Complex-valued Burgers and KdV-Burgers equations

0.2 MB·English

by  Netra Khanal

Complex-valued Burgers and KdV-Burgers equations
Netra Khanal Jiahong Wu Juan-Ming Yuan
∗ † ‡
Bing-Yu Zhang
§
9
0
0
2
Abstract
n
a
Spatially periodic complex-valued solutions of the Burgers and KdV-Burgers
J
equations are studied in this paper. It is shown that for any sufficiently large time
4
1 T, there exists an explicit initial data such that its corresponding solution of the
BurgersequationblowsupatT. Inaddition,theglobalconvergenceandregularity
]
P of series solutions is established for initial data satisfying mild conditions.
A
. Key Words:
h
t
a
Complex Burgers equation, complex KdV-Burgers equation, finite-time singularity,
m
global regularity
[

1
Mathematics Subject Classification:
v
2
35Q53, 35B35, 35B65, 76D03
3
1
2
.
1
0
9
0
:
v
i
X
r
a
∗Department of Mathematics, Oklahoma State University, 401 Mathematical Sciences, Stillwater,
OK 74078,USA. Email: [email protected].
†Department of Mathematics, Oklahoma State University, 401 Mathematical Sciences, Stillwater,
OK 74078,USA. Email: [email protected].
‡Department of Applied Mathematics, Providence University, Shalu, Taichung 433, Taiwan. Email:
[email protected]
§Department of MathematicalSciences, University of Cincinnati, Cincinnati, OH 45221-0025,USA.
Email: [email protected]
1
1 Introduction
This work addresses the global regularity issue on solutions of the complex Burgers and
KdV-Burgers equations
u 6uu +αu νu = 0, (1.1)
t x xxx xx
− −
where ν 0 and α 0 are parameters and u = u(x,t) is a complex-valued function.
≥ ≥
Attention will be focused on the spatially periodic solutions, namely x T = R/(2π),
∈
the one-torus and we supplement (1.1) with a given initial data
u(x,0) = u (x), x T. (1.2)
0
∈
Our first major result is for the complex Burgers equation ((1.1) with α = 0) and
it asserts that for any sufficiently large time T, there exists an explicit smooth initial
data u such that its corresponding solution blows up at t = T (Theorem 2.1). This
0
result was partially motivated by a recent paper of Pol´aˇcik and Sˇver´ak [9], in which
the complex-valued Burgers equation on the whole line was shown to develop finite-
time singularities for compactly supported smooth data. Their proof takes advantage
of the explicit solution formula obtained via the Hopf-Cole transform. By contrast, the
finite-time singular solutions constructed in this paper assume the form
u(x,t) = ∞ a (t)eikx (1.3)
k
k=1
X
andcorrespondto theinitialdatau (x) = aeix. Weemphasize thatsolutions oftheform
0
(1.3)arelocallywell-posed intheusualSobolevspaceHs := Hs(T)withasuitable index
s (see Theorem 2.5 for more details). For any T T (a fixed number depending on ν
0
≥
only), we obtain a lower bound for a (T) through a careful observation of the pattern
k
| |
that a (t)’s exhibit and the finite time singularity of (1.3) in L2 then follows if we take
k
a in u to be sufficiently large. This result reveals a fundamental difference between
0
the real-valued solutions of the Burgers equation and their complex counterparts. The
diffusion in the case of complex-valued solutions no longer dissipates the L2-norm, which
can blow up in a finite time. However, if we know the L2-norm of a complex-valued
solution is bounded, then there would be no finite-time singularity (Theorem 2.6).
We also explore the conditions under which solutions of (1.1) are global in time. A
simple example of the global solutions of (1.1) corresponds to the initial data u (x) =
0
a eix with a < 1 provided ν and α satisfy a suitable condition, say ν2 +4α2 9 (see
0 0
| | ≥
Theorem 3.5). For general initial data of the form
u (x) = ∞ a eikx
0 0k
k=1
X
with a < 1, (1.1) possesses a unique local solution (1.3) with a (t) given by a finite
0k k
| |
sum of terms that can be made explicit through an inductive relation. To show the
convergence of (1.3) for large time, it is necessary to estimate a (t) and our approach
k
| |
2
is to count the total number of terms that it contains. This counting problem is closely
related to the number of nonnegative integer solutions to the equation
j +2j +3j + +kj = k
1 2 3 k
···
for a fixed integer k > 0. Using a result by Hardy and Ramanujan [3], we are able to
establish the global regularity of (1.3) under a mild assumption (see Theorem 3.3). In
addition, u( ,t) for any s 0 decays exponentially in t for large t.
Hs
k · k ≥
We remark that the study of complex-valued Burgers and KdV-Burgers equations
can be justified both physically and mathematically. Physically these complex equations
doariseinthemodelingofseveral physicalphenomena ([4],[5],[6]). Mathematicallythese
equations exhibit some remarkable features and admit solutions with much richer struc-
tures than those of their real-valued ones. In fact, these equations and other complex-
valued partial differential equations have attracted quite some attention recently. A lot
of efforts have been devoted to the important issue of whether or not their solutions
can blow up in a finite time. In [1] Birnir considered the complex KdV equation and
constructed a family of singular solutions represented by the Weierstrass function. Very
recently Y. Li [8] obtained simple explicit formulas for finite-time blowup solutions of
the complex KdV equation through Darboux transform. In [2] Bona and Weissler ad-
dressed the blowup issue for a family of complex-valued nonlinear dispersive equations.
The papers of Yuan and Wu ([11],[12],[13]) treated the complex KdV and KdV-Burgers
equations as systems of two nonlinearly coupled equations and clarified how the poten-
tial singularities of the real part are related to those of the imaginary part. In addition,
extensive numerical experiments were performed to reveal the blowup structures. An-
other important example that shows significant differences between the real-valued and
complex-valued solutions is the Navier-Stokes equations. It remains open whether or not
classical solutions of the 3D incompressible Navier-Stokes equations can develop finite-
time singularities. However, Li and Sinai [7] recently showed that the complex solutions
of the 3D Navier-Stokes equations corresponding to large parameter family of initial
data blow up in finite time. Their work motivated the study of Pol´aˇcik and Sˇver´ak on
the complex-valued solutions of the Burgers equation, as we mentioned earlier.
The rest of this paper is divided into three sections. The second section focuses on
the complex Burgers equation and presents Theorems 2.1, 2.5 and 2.6. The third section
details the global regularity results concerning the complex KdV-Burgers equations.
2 Blowup for the complex Burgers equation
Thissectionpresents threemajorresults. Thefirst oneisablowupresult forthecomplex
Burgers equation in a periodic domain T= [0,2π], namely
u 6uu νu = 0, x T, t > 0,
t − x− xx ∈ (2.1)
u(x,0) = u (x), x T
( 0
∈
3
It states that for any sufficiently large T > 0, there exists an initial data u such that
0
its corresponding solution u blows up at t = T. This solution can be represented by
u(x,t) = ∞ a (t)eikx (2.2)
k
k=1
X
and the blowup is in the L2 sense.
For the sake of completeness of our theory on (2.1), we also present a local existence
and uniqueness result on solutions of the form (2.2) to the complex-valued KdV-Burgers
type equation
u 6uu +ν( ∆)γu+αu = 0, (2.3)
t x xxx
− −
which reduces to the complex Burgers equation when γ = 1 and α = 0. The fractal
Laplacian ( ∆)γ is defined through Fourier transform,
−
( ∆)γu(ξ) = ξ 2γu(ξ).
− | |
The third result asserts that if thedL2-norm of a soblution of (2.3) is bounded on [0,T],
then all higher derivatives are bounded and no singularity is possible on [0,T].
We divide the rest of this section into two subsections with the first devoted to the
blowup result and the second to the local existence uniqueness.
2.1. Finite-time blowup
Theorem 2.1 For every sufficiently large T > 0, there exists an initial data u of the
0
form
u (x) = aeix (2.4)
0
such that the corresponding solution u of (2.1) blows up at t = T in the L2-norm, namely
u( ,T) L2(T) = . (2.5)
k · k ∞
For any s R, the homogeneous Sobolev space H˚s(T) and the inhomogeneous
∈
Sobolev space Hs(T) are defined in the standard fashion. In particular, a function of
the form
u(x,t) = ∞ a eikx
k
k=1
X
is in H˚s(T) if
u 2 ∞ k2s a 2 < ,
k kH˚s(T) ≡ | k| ∞
k=1
X
and in Hs(T) if
u 2 ∞ (1+k2)s a 2 < .
k kHs(T) ≡ | k| ∞
k=1
X
4
Clearly, L2(T) can be identified with H0(T).
For u given by (2.4), the local existence and uniqueness result of the next subsection
0
asserts that the corresponding solution u can be written as
u(x,t) = ∞ a (t)eikx
k
k=1
X
before it blows up. The idea is to choose large a such that
u( ,T) 2 = ∞ a (T) 2 = .
k · kL2 | k | ∞
k=1
X
We attempt to find an explicit representation for a (t). It is easy to verify the following
k
iterative formula
t
a (t) = ae νt, a (t) = 3ike νk2t eνk2τ a (τ)a (τ)dτ, k = 2,3, .
1 − k − k1 k2 ···
Z0 k1+Xk2=k
(2.6)
To see the pattern in a (t), we calculate the first few of them explicitly:
k
a (t) = ae νt, (2.7)
1 −
a (t) = ia2ν 1 3e 2νt +3e 4νt , (2.8)
2 − − −
− −
h 27 i 9
a (t) = a3ν 2 9e 3νt e 5νt + e 9νt , (2.9)
3 − − − −
− − 2 2
(cid:20) (cid:21)
27 9
a (t) = ia4ν 3 27e 4νt +54e 6νt e 8νt 18e 10νt + e 16νt , (2.10)
4 − − − − − −
− − 2 − 2
(cid:20) (cid:21)
405 405 135 135
a (t) = a5ν 4 81e 5νt e 7νt + e 9νt + e 11νt e 13νt
5 − − − − − −
− 2 4 2 − 4
(cid:20)
135 27
e 17νt + e 25νt ,
− −
− 8 8
(cid:21)
2187 729
a (t) = ia6ν 5 243e 6νt +729e 8νt e 10νt e 12νt
6 − − − − −
− − − 4 − 4
(cid:20)
81 243 243 81
+243e 14νt + e 18νt e 20νt e 26νt + e 36νt .
− − − − −
2 − 8 − 20 40
(cid:21)
The following lemma summarizes the pattern exhibited by a (t)’s.
k
Lemma 2.2 For any t > 0,
a (t) = ab (t), a (t) = ia2b (t), a (t) = a3b (t), a (t) = ia4b (t) (2.11)
1 1 2 2 3 3 4 4
− −
and more generally, for k = 4n+j with n = 0,1,2, and j = 1,2,3,4,
···
a (t) = a (t) = ij 1a4n+j b (t), (2.12)
k 4n+j − 4n+j
where b (t) > 0 for any t > 0.
4n+j
5
Remark. A special consequence of this lemma is that all terms in the summation in
(2.6) have the same sign and thus
t
a (t) = 3ke νk2t eνk2τ a (τ) a (τ) dτ. (2.13)
| k | − | k1 || k2 |
Z0 k1+Xk2=k
Proof of Lemma 2.2. (2.12) can be shown through induction. For n = 0, (2.12) is just
(2.11). By (2.6), a (t) = ae νt and
1 −
t
a (t) = 6ia2e 4νt e4ντ b2(τ)dτ = ia2b (t),
2 − 1 2
Z0
where b (t) = 6e 4νt te4ντ b2(τ)dτ > 0. Similarly, a (t) = a3b (t) and a (t) =
2 − 0 1 3 − 3 4
ia4b (t) for some b (t) > 0 and b (t) > 0.
4 3 R 4
−
We now consider the general case. Without loss of generality, we prove (2.12) with
k = 4n+1. Assume (2.12) is true for all k < 4n+1. By (2.6),
t
a (t) = 3ike νk2t eνk2τ a (τ)a (τ)dτ.
k − k1 k2
Z0 k1+Xk2=k
Noticing that a (τ)a (τ) with k +k = 4n+1 assumes two forms
k1 k2 1 2
a (τ)a (τ) and a (τ)a (τ)
4n1 4n2+1 4n1+2 4n2 1
−
where n 0, n 0 and n +n = n, we conclude by the inductive assumptions that
1 2 1 2
≥ ≥
a (τ)a (τ) must be of the form iakb (τ) for some positive function b (τ) > 0.
k1 k2 − k1,k2 k1,k2
Therefore,
a (t) = a (t) = akb (t)
k 4n+1 k
with
t
b (t) = 3ke νk2t eνk2τ b (τ)dτ > 0 for any t > 0.
k − k1,k2
Z0 k1+Xk2=k
This completes the proof of Lemma 2.2.
Proof of Theorem 2.1. Without loss of generality, we set ν = 1. Assume
1 3k 3
∞
T T ln − (2.14)
≥ 0 ≡ k2 2k 3
kX=2 −
and choose a such that
A ae T 1
−
≡ ≥
We prove by induction that
a (T) Ak for k = 1,2,3, (2.15)
k
| | ≥ ···
which, in particular, yields (2.5). Obviously, for any 0 t T,
≤ ≤
a (t) a (T) = ae T = A 1.
1 1 −
| | ≥ | | ≥
6
To prove (2.15) for k 2, we recall (2.13), namely
≥
t
a (t) = 3ke k2t ek2τ a (τ) a (τ) dτ.
| k | − | k1 || k2 |
Z0 k1+Xk2=k
Therefore, for T t t 1 ln3,
≥ ≥ 2 ≡ 4
t 3
a (t) = 6e 4t e4τa2(τ)dτ = A2(1 e 4t) A2.
| 2 | − 1 2 − − ≥
Z0
For k = 3, if T t t t + 1 ln 2,
≥ ≥ 3 ≡ 2 9
t
a (t) = 9e 9t e9τ2 a (τ) a (τ) dτ
3 − 1 2
| | | || |
Z0
t
9e 9t e9τ2 a (τ)a (τ) dτ
− 1 2
≥ | |
Zt2
2A3(1 e 9(t t2)) A3.
− −
≥ − ≥
More generally, for any t t = t + 1 ln 3k 3,
≥ ≥ k k−1 k2 2k−−3
t
a (t) = 3ke k2t ek2τ ( a (τ) a (τ) + a (τ) a (τ)
k − 1 k 1 2 k 2
| | Z0 | || − | | || − |
+ + a (τ) a (τ) + a (τ) a (τ) )dτ
k 2 2 k 1 1
··· | − || | | − || |
t
3ke k2t ek2τ ( a (τ) a (τ) + a (τ) a (τ)
− 1 k 1 2 k 2
≥ Ztk−1 | || − | | || − |
+ + a (τ) a (τ) + a (τ) a (τ) )dτ
k 2 2 k 1 1
··· | − || | | − || |
3k(k 1)
− (1 e−νk2(t−tk−1))Ak Ak.
≥ k2 − ≥
If T T as defined in (2.14), then t < T for any integer k 1 and thus
0 k
≥ ≥
a (T) Ak.
k
| | ≥
This completes the proof of Theorem 2.1.
We state and prove a few specific properties for a (t).
k
Proposition 2.3 Assume u is given by (1.2). For each k 1, a (t) is of the form
0 k
≥
k2
a (t) = α e mνt (2.16)
k k,m −
m=k
X
where the complex-valued coefficients α satisfy
k,m
k2
α = 0 for k 2, (2.17)
k,m
≥
m=k
X
3ik
α = α α for k m < k2. (2.18)
k,m k2 m k1,m1 k2,m2 ≤
− k1+Xk2=km1+Xm2=m
The indices k , k , m and m in the summation above obey
1 2 1 2
1 k k 1, 1 k k 1, k m k2 and k m k2.
≤ 1 ≤ − ≤ 2 ≤ − 1 ≤ 1 ≤ 1 2 ≤ 2 ≤ 2
7
Proof. (2.17) is a consequence of the fact that a (0) = 0 for k 2. (2.16) follows
k
≥
from a simple induction. Obviously, a (t) = ae νt. Fix k and assume (2.16) is valid
1 −
for all integers up to k. Then, for k 1,k 1,k + k = k + 1,k m k2 and
1 ≥ 2 ≥ 1 2 1 ≤ 1 ≤ 1
k m k2,
2 ≤ 2 ≤ 2
t
a (t) = 3i(k +1) α α e ν(k+1)2t eν((k+1)2 (m1+m2))τ dτ
k+1 k1,m1 k2,m2 − −
k1+kX2=k+1mX1,m2 Z0
3i(k +1)α α
= k1,m1 k2,m2 e ν(m1+m2)t e ν(k+1)2t .
− −
ν((k +1)2 (m +m )) −
k1+kX2=k+1mX1,m2 − 1 2 (cid:16) (cid:17)
Since m +m k2 +k2 (k +k )2 = (k +1)2, this proves (2.16) with (2.18).
1 2 ≤ 1 2 ≤ 1 2
Proposition 2.4 Assume that u is given by (1.2).
0
1) Let k 1 be an integer. Then
≥
3i k 1 k
α = − ak and α = α ; (2.19)
k,k k,k+2 k,k
ν −2
(cid:18) (cid:19)
2) Let k 1 be an integer. Then, for n = 1,3,5, ,
≥ ···
α = 0;
k,k+n
3) Let k 1 be an integer and let k2 > m > U(k) k2 2k+2. Then
≥ ≡ −
α = 0. (2.20)
k,m
Proof. Letting m = k and m = k in (2.18), we find
1 1 2 2
3ik 3i
α = α α = α α .
k,k k1,k1 k2,k2ν(k2 k) ν(k 1) k1,k1 k−k1,k−k1
k1+Xk2=k − − k1+Xk2=k
Asimple induction allows us toobtainthe expression forα . To show α = k α ,
k,k k,k+2 −2 k,k
we set m = k +2 in (2.18) to get
3ik
α = (α α +α α +α α
k,k+2 ν(k2 k 2) 1,1 k−1,k+1 2,2 k−2,k 2,4 k−2,k−2
− −
+ +α α +α α +α α ). (2.21)
k 2,k 2 2,4 k 2,k 2,2 1,1 k 1,k+1
··· − − − −
Inserting the inductive assumptions such as
k 1 k 2
α = − α , α = − α , α = α
k 1,k+1 k 1,k 1 k 2,k k 2,k 2 2,4 2,2
− − 2 − − − − 2 − − −
8
in (2.21), we obtain
3ik k k 1
−
α = α α +α α
k,k+2 ν(k2 k 2) −2 k1,k1 k−k1,k−k1 1,1 k−1,k−1
− − kX1=1
k k2 k  3ik k 1 
−
= − α α
−2k2 k 2ν(k2 k) k1,k1 k−k1,k−k1
− − − kX1=1
3ik
+ α α
ν(k2 k 2) 1,1 k−1,k−1
− −
k k2 k k 2 k
= − α − α = α .
−2k2 k 2 k,k − 2k2 k 2 k,k −2 k,k
− − − −
To show α = 0, we set m = k +1 to obtain
k,k+1
3ik
α = (α α +α α + +α α ),
k,k+1 ν(k2 (k +1)) 1,1 k−1,k 2,2 k−2,k−1 ··· k−1,k 1,1
−
which can be seen to be zero after inserting the inductive assumptions.
To prove (2.20), it suffices to notice in (2.18) that the second summation is over
m +m = m with k m k2 and k m k2. Thus, m = m +m k2 +k2 =
1 2 1 ≤ 1 ≤ 1 2 ≤ 2 ≤ 2 1 2 ≤ 1 2
(k +k )2 2k k k2 2(k 1) and α with U(k) < m < k2 is equal to zero. This
1 2 1 2 k,m
− ≤ − −
completes the proof of Proposition 2.4.
2.2 Local well-posedness
This subsection establishes the following two major results.
Theorem 2.5 Consider (2.3) with γ > 1. Let s > 1. Assume u Hs(T) has the form
2 2 0 ∈
u (x) = ∞ a eikx. (2.22)
0 0k
k=1
X
Then there exists T = T( u ) such that (2.3) with the initial data u has a unique
0 Hs 0
solution u C([0,T);Hs)k Lk2([0,T);H˚s+γ) that assumes the form
∈ ∩
u(x,t) = ∞ a (t)eikx.
k
k=1
X
In the case when γ 1, we can actually show that no finite-time singularity is
≥
possible if we know that the L2-norm is bounded a priori. In fact, the following theorem
states that the L2-norm controls all higher-order derivatives.
Theorem 2.6 Let T > 0 and let u be a weak solution of (2.3) with γ 1 on the time
≥
interval [0,T]. If we know a priori that u L ([0,T];L2) L2([0,T];H˚γ), namely
∞
∈ ∩
T
M sup u( ,t) 2 +ν Λγu( ,t) 2 dt < , (2.23)
0 ≡ k · kL2 k · kL2 ∞
t [0,T] Z0
∈
9
then, for any integer k > 0,
T
M sup u(k)( ,t) 2 +ν Λk+γu( ,t) 2 dt < .
k ≡ k · kL2 k · kL2 ∞
t [0,T] Z0
∈
where Λ = ( ∆)12 and u(k) denotes any partial derivative of order k.
−
We first prove Theorem 2.5.
Proof of Theorem 2.5. The existence of such a solution follows from the Galerkin ap-
proximation. LetN 1anddenotebyP theprojectiononthesubspace eix,e2ix, ,eiNx .
N
≥ { ··· }
Let
N
uN(x,t) = aN(t)eikx
k
k=1
X
where a (t) satisfies
k
d
aN(t) = 3ik aN(t)aN(t)+iαk3aN(t) νk2γ aN(t),
dt k k1 k2 k − k
k1+Xk2=k
aN(0) = aN a . (2.24)
k 0k ≡ 0k
Here 1 k N and 1 k N. From the theory of ordinary differential equations,
1 2
≤ ≤ ≤ ≤
we know that (2.24) has a unique local solution aN(t) on [0,T]. We derive some a priori
k
bounds for uN(x,t). Clearly, uN(x,t) solves
∂ uN = 6P (uNuN)+αuN ν( ∆)γuN, uN(x,0) = P u .
t N x xxx − − N 0
We now show that
d 6γ−2
uN 2 +ν uN 2 C(ν,s) uN 2γ−1. (2.25)
dtk kHs k kHs+γ ≤ k kHs
It follows from the equation
d
aN(t)+νk2γaN(t) iαk3aN(t) = 3ik aN(t)aN(t)
dt k k − k k1 k2
k1+Xk2=k
that, after omitting the upper index N for notational convenience,
d N N N
k2s a (t) 2 = 2ν k2(s+γ) a (t) 2 6 k2s+1 a¯ a a ,
dt | k | − | k | − I k k1 k2
kX=1 kX=1 kX=1 k1+Xk2=k
 
where denotes the imaginary part. To bound the nonlinear term on the right (denoted
I
by J), we first notice that the summation over k + k = k is less than twice the
1 2
summation over k +k = k with k k and 2k k. Thus,
1 2 1 2 2
≤ ≥
N
J 6 k2s+1 a a a
≤ | k| | k1|| k2|
kX=1 k1+Xk2=k
N
≤ 12 ks+21|ak| (2k2)s+12|ak1||ak2|.
kX=1 k/2X≤k2≤k
10