Electronic Journal of Differential Equations, Vol. 2004(2004), No. 113, pp. 1–9.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu (login: ftp)
THE CAUCHY PROBLEM AND STEADY STATE SOLUTIONS
FOR A NONLOCAL CAHN-HILLIARD EQUATION
JIANLONG HAN
Abstract. We study the existence, uniqueness, and continuous dependence
on initial data of the solution to the Cauchy problem and steady state solutions
of a nonlocal Cahn-Hilliard equation on a bounded domain.
1. Introduction
We are concerned with two different problems, the first being the Cauchy problem
for a nonlocal Cahn-Hilliard equation
∂u
= 4(ϕ(u) − J ∗ u) in Rn × (0, T ),
∂t
(1.1)
u(x, 0) = u0 (x),
where ϕ(u) = u +Rf (u), f is bistable (e.g. f (u) = au(u2 − 1) for some a > 0), ∗ is
convolution, and Rn J = 1.
The second problem is for the steady state equation
Z
Z
J(x − y)dyu(x) −
J(x − y)u(y)dy + f (u) = C in Ω,
Ω
ZΩ
(1.2)
u(x)dx = 0,
Ω
where Ω is a bounded domain, C is a constant. The case when Ω = R or Rn has
been treated in [3, 5, 10, 11] and references therein.
To derive equations (1.1) and (1.2), we consider the free energy
ZZ
Z
E(u) = C
J(x − y)(u(x) − u(y))2 dxdy + F (u(x))dx,
(1.3)
where C is a constant, F is the primitive of f , and u represents the concentration
of one of the species of a binary material.
Following [16], we consider the gradient flow for (1.3) in H0−1 (Ω), where H0−1 is
the space of distributions in the dual space of H 1 and with mean value zero. We
do this since the total energy, E, decreases along the trajectories, and the average
of u should be conserved. We have
ut = − grad H −1 E(u).
0
2000 Mathematics Subject Classification. 35A05, 35M99.
Key words and phrases. Phase transition; long range interaction; steady state solution.
c
2004
Texas State University - San Marcos.
Submitted July 18, 2004. Published September 29, 2004.
1
(1.4)
2
JIANLONG HAN
EJDE-2004/113
Since the representative of grad E(u) in H0−1 is
Z
grad H −1 E(u) = −4(J(x − y)dyu(x) −
0
J(x − y)u(y)dy + f (u)),
(1.4) gives
Z
Z
∂u
= 4( J(x − y)dyu(x) −
J(x − y)u(y)dy + f (u)).
∂t
Ω
Ω
(1.5)
In (1.3), making the approximation
u(x) − u(y) ' ∇u(x) · (x − y),
and assuming J to be isotropic, equation (1.4) leads to
∂u
= −4(d4u − f (u)),
(1.6)
∂t
which is the classical Cahn-Hilliard equation.
Equations (1.5) and (1.6) are important in the study of materials science for
modelling certain phenomena such as spinodal decomposition, Ostwald ripening,
and grain boundary motion.
There is a lot of work on equation (1.6) (see for example [1, 2, 4, 9, 12, 13, 15,
14, 20] and references therein). For equation (1.5), there are very few results. In
[6] and [7], we discussed the Neumann and Dirichlet boundaryR problems for (1.5).
Here, we consider the Cauchy problem, where Ω = Rn and J = 1. Note that
the steady state solutions for (1.5) in a bounded domain with no flux boundary
condition satisfy the equation in (1.2) without the constraint.
In this paper, we prove the global existence and uniqueness of solutions for
equation (1.1). Also we prove the existence of nonconstant solutions for equation
(1.2). The techniques used in the proof of the latter result can also be applied to
the nonlocal phase field system discussed in [8].
We organize this paper as follows. In section 2, we establish the existence,
uniqueness and continuous dependence on initial values for classical solutions of
equation (1.1). In section 3, we prove that under certain conditions, there exists a
discontinuous steady state solution for equation (1.2).
2. The Cauchy problem for the nonlocal Cahn-Hilliard equation
For T > 0, let QT = Rn × (0, T ). We make the following assumptions:
0
(D1) f ∈ C 2+β (R) and ϕ (u) ≥ c for some positive Rconstants c and β,
(D2) J ∈ C 2+β (Rn ), 4J ∈ L1 (Rn ) ∩ L∞ (Rn ), and Rn J = 1.
First, we prove the uniqueness and continuous dependence of solutions on initial
data. We have
Proposition 2.1. Let ui (i = 1, 2) be two solutions of (1.1) with initial data ui0
(i = 1, 2). If conditions (D1)–(D2) are satisfied, if ui ∈ C([0, T ], L1 (Rn ))∩L∞ (QT ),
and if ui0 ∈ L1 (Rn ) ∩ L∞ (Rn ) (i = 1, 2), then
Z
Z
sup
|u1 − u2 |dx ≤ C(T )
|u10 − u20 |dx
(2.1)
0≤t≤T
for some constant C(T ).
Rn
Rn
EJDE-2004/113
THE CAUCHY PROBLEM
3
Proof. For any τ ∈ (0, T ), and ψ ∈ C 2,1 (Qτ ), with ψ = 0 for |x| large enough, after
multiplying (1.1) by ψ, integrating over [0, τ ] × Rn , we have
Z
ui (x, τ )ψ(x, τ )dx
Rn
Z
Z τZ
Z τZ
=
ui (x, 0)ψ(x, 0)dx +
(ui ψt + ϕ(ui )4ψ)dx dt −
ψ4J ∗ ui dx dt.
Rn
Rn
0
0
Rn
Set z = u1 − u2 , z0 = u10 − u20 , then the above equality gives
Z
Z
Z τZ
z(x, τ )ψ(x, τ )dx =
z0 (x)ψ(x, 0)dx +
z(x, t)(ψt + b(x, t)4ψ)dx dt
Rn
Rn
0
Rn
Z τZ
−
ψ4J ∗ z(x, t)dx dt,
Rn
0
(2.2)
where
(
ϕ(u1 )−ϕ(u2 )
u1 −u2
ϕ0 (u1 )
b(x, t) =
for u1 6= u2 ,
for u1 = u2 .
(2.3)
Let g(x) ∈ C0∞ (Rn ) have compact support, 0 ≤ g(x) ≤ 1, and take λ > 0.
We will choose ψ, above, to satisfy certain conditions. First, consider the following final value problem on a large ball BR (0)
∂ψ
= −b(x, t)4ψ + λψ for |x| < R, 0 < t < τ
∂t
ψ = 0 on |x| = R, 0 < t < τ
(2.4)
ψ(x, τ ) = g(x) |x| ≤ R.
There exists a unique solution ψ ∈ C 2,1 (BR (0) × (0, τ )) of of (2.4) which satisfies
the following properties:
0 ≤ ψ ≤ eλ(t−τ ) ,
Z
0
τ Z
b(x, t)|4ψ|2 dx dt ≤ C,
(2.5)
(2.6)
BR (0)
Z
|∇ψ|2 dx ≤ C,
sup
0≤t≤τ
(2.7)
BR (0)
where the constant C depends only on g. To extend ψ to be zero outside of BR (0),
we define ξR ∈ C0∞ (Rn ) such that
0 ≤ ξR ≤ 1,
if |x| < R − 1,
1
ξR = 0 if |x| > R − ,
2
|∇ξR (x)|, |4ξR (x)| ≤ C
ξR = 1
for some constant C which does not depend on R.
(2.8)
4
JIANLONG HAN
EJDE-2004/113
Let γ = ξR ψ, where ψ satisfies (2.4) in BR (0) and is zero outside. Using γ
instead of ψ in (2.2), we have
Z
Z
Z Z
z(x, τ )gξR dx −
ξR (x)z0 (x)ψ(x, 0)dx +
(4J ∗ z − λz)ξR ψdx dt
Rn
Rn
Qτ
Z Z
=
b(x, t)z(x, t)(2∇ξR · ∇ψ + ψ4ξR )dx dt ≡ G(z, R).
Qτ
Since u1 and u2 belong to L∞ (QT ), and since b is positive, from estimates (2.5)-(2.7)
and (2.8), we have
Z τZ
|G(z, R)| ≤
(b|u1 − u2 |((2|∇ξR k∇ψ| + |ψk4ξR |))
BR \BR−1
τ Z
0
Z
≤C
b(|u1 | + |u2 |)(|∇ψ| + 1)dx dt
(2.9)
BR \BR−1
0
τ Z
Z
≤C
(|u1 | + |u2 |)dx dt.
BR \BR−1
0
Since u1 and u2 belong to L1 (QT ), letting R → ∞ we have G(z, R) → 0. This
implies
Z
Z
Z τZ
z(x, τ )g(x)dx ≤
|z0 (x)|e−λτ dx +
(|4J ∗ z − λz|eλ(t−τ ) dx dt.
Rn
Rn
Rn
0
+
Letting λ → 0 and g(x) → sign z (x, τ ), we obtain
Z
+
Z
Z
(u1 − u2 ) dx ≤
Rn
τ
Z
|u10 − u20 |dx + C
Rn
|u1 − u2 |dx dt.
Interchanging u1 and u2 yields
Z
Z
Z
|u1 − u2 |dx ≤
|u10 − u20 |dx + C
Rn
Rn
(2.10)
Rn
0
τ
0
Z
|u1 − u2 |dx dt.
Rn
Inequality (2.1) follows from the above inequality and Gronwall’s inequality.
Next we prove the existence of a solution to (1.1).
Theorem 2.2. For any T > 0, if u0 (x) ∈ C02+β (Rn ), and if ϕ and J satisfy
assumptions (D1 ) − (D2 ), then there exists a unique solution of (1.1) which belongs
2+β
to C 2+β, 2 (QT ) ∩ L1 (QT ) ∩ L∞ (QT ).
Proof. Since u0 (x) = 0 for |x| large enough, we consider
∂u
= 4(ϕ(u) − J ∗ u) in BR (0) × (0, T ),
∂t
u(x, t) = 0 on ∂BR (0) × (0, T ),
(2.11)
u(x, 0) = u0 (x).
From [7, Theorem 2.4], there exists a unique solution u(x, t) ∈ C 2+β,
(0, T )) of (2.11). Let u(x, t) = vet in (2.11), we have
et vt + vet = ϕ0 (u)et 4v + ϕ00 (u)|∇v|2 e2t − et 4J ∗ v.
2+β
2
(BR (0) ×
(2.12)
EJDE-2004/113
THE CAUCHY PROBLEM
5
Multiplying (2.12) by v and using v4v = 12 4v 2 − |∇v|2 , we obtain
1 2
1
1
(v )t + v 2 = ϕ0 (u)4v 2 + ϕ00 (u)∇v · ∇v 2 et − ϕ0 (u)|∇v|2 − v4(J ∗ v). (2.13)
2
2
2
If there exists (P0 , t0 ) ∈ BR (0) × (0, T ] such that v 2 (P0 , t0 ) = max v 2 , then we have
4v 2 (P0 , t0 ) ≤ 0, ∇v 2 (P0 , t0 ) = 0, ∇v(P0 , t0 ) = 0, vt2 (P0 , t0 ) ≥ 0, and (2.13) yields
Z
2
v (P0 , t0 ) ≤ −
4J(P0 − y)v(y, t0 )dyv(P0 , t0 ).
(2.14)
BR
This yields
Z
max |v| ≤ M
|v(y, t0 )|dy
(2.15)
BR
for some constant M which does not depend on R.
Since u = 0 is also a solution of (2.11) with initial data u0 = 0, by [7, Theorem
2.5], we have
Z
Z
|u(x, t) − 0|dx ≤ C(T )
BR
|u0 − 0|dx
(2.16)
BR
for some constant C(T ) which does not depend on R. Inequalities (2.15) and (2.16)
imply
Z
max |v| ≤ C(T )
|u0 |dx.
(2.17)
BR
1
n
Since u0 ∈ L (R ), we have
max |v| ≤ B(T )
(2.18)
for some constant B(T ) which does not depend on R. This yields
max |u| ≤ B(T )eT
(2.19)
for some constant B(T ) which does not depend on R. We have proved the solution
of (2.11) is uniformly bounded, i.e.,
max |u(x, t)| ≤ C
BR ×[0,T ]
for any R > 0, where C does not depend on R. A similar argument to that in the
proof in [7, Theorem 2.2] yields
kuR k2+β ≤ C(K, T )
(2.20)
for any R > K ≡ constant, where uR is a solution of (2.11) in BR × (0, T ) and
C(K, T ) is a constant which does not depend on R (k · k2+β is a H ölder norm
defined in [19]).
By employing the usual diagonal process, we can choose a sequence {Ri } such
that uRi , DuRi , and D2 uRi converge to u, Du, and D2 u pointwise, and u satisfies
equation (1.1). From (2.16) and (2.19), we also have u ∈ L1 (QT ) ∩ L∞ (QT ).
Uniqueness follows from Proposition 2.1.
6
JIANLONG HAN
EJDE-2004/113
3. Steady state solutions for the nonlocal Cahn-Hilliard equation
In this part, we study equation (1.2).
Proposition 3.1. Suppose Ω ⊂ Rn is a closed and bounded set, J(x) ≥ 0 and is
continuous on Rn , suppJ ⊃ Bδ (0) for some positive constant δ, and f is nondecreasing. Then the only continuous solution of equation (1.2) is zero.
Proof. Without loss of generality, we assume that f (0) = 0. If f (0) 6= 0, we may
use f (u) − f (0) instead of f (u) in (1.2).
R
Case 1: C ≤ 0 in equation (1.2). If the conclusion is not true, since udx = 0, and
u is continuous on Ω, there exists P0 ∈ Ω such that u(P0 ) = max u(x) > 0. Let
A = {y ∈ Ω : u(y) = max u(x)}.
We claim: There exist P0 ∈ ∂A and r > 0 such that K := (Ω \ A) ∩ Br (P0 )
has positive measure. If this
is not
R
R true, we have meas(ΩR\ A) = 0. This and
u(x) = max u on A imply Ω u = A u > 0. This contradicts Ω u = 0.
Since supp J ⊃ Bδ (0) implies supp J(P0 − ·) ⊃ Bδ (P0 ), choosing r1 = min{δ, r}
gives
meas(K ∩ Br1 (P0 )) > 0,
J(P0 − y) > 0
u(P0 ) − u(y) > 0
on K ∩ Br1 (P0 ),
on K ∩ Br1 (P0 ).
Inequalities (3.1)-(3.3) imply
Z
Z
J(P0 − y)(u(P0 ) − u(y))dy ≥
Ω
(3.1)
(3.2)
(3.3)
J(P0 − y)(u(P0 ) − u(y))dy > 0.
K∩Br1 (P0 )
This and f (u(P0 )) ≥ 0 imply
Z
Z
J(P0 − y)u(P0 )dy −
J(P0 − y)u(y)dy + f (u(P0 )) > 0,
Ω
(3.4)
Ω
contradicting (1.2).
Case 2: C > 0 in (1.2). In this case, taking P0 such that u(P0 ) = min u < 0 leads
to a contradiction in a similar way.
If f 0 (u) changes sign, we make the following assumptions:
(E1) Ω = (−1, 1) if dim Ω = 1, Ω = (−1, 1) × Ω0 if dim Ω > 1.
(E2) J(x) = J(|x|), J(x) ≥ 0, and
Z
Z
M ≥ sup
J(x − y)dy ≥ inf
J(x − y)dy ≥ m > 0
x∈Ω
Ω
x∈Ω
Ω
for positive constants M and m.
(E3) f ∈ C 1 (R) is odd, f (1) = 0, there exist δ > 0 and a ∈ (0, 1) such that
f 0 (u) ≥ δ on [a, ∞), and f (−a) ≥ (1 + a)M .
(E4) C = 0 in (1.2).
Remark 3.2. Condition (E3) implies that f (−1) = 0, f 0 (u) ≥ δ on (−∞, −a], and
−f (a) ≥ (1 + a)M .
R
Let j(x) = Ω J(x − y)dy. From (E2 ), we have
m ≤ j(x) ≤ M.
(3.5)
EJDE-2004/113
THE CAUCHY PROBLEM
Dividing (1.2) by j(x), we consider
Z
1
f (u(x))
J(x − y)u(y)dy +
= 0,
u(x) −
j(x) Ω
j(x)
Z
u(x)dx = 0.
7
(3.6)
Ω
Theorem 3.3. If assumptions (E1 )−(E4 ) are satisfied, then there exists a solution
of equation (3.6) such that
(
≥a
for x ∈ M1 ≡ (0, 1) × Ω0 ,
u(x)
(3.7)
≤ −a for x ∈ M2 ≡ (−1, 0) × Ω0 .
Moreover, we have
−1 ≤ u(x) ≤ 1.
(3.8)
Proof. Following [3], we let
B = {u ∈ L∞ (Ω) : u(−x1 , x0 ) = −u(x1 , x0 ), u(x) ∈ [a, 1] for x ∈ M1 }.
The definition of B implies that u(x) ∈ [−1, −a] for x ∈ M2 . Define
Z
1
1
T u(x) = u(x) + h[
J(x − y)u(y)dy − u(x) −
f (u(x))].
j(x) Ω
j(x)
We want to show
T : B → B is a contraction map if h is small enough. In fact,
R
since j(x) = Ω J(x − y)dy, with assumption (E2), we have j(−x1 , x0 ) = j(x1 , x0 ).
And if u(x) ∈ B, we have
T (u(−x1 , x0 ))
Z 1Z
h
J(−x1 − y1 , x0 − y 0 )u(y1 , y 0 )dy1 dy 0
= u(−x1 , x ) +
j(−x1 , x0 ) −1 Ω0
h
− hu(−x1 , x0 ) +
f (u(−x1 , x0 ))
j(−x1 , x0 )
Z 1Z
h
0
= −u(x1 , x ) −
J(−x1 + z1 , x0 − y 0 )u(z1 , y 0 )dz1 dy 0
j(x1 , x0 ) −1 Ω0
h
f (u(x1 , x0 ))
+ hu(x1 , x0 ) −
j(x1 , x0 )
Z 1Z
h
0
= −u(x1 , x ) −
J(x1 − z1 , x0 − y 0 )u(z1 , y 0 )dz1 dy 0
j(x1 , x0 ) −1 Ω0
h
f (u(x1 , x0 ))
+ hu(x1 , x0 ) −
j(x1 , x0 )
Z 1Z
h
= −(u(x1 , x0 ) +
J(x1 − z1 , x0 − y 0 )u(z1 , y 0 )dz1 dy 0
j(x1 , x0 ) −1 Ω0
h
− hu(x1 , x0 ) +
f (u(x1 , x0 )))
j(x1 , x0 )
= −T (u(x1 , x0 )).
0
(3.9)
Choose h small enough such that
h
1 0
f (u) < 1 − h
j(x)
(3.10)
8
JIANLONG HAN
EJDE-2004/113
for u ∈ [−1, −a] ∪ [a, 1] and x ∈ Ω.
1
This implies that u − h[u + j(x)
f (u)] is increasing in u on [a, 1]. Since u(y) ≥ a
for y ∈ M1 , and u(y) ≥ −1 for y ∈ M2 , we have for x ∈ M1
T u(x)
Z
1
1
=h
J(x − y)u(y)dy + u − h[u +
f (u)]
j(x) Ω
j(x)
Z
1
1
≥h
J(x − y)u(y)dy + a − ha − h
f (a)
j(x) Ω
j(x)
Z
Z
1
1
1
=h
J(x − y)u(y)dy + h
J(x − y)u(y)dy + a − ha − h
f (a)
j(x) M1
j(x) M2
j(x)
Z
Z
1
1
1
≥ ha
J(x − y)dy − h
J(x − y)dy + a − ha − h
f (a)
j(x) M1
j(x) M2
j(x)
Z
Z
1
1
h
= a − ha
J(x − y)dy − h
J(x − y)dy −
f (a)
j(x) M2
j(x) M2
j(x)
Z
h
≥a−
[(1 + a)
J(x − y)dy + f (a)] ≥ a
j(x)
M2
by (E3). Also,
Z
1
1
f (u)]
J(x − y)u(y)dy + u − h[u +
j(x) Ω
j(x)
Z
1
1
≤h
J(x − y)u(y)dy + 1 − h − h
f (1) ≤ 1
j(x) Ω
j(x)
T u(x) = h
(3.11)
for x ∈ M1 . Estimates (3.9)-(3.11) imply that T maps B to B.
1
) < 1, we have
For u, v ∈ B, choosing h small enough so that 0 < 1 − h(1 + δ M
kT u − T vk∞
= k(u − v) +
h
j(x)
Z
J(x − y)(u(y) − v(y))dy
Ω
h
(f (u) − f (v))k∞
j(x)
Z
hf 0 (θu + (1 − θ)v)
h
= k(1 − h −
)(u − v) +
J(x − y)(u(y) − v(y))dyk∞
j(x)
j(x) Ω
1
≤ (1 − h(1 + δ ))ku − vk∞ + hk(u − v)k∞
M
1
≤ (1 − hδ )ku − vk∞ ,
M
− h(u(x) − v(x)) −
where θ(x) ∈ (0, 1) for all x ∈ Ω. Here we used (E3) and the fact that for any x ∈ Ω
either u(x), v(x) ≥ a or u(x), v(x) ≤ −a.
Therefore, T is a contraction map from B to B. There exists a unique fixed
point u(x) such that T u = u. Estimates (3.8) follows from the definition of B. Remark 3.4. If we just consider the solution to
Z
Z
J(x − y)u(x)dy −
J(x − y)u(y)dy + f (u) = 0
Ω
Ω
in Ω
(3.12)
EJDE-2004/113
THE CAUCHY PROBLEM
9
R
without the condition Ω udx = 0, then the conditions that f is odd and J(x) =
J(|x|) are not necessary. In this case, we can use a similar method to that in [3] to
prove the existence of a discontinuous solution under conditions (E2), (E3)’, and
(E4), where
(E3)’ f ∈ C 1 (R), f (−1) = f (c) = f (1) = 0 for c ∈ (−1, 1), there exist δ > 0,
a ∈ (0, 1), b ∈ (−1, 0) such that f 0 (x) ≥ δ on [a, ∞) ∪ (−∞, b), f (a) ≤
−(1 + a)M , and f (b) ≥ (1 + b)M , where M is defined in (E2).
Acknowledgement. The author would like to thank Professor Peter W. Bates for
his helpful discussions.
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Jianlong Han
Department of Mathematics, Michigan State University, East Lansing, MI 48824, USA
E-mail address: hanjianl@math.msu.edu