AARMS SUMMER SCHOOL–LECTURE VII:
INTRODUCTION TO INFINITE DIMENSIONAL
REDUCTION METHODS FOR SOLVING PDE’S
MANUEL DEL PINO AND JUNCHENG WEI
1. Back to Allen Cahn in R2
We consider the functional
¶
Z µ
2
(1 − u2 )2
2 |∇u|
J(u) =
ε
a(x)dx.
+
2
4
R2
Critical points of J are solutions of
ε2 div(a(x)∇u) + a(x)(1 − u2 )u = 0,
where we suppose 0 < α ≤ a(x) ≤ β. This equation is equal to
∇a
(1.1)
ε2 ∆u + ε2
(x)∇u + (1 − u2 )u = 0.
a
Using the change of variables v(x) = u(εx), we find the equation
∇a
(1.2)
∆v + ε
(x)∇v + (1 − v 2 )v = 0.
a
We will study the problem: Given a curve Γ in R2 we want to find a
solution uε (x) to (1.1) such that uε (x) ≈ w( zε ), for points x = y +zν(y),
y ∈ Γ, |z| < δ, where ν(y) is a vector perpendicular to the curve and
w(t) = tanh( √t2 ), which solves the problem
w00 + (1 − w2 )w = 0,
w(±∞) = ±1.
First issue: Laplacian near Γ, which we will consider as smooth as we
need.
Assume: Γ is parametrized by arc-length
γ : [0, l] → R2 , s → γ(s), |γ̇(s)| = 1, l = |Γ|.
Convention: ν(s) inner unit normal at γ(s). We have that |ν(s)|2 = 1,
which implies that 2ν ν̇ = 0, so we take ν̇(s) = −k(s)γ̇(s), where k(s)
is the curvature.
Coordinates: x(s, t) = γ(s) + zν(s), s ∈ (0, l) and |z| < δ. If we
take a compact supported function ψ(x) near Γ, and we call ψ̃(s, z) =
ψ(γ(s) + zν(s)), then ∂∂sψ̃ = ∇ψ · [γ̇ + z ν̇] = (1 − kz)∇ψ · γ̇ and ∂∂tψ̃ =
1
2
MANUEL DEL PINO AND JUNCHENG WEI
∇ψ · ν. Observe that ∇ψ = (∇ψ · γ̇)γ̇(∇ · ν)ν. This means that
1
1 ∂ ψ̃
2
2
γ̇ + ∂∂zψ̃ ν, and |∇ψ|2 = (1−kz)
∇ψ = 1−kz
2 |ψ̃s | + |ψ̃z | . Then
∂s
¶
Z
ZZ µ
1
2
2
2
|∇ψ(x)| dx =
|ψ̃s | + |ψ̃z | (1 − kz)dsdz
(1 − kz)2
R2
ψ → ψ + tϕ and differentiating at t = 0 we get
Z
ZZ
1
∇ψ∇ϕdx =
ψ̃s ϕ̃s + ψ̃z ϕ̃z (1 − kz)dsdz
(1 − kz)
So
Z
ZZ
− ∆ψϕdx = −
then
∆ψ̃ =
1
(1 − kz)
µµ
1
ψ̃s
(1 − kz)
¶
¶
+ (ψ̃z (1 − lz))z ϕ̃(1−kz)dsdz
s
∂
1
k
1
(
ψ̃s ) + ψ̃zz −
ψ˜z
(1 − kz) ∂s 1 − kz
1 − kz
We just say
∆ψ̃ =
1
1
k
(
ψs )s + ψzz −
ψz
1 − kz 1 − kz
1 − kz
Near Γ (x = γ(s) + zν(s)), we have the new equation for u → ũ(s, z)
S[u] = ε2
1
1
ε2 k
ε2 a s
ε2 az
(
us )s +ε2 uzz +(1−u2 )u−
uz +
us +
uz = 0
1 − kz 1 − kz
1 − kz
1 − kz a
1 − kz a
we want a solution u(s, z) ≈ w( zε ).
z
az
k(s)
z
S[w( )] = ε[ −
]w0 ( )
ε
a
1 − k(s)z
ε
The condition we ask (geodesic condition) is
language we want
az
(s, 0)
a
= k(s). In v
∇a
(εx) · ∇v + f (v) = 0
a
transition on Γε = 1ε Γ. we use coordinates relative to Γε rather than Γ
∆v + ε
1
Xε (s, z) = γ(εs) + zν(εs), |z| < δ/ε
ε
Laplacian for coordinates relative to Γε are
¶
µ
εk(εs)
as
az
1
1
1
vs +ψzz −
+ε
vs +ε vz +
∆ψ =
2
(1 − εk(εs)z) (1 − εk(εs)z)
(1 − εk(εs)z)
a (1 − εk(εs)z)
a
s
where we use the computation
∂γ(εs)
∂s
= −k(ε)γ˙ε (s), where kε = εk(εs)
LYAPUNOV SCHMIDT REDUCTION METHOD
3
Hereafter we use s̃ instead of s and z̃ instead of z̃. Observation: The
operator is closed to the Laplacian on (s̃, z̃) variables, at least on the
curve Γ, if we assume the validity of the relation
az̃ (s̃, 0) = k(s̃)a(s̃, 0),
∀s̃ ∈ (0, l).
We can write this relation also like ∂ν a = ka on Γ (Geodesic condition).
This relation meansRthat Γ is a critical point of curve length weighted
by a. Let La [Γ] = Γ adl. Consider a normal perturbation of Γ, say
Γh := {γ(s̃) + h(s̃)ν(s̃)|s̃ ∈ (0, l)}, khkC 2 (Γ) ¿ 1. We want: first
variation along this type of perturbation be equal to zero. This is
DLa [Γh ]|h=0 = 0
This means
∂
L[Γλh ]|h=0 = 0
∂λ
or just hDL(Γ), hi = 0 for all h. Observe that
Z l
a(γ(s̃) + h(s̃)ν(s̃)) · |γ̇(s̃)λh |ds̃
L(Γλh ) =
0
and also γ̇λh (s̃) = γ̇(s̃) + λḣν + λhν̇, and ν̇ = −k γ̇. With the taylor
expansion
1
1
(1−2kλh+λ2 k 2 h2 +λ2 ḣ2 )1/2 = 1+ (−2kλh+λ2 k 2 h2 +λ2 ḣ2 )− 4k 2 λ2 h2 +O(λ2 h3 )
2
8
and
1
˜
a(γ((s))+λh(s̃ν(s̃))
= a(s̃, λh(s̃)) = a(s̃, 0)+λaz̃ (s̃, 0)h(s̃)+ λ2 az̃z̃ (s̃, 0)h(s̃)2 +O(λ3 h3 ).
2
we conclude
Z l
Z l
ḣ2
1
2
Lh [Γλh ] = La (Γ) = λ (−ka+az̃ )(s̃, 0)h(s̃)ds̃+λ
(a +az̃ k 2 h2 + az̃z̃ h2 )+O(λ3 h3 )
2
2
0
0
This tells us:
∂
Lh [Γλh ]|λ=0 = 0 ⇔ k(s̃)a(s̃, 0) = az̃ (s̃, 0),
∂λ
the geodesic condition. Also we conclude that
Z l
Z l
∂2
2
2
2
L(Γλh )|λ=0 =
(aḣ −2k a+az̃z̃ h )ds̃ = − (a(s̃, 0)ḣs̃)0 h+(2a(s̃, 0)k 2 −az̃z̃ (s̃, 0)h)h
2
∂λ
0
0
This can be expressed as D2 L(Γ) = Ja , which means D2 L(Γ)[h]2 =
Rl
− 0 Ja [h]h. Ja [h] is called the Jacobi operator of the geodesic Γ. Assumption: Ja is invertible.
4
MANUEL DEL PINO AND JUNCHENG WEI
We assume that if h(s̃), s̃ ∈ (0, l) is such that h(0) = h(l), ḣ(0) = ḣ(l)
and Ja [h] = 0 then h ≡ 0. Ker(Ja ) = {0}, in the space of l−periodic
C 2 functions. This implies (exercise) that the problem
Ja [h] = g, g ∈ C(0, l), g(0) = g(l), h(0) = h(l), ḣ(0) = ḣ(l)
has a unique solution φ. Moreover kφkC 2,α (0,l) ≤ CkgkC α (0,l) .
Remember that the equation in coordinates (s, z) is
µ
¶
1
1
εk(εs)
E(v) =
vs + vzz −
vz +
(1 − εk(εs)z) (1 − εk(εs)z)
(1 − εk(εs)z)
s
as̃
1
az̃
vs + ε vz + f (v) = 0
2
a (1 − εk(εs)z)
a
Change of variables: Fix a function h ∈ C 2,α (0, l) with khk ≤ 1 and do
the change of variables z − h(εs) = t and take as first approximation
v0 ≡ w(t). Let us see that v0 (s, z) = w(z − h(εs)) so
1
1
(
w0 (−ḣ(εs, εz))s + w00 + f (w)
E(v0 ) =
1 − εkz 1 − εkz
ε
az̃
k(εs)
ε
as̃ 0
(εs, εz) −
)w0 − εḣ
w
a
1 − k(εs)εz
(1 − εkz)2 a
Error in terms of coordinates (s, t) z = t + h(εs):
¸
·
az̃
k(εs)
ε2 w 0
0
(εs, ε(t + h)) −
−
h00
E(v0 )(s, t) = εw (t)
2
a
1 − k(εs)(t + h)ε
(1 − kε(t + h))
+ε(
1
1
ε
as̃ 0
00 2 2
2
0
w
ḣ
ε
−
ε
k̇(t+h)
ḣw
(t)−ε
ḣ
w
(1 − kε(t + h))2
(1 − εk(t + h))3
(1 − εkz)2 a
In fact
|E(v0 )(t, s)| ≤ Cε2 e−σ|t|
σ < 1, and
keσ|t| E(v0 )kC 0,α (|t|< δ ) ≤ Cε2
ε
R δ/ε
Formal computation: We would like −δ/ε E(v0 )(s, y)w0 (t)dt ≈ 0. Observe that
Z
Z
w02
2 00
2 00
w02 dt + O(ε3 )
= −ε h
−ε h (εs)
(1
−
kε(t
+
h))
R
|t|<δ/ε
+
Also
Z
1
w00 w0 dt = 0 + O(ε3 ).
1 − εk(t + h)
Z
Z
as
2
02
2
2 as̃
ε ḣ
(εs, ε(t + h))w /(1 + kε(t + h)) = ε ḣ (εs, 0) w02 + O(ε3 )
a
a
2 2
ḣ ε
LYAPUNOV SCHMIDT REDUCTION METHOD
5
and finally
Z
Z
az̃
k(εs)
02 az̃
2
ε
w ( (εs, ε(t+h))−
w0 (t)2 (ε2 )(( )(εs, 0)−k 2 )h(εs)+O(ε
)=ε
a
1 − k(εs)(t + h)ε
a
|t|<δ/ε
R
Then
R
³a ´
− Ew0 dt
z̃
00
0 as̃
R
=
h
+
h
−
(
(εs, 0) − k 2 )h + O(ε)
a
a z̃
ε2 w02
we call s̃ = εs, and we conclude that the right hand side of the above
equality is equal to
1
((a(s̃, 0))h0 (s̃)0 + (2k 2 a(s̃, 0) − az̃z̃ (s̃, 0))h) + O(ε)
a(s̃, 0)
and this is equal to
1
(Ja [h] + O(ε))
a(s̃, 0)
We need the equation for v(s, z) = ṽ(s, z − h(εs)). We have
∂v
∂ṽ ∂ṽ
=
−
ḣε
∂s
∂s
∂t
We write z = t + h, so we have
S(ṽ) =
1
∂
∂
1
∂
∂
( − εḣ )[
( − εḣ )]ṽ + ṽtt
(1 − εkz) ∂s
∂t 1 − εk(t + h) ∂s
∂t
k
az̃
as̃
1 2
+ ]ṽt + ε
[ṽs − εḣṽt ] + f (ṽ) = 0
1 − εkz
a
a 1 − kεz
The first term of this equation is equal to
ε[−
1
ε(εk̇(t + h) + εk ḣ)
1
1
{
(ṽs −εḣvt )+
(−ε2 h00 vt −2εḣṽt s)+
ṽs s}
2
1 − εkz (1 − εk(t + h))
1 − kε(t + h)
1 + εk(t + h)
εk
1
(ṽs − εḣṽt ) +
(−εḣṽtt )} + f (ṽ) = 0
2
(1 − εk(t + h))
1 − εk(t + h)
Let us observe that for |t| < δ/ε, δ ¿ 1
−εḣ{
S[ṽ](s, t) = ṽss +ṽtt +O(ε)∂ts ṽ+O(ε̃)∂tt ṽ+O(εk(|t|+1))∂ss ṽ+O(ε)∂t ṽ+O(ε)∂s ṽ+f (v) = 0
We will call the operator that appears in the equation B[ṽ]. We look
for a solution of the form ṽ(s, t) = w(t) + φ(s, t). The equation for φ is
φss + φtt + f 0 (w(t))φ + E + B(φ) + N (φ) = 0,
|t| < δ/ε
where E = S(w(t)) = O(ε2 e−σt ), N (φ) = f (w + φ) − f (w) − f 0 (w)φ,
s ∈ (0, l/ε). We use the notation L(φ) = φss + φtt + f 0 (w(t))φ. We also
need the boundary condition φ(0, t) = φ(l/ε, t) and φs (0, t) = φs (l/ε, t).
6
MANUEL DEL PINO AND JUNCHENG WEI
It is natural to study the linear operator in R2 and the linear projected problem
φss + φtt + f 0 (w(t))φ + g(t, s) = c(s)w0 (t)
R
where c(s) =
0
RRg(t,s)w (t)dt
0 (t)2 dt
w
R
Z
∞
and under the orthogonally condition
φ(s, t)w0 (t)dt = 0,
∀s ∈ R
−∞
Basic ingredient: (Even more general) Consider the problem in Rm ×
R, with variables (y, t):
∆y φ + φtt + f 0 (w(t))φ = 0,
φ ∈ L∞ (Rm × R)
If φ is a solution of the above Rproblem, then φ(y, t) = αwR0 (t) some
α ∈ R. Ingredient: ∃γR > 0 :
p0 (t)2 − fR0 (w(t))p(t)2 ≥ γ R p2 (t)dt
R
1
0
for all p ∈ H with R pw = 0. ψ(y) = R φ2 (y, t)dt. This is well
defined
out that |φ(y, t)| ≤ Ce−σt ,
√ (as we will see) Indeed: It turns
σ < 2, thanks to the fact that φ ∈ L∞ . We use x = (y, t) and we
obtain
∆x φ − (2 − 3(1 − w(t)2 ))φ = 0
√
Observe that 1 − w(t)2 is small if |t| À 1. Fix 0 < σ < 2, for |t| > R0
we have 2 − 3(1 − w2 (t)) > σ 2 . Let
φ̄ρ (y, t) = ρ
n
X
cosh(σyi ) + ρ cosh(σt) + kφk∞ eσR0 e−σ|t| .
i=1
We have that
φ(y, t) ≤ φ̄ρ (y, t), for |t| = R0
also true that for |t| + |y| > Rρ À 1, φ(y, t) ≤ φ̄ρ .
−∆x φ + (2 − 3(1 − w(t)2 ))φ̄ = (2 − σ 2 − 3(1 − w(t)2 )φ̄ρ ) > 0
for |t| > R0 . So is a supersolution of the operator
−∆x φ + (2 − 3(1 − w(t)2 ))φ
in Dρ , which implies that φ ≤ φ̄ρ for |t| > R0 . This implies that
|φ(x)| ≤ C φ̄ρ for all x, and we conclude the assertion taking ρ → 0. If
φ solves −∆φ+(1−3w2 )φ = 0, then kφkC 2,α (B1 (x0 )) ≤ CkφkL∞ (B2 (x0 )) .
This implies that also
|φy | + |φyy | ≤ Ce−σt .
˜ t) = φ(y, t) −
Let φ(y,
R
0
φ(y,τ
R )w (τ )dτ w 0 .
w02
R
We call β(y) =
0
φ(y,τ
R )w (τ )dτ
w02
∆φ̃ + f 0 (w)φ̃ = ∆φ + f 0 (w)φ + (∆y β)w0 + β(∆w0 + f 0 (w))w0 = 0
LYAPUNOV SCHMIDT REDUCTION METHOD
7
R
because ∆y β = 0 by integration by parts. Let ψ(y) = R φ̃2 dt.
Z
Z
Z
Z
Z
2
2
∆y ψ =
∇y (2φ̃∇y φ̃)dt = 2 |∇y φ̃| dt+2 φ̃∆y φ̃ = 2 |∇y φ̃| −2 φ̃[φ̃tt +f 0 (w)φ̃]dt
R
R
R
Using 2 |∇y φ̃|2 dt + 2 (φ̃2t − f 0 (w)φ̃2 ) This implies that ∆ψ ≥ 2γψ
which implies −∆ψ + 2γψ ≤ 0, 0 ≤ ψ ≤ c.
We obtain
that ψ ≡ 0 and this implies φ̃ = 0. This implies that
R
φ(t) = ( φw0 )w0 = β(y)w0 and ∆β = 0, β ∈ L∞ . Liouville implies
that β = constant so φ = constantw0 .
Lemma: L∞ a priori estimates for the linear projected problem:
∃C : kφk∞ ≤ Ckgk∞ .
Proof: If not exists kgn k∞ → 0 and kφn k∞ = 1.
L[φn ] = −gn + cn (t)w0 (t) = hn (t)
and hn → 0 in L∞ . kφn k = 1 which implies that ∃(yn , tn ): |φ(yn , tn )| ≥
γ > 0. Assume that |tn | ≤ C and define φ̃(y, t) = φn (yn + y, t). Then
∆φ̃n + f 0 (w(t))φ̃n = h̃n
but f 0 (w(t))φ̃n is uniformly bounded and the right hand side goes to 0.
This implies that kφkC 1 (Rm+| ) ≤ C This implies that φ̃n → φ̃ passing
to subsequence, and the convergence is uniformly on compacts, where
∆φ̃ + f 0 (w)φ̃ = 0, φ̃ ∈ L∞ . We conclude after a classic argument
that
√
σ|t|
σ|t|
φ̃ = 0. We have also that ke φk∞ ≤ Cke gk∞ , 0 < σ < 2. Elliptic
regularity implies that keσ|t| φkC 2,σ ≤ keσ|t| gkC 0,σ .
Existence: Assume g has compact
forR support and 0take the weak
R
mulation: Find φ ∈ H such that Rm+1 ∇φ∇ψ
−
f
(w)φψ
=
gy,
for
R
1
m+1
0
m
all ψ ∈ H, where H = {f R∈ H (R
)| R ψw dt =R 0, ∀y ∈ R }.
2
0
Let us see that a(ψ, ψ) = |∇ψ| − f (w)ψ 2 ≥ γ ψ 2 + ψ 2 . So
a(ψ, ψ) ≥ Ckψk2H 1 (Rm+1 ) This implies the unique existence solution.
Observe that
Z
(∆φ + f 0 (w)φ + g)ψ = 0
R
ψ̃w0 dt
for all ψ ∈ H. Let ψ ∈ H 1 and ψ = ψ̃ − R w02 w0 = Π(ψ̃). We have
that
Z
Z
Z
dy gΠ(ψ̃)dt = Π(g)ψ
which implies that
+ f 0 (w)φ + g) = 0 if and only if ∆φ +
R Π(∆φ
0
(∆φ+f
R (w)+g) w 0 Regularity implies that φ ∈ L∞ and
f 0 (w) + φ + g =
w02
kφk∞ ≤ Ckgk∞ . Approximating g ∈ L∞ by gR ∈ Cc∞ (RN ) locally over
compacts. This implies existence result.
8
MANUEL DEL PINO AND JUNCHENG WEI
We can bound φ in other norms. For example if 0 < σ <
√
2, then
keσ|t| φk∞ ≤ Ckeσ|t| gk∞ .
Indeed, f 0 (w) < −σ 2 − η if |t| > R, with η = (2 − σ 2 )/2. We set
n
X
−σ|t|
φ̄ = M e
+ρ
cosh(σyi ) + ρ cosh(σt).
i=1
Therefore
−∆φ̄ + (−f 0 (w))φ̄ ≥ −δ φ̄ + (σ 2 + η)φ̄ = η φ̄ > g̃ = −g + c(y)w0 (t)
if M ≥ Aη keσ|t| gk∞ . In addition we have φ̄ ≥ φ on |t| = R if M ≥
kφk∞ eσR . By an standard argument based on maximum principle, we
conclude that φ ≤ φ̄. This means, letting ρ → 0, φ ≤ M e−σ|t| , where
M ≥ C max{kφk∞ , kgeσ|t| k∞ }. Since kφk∞ ≤ Ckgk∞ ≤ Ckgeσ|t| k∞ ,
we can take M = Ckgeσ|t| k∞ . Finally, we conclude kφeσ|t| k∞ ≤ kgeσ|t| k∞ .
Reminder: If ∆φ = p implies that
k∇φkL∞ (B1 (0)) ≤ C[kφkL∞ B2 (0) + kpkL∞ (B1 (0)) ].
Remember that
kpkC 0,α (A) = kpk∞ + [φ]0,α,A
1 )−p(x2 )|
where [φ]0,α,A = supx1 ,x2 ∈A,x1 6=x2 |p(x
. Also we have the following
|x1 −x2 |α
interior Schauder estimate: for 0 < α < 1
kφkC 2,σ (B1 ) ≤ C[kφkL∞ (B2 (0)) + kpkC 0,α (B2 (0)) ].
Conclusion: If φ solves the equation in Rn+1 then
kφkC 2,α (Rn+1 ) ≤ CkgkC 0,α (Rn+1 ) .
Sketch of the proof of this fact: Fix x0 ∈ Rn+1 , then
C[φ]0,α,B1 (x0 ) ≤ k∇φkL∞ (B1 (x0 )) ≤ C[kφk∞ + kgk∞ ] ≤ Ckgk∞
This implies that kφkC 0,α (B1 (x0 )) ≤ Ckgk∞ , which implies kφkC 0,α (Rn ) ≤
Ckgk∞ . Clearly kpkC 0,α (B2 (x0 )) ≤ Ckgk∞ , so kφkC 0,α (B1 (x0 )) ≤ CkgkC 0,α (Rn+1 ) ,
from where we deduce the estimate.
We also get
keσ|t| φkC 2,α (Rn+1 ) ≤ Ckeσ|t| gkC 0,α (Rn+1 ) .
The proof of this fact is very similar to the previous one (use that
σ|t|
g ≤ e−σ|t0 |kge k , for |t0 | À 1).
Another result is the following
k(1 + |y|2 )µ/2 φk∞ ≤ Ck(1 + |y|2 )µ/2 gk∞
In order to prove this result we define ρ(y) = (1 + |y|µ ) and we consider
φ̃ = ρ(δy)φ. Observe that
LYAPUNOV SCHMIDT REDUCTION METHOD
9
∆φ = ρ−1 ∆φ̃ − 2δ∇φ̃∇(ρ−1 (δy)) + φ̃δ 2 ∆(ρ−1 )(δy) = f 0 (w)φ + g − cw0
We get L[φ̃] + O(δ 2 )φ̃ + O(δ)∇φ̃ = ρ(g − cw0 ). We get
k∇φ̃k∞ + kφ̃k∞ ≤ C[δ 2 kφ̃k∞ + δk∇φ̃k∞ + kρgk∞ ].
If δ is small we conclude that
kφ̃k∞ + k∇φ̃k∞ ≤ Ckρgk∞
and we obtain
kρφkC ≤ kρgk.
Our setting:
∇a
· ∇u] + f (u) = 0
a
We want a solution to (1.3) uε (x) ≈ W (z/ε). Writing x = y + zγ(y),
|z| < δ, we have
(1.3)
ε2 [δu +
∆v + ∇a(εx)/a · ∇v + f (v) = 0,
in Γε = 1ε Γ: x = y + zν(εy), which means x = 1ε γ(εs) + zν(εs).
Remember that |γ̇(s̃)| = 1 which implies ν̇(s̃) = −k(s̃)γ̇(s̃). We also set
z = h(εs) + t. x = 1ε γ(εs) + (t + h(εs))ν(εs). We assume khkα,(0,l) ≤ 1,
for 0 < α < 1. We wrote ∆x in terms of this coordinates (t, s) and the
equations S(v) = 0 is rewritten taking as first approximation w(t). We
evaluated S(w(t)) and got that S(w(t)) = 0.
From the expression of ∆x we get (x = 1ε γ(εs) + (t + h(εs))ν(εs))
∆x v = ∂ss + ∂tt + ε[bε1 (t, s)∂ss + bε2 ∂tt + bε3 ∂st + bε4 ∂t + bε5 ∂s ]
|εbi | ≤ Cδ in the region |t| < δ/ε. The coefficients are periodic (same
values at s = 0 and s = l/ε). Our equation reads
∂ss v + ∂tt v + Bε [v] + f (v) = 0,
for s ∈ (0, l/ε), |t| < δ/ε.
This expression does not make sense globally. We consider δ ¿ 1. We
define
½
−1 in Ωε−
H(x) =
+1 in Ωε+
where Ωε+ is a bounded component of R2 \ Γ, and Ωε− the other. For
the equation
∇a
∆v + ε
· ∇v + f (v) = 0
a
we take as first (global) approximation
v0 (x) = w(t)η3 + (1 − η4 )H(x)
10
MANUEL DEL PINO AND JUNCHENG WEI
where
(
ηl (x) =
³
η
ε|t|
lδ
0
´
if |t| < 2δl/ε
otherwise
Look for a solution of the form v = v0 + φ̃, so
∇a
∆x φ̃ + ε
· ∇φ̃ + f 0 (vo )φ̃ + E + N (φ̃) = 0
a
where E = S(v0 ) and N (φ̃) = f (v0 + φ̃) − f (v0 ) − f 0 (v0 )φ̃.
We write φ̃ = η3 φ + ψ. We require that φ and ψ solve the system
∇a
∇
∆x ψ−2ψ+(2+f 0 (v0 ))(1−η1 )ψ+ε
∇ψ+(1−η1 )E+(1−η1 )N (η3 φ+ψ)+∇η3 ∇φ+∇η3 ∇φ+ε
a
a
·
¸
∇a
η3 ∆x φ + f 0 (w(t))φ + η1 (2 + f 0 (w(t)))ψ + η1 E + η1 N (φ + ψ) + ε
· ∇φ = 0.
a
We need that the φ above satisfies the equation just for |t| < 6δ/ε. We
assume that φ(s, t) is defined for all s and t (and it is l/ε- periodic in
s). We require that φ satisfies globally
φtt + φss + η6 Bε [φ] + f 0 (w(t))φ + η1 E + η1 N (φ + ψ) + η1 (2 + f 0 (w))ψ = 0
and φ ∈ L∞ (Rn + 1) and periodic in s. Notice that φtt +φss +η6 Bε [φ] =
∆x φ inside the support of η3 . Rather than solving this problem directly
we solve the projected problem
(1.4)
φtt +φss +η6 Bε [φ]+f 0 (w(t))φ+η1 E+η1 N (φ+ψ)+η1 (2+f 0 (w))ψ = c(s)w0 (t)
R
and R φw0 (t)dt = 0. We solve (1)-(1.4) first, then we find h such
that c(s) ≡ 0. We consider φ with kφk∞ + k∇φk∞ ≤ ε. The operator
−∆ψ +2ψ is invertible L∞ (R3 ) → C 1 (R2 ). We conclude that if g ∈ L∞
the exist a unique solution ψ = T [g] ∈ C 1 (R2 ) with kφkC 1 ≤ Ckgk∞ of
equation −∆ψ + 2ψ = g in R2 . Observe that (1) is equivalent to
∇a
∇a
ψ = T [(2+f 0 (v0 ))(1−η1 )ψ+ε
∇ψ+(1−η1 )E+(1−η1 )N (η3 φ+ψ)+∇η3 ∇φ+∇η3 ∇φ+ε
∇
a
a
Using contraction mapping in C 1 on kψkC 1 ≤ Cε, we conclude that
there exist a unique solution of the this problem ψ = ψ(φ, h) such that
kψk ≤ C[ε2 + εkφkC 1 ].
Even more, kψ(φ1 , h) − ψ(φ2 , h)kC 1 ≤ Cεkφ1 − φ2 kC 1 .