INTRODUCTION TO LYAPUNOV SMICHDT
REDUCTION METHODS FOR SOLVING PDE’S
MANUEL DEL PINO AND JUNCHENG WEI
1. Allen Cahn Equation
Energy: Phase transition model.
Let Ω ⊆ RN of a “binary mixture”: Two materials coexisting (or one
material in two phases). We can take as an example of this: Water in
solid phase (+1), and water in liquid phase (−1). The configuration of
this mixture in Ω can be described as a function
½
+1 in Λ
∗
u (x) =
−1 in Ω \ Λ
where Λ is some open subset of Ω. We will say that u∗ is the phase
function.
Consider the functional
Z
1
(1 − u2 )2
4 Ω
minimizes if u = 1 or u = −1. Function u∗ minimize this energy
functional. More generally this well happen for
Z
W (u)dx
Ω
where W (u) minimizes at 1 and −1, i.e. W (+1) = W (−1) = 0,
W (x) > 0 if x 6= 1 or x 6= −1,W 00 (+1), W 00 (−1) > 0.
1.1. The gradient theory of phase transitions. Possible configurations will try to make the boundary ∂Λ as nice as possible: smooth
and with small perimeter. In this model the step phase function u∗ is
replaced by a smooth function uε , where ε > 0 is a small parameter,
and
½
+1 inside Λ
uε (x) ≈
−1 inside Ω \ Λ
and uε has a sharp transition between these values across a “wall” of
width roughly O(ε): the interface (thin wall).
1
2
MANUEL DEL PINO AND JUNCHENG WEI
In grad theory of phase transitions we want minimizers, or more
generally, critical points uε of the functional
Z
Z
|∇u|2 1
(1 − u2 )2
Jε (u) = ε
+
2
ε Ω
4
Ω
Let us observe that the region where (1 − u2ε ) > γ > 0 has area of order
O(ε) and the size of the gradient of uε in the same region is O(ε2 ) in
such a way J(uε ) = O(1). We will find critical points uε to functionals
of this type so that J(uε ) = O(1).
Let us consider more generally the case in which the container isn’t
homogeneous so that distinct costs are paid for parts of the interface
in different locations
¶
Z µ
|∇u|2 1 (1 − u2 )2
Jε (u) =
ε
a(x)dx
+
2
ε
4
Ω
a(x) non-constant, 0 < γ ≤ a(x) ≤ β and smooth.
1.2. Critical points of Jε . First variation of Jε at uε is equal to zero.
¯
¯
∂
Jε (uε + tϕ)¯¯ = DJε (uε )[ϕ] = 0, ∀ϕ ∈ Cc∞ (Ω)
∂t
t=0
We have
Jε (uε + tϕ) =
i.e. ∀ϕ ∈
Cc∞ (Ω)
Z
1
0 = DJε (uε )[ϕ] = ε (∇uε ∇ϕ)a +
ε
Ω
2
Z
W 0 (uε )φa.
Ω
If uε ∈ C (Ω)
Z ³
´
a
−ε∇ · (a∇uε ) + W 0 (uε ) ϕ = 0, ∀ϕ ∈ Cc∞ (Ω)
ε
Ω
This give us the weighted Allen Cahn equation in Ω
a
−ε∇ · (a∇u) + u(1 − u2 ) = 0 in Ω.
ε
We will assume in the next lectures Ω = RN , where N = 1 or N = 2.
If N = 1 weight Allen Cahn equation is
a0
(1.1)
ε2 u00 + ε2 u0 + (1 − u2 )u = 0, in (−∞, ∞).
a
Look for uε that connects the phases −1 and +1 from −∞ to ∞.
Multiplying (1.1) against u0 and integrating by parts we obtain
µ 02
¶ Z ∞ 0
Z ∞
(1 − u2 )2
d
u
a 02
−
u =0
(1.2)
ε
+
2
4
−∞ dx
−∞ a
LYAPUNOV SCHMIDT REDUCTION METHOD
3
Assume that u(−∞) = −1, u(∞) = 1, u0 (−∞) = u0 (∞) = 0, a > 0,
then (1.2) implies that
Z ∞ 0
(1 − u2 )2
a 02
+
u =0
4
−∞ a
from which we conclude that unless a is constant, we need a0 to change
sign. So: if a is monotone and a0 6= 0 implies the non-existence of
solutions as we look for. We need the existence (if a0 6= 0) of local
maximum or local minimum of a. We will prove that under some
general assumptions on a(x), given a local max. or local min. x0 of
a non-degenerate (a00 (x0 ) 6= 0), then a solution to (1.1) exists, with
transition layer.
We consider first the problem with a ≡ 1, ε = 1:
(1.3)
W 00 + (1 − W 2 )W = 0,
The solution of this problem is
W (−∞) = −1, W (∞) = 1.
µ
¶
t
W (t) = tanh √
2
This solution is called “the heteroclinic solution”, and it’s the unique
solution of the problem (1.3)up to translations.
Observation 1.1. This solution exists also for the problem
(1.4)
w00 + f (w) = 0,
w(−∞) = −1, w(∞) = 1
02
where f (w) = −W 0 (w). Solutions satisfies w2 − W (w) = E, where E
is constant, and w(−∞) = −1 and w(∞) = 1 if and only if E = 0.
This implies
Z w
ds
p
=t
2w(s)
0
t(w) → ∞ if w → 1, and t(w) → −∞ if w → −1, so the previous
relation defines a solution w such that w(0) = 0, and w(−∞) = −1,
w(∞) = 1.
If we wright the Hamiltonian system associated to the problem we
have:
p0 = −f (q), q 0 = p.
Trajectories lives on level curves of H(p, q) =
(1−q 2 )2
.
4
p2
2
−W (q), where W (q) =
Let x0 ∈ R (we will make assumptions on this point). Fix a number
h ∈ R and set
v(t) = u(x0 + ε(t + h)),
v 0 (t) = εu0 (x0 + ε(t + h))
4
MANUEL DEL PINO AND JUNCHENG WEI
Using (1.1), we have
a0
ε2 u00 (x0 + ε(t + h)) = −ε2 u0 (x0 + ε(t + h)) − (1 − v 2 (t))v(t)
a
so we have the problem
(1.5)
a0
v 00 (t)+ε (x0 +ε(t+h))v 0 (t)+(1−v(t)2 )v(t)2 = 0, w(−∞) = −1, w(∞) = 1.
a
Let us observe that if ε = 0 the previous problem becomes formally in
(1.3), so is natural to look for a solution v(t) = W (t) + φ, with φ a
small error in ε.
Assumptions:
(1) There exists β, γ > 0 such that γ ≤ a(x) ≤ β, ∀x ∈ R
(2) ka0 kL∞ (R) , ka00 kL∞ (R) < +∞
(3) x0 is such that a0 (x0 ) = 0, a00 (x0 ) 6= 0, i.e. x0 is a non-degenerate
critical point of a.
Theorem 1.1. ∀ε > 0 sufficiently small, there exists a solution v = vε
to (1.5) for some h = hε , where |hε | ≤ Cε and vε (t) = w(t) + φε (t) and
kφε k ≤ Cε
Proof. We write in (1.5) v(t) = w(t) + φ(t). From now on we write
f (v) = v(1 − v 2 ). We get
a0
a0
w00 +φ00 +ε (x0 +ε(t+h))φ0 +ε (x0 +ε(t+h))w0 +f (w+φ)−f (w)−f 0 (w)φ+f (w)+f 0 (w)φ = 0
a
a
φ(−∞) = φ(∞) = 0.
It can be written in the following way
(1.6) φ00 + f 0 (w(t))φ + E + B(φ) + N (φ) = 0,
φ(−∞) = φ(∞) = 0
where
a0
B(φ) =ε (x0 + ε(t + h))φ0 ,
a
N (φ) =f (w + φ) − f (w) − f 0 (w) = −3wφ2 − φ3 ,
a0
E =ε (x0 + ε(t + h))w0 .
a
We consider the problem
(1.7)
φ00 + f 0 (w(t))φ + g(t) = 0,
φ ∈ L∞ (R),
and we want to know when (1.7) is solvable. We will assume g ∈
L∞ (R). Multiplying (1.7) against w0 we get
Z ∞
Z ∞
000
0
0
(w + f (w)w )φ +
gw0 = 0
−∞
−∞
LYAPUNOV SCHMIDT REDUCTION METHOD
5
the first integral is zero because (1.4). We conclude that a necessary
condition is
Z ∞
gw0 = 0.
−∞
This condition is actually sufficient for solvability. In fact, we write
φ = w0 Ψ, we have
φ00 + f 0 (w)φ = g ⇔ w0 Ψ + 2w00 Ψ0 = −g
Multiplying this last expression by w0 (integration factor), we get
Z ∞
02 0 0
0
0
0
(w Ψ ) = gw ⇒ w 2Ψ (t) = −
g(s)w0 (s)ds
−∞
Let us choose
Z
t
Ψ(t) = −
0
Then the function
dτ
02
w (τ )
Z
t
0
φ(t) = −w (t)
0
Recall that
Claim: if
R∞
−∞
Z
τ
g(s)w0 (s)ds
−∞
dτ
02
w (τ )
Z
τ
g(s)w0 (s)ds
−∞
√ √
w0 (t) ≈ 2 2e− 2|t|
gw0 = 0 then we have
kφk∞ ≤ Ckgk∞ .
In fact, if t > 0
Z
0
|φ(t)| ≤ |w (t)|
t
C
√
¯Z
¯
¯
¯
∞
¯
Z t √
√
¯
0
− 2t
¯
e 2τ dτ ≤ Ckgk∞ .
gw ds¯ dτ ≤ Ckgk∞ e
e−2 2τ τ
For t < 0 a similar estimate yields, so we conclude
0
0
|φ(t)| ≤ Ckgk∞ .
¤
The solution of (1.7) is not unique because if φ1 is a solution implies
that φ2 = φ1 + Cw0 (t) is also a solution. The solution we found is
actually the only one with φ(0) = 0. For g ∈ L∞ arbitrary we consider
the problem
φ00 + f 0 (w)φ + (g − cw0 ) = 0, in <,
R∞
gw0
−∞
R
where C = C(g) = ∞ 02 .
w
−∞
(1.8)
φ ∈ L∞ (R)
6
MANUEL DEL PINO AND JUNCHENG WEI
Lemma 1.1. ∀g ∈ L∞ (R) (1.8) has a solution which defines a operator
φ = T [g] with
kT [g]k∞ ≤ Ckgk∞ .
In fact if T̂ [ĝ] is the solution find in the previous step then φ = T̂ [g −
C(g)w0 ] solves (1.8) and
(1.9)
kφk∞ ≤ Ckgk∞ + |C(g)|C ≤ Ckgk∞
Proof. Back to the original problem: We solve first the projected problem
φ00 + f 0 (w)φ + E + B(φ) + N (φ) = Cw0 ,
where
φ ∈ L∞ (R)
R
(E + B(φ) + N (φ))w0
R
.
w02
R
We solve first (1.9) and then we find h = hε such that in (1.9) C=0
in such a way we find a solution to the original problem. We assume
|h| ≤ 1. It’s sufficient to solve
C=
R
φ = T [E + B(φ) + N (φ)] := M [φ].
We have the following remark
|E| ≤ Cε2 ,
kB(φ)k∞ ≤ Cεkφ0 k∞ ,
kN (φ)k ≤ C(kφ2 k∞ + kφ3 k∞ )
where C is uniform on |h| ≤ 1. We have
d
M k∞ ≤ C(kEk∞ +kB(φ)k∞ +kN (φ)k∞ ≤ C(ε2 +εkφ0 k∞ +kφ2 k∞ +kφ3 k∞ )
dt
then if kφk∞ + kφ0 k∞ ≤ M ε2 we have
kM k∞ +k
d
M k∞ ≤ C ∗ ε2 .
dt
We define the space X = {φ ∈ C 1 (R) : kφk∞ + kφ0 k∞ ≤ C ∗ ε2 }. Let us
observe that M (X) ⊂ X. In addition
kM k∞ + k
d
(M (φ1 )−M (φ2 ))k∞ ≤ Cε(kφ1 −φ2 k∞ +kφ01 −φ02 k∞ ).
dt
So if ε is small M is a contraction mapping which implies that there
exists a unique φ ∈ X such that φ = M [φ].
¤
kM (φ1 )−M (φ2 )k∞ +k
In summary: We found for each |h| ≤ 1
φ = Φ(h), solution of1.7
. We recall that
h → Φ(h)
LYAPUNOV SCHMIDT REDUCTION METHOD
7
is continuous (in kkC 1 ) . Notice that from where we deduce that M is
continuous in h.
The problem is reduced to finding h such that C = 0 in (1.7) for
φΦ(h) =. Let us observe that
Z
C = 0 ⇔ αε (h) := (Eh + B[Φ(h)]) + N [Φ(h)])w0 = 0.
R
Let us observe that if we call ψ(x) =
a0
(x),
a
Z
1
0
ψ(x0 +ε(t+h)) = ψ(x0 )+ψ (x0 )ε(t+h)+
then
(1−s)ψ 00 (x0 +sε(t+h))ε2 (t+h)2 ds
0
000
∞
We add the assumption a ∈ L (R) in order to have ψ 00 ∈ L∞ (R). We
deduce that
Z
Z
Z Z 1
0
2 0
0
2
3
( (1−s)ψ 00 (x0 +sε(t+h))ds)(t+h)2 w0 (t)dt
Eh w = ε ψ (x0 ) (t+h)w (t) +ε
R
R
0
We recall that: R tw0 (t)2 and
Z
| (B[φ(h)] + N [φ(h)])w0 | ≤ C(εkΦ(h)kC 1 + kΦ(h)kL∞ ) ≤ Cε3 .
R
So, we conclude that
αε (h) = ψ 0 (x0 )ε2 (h + O(ε))
and the term inside the parenthesis change sign. This implies that
∃hε : |hε | ≤ M ε such that αε (h) = 0, so C = 0.
Observe that
√
1
L(φ) = φ00 −2φ+εψ+3(1−w2 )φ+ f 00 (w+sφ)φφ+O(ε2 )e− 2|t| = 0, |t| > R
2
We consider t > R. Notice that 12 f 00 (w + sφ)φ = O(ε2 ). Then using
φ̂ = εe−|t| + δe|t| . Then using maximum principle and after taking
δ → 0, we obtain φ ≤ εe−|t| .
A property: We call
L(φ) = φ00 + f 0 (w)φ,
φ ∈ H 2 (R).
We consider the bilinear form associated
Z
Z
B(φ, φ) = − L(φ)φ =
φ02 − f 0 (w)2 φ2 ,
R
φ ∈ H 1 (R).
R
1
Claim: B(φ, φ) ≥ 0, ∀φ ∈ H (R) and B(φ, φ) = 0 ⇔ φ = cw0 (t).
In fact: J 00 (w)[φ, φ] = B(φ, φ). We give now the proof of the claim:
8
MANUEL DEL PINO AND JUNCHENG WEI
Take φ ∈ Cc∞ (R). Write φ = w0 Ψ =⇒ Ψ ∈ Cc∞ (R). Observe that
L[w0 Ψ] = w10 (w02 Ψ0 )0 and
Z
Z
1 02 0 0 0
w02 Ψ02 , ∀φ ∈ Cc∞ (R)
B(φ, φ) = −
(w Ψ ) w Ψ =
0
w
R
R
1
|φ0 |2 −
Same is valid
for
all
φ
∈
H
(R),
by
density.
So
B(φ,
φ)
=
R
R 0
0
2
0 2
0
f (w)φ = R w 2|Ψ | ≥ 0 and B(φ, φ) = 0 ⇔ Ψ = 0 which implies
φ = cw0 .
Corollary 1.1. Important
for later porpuses There exists r > 0 such
R
that if φ ∈ H 1 (R) and R φw0 = 0 then
Z
B(φ, φ) ≥ γ φ2
R
R 2
1
φ .
Proof. If not there exists φn H 1 (R) such that
0
≤
B(φ
,
φ
)
<
n
n
n R n
R
We may assume without loss of generality φ2n = 1 which implies that
up to subsequence
φn * φ ∈ H 1 (R)
R
and φn → φ uniformly and in L2 sense on bounded intervals. This
implies that
Z
Z
0
0 = lim
φn w =
φw0
n→∞
R
R
On the other hand
Z
Z
Z
0 2
2
|φn | + 2 φn − 3 (1 − w2 )φ2n → 0
R
R
R
R
R
R
and also |φ0n |2 +2 Rφ2n −3 (1−w2 )φ2n → |φ0 |2 +2 φ2 −3 (1−w2 )φ2 ,
so B(φ, φ) = 0, and w0 φ = 0 so φ = 0. But also
Z
2 ≤ 3 (1 − w2 )φ2n + o(1)
R
which implies that 2 ≤ 3 (1 − w2 )φ2 and this means that φ 6= 0, so
we obtain a contradiction.
¤
Observation 1.2. If we choose δ = 2kfγ0 k∞ then
Z
φ02 − (1 + δ)f 0 (w)φ2 ≥ 0.
This implies in fact that
Z
B(φ, φ) ≥ α
φ02 .
LYAPUNOV SCHMIDT REDUCTION METHOD
9
2. Nonlinear Schrödinger eqution (NLS)
εiΨt = ε2 ∆Ψ − W (x)Ψ + |Ψ|p−1 Ψ.
R
A first fact is that RN |Ψ|2 = constant. We are interested into study
solutions of the form Ψ( x, t) = e−iEt u(x) (we will call this solutions
standing wave solution). Replacing this into the equation we obtain
εEu = ε2 ∆u − W u − |u|p−1 u
whose transforms into
ε2 ∆u − (W − λ)u + |u|p−1 u = 0,
u(x) → 0, as |x| → ∞
choosing E = λε . We define V (x) = (W (x) − λ)
2.1. The case of dimension 1.
(2.1)
ε2 u00 − V (x)u + up = 0, x ∈ R,
0
00
Assume: V ≥ γ > 0, V, V , V , V
point
(2.2)
w00 − w + wp = 0,
0 < u(x) → 0, as |x| → ∞, p > 1.
000
∈ L∞ , and V ∈ C 3 (R). Starting
w > 0,
w(±∞) = 0, p > 1
There exists a homoclinic solution
Cp
w(t) =
cosh
2 ,
¡ p−1 ¢ p−1
t
2
µ
Cp =
p+1
2
1
¶ p−1
Let us observe that w(t) ≈ 22/(p−1) Cp e−|t| as t → ∞ and also that
W (t + c) satisfies same equation.
¡
¢
0
Staid at x0 with V (x0 ) = 1 we want uε (x) ≈ w x−x
of the problem
ε
(2.1).
Observation 2.1. Given x0 we can assume V (x0 ) = 1. Indeed writing
2
u(x) = λ p−1 v(λx0 + (1 − λ)x0 )
we obtain the equation
ε2 v 00 (y) − V̂ (y)v + v p = 0
0
where y = λx0 + (1 − λ)x0 , and V̂ (y) = V ( y−(1−λ)x
). Then choosing
λ
p
λ = V (x0 , we obtain V̂ (x0 ) = 1.
Theorem 2.1. We assume V (x0 ) = 1, V 0 (x0 ) = 0, V 00 (x0 ) 6= 0. Then
there exists a solution to (2.1) with the form
µ
¶
x − x0
uε (x) ≈ w
.
ε
10
MANUEL DEL PINO AND JUNCHENG WEI
We define v(t) = u(x0 + ε(t + h)), with |h| ≤ 1. Then v solves the
problem
(2.3)
v 00 − V (x0 + ε(t + h)v + v p = 0,
v(±∞) = 0.
We define v(t) = w(t) + φ(t), so φ solves
(2.4)
φ00 − φ + pwp−1 φ − (V (x0 + ε(t + h)) − V (x0 ))φ + (w + φ)p − wp − pwp−1 φ
(2.5)
−(V (x0 + ε(t + h)) − V (x0 ))w(t) = 0
So we want a solution of
(2.6)
φ00 − φ + pwp−1 φ + E + N (φ) + B(φ) = 0,
φ(±) = 0.
Observe that
1
E = V 00 (x0 + ξε(t + h))ε2 (t + h)2 w(t),
2
2 2
so |E| ≤ Cε (t + 1)e−|t| ≤ Ce−σt for 0 < σ < 1.
We won’t have a solution unless V 0 doesn’t change sign and V 6= 0.
For instance consider V 0 (x) ≥ 0, and after Rmultiplying the equation by
2
u0 and integrating by parts, we see that R v 0 u2 = 0, which by ODE
implies that u ≡ 0, because u and u0 equals 0 on some point.
2.2. Linear projected problem.
L(φ) = φ00 − φ + pwp−1 φ + g = 0,
φ ∈ L∞ (R)
R
For solvability
we
have
the
necessary
condition
L(φ)w0 = 0. Assume
R
0
0
g such that R gw = 0. We define φ = w Ψ, but we have the problem
that w0 (0) = 0. We conclude that (w02 Ψ0 )0 + w0 g = 0 for t 6= 0. We
take for t < 0
Z −1
Z τ
dτ
0
g(s)w0 (s)ds
φ(t) = w (t)
0 (τ )2
w
t
−∞
and for t > 0
Z
t
0
φ(t) = w (t)
1
dτ
0
w (τ )2
Z
∞
g(s)w0 (s)ds
τ
In order to have a solution of the problem we need φ(0− ) = φ(0+ ).
R t dτ R τ
Rt
1
0
−
g(s)w
(s)ds
gw0
−
0 (τ )2
0 (t)2
1
w
w
−1
−∞
−∞
−
φ(0 ) = lim−
= lim−
=
R0
1
1
00
00
t→0
t→0
− w0 (t)2 w (t)
w (0) −∞ gw0
w0 (t)
and
φ(0+ ) = −
w00 (0)
1
R∞
0
gw0
LYAPUNOV SCHMIDT REDUCTION METHOD
11
and the condition is satisfies because of the assumption of orthogonality
condition.
We get kφk∞ ≤ Ckgk∞ . In fact we get also: ∀ 0 < σ < 1, ∃C > 0 :
kφeσt kL∞ + kφ0 eσt kL∞ ≤ Ckgeσt k
Observation: We use g = g − cw0 .(Correct this part!!!!)
2.3. Method for solving. In this section we consider a smooth radial
cut-off function η ∈ C ∞ (R), such that η(s) = 1 for s³< 1´ and η(s) = 0
if s > 2. For δ > 0 small fixed, we consider ηk,ε = η ε|t|
, k ≥ 1.
kδ
2.3.1. The gluing procedure. Write φ̃ = η2,ε φ + Ψ, then φ solves (2.5)
if and only if
£
¤
0
(2.7)
η2,ε φ00 + (pwp−1 − 1)φ + B(φ) + 2φ0 η2,ε
(2.8)
£
¤
+ Ψ00 + (pwp−1 − 1)Ψ + BΨ + E + N (η2,η φ + Ψ) = 0.
(φ, Ψ) solves (2.8) if is a solution of the system
(2.9)
φ00 − (1 − pwp−1 )φ + η1,ε E + η3,ε B(φ) + η1,ε pwp−1 Ψ + η1,η N (φ + Ψ) = 0
(2.10)
(2.11)
Ψ00 − (V (x0 + ε(t + h)) − pwp−1 (1 − η1,ε ))Ψ
0
00
+(1 − η1,ε )E + (1 − η1,ε )N (η2,ε φ + Ψ) + 2φ0 η2,ε
+ η2,ε
φ=0
We solve first (2.11). We look first the problem
Ψ00 − W (x)Ψ + g = 0
where 0 < α ≤ W (x) ≤ β, W continuous and g ∈ C(R) ∩ L∞ (R).
We claim that (2.3.1) has a unique solution φ ∈ L∞ (R). Assume first
that g has compact support and consider the well defined functional in
H 1 (R)
Z
Z
Z
1
1
0 2
2
|Ψ | +
wΨ − gΨ.
J(Ψ) =
2 R
2 R
R
Also, this functional is convex and coercive. This implies that J has a
minimizer, unique solution of (2.3.1) in H 1 (R) and it is bounded. Now
we consider the problem
µ ¶
|t|
00
ΨR − W ΨR + gη
=0
R
12
MANUEL DEL PINO AND JUNCHENG WEI
Let us see that ΨR has a uniform bound. Take ϕ(t) =
for ρ > 0 very small. Since ΨR ∈ L∞ (R) we have
ΨR ≤ ϕ(t),
kgk∞
+ρ cosh
α
³√
α
|t|
2
for |t| > tρ,R .
Let us observe that in [−tρ,R , tρ,R ]
00
ϕ − W ϕ + gη
µ
|t|
R
¶
< 0.
From (2.3.1), we see that γ = (ΨR − ϕ) satisfies
γ 00 − W γ > 0.
(2.12)
Claim: γ ≤ 0 on R. It’s for |t| > tρ,R if γ(t̄) > 0 there is a global maximum positive γ ∈ [−tρ,R , tρ,R ]. This implies that γ 00 (t) ≤ 0 which
is a contradiction
with (2.12). This implies that ΨR (t) ≤ kgkα∞ +
³
´
√
ρ cosh 2α t . Taking the limit ρ going to 0 we get ΨR ≤
similarly we can conclude that
kgk∞
,
α
and
kgk∞
, ∀R
α
Passing to a subsequence we get a solution Ψ = limR→∞ ΨR , and the
convergence is uniform over compacts sets, to (2.3.1) with
kΨR kL∞ ≤
kgk∞
α
∞
. Also, the same argument shows that the solution is unique
√ (in L
σ|t|
sense). Besides: We observe that if ke gk∞ < ∞, 0 < σ < α then
kΨk∞ ≤
keσ|t| Ψk∞ ≤ Ckeσ|t| gk
The proof of this fact is similar to the previous one. Just take as the
function ϕ as follows
µ√
¶
keσ|t| gk∞ −σ|t|
α
ϕ=M
e
+ ρ cosh
|t| .
α
2
Observe now that Ψ satisfies (2.11) if and only if
µ
¶−1
d2
Ψ= − 2 +W
[F [Ψ, φ]]
dt
where W (x) = V (x0 +ε(t+h))−pwp−1 (1−η1,ε ) and F [φ] = (1−η1,ε )E+
0
+ η 00 φ. The previous result tell us that
(1 − η1,ε )N (η2,ε φ + Ψ) + 2φ0 η2,ε
³ 2 2,ε ´
the inverse of the operator − dtd 2 + W is well define. Assume that
kφkC 1 := kφk∞ + kφ0 k∞ ≤ 1, for some σ < 1 and kΨk∞ ≤ ρ, where ρ
´
LYAPUNOV SCHMIDT REDUCTION METHOD
13
is a very small positive number. Observe that k(1 − η1,ε )Ek∞ ≤ e−cδ/ε .
Furthermore, we have
|F (Ψ, φ)| ≤ e−cδ/ε + cεkφkC 1 + kφk2∞ + kΨk2∞
This implies that
kM [Ψ]k ≤ C∗ [µ + kΨk2∞ ]
where µ = e−cδ/ε + cεkφkC 1 + kφk2∞ . If we assume µ <
choosing ρ = 2C∗ µ, we have
1
,
4C∗ 2
and
kM [Ψ]k < ρ.
If we define X = {Ψ|kΨk∞ < ρ}, then M is a contraction mapping
in X. We conclude that
kM [Ψ1 ] − M [Ψ2 ]k ≤ C∗ CkΨ1 − Ψ2 k,
where C∗ C < 1.
Conclusion: There exists a unique solution of (2.11) for given φ (small
in C 1 -norm) such that
kΨ(φ)k∞ ≤ [e−cδ/ε + εkφkC 1 + kφk2∞ ]
Besides: If kφk ≤ ρ, independent of ε, we have
kΨ(φ1 ) − Ψ(φ2 )k∞ ≤ o(1)kφ1 − φ2 k.
Next step: Solver for (2.9), with kφk very small, the problem
(2.13)
φ00 −(1−pwp−1 )φ+η1,ε E+η3,ε B(φ)+η1,ε pwp−1 Ψ+η1,η N (φ+Ψ)−cw0 = 0
R
where c = R w1 02 R (η3,ε B(φ) + η1,ε pwp−1 Ψ + η1,η N (φ + Ψ))w0 . To solve
(2.13) we write it as
φ = T [η3,ε Bφ] + T [N (φ + Ψ(φ)) + pwp−1 Ψ(φ)] + T [E] =: Q[φ]
Choosing δ sufficiently small independent of ε we conclude that Q(x) ⊆
X, and Q is a contraction in X for k · kC 1 . This implies that (2.13) has
a unique solution φ with kφkC 1 < M ε2 . Also the dependence φ = Φ(h)
is continuous. Now we only need to adjust h in such a way that c = 0.
After some calculations we obtain
0 = Kε2 V 00 (x0 )h + O(ε3 ) + O(δε2 ).
So we can find h = hε and |hε | ≤ Cε, such that c = 0.
14
MANUEL DEL PINO AND JUNCHENG WEI
3. Schrodinger equation in dimension N
½
(3.1)
ε2 ∆u − V (y)u + up = 0
in RN
n
0 < u in R
u(x) → 0, as|x| → ∞
+2
We consider 1 < p < ∞ if N ≤ 2, and 1 < p < N
if N ≥ 3. The
N −2
basic problem that we consider is
½
∆w − w + wp = 0
in RN
0 < u in Rn
w(x) → 0, as|x| → ∞
We look for a solution w = w(|x|), a radially symmetric solution. w(r)
satisfies the ordinary differential equation
½ 00 N −1 0
w + r w − w + wp = 0
r ∈ (0, ∞)
(3.2)
w0 (0) = 0, 0 < w in (0, ∞) w(|x|) → 0, as|x| → ∞
Proposition 3.1. There exist a solution to (3.2).
Proof. Let us consider the space
Hr1 = {u = u(|x|) : u ∈ H 1 (RN )}
R∞
with the norm kukH 1 = 0 (|u0 |2 + |u|2 )rN −1 dr. Let
R
2
2
N |∇u| + u
S = inf 1 R R
u6=0,u∈Hr ( N |u|p+1 )2/(p+1)
R
We recall that H 1 (RN ) → Lp+1 (RN ) continuously, which means that
S > 0 (the larger constant such that ckukLp+1 ≤ kukH 1 ). Strategy:
Take un ≥ 0a minimizing sequence for S. We may assume kun kLp+1 =
1. This means that kun k2H 1 → S. This means that the sequence is
bounded in H 1 1. We may assume un * u ∈ H 1 . We have by lower
weak s.c.i.
Z
Z
2
2
|∇u| + u ≤ lim |∇un |2 + u2n = S.
n
We could get existence of a minimizer for S if we prove that kukLp+1 =1 .
This is indeed the case thanks to:
Strauss Lemma: There exist a constant C such that ∀u ∈ Hr1 (RN ):
|u(|x|)| ≤
C
|x|
N −1
2
kukH 1
The proof of this fact is the following: Let u ∈ Cc∞ (RN ), u = u(|x|).
Z ∞
Z ∞
sN −1
2
0
u (r) = −2
u(s)u (s)ds ≤ 2
|u(s)||u0 (s)| N −1 ds
r
r
r
LYAPUNOV SCHMIDT REDUCTION METHOD
15
Z ∞
Z ∞
C
2 n−1
1/2
(3.3) ≤ N −1 (
|u| s ds) (
|u0 |2 sN −1 ds)1/2 ≤ N −1 kuk2H 1
r
r
0
0
By density we concludes the proof.
Let us observe that
2
p+1
kun kp+1
= kun kp+1
Lp+1 (BR ) + kun kLp+1 (B c )
Lp+1 (RN )
R
and
Z
kun kp+1
c)
Lp+1 (BR
≤
kun kp−1
L∞ (|x|>R)
RN
u2n ≤ ε
if R ≥ R0 (ε) (here we use the lemma of Strauss). On the other hand:
un → u
strong in Lp+1 (BR ) since H 1 (BR ) → Lp+1 (BR ) compactly. This implies
p+1 (B ) + ε ≤ kukLp+1 (RN ) + ε
that 1 ≤ limn→∞ kun kp+1
R
Lp+1 (BR ) + ε = kukL
This implies that kukLp+1 ≥ 1 and we conclude kukLp+1 = 1. R
u is a minimizer for S, u ≥ 0, u 6= 0. We define Φ(v) = kvkH 1 /( |v|p+1 )2 /p+
1. So u is a minimizer for Φ. This means that u is a weak solution of
the problem
−∆u + u = αup
−1
where α = kukH 1 . We define u = α p−1 ũ, then ũ is a solution of
−∆ũ + ũ = ũp
And, with the aid of maximum strong principle we can conclude that
ũ is in fact strictly positive everywhere. This concludes the proof ¤
Observation 3.1. There no exist a solution of class C 2 for p ≥
The proof of this fact is an application of Pohozaev identity.
N +2
.
N −2
N −1
We claim that w(r) ≈ Cr− 2 e−r . This can be proved with the
N −1
change of variables k = r− 2 h. The equation that satisfies h is like
h00 − h(1 + rc2 ) = 0, and the solution of this equation is like e−r .
Theorem 3.1. Kwong, 1989 The radial solution of (3.2) is unique.
3.1. Linear problem. Consequence of the proof of Kwong: We define
L(φ) = ∆φ + pw(x)p−1 φ − φ.
Let us consider the problem
L(φ) = 0,
φ ∈ L∞ (RN )
A known fact is that if φ is a solution of this problem, then φ is a linear
∂w
combination of the functions ∂x
(x), j = 1, . . . , N . This is known as
j
non degeneracy of w.
16
MANUEL DEL PINO AND JUNCHENG WEI
We assume as always 0 < α ≤ V ≤ β. We want to solve the problem
½ 2
ε ∆ũ − V (y)ũ + ũp = 0
in RN
(3.4)
n
0 < ũ in R
ũ(x) → 0, as|x| → ∞
´
³p
1
We fix a point ξ ∈ RN . Observe that Uε (y) = V (ξ) p−1
V (ξ) y−ξ
,
ε
is a solution of the problem equation
ε2 ∆u − V (ξ)u + up = 0.
We will look for a solution of (3.4) uε (x) ≈ Uε (y). We define wλ =
√
1
λ p−1 w( λx).
Let us observe that if ũ satisfies (3.4), then u(x) = ũ(εz) satisfies
the problem
½
∆u − V (εz)u + up = 0
in RN
(3.5)
n
0 < u in R
u(x) → 0, as|x| → ∞
Let ξ 0 = ξε . We want a solution of (3.5) with the form u(z) = wλ (z −
ξ 0 ) + φ̃(z), with λ = V (ξ) and φ̃ small compared with wλ (z − ξ 0 ).
3.2. Equation in terms of φ. φ(x) = φ̃(ξ 0 − x). Then φ satisfies the
equation ∆x [wλ (x)+φ(x)]−V (ξ+εx)[wλ (x)+φ(x)]+[wλ (x)+φ(x)]p = 0.
We can write this equations as
∆φ − V (ξ)φ + pwλp−1 (x)φ − E + B(φ) + N (φ) = 0
where E = (V (ξ + εx) − V (ξ))wλ (x), B(φ) = (V (ξ) − V (ξ + εx))φ and
N (φ) = (wλ + φ)p − wλp − pwλp−1 φ. We consider the linear problem for
λ = V (ξ),
(
P
∂w
L(φ) = ∆φ − V (ξ + εx)φ + pwλ (x)φ = g − N
i=1 ci ∂xi
R
(3.6)
λ
φ ∂w
= 0, i = 1, . . . , N
∂xi
RN
The c0i s are defined as follows
Z
Z
Z
L(φ)(wλ )xi = L0 (φ)(wλ )xi + (V (ξ) − V (ξ + εx))φ(wλ )xi
xi
. This implies that
w = w(|x|). (wλ )xi (x) = wλ0 |x|
Z
Z
1
(wλ )xi (wλ )xj = wλ0 (|x|)2 xi xj 2
|x|
R
R
0
2 2 1
This integral is 0 if i 6= j and equals to RN wλ (|x|) xi |x|2 dx = 1/N |∇wλ |2 =
R
R
γ. Then ci = g(wλ )xi + RN [V (ξ + εx) − V (ξ)]φ(wλ )xi R (wλ1)xi 2 .
Problem: Given g ∈ L∞ (RN ) we want to find φ ∈ L∞ (RN ) solution
to the problem (3.6). Assumptions: We assume V ∈ C 1 (RN ), kV kC 1 <
∞. We assume in addition that |ξ| ≤ M0 and 0 < α ≤ V .
LYAPUNOV SCHMIDT REDUCTION METHOD
17
Proposition 3.2. There exists ε0 , C0 > 0 such that ∀0 < ε ≤ ε0 ,
∀|ξ| ≤ M0 , ∀g ∈ L∞ (RN ) ∩ C(RN ), there exist a unique solution φ ∈
L∞ (RN ) to (3.6), φ = T [g] satisfies
kφkC 1 ≤ C0 kgk∞
Proof. Step 1: A priori estimates on bounded domains: There exist
∞
R0 , ε0 , C0 such thatP
∀ε < ε0 , R > R0 , |ξ|
R ≤ M0 such that ∀φ, g ∈ L
solving L(φ) = g − i ci (wλ )xi in BR , BR φ(wλ )xi = 0 and φ = 0 on
∂BR , we have
kφkC 1 (BR ) ≤ C0 kgk∞
We prove first kφk∞ ≤ C0 kgk∞ . Assume the opposite, then there exist
sequences φn , gn , ε → 0, Rn → ∞, |ξn | ≤ M0 such that
L(φn ) = gn − cni
∂wλ
∂xi
. The first fact is that cni → 0 as n → ∞. This fact follows just after
multiplying the equation against (wλ )xi and integrating by parts.
Observation: If ∆φ = g in B2 then there exist C such that
k∇φkL∞ (B1 ) ≤ C[kgkL∞ (B2 ) + kφkL∞ (B2 ) ]
Where B1 and B2 are concentric balls. This implies that k∇φn kL∞ (B) ≤
C a given bounded set B, ∀n ≥ n0 . This implies that passing to a
subsequence φn → φ uniformly on compact sets, and φ ∈ L∞ (RN ).
Observe that kφn k∞ = 1, and this implies that kφk∞ ≤ 1. We can
assume that ξn → ξ0 .
φ satisfies the equation ∆φ − V (ξ0n
)φ + pwλp−1
(x)φo= 0, where λ0 =
0
R
∂wλ0
∂wλ0
V (ξ0 ), and this implies that φ ∈ Span x1 , . . . , xN , but also RN φ(wλ0 )xi =
0, i = 1, . . . , N . Then φ = 0 and this implies that kφn kL∞ (BM (0)) →
0, ∀M < ∞. Maximum principle implies that kφn kL∞ (BRn \BM0 → 0,
just because |φn | = o(1) on ∂BRn \ BM0 and kgn k∞ → 0. Therefore kφn k∞ → 0, a contradiction. This implies that kφkL∞ (BR ) ≤
C0 kgkL∞ (BR ) uniformly on large R. The C 1 estimate follows from elliptic local boundary estimates for ∆.
Step 2: Existence: g ∈ L∞ . We want to solve (3.6). WeR claim that
solving (3.6) is equivalent to finding φ ∈ X = {ψ ∈ H01 : ψ(wλ )xi =
0, i = 1, . . . , N } such that
Z
Z
Z
p−1
∇φ∇ψ + V (ξ + εx)φψ − pw φψ + gψ = 0, ∀ψ ∈ X.
18
MANUEL DEL PINO AND JUNCHENG WEI
R
P
Ψ(w )
Take general Ψ ∈ H01 , Ψ = ψ + i αi (wλ )xi , with αi = R (wλλ)xxi . We
i
have
Z
Z
X
X
X
− ∆(
αi (wλ )xi )∇φ+ V (ξ)(
αi (wλ )xi )φ−pwp−1 (
αi (wλ )xi )φ = 0
i
i
Which implies that
Z
Z
V (ξ)φΨ − pwp−1 φΨ
∇φ∇Ψ +
Z
−
i
(V (ξ) − V (ξ + εx))(Ψ −
X
Z
αi (wλ )xi ) +
g(Ψ −
i
Then
Z
[(V (ξ + εx) − V (ξ))φ + g](Ψ −
and ΠX (Ψ) =
X
αi (wλ )xi )
i
X
αi (wλ )xi )
i
P
i αi (wλ )xi , then the previos integral is equal to
Z
ΠX ([(V (ξ + εx) − V (ξ))φ + g]φ)Ψ
¤
This implies that
−∆φ + V (ξ)φ − pwp−1 φ + ΠX ([(V (ξ + εx) − V (ξ))φ + g]φ) = 0.
The problem is formulated weakly as
Z
Z
Z
p−1
∇φ∇ψ + (V (ξ + εx) − pw )φψ + gψ = 0, φ ∈ X, ∀ψ ∈ X
This can be written as φ = A[φ] + g̃, where A is a compact operator.
The a priori estimate implies that the only solution when g = 0 of this
equation is φ = 0. We conclude existence by Fredholm alternative.
We look for a solution which near xj = ξj0 = ξj /ε, j = 1, . . . , k looks
like v(x) ≈ Wλj (x − ξj0 ), λj = V (ξj ), where Wλ solves
∆Wλ − λW + W p = 0,
Wλ radial, Wλ (|x|) → 0, as |x| → ∞
√
Observe that Wλ (y) = λ1/(p−1) w( λy), where w solves the equation
∆w − w + wp = 0. The equation
∆v − V (εx)v + v p = 0
looks like ∆v −V (ξj )v +v p = 0, where ξ1 , ξ2 , . . . ξk ∈ RN and we assume
P
also |ξj0 − ξl0 | À 1, if j 6= l. We look for a solution v(x) ≈ kj=1 Wλj (x −
ξj0 ), λj = V (ξj ). We assume V ∈ C 2 (RN ) and kV kC 2 < ∞, 0 < α ≤ V .
LYAPUNOV SCHMIDT REDUCTION METHOD
We use the notation Wj = Wλj (x − ξj0 ), λj = V (ξj ) and W =
Look for a solution v = W + φ, so φ solves the problem
19
Pn
j=1
Wj .
∆φ − V (εx)φ + pW p−1 φ + E + N (φ) = 0
where
E = ∆W − V W + W p , N (φ) = (W + φ)p − W p − pW p−1 φ.
P
P
Observe that ∆W = j ∆Wj = j λj Wj − Wjp . So we can write
X
X
X p
E=
(λj − V (εx))Wj + (
Wj )p −
Wj .
j
j
j
3.3. Linearized (projected) problem. We use the following nota∂W
tion Zji = ∂xij . The linearized projected problem is the following
X
cij Zji ,
∆φ − V (εx)φ + pW p−1 φ + g =
i,j
R
with the orthogonality condition φZji = 0, ∀i, j. The Zji ’s are “nearly
orthogonal” if the centers ξj0 are far away one to each other. The cij ’s
are, by definition, the solution of the linear system
Z
X Z
i0
p−1
(∆φ − V (εx)φ + pW φ + g)Zj0 =
cij
Zji Zji00 ,
RN
i,j
RN
for i0 = 1, . . . , N , j0 = 1, . . . , k. The cij ’s are indeed uniquely determined provided that |ξl0 − ξj0 | > R0 À 1, because the matrix with
R
coefficients αi,j,i0 ,j0 = Zji Zji00 is “nearly diagonal”, this means
½ 1 R
|∇Wj |2 if (i, j) = (i0 , j0 ),
N
αi,j,i0 ,j0 =
o(1)
if not
Moreover:
Z
XZ
p−1
i0
p−1
i
|cj0 | ≤ C
|φ|[|λj −V |+p|W −Wj |]|Zj |+ |g||Zji | ≤ C(kφk∞ +kgk∞ )
i,j
with C uniform in large R0 . Even more, if we take x = ξ 0 + y
|(λj − V (εx))Zji | ≤ |(V (ξj ) − V (ξj + εy))||
because |
∂Wλj
∂yi
√
∂Wλj
α
| ≤ Cεe− 2 |y| ,
∂yi
√
| ≤ Ce−|y| λj |y|−(N −1)/2 . Observe also that
|(W p−1 − Wjp−1 )Zji | = |((1 −
X Wl
)p−1 − 1)|Wjp−1 Zji .
Wj
l6=j
20
MANUEL DEL PINO AND JUNCHENG WEI
Observe that if |x − ξj0 | < δ0 minj1 6=j2 |ξj0 1 − ξj0 2 |, then
√
√
0
0
Wl (x)
e− λl |x−ξl |
e− λl |x−ξl |
√
√
≈
<
0
0
0
Wj (x)
e− λj |x−ξj |
e− λj δ0 minj1 6=j2 |ξj1 −ξj2 |
√
√
√
0
0
0
0
If δ0 ¿ 1 but fixed, we conclude that e− λl |ξj −ξl |+δ0 ( λl − λj ) minj1 6=j2 |ξj1 −ξj2 | <
0
0
e−ρ minj1 6=j2 |ξj1 −xij2 |¿1 . Conclusion: if |x − ξj0 | < δ0 minj1 6=j2 |ξj0 1 − xi0j2 |
implies that
0
0
α
0
|(W p−1 − Wjp−1 )Zji | ≤ e−ρ minj1 6=j2 |ξj1 −xij2 | e− 2 |x−ξj | .
If |x − ξj0 | > δ0 minj1 6=j2 |ξj0 1 − xi0j2 |, then
0
0
α
0
|(W p−1 − Wjp−1 )Zji | ≤ C|Zji | ≤ Ce−ρ minj1 6=j2 |ξj1 −xij2 | e− 2 |x−ξj |
As a conclusion we get
0
0
|cij00 | ≤ C(ε + e−ρ minj1 6=j2 |ξj1 −ξj2 | )kφk∞ + kgk∞
Lemma 3.1. Given k ≥ 1, there exist R0 , C0 , ε0 such that for all points
ξj0 with |ξj0 1 − ξj0 2 | > R0 , j = 1, . . . , k and all ε < ε0 then exist a unique
solution φ to the linearized projected problem with
kφk∞ ≤ C0 kgk∞ .
Proof. We first prove the a priori estimate kφk∞ ≤ C0 kgk∞ . If not
there exist εn → 0, kφnP
k∞ = 1, kgn k → 0, ξj0n with minj1 6=j2 |ξj0n1 −ξj0n2 | →
∞. We denote Wn = j Wjn , and we have
X
∆φn − V (εn x)φn + pWnp−1 φn + gn =
(cij )n (zji )n
i,j
First observation:
→ 0 (follows from estimate for cij00 ). Second:
∀R > 0 kφn kL∞ (B(ξj0n ,R)) → 0, j = 1, . . . , k. If not, there exist j0
kφn kL∞ (B(ξj0n ,R)) ≥ γ > 0. We denote φ˜n (y) := φn (ξj0n0 + y). We have
0
kφ˜n kL∞ (B(0,R)) ≥ γ > 0. Since |∆φ˜n | ≤ C, kφ̃n k∞ ≤ 1. This implies
(cij )n
that k∇φ̃n k ≤ C. Passing to a subsequence we may assume φ̃n →
φ̃ uniformly on compacts sets. Observe that also V (εn (ξj0n0 + y)) =
V (εn ξj0n0 )+O(εn |y|) → λj0 over compact sets and Wn (ξj0n0 +y) → Wλj0 (y)
uniformly on compact sets. This implies that φ̃ is a solution of the
problem
Z
∂Wλj0
p−1 ˜
dy = 0, i = 1, . . . , N
∆φ̃ − λj0 φ̃ + pwλ0 p − 1 = 0,
φ̃
∂yi
P ∂wλ
Non degeneracy of wλj0 implies that φ̃ = i αi ∂yij0 . The orthogonality condition implies that αi = 0, ∀i = 1, . . . , N . This implies that
LYAPUNOV SCHMIDT REDUCTION METHOD
21
φ̃ = 0 but kφ̃kL∞ (B(0,R)) ≥ γ > 0, a contradiction. Now we prove:
kφn kL∞ (RN \ ∪n B(ξj0n , R)) → 0, provided that R À 1 and fixed so that
φn → 0 in the sense of kφn k∞ (again a contradiction). We will denote
Ωn = RN \ ∪n B(ξj0n , R). For R À 1 the equation for φn has the form
∆φn − Qn φn + gn = 0
where Qn = V (εx) − pWnp−1 ≥ α2 > 0 for some R sufficiently large (but
fixed). Let’s take for σ 2 < α/2
X
0n
φ̄ = δ
eσ|x−ξj | + µn .
j
We denote ϕ(y) = eσ|y| , r = |y|. Observe that ∆ϕ − α/2ϕ = eσ|y| (σ 2 +
N −1
− α/2) < 0 if |y| > R À 1. Then
|y|
α
α
φ̄ − kgn k∞ > µn − kgn k∞ > 0
2
2
P
if we choose µn ≥ kgn k∞ α2 . In addition we take µn = j kφn kL∞ (B(ξjn ,R)) +
kgn k∞ α2 . Maximum principle implies that φn (x) ≤ φ̄ for all x ∈ Ωn .
Taking δ → 0 this implies that φn (x) ≤ µn , for all x ∈ Ωn . Also true
that |φn (x)| ≤ µn for all x ∈ Ωnc , and this implies that kφn kL∞ (RN ) →
0.
¤
³P
´
−1
−ρ|x−ξj0n |
k∞ →
Observation 3.2. If in addition we have θn = kgn
je
0 with ρ < α/2. Then we can use as a barrier
X
X
0n
0n
φ̄ = δ
eσ|x−ξj | + µn
e−ρ|x−ξj |
−∆φ̄ + Qn φ̄ − gn > −∆φ̄ +
j
j
P
with µn = eρR j kφn kL∞ (B(ξj0n ,R)) + θn , then φ̄ is a super solution of
the equation and we have |φn | ≤ φ̄, and letting δ → 0 we get |φn (x)| ≤
P
0n
µn j e−ρ|x−ξj | . As a conclusion we also get the a priori estimate
à k
!−1
à k
!−1
X
X
−ρ|x−ξj0 |
−ρ|x−ξj0 |
kφ
e
k∞ ≤ Ckg
e
k∞
j=1
j=1
provided that 0 ≤ ρ < α/2, |ξj0 1 − ξj0 2 | > R0 À 1, ε < ε0 .
We now give the proof of existence
Proof. Take g compactly supported. The weak formulation for
Z
X
i i
p−1
cj Zj ,
φZji , ∀i, j
(3.7)
∆φ − V (εx)φ + pW φ + g =
i,j
22
MANUEL DEL PINO AND JUNCHENG WEI
R
is find φ ∈ X = {φ ∈ H 1 (RN ) : φZji = 0, ∀i, j} such that
Z
(3.8)
∇φ∇ψ + V φψ − pwp−1 φψ − gψ = 0, ∀ψ ∈ X.
RN
R
Assume φ solves (3.7). For g ∈ L2 , write g = g̃ + Π[g] where g̃Zji = 0,
for all i, j. Π is the orthogonal projection of g onto the space spanned
by the Zji ’s. Take ψ ∈ H 1 (RN ) arbitrary and use ψ − Π[ψ] as a test
function in (3.8). Then if ϕ ∈ Cc∞ (RN ), then
Z
Z
Z
∇ϕ∇(Π[ψ]) = −
∆ϕΠ[ψ] = −
Π[∆ϕ]ψ.
RN
RN
P
RN
αi,j Zji , where
Z
Z
Z
X
j0
αi,j Zi,j Zi0 ,j0 = ∆ϕZi0 = ϕ∆Zij00
But Π[∆ϕ] =
i,j
Then kΠ[∆ϕ]kL2 ≤ CkϕkH 1 . By density is true also for ϕ ∈ H 1 where
∆ϕ ∈ H −1 . Therefore
Z
Z
Z
p−1
∇φ∇ψ + (V φ − pW φ − g)ψ = Π(V φ − pW p−1 φ + g)ψ
then φ solves in weak sense
−∆φ + V φ − pW p−1 φ − g = Π[−∆φ + V φ − pW p−1 φ − g]
P
and Π[−∆φ + V φ − pW p−1 φ − g] = i,j cji Zi j. Therefore by definition
φ solves (3.8) implies that φ solves (3.8). Classical regularity gives
that this weak solution is solution of (3.7) in strong sense, in particular
φ ∈ L∞ so that
kφk∞ ≤ Ckgk∞
. Now we give the proof of existence for (3.7). We take g compactly
supported. The equation (3.8) can be written in the following way
(using Riesz theorem):
hφ, ψiH 1 + hB[φ], ψiH 1 = hg̃, ψiH 1
or φ + B[φ] = g̃, φ ∈ X. We claim that B is a compact operator.
Indeed if φn * 0 in X, then φn → 0 in L2 over compacts.
Z
Z
Z
p−1
p−1 2 1/2
|hB[φn ], ψi| ≤ | pW φn ψ| ≤ ( pw φn ) ( pW p−1 ψ 2 )1/2
then
Z
|hB[φn ], ψi| ≤ c( pW p−1 φ2n )1/2 kψkH 1
LYAPUNOV SCHMIDT REDUCTION METHOD
23
Take ψ = B[φn ], which implies
kB[φn ]kH 1
Z
≤ c( pW p−1 φ2n )1/2 → 0.
This implies that B is a compact operator. Now we prove existence
with the aid of fredholm alternative. Problem is solvable if for g̃ = 0
implies that φ = 0. But φ + B[φ] = 0 implies solve (3.7)(strongly) with
g = 0. This implies φ ∈ L∞ , and the a priori estimate implies φ = 0.
Considering gΞBR (0) we conclude that
kφR k∞ ≤ kgk∞
Taking R → ∞ then along a subsequence φR → φ uniform over compacts.
¤
We take g ∈ L∞ . We have φ = Tξ0 [g], where ξ 0 = (ξ10 , . . . , ξk0 ).
We want to analyze derivatives ∂ξji0 Tξ0 [g]. We know that kTξ0 [g]k ≤
C0 kgk∞ . First we will make a formal differentiation. We denote Φ =
∂φ
.
∂ξi0 j
0 0
R
P
We have ∆φ − V φ + pW p−1 φ + g = i,j cij Zji and φZji = 0, for all
i, j. Formal differentiation yields
X
X
∆Φ − V Φ + pW p−1 Φ + +∂ξi0 j0 (W p−1 )φ −
cij ∂ξi0 j0 Zij =
c̃ij Zji
i,j
i,j
where formally c̃ji = ∂ξi0 j0 cji . The orthogonality conditions traduces
into
½
Z
0
if j 6= j0
i
ΦZj = − R φ∂
i
if j = j0
ξi0 j0 Zj
RN
0
R
P
Let us define Φ̃ = Φ − i,j αi,j Zji . We want Φ̃Zji = 0, for all i, j. We
need
½
Z
X
0
if j̄ 6= j0
i ī
αi,j Zj Zj̄ = − R φ∂
i
if j̄ = j
ξi0 j0 Zj
i,j
0
0
The system has a unique solution and |αi,j | ≤ Ckφk
R ∞ (since the system
is almost diagonal). So we have the condition Φ̃Zji = 0, for all i, j.
P
We add to the equation the term i,j αi,j (∆ − V + pW p−1 )Zji , so Φ̃
P
satisfies the equation∆φ − V φ + pW p−1 φ + g = i,j cij Zji
X
X
X
αi,j (∆−V +pW p−1 )Zji
cij ∂ξi0 j0 Zij =
c̃ij Zji −
∆Φ̃−V Φ̃+pW p−1 Φ̃+∂ξi0 j0 (W p−1 )φ−
i,j
i,j
i,j
This implies kΦ̃k ≤ C(khk + kgk) ≤ Ckgk∞ . This implies kΦk ≤
Ckgk∞ . We do this in a discrete way, and passing to the limit all
these calculations are still valid. Conclusion: The map ξ → ∂ξ φ is well
24
MANUEL DEL PINO AND JUNCHENG WEI
defined and continuous (into L∞ ). Besides k∂ξ φk∞ ≤ Ckgk∞ , and this
implies
k∂ξ Tξ [φ]k ≤ Ckgk
3.4. Nonlinear projected problem. Consider now the nonlinear projected problem
Z
X j
i
p−1
ci Zj ,
φZij = 0, ∀i, j
∆φ − V φ + pw φ + E + N (φ) =
i,j
We solve this by fixed point. We have φ = T (E + N (φ)) =: M (φ). We
N
∞
N
define Λ =
≤ M kEk∞ }. Remember
∞P
P{φ ∈ C(R ) ∩ L (RP ) : kφk
that E = i (λj − V (εx))Wj + ( j Wj )p − j Wjp . Observe that
X
X
0
0
0
0
e−σ|x−ξj |
|E| ≤ ε
e−σ|x−ξj | + ce−δ0 minj1 6=j2 |ξj1 −ξj2 |
j
i
−δ0 minj1 6=j2 |ξj0 −ξj0 |
1
2 ] =: ρ (see that
so, for existence we have kEk ≤ C[ε+e
ρ is small). Contraction mapping implies unique existence of φ = Φ(ξ)
and kΦ(ξ)k ≤ M ρ.
3.5. Differentiability in ξ 0 of Φ(ξ 0 ). We have
Φ − Tξ0 (Eξ0 + Nξ0 (φ)) = A(Φ, ξ 0 ) = 0
If (DΦ A)(Φ(ξ 0 ), ξ 0 ) is invertible in L∞ , then Φ(ξ 0 ) turns out to be of
class C 1 . This is a consequence of the fixed point characterization, i.e.,
DΦ A(Φ(ξ 0 ), ξ 0 ) = I + o(1) (the order o(1) is a direct consequence of
fixed point characterization). Then is invertible. Theorem and the C 1
derivative of A(Φ, ξ 0 ) in (φ, ξ 0 ). This implies Φ(ξ 0 ) is C 1 . kDξ0 Φ(ξ 0 )k ≤
Cρ (just using the derivate given by the implicit function theorem).
3.6. Variational reduction. We want to find ξ 0 such that the cij = 0,
for all i, j, to get a solution to the original problem. We use a procedure
that we call Variational Reduction in which the problem of finding ξ 0
with cij = 0, for all i, j, is equivalent to finding a critical point of a
functional of ξ 0 . Recall:
Z
Z
1
1
2
2
J(v) :=
|∇v| + V (εx)v −
v p+1
2 RN
p + 1 RN +1 +
+2
is defined in H 1 (RN ), since 1 < p < N
. v is a solution of ∆v − V v +
N −2
p
1
N
v = 0, v → R0 if and only if v ∈ H (R ) and J 0 (v) = 0. Observe that
p
hJ 0 (v), ϕi = ∇v∇ϕ + V vϕ − v+
ϕ.
The following fact happens: v = Wξ∗0 + φ(ξ 0 ) is a solution of the
original problem (for ρ ¿ 1) if and only if
∂ξ0 J(Wξ0 + φ(ξ 0 ))|ξ0 =ξ∗0 = 0.
LYAPUNOV SCHMIDT REDUCTION METHOD
25
0
Indeed, observe that v(ξ
Wξ0 + φ(ξ 0 ) solves the problem ∆v(ξ 0 ) −
P ) :=
0
0 p
i i
V (εx)v(ξ ) + v(ξ ) = i,j cj Zj and also that
X Z
X Z j
0
0
0
0
i
i
∂ξj0 i J(v(ξ )) = hJ (v(ξ )), ∂ξj0 i v(ξ )i = −
cj Zj ∂ξj0 i v = −
cij Zi (∂ξj0
0 0
0 0
Remember that Wξ0 =
∂ξj0
0 i0
Wξ0 = ∂ξj0
Pk
j=1
j,i
0 0
i,j
0
0 i0
0
0 i0
wλj0 (ξ0 ) (x−ξj0 ) = (∂λ wλ (x−ξj0 0 ))|λ=λj0 −∂xi0 wλj0 (x−ξj0 0 ) = O(e−δ|x−ξ0 | )o(ε)−Zj
0
Cρe
Wξ0 +∂ξj0
wλj (x − ξj0 ),
This because ∂λ wλ = O(e−δ|x−ξ0 | ). On the other hand |∂ξj0
−δ|x−ξj0 |
0 i0
. Finally, observe that
Z
− Zji (∂ξj0 i Wξ0 + ∂ξj0
0 0
0 i0
φ| ≤
Z
0 i0
Zji Zji00 + O(ρ)
φ) =
The matrix of these numbers is invertible provided ρ ¿ 1.
A consequence (D, Felmer 1996): Assume j = 1 and that there exist
an open, bounded set Λ ⊂ RN such that
inf V > inf V,
∂Λ
Λ
then there exist a solution to the original problem, vε with vε (x) =
WV (ξε ) ((x − ξε )/ε) + o(1) and V (ξε ) → minΛ V , ξ = ξε .
Another consequence (D, Felmer 1998): Λ1 , . . . , Λk disjoint bounded
with inf Λj V < inf ∂Λj V , for all j. For the problem ε2 ∆u − V (x)u +
up = 0, 0 < u → 0 at ∞, there exist a solution uε with uε (x) ≈
Pk
ε
ε
ε
ε
j=1 WV (ξj ) (x − ξj /ε), ξj ∈ Λj and V (ξj ) → inf Λj V (in the case
of non-degeneracy minimal or more generally non-degenerate critical
points the result is due to Oh (1990))
Proof. First result: j = 1. v(ξ 0 ) = Wξ0 + φ(ξ 0 ). Then
1
J(W( ξ 0 )) = J(Wξ0 + φ(ξ 0 )) + hJ 0 (Wξ0 + φ), −φi + J 00 (Wξ0 + (1 − t)φ)[φ]2
2
0
(Taylor expansion of the function α(t) = J(Wξ + (1 − t)φ)). Observe
R
P
that hJ 0 (Wξ0 + φ), −φi = i,j cij Zij φ = 0. Also observe that
Z
00
0
2
J (Wξ +(1−t)φ)[φ] = |∇φ|2 +V (εx)φ2 −p(Wξ0 +(1−t)φ)φ2 = O(ε2 )
0
uniformly on ξ 0 because ∇φ, φ = O(εe−δ|x−ξ | ). We call Φ(ξ) := J(v(ξ 0 )) =
J(Wξ0 ) + O(ε2 ), and
Z
Z
Z
1
1
0p+1
02
0 2
0
|∇Wξ | +V (ξ)Wξ −
Wξ + (V (εx)−V (ξ 0 ))Wξ2
J(Wξ ) =
2
p+1
26
MANUEL DEL PINO AND JUNCHENG WEI
Taking λ = V (ξ), we have that
Z
Z
2
−N/2
|∇wλ (x)| = λ
|∇w(λ1 /2x)|2 λ1+2/(p−1) λN/2 dx = λ−N/2+p+1/p−1 |∇w(y)|2 dy
and
Z
Z
wλ2 (x)
λ
=λ
−N/2p+1/p−1
w(y)p+1 dy
This implies that
Z
Z
1
1
02
0 2
|∇Wξ | + V (ξ)Wξ −
Wξ0p+1 = V (ξ)p+1/p−1−N/2 cp,N .
2
p+1
also
Z
(V (εx) − V (ξ 0 ))wλ (x − ξ 0 )2 = O(ε)
uniformly on ξ. In summary Φ(ξ) = J(v(ξ 0 )) = V (ξ)p+1/p−1−N/2 cp,N +
− N2 > 0. Then ∀ε ¿ 1 we have
O(ε) and p+1
p−1
inf Φ(ξ) < inf Φ(ξ)
ξ∈Λ
ξ∈∂Λ
therefore Φ has a local minimum ξε ∈ Λ and V (ξε ) → minΛ V . Same
thing works at a maximum.
For several spikes separated: |ξj1 − ξj2 | > δ, for all j1 6= j2 . ρ =
−δ0 minj1 6=j2 |ξj0 −ξj0 |
−δ0 δ/ε
1
2 + ε ≤ e
e
+ ε < 2ε, so we have
X
0
|∇x φ(ξ 0 )| + |φ(ξ 0 )| ≤ Cε
e−δ0 |x−ξj |
j
Now we get
J(v(ξ 0 )) =
X
V (εj )p+1/p−1−N/2 cp,N + O(ε)
j
ξ 0 = 1/ε(ξ1 , . . . , ξk ) implies for several minimal on the Λj we have the
result desired.
¤
Result at one non-degenerate critical point: if ξ0 is a non-degenerate
critical point of V (V 0 (ξ0 ) = 0 and V 00 (ξ0 ) invertible), then there exist
a solution uε (x) such that
uε (x) ≈ WV (ξε ) (x − ξε )/ε,
ξε → ξ0 .
For small δ we have that J(v) has degree different from 0 in a ball
centered at x0 and of radius δ.
LYAPUNOV SCHMIDT REDUCTION METHOD
27
4. Back to Allen Cahn in R2
We consider the functional
¶
Z µ
2
(1 − u2 )2
2 |∇u|
ε
+
a(x)dx.
J(u) =
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
(x)∇u + (1 − u2 )u = 0.
a
Using the change of variables v(x) = u(εx), we find the equation
(4.1)
ε2 ∆u + ε2
∇a
(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 (4.1) 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
(4.2)
∆v + ε
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 =
∇ψ · ν. Observe that ∇ψ = (∇ψ · γ̇)γ̇(∇ · ν)ν. This means that
1 ∂ ψ̃
1
2
2
∇ψ = 1−kz
γ̇ + ∂∂zψ̃ ν, and |∇ψ|2 = (1−kz)
2 |ψ̃s | + |ψ̃z | . Then
∂s
¶
Z
ZZ µ
1
2
2
2
|ψ̃s | + |ψ̃z | (1 − kz)dsdz
|∇ψ(x)| dx =
(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)
28
MANUEL DEL PINO AND JUNCHENG WEI
So
Z
ZZ
− ∆ψϕdx = −
then
∆ψ̃ =
1
(1 − kz)
µµ
1
ψ̃s
(1 − kz)
¶
¶
+ (ψ̃z (1 − lz))z ϕ̃(1−kz)dsdz
s
1
1
∂
k
(
ψ̃s ) + ψ̃zz −
ψ˜z
(1 − kz) ∂s 1 − kz
1 − kz
We just say
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)
∆ψ̃ =
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 ).
S[u] = ε2
z
az
k(s)
z
S[w( )] = ε[ −
]w0 ( )
ε
a
1 − k(s)z
ε
The condition we ask (geodesic condition) is aaz (s, 0) = k(s). In v
language we want
∇a
∆v + ε
(εx) · ∇v + f (v) = 0
a
transition on Γε = 1ε Γ. we use coordinates relative to Γε rather than Γ
1
Xε (s, z) = γ(εs) + zν(εs), |z| < δ/ε
ε
Laplacian for coordinates relative to Γε are
µ
¶
1
1
1
εk(εs)
as
az
∆ψ =
vs +ψzz −
+ε
vs +ε vz +
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)
= −k(ε)γ˙ε (s), where kε = εk(εs)
∂s
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
LYAPUNOV SCHMIDT REDUCTION METHOD
29
This means
∂
L[Γλh ]|h=0 = 0
∂λ
or just hDL(Γ), hi = 0 for all h. Observe that
Z l
L(Γλh ) =
a(γ(s̃) + h(s̃)ν(s̃)) · |γ̇(s̃)λh |ds̃
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
(a +az̃ k 2 h2 + az̃z̃ h2 )+O(λ3 h3 )
Lh [Γλh ] = La (Γ) = λ (−ka+az̃ )(s̃, 0)h(s̃)ds̃+λ
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
∂λ2
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.
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
εk(εs)
1
E(v) =
vs + vzz −
vz +
(1 − εk(εs)z) (1 − εk(εs)z)
(1 − εk(εs)z)
s
ε
as̃
az̃
1
vs + ε vz + f (v) = 0
2
a (1 − εk(εs)z)
a
30
MANUEL DEL PINO AND JUNCHENG WEI
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
E(v0 ) =
(
w0 (−ḣ(εs, εz))s + w00 + f (w)
1 − εkz 1 − εkz
az̃
k(εs)
ε
as̃ 0
(εs, εz) −
)w0 − εḣ
w
2
a
1 − k(εs)εz
(1 − εkz) a
Error in terms of coordinates (s, t) z = t + h(εs):
¸
·
k(εs)
ε2 w 0
az̃
0
E(v0 )(s, t) = εw (t)
(εs, ε(t + h)) −
−
h00
a
1 − k(εs)(t + h)ε
(1 − kε(t + h))2
+ε(
1
ε
as̃ 0
1
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
−ε h (εs)
= −ε h
w02 dt + O(ε3 )
|t|<δ/ε (1 − kε(t + h))
R
+
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
and finally
Z
Z
k(εs)
az̃
02 az̃
2
ε
w ( (εs, ε(t+h))−
)=ε
w0 (t)2 (ε2 )(( )(εs, 0)−k 2 )h(εs)+O(ε
a
1 − k(εs)(t + h)ε
a
R
|t|<δ/ε
2 2
ḣ ε
Then
R
³a ´
− Ew0 dt
z̃
00
0 as̃
R
=h +h
−(
(εs, 0) − k 2 )h + O(ε)
2
02
a
a z̃
ε w
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)
LYAPUNOV SCHMIDT REDUCTION METHOD
31
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
1
∂
∂
1
∂
∂
S(ṽ) =
( − 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
2 00
{
(ṽ
(−ε
h
v
ṽs s}
−ε
ḣv
)+
−2ε
ḣṽ
s)+
s
t
t
t
1 − εkz (1 − εk(t + h))2
1 − kε(t + h)
1 + εk(t + h)
1
εk
(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).
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: ∃γ > 0 :
p0 (t)2 − f 0 (w(t))p(t)2 ≥ γ R p2 (t)dt
R
32
MANUEL DEL PINO AND JUNCHENG WEI
R
R
for all p ∈ H 1 with R pw0 = 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 .
02
w
R
We call β(y) =
0
φ(y,τ
R )w (τ )dτ
02
w
∆φ̃ + f 0 (w)φ̃ = ∆φ + f 0 (w)φ + (∆y β)w0 + β(∆w0 + f 0 (w))w0 = 0
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)
LYAPUNOV SCHMIDT REDUCTION METHOD
33
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 }.
Let us see that a(ψ, ψ) = |∇ψ|2 − f 0 (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.
√
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
φ̄ = M e−σ|t| + 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∞ .
34
MANUEL DEL PINO AND JUNCHENG WEI
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
∆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:
(4.3)
ε2 [δu +
∇a
· ∇u] + f (u) = 0
a
LYAPUNOV SCHMIDT REDUCTION METHOD
35
We want a solution to (4.3) uε (x) ≈ W (z/ε). Writing x = y + zγ(y),
|z| < δ, we have
∆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 + f (v) = 0
∆v + ε
a
we take as first (global) approximation
v0 (x) = w(t)η3 + (1 − η4 )H(x)
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
36
MANUEL DEL PINO AND JUNCHENG WEI
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
(4.4)
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 (4)-(4.4) 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 (4) 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 .