Convergence to equilibrium of gradient flows defined on planar curves

Matteo Novaga; Shinya Okabe

doi:10.1515/crelle-2015-0001

Article

Convergence to equilibrium of gradient flows defined on planar curves

Published/Copyright: June 5, 2015

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From the journal Journal für die reine und angewandte Mathematik (Crelles Journal) Volume 2017 Issue 733

Abstract

We consider the evolution of open planar curves by the steepest descent flow of a geometric functional, with different boundary conditions. We prove that, if any set of stationary solutions with fixed energy is finite, then a solution of the flow converges to a stationary solution as time goes to infinity. We also present a few applications of this result.

Funding statement: The second author was partially supported by the JSPS Strategic Young Researcher Overseas Visits Program for Accelerating Brain Circulation and by Grant-in-Aid for Young Scientists (B) (No. 24740097).

A Appendix

Lemma A.1.

Let γ⁢(s):[0,∞)→R2 be a planar open curve with curvature satisfying

(d⁢κd⁢s)2+F⁢(κ)=E.

Then the function

L⁢(E)=2⁢∫κm⁢(E)κM⁢(E)d⁢κE-F⁢(κ)

is analytic on (-λ4/4,0)∪(0,∞). Furthermore, it holds that

(A.1)L⁢(E)→∞ as ⁢E→0.

Proof.

To begin with, we show that L⁢(E) is analytic on (0,∞) and that L⁢(E)→∞ as E↓0. Recall that L⁢(E) is written as

L⁢(E)=4⁢∫0κM⁢(E)d⁢κE-F⁢(κ) for E∈(0,∞).

Since F is analytic, it is clear that κM⁢(E) is analytic. Moreover, the definition of κM⁢(E) implies that F′⁢(κ⁢(E))≠0. The Taylor expansion of F at κ=κM⁢(E) is expressed as

F⁢(κ)=F⁢(κM)+F′⁢(κM)⁢(κ-κM)+F′′⁢(κM)2!⁢(κ-κM)2

+F(3)⁢(κM)3!⁢(κ-κM)3+F(4)⁢(κM)4!⁢(κ-κM)4.

It follows from F′⁢(κM)≠0 that

E-F⁢(κ)=F′⁢(κM)⁢(κM-κ)⁢1+∑n=13an⁢(E)⁢(κM-κ)n for any κ∈[0,κM],

where an⁢(E) is given by

an⁢(E)=(-1)n⁢F(n+1)⁢(κM⁢(E))(n+1)!⁢F′⁢(κM⁢(E)).

Since it holds that

|an⁢(E)|≤C⁢|λ|-n for any E>0,

we see that

1E-F⁢(κ)=∑k=0∞bk⁢(κM⁢(E)-κ)k-12 for any κ∈(κM⁢(E)-ε,κM⁢(E)),

where ε is a positive constant satisfying

max1≤n≤3⁡|an|⁢ε⁢(1+ε+ε2)<1.

Remark that bk=bk⁢(E) is analytic on (0,∞). In the following let us set

L⁢(E)4=𝔏1⁢(E)+𝔏2⁢(E),

which are written as

𝔏1⁢(E)=∫0κ0⁢(E)d⁢κE-F⁢(κ),𝔏2⁢(E)=∫κ0⁢(E)κM⁢(E)d⁢κE-F⁢(κ),

where κ0⁢(E)=κM⁢(E)-ε/2. First we check that 𝔏1⁢(E) is analytic. Let us write 𝔏1⁢(E) as

𝔏1⁢(E)=1E⁢∫0κ0(1-F⁢(κ)E)-12⁢𝑑κ.

Notice that the function (1-y)-12 is analytic on (-∞,1), and for any y0<1 one can write

(1-y)-12=∑k=0∞ck⁢(y-y0) for all ⁢y∈(y0,1),

where the coefficients ck depend on y0. Letting κ¯=|λ| which is a minimum point of F and setting y=F⁢(κ)/E, y0=F⁢(κ¯)/E, we have

(A.2)𝔏1⁢(E)=1E⁢∑k=0∞ck⁢∫0κ0(F⁢(κ)E-F⁢(κ¯)E)k⁢𝑑κ

=∑k=0∞ck⁢gk⁢(κ0⁢(E))Ek+12,

where

gk⁢(x)=∫0x(F⁢(κ)-F⁢(κ¯))k⁢𝑑κ.

Since it holds that

(∫0κ0(F⁢(κ)E-F⁢(κ¯)E)k⁢𝑑κ)1k→supκ∈(0,κ0)⁡F⁢(κ)-F⁢(κ¯)E<1-y0 as k→∞,

we see that the series in (A.2) converges for each E>0. Recalling κM⁢(E) is analytic, all the functions gk⁢(κ0⁢(E)) are also analytic. This implies that 𝔏1⁢(E) is analytic for E>0.

Regarding 𝔏2⁢(E), we have

𝔏2⁢(E)=∫κ0κM⁢(E)d⁢κE-F⁢(κ)=-∑k=0∞bk⁢(E)(k+1/2)⁢(ε2)k+12.

Since bk⁢(E) is analytic, this implies that 𝔏2⁢(E) is also analytic for E>0. Therefore we observe that L⁢(E) is analytic on (0,∞). On the other hand, it follows from (A.2) that

(A.3)L⁢(E)→∞ as ⁢E↓0.

Next we prove that the function L⁢(E) is analytic on (-λ4/4,0). Along the same line as above, we see that

1E-F⁢(κ)=∑k=0∞bk⁢(κM⁢(E)-κ)k-12 for any κ∈(κM-ε,κM)

and

1E-F⁢(κ)=∑k=0∞b~k⁢(κm⁢(E)-κ)k-12 for any κ∈(κm,κm+ε),

where ε is an appropriate small number. Set

L⁢(E)=𝔏~1⁢(E)+𝔏~2⁢(E)+𝔏~3⁢(E)+𝔏~4⁢(E),

where

𝔏~1⁢(E)=∫κmκm+ε2d⁢κE-F⁢(κ),𝔏~2⁢(E)=∫κm+ε2κ¯d⁢κE-F⁢(κ),

𝔏~3⁢(E)=∫κ¯κM-ε2d⁢κE-F⁢(κ),𝔏~4⁢(E)=∫κM-ε2κMd⁢κE-F⁢(κ).

Regarding 𝔏~1⁢(E) and 𝔏~4⁢(E), we can verify that 𝔏~1⁢(E) and 𝔏~4⁢(E) are analytic on (-λ4/4,0) along the same argument for 𝔏2⁢(E). Next we turn to 𝔏~2⁢(E). Along the same line as the argument for 𝔏1⁢(E), we have

(A.4)𝔏~2⁢(E)=∫κm+ε2κ¯1-F⁢(κ)⁢d⁢κ1-E/F⁢(κ)

=∑k=0∞c~k⁢∫κm+ε2κ¯1-F⁢(κ)⁢(EF⁢(κ)-EF⁢(κ¯))k⁢𝑑κ

=∑k=0∞c~k⁢g~k⁢(κm⁢(E)+ε2)⁢Ek,

where

g~k⁢(x)=∫xκ¯(F⁢(κ)-1-F⁢(κ¯)-1)⁢(-F⁢(κ))-12⁢𝑑κ.

Since it holds that

𝔏~2⁢(E)≤1-E⁢∑k=0∞c~k⁢∫κm+ε2κ¯(EF⁢(κ)-EF⁢(κ¯))k⁢𝑑κ

and

(∫κm+ε2κ¯(EF⁢(κ)-EF⁢(κ¯))k⁢𝑑κ)1k→supκ∈(κm+ε2,κ¯)⁡EF⁢(κ)-EF⁢(κ¯)<1-y0 as k→∞,

we observe that the series in (A.4) converges for E∈(-λ4/4,0). Recalling κm⁢(E) is analytic, all the functions g~k⁢(κm⁢(E)+ε/2) are also analytic. This implies that 𝔏~2⁢(E) is analytic for E∈(-λ4/4,0). A similar argument gives us that 𝔏~3⁢(E) is also analytic for E∈(-λ4/4,0).

Finally, we prove that L⁢(E)→∞ as E↑0. Regarding 𝔏~1⁢(E), it holds that

(A.5)𝔏~1⁢(E)>∫κmκm+ε2d⁢κ-F′⁢(κm)⁢(κ-κm)-F′′⁢(κm)⁢(κ-κm)2

=12⁢-F′′⁢(κm)⁢log⁡1+εε+2⁢a1-εε+2⁢a,

where a=F′⁢(κm)/F′′⁢(κm). Since F′⁢(κm⁢(E))→0 and F′′⁢(κm⁢(E))→-λ2 as E↑0, it follows from (A.5) that

𝔏~1⁢(E)→∞ as ⁢E↑0.

This clearly implies that L⁢(E)→∞ as E↑0. ∎

The arguments in the proof of Lemma A.1 also implies the analyticity of Li⁢(E) which are defined by (4.36)–(4.37).

Lemma A.2.

Let α>0. If F⁢(α)≥0, then the function L1⁢(E) is analytic on (F⁢(α),∞). If F⁢(α)<0, then the function L1⁢(E) is analytic on (F⁢(α),0)∪(0,∞). Moreover, as E→0, it holds that

(A.6)L1⁢(E)→∞ as ⁢E→0.

Proof.

The proof of Lemma A.1 gives us the conclusion. ∎

Lemma A.3.

For each α>0, the function L2⁢(E) is analytic on (F⁢(α),∞). Moreover, for each α∈(0,2⁢|λ|), it holds that

(A.7)L2⁢(E)→0 as ⁢E→∞.

Proof.

The analyticity of L2⁢(E) follows from the same argument of the proof of Lemma A.1. We shall prove (A.7). Since 0⁢<α⁢<2|⁢λ|, we divide L2⁢(E) into two part as follows:

(A.8)∫ακMd⁢κE-F⁢(κ)=∫α2⁢|λ|d⁢κE-F⁢(κ)+∫2⁢|λ|κMd⁢κE-F⁢(κ).

Recalling F⁢(2⁢|λ|)=0, we have

∫α2⁢|λ|d⁢κE-F⁢(κ)≤1E⁢∫α2⁢|λ|𝑑κ→0 as ⁢E→∞.

Thus it is sufficient to estimate the second term of the right-hand side of (A.8). By changing the variable

κ(4⁢E)14=x,

we have

∫2⁢|λ|κMd⁢κE-F⁢(κ)≤1E⁢∫2⁢|λ|κMd⁢κ1-κ44⁢E

=2E1/4⁢∫2⁢|λ|(4⁢E)1/4κM(4⁢E)1/4d⁢x1-x4.

Then, the conclusion is obtained from the following calculation:

2E1/4⁢∫2⁢|λ|(4⁢E)1/4κM(4⁢E)1/4d⁢x1-x4≤2E1/4⁢∫2⁢|λ|(4⁢E)1/4κM(4⁢E)1/4d⁢x1-x2

=2E1/4⁢{sin-1⁡κM(4⁢E)1/4-sin-1⁡2⁢|λ|(4⁢E)1/4}→0

as E→∞. ∎

B Appendix

The scope of this appendix is to prove that (4.31) has a unique smooth solution defined for all times.

Let us first show that the L2-gradient flow for the functional ℰλ,α under (4.2) and (4.26) can be written as (4.1). Indeed, let γ:[0,1]→ℝ2 be a smooth planar curve satisfying the symmetric Navier boundary conditions

(B.1)γ⁢(0)=(0,0),γ⁢(1)=(R,0),κ⁢(0)=κ⁢(1)=α.

We consider a variation of γ defined as follows:

γ⁢(x,ε)=γ⁢(x)+ϕ⁢(x,ε)⁢𝝂⁢(x),

where 𝝂 is the unit normal vector, pointing in the direction of the curvature, given by

𝝂=(0-110)⁢γx|γx|:=ℛ⁢γx|γx|,

and ϕ⁢(x,ε)∈C∞⁢((-ε0,ε0);C∞⁢(0,1)) is an arbitral smooth function with

ϕ⁢(x,0)≡ϕ⁢(0,ε)≡ϕ⁢(1,ε)≡0.

In the following we shall derive a first variational formula for the functional ℰλ,α⁢(γ). Put

τ=γx|γx|.

Since the curvature of γ is expressed as

(B.2)κ=γx⁢x⋅ℛ⁢γx|γx|3,

we have

κ=γx⁢x⋅𝝂⁢|γx|-2,

and then Frenet–Serret’s formula ∂s⁡𝝂⋅τ=-κ yields that

𝝂x⋅τ=-κ⁢|γx|.

To begin with, we derive useful variational formulae. First we find the first variational formula of the local length:

(B.3)dd⁢ε⁢|γx⁢(x,ε)||ε=0=γx⋅(ϕε⁢𝝂)x|γx|=τ⋅ϕε⁢𝝂x=-κ⁢|γx|⁢ϕε,

where ϕε⁢(⋅)=(∂⁡ϕ/∂⁡ε)⁢(⋅,0). Next we find the first variation formula of the curvature. From formula (B.2) and

𝝂⋅𝝂x⁢x=-|𝝂x|2=-κ2⁢|γx|2,

γx⁢x⋅ℛ⁢𝝂x=γx⁢x⋅ℛ⁢(-κ⁢γx)=-κ2⁢|γx|3,

(|γx|-1)x=γx⁢x⋅ℛ⁢𝝂|γx|2,

it follows that

(B.4)dd⁢ε⁢κ⁢(x,ε)|ε=0=ϕε⁢x⁢x|γx|2+κ2⁢ϕε+(|γx|-1)x⁢ϕε⁢x|γx|.

Using (B.3), we obtain

dd⁢ε⁢ℰλ,α⁢(γ⁢(⋅,ε))|ε=0=∫01{2⁢(κ-α)⁢dd⁢ε⁢κ|ε=0-(κ3-2⁢α⁢κ2+λ2⁢κ)⁢ϕε}⁢|γx|⁢𝑑x.

Using (B.4) and integrating by parts, we get

∫01(κ-α)⁢dd⁢ε⁢κ|ε=0⁢|γx|⁢d⁢x

=∫01(κ-α)⁢{ϕε⁢x⁢x|γx|2+κ2⁢ϕε+(|γx|-1)x⁢ϕε⁢x|γx|}⁢|γx|⁢𝑑x

=∫01-(κ-α|γx|)x⁢ϕε⁢x+(κ3-α⁢κ2)⁢ϕε⁢|γx|+(|γx|-1)x⁢(κ-α)⁢ϕε⁢x⁢d⁢x+[κ-α|γx|⁢ϕε⁢x]01

=∫01-κx|γx|⁢ϕε⁢x+(κ3-α⁢κ2)⁢ϕε⁢|γx|⁢d⁢x

=∫01(κx|γx|)x⁢ϕε+(κ3-α⁢κ2)⁢ϕε⁢|γx|⁢d⁢x

=∫01{(∂x|γx|)2⁢κ+(κ3-α⁢κ2)}⁢ϕε⁢|γx|⁢𝑑x.

Here we use κ⁢(0)=κ⁢(1)=α. Thus we find

(B.5)dd⁢ε⁢ℰλ,α⁢(γ⁢(⋅,ε))|ε=0=∫ab{2⁢(∂x|γx|)2⁢κ+κ3-λ2⁢κ}⁢ϕε⁢|γx|⁢𝑑x.

Parameterizing by the arclength, formula (B.5) is written as

dd⁢ε⁢ℰλ,α⁢(γ⁢(⋅,ε))|ε=0=∫01{2⁢κs⁢s+κ3-λ2⁢κ}⁢ϕε⁢𝑑s.

Therefore we see that the flow (4.1) is the L2-gradient flow for the functional ℰλ,α under the symmetric Navier boundary conditions (B.1).

Since (4.31) is a nonlinear boundary value problem for a quasi-linear parabolic equation, a short time existence is a standard matter. In what follows we shall prove a long time existence of solutions to (4.31). Throughout the section, put

Vλ=2⁢∂s2⁡κ+κ3-λ2⁢κ.

Then the equation in (4.31) is written as

(B.6)∂t⁡γ=-Vλ⁢𝝂.

Since s depends on t, remark that the following holds.

Lemma B.1.

Under (B.6), the following commutation rule holds:

∂t⁡∂s=∂s⁡∂t-κ⁢Vλ⁢∂s.

Lemma B.1 gives us the following:

Lemma B.2.

Let γ⁢(x,t) satisfy (B.6). Then it holds that

(B.7)∂t⁡κ=-∂s2⁡Vλ-κ2⁢Vλ.

Furthermore, the line element ds of γ⁢(x,t) satisfies

(B.8)∂t⁡d⁢s=κ⁢Vλ⁢d⁢s.

The boundary conditions in (4.31) imply that several terms vanish on the boundary.

Lemma B.3.

Suppose that γ satisfies (4.31). Then it holds that

(B.9)∂t⁡γ=0on ⁢∂⁡I×[0,∞),

(B.10)∂t⁡κ=0on ⁢∂⁡I×[0,∞),

(B.11)Vλ=0on ⁢∂⁡I×[0,∞),

(B.12)∂s2⁡Vλ=0on ⁢∂⁡I×[0,∞),

(B.13)∂t⁡Vλ=0on ⁢∂⁡I×[0,∞),

(B.14)∂t⁡∂s2⁡Vλ=0on ⁢∂⁡I×[0,∞),

(B.15)∂t⁡∂s=∂s⁡∂ton ⁢∂⁡I×[0,∞).

Proof.

Since both γ⁢(t) and κ⁢(t) are fixed on ∂⁡I, we observe (B.9)–(B.10). It follows from (B.6) and (B.9) that (B.11) holds. By virtue of (B.2), (B.10), and (B.11), we obtain (B.12). Then (B.11) and (B.12) implies (B.13) and (B.14), respectively. Finally, (B.15) follows from Lemma B.1 and (B.11) ∎

Here we introduce interpolation inequalities for open curves, which has been inspired by [6] for closed curves and given in [9]. The interpolation inequalities are written in terms of the following the scale invariant Sobolev norms:

∥κ∥k,p:=∑i=0k∥∂si⁡κ∥p,

∥∂si⁡κ∥p:=ℒ⁢(γ)i+1-1p⁢(∫I|∂si⁡κ|p)1p.

Lemma B.4 ([9]).

Let γ:I→R2 be a smooth curve. Then for any k∈N∪{0}, p≥2, and 0≤i<k, we have

∥∂si⁡κ∥p≤c⁢∥κ∥21-α⁢∥κ∥k,2α,

where α=(i+12-1p)⁢1k and c=c⁢(n,k,p).

In order to prove a long time existence of solutions to (4.31), we make use of the following lemma, which is a modification of [6, Lemma 2.2].

Lemma B.5.

Let γ:I×[0,T)→R2 satisfy equation (B.6) and let ϕ:I×[0,T)→R be a scalar function defined on γ satisfying

(B.16){∂t⁡ϕ=-2⁢∂s4⁡ϕ+Yin ⁢I×[0,T),ϕ=0,∂s2⁡ϕ=0on ⁢∂⁡I×[0,T).

Then it holds that

(B.17)dd⁢t⁢14⁢∫γϕ2⁢𝑑s+∫γ(∂s2⁡ϕ)2⁢𝑑s=12⁢∫γϕ⁢Y⁢𝑑s+14⁢∫γϕ2⁢κ⁢Vλ⁢𝑑s.

Proof.

It follows from the equation in (B.16) and Lemma B.2 that

dd⁢t⁢14⁢∫γϕ2⁢𝑑s=12⁢∫γϕ⁢∂t⁡ϕ⁢d⁢s+14⁢∫γϕ2⁢∂t⁡(d⁢s)

=12⁢∫γϕ⁢(-2⁢∂s4⁡ϕ+Y)⁢𝑑s+14⁢∫γϕ2⁢κ⁢Vλ⁢𝑑s.

With the aid of the boundary conditions in (B.16), we obtain

∫γϕ⁢∂s4⁡ϕ⁢d⁢s=-∫γ∂s⁡ϕ⁢∂s3⁡ϕ⁢d⁢s=∫γ(∂s2⁡ϕ)2⁢𝑑s.

Then we observe (B.17). ∎

By virtue of Lemma B.3, we observe that ∂tm⁡Vλ=0 and ∂s2⁡∂tm⁡Vλ=0 hold on ∂⁡I for any m∈ℕ∪{0}. The fact implies that we can apply Lemma B.5 to ϕ=∂tm⁡Vλ. To do so, first we introduce the following notation for convenience.

Definition B.1 ([2]).

We use the symbol 𝔮r⁢(∂sl⁡κ) for a polynomial with constant coefficients such that each of its monomials is of the form

∏i=1N∂sji⁡κ with ⁢0≤ji≤l⁢ and ⁢N≥1

with

r=∑i=1N(ji+1).

Making use of the notation, we obtain the following:

Lemma B.6.

Suppose that γ:I×[0,∞)→R2 satisfies (4.31). Let ϕ be a scalar function defined on γ. Then the following formulae hold for any m,l∈N:

(B.18)∂sm⁡Vλ=𝔮3+m⁢(∂s2+m⁡κ)-λ2⁢∂sm⁡κ,

(B.19)∂t⁡∂sm⁡ϕ=∂sm⁡∂t⁡ϕ+∑i=0m-1(𝔮4+i⁢(∂s2+i⁡κ)+𝔮2+i⁢(∂si⁡κ))⁢∂sm-i⁡ϕ,

(B.20)∂t⁡∂sm⁡κ=-2⁢∂sm+4⁡κ+𝔮m+5⁢(∂sm+2⁡κ)+𝔮m+3⁢(∂sm+2⁡κ),

(B.21)∂t⁡𝔮l⁢(∂sm⁡κ)=𝔮l+4⁢(∂sm+4⁡κ)+𝔮l+2⁢(∂sm+2⁡κ).

Proof.

Since Vλ=𝔮3⁢(∂s2⁡κ)-λ2⁢κ, assertion (B.18) follows from a simple calculation. Regarding (B.19), we proceed by induction on m. For m=1, we have

∂t⁡∂s⁡ϕ=∂s⁡∂t⁡ϕ-κ⁢Vλ⁢∂s⁡ϕ

=∂s⁡∂t⁡ϕ-(𝔮4⁢(∂s2⁡κ)+𝔮2⁢(κ))⁢∂s⁡ϕ.

Assuming that (B.19) is true for some m≥1, we obtain

∂t⁡∂sm+1⁡ϕ=∂s⁡∂t⁡∂sm⁡ϕ+(𝔮4⁢(∂s2⁡κ)+𝔮2⁢(κ))⁢∂sm+1⁡ϕ

=∂s⁡{∂sm⁡∂t⁡ϕ+∑i=0m-1(𝔮4+i⁢(∂s2+i⁡κ)+𝔮2+i⁢(∂si⁡κ))⁢∂sm-i⁡ϕ}

+(𝔮4⁢(∂s2⁡κ)+𝔮2⁢(κ))⁢∂sm+1⁡ϕ

=∂sm+1⁡∂t⁡ϕ+∑i=0m(𝔮4+i⁢(∂s2+i⁡κ)+𝔮2+i⁢(∂si⁡κ))⁢∂sm+1-i⁡ϕ.

Assertion (B.20) follows from (B.2) and (B.19) directly. Finally, we obtain (B.21) for m,l∈ℕ fixed arbitrarily as follows:

∂t⁡𝔮l⁢(∂sm⁡κ)=∑j=0m𝔮l-j-1⁢(∂sm⁡κ)⋅∂t⁡∂sj⁡κ

=∑j=0m𝔮l-j-1⁢(∂sm⁡κ)⋅{-2⁢∂sj+4⁡κ+𝔮j+5⁢(∂sj+2⁡κ)+𝔮j+3⁢(∂sj+2⁡κ)}

=∑j=0m𝔮l+4⁢(∂smax⁡{m,j+4}⁡κ)+∑j=0m𝔮l+2⁢(∂smax⁡{m,j+2}⁡κ)

=∑j=04𝔮l+4⁢(∂sm+j⁡κ)+∑j=02𝔮l+2⁢(∂sm+j⁡κ)

=𝔮l+4⁢(∂sm+4⁡κ)+𝔮l+2⁢(∂sm+2⁡κ).∎

With the aid of Lemma B.6, we obtain a representation of ∂tm⁡Vλ.

Lemma B.7.

For each m∈N, it holds that

(B.22)∂tm⁡Vλ=(-1)m⁢2m+1⁢∂s4⁢m+2⁡κ+𝔮4⁢m+3⁢(∂s4⁢m⁡κ)

+∑j=1m𝔮4⁢m+3-2⁢j⁢(∂s4⁢m+2-2⁢j⁡κ).

Proof.

We proceed by induction on m. For m=1, we have

∂t⁡Vλ=∂t⁡(2⁢∂s2⁡κ+κ3-λ2⁢κ)

=2⁢(-2⁢∂s6⁡κ+𝔮7⁢(∂s4⁡κ)+𝔮5⁢(∂s4⁡κ))+3⁢κ2⁢∂t⁡κ-λ2⁢∂t⁡κ

=-22⁢∂s6⁡κ+𝔮7⁢(∂s4⁡κ)+𝔮5⁢(∂s4⁡κ).

Suppose that (B.22) holds for m=k. Then we have

(B.23)∂tk+1Vλ=∂t{(-1)k2k+1∂s4⁢k+2κ+𝔮4⁢k+3(∂s4⁢kκ)

+∑j=1k𝔮4⁢k+3-2⁢j(∂s4⁢k+2-2⁢jκ)}

=(-1)k⁢2k+1⁢{-2⁢∂s4⁢k+6⁡κ+𝔮4⁢k+7⁢(∂s4⁢k+6⁡κ)+𝔮4⁢k+5⁢(∂s4⁢k+6⁡κ)}

+∂t⁡{𝔮4⁢k+3⁢(∂s4⁢k⁡κ)+∑j=1k𝔮4⁢k+3-2⁢j⁢(∂s4⁢k+2-2⁢j⁡κ)}.

By virtue of (B.21), the last term in (B.23) is reduced to

∂t⁡{𝔮4⁢k+3⁢(∂s4⁢k⁡κ)+∑j=1k𝔮4⁢k+3-2⁢j⁢(∂s4⁢k+2-2⁢j⁡κ)}

=𝔮4⁢k+7⁢(∂s4⁢k+4⁡κ)+𝔮4⁢k+5⁢(∂s4⁢k+2⁡κ)

+∑j=1k{𝔮4⁢k+7-2⁢j⁢(∂s4⁢k+6-2⁢j⁡κ)+𝔮4⁢k+5-2⁢j⁢(∂s4⁢k+4-2⁢j⁡κ)}

=𝔮4⁢(k+1)+3⁢(∂s4⁢(k+1)⁡κ)+∑j=1k+1𝔮4⁢(k+1)+3-2⁢j⁢(∂s4⁢(k+1)+2-2⁢j⁡κ).

This implies that (B.22) holds for any m∈ℕ. ∎

We are in a position to prove the main result of this section.

Theorem B.1.

Let λ∈R be a non-zero constant. Let γ0:I→R2 be a smooth open curve satisfying

γ0⁢(0)=(0,0),γ0⁢(1)=(R,0),κ0⁢(0)=κ0⁢(1)=α,

where α∈R is a given constant with |α|<|λ|. Then there exists a unique family of smooth open planar curves γ⁢(x,t) satisfying (4.31) for any finite time t>0.

Proof.

Suppose not; then there exists a time t1>0 such that the smooth solution γ⁢(x,t) of (4.31) remains up to t=t1. Setting

ϕ=∂tm⁡Vλ,

Lemma B.5 implies that

(B.24)dd⁢t⁢14⁢∫γ(∂tm⁡Vλ)2⁢𝑑s+∫γ(∂s2⁡∂tm⁡Vλ)2⁢𝑑s

=12⁢∫γ∂tm⁡Vλ⁢Y⁢d⁢s+14⁢∫γ(∂tm⁡Vλ)2⁢κ⁢Vλ⁢𝑑s.

Regarding the integral of (∂s2⁡∂tm⁡Vλ)2, we have

(∂s2⁡∂tm⁡Vλ)2≥(22⁢(m+1)-ε)⁢(∂s4⁢m+4⁡κ)2

+{𝔮4⁢m+5⁢(∂s4⁢m+2⁡κ)+∑j=1m𝔮4⁢m+5-2⁢j⁢(∂s4⁢m+4-2⁢j⁡κ)}2

=cm⁢(∂s4⁢m+4⁡κ)2+∑j=0m𝔮8⁢m+10-2⁢j⁢(∂s4⁢m+2⁡κ)

+∑j,l=1m𝔮8⁢m+10-2⁢(j+l)⁢(∂s4⁢m+4-2⁢min⁡{j,l}⁡κ).

Regarding the integral of ∂tm⁡Vλ⁢Y, setting

∂tm⁡Vλ⁢Y=∂tm⁡Vλ⁢𝔮4⁢m+7⁢(∂s4⁢m+4⁡κ)+∂tm⁡Vλ⁢𝔮4⁢m+5⁢(∂s4⁢m+4⁡κ)

+∂tm⁡Vλ⁢∑j=2m+1𝔮4⁢m+7-2⁢j⁢(∂s4⁢m+6-2⁢j⁡κ)

=:I1+I2+I3,

and integrating by part once the highest order term, we find

∫γI1⁢𝑑s=-∫γ∂s⁡∂tm⁡Vλ⁢{𝔮4⁢m+7⁢(∂s4⁢m+3⁡κ)+𝔮4⁢m+6⁢(∂s4⁢m+3⁡κ)}⁢𝑑s

=-∑j=0m∫γ{𝔮8⁢m+11-2⁢j⁢(∂s4⁢m+3⁡κ)+𝔮8⁢m+10-2⁢j⁢(∂s4⁢m+3⁡κ)}⁢𝑑s

and

∫γI2⁢𝑑s=-∫γ∂s⁡∂tm⁡Vλ⁢{𝔮4⁢m+5⁢(∂s4⁢m+3⁡κ)+𝔮4⁢m+4⁢(∂s4⁢m+3⁡κ)}⁢𝑑s

=-∑j=0m∫γ{𝔮8⁢m+9-2⁢j⁢(∂s4⁢m+3⁡κ)+𝔮8⁢m+8-2⁢j⁢(∂s4⁢m+3⁡κ)}⁢𝑑s.

Hence we see that

∫γ∂tmVλYds=∫γ∑j=0m+1{𝔮8⁢m+11-2⁢j(∂s4⁢m+3κ)+𝔮8⁢m+10-2⁢j(∂s4⁢m+3κ)

+𝔮8⁢m+8-2⁢j(∂s4⁢m+2κ)+𝔮8⁢m+6-2⁢j(∂s4⁢mκ)}

+∑l=1,j=1m𝔮8⁢m+8-2⁢(j+l)⁢(∂s4⁢m+2-2⁢min⁡{j,l}⁡κ)⁢d⁢s.

Since it holds that

∫γ(∂tm⁡Vλ)2⁢κ⁢Vλ⁢𝑑s=∫γ∑j=0m+1{𝔮8⁢m+10-2⁢j⁢(∂s4⁢m+2⁡κ)+𝔮8⁢m+10-2⁢j⁢(∂s4⁢m⁡κ)}

+∑l=1,j=1m𝔮8⁢m+10-2⁢(j+l)⁢(∂s4⁢m+2-2⁢min⁡{j,l}⁡κ)⁢d⁢s,

equality (B.24) is reduced to

(B.25)dd⁢t⁢14⁢∫γ(∂tm⁡Vλ)2⁢𝑑s+cm⁢(ε)⁢∫γ(∂s4⁢m+4⁡κ)2⁢𝑑s

=∫γ[∑j=0m+1{𝔮8⁢m+11-2⁢j(∂s4⁢m+3κ)+𝔮8⁢m+10-2⁢j(∂s4⁢m+3κ)

+𝔮8⁢m+10-2⁢j(∂s4⁢m+2κ)+𝔮8⁢m+10-2⁢j(∂s4⁢mκ)}

+∑l=1,j=1m𝔮8⁢m+10-2⁢(j+l)(∂s4⁢m+2-2⁢min⁡{j,l}κ)]ds.

We estimate the integral of 𝔮8⁢m+11⁢(∂s4⁢m+3⁡κ) which is the highest order term on the right-hand side of (B.25). By Definition B.1, this term can be written as

𝔮8⁢m+11⁢(∂s4⁢m+3⁡κ)=∑j∏i=1Nj∂scji⁡κ

with all the cji less than or equal to 4⁢m+3, and

∑i=1Nj(cji+1)=8⁢m+11 for every j.

Hence we have

|𝔮8⁢m+11⁢(∂s4⁢m+3⁡κ)|≤∑j∏i=1Nj|∂scji⁡κ|.

Putting

Qj=∏i=1Nj|∂scji⁡κ|,

it holds that

∫γ|𝔮8⁢m+11⁢(∂s4⁢m+3⁡κ)|⁢𝑑s≤∑j∫γQj⁢𝑑s.

After collecting the derivatives of the same order in Qj, we can write

Qj=∏l=04⁢m+3|∂sl⁡κ|αjl

with

∑l=04⁢m+3αjl⁢(l+1)=8⁢m+11.

Using Hölder’s inequality, we get

∫γQj⁢𝑑s≤∏l=04⁢m+3(∫γ|∂sl⁡κ|αjl⁢λl)1λl=∏l=04⁢m+3∥∂sl⁡κ∥αjl⁢λlαjl,

where the value λl is chosen as follows: λl=0 if αjl=0 (in this case the corresponding term is not present in the product) and λl=(8⁢m+11)/αjl⁢(l+1) if αjl≠0. Clearly,

αjl⁢λl=8⁢m+11l+1≥8⁢m+114⁢m+4>2

and

∑l=0λl≠04⁢m+31λl=∑l=0λl≠04⁢m+3αjl⁢(l+1)8⁢m+11=1.

Let kl=αjl⁢λl-2. The fact αjl⁢λl>2 implies that kl>0. Then we obtain

∥∂sl⁡κ∥αjl⁢λl≤c⁢∥κ∥21-σjl⁢∥κ∥4⁢m+4,2σjl,

where σjl=(l+12-1αjl⁢λl)⁢14⁢m+4 and c=c⁢(j,l,m). Since

∥κ∥4⁢m+4,22≤C⁢(m)⁢(∥∂s4⁢m+4⁡κ∥22+∥κ∥22),

we observe that

∥∂sl⁡κ∥αjl⁢λl≤C⁢∥κ∥21-σjl⁢(∥∂s4⁢m+4⁡κ∥22+∥κ∥22)σjl.

Multiplying together all the estimates, we obtain

(B.26)∫γQj⁢𝑑s≤C⁢∏l=04⁢m+3∥κ∥2(1-σjl)⁢αjl⁢(∥∂s4⁢m+4⁡κ∥2+∥κ∥2)σjl⁢αjl

=C⁢∥κ∥2∑l=04⁢m+3(1-σjl)⁢αjl⁢(∥∂s4⁢m+4⁡κ∥2+∥κ∥2)∑l=04⁢m+3σjl⁢αjl.

Then the exponent in the last term of (B.26) is written as

∑l=04⁢m+3σjl⁢αjl=∑l=04⁢m+3αjl⁢(l+12-1αjl⁢λl)4⁢m+4=∑l=04⁢m+3αjl⁢(l+12)-14⁢m+4,

and hence by using the rescaling condition we have

∑l=04⁢m+3σjl⁢αjl=∑l=04⁢m+3αjl⁢(l+1)-12⁢∑l=04⁢m+3αjl-14⁢m+4

=8⁢m+11-12⁢∑l=04⁢m+3αjl-14⁢m+4

=16⁢m+20-∑l=04⁢m+3αjl2⁢(4⁢m+4).

Noting that

∑l=04⁢m+3αjl≥∑l=04⁢m+3αjl⁢l+14⁢m+4=8⁢m+114⁢m+4,

we see that

∑l=04⁢m+3σjl⁢αjl≤16⁢m+20-8⁢m+104⁢m+42⁢(4⁢m+4)=2-1(4⁢m+4)2<2.

Hence we can apply the Young inequality to the product in the last term of (B.26), in order to get the exponent 2 on the first quantity, that is,

∫γQj⁢𝑑s≤δj2⁢(∥∂s4⁢m+4⁡κ∥2+∥κ∥2)2+Cj⁢∥κ∥2β

≤δj∥∂s4⁢m+4κ∥22+∥κ∥22++Cj∥κ∥2βj

for arbitrarily small δj>0 and some constant Cj>0 and exponent βj>0. Hence we get

dd⁢t⁢14⁢∫γ(∂tm⁡Vλ)2⁢𝑑s+12⁢∫γ(∂s2⁡∂tm⁡Vλ)2⁢𝑑s+cm⁢(ε)2⁢∫γ(∂s4⁢m+4⁡κ)2⁢𝑑s

≤∑j=0m+1δj⁢∥∂s4⁢m+4⁡κ∥22+C⁢∑j=0m+1∥κ∥2βj.

Letting δj>0 be sufficiently small, we obtain

(B.27)dd⁢t⁢14⁢∫γ(∂tm⁡Vλ)2⁢𝑑s≤C⁢∑j=0m+1∥κ∥2βj.

Since Lemma 4.4 gives us

∥κ∥22≤C⁢(α,λ)⁢ℰλ,α⁢(γ0),

estimate (B.27) implies that

(B.28)∥∂tm⁡Vλ⁢(t)∥L22≤C1⁢t+∥∂tm⁡Vλ⁢(0)∥L22

for any time t∈[0,t1). Using (B.22) and the interpolation inequality, we reduce (B.28) to

(B.29)∥∂s4⁢m+2⁡κ∥L22≤C1⁢t+∥∂tm⁡Vλ∥L22⁢(0)+C2,

where C2 depends on ℰλ,α⁢(γ0). Combining (4.30) and (B.29) with the interpolation inequality, we observe that there exists a positive constant depending only on ℰλ,α⁢(γ0) such that

(B.30)∥∂sl⁡κ∥L2≤C1⁢t+∥∂tm⁡Vλ∥L22⁢(0)+C3

for any 0≤l<4⁢m+2. For each l∈ℕ, it is easy to obtain that

(B.31)∥∂sl-1⁡κ∥L∞≤C⁢∥∂sl⁡κ∥L1+ℒ⁢(γ)-1⁢∥∂sl-1⁡κ∥L1.

Applying Hölder’s inequality to (B.31), we obtain

(B.32)∥∂sl-1⁡κ∥L∞≤ℒ⁢(γ)12⁢∥∂sl⁡κ∥L2+ℒ⁢(γ)-12⁢∥∂sl-1⁡κ∥L2.

Then it follows from (B.30) and (B.32) that there exists a constant C=C⁢(γ0,t1,α,λ) such that

(B.33)∥∂sl-1⁡κ⁢(t)∥L∞≤C

for each l∈ℕ and any t∈[0,t1). This contradicts that the solution of (4.31) remains smooth to t=t1. We have thus completed the proof. ∎

Acknowledgements

This work was done while the second author was visiting Centro di Ricerca Matematica Ennio De Giorgi and Department of Mathematics, University of Padova, whose hospitality he gratefully acknowledges.

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Received: 2013-11-8

Revised: 2014-12-26

Published Online: 2015-6-5

Published in Print: 2017-12-1

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https://doi.org/10.1515/crelle-2015-0001