Blow-up of solutions to a nonlinear dispersive rod
equation
Yong Zhou
Department of Mathematics, East China Normal University
Shanghai 200062, China
and
The Institute of Mathematical Sciences, The Chinese University of Hong Kong
Shatin, N.T., Hong Kong
[email protected]
Abstract: In this paper, firstly we find the best constant for a convolution problem on the unit circle via a variational method. Then we apply the best constant
on a nonlinear rod equation to give sufficient conditions on the initial data, which
guarantee finite time singularity formation for the corresponding solutions.
Mathematics Subject Classification(2000): 30C70, 37L05, 35Q58, 58E35
Key words: best constant, convolution problem, rod equation, singularity
1
Introduction
Although a rod is always three-dimensional, if its diameter is much less than the
axial length scale, one-dimensional equations can give a good description of the
motion of the rod. Recently Dai [6] derived a new (one-dimensional) nonlinear
dispersive equation including extra nonlinear terms involving second-order and
third-order derivatives for a compressible hyperelastic material. The equation
1
reads
vτ + σ1 vvξ + σ2 vξξτ + σ3 (2vξ vξξ + vvξξξ ) = 0,
where v(ξ, τ ) represents the radial stretch relative to a pre-stressed state, σ1 6= 0,
σ2 < 0 and σ3 ≤ 0 are constants determined by the pre-stress and the material
parameters. If one introduces the following transformations
√
√
3 −σ2
t, ξ = −σ2 x,
τ=
σ1
then the above equation turns into
ut − utxx + 3uux = γ(2ux uxx + uuxxx ),
(1.1)
where γ = 3σ3 /(σ1 σ2 ). In [7], the authors derived that value range of γ is from 29.4760 to 3.4174 for some special compressible materials. From the mathematical
view point, we regard γ as a real number.
When γ = 1 in (1.1), we recover the shallow water (Camassa-Holm) equation derived physically by Camassa and Holm in [2] (found earlier by Fokas and
Fuchssteiner [8] as a bi-Hamiltonian generalization of the KdV equation) by approximating directly the Hamiltonian for Euler’s equations in the shallow water
regime, where u(x, t) represents the free surface above a flat bottom. Some satisfactory results have been obtained for this shallow water equation recently. Local
well-posedness for the initial datum u0 (x) ∈ H s with s > 3/2 was proved by
several authors, see [12, 14, 16]. On the other hand, Himonas and Misiole [10]
showed that the Camassa-Holm equation is not well-posed for initial data in H s
for s < 3/2. These results suggest that s = 3/2 is the critical index for wellposedness. Moreover, wave breaking for a large class of initial data has been
established in [3, 4, 12, 13, 20]. However, In [18, 19], Xin and Zhang showed
global existence and uniqueness for weak solutions with u0 (x) ∈ H 1 . The solitary
waves of Camassa-Holm equation are peaked solitons [2] with u(x, t) = e−|x−ct| ,
where c ∈ R is the wave speed. The orbital stability of the peakons was showed
by Constantin and Strauss [5] 4 years ago.
If γ = 0, (1.1) is the BBM equation, a well-known model for surface waves in
a canal [1], and its solutions are global.
2
For general γ ∈ R, the rod equation (1.1) was studied systemically by the
author firstly in [22]. Local well-posedness of strong solutions to (1.1) was established by applying Kato’s theory [11] and various sufficient conditions on the
initial data were found to guarantee the finite blow-up of the corresponding solutions for both periodic and nonperiodic cases. Later, the refined blow-up criteria
R
for a special class of initial data S u0 = 0 for the periodic rod equation was
presented in [23], where S = R/Z is the unit circle. It should be mentioned that
for γ < 1, (1.1) admits smooth solitary waves observed by Dai and Huo [7]. Let
u(x, t) = φ(ξ), ξ = x − ct be the solitary wave to (1.1). It was showed that φ(ξ)
satisfies
√
µ
±ξ = − −γ
for γ < 0 and
±ξ =
√
1
2γφ − (γ + 1)c
π + arcsin
2
(1 − γ)c
¶
p
p
( c(c − φ) + c(c − γφ))2
− ln
(1 − γ)cφ
p
p
√
√
c − γφ) − γ(c − φ))2
( c − γφ + c − φ)2
γ ln
− ln
(1 − γ)c
(1 − γ)φ
(
for 0 < γ < 1. Recently, the author [21] proved the stability of these solitary waves
by applying a general theorem established by Grillakis, Shatah and Strauss [9].
We now finish this introduction by outlining the rest of the paper. In section
2, we recall the local well-posedness for (1.1) with initial datum u0 ∈ H s , s > 3/2,
and the lifespan of the corresponding solution is finite if and only if its first-order
derivative blows up. In section 3, we find the best constant for a convolution
problem by a variational method described in Struwe’s book [17]. Then we show
blow-up of solutions to the nonlinear dispersive rod equation by applying the
best constant for the convolution problem. Finally, in section 5, we discuss the
analogous problem for the nonperiodic case.
2
Preliminaries
In this paper, we concentrate on the periodic case. In [22], it is proved that
Theorem 2.1 [22] Let the initial datum u0 (x) ∈ H s (S), s > 3/2. Then there
exists T = T (ku0 kH s ) > 0 and a unique solution u, which depends continuously
on the initial datum u0 , to (1.1) such that
¡
¢
u ∈ C ([0, T ); H s (S)) ∩ C 1 [0, T ); H s−1 (S) .
3
Moreover, the following two quantities E and F are invariants with respect to
time t for (1.1).
Z

¢
¡ 2
2


E(u)(t)
=
u
(x,
t)
+
u
(x,
t)
dx

x


S
Z


¡ 3
¢


 F (u)(t) =
u (x, t) + γu(x, t)u2x (x, t) dx
S
Actually, the local well-posedness was proved for both periodic and nonperiodic
case in the above paper.
The maximum value of T in Theorem 2.1 is called the lifespan of the solution,
in general. If T < ∞, that is lim supt↑T ku(., t)kH s = ∞, we say that the solution
blows up in finite time. The following theorem tells us that the solution blows
up if and only if the first-order derivative blows up. This phenomenon coincides
physically with the rod breaking.
Theorem 2.2 [22] Let u0 (x) ∈ H s (S), s > 3/2, and u be the corresponding
solution to problem (1.1) with lifespan T . Then
sup
x∈S,0≤t<T
|u(x, t)| ≤ C(ku0 kH 1 ).
(2.1)
T is bounded if and only if
lim inf inf {γux (x, t)} = −∞
(2.2)
m(t) := inf (ux (x, t)sign{γ}) , t ≥ 0,
(2.3)
t↑T
x∈S
For γ 6= 0, we set
x∈S
where sign{a} is the sign function of a ∈ R and we set m0 := m(0). Then for
every t ∈ [0, T ) there exists at least one point ξ(t) ∈ S with m(t) = ux (ξ(t), t).
Just as the proof given in [4], one can show the following property of m(t).
Lemma 2.3 Let u(t) be the solution to (1.1) on [0, T ) with initial data u0 ∈
H s (S), s > 3/2, as given by Theorem 2.1. Then the function m(t) is almost
everywhere differentiable on [0, T ), with
dm(t)
= utx (ξ(t), t), a.e. on (0, T ).
dt
4
To consider the quantity m(t) for wave breaking comes from an idea of Seliger [15]
originally, the rigorous regularity proof is given in [4] for Camassa-Holm equation.
Set Qs = (1 − ∂x2 )s/2 , then the operator Q−2 can be expressed by
Z
−2
Q f = G ∗ f = G(x − y)f (y)dy
T
2
for any f ∈ L (S) with
G(x) =
cosh(x − [x] − 1/2)
,
2 sinh(1/2)
(2.4)
where [x] denotes the integer part of x. Then equation (1.1) can be rewritten as
µ
¶
3−γ 2 γ 2
−2
ut + γuux + ∂x Q
u + ux = 0.
(2.5)
2
2
Just as in [22], it is easy to derive a equation for m(t) from (2.5) as
·
µ
¶¸
3−γ 2 γ 2
dm
γ 2 3−γ 2
=− m +
u (ξ(t), t) − G ∗
u + ux (ξ(t), t)
dt
2
2
2
2
(2.6)
a.e. on (0, T ), where m(t) and ξ(t)was defined in (2.3) and Lemma 2.3.
If γ = 3, it turns out that (2.6) is a Riccati type equation with negative initial
data for any nonconstant u0 . So the solutions to (1.1) in periodic case definitely
blow up in finite time with arbitrary nonconstant initial data u0 .
In what follows, we assume that 0 < γ < 3.
3
The best constant for a convolution problem
In this section, we consider the following convolution problem
µ
¶
1 2
2
G ∗ f + fx (x),
2
where G is the Green function for Q−2 in the unit circle, defined by (2.4) and
function f ∈ H 1 (S).
Direct computation yields
1
[G ∗ (f 2 + fx2 )](x)
2
¶
µ
Z x x−η− 1
1
+η−x
2 + e2
1
e
1 2
2
=
f (η) + fx (η) dη
2
2
2 sinh( 21 ) 0
¶
Z 1 x−η+ 1
1 µ
η−x− 2
2 + e
1
1 2
e
2
+
f (η) + fx (η) dη.
2
2
2 sinh( 12 ) x
5
(3.1)
Due to Cauchy-Schwartz inequality, we have
Z x
¡
¢
e−η f 2 (η) + fx2 (η) dη
0
Z x
Z
−η
−η 2
x
≥ −2
e f (η)fx (η)dη = −e f (η)|0 −
0
which implies that
Z x
µ
−η
e
0
x
0
Similarly we have
Z 1
−η
Z
0
e
(3.3)
¶
¯1
1 2
1
f (η) + fx (η) dη ≥ − e−η f 2 (η)¯x
2
2
2
µ
1
η
x
(3.2)
eη f 2 (η)dη,
µ
¶
¯x
1 2
1
2
e f (η) + fx (η) dη ≥ eη f 2 (η)¯0 .
2
2
x
and
x
η
µ
e
0
¶
¯x
1
1 2
f (η) + fx (η) dη ≥ − e−η f 2 (η)¯0 .
2
2
On the other hand, we compute
Z x
¡
¢
eη f 2 (η) + fx2 (η) dη
0
Z x
Z
η
η 2
x
≥ 2
e f (η)fx (η)dη = e f (η)|0 −
Z
e−η f 2 (η)dη,
2
0
which implies
x
¶
¯1
1 2
1
f (η) + fx (η) dη ≥ eη f 2 (η)¯x .
2
2
2
(3.4)
(3.5)
Combining (3.1)–(3.5), we get
1
1
G ∗ (f 2 + fx2 )(x) ≥ f 2 (x).
2
2
The question is whether
1
2
is the best constant for the problem
¶
µ
1 2
2
G ∗ f + fx (x) ≥ Cf 2 (x),
2
(3.6)
for f ∈ H 1 (S).
From the above computation, it is easy to find that
1
2
is the best constant
1
if and only if f ∈ H (S) is a solution to f = −fx and f = fx . Obviously it is
impossible. Hence
1
2
is not the best constant to problem (3.6). The goal of this
section is to find the best constant C via a variational method. Actually we will
prove
6
Theorem 3.1 For all f ∈ H 1 (S), the following inequality holds
µ
¶
1 2
2
G ∗ f + fx (x) ≥ C0 f 2 (x),
2
with
C0 =
1
arctan(sinh(1/2))
≈ 0.869.
+
2 2 sinh(1/2) + 2 arctan(sinh(1/2)) sinh2 (1/2)
Moreover, C0 is the optimal constant obtained by the function
f0 =
1 + arctan(sinh(x − [x] − 1/2)) sinh(x − [x] − 1/2)
.
1 + arctan(sinh(1/2)) sinh(1/2)
Proof: Let
A = {f ∈ H 1 (S) | kf kL∞ = 1}
and
µ
¶
µ
¶
Z
1 2
1 2
2
I[f ](x) = G ∗ f + fx (x) = G(x − y) f (y) + fx (y) dy.
2
2
S
2
Since A and I[f ] defined as above is translation invariant on the unit circle S,
we can assume A is defined on the interval [0, 1] with f ≥ 0 and f (0) = f (1) = 1
without loss of generality. Hence to find the best constant for problem (3.6) is
equivalent to fine the minimum value for
µ
¶
Z 1
1
1 2
2
I[f ](0) =
cosh(x − 1/2) f (x) + fx (x) dx.
2 sinh(1/2) 0
2
From now on, we follow the variational method discussed in a comprehensive
book written by Struwe [17].
It is clearly that
Z 1
Z 1
¡ 2
¢
¡ 2
¢
cosh(1/2)
1
2
f (x) + fx (x) dx ≤ I[f ](0) ≤
f (x) + fx2 (x) dx,
4 sinh(1/2) 0
2 sinh(1/2) 0
for any f ∈ A. The above inequality means that I[f ](0) is equivalent to the
H 1 -norm of f .
Suppose {fk }∞
k=1 is a minimizing sequence, i.e., I[fk ](0) → inf f ∈A I[f ](0), as
k → ∞. Hence it is easy to show that there exists a subsequence {fkj }∞
j=1 ⊂
∞
{fk }∞
k=1 , denoted it by {fk }k=1 also, and a function g ∈ A with fk → g as
k → ∞. For the details we refer to [20].
7
Due to the identity cosh(3x) = cosh3 (x) + 3 cosh(x) sinh2 (x), we have
I [cosh(x − 1/2)/ cosh(1/2)] (0)
¶
Z 1µ
1
1
3
2
=
cosh (x − 1/2) + cosh(x − 1/2) sinh (x − 1/2) dx
2
2 sinh(1/2) cosh2 (1/2) 0
¶
Z 1/2 µ
1
5
2
=
cosh(3x) − cosh(x) sinh (x) dx
2
2 sinh(1/2) cosh2 (1/2) −1/2
2 sinh(3/2) − 5 sinh3 (1/2)
2 + sinh2 (1/2)
=
6 sinh(1/2) cosh2 (1/2)
2 cosh2 (1/2)
sinh2 (1/2)
< 1 = I[1](0),
= 1−
2 cosh2 (1/2)
=
where we used the identity sinh(3x) = 4 sinh3 (x) + 3 sinh(x).
The above inequality implies that 1 is not the minimizer for I[f ](0), in other
words, there exists region U where the value of g is strictly less than 1.
Let φ be a smooth function with compact support in U . One can choose ² is
sufficient small such that g + ²φ ∈ A. Now set
µ
¶
Z 1
1
2
2
i(t) = I[g + t²φ](0) =
G(x) (g + t²φ) + (gx + t²φx ) dx,
2
0
where t ∈ R such that g + t²φ ∈ A. Since g is the minimizer, we have
Z 1
Z 1
0
0 = i (0) = ²
(2Ggφ + Ggx φx )dx = ²
(2Gg − (Ggx )x )φdx.
0
0
Therefore the equation for g in the region g < 1 reads
(Ggx )x = 2Gg, with G(x) =
cosh(x − 1/2)
.
2 sinh(1/2)
Just as what done in [22], we have the following claim that g < 1 at all points
except 0 and 1. So the equation for g is
(Ggx )x = 2Gg, in (0, 1), with g(0) = g(1) = 1.
(3.7)
After changing variable we can rewrite (3.7) to
cosh(x)g 00 (x) + sinh(x)g 0 − 2 cosh(x)g(x) = 0,
where prime means taking derivative with respect to x.
8
(3.8)
Equation (3.8) is a second order ordinary equation with variable coefficients.
In general, it is difficult to find the explicit solution for this type equation. However, if one can find one special solution, then the other one can be reduced to
solve a first order ordinary differential equation. Fortunately, it is easy to observe
that g1 (x) = sinh(x) is a special solution to (3.8).
Letting g2 = v(x)g1 (x) = v(x) sinh(x) and putting it into (3.8), after simple
computation we obtain the equation for v
cosh(x) sinh(x)v 00 (x) + (2 cosh2 (x) + sinh2 (x))v 0 (x) = 0.
Rewrite (3.9) as
2 cosh(x) sinh(x)
(ln(v )) = −
−
=
sinh(x)
cosh(x)
0
µ
0
(3.9)
¶0
1
ln(
) .
sinh2 (x) cosh(x)
So up to a constant, we have
Z
Z
1
d(sinh(x))
v(x) =
dx =
2
sinh (x) cosh(x)
sinh2 (x) cosh2 (x)
Z
d(sinh(x))
d(sinh(x))
1
=
−
=−
− arctan(sinh(x)).
2
2
sinh(x)
sinh (x)
1 + sinh (x)
Therefore the general solution to (3.8) reads
g = A sinh(x) + B (1 + arctan(sinh(x)) sinh(x)) .
Changing the variable back and using the boundary condition, we have
1 + arctan(sinh(x − 1/2)) sinh(x − 1/2)
g=
.
1 + arctan(sinh(1/2)) sinh(1/2)
Using the equation satisfied by g, we compute
G(x)gx2 (x) = G(x)(g(x)gx (x))x − G(x)gxx (x)g(x)
= G(x)(g(x)gx (x))x + Gx (x)gx (x)g(x) − 2G(x)g 2 (x).
Hence
1
1
(G(x)(g(x)gx (x))x + Gx (x)gx (x)g(x))
G(x)g 2 (x) + G(x)gx2 (x) =
2
2
1
=
(G(x)g(x)gx (x))x .
2
Hence C0 can be obtained easily by
¯1−
1
C0 = G(x)g(x)gx (x)¯0+ .
2
This finishes the proof.
9
2
4
Blow-up criteria
After local well-posedness of strong solutions (see Theorem 2.1) was established,
the next question is whether this local solution can exist globally. If not, how
about the behavior of the solution when it blows up? What induces the blowup? On the other hand, to find sufficient conditions to guarantee the finite time
blow-up or global existence is of great interest, especially for sufficient conditions
added on the initial data.
The main theorem of this section is as follows.
Theorem 4.1 Let 0 < γ < 3. Assume that u0 ∈ H 2 (S) satisfies m0 < 0 and
p

(6
−
γ
−
12γ − 3γ 2 ) cosh(1/2)

2


m
>
ku0 k2H 1 (S) , if 0 < γ ≤ γ0 ,
0


4γ
sinh(1/2)









(3 − (1 + 2C0 )γ) cosh(1/2)


ku0 k2H 1 (S) ,
if γ0 < γ ≤ 1,
m20 >


2γ sinh(1/2)




(3 − γ)(1 − C0 ) cosh(1/2)

2

m
ku0 k2H 1 (S) ,
if 1 < γ ≤ γ1 ,
>

0

2γ
sinh(1/2)






p



(6
−
γ
−
12γ − 3γ 2 ) cosh(1/2)


 m20 >
ku0 k2H 1 (S) , if γ1 < γ < 3,
4γ sinh(1/2)
where
γ0 =
3
3C02
,
γ
=
,
1
4C02 + 2C0 + 1
C02 − C0 + 1
while C0 is the optimal constant in Theorem 3.1. Then the life span T > 0 of the
corresponding solution to (1.1) is finite, i.e., rod breaking occurs.
Remark 4.1 If γ = 1, (1.1) is reduced to the Camassa-Holm equation, while
the condition is m0 >
cosh(1/2)
ku0 kH 1 ,
2 sinh(1/2)
(1−C0 ) cosh(1/2)
ku0 kH 1 ,
sinh(1/2)
which is an improvement of m0 >
which was proved in [20]. Where γ0 ≈ 0.521 and γ1 ≈ 2.555 are
solutions respectively to
r
r
µ
¶
1 1 12 − 3γ
1 1 12 − 3γ
− +
= 2C0 and − +
γ = (3 − γ)C0 .
2 2
γ
2 2
γ
The conditions for 0 < γ ≤ γ0 and γ1 < γ < 3 are established in [22] first. The
cases for γ < 0 and γ > 3 were discussed in [22, 23].
10
First, we have the following blow-up result for a Riccati type ordinary differential
equation.
Lemma 4.2 Assume that a differentiable function y(t) satisfies
y 0 (t) ≤ −Cy 2 (t) + K,
(4.1)
q
with constants C, K > 0. If the initial datum y(0) = y0 < −
K
,
C
then the
solution to (4.1) goes to −∞ in finite time.
Proof: First, we claim that y 0 (t) < 0 for all t ≥ 0.
Suppose not, by the continuity of y(t), there exists a time t0 , such that y 0 (t) <
0
0 for all t ∈ [0, t0 ] and
q y (t0 ) = 0. Then by the decreasing property of y(t), we
have y(t0 ) ≤ y0 < − K
. But from the equation (4.1), we have
C
0 ≤ −Cy 2 (t0 ) + K.
This is a contradiction. Therefore, the claim is true.
From the equation again, one has
µ
K
y (t) ≤ −Cy (t) + K = −C 1 −
Cy02
µ
¶
K
≤ −C 1 −
y 2 (t).
2
Cy0
0
2
¶
y 2 (t) −
K 2
y (t) + K
y02
Solving the above ordinary equation, we obtain that
µ
µ
¶ ¶−1
1
K
y(t) ≤
+C 1−
t
.
y0
Cy02
Hence the solution to (4.1) goes to −∞ before t tends to
1
−Cy0 +
K
y0
.
The proof is complete.
2
Secondly, let us recall the best constant for a Sobolev inequality proved in [20].
kf k2L∞ (S) ≤
cosh(1/2)
kf k2H 1 (S) ,
2 sinh(1/2)
(4.2)
for f ∈ H 1 (S). Moreover, it is an optimal constant for the Sobolev imbedding
H 1 ⊂ L∞ in the sense that (4.2) holds if and only if f (x) = λG(x − y) for some
λ, y ∈ R.
11
We start the proof for the main theorem from (2.6). We will treat it case by
case.
(i) 0 < γ ≤ γ0 .
By the representation of G, we have
·
¸
3−γ 2 γ 2
G∗(
u + ux ) (x, t)
2
2
µ
¶
Z x x−η− 1
1
+η−x
2 + e2
1
e
3−γ 2
γ 2
=
u (η, t) + ux (η, t) dη
2
2
2
2 sinh( 12 ) 0
¶
Z 1 x−η+ 1
1 µ
η−x− 2
2 + e
1
e
3−γ 2
γ 2
+
u (η, t) + ux (η, t) dη. (4.3)
2
2
2
2 sinh( 12 ) x
Direct computation yields that
Z x
³γ
´
γ
e−η
α2 u2 (η, t) + u2x (η, t) dη
2
2
0
Z x
Z x
γα
−η
−η γα 2
x
≥ −
e γαu(η, t)ux (η, t)dη = −e
u (η, t)|0 −
e−η u2 (η, t)dη
2
2
0
0
holds for any α > 0. We have
Z x
³
´
¯x
γ
γ
αγ
e−η (α2 + α) u2 (η, t) + u2x (η, t) dη ≥ − e−η u2 (η, t)¯0 .
2
2
2
0
Now we let
α2 + α =
3−γ
,
γ
which has one positive root α0 with
1 1
α0 = − +
2 2
Therefore
Z
x
µ
−η
e
0
r
12 − 3γ
.
γ
(4.4)
¶
¯x
α0 γ −η 2
γ 2
3−γ 2
u (η, t) + ux (η, t) dη ≥ −
e u (η, t)¯0 .
2
2
2
Moreover, from (4.3), just use the above trick for each term, then one obtains
that
[G ∗ (
3−γ 2 γ 2
α0 γ 2
u + ux )](ξ(t), t) ≥
u (ξ(t), t).
2
2
2
12
(4.5)
Now combining (2.6) and (4.5) together, we have
p
dm
γ 2 6 − γ − 12γ − 3γ 2 2
≤ − m +
u (ξ(t), t)
dt
2
4
p
(6 − γ − 12γ − 3γ 2 ) cosh(1/2)
γ
ku0 k2H 1 ,
≤ − m2 +
2
8 sinh(1/2)
(4.6)
where we used (4.2) and the conservation of H 1 -norm.
If
Ã
m0 < −
6−γ−
p
12γ − 3γ 2
4γ
!1/2
ku0 kH 1 ,
then the solution m(t) to (4.6) goes to −∞ in finite time by applying Lemma 4.2.
(ii) γ0 < γ ≤ 1
By direct computation, we have
¶
µ
µ
3−γ 2 γ 2
u + ux (x, t) = γG ∗ u2 +
G∗
2
2
µ
≥ γG ∗ u2 +
¶
1 2
3(1 − γ)
ux (x, t) +
G ∗ u2 (x, t)
2
2
¶
1 2
u (x, t)
2 x
≥ γC0 u2 (x, t),
(4.7)
where C0 is the constant in Theorem 3.1.
Putting (4.7) into (2.6), one has
µ
¶
dm
3−γ
γ 2
≤ − m +
− γC0 u2 (ξ(t), t)
dt
2
2
(3 − (1 + 2C0 )γ) cosh(1/2)
γ
ku0 k2H 1 .
≤ − m2 +
2
4γ sinh(1/2)
Due to Lemma 4.2, it is easy to see that the condition given in Theorem 4.1
guarantees the blow-up of solutions.
(iii) 1 < γ ≤ γ1
Similar to the case (ii),
µ
µ
¶
¶
3−γ 2 γ 2
1 2
3−γ
3(γ − 1)
2
G∗
u + ux (x, t) =
G ∗ u + ux (x, t) +
G ∗ u2x (x, t)
2
2
2
2
4
µ
¶
3−γ
1 2
2
≥
G ∗ u + ux (x, t)
2
2
3−γ
≥
C0 u2 (x, t),
2
13
which reduces (2.6) to
dm
γ
(3 − γ)(1 − C0 ) cosh(1/2)
≤ − m2 +
ku0 k2H 1 .
dt
2
4γ sinh(1/2)
we can get the blow-up result in this case.
(iv) γ1 < γ < 3
Similar argument for the first case 0 < γ < γ0 can be used here.
The proof of Theorem 4.1 is complete.
Remark 4.2 From the proof, the reader may ask why we do not find the optimal
constant C for the following convolution problem directly.
µ
¶
3−γ 2 γ 2
G∗
u + ux (x) ≥ Cu2 (x).
2
2
The question is that we do not have any effective method to solve this problem at
present. We hope we can deal with it in the near future.
5
The nonperiodic case
To find the best constant in R is different from the periodic case. In this case the
Green function for Q−2 is G(x) = 21 e−|x| . We can compute
µ
G∗
µ
¶
¶
Z
γα02 2 γ 2
1 0 x γα02 2
γ 2
e
u + ux (0) =
u (x) + ux (x) dx
2
2
2 −∞
2
2
µ 2
¶
Z ∞
γα0 2
1
γ 2
−x
e
+
u (x) + ux (x) dx
2 0
2
2
Z 0
Z
1
1 ∞ −x
x
≥
e γα0 u(x)ux (x)dx −
e γα0 u(x)ux (x)dx
2 −∞
2 0
¯∞ ´ γα0
γα0 ³ x ¯¯0
e −∞ − e−x ¯0 −
G ∗ u2 (0),
=
4
2
where α0 is the number in (4.4). Hence we have
µ
¶
3−γ 2 γ 2
γα0
G∗
u + ux (0) ≥
.
2
2
2
Moreover,
γα0
2
is the best constant if and only if
α0 u = ux , in (−∞, 0) and − α0 u = ux , in (0, ∞),
14
which can be solved as u = e−α0 |x| . Therefore we have
µ
¶
3−γ 2 γ 2
γα0 2
G∗
u + ux (x) ≥
u (x).
2
2
2
Moreover,
6
γα0
2
is the best constant obtained by u = λe−α0 |x−y| for some λ, y ∈ R.
Acknowledgment
The author thanks The Institute of Mathematical Sciences for the financial support and the warm hospitality during his visit in December 2004. This work is
partially supported by Hong Kong RGC Earmarked Grants CUHK-4028-04P and
Shanghai Leading Academic Discipline.
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