Stability of Cahn-Hilliard Fronts in Three Dimensions

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Stability of Cahn–Hilliard Fronts in Three Dimensions

Timo Korvola

Academic dissertation

To be presented, with the permission of the Faculty of Science of the University of Helsinki, for public criticism in Auditorium XII of the Main

Building of the University, on September 20th, 2003, at 10 o’clock am.

Department of Mathematics Faculty of Science University of Helsinki

Helsinki 2003

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ISBN 952-91-6289-8 (paperback) ISBN 952-10-1351-6 (PDF) Otamedia Oy

Espoo 2003

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Preface

The topic of this work was suggested to me in 1997 by Professor Kupiainen, who had been working on the one-dimensional case and was then finishing [BKT99] with the other authors. Thus began a joint project where I worked on detailed analysis of the linearised problem, while he and Jari Taskinen were looking at the nonlinear analysis. Kupiainen and Taskinen were trying to formulate assumptions on the linear semigroup that would be sufficient for carrying out the nonlinear analysis, I was trying to find out what I could actually prove about the linear semigroup, and many meetings were held trying to fit the two pieces together.

The bulk of the present work consists of the linear analysis, which borrows heavily from [BKT99] but is otherwise my work. The nonlinear analysis of Chapter 6 and parts of the introduction were written jointly with Kupiainen and Taskinen.

I wish to thank Professors Kupiainen and Taskinen for their help in creat- ing this work and Alain Schenkel for suggesting many improvements. I would also like to thank Mikko Stenlund, Ville Hakulinen, all my other friends and colleagues in our research group and my parents.

Some parts of this work, Chapter 3 in particular, involve lengthy routine computations. Mathematica and Maxima were used for these. I would like to extend my thanks to the people who have contributed to the development of Maxima and other free software, which provides for most of my computing needs.

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Chapter 1 Introduction

1.1 The equation

In this work we consider the Cahn–Hilliard equation, which we write

∂ψ

∂t =4(− 4ψ− ¹₂ψ+ ¹₂ψ³). (1.1) Denote the spatial coordinates by

x= (x, ~x)∈R^d, x∈R, ~x∈R^d⁻¹. (1.1) has the time-independent solution

ψ0(x) := tanh(x

2), (1.2)

which is also independent of ~x. Our aim is to establish the stability of this solution. This has been done in one dimension in [BKT99], where the following asymptotic behaviour was established:

ψ(x, t) =ψ₀(x−a) + A

√t d

dx ψ₀(x)e⁻^x^4t²

− B

√t d

dxe⁻^x^4t² +o(1 t).

Thus at large times there is a translated front and two perturbation terms near the origin: one of magnitude proportional to 1/√

t and constant width and another of magnitude proportional to 1/t and width to √

t.

The proof in [BKT99] is not directly applicable to several dimensions although we shall use rather similar techniques. One would also expect the equation to behave somewhat differently: in several dimensions an initial perturbation of finite mass should not be able to translate the whole front.

We shall prove the following:

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Theorem 1.1. Let d≥3, r > d+ 1 and assume

|η0(x)| ≤ δ (1 +|x|)^r

for a small enoughδ. Then (1.1) with the initial conditionψ(0, x) =ψ0(x) + η0(x) has a unique classical solution, which for t ≥1 satisfies

ψ(t, x) =ψ₀(x) + A

2∂_xψ₀(x)φ(t, ~x) + ˜ψ(t, x) (1.3) where

A= Z

R^d

η0dx, φ(t, ~k) =ˆ e⁻¹³^|^~k^|³^t, |ψ(t, x)˜ | ≤ Cδ t^d−1³ ⁺¹²¹ . (1.3) describes by linear approximation a local translation of the front by ^A₂φ(t, ~x) at each~x. Because φ(t, ~x) =t⁻^d+1³ φ(1, ~x/t^1/3), as t increases the perturbation grows wider but reduces in magnitude appropriately to preserve mass. The decay rate is different from normal diffusion where one would have

√t instead of t^1/3. Such scaling was to be expected in light of the work on the linear problem in [SO93] and [BR96].

Unfortunately technical problems prevent us from establishing this result in two dimensions. We do not know whether it holds there. We will try to point out the problems by generally assumingd >1 and indicating explicitly whend >2 is needed.

In bounded domains quite a bit more is known about (1.1) because with appropriate boundary conditions the free energy

Z

(¹₂|∇ψ|²+¹₈(ψ²−1)²) dx

is a Lyapunov function. It can also be useful inRas demonstrated in [CCO01]

because in one dimensionψ0 has finite free energy. That, however, is not true in higher dimensions.

Writing ψ =ψ0 +η we get an equation for η:

∂η

∂t =4(Hη+ ³₂ψ₀η²+¹₂η³), (1.4) whereH :=− 4+1 +V and

V(x) :=−³₂cosh(^x₂)⁻². This is equivalent to the integral equation

η(t) =e^t⁴^Hη(0) + Z t

0

e^(t⁻^s)⁴^H4(³₂ψ0η²+ ¹₂η³) ds. (1.5)

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1.1. THE EQUATION 3 The idea is to solve this using the contraction mapping principle. That requires estimates for e^t⁴^H and e^t⁴^H4. Hence we need to study the semigroup which solves the linearised equation

∂tu=4Hu. (1.6)

The independence of V of the d−1 spatial directions can be exploited by applying the Fourier transform in these directions. We denote, somewhat unconventionally

fˆ(x, ~k) :=

Z

R^d−1

e⁻^i~k^·^~^xf(x, ~x) d~x (1.7) and hope that this does not cause too much confusion.¹ Then

4\Hu=−DkHkuˆ (1.8)

where k=|~k|,

Dk =−∂_x²+k², Hk =Dk+ 1 +V.

− 4H is quasi-accretive² in L²(R^d):

Rehu,− 4Hui=k 4uk²−Reh4u,(1 +V)ui

=k 4u− ¹₂(1 +V)uk²− ¹₄k(1 +V)uk² ≥ −¹₄kuk². In fact using the second Neumann series

(ζ+4H)⁻¹ = X∞

k=0

((ζ− 4²+4)⁻¹4V)^k(ζ− 4²+4)⁻¹,

which converges for ζ far enough from the positive real axis, we see that

− 4H is sectorial and generates a quasi-bounded analytic semigroup. That means we can write the integral kernel of the semigroup as a Dunford–Cauchy integral, using the integral kernel of the resolvent:

e^t⁴^H(x, ξ) = Z

−Γ

e^ζt(ζ− 4H)⁻¹(x, ξ) dζ 2πi

= Z

−Γ

e^ζt Z

R^d−1

e^i~k^·^(~^x⁻^~^ξ)(ζ+DkHk)⁻¹(x, ξ) d~k 2π

dζ 2πi

= Z

R^d−1

e^i~k^·^(~^x⁻^ξ)^~ Z

Γ

e⁻^ζt(ζ−DkHk)⁻¹(x, ξ) dζ 2πid ~k

2π (1.9)

1Unfortunately we shall sometimes need Fourier transform in the remainingxvariable, which will also be denoted byˆ. Sorry about that.

2See [Kat66] for an introduction to semigroup theory.

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Γ

σ

(a) Generic (b)k-dependent

Figure 1.1: Integration paths for the analytic semigroup. Due to analyticity the precise shape of the path does not matter.

where Γ is a suitable curve around the spectrum of− 4H as in Figure 1.1(a).

However, after reaching the last line above we can replace Γ with a k- dependent path tailored to fit the spectrum of D_kH_k. This will of course require some knowledge about the spectrum. For most interesting values of k the path will look like Figure 1.1(b), where the spectrum is represented by the thick parts.

1.2 On the spectrum

The problem with integrating around the spectrum ofDkHkis that one would like to stay in the right half plane in order to keepe⁻^ζt bounded (rather than just quasi-bounded). AlthoughDkHkis positive there is not much room when k is small. Here we present some coarse estimates which serve to illustrate the structure of the spectrum. These are based on [SO93].

The spectrum of H0 is known [LL81, page 79]: there are two isolated eigenvalues at 0 and 3/4 and a continuous spectrum [1,∞). The eigenfunc- tions are V and x7→sinh(x/2)/cosh(x/2)².

For k > 0 Dk is positive and self-adjoint. Its inverse can be represented with the convolution kernel

Gk(x) = 1

2ke⁻^k^|^x^|.

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1.2. ON THE SPECTRUM 5 Alternatively, on the Fourier side we have

Gˆk(p) = 1 k²+p², which is bounded by 1/k².

To overcome the lack of self-adjointness of DkHk we study instead the positive semidefinite and self-adjoint operatorAk :=D^1/2_k HkD^1/2_k .

Lemma 1.2. When k > 0 Ak and DkHk have the same eigenvalues with same multiplicities.

Proof. Let DkHku =ζu. Set v =G^1/2_k u. Now DkHkD_k^1/2v =ζD_k^1/2v and v is an eigenvector of Ak.

Any eigenvector v of Ak obviously has to be in the domain ofD^1/2_k , thus we can set u=D^1/2_k v and reverse the calculation above.

The bottom of the spectrum of A_k can be obtained using the Rayleigh quotient. We get

ζ0 = inf

u

hu, Hkui

hu, Gkui (1.10)

(by substituting G^1/2_k u for the variable v in the usual Rayleigh quotient for Ak, which is permissible since any v in the domain of Ak is of such form).

Plugging the normalised eigenfunction u0 of H0 (corresponding to the zero eigenvalue) into (1.10) we immediately get an upper bound:

ζ0≤ k²

hu₀, G_ku₀i = 2k³ R_∞

−∞

R_∞

−∞u0(x)e⁻^k^|^x⁻^y^|u0(y) dxdy.

The denominator can be bounded from below by a constant C⁻¹ for, say, k ≤1, and we getζ0 ≤Ck³ for smallk. We also get a lower bound:

ζ0 ≥inf

u

k⁴kuk²

hu, k²Gkui ≥k⁴.

To see that ζ0 is a discrete eigenvalue of multiplicity one we proceed further with the minimax principle:

ζ₁ = sup

v uinf∈v^⊥

hu, Akui

hu, ui = sup

v

infu hG^1/2_k u,vi=0

hu, Hkui hu, Gkui

= sup

v inf

u∈v^⊥

hu, Hkui

hu, Gkui ≥sup

v inf

u∈v^⊥

hu, H0ui kuk²

k²kuk²

hu, k²Gkui ≥ ³₄k²,

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which for small k is larger than ζ0.

D_kH_khas asymptotically constant coefficients, thus its essential spectrum is the same as that of D²_k+Dk [CE90, Proposition 26.2], which is real and bounded from below by k²+k⁴.

We have gathered these facts:

Lemma 1.3. For anykthe spectrum ofDkHkis real and bounded from below by k⁴.

For smallk the bottom of the spectrum is an isolated eigenvalueζ0 of multiplicity one with ζ0 ≤ Ck³. The remaining part of the spectrum is bounded from below by ³₄k².

1.3 Notation

C will be used as generic notation for constants. The actual values ofC may vary (typically increasing) from expression to expression. For clarity, one might label every C with a distinct subscript. For brevity, we usually don’t.

When we do equip C with a subscript it denotes a specific constant.

R will be used to denote the resolvent (ζ −DkHk)⁻¹. K will be used to denote the semigroup e⁻^tD^k^H^k. Operators and their integral kernels are identified, e.g., we may write (RDk)(x, ξ).

The O(·) notation will be used in the usual sense, except that we sometimes don’t bother writing absolute value signs: O(f(λ)) denotes an expression which is bounded by|f(λ)|whenλis in some region, which hopefully will be clear in each context. Our laxness with absolute values extends to writing O(λⁿ) whenλis a finite-dimensional vector, in which case we naturally mean O(|λ|ⁿ).

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Chapter 2 The Pego–Weinstein formalism

2.1 Introduction

The linear analysis is based on an approach used in [PW92]. Let us study a system of ordinary linear differential equations in Cⁿ:

y⁰(x) =A(λ, x)y(x) +b(x), (2.1) where the coefficient matrix depends analytically on one or more complex parameters λ.¹ Assume A_∞(λ) := limx→±∞A(λ, x), i.e., both limits exist and are the same and that R(λ, x) := A(λ, x)−A_∞(λ) can be bounded by Ce⁻^ρ^|^x^|uniformly inλ. We will usually not bother to write theλdependence explicitly. Here is a summary of relevant notation:

Definition 2.1 (notation). µi, i ∈ {1, . . . , n}, are the eigenvalues of A_∞. v_i andw_i are the corresponding right and left eigenvectors, i.e., A_∞v_i =µ_iv_i, wiA_∞=µiwi.

Some of theµi and their eigenvectors may be equal to each other for some values ofλ.

It proves useful to study the homogeneous equation corresponding to (2.1) along with the so-called transposed equation:

y⁰(x) =A(x)y(x) (2.2)

z⁰(x) =−z(x)A(x). (2.3)

The motivation for (2.3) is that ify andz are solutions of (2.2) and (2.3) the product z(x)y(x) is independent of x.

1Thusλis a complex vector of unspecified dimension and in some unspecified region.

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As the coefficient matrix A(x) tends to the constant A_∞ when x→ ±∞

one would also expect the solutions of (2.2) and (2.3) to tend to the solutions of the corresponding constant coefficient equations. Indeed, Pego and Wein- stein show in [PW92] that if there is a simple eigenvalue µ1 with a smaller real part than any other eigenvalue (2.2) admits a unique solution y⁺₁ with the property thate⁻^µ¹^xy₁⁺(x)→v1 asx→ ∞. Similarly (2.3) has an unique solution z₁⁻ with e^µ¹^xz⁻₁(x)→w1 as x→ −∞.

2.2 More solutions of the homogeneous equa- tion

For our purposes it is useful to extract solutions of (2.2) and (2.3) corresponding to each eigenvalue µ_j, which is possible under suitable assumptions. We prove things for y_j⁺ but the z_j⁻ case is similar.

Choose some λ-dependent µ and substitute y(x) =: e^µxv(x) in (2.2) to get

v⁰(x) = (B+R(x))v(x), (2.4)

where B := A_∞ − µ and R(x) := A(x) − A_∞. We want to divide the eigenvalues of A_∞ into two sets: those with smaller real part thanµand the rest. However, we need some margins in order to keep everything uniform in λ because the eigenvalues move around.

Let Ei := S_∞

n=1ker(µi −A_∞)ⁿ. Assume that there are constants α and β and a set of indices I such that Re(µi − µ) ≤ α < 0 when i ∈ I and Re(µi−µ)≥β >−ρ whenever i6∈I. α, β and I are assumed independent of λ. Also, if some eigenvalues of A_∞ coincide for some λ, such a group of eigenvalues shall not be split betweenI and its complement: they must all be in one or the other. Let P be the projection onto L

i∈IE_i which commutes with A_∞ and let Q:= 1−P. Fix some positivex0 and define

(FPv)(x) :=

Z x x0

e^(x⁻^ξ)BP R(ξ)v(ξ) dξ, (FQv)(x) := −

Z _∞

x

e^(x⁻^ξ)BQR(ξ)v(ξ) dξ, F :=FP +FQ

(2.5)

for bounded continuous functions v on [x0,∞].

Lemma 2.2. When x0 is sufficiently large F is a contraction in the norm kvk= sup_x_∈_[x₀_,_∞₎|v(x)|.

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2.2. MORE SOLUTIONS OF THE HOMOGENEOUS EQUATION 9 Proof. Some eigenvalues of B may be defective and solutions corresponding to them may have additional polynomial growth rather than just the usual exponential behaviour. Fix a small to absorb that. Choose ∈(0,−α−ρ) if possible, otherwise just ∈ (0,−α). Letting C denote a generic constant independent ofx₀ (but dependent on ) we have

|FPv(x)|=

Z x x0

e^(x⁻^ξ)BP R(ξ)v(ξ) dξ

≤C Z x

x0

e^(α+)(x⁻^ξ)e⁻^ρξkvkdξ

≤Ce^(α+)xkvk Z x

x0

e⁻^(ρ+α+)ξdξ

≤

(Ce⁻^ρxkvk if α <−ρ, Ce⁻^(ρ+α+)x⁰e^(α+)xkvk otherwise and, taking < ρ+β,

|F^Qv(x)|=

Z _∞

x

e^(x⁻^ξ)BQR(ξ)v(ξ) dξ

≤C Z _∞

x

e^(β⁻^)(x⁻^ξ)e⁻^ρξkvkdξ

=Ce^(β⁻^)xkvk Z _∞

x

e⁻^(β+ρ⁻^)ξdξ

≤Ce⁻^ρxkvk,

Thus we can make the operator norm ofF arbitrarily small by taking a large enough x0.

Corollary 2.3. (Fv)(x) =O(e^max(⁻^ρ,α+)x) for large x and any >0.

Hence v = ˜v+Fv can be solved forv given any ˜v. For such a solution (v−˜v)⁰ =P Rv+BFPv+QRv+BFQv

=B(v−v˜) +Rv.

In particular if ˜v is a bounded solution of ˜v⁰ = B˜v (a constant coefficient equation) for large xthen v will be a solution of (2.4) with the same asymptotic behaviour as ˜v in the sense thatv−v˜=Fv tends to zero exponentially fast as x→ ∞.

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Corollary 2.4. For each eigenvalue µi there is a solution yi of (2.2) which behaves asymptotically (asx → ∞) like e^µⁱ^xv_i:

|e⁻^µⁱ^xy_i⁺(x)−v_i|< Ce⁻^ρ²^x for someC and all x >0.

Proof. Pick µ= µi, α = −⁵₈ρ, β = −⁷₈ρ, = ₁₆¹ρ and ˜v(x) = vi. It is clear that a suitableI can be found andF can be used to get a solution forx > x0, which can then be extended to the whole real line.

In general the solutions of Corollary 2.4 are not unique even after fixing normalisation of vi because one can add similar solutions corresponding to µj with smaller real parts thanµi.

Corollary 2.5. If the µi and vi are analytic functions of λ in some domain and we can fix theI in the previous proof uniformly for allλthen the solutions in Corollary 2.4 are analytic in this domain when evaluated at some fixed x.

If yi is such a solution then |e⁻^µⁱ^(λ)xyi(λ;x)−vi(λ)|< Ce⁻^ρ²^x for x >0 and λ in compact subsets of the domain (C depends on the subset).

2.3 Application to the resolvent of D

k

H

k Write (ζ−DkHk)u=f in the form y⁰ =Ay+b by setting

y= (u, u⁰, u⁰⁰, u⁰⁰⁰)^T, b = (0,0,0,−f)^T (2.6) and

A =







0 1 0 0

0 0 1 0

0 0 0 1

ζ−k²−k⁴−k²V +V⁰⁰ 2V⁰ 1 + 2k²+V 0







. (2.7)

The eigenvalues of A_∞ are the zeros of the polynomial ζ−(−µ²+k²)²−(−µ² +k²) which are

µ_j =± q

1

2 +k²± ¹₂p

1 + 4ζ. (2.8)

Unfortunately they have somewhat poor analyticity for k and ζ near zero.

This problem can be overcome by writing

ζ =k²+k⁴+ (1 + 2k²)²τ² (2.9)

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2.3. APPLICATION TO THE RESOLVENT OF DKHK 11 and using λ= (k, τ) as the parameters. Then (2.7) and (2.8) become

A=







0 1 0 0

0 0 1 0

0 0 0 1

(1 + 2k²)²τ²−k²V +V⁰⁰ 2V⁰ 1 + 2k²+V 0







and

µj =±√

1 + 2k² q1

2 ± ¹₂√

1 + 4τ². (2.10)

To fix the branches when the second sign is negative set

± q1

2 − ¹₂√

1 + 4τ² = ±iτ q1

2 + ¹₂√

1 + 4τ²

(2.11)

and choose µ2 to have the positive sign with the principal branch in the denominator. Now the µ_i are analytic functions of k and τ when k ∈ C\

±i[1/√

2,∞) and τ ∈ C \ ±i[¹₂,∞). We are mostly interested in real k, considering that k used to be the length of something. Then the cuts in the τ plane correspond to ζ ∈ (−∞,−¹₄]. τ ∈ ±i[0,¹₂k] ∪R is mapped onto [³₄k² − k⁶,∞) which by Lemma 1.3 contains all of the spectrum of DkHk except the lowest eigenvalue, which should be somewhere around τ ≈ ik when k is small. For larger k it is more useful to use the k⁴ bound:

τ ∈ ±i[0, k/(1 + 2k²)]∪R is mapped onto [k⁴,∞) which contains the entire spectrum.

τ is mapped two-to-one to ζ and will usually be kept in the upper half plane. There Reµ2 <0 when k is real. However, we need some results to be valid also in a small complex neighbourhood of k = τ = 0 because we will develop a power series there later.

Set µ1 :=−√

1 + 2k² q1

2 +¹₂√

1 + 4τ², µ4 :=−µ1 and µ3 :=−µ2. These equations are sometimes useful:

µ²₁+µ²₂ = 1 + 2k², µ₁µ₂ =−iτ(1 + 2k²), µ²₁ −µ²₂ = (1 + 2k²)√

1 + 4τ² =p

1 + 4ζ. (2.12)

The eigenvectors vj and wj are easy to express in terms of µj and are thus also analytic. Normalise the eigenvectors so that (vj)1 = (wj)4 = 1. This results in

vj = 1, µj, µ²_j, µ³_jT

,

wj = µ_j(µ²_j −1−2k²), µ²_j −1−2k², µ_j, 1 .

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According to Corollary 2.4 for j ∈ {1,2,4} there is a y⁺_j which solves the homogeneous equation ∂_xy⁺_j = Ay_j⁺ and behaves like e^µ^j^xv_j as x → ∞. There is also a z_j⁻ which solves the transposed equation ∂xz_j⁻ =−z_j⁻A and behaves like e⁻^µ^j^xwj asx→ −∞. It would also be possible to define y⁺₃ and z₃⁻ in this way but that would not be very useful because A_∞ is defective at τ = 0: the eigenvectors v2 and v3 collide and we would not have a set of four linearly independent solutions. Thus with some abuse of notation we require instead

y₃⁺(x)∼ e^µ³^xv3−e^µ²^xv2

µ₃−µ₂ and z₃⁻(x)∼ e⁻^µ³^xw3−e⁻^µ²^xw2

µ₃−µ₂

as x → ∞ and x → −∞, respectively. As µ₃ − µ₂ → 0 these converge (pointwise in x) to solutions with linear asymptotes.

Our expression for the integral kernel of (ζ−DkHk)⁻¹ will contain only y_j⁺ and z_j⁻ for j ∈ {1,2}. However, the other two values of j are needed for understanding the behaviour of the kernel.

Theorem 2.6. Assume

Re(µ1±µ2)<−⁵₈ and Re(µ2)< ₃₂¹ . (2.13) Then(2.2) with (2.7)has solutionsy⁺_i such that|y_i⁺(x)−e^µⁱ^xvi|< C|e^(µⁱ⁻¹²^)x| for i∈ {1,2,4} and

|y₃⁺(x)−e^µ³^xv3−e^µ²^xv2

2µ3 |< C|e^(µ³⁻¹²^)x| when x is bounded from below. Furthermore,

|∂_xⁿy_i⁺(x)−µⁿ_ie^µⁱ^xvi|< C|e^(µⁱ⁻¹²^)x|, n ∈ {1,2}, i6= 3

|∂xy₃⁺(x)− ¹₂(e^µ³^xv3+e^µ²^xv2)|< C|e^(µ³⁻¹²^)x|,

|∂_x²y₃⁺(x)−¹₂(µ3e^µ³^xv3+µ2e^µ²^xv2)|< C|e^(µ³⁻¹²^)x|.

Similarly, (2.3) has solutionsz_i⁻ such that |z_i⁻(x)−e⁻^µⁱ^xwi|< Ce⁽¹²⁻^µⁱ^)x for i∈ {1,2,4} and

|z₃⁻(x)− e⁻^µ³^xw3−e⁻^µ²^xw2

2µ3 |< C|e⁽¹²⁻^µ³^)x|

when x is bounded from above. The corresponding estimates for the derivatives hold.

For i∈ {1,2,4}y_i⁺(x) and z_i⁻(x) are analytic in λ wherever the assumptions hold. y⁺₃(x) and z₃⁻(x) are continuous and when

Re(µ2)>−¹₄ (2.14)

also analytic. C above depends on λ but can be fixed in any compact subset.

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2.3. APPLICATION TO THE RESOLVENT OF DKHK 13 Before the proof we would like to comment on the assumptions of the theorem. Specifying them in terms of µ₁ and µ₂ was perhaps a bit opaque.

However, there are two cases which interest us. The first isλ≈0. In a small enough neighbourhood of the origin both (2.13) and (2.14) clearly hold. The second case is whenk is real andτ in the upper half plane. Thenµ₂ is in the left half plane. For the first inequality of (2.13) use (2.12) to get (µ1±µ2)² = (1 + 2k²)(1∓2iτ). From this we see that the inequality is equivalent to τ lying beneath a parabola whose apogee is at i(128k² + 39)/(256k² + 128).

This is always above the values corresponding to the spectrum.

Proof. We prove this for they_i⁺, thez_i⁻ case is similar enough. ρ= 1. Apply Corollary 2.5 with

I =







∅, wheni= 1 {1}, when i= 2 {1,2,3}, when i= 4.

When i= 3 choose µ=µ3+₁₆¹ and

˜

v(x) =e⁻¹⁸^xv3−e^2µ²^xv2

2µ3

which is bounded even when µ3 → 0. Unfortunately we have to split this up as follows: when Re(µ2) >−³₈ choose I = {1} and when Re(µ2) < −¹¹₃₂ choose I = {1,2}. α = −⁵₈, β = −⁷₈ will do for both regions just as in Corollary 2.4. Solve v = ˜v+Fv forv in each region separately, glue together with a partition of unity and define y₃⁺(x) = e^µxv(x). As α+ ₁₆¹ < −¹₂ we still get the claimed estimate.

That gave us the solutions and the basic estimates. For the derivatives we need to use (2.2). Wheni6= 3y_i⁺(x) =e^µⁱ^xv_i+r_i(x) and|r_i(x)| ≤C|e^(µⁱ⁻¹²^)x|. Thus

∂_xy⁺_i (x) =A(x)y_i⁺(x) = (A_∞+R(x))(e^µⁱ^xv_i+r_i(x))

=µie^µⁱ^xvi+e^µⁱ^xR(x)vi+A(x)ri(x)

=µie^µⁱ^xvi+O(e^(µⁱ⁻¹²^)x),

∂_x²y⁺_i (x) =A(x)∂_xy⁺_i (x) +∂_xR(x)y_i⁺(x)

=µ²_ie^µⁱ^xvi+O(e^(µⁱ⁻¹²^)x).

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The i= 3 is slightly but not substantially different:

y⁺₃(x) = e^µ³^xv3−e^µ²^xv2

2µ3

+r₃(x) where |r₃(x)| ≤C|e^(µ³⁻¹²^)x|,

∂xy⁺₃(x) = ¹₂(e^µ³^xv3+e^µ²^xv2) +R(x)e^µ³^xv3−e^µ²^xv2

2µ3

+A(x)r3(x)

= ¹₂(e^µ³^xv3+e^µ²^xv2) +O(e^(µ³⁻¹²^)x),

∂_x²y⁺₃(x) =A(x)∂xy⁺₃(x) +∂xR(x)y₃⁺(x)

= ¹₂(µ₃e^µ³^xv₃+µ₂e^µ²^xv₂) +O(e^(µ³⁻¹²^)x).

Lemma 2.7. Under the assumptions (2.13) there are also solutions y_j⁻ and z_j⁺ of the homogeneous (2.2) and transposed equation (2.3), respectively, with prescribed behaviour forx→ −∞andx→ ∞: (e.g. y⁻₁(x)∼e⁻^µ¹^xv4 asx→

−∞—note that µ1 = −µ4). These can be expanded as linear combinations of y_j⁺ and z_j⁻. The coefficients of this expansion are continuous and when (2.14) holds also analytic.

Proof. This is an immediate consequence of the symmetry of (1.6) under the reflection x7→ −x: we have y_j⁻(x) = Jy_j⁺(−x) and z_j⁺(x) =−z_j⁻(−x)J, where J = diag(1,−1,1,−1). The coefficients of the expansion can be de- termined at some fixed point, say x = 0; the y^±_j (0) are continuous in the parameters and linearly independent.

The solutions y_j^± can of course naturally be written in terms of a scalar valued function Y_j^±. Similarly the solutions of the transposed equation can be written in terms of a single function. A straightforward computation yields:

Lemma 2.8.

y_j^±= Y_j^±, ∂xY_j^±, ∂_x²Y_j^±, ∂_x³Y_j^±T

,

z^∓_j = (Hk∂x+k²∂x−V⁰)Z_j^∓, −(Hk+k²)Z_j^∓, −∂xZ_j^∓, Z_j^∓ with (ζ−DkHk)Y_j^± = 0and (ζ−HkDk)Z_j^∓= 0. For j ∈ {1,2,4} we have

xlim→∞e⁻^µ^j^xY_j⁺(x) = lim

x→−∞e^µ^j^xZ_j⁻(x) = 1 while

xlim→∞e⁻^µ³^xY₃⁺(x)− e^2µ²^x−1 2µ2

= lim

x→−∞e^µ³^xZ₃⁻(x)− e⁻^2µ²^x−1 2µ2

= 0 (when µ2 = 0, replace the fractions by their limits, i.e., x and −x).

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2.3. APPLICATION TO THE RESOLVENT OF DKHK 15 We want to write a solution of y⁰ =Ay+b in the form

y(x) = Z x

−∞

y⁺(x)(Ω⁺)⁻¹z⁻(ξ)b(ξ) dξ+ Z x

∞

y⁻(x)(Ω⁻)⁻¹z⁺(ξ)b(ξ) dξ, (2.15) where y^± = (y^±₁, y₂^±), z^∓ = _z∓

1

z₂^∓

and Ω^± = z^∓y^±. These products are independent of x and it is also easy to see that Ω⁻ = −Ω⁺. For simplicity we shall from now on denote Ω⁺ just by Ω. For this formula to make sense Ω of course needs to be invertible.

Theorem 2.9. Assume Reµ1 <Reµ2 <0 and ζ ∈ρ(DkHk) (the resolvent set). Then Ω =z⁻y⁺ is invertible.

Proof. Assume the contrary: let Ω(α1, α2)^T = 0, i.e., z⁻_i y= 0 for i∈ {1,2}, y:=α1y⁺₁ +α2y₂⁺ 6= 0. y can also be written as P4

j=1βjy_j⁻. We have e^µⁱ^xz_i⁻(x)→wi, e⁻^µⁱ^xy₅⁻₋_i(x)→

( 1

2µ3v2 when i= 2, vi otherwise

as x → −∞. Thus z⁻₁(x)y_j⁻(x) → δ4jw1v1 but the left side is actually independent of x. Since µ1 is a simple eigenvalue we must have w1v1 6= 0, consequently z₁⁻y = 0 implies β₄ = 0. In a similar vein z₂⁻y = 0 implies β3 = 0. Hence y = β1y₁⁻+β2y⁻₂ decreases exponentially at ±∞, giving a nontrivialL² solution to (ζ−DkHk)u= 0. Such a solution was assumed not to exist.

Lemma 2.10. Under the assumptions of the previous theorem (2.15) solves y⁰ =Ay+b.

Proof. Differentiating (2.15) and recalling that y^± solve the homogeneous equation we get

y⁰(x) =A(x)y(x) + y⁺(x)Ω⁻¹z⁻(x)−y⁻(x)Ω⁻¹z⁺(x) b(x).

Thus we need to show that the matrix inside the parentheses is the identity.

y⁺(x)Ω⁻¹z⁻(x) is a projection onto the space spanned by{y⁺₁(x), y₂⁺(x)}. These two vectors are linearly independent. Similarly −y⁻(x)Ω⁻¹z⁺(x) is a projection onto the two-dimensional space spanned by {y₁⁻(x), y⁻₂(x)}. The two projections are orthogonal to each other asz^±(x)y^±(x) = 0: the product would have exponential decay as x→ ±∞but is actually independent of x.

Thus their sum is the identity.

Recalling (2.6) we get for the original equation:

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Theorem 2.11. Under the assumptions of Theorem 2.9 the integral kernel of the resolvent is given by

R(x, ξ) := (ζ−DkHk)⁻¹(x, ξ) =

(−Y⁺(x)Ω⁻¹Z⁻(ξ) for ξ < x,

−Y⁺(−x)Ω⁻¹Z⁻(−ξ) for ξ > x, (2.16) where Y⁺ = (Y₁⁺, Y₂⁺) and Z⁻ = (Z₁⁻, Z₂⁻)^T.

Note that from y⁺(x)Ω⁻¹z⁻(x)−y⁻(x)Ω⁻¹z⁺(x) = 1 it follows in particular that the (i, j) component of the left hand side is 0 for i < j. From this we see that our resolvent kernel has continuous derivatives with respect tox orξ up to total order two.

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Chapter 3 Leading terms of the resolvent

3.1 Outline

Here we estimate the resolvent for very small k and |ζ|. We use (2.9) and assume k < , |τ| < for a small enough .¹ Then the assumptions of Theorem 2.6 are satisfied, including the analyticity condition (2.14). We shall expand Ω of (2.16) into a power series to get some explicit leading terms for the resolvent. Thus this chapter is mostly about computing derivatives of things at k =τ = 0.

3.2 Solutions of the homogeneous equation

We’ll denote the solutions atk =τ = 0 with a r˚ıng. ˚µ1 =−1, ˚µ2 = 0 and Y˚₁⁺(x) = 1

4 cosh(^x₂)²

Y˚₂⁺(x) = −1−6e^x+ 5e^2x+ 2e^3x+ 6e^2xx 2e^x(1 +e^x)²

Z˚₁⁻(x) = log(e^x+ 1) Z˚₂⁻(x) = 1.

(3.1)

By a mechanical computation we get

˚Ω = 0 1

0 0

.

1We start with somebut Cauchy’s estimates and such nibble at it, and we shall end up with a somewhat smallerthan we started with.

17

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Although they do not appear in Ω, we also list for later use the solutions which are asymptotically equal tox7→ |x|, i.e., ˚Y₃⁺(x)/x→1 asx→ ∞and Z˚₃⁻(x)/x→ −1 asx→ −∞. After arduous but routine calculations² we find these:

˚Y₃⁺(x) = 1 2(1 +e^x)²

1 + 4e^x−e^2x+ 16e^xlog 2−6x−xe⁻^x+ 7xe^x+ 2xe^2x + 3x²e^x+ (−8−e⁻^x+ 8e^2x+e^3x) log(1 +e⁻^x) + 12e^xLi2(−e⁻^x) Z˚₃⁻(x) =−x

where Li2 is the dilogarithm:

Li2(z) :=

X∞

k=1

z^k k² =

Z 0 z

log(1−s)

s ds.

These products will be needed later:





˚z₁⁻

˚z₂⁻

˚z₃⁻



˚y₃⁺=



 0

−1 0



 and ˚z⁻₃ ˚y₁⁺ ˚y⁺₂ ˚y₃⁺

= 0 −1 0

. (3.2)

We shall also also need the rapidly growing solution

˚Y₄⁺(x) = 4 cosh(x 2)².

3.3 First order

Define v_j⁺(x) = e⁻^µ^j^xy_j⁺(x) and w⁻_j (x) = e^µ^j^xz⁻_j (x) for j ∈ {1,2}. By Corollary 2.5 they are analytic functions of λ for each fixed x and satisfy

|v_j⁺(x)−vj|< Ce⁻^x/2 for x > 0 and |w_j⁻(x)−wj| < Ce^x/2 for x < 0 where C can be fixed independently ofk and τ. They will also satisfy the obvious differential equations, which we now differentiate, then setk =τ = 0:

∂x∂λv⁺_j = (A−µj)∂λv_j⁺−∂λµjv⁺_j

∂x∂λw_j⁻=−∂λw⁻_j (A−µj) +∂λµjw_j⁻

(note that ∂λA = 0). These are just inhomogeneous versions of the equations satisfied by v⁺_j and w_j⁻. The “initial conditions” are again asymptotic:

2Admittedly somewhat less arduous for ˚Z₃⁻ than for ˚Y₃⁺.

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3.4. SECOND ORDER 19

|∂λv_j⁺(x)−∂λvj|< Ce⁻^x/2 and |∂λw⁻_j (x)−∂λwj| < Ce^x/2 by Cauchy’s estimates. The first component of ∂_λv_j and the last component of∂_λw_j are zero because of the normalisation used.

The j = 1 case is trivial: we have ∂λµ1 = 0, ∂λv1 = 0 and the equations above are actually homogeneous. They have no nontrivial solutions with the required asymptotic behaviour, and we conclude ∂λv⁺₁ = 0, ∂λw⁻₁ = 0.

For j = 2 we easily find special solutions: just take −∂λµ2xv₂⁺ and

∂_λµ₂xw⁻₂. These of course grow linearly, which can be countered by adding some ˚y₃⁺ and ˚z₃⁻. When we do that we end up with solutions whose first and last components, respectively, decrease exponentially. We are happy with that:

∂λv⁺₂ =∂λµ2(˚y₃⁺−x˚y⁺₂)

∂λw⁻₂ =∂λµ2(˚z⁻₃ +x˚z⁻₂).

Note thaty₂⁺and z₂⁻ are only unique up to multiples ofy₁⁺ andz⁻₁. We fixed the values at λ= (0,0) earlier and now we fix theirλ-derivatives so that no

˚y⁺₁ or ˚z₁⁻ appears in the expressions above.

It is also worth noting that ˚Z₃⁻+xZ˚₂⁻ = 0. Not all components of the vector ∂λw⁻₂ vanish, though.

Thus we end up with these reasonably simple expressions:

∂λy_j⁺=∂λµj˚y₃⁺, ∂λz⁻_j =∂λµj˚z₃⁻ (3.3) and, using (3.2),

∂λΩ = ∂λz⁻y⁺+z⁻∂λy⁺=

0 0 0 −2∂_λµ₂

, with ∂λµ1 =∂kµ2 = 0 and ∂τµ2 = i.

3.4 Second order

So now we have come up with some non-zero leading terms for the right column of Ω. Let us also try to get something to the left column in order to eventually compute some non-zero terms for det Ω. Here we shall deal with

∂_λ²(z_i⁻y₁⁺) =∂_λ²z_i⁻y₁⁺+z_i⁻∂_λ²y₁⁺

—remember that ∂λy₁⁺= 0 at λ = 0. It is convenient to estimate the terms on the right hand side at separate values of x, which is possible with the following trick:

∂x(∂²_λz⁻_i y₁⁺) =−∂_λ²(z_i⁻A)y₁⁺+∂_λ²z_i⁻Ay⁺₁ =−z_i⁻∂_λ²Ay₁⁺

Stability of Cahn-Hilliard Fronts in Three Dimensions