EXISTENCE AND UNIQUENESS OF ODD INCREASING
SOLUTIONS FOR A CLASS OF INHOMOGENEOUS
ALLEN-CAHN EQUATIONS IN R
C. SOURDIS
Abstract. We prove existence and uniqueness of odd increasing solutions to
a class of inhomogeneous Allen-Cahn type equations. A typical example is
u00 + |x|α (u − u3 ) = 0, α > 0. We use the fact that this is well known when
α = 0 and a continuation argument. Our motivation is the study of layered
solutions for ε2 u00 + h(x)(u − u3 ) = 0, h(x) = |x − x1 |α + o(|x − x1 |α ), x → x1 ,
h > 0, x 6= x1 in a bounded interval with Neumann boundary conditions and
ε > 0 small (see [3]).
1. Introduction and main result
We consider the problem of finding increasing solutions to the problem
 00
α
x∈R

 u + |x| f (u) = 0



u(x) → −1 as x → −∞ u(x) → 1 as x → ∞,
(1.1)





u odd,
where α > 0 and f ∈ C 2 (R) is odd, f (±1) = 0, fu (1) < 0, f (u) > 0, u ∈ (0, 1),
f (δu) ≥ δf (u) ∀u ∈ (0, 1), δ ∈ (0, δ0 ],
(1.2)
for some small δ0 > 0. A typical example is f (u) = u − u|u|p , p > 0.
Remark 1.1. In [3] we did not require (1.2) but fu (0) > 0. Existence and asymptotic stability of a solution were established by topological arguments.
Let u0 be the unique odd solution of
 00
 u + f (u) = 0
(1.3)

u(x) → −1 as x → −∞
x∈R
u(x) → 1 as x → ∞.
u00
We have
> 0.
By odd reflection, (1.1) is equivalent to
 00
 u + xα f (u) = 0 x > 0
(1.4)

u(0) = 0
u(x) → 1 as x → ∞.
Proposition 1.1. There exists a solution u of (1.4) such that u0 (x) > 0, x ≥ 0.
Partly supported by grant FONDECYT 3085026.
1
2
C. SOURDIS
Proof. Note that
(1.5)

 δu0 (x − 1) x ≥ 1
uδ (x) =

0 ≤ x ≤ 1,
0
δ ≥ 0 is a nontrivial subsolution of (1.4) if δ ∈ (0, δ0 ].
Indeed, in [0,1) we have uδ = 0. So,
u00δ + xα f (uδ ) = xα f (0) = 0, x ∈ [0, 1).
In (1, ∞) we have, for small δ ∈ (0, δ0 ],
u00δ + xα f (uδ )
= δu000 (x − 1) + xα f (δu0 (x − 1))
(by (1.2)) ≥ δu000 (x − 1) + δxα f (u0 (x − 1))
≥ δu000 (x − 1) + δf (u0 (x − 1)) = 0.
Moreover, uδ (0) = 0, uδ (∞) = δ < 1. Finally, the function uδ is the maximum of
two regular sub-solutions, and therefore uδ itself must be a subsolution.
Let φε solve
2
x > ε− α

 −φ00ε − fu (1)|x|α φε = −φε

2
φε (ε− α ) = 1 − u0 (ε−
2+α
α
2
) > 0 φ0ε (ε− α ) = −ε−1 u00 (ε−
2+α
α
) < 0.
2
We have φε > 0, φ0ε < 0, φ00ε > 0, x > ε− α , and a standard barrier argument yields
−
0 < φε (x) ≤ Ce
Cε
1
2+α
α
2 − x−ε− α
e
,
2
x > ε− α ,
provided ε > 0 is small (recall that fu (1) < 0 and 1−u0 (x) ≤ Ce−|x|/C for a generic
constant C).
We claim that, for small ε,
vε =

2
 1 − φε (x) x > ε− α

u0
x
ε
2
0 < x ≤ ε− α
is a supersolution of (1.4). Note that vε > uδ if δ ≤ δ0 and ε > 0 is small. We have
2
vε ∈ C 1 (R) and vε (0) = 0, vε (∞) = 1. In (0, ε− α ),
vε00
2
In (ε− α , ∞),
α
+ |x| f (vε ) =
1
α
− 2 + |x| f (vε ) ≤ 0.
ε
ODD INCREASING SOLUTIONS FOR INHOMOGENEOUS ALLEN-CAHN EQUATIONS
vε00 + |x|α f (vε )
3
= −φ00 + |x|α f (1 − φ)
= −φ00 − fu (1)|x|α φ + |x|α O(φ2 )
=
(−1 + |x|α O(|φ|)) φ
≤
1
− 2+α
2 2
− x−ε− α
−1 + C |x − ε− α |α + ε−2 e Cε α e
φ<0
if ε is small.
Hence, there exists a solution u of the differential equation of (1.4) such that
uδ ≤ u ≤ vε , x ≥ 0.
0
It is easy to see that u (x) > 0, x ≥ 0 and, thus, u solves (1.4). The proof of the
proposition is complete.
Remark 1.2. Alternatively, one can get existence using the continuation argument
we will use for uniqueness (starting from α = 0).
Lemma 1.1. Every increasing solution u of (1.4) satisfies
u(x) ≥ δ0 u0 (x − 1),
x ≥ 1.
Proof. We prove the result by using the subsolutions uδ defined in (1.5) and Serrin’s
sweeping technique (cf. [2]).
By the proof of Proposition 1.1, uδ is a subsolution of (1.4) if δ ∈ [0, δ0 ]. Suppose
that u is an increasing solution of (1.4). Then, if δ = 0, u ≥ uδ in (0, ∞). By the
Serrin sweeping principle, and since u(0) = 0, u(∞) = 1 while uδ (0) = 0, uδ (∞) =
δ < 1, u0δ (0) = 0, we see that
u(x) ≥ uδ0 (x) = δ0 u0 (x − 1)+ in (0, ∞).
To prove this, we let δ̃ = sup {δ ∈ [0, δ0 ] : u ≥ uδ in (0, ∞)}, note that u ≥ uδ̃ in
(0, ∞), and apply the maximum principle to u − uδ̃ to deduce that this function has
a positive lower bound in (0, ∞). This contradicts the maximality of δ̃ if δ̃ < δ0 .
The proof of the lemma is complete.
Lemma 1.2. There exists α0 > 0 such that (1.4) has a unique increasing solution
for each 0 < α ≤ α0 .
Proof. The idea is to obtain good asymptotics for the solution and then use this to
prove uniqueness.
Step 1 Suppose that ui are increasing solutions of (1.4) for αi where αi > 0 and
αi → 0 as i → ∞.
We have 0 < ui < 1 in (0, ∞) and by equation
(1.6)
u00i + xαi f (ui ) = 0, x ∈ (0, ∞),
we see that
(1.7)
kui kC 2 [0,L] ≤ C(L), L > 0, i ≥ 1.
Using the Arczela-Ascoli theorem and the standard diagonal argument we obtain
that, for a subsequence,
2
ui → u∗ in Cloc
[0, ∞),
4
C. SOURDIS
for some nondecreasing 0 ≤ u∗ ≤ 1 bounded solution of
u00 + f (u) = 0, x > 0, u(0) = 0,
(we also used that xα → 1 in Cloc [0, ∞) as α → 0).
By Lemma 1.1, we also have u∗ ≥ δ0 u0 (x − 1), x ≥ 1, i.e, u∗ is nontrivial. Hence,
u∗ ≡ u0 , By the uniqueness of the limit, we conclude that the every subsequence of
ui satisfies
(1.8)
2
ui → u0 in Cloc
[0, ∞) as i → ∞.
Step 2 Suppose that the lemma is false and that ui , vi are two distinct increasing
solutions of (1.4) with α = αi and αi → 0 as i → ∞. Let
ui − vi
.
φi =
kui − vi kL∞ (0,∞)
Then φi satisfies
 00
 φi + Vi (x)φi = 0 x ∈ (0, ∞)
(1.9)

φi (0) = 0
kφi kL∞ (0,∞) = 1, φi (∞) = 0,
where
Vi (x) =

f (u )−f (v )
 xαi ui i −vi i
if ui (x) 6= vi (x)
xαi fu (ui )
if ui (x) = vi (x).
which converges in Cloc [0, ∞) to fu (u0 ) by Step 1. Using once more the ArczelaAscoli theorem and the standard diagonal argument we see that, for a subsequence,

(1.10)
2
φi → φ∗ in Cloc
[0, ∞) as i → ∞,
where φ∗ is bounded in (0, ∞) and solves
φ00 + fu (u0 )φ = 0, x > 0, u(0) = 0,
Hence,
(1.11)
φ∗ ≡ 0.
(To see this note that the boundedness of φ and fu (1) < 0 imply that φ∗ ∈ H 2 (0, ∞).
So, its odd reflection is in the kernel of the operator Bφ = φ00 +fu (u0 )φ, φ ∈ H 2 (R).
Thus, φ∗ = λu00 , and by setting x = 0 we get λ = 0).
There exist c, d > 0 such that fu (u) < −c if |u − 1| ≤ d. Moreover, there is
an L0 > 0 such that 0 < 1 − u0 (x) < d if x ≥ L0 . Since ui , vi are increasing and
converge to u0 uniformly in compact intervals, we get that 1 − d < ui (x), vi (x) <
1, x ≥ L0 if i is sufficiently large. By the mean value theorem,
Vi (x) = xαi fu (θui + (1 − θ)vi ) ,
θ ∈ [0, 1].
Hence,
(1.12)
Vi (x) ≤ −cxαi ,
x ≥ L0 .
There exist xi > 0 such that, without loss of generality, φi (xi ) = 1, φ0i (xi ) =
0, φ00i (xi ) ≤ 0, i ≥ 1. We claim that the sequence {xi } is bounded. Indeed, if not,
by (1.9), (1.12) we find that φ00i (xi ) → ∞ as i → ∞, a contradiction.
By passing to a subsequence we may assume that xi → x∗ ≥ 0. In view of (1.10),
we get
φ∗ (x∗ ) = 1,
which contradicts (1.11). The proof of the lemma is complete.
ODD INCREASING SOLUTIONS FOR INHOMOGENEOUS ALLEN-CAHN EQUATIONS
5
Lemma 1.3. If u is an increasing solution of (1.1) then u is nondegenerate. In
particular it is asymptotically stable, i.e, if
−ψ100 − |x|α fu (u)ψ1 = µ1 ψ1 in R,
(1.13)
ψ1 ∈ L2 (R), ψ1 > 0 in R, then µ1 > 0.
Proof. We will show that the spectrum, in L2 (R), of the operator defined by the
left-hand side of (1.13) is strictly positive. Since −|x|α fu (u) → ∞ as |x| → ∞, its
spectrum consists of discrete simple eigenvalues µ1 < µ2 < · · · with µi → ∞ as
i → ∞ and corresponding eigenfunctions ψ1 , ψ2 , · · · . Each ψi , i ≥ 1 has exactly
i − 1 nodes. Without loss of generality we can assume that ψ1 > 0 is an even
function of x. It is standard to show that 1 − u, ψ1 tend to 0 super-exponentially
as |x| → ∞ (see [3]). Note that w = u0 > 0 solves
−w00 − |x|α fu (u)w = α|x|α−2 xf (u),
(1.14)
x ∈ R.
By multiplying (1.13) with w, (1.14) with ψ1 and subtracting, we find that
Z ∞
Z ∞
Z ∞
µ1
ψ1 wdx = α
|x|α−2 xf (u)ψ1 dx = 2α
xα−1 f (u)ψ1 dx > 0.
−∞
−∞
0
The proof is complete.
Lemma 1.4. If U is an increasing solution of (1.1)A (this means α = A) then there
exists a δ > 0 and an increasing solution uα of (1.1)α for |α − A| ≤ δ. Moreover,
(1.15)
kuα − uA kL∞ (0,∞) → 0
as α → A.
Proof. We seek a solution u of (1.1)α in the form u = uA + ϕ, ϕ ∈ H 2 (R) odd. In
terms of ϕ, (1.1)α becomes
L(ϕ) = |x|α N (ϕ) + E,
where
L(ϕ) = −ϕ00 − |x|α fu (U )ϕ
N (ϕ) = f (U + ϕ) − f (U ) − fu (U )ϕ
E = U 00 + |x|α f (U ).
We introduce the norm kϕkw = ke|x| ϕkL∞ (R) and the Banach space
X = {ϕ : ϕ is odd, kϕkw < ∞}.
Arguing as in Lemma 1.2 we see that L is nonsingular if α is close to A. By rather
standard arguments we see that if f ∈ X then there exists a unique ϕ ∈ H 2 (R) ∩ X
such that L(ϕ) = f . Moreover kϕkw ≤ Ckf kw (C is a generic constant). Note that
(if kϕkL∞ ≤ 1) we have
|x|α |N (ϕ)| ≤ C|x|α ϕ2 ≤ C|x|α e−2|x| kϕk2w
1+α/2
|E| = |x|α − |x|A f (U ) ≤ C |x|α − |x|A e−|x|
,
(1 − U decays super-exponentially). Now the existence of a solution u of (1.1)α with
|α − A| small satisfying (1.15) follows from Banach’s fixed point theorem. We can
show that |u0α (x) − U 0 (x)| → 0 as α → A. It follows that u ≥ 0 in (0, ∞) and, thus,
u0 > 0 in R provided α is close to A.
6
C. SOURDIS
Lemma 1.5. If uα are increasing solutions of (1.1)α , α > α∗ then there exists
δ > 0 and increasing solutions uα of (1.1)α for α∗ − δ ≤ α ≤ α∗ . Moreover uα are
continuous functions of [α∗ − δ, α] to L∞ (R) (assumming that they are for α > α∗ ).
Proof. We know that there exists a sequence of increasing solutions ui of (1.1) with
αi → α∗ as i → ∞ and αi > α∗ . Since |ui | ≤ 1, we can employ Arczela-Ascoli’s
theorem and the standard diagonal argument again to show that there exists a
nondecreasing solution u∗ of the equation of (1.1) with α = α∗ . By Lemma 1.1 we
get that u∗ is nontrivial and, thus, is an increasing solution of (1.1). By the Cloc
convergence and the fact that ua are increasing we have that the convergence is
uniform in R. This establishes continuity from [α∗ , α] to L∞ (R). By Lemma 1.4 we
obtain increasing solutions for α∗ − δ ≤ α ≤ α∗ . The continuity assertion follows by
substracting two different equations and working as in the proof of Lemma 1.4. Theorem 1.1. If α > 0 then there exists a unique increasing solution of (1.1).
Proof. Existence was established in Proposition 1.1. For the uniqueness we will
argue by contradiction. Suppose that u1 , u2 are two distinct increasing solutions
of (1.1)α0 for some α0 > 0. By Lemmas 1.4, 1.5 we can construct increasing
solutions ui (α), i = 1, 2 of (1.1)α for 0 < α < α0 such that ui : (0, α0 ] → L∞ (R)
is continuous. Lemma 1.2 implies that there exists an α1 such that u1 (α1 ) =
u2 (α1 ). The continuity of the branches yields that the linearization of (1.1) at
u1 (α1 ) is singular (as in 1.2). On the other hand, since u1 (α1 ) is increasing, we get
a contradiction by Lemma 1.3. Therefore, the proof of Theorem 1.1 is complete. Remark 1.3. Our approach for uniqueness is motivated from [1].
Remark 1.4. If instead of (1.2) we had the stronger
f (δu) ≥ δf (u),
u ∈ [0, 1], δ ∈ [0, 1],
p
as in the case f (u) = u − u|u| , p > 0, the uniqueness proof is very simple. Suppose
that u, v are two increasing solutions of (1.4). Then δu, δ ∈ [0, 1] is a family of
subsolutions and δu ≤ v in (0, ∞) if δ = 0. The Serrin sweeping principle yields
u ≤ v in (0, ∞). Similarly, v ≤ u in (0, ∞).
References
[1] C.S. LIN, Uniqueness of least energy solutions to a semilinear elliptic equation in R2 ,
Manuscripta Math 84 (1994) 13-19.
[2] D. SATTINGER, Topics in stability and bifurcation theory, Springer Lectures Notes in Mathematics 309 (1973).
[3] C. SOURDIS, Transition layer solutions in a critical case, preprint (2008).
C. Sourdis, Departamento de Ingenieria Matematica, Universidad de Chile, Santiago,
Chile
E-mail address: [email protected]