2
j 2
= j+1/2 ∂x w ·, ·, 3/2 t 0
(4.67)
2
2
≤ j+1/2 · M j 3/2 t
= j+5/2 M j 3/2 t
for 0 ≤ 3/2 t ≤ 1. Hence if 0 ≤ t ≤ −3/2 , then
j
∂x η(·, ·,t) − ∂xj ξ(·, ·,t) ≤ j/2+5/4 M j 3/2 t .
0
(4.68)
Similarly,
j
j
j
j
∂ y η(x, y,t) = j+1 ∂ y u 1/2 x − 3/2 t, y, 3/2 t ,
(4.69)
∂ y ξ(x, y,t) = j+1 ∂ y v 1/2 x − 3/2 t, y, 3/2 t ,
so that
j
∂ y η(·, ·,t) − ∂ yj ξ(·, ·,t)2 = 2 j+2
0
R2
j
∂ y u 1/2 x − 3/2 t, y, 3/2 t
2
j − ∂ y v 1/2 x − 3/2 t, y, 3/2 t dx d y
= 2 j+1/2
R2
j
j
∂ y u(z1 ,z2 , 3/2 t − ∂ y v z1 ,z2 , 3/2 t
2
dz1 dz2
j 2
= 2 j+1/2 ∂ y w ·, ·, 3/2 t 0
2
2
≤ 2 j+1/2 · M j 3/2 t
= 2 j+5/2 M j 3/2 t
(4.70)
for 0 ≤ 3/2 t ≤ 1. Hence if 0 ≤ t ≤ −3/2 , then
j
∂ y η(·, ·,t) − ∂ yj ξ(·, ·,t) ≤ j+5/4 M j 3/2 t ,
0
where M j is of order one.
(4.71)
Corollary 4.3. Assume the hypothesis in Theorem 4.1. Then, there are constants K j such
that
j
∂x η(x, y,t) − ∂xj ξ(x, y,t) ≤ 2+ j/2 K j 3/2 t
∞
for 0 ≤ j ≤ k − 2 and 0 ≤ t ≤ −3/2 min{T,T0 }, and j ≤ k/2.
(4.72)
G. P. Daspan and M. M. Tom 19
Proof. Note that
j
∂x η(x, y,t) − ∂xj (x, y,t) ≤ 2∂xj (η − ξ)1/4 ∂xj η y − ξ y 1/2 ∂xj+2 (η − ξ)1/4 .
0
0
0
∞
(4.73)
Now observe that
j
∂x ∂ y (η − ξ) ≤ ∂2x j (η − ξ)1/2 ∂2 (η − ξ)1/2 ,
y
0
0
0
(4.74)
so that
j
∂x (η − ξ) ≤ 2∂xj (η − ξ)1/4 ∂2x j (η − ξ)1/4 1/4 ∂2 (η − ξ)1/4 ∂xj+2 (η − ξ)1/4
y
0
0
0
0
0
∞
1/4 2 j/2+5/4
1/4 13/4 1/4 ( j+2)/2+5/4 1/4
≤ j/2+5/4 m j
m2 j
mj
m1
= j/8+5/16+2 j/8+5/16+13/16+( j+2)/8+5/16 m
= j/8+5/16+2 j/8+5/16+13/16+( j+2)/8+5/16 m
= 2+ j/2 m
3/2 t .
(4.75)
From which, with j = 0, we get the following interesting case:
η(x, y,t) − ξ(x, y,t) ≤ K0 7/2 t
∞
(4.76)
holding for 0 ≤ t ≤ −3/2 . Inequality (4.76) is exactly what was obtained when comparing
solutions of the two linearized model evolution equations. Note in particular that at t =
t1 = −3/2
η x, y,t1 − ξ x, y,t1 ≤ K0 2 ,
∞
(4.77)
showing explicitly that the two solutions are the same in the formal order of accuracy 2
achieved by either model. Similarly, we state without proof the following corollary.
Corollary 4.4. Assume the hypothesis in Theorem 4.1. Then, there are constants K j such
that
j
∂ y η(x, y,t) − ∂ yj ξ(x, y,t) ≤ 2+ j K j 3/2 t
∞
(4.78)
for 0 ≤ j ≤ k − 2 and 0 ≤ t ≤ −3/2 min{T,T0 }, and j ≤ k/2 − 2.
5. Remark
The same result is obtained by analyzing the solutions of the initial-value problems for
the KP II and the BBM KP II equations. Indeed, that analysis was carried out by Mammeri
in [11]. This was made known to us after the submission of this paper.
20
International Journal of Mathematics and Mathematical Sciences
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Gideon P. Daspan: Department of Mathematics, Louisiana State University, Baton Rouge,
LA 70803, USA
Email address: daspan2@math.lsu.edu
Michael M. Tom: Department of Mathematics, Louisiana State University, Baton Rouge,
LA 70803, USA
Email address: tom@math.lsu.edu