Local well-posedness and blow-up criteria of solutions for a rod equation

Math. Nachr. 278, No. 14, 1726 – 1739 (2005) / DOI 10.1002/mana.200310337

Local well-posedness and blow-up criteria of solutions for a rod equation

Yong Zhou∗ 1

1 Department of Mathematics, East China Normal University, Shanghai 200062

Received 14 April 2003, revised 29 December 2003, accepted 17 January 2004Published online 6 October 2005

Key words Well-posedness, blow-upMSC (2000) 49K40, 37L05

In this paper we consider a new rod equation derived recently by Dai [Acta Mech. 127 No. 1–4, 193–207(1998)] for a compressible hyperelastic material. We establish local well-posedness for regular initial dataand explore various sufficient conditions of the initial data which guarantee the blow-up in finite time bothfor periodic and non-periodic case. Moreover, the blow-up time and blow-up rate are given explicitly. Someinteresting examples are given also.

c© 2005 WILEY-VCH Verlag GmbH & Co. KGaA, Weinheim

1 Introduction

Although a rod is always three-dimensional, if its diameter is much less than the axial length scale, one-dimensionalequations can give a good description of the motion of the rod. Recently Dai [7] derived a new (one-dimensional)nonlinear dispersive equation including extra nonlinear terms involving second-order and third-order derivativesfor a compressible hyperelastic material. The equation reads

vτ + σ1vvξ + σ2vξξτ + σ3(2vξvξξ + vvξξξ) = 0

where u(ξ, τ) represents the radial stretch relative to a pre-stressed state, and σ1 �= 0, σ2 < 0 and σ3 ≤ 0 areconstants determined by the pre-stress and the material parameters. If one introduces the following transforma-tions

τ =3√−σ2

σ1t , ξ =

√−σ2 x ,

then the above equation turns into

ut − utxx + 3uux = γ(2uxuxx + uuxxx) (1.1)

where γ = 3σ3/(σ1σ2). In [8], the authors derived that value range of γ is from −29.4760 to 3.4174 for somespecial compressible materials. In this paper, from the mathematical view point, we regard γ as a real number.

When γ = 1 in Eq. (1.1), we recover the shallow water (Camassa–Holm) equation derived physically byCamassa and Holm in [2] by approximating directly the Hamiltonian for Euler’s equations in the shallow waterregime, where u(x, t) represents the free surface above a flat bottom. Some satisfactory results have been obtainedfor this shallow water equation recently. Local well-posedness for the initial datum u0(x) ∈ Hs with s > 3/2was proved by several authors, see [14, 16, 18]. On the other hand, Himonas and Misiole [10] showed thatthe Camassa–Holm equation is not well-posed for initial data in Hs for s < 3/2. These results suggest thats = 3/2 is the critical index for well-posedness. Moreover, wave breaking for a large class of initial data hasbeen established in [3, 4, 5, 14, 15, 21, 22]. However, in [19, 20], Xin and Zhang showed global existence anduniqueness for weak solutions with u0(x) ∈ H1. The solitary waves of Camassa–Holm equation are peakedsolitons [2] and are orbitally stable [6].

∗ e-mail: [email protected], Phone: +21 6223 3050 (O), +21 6245 7134 (H), Fax: +21 6223 8105


Math. Nachr. 278, No. 14 (2005) / www.mn-journal.com 1727

If γ = 0, Eq. (1.1) is the BBM equation, a well-known model for surface waves in a canal [1], and its solutionsare global.

We now conclude this introduction by outlining the rest of the paper. In Section 2, we establish local well-posedness for Eq. (1.1) with initial datum u0 ∈ Hs, s > 3/2, and that the lifespan of the corresponding solutionis finite if and only if its first-order derivative blows up. In Section 3, we investigate various sufficient conditions(which are different for different γ) of the initial datum to guarantee the finite time blow-up for periodic case andsome examples are given at the end of that section. Blow-up criteria are established for the non-periodic case inSection 4. Finally we compare the rod equation with Camassa–Holm equation and some questions are proposedin Section 5.

2 Local well-posedness

In this section, we consider both periodic and nonperiodic case for γ �= 0. Let S := R/Z be the unit circle andT = R or S.

Set Qs =(1 − ∂2

x

)s/2, then the operator Q−2 can be expressed by

Q−2f = G ∗ f =∫

T

G(x − y)f(y) dy

for any f ∈ L2(T ) with

G(x) =

⎧⎪⎪⎨⎪⎪⎩

12

e−|x| , if T = R ,

cosh(x − [x] − 1/2)2 sinh(1/2)

, if T = S ,

where [x] denotes the integer part of x. Then Eq. (1.1) can be written as

ut + γuux + ∂xQ−2„

3 − γ

2u2 +

γ

2u2

x

«= 0 . (2.1)

The local well-posedness theorem for Eq. (1.1) reads

Theorem 2.1 Let u0(x) ∈ Hs(T), s > 3/2, be the initial datum. Then there exists T = T (‖u0‖Hs) > 0 anda unique solution u, which depends continuously on the initial datum u0, to Eq. (1.1) (or Eq. (2.1)) such that

u ∈ C([0, T ); Hs(T)

) ∩ C1([0, T ); Hs−1(T)

).

Moreover, the following two quantities E and F , defined by⎧⎪⎪⎨⎪⎪⎩

E(u)(t) =∫

T

(u2(x, t) + u2

x(x, t))dx ,

F (u)(t) =∫

T

(u3(x, t) + γu(x, t)u2

x(x, t))dx ,

are invariants with respect to time t for Eq. (1.1).

This result can be showed by applying Kato’s theorem [11, 13], just as what was done for the Camassa–Holmequation (γ = 1 in Eq. (2.1)) [4, 16]. So we would like to skip the detailed proof here, for concise.

The maximum value of T in Theorem 2.1 is called the lifespan of the solution, in general. If T < ∞, that islim supt↑T ‖u(., t)‖Hs = ∞, we say that the solution blows up in finite time. The following theorem tells us thatthe solution blows up if and only if the first-order derivative blows up. This phenomenon coincides physicallywith the rod breaking.

Theorem 2.2 Let u0(x) ∈ Hs(T), s ≥ 2, and u be the corresponding solution to problem (1.1) with lifespanT . Then

supx∈T, 0≤t<T

|u(x, t)| ≤ C (‖u0‖H1) . (2.2)


1728 Zhou: Well-posedness and blow-up for a rod equation

T is bounded if and only if

lim inft↑T

infx∈T

{γux(x, t)} = −∞ . (2.3)

P r o o f. First, we have the following Sobolev embedding inequality

‖f‖2L∞ ≤ C ‖f‖2

H1 ,

where C is an absolute constant. More precisely,

C =

⎧⎪⎪⎨⎪⎪⎩

e1/2 + e−1/2

2(e1/2 − e−1/2), if T = S ,

12

, if T = R ,

see Theorem 3.1 and Eq. (4.1) below, respectively. Then Eq. (2.2) follows from the conservation of H1-norm ofthe solution.

Taking y = u − uxx, then from Eq. (1.1) one can get the equation for y

yt + γyxu + 2γyux + 3(1 − γ)uux = 0 . (2.4)

Applying y on Eq. (2.4) and integration by parts, we obtain

d

dt

∫T

y2 dx = − 3γ

∫T

y2ux dx + 6(γ − 1)∫

T

yuux dx

≤ − 3γ

∫T

y2ux dx + 6C |γ − 1|∫

T

yux dx

≤ − 3γ

∫T

y2ux dx + 3C |γ − 1|∫

T

(y2 + u2

x

)dx

≤ − 3γ

∫T

y2ux dx + 6C |γ − 1|∫

T

y2 dx ,

(2.5)

where we use the bounds of u and that ‖y‖L2 is equivalent to ‖u‖H2 .Due to Gronwall’s inequality, it is clear that, from Eq. (2.5), γux is bounded from below on [0, T ), then the

H2-norm of the solution is also bounded on [0, T ).On the other hand,

u = Q−2y =∫

T

G(x − ξ)y(ξ) dξ ,

therefore

‖ux‖L∞ ≤ ‖Gx‖L2‖y‖L2 . (2.6)

Eq. (2.6) tells us if the H2-norm of the solution is bounded then the L∞-norm of the first derivative is bounded.This finishes the proof.

For γ �= 0, we set

m(t) := infx∈T

ux(x, t)sign{γ} , t ≥ 0 , (2.7)

where sign{a} is the sign function of a ∈ R and we set m0 := m(0). Then for every t ∈ [0, T ) there existsat least one point ξ(t) ∈ T with m(t) = ux(ξ(t), t). Just as the proof given in [5], one can show the followingproperty of m(t).

Lemma 2.3 Let u(t) be the solution to Eq. (1.1) on [0, T ) with initial data u0 ∈ H2(T), as given by Theorem2.1. Then the function m(t) is almost everywhere differentiable on [0, T ), with

dm(t)dt

= utx(ξ(t), t) a.e. on (0, T ) .

To consider the quantity m(t) for wave breaking comes from an idea of Seliger [17] originally, the rigorousregularity proof is given in [5] for Camassa–Holm equation.



3 Blow-up criteria for periodic case

Before give the main theorems of this section, we would like to prove the following theorem concerning optimalconstant for Sobolev embedding.

Theorem 3.1 For all f ∈ H1(S), the following inequality

maxx∈[0,1]

f2(x) ≤ C0‖f‖2H1(S)

holds, where C0 = e1/2+e−1/2

2(e1/2−e−1/2).

Moreover, C0 is the minimum value, so in this sense, C0 is the optimal constant which is obtained by the

associated Green function G = cosh(x−[x]− 12 )

2 sinh( 12 )

.

P r o o f. The aim of the proof is to find the optimal constant c such that

‖u‖2L∞(S) ≤ c ‖u‖2

H1(S) for all u ∈ H1(S) .

Let

A ={u ∈ H1(S)

∣∣ ‖u‖L∞(S) = 1}

and

I[u] = ‖u‖2H1(S) =

∫S

(u2 + u2

x

)dx , u ∈ A .

Step 1. We need to find the minimizer of the functional I[u] in the space A.Let {uk}∞k=1 ⊂ A be a minimizing sequence, i.e., I[uk] → infu∈A I[u] as k → ∞.

Since I[1] = 1, then 0 < infu∈A I[u] ≤ 1. Therefore {uk}∞k=1 is bounded in H1(S), so there exists asubsequence {ukj}∞k=1 ⊂ {uk}∞k=1, which we also denote by {uk}∞k=1, and a u0 ∈ H1(S) with

uk ⇀ u0 in H1(S) .

On the other hand,

I[u] = ‖u‖2H1(S) and I[u0] ≤ lim inf

k→∞I[uk] ,

so I[u0] = infu∈A I[u] and we can get uk → u0 in H1(S). By the Sobolev embedding, we know that

‖uk − u0‖2L∞(S) ≤ c ‖uk − u0‖2

H1(S) ,

so ‖u0‖L∞(S) = 1. Hence we have u0 ∈ A and uk → u0 in A.

Step 2. By step 1, we know that I[u] has a minimizer in A, say g. In the following steps, the property of g willbe explored. Let

f(x) =4π2 + 1 + 2 cos(2πx)

4π2 + 3.

By direct computation, it is easy to obtain that f ∈ A and

I[f ] =4π2 + 14π2 + 3

.

Therefore, the steady state 1 is not a minimizer and consequently, for g, there exists some point, say x0, such thatg(x0) < 1.

Now let ϕ be any C∞ function with compact support in a small neighborhood U(x0, δ) = {x ∈ S ||x − x0| < δ}. Since g is continuous

(H1(S) ⊂ C(S)

), one can choose δ and ε sufficiently small such that

(εϕ + g)(x) < 1 for all x ∈ U(x0, δ) .

Then ‖g + εϕ‖L∞(S) = 1, since εϕ ≡ 0 outside U(x0, δ).



Set

i(t) = I[g + tεϕ] =∫

S

((g + tεϕ)2 + (gx + tεϕx)2

)dx ,

where t ∈ R satisfies ‖g + tεϕ‖L∞(S) = 1.By g being a minimizer,

0 = i′(0) = 2ε

∫S

(gϕ + gxϕx) dx = 2ε

∫U(x0,δ)

g(ϕ − ϕxx) dx .

So in the sense of distribution, g is a weak (smooth) solution to the following equation

u − uxx = 0 in {x ∈ S | g(x) < 1} .

Step 3. We will show that g must have only one maximum point.Since g is continuous on S, ‖g‖L∞(S) = maxx∈S |g(x)|.Claim 1 The maximum points cannot be dense in any interval contained in S.

Suppose otherwise, then by the continuity of g, we get g ≡ 1 on some interval, say [a, b] ⊂ [0, 1].Let

g(x) =

⎧⎪⎨⎪⎩

4π2 + 1 + 2 cos(

2π(x−a)b−a

)4π2 + 3

, if x ∈ [a, b] ,

g(x), otherwise .

It is easy to check that g(x) ∈ A. But

I[g] =∫

S

(g2 + g2

x

)dx =

∫[a,b]

(g2 + g2

x

)dx +

∫S\[a,b]

(g2 + g2

x

)dx

= (b − a)4π2 + 14π2 + 3

+∫

S\[a,b]

(g2 + g2

x

)dx

< I[g] .

This is a contradiction, since g is a minimizer.

Claim 2 The number of maximal points cannot exceed 1.

Suppose that the number of maximal points is larger than or equal to 2. By Claim 1, we can suppose that x0

is a maximal point and there exist 0 < ε0 < 1 and 0 < δ < 1/2 such that

1 − ε0 < |g(x)| < 1 for all x ∈ {U(x0, δ) \ x0} .

Let M0 = max{|g(x0 − δ)|, |g(x0 + δ)|} and U(x0) be a neighborhood of x0 defined as follows

U(x0) = {x ∈ U(x0, δ) | M0 ≤ |g(x)| } .

Then set

g1(x) =

{M0 sign{g(x0)} , if x ∈ U(x0) ,

g(x) , otherwise .

It is not difficult to check that g1 ∈ A, and

I[g1] =∫

S

(g21 + g2

1x

)dx =

∫U(x0)

M20 dx +

∫S\U(x0)

(g2 + g2

x

)dx

<

∫U(x0)

g2 + g2x dx +

∫S\U(x0)

(g2 + g2

x

)dx

= I[g] .



This contradicts the fact that g is the minimizer. Hence Claim 2 is true.

Step 4. By step 3 we know that g has only one maximal point. After choosing a suitable coordinate, we canassume that

g < 1 in (0, 1) and g(0) = g(1) = 1 .

It follows from step 2 that g is a weak solution of the following equation in (0, 1):

g − gxx = 0 , x ∈ (0, 1) , g(0) = g(1) = 1 . (3.1)

For the above equation, we know the nontrivial classical solution has the form aex + ce−x. Thus by the boundarycondition, after calculation we get

a =1 − e−1

e − e−1and c =

e − 1e − e−1

.

Then, by the uniqueness of solutions to Eq. (3.1), one obtains

g =cosh

(x − 1

2

)cosh

(12

) , 0 ≤ x ≤ 1 .

Finally, we can get the constant C0 = e1/2+e−1/2

2(e1/2−e−1/2).

Let m(t) be defined by Eq. (2.7), we have the following blow-up criterion first.

Theorem 3.2 Let γ ∈ R, γ �= 0, 3. Assume that u0 ∈ H2(S) satisfies m0 < 0 and

m20 >

⎧⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎩

(3 − γ)C0

−γ‖u0‖2

H1(S) , if γ < 0 ,

(6 − γ −

√12γ − 3γ2

)C0

2γ‖u0‖2

H1(S) , if 0 < γ < 3 ,

(γ − 3)C0

γ‖u0‖2

H1(S) , if γ > 3 ,

where m0 = infx∈S u0x(x), and C0 is given in Theorem 3.1. Then the life span T > 0 of the correspondingsolution to Eq. (1.1) is finite, i.e., rod breaking occurs. If γ = 3, all solutions with non-constant initial datau0 ∈ H2(S) blow up in finite time.

P r o o f. Differentiating Eq. (2.1) with respect to x, one obtains

uxt + γuuxx +γ

2u2

x =3 − γ

2u2 − G ∗

(3 − γ

2u2 +

γ

2u2

x

). (3.2)

Then by Lemma 2.5 and the definition of m(t) (see Eq. (2.7)), we have the equation for m(t), from Eq. (3.2)

sign{γ} dm

dt= −γ

2m2 +

3 − γ

2u2(ξ(t), t) −

[G ∗

(3 − γ

2u2 +

γ

2u2

x

)](ξ(t), t) (3.3)

a.e. on (0, T ).If γ < 0, from Eq. (3.3), we get

dm

dt=

γ

2m2 − 3 − γ

2u2(ξ(t), t) +

[G ∗

(3 − γ

2u2 +

γ

2u2

x

)](ξ(t), t)

≤ γ

2m2 +

[G ∗ (3 − γ)u2

2

](ξ(t), t)

≤ γ

2m2 +

3 − γ

2‖G‖L∞‖u2‖L1

≤ γ

2m2 +

(3 − γ)C0

2‖u0‖2

H1 .

(3.4)



It is easy to prove that if

m0 < 0 and m20 >

(3 − γ)C0

−γ‖u0‖2

H1 ,

then the solution m(t) of the Riccati type equation (3.4) goes to −∞ in finite time. More precisely, if we set aconstant δ ∈ (0, 1) satisfying

δm20 =

(3 − γ)C0

−γ‖u0‖2

H1 ,

then the lifespan of the solution to Eq. (1.1) is less than 2/(m0(1 − δ)γ).Now we go to the case 0 < γ < 3. By the representation of G, it follows[

G ∗(

3 − γ

2u2 +

γ

2u2

x

)](x, t)

=1

2 sinh(

12

) ∫ x

0

ex−η− 12 + e

12+η−x

2

(3 − γ

2u2(η, t) +

γ

2u2

x(η, t))

dη

+1

2 sinh(

12

) ∫ 1

x

ex−η+ 12 + eη−x− 1

2

2

(3 − γ

2u2(η, t) +

γ

2u2

x(η, t))

dη .

(3.5)

Since ∫ x

0

e−η(γ

2α2u2(η, t) +

γ

2u2

x(η, t))

dη ≥ −∫ x

0

e−ηγαu(η, t)ux(η, t) dη

= − e−η γα

2u2(η, t)

∣∣x0−∫ x

0

e−η γα

2u2(η, t) dη

holds for any α > 0, we have∫ x

0

e−η((

α2 + α) γ

2u2(η, t) +

γ

2u2

x(η, t))

dη ≥ −αγ

2e−ηu2(η, t)

∣∣x0

.

Now we let

α2 + α =3 − γ

γ. (3.6)

Eq. (3.6) has one positive root α0 with

α0 = −12

+12

√12 − 3γ

γ.

Therefore ∫ x

0

e−η

(3 − γ

2u2(η, t) +

γ

2u2

x(η, t))

dη ≥ −α0γ

2e−ηu2(η, t)

∣∣x0.

Moreover, from Eq. (3.5), just use the above trick for each term, then one obtains[G ∗(

3 − γ

2u2 +

γ

2u2

x

)](ξ(t), t) ≥ α0γ

2u2(ξ(t), t) . (3.7)

Now combining Eqs. (3.3) and (3.7), we have

dm

dt≤ − γ

2m2 +

6 − γ −√

12γ − 3γ2

4u2(ξ(t), t)

≤ − γ

2m2 +

6 − γ −√

12γ − 3γ2

4C0 ‖u0‖2

H1 .

(3.8)



Just as the case γ < 0, we can get that if

m0 < 0 and m20 ≥ 6 − γ −

√12γ − 3γ2

2γC0 ‖u0‖2

H1 ,

then the solution to Eq. (1.1) (or Eq. (3.8)) blows up in finite time. And we can give an explicit estimate for thelifespan of the corresponding solution. Note that

6 − γ −√12γ − 3γ2

4<

3 − γ

2for 0 < γ < 3 .

If γ > 3, it follows from Eq. (3.3) that m(t) satisfies

dm

dt≤ − γ

2m2 +

[G ∗ (γ − 3)u2

2

](ξ(t), t)

≤ − γ

2m2 +

γ − 32

‖G‖L∞‖u2‖L1

≤ − γ

2m2 +

(γ − 3)C0

2‖u0‖2

H1 .

(3.9)

Eq. (3.9) is similar to Eq. (3.4) and a corresponding condition for blow-up is easy to be obtained.If γ = 3, Eq. (3.3) reduces to

dm

dt= −3

2m2 −

[G ∗ 3

2u2

x

](ξ(t), t) ≤ −3

2m2. (3.10)

For any non-constant initial datum u0, we have m0 < 0. Hence the solution to Eq. (3.10) (also Eq. (1.1)) blowsup in finite time.

Theorem 3.3 Suppose u0 ∈ H2(S) satisfies m0 < 0 and

m20 >

(6 − γ −

√12γ − 3γ2

)2γ

(‖u0‖2

H1(S) +(∫

S

u0 dx

)2)

,

for 0 < γ < 3. Then the life span of the corresponding solution to Eq. (1.1) is finite.

P r o o f. First we prove the following inequality for any function f ∈ H2(S):

‖f‖2L∞ ≤ ‖f‖2

H1 +(∫

S

f dx

)2

. (3.11)

Indeed, since f is continuous on S, there exists a point x0 ∈ S such that∫

Sf dx = f(x0). For any x ∈ S, we

have

f2(x) −(∫

S

f dx

)2

= f2(x) − f2(x0) =∫ x

x0

2ffx dx ≤∫ x

x0

(f2 + f2

x

)dx ≤ ‖f‖2

H1(S) ,

which yields Eq. (3.11).On the other hand, it is obvious that∫

S

u(x, t) dx

is unchanged for solutions u(x, t) with respect to time for initial data u0 ∈ H2(S).Therefore Theorem 3.3 can be proved if one uses

‖u(., t)‖2L∞(S) ≤ ‖u0‖2

H1(S) +(∫

S

u0 dx

)2

, for all t ≥ 0 ,

in Eq. (3.8) instead of ‖u(., t)‖2L∞(S) ≤ C0‖u0‖2

H1(S).



Remark 3.4 Even for γ = 1 (Camassa–Holm equation), Theorem 3.2 is a significant improvement of theprevious result (see [3]). Theorem 3.3 is similar to the criterion established by the author in [21] for Camassa–Holm equation.

When we study the blow-up problems for differential equations, according to [9], the basic questions includeswhen, where, and how. Theorems 3.2 and 3.3 give answer to the first two questions. The following theoremanswers one aspect of the third question, the rate of blow-up.

Theorem 3.5 Suppose u0(x) ∈ H2(S), and that u(x, t) is the corresponding solution. If the lifespan of thesolution u(x, t) is finite, then

limt→T

{(T − t)m(t)} = − 2|γ| .

P r o o f. The conclusion follows from the theory of ordinary differential equation corresponding to Eq. (3.3).Indeed, for any γ �= 0, we have

0 ≤ dm

dt+

|γ|2

m2 ≤ C , (3.12)

where C > 0 is a constant depending on γ, C0 and ‖u0‖H1 . Since m(t) → −∞ as t → T , from Eq. (3.3) itfollows that for any 0 < ε < |γ|/2 there exists a t0 such that m(t) < −C/ε for all t ∈ (t0, T ). Then we obtain

−|γ|2

≤ 1m2

dm

dt≤ −|γ|

2+ ε . (3.13)

For t ∈ (t0, T ), integrating Eq. (3.13) on (t, T ) gives( |γ|2

− ε

)(T − t) ≤ − 1

m(t)≤ |γ|

2(T − t) .

The proof is complete from the arbitrariness of ε.

At the end of this section, some examples are given to illustrate the applications of Theorem 3.2 and The-orem 3.3, respectively. We just take γ = 1, for simplicity. Of course, we can give examples for any fixed0 < γ < 3.

Example 3.6 u0(x) = sin(2πx) + 4, x ∈ [0, 1].Direct computation yields

‖u0‖2H1(S) = 2π2 +

12

+ 16

and

minx∈S

u0x(x) = −2π > −(‖u0‖2

H1(S) +(∫ 1

0

u0 dx

)2)1/2

(≈ −7.22767) .

So the classical solution u(x, t) to Eq. (1.1) corresponding to u0(x) = sin(2πx) + 4 only exists for finite timedue to Theorem 3.2 (γ = 1). But Theorem 3.3 does not work. Note that also none of theorems in [3, 4] can beapplied in this case.

Example 3.7 We would like to give an example, which illustrates the applicability of Theorem 3.3 as follows.Let

u0(x) =

⎧⎪⎨⎪⎩

x − a , 0 ≤ x ≤ a ,

0 , a ≤ x ≤ 1 − a ,

1 − a − x , 1 − a ≤ x ≤ 1 ,

where 0 < a < 12 is to be determined later. By simple computations, we get

‖u0‖2H1(S) =

2a3

3+ 2a and

(∫ 1

0

u0(x) dx

)2

= a4 .



Since minx∈S u0(x) = −1, we want a to satisfy

1 < C0

(2a3

3+ 2a

)and 1 >

2a3

3+ 2a + a4 .

By Mathematica, one obtains that when a = 0.44,

−1 > −√

C0

(2a3

3+ 2a

)(≈ −1.01358) (3.14)

and

−1 < −√

2a3

3+ 2a + a4 (≈ −0.97427) . (3.15)

It is easy to show that by standard mollification at the points 0, a and 1−a (a = 0.44), we can get a new functionu0 ∈ H2(S) that also satisfies Eqs. (3.14) and (3.15).

Let u0 be the initial data to Eq. (1.1). By the above analysis, we can see in this case, only Theorem 3.3 works.

4 Blow-up criteria for non-periodic case

Theorem 4.1 Assume that the initial profile u0 ∈ H2(R) satisfies m0 < 0 and

m20 >

⎧⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎩

3 − γ

−2γ‖u0‖2

H1(R) , if γ < 0 ,

6 − γ −√

12γ − 3γ2

4γ‖u0‖2

H1(R) , if 0 < γ < 3 ,

γ − 32γ

‖u0‖2H1(R) , if γ > 3 .

Then the life span T > 0 of the corresponding solution to Eq. (1.1) is finite, i.e., rod breaking occurs. If γ = 3,all solutions with nonzero initial data u0 ∈ H2(R) blow up in finite time.

P r o o f. First, we have the following inequality

‖f‖L∞(R) ≤ 1√2‖f‖H1(R) for all f ∈ H1(R) . (4.1)

Indeed, for any x ∈ R, we have

2f2(x) = 2(∫ x

−∞ffx dx −

∫ ∞

x

ffx dx

)≤∫

R

(f2 + f2

x

)dx = ‖f‖H1(R) ,

which yields Eq. (4.1).For the rod equation, since the H1-norm is conserved, one has

supx∈R,t∈[0,T )

|u(x, t)| ≤ 1√2‖u0‖H1 .

Just as the proof for Theorem 3.2, we have Eq. (3.3) for m(t) and the remaining proof is similar. For example, inEq. (3.4) we use ‖G‖L∞(R) ≤ 1/2 instead. So for γ < 0, we have

dm

dt≤ γ

2m2 +

3 − γ

4‖u0‖2

H1 ,

and the case γ > 3 is similar to this case.



For 0 < γ < 3, we obtain by the representation of G, that

G ∗(

3 − γ

2u2 +

γ

2u2

x

)(x) =

12

∫ x

−∞e−x+y

(3 − γ

2u2 +

γ

2u2

x

)(y) dy

+12

∫ ∞

x

ex−y

(3 − γ

2u2 +

γ

2u2

x

)(y) dy .

(4.2)

The following inequality∫ x

−∞ey(α2

0

γ

2u2 +

γ

2u2

x

)(y) dy ≥

∫ x

−∞eyαγ(uux)(y) dy

=αγ

2exu2(x) +

∫ x

−∞

αγ

2eyu2(y) dy

implies ∫ x

−∞e−x+y

((α2 + α)

γ

2u2 +

γ

2u2

x

)(y) dy ≥ αγ

2u2(x) . (4.3)

Then we take α2 + α = 3−γγ , that is

α = −12

+12

√12 − 3γ

γ.

Similarly we get the estimate of the second term in Eq. (4.2) as∫ ∞

x

ex−y

(3 − γ

2u2 +

γ

2u2

x

)(y) dy ≥ αγ

2u2(x) . (4.4)

Combining Eqs. (4.2), (4.3) and (4.4), one has[G ∗(

3 − γ

2u2 +

γ

2u2

x

)](ξ(t), t) ≥ αγ

2u2(ξ(t), t) =

−γ +√

12γ − 3γ2

4u2(ξ(t), t) . (4.5)

Hence Eq. (3.3) reduces to

dm

dt≤ −γ

2m2 +

6 − γ −√12γ − 3γ2

8‖u0‖2

H1 .

For γ = 3, we have the following equation for m(t)

dm

dt= −3

2m2 − G ∗

(32

u2x

)≤ −3

2m2 .

For any nonzero u0 ∈ H2(R), it is clearly that m(0) is negative, hence the lifespan of the solution is finite.The proof is complete.

Theorem 4.2 Assume that u0 ∈ H2(R) and that u(x, t) is the corresponding solution. Set

q(t) := sign{γ}∫

R

u3x(x, t) dx and q0 := q(0) .

If q0 < 0 the following conditions are satisfied:

q20 >

⎧⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎩

3(3 − γ)−2γ

‖u0‖6H1(R) , if γ < 0 ,

3(6 − γ −

√12γ − 3γ2

)4γ

‖u0‖6H1(R) , if 0 < γ < 3 ,

3(γ − 3)2γ

‖u0‖6H1(R) , if γ > 3 .

Then the u(x, t) to Eq. (1.1) blows up in finite time.



P r o o f. First, for any f ∈ H2(R), since∣∣∣∣∫

R

f3x dx

∣∣∣∣ ≤(∫

R

f4x dx

)1/2(∫R

f2x dx

)1/2

,

it follows that∫R

f4x dx ≥ 1

‖f‖2H1

(∫R

f3x dx

)2

. (4.6)

Multiplying Eq. (3.2) by u2x, after integration by parts, one has

sign{γ}dq

dt= −γ

2

∫R

u4x dx +

3(3 − γ)2

∫R

u2xu2 dx − 3

∫R

u2xG ∗

(3 − γ

2u2 +

γ

2u2

x

)dx . (4.7)

If γ < 0, by Eqs. (4.1), (4.6) and (4.7), one obtains

dq

dt≤ γ

2‖u0‖2H1

q2 +3(3 − γ)

2

∫R

u2x

(G ∗ u2

)dx

≤ γ

2 ‖u0‖2H1

q2 +3(3 − γ)

2‖u0‖2

H1 ‖G ∗ u2‖L∞ (4.8)

≤ γ

2 ‖u0‖2H1

q2 +3(3 − γ)

4‖u0‖4

H1 .

It is easy to see that under the corresponding condition of Theorem 4.2, the solution of Eq. (4.8) blows up in finitetime.

If 0 < γ < 3, from Eqs. (4.5) and (4.7), we obtain,

dq

dt≤ − γ

2‖u0‖2H1

q2 +3(3 − γ)

2

∫R

u2u2x dx − 3α0γ

2

∫R

u2u2x dx

≤ − γ

2‖u0‖2H1

q2 +3(6 − γ −

√12γ − 3γ2

)4

‖u‖2L∞

∫R

u2x dx (4.9)

≤ − γ

2 ‖u0‖2H1

q2 +3(6 − γ −

√12γ − 3γ2

)8

‖u0‖4H1 .

Similarly for γ > 3, one has

dq

dt≤ − γ

2 ‖u0‖2H1

q2 +3(γ − 3)

2

∫R

u2x

(G ∗ u2

)dx

≤ − γ

2 ‖u0‖2H1

q2 +3(γ − 3)

4‖u0‖4

H1 .

(4.10)

From standard argument for Riccati type equations, it is clear that under the condition of Theorem 4.2 the solutionof Eqs. (4.9) and (4.10) blow up in finite time. Therefore, for any γ ∈ R, γ �= 0, we have

limt→T

[sign{γ}

∫R

u3x(x, t) dx

]= −∞ for some T > 0 . (4.11)

Eq. (4.11) and the estimate∣∣∣∣∫

R

u3x dx

∣∣∣∣ ≤ C(s) ‖u‖2H1‖u‖Hs = C(s) ‖u0‖2

H1‖u‖Hs , for any s ∈ (3/2, 2] ,

show that limt→T ‖u‖H2 = ∞. Moreover,

sign{γ}∫

R

u3x(x, t) dx ≥ 1

|γ| infx∈R

{γux} ‖u0‖2H1

yields that limt→T infx∈R{γux(x, t)} = −∞.



From the above argument, we show that if γ = 3 then any solution with nonzero initial datum blows up infinite time again.

Remark 4.3 Theorem 4.1 is a similar result as that in [5], where γ = 1 for Camassa–Holm equation, whileTheorem 4.2 is a improvement of that in [14], even for γ = 1.

Just as the periodic case, we have the blow-up rate for non-periodic case

limt→T

{(T − t)m(t)} = − 2|γ| and lim

t→T

{(T − t)

∫R

u3(x, t) dx

}= −2 ‖u0‖2

H1

|γ| .

5 Comparison with Camassa–Holm equation

In [15], McKean gave a sufficient and necessary criterion on y0 = Q2u0 (initial potential) for the finite timeformation of singularities in a smooth solution to Camassa–Holm equation

ut − utxx + 3uux = 2uxuxx + uuxxx . (5.1)

He showed that the solution to Eq. (5.1) with u0 ∈ H3(T) blows up in finite time if and only if there exist pointsξ < ζ such that y0(ξ) > 0 and y0(ζ) < 0. In other words, the smooth solution exists globally if and only if

i) y0 does not change sign for periodic case,ii) y0 does not change sign or the set {x | m0(x) > 0} locates right to the set {x | m0(x) < 0} for

non-periodic case.Let u(x, t) solve Eq. (5.1) and let q(x, t) be the particle line along u(x, t), i.e., q(x, t) satisfies the following

equation {qt = u(q, t) , 0 < t < T , x ∈ R ,

q(x, 0) = x , x ∈ R ,(5.2)

where T is the life span of the solution, then q is a diffeomorphism of the line.Set the potential y := u − uxx, then

yt = −yxu − 2yux . (5.3)

From Eqs. (5.2) and (5.3), one obtains⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩

qx(x, t) = eR t0 ux(q,s) ds , y(q)q2

x = y0(x) ,

d

dt[eq(ux − u)] (q) = −1

2eq(ux − u)2(q) − 1

2

∫ q

−∞es(ux − u)2(s) ds ,

d

dt

[e−q(ux + u)

](q) = −1

2e−q(ux + u)2(q) − 1

2

∫ ∞

q

e−s(ux + u)2(s) ds ,

d

dtux(q) =

12

u2(q) − 12

u2x(q) − 1

4

∫ ∞

−∞e−|q−s|(ux ± u)2(s) ds .

The above identities are the most important ingredients of the whole paper [15].But for the rod equation Eq. (1.1), y := u − uxx satisfies

yt = −γyxu − 2γyux + 3(γ − 1)uux ,

which destroys all the above important and beautiful identities, if γ �= 1.So far, the sufficient conditions to guarantee the global existence of smooth solution to Eq. (1.1) are not clear

for general γ ∈ R, γ �= 1. We hope to report it later.Dai and Hou investigated the solitary waves for the rod equation in [8] for γ ∈ R. They showed that only for

γ < 1, there are smooth solitary waves such that

u(x, t) = u(ζ) ∈ H1(R) , ζ = x − V t with V a speed parameter ,



and only for γ = 1, the solitary waves u(x, t) = u(ζ) ∈ H1(R) are peaked solitons (see also [2]). For γ = 1,Constantin and Strauss [6] proved the orbital stability in H1-norm of peakons by a method which depends on thespecial structure of Camassa–Holm equation. We hope to discuss the stability of the smooth solitary waves forγ < 1 in a forthcoming paper, since it is very interesting and useful both mathematically and for applications.

On the other hand, one can establish like in [19, 20] the well-posedness of weak solutions for the problem(1.1).

Acknowledgements The author would like to express sincere gratitude to his supervisor Professor Zhouping Xin for en-thusiastic guidance and constant encouragement. Thanks also to Professor Shing-Tung Yau and Professor Zhouping Xin forproviding an excellent study and research environment in The Institute of Mathematical Sciences. This work is partiallysupported by NSFC and Shanghai Rising-Star Program 05QMX1417.

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Local well-posedness and blow-up criteria of solutions for a rod equation

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