THE FOURIER SPECTRAL METHOD FOR THE CAHN-HILLIARD EQUATION
1Ye Xingde Cheng Xiaoliang
Department of Mathematics, Zhejiang University Hangzhou , 310028, P. R. China
[email protected] [email protected]
Abstract In this paper, a Fourier spectral method for numerically solving Cahn-Hilliard equation with periodic boundary conditions is developed. We establish their semi-discrete and fully discrete schemes that inherit the energy dissipation property and mass conservation property from the associated continuous problem. we prove existence and uniqueness of the numerical solution and derive the optimal error bounds.
we perform some numerical experiments which confirm our results.
Keywords: Cahn-Hilliard Equation, Spectral Method.
1 Introduction
The purpose of this paper is to develop a Fourier spectral method for numerically solving Cahn-Hilliard (C-H) equation with periodic boundary conditions. We consider the following nonlinear evolution equation in one space dimension, for real u,
ut+ γD4u = D2φ(u), in x ∈ Ω ≡ (0, 2π), t > 0 (1.1) where
φ(u) = r2u3+ r1u2+ r0u, r2> 0 (1.2) subject to the boundary condition which can be of either periodic or Neumann type:
u(x + 2π, t) = u(x, t), x ∈ Ω, t > 0 (1.3)
Du = γD3u − Dφ(u) = 0, x = 0, 2π (1.4)
and initial condition
u(x, 0) = u0(x), x ∈ Ω (1.5)
Here, D ≡ ∂x∂ , ut≡ ∂u∂t, γ is a prescribed positive constant and r0, r1, r2 are given constants. This problem arises in the study of phase separation in binary alloys(see [1, 2, 3, 4]).
The study of the problem (1.1)-(1.5) can be found in many references (cf.[5, 6, 7]). The existence of a solution locally in time is proved by the standard Picard iteration. Global existence results are obtained by proving a priori estimate for the appropriate norms of u. Adjusted to our needs, the results can be given in the following form:
1This work was supported by National Science Foundation of China(No. 10471129).
Theorem 1.1 Suppose u0 ∈ L2(Ω),the problem (1.1)-(1.5) possesses a unique solution u defined for all t ≥ 0, such that
u ∈ C(R+, L2(Ω)) ∩ L2(0, T ; H2(Ω)) ∩ L4(0, T ; L4(Ω)), ∀T > 0.
if u0∈ Hb2(Ω) ≡ {u ∈ H2(Ω);the considered boundary conditions hold}, then u ∈ C(R+, Hb2(Ω)) ∩ L2(0, T ; H4(Ω)), ∀T > 0.
if u0 ∈ Hk(Ω), k ≥ 1, then u ∈ L∞(R+, Hk(Ω)). For the case of periodic boundary condition (1.3),if u0∈ C∞(Ω), then u(x, t) ∈ C∞(Ω × R+).
In this paper, we only consider the case of periodic boundary condition (1.3). The case of Neumann boundary condition (1.4) can be found in [8, 9, 10].
The important features of C-H equations are as follows.
(i) the extended Ginzberg-Landau free energy functional F(u) =
Z
Ω
[γ
2(Du)2+ ψ(u)]dx (1.6)
where
ψ(u) = r2
4 u4+r1
3u3+r0
2 u2, ψ0(u) = φ(u), (1.7)
is a Lyapunov functional (see [5, 7, 11, 12]), dtdF(u) ≤ 0, i.e. the energy is dissipative as time passes.
(ii) The total mass remains constant 1
|Ω|
Z
Ω
u(x, t)dx = 1
|Ω|
Z
Ω
u0dx = M t > 0. (1.8)
These properties play an important role both in C-H’s mathematical theoretical analysis and its numerical analysis. They are used to estimate the absolute pointwise maximum value of the solution ,that is ,for continuous problem (1.1)-(1.5),
kuk∞≤ c, t > 0. (1.9)
Numerical methods for C-H equations can be found in many references (see [5, 11, 12, 13, 14, 15, 16, 17]
and the references cited therein). From these references, we know that it is the key of the numerical analysis of C-H equations to construct discrete schemes that possess a Lyapunov functional and remain mass constant.
For example, in [5, 11], the problem (1.1) (1.2) (1.4) (1.5) is discretized using conforming finite schemes with an implicit time discretization. In their fully discrete scheme, no Lyapunov function is known at present, the pointwise estimate of the numerical solution can not be obtained, so such a hypothesis is required to avoid this difficulty. In [15], semi-discrete schemes which can define a Lyapunov functional and remains mass constant are used for a mixed formulation of the governing equation. In [13], a mixed finite element formulation with an implicit time discretization was presented for the C-H equation (1.1) with Dirichlet boundary conditions.
Both their semi-discrete schemes and fully discrete schemes have the Lyapunov functional. In [17], a finite difference scheme that inherit the energy dissipation property and the mass conservation property was used.
Compared to large amount of studies in finite element methods and finite difference methods, there are no numerical results on C-H equations by spectral methods to our knowledge. But spectral methods have the natural advantage in keeping the physical properties of primitive problems. In [8], we considered a Fourier collocation spectral method for periodic C-H equations. In this paper, we develop a Fourier Galerkin method for C-H equation (1.1) (1.2) (1.3) (1.5) . We establish their semi-discrete and fully discrete schemes that preserve the property (i)(ii), and give their numerical analysis. The layout of the paper is as follows: In Section 2,we consider a semi-discrete Fourier spectral approximation; prove its existence and uniqueness of the numerical solution and derive the error bound. In Section 3, we consider a fully discrete implicit scheme and prove existence and uniqueness of the numerical solution under the condition ∆tN2 < β, where ∆t is time step, N is the dimension of approximation space, β is a constant. Further, we prove convergence to the solution of the associated continuous problem. In Section 4, we perform some numerical experiments which confirm our results.
2 Semi-discrete Approximation
Let Ω = (0, 2π), L2(Ω) denote the set of all square integrable functions with the inner product (u, v) = R
Ωu(x)v(x)dx and the norm kuk20= (u, u). Let L∞(Ω) denote the Lebesgue space with the norm kuk∞ = ess supx∈Ω|u(x)| and Hpm(Ω) denote the periodic Sobolev space with the norm kukm= (P
|α|≤ m|Dαuk20)12. Denote
L2(0, T ; Hpm(Ω)) = {u(x, t) ∈ Hpm(Ω);
Z T
0
kuk2mdt < +∞}
L∞(0, T ; Hpm(Ω)) = {u(x, t) ∈ Hpm(Ω); sup
0≤t≤T
kuk2mdt < +∞}
Denote SN = span{eikx; −N/2 ≤ k ≤ N/2}, let PN : L2(Ω) → SN be an orthogonal projecting operator :
(u − PNu, φ) = 0, ∀φ ∈ SN (2.1)
For operator PN and functions in SN, the following results hold: (cf.[19, 18]) 1) PN commutes with derivation on Hp1(Ω), i.e.
PNdu = DPNu ∀u ∈ Hp1(Ω) (2.2)
2) For any real 0 ≤ µ ≤ σ, there exists a constant c, such that
ku − PNukµ ≤ cNµ−σ|u|σ ∀u ∈ Hpσ(Ω) (2.3)
3) (inverse inequality) For any real 1 ≤ p ≤ q ≤ ∞,
kukLq(Ω)≤ cN1p−q1kukLp(Ω) ∀u ∈ SN (2.4)
For all real p, 1 ≤ p ≤ ∞ and for all integer r ≥ 1,
ku(r)kLp(Ω)≤ NrkukLp(Ω) ∀u ∈ SN (2.5) where u(r) denotes the derivation of order r of u.
Now we construct an approximation solution for periodic initial value problem (1.1) (1.2) (1.3) (1.5) as following: Find uN(t) ∈ SN, such that
(uNt + γD4uN − D2φ(uN), vN) = 0 ∀vN ∈ SN (2.6)
uN(0) = PNu0 (2.7)
From (2.1)(2.2), we know (2.6) is equivalent to (2.8):
uNt + γD4uN− D2PNφ(uN) = 0 (2.8)
We can easily obtain local in time existence and uniqueness results by Picard iteration as the continuous problem (1.1). Therefore in order to obtain existence on [0,T] for any T > 0, we need a priori estimates on uN .
Lemma2.1 Let u0∈ Hp1(Ω), then
d
dtF(uN(t)) ≤ 0 (2.9)
Proof:
d
dtF(uN(t)) = γ(DuN, DuNt ) + (φ(uN(t)), uNt )
= (−γD2uN, uNt ) + (φ(uN(t)), uNt )
= (−γD2uN+ φ(uN(t)), uNt )
= (−γD2uN+ φ(uN(t)), −γD4uN+ D2PNφ(uN(t))) (by (2.8))
= (−γD2uN + φ(uN(t)), D2(−γD2uN+ PNφ(uN(t))))
= −(D(−γD2uN+ φ(uN(t))), D(PN(−γD2uN+ φ(uN(t)))))
= −(D(−γD2uN+ φ(uN(t))), PND(−γD2uN+ φ(uN(t))))
= −kPND(−γD2uN + φ(uN(t))k2≤ 0
Lemma2.2
(uN(t), 1) = (u0, 1) ∀t > 0 (2.10)
Proof: It can be easily obtained from (2.6) if only we choose vN = 1.
Lemma2.3 Let u0∈ Hp1(Ω), then there exists a constant c, independent of N , such that
kuN(t)k1≤ c, ∀t ≥ 0 (2.11)
Consequently,
kuN(t)k∞≤ c, ∀t ≥ 0 (2.12)
Proof: By Lemma 2.1, we have
F(uN(t)) ≤ F(uN(0))
i.e. γ
2kDuNk20+ Z
Ω
Ψ(uN)dx ≤ γ
2kDPNu0k20+ Z
Ω
Ψ(PNu0)dx (2.13)
Using the Young inequality, we infer the existence of two constants c1, c2 (depending only on the coefficients of Ψ), such that
r2
8s4− c1≤ Ψ(s) ≤ r2
2s4+ c2 (2.14)
Thus, we have from (2.2)(2.13)(2.14) γ
2kDuNk20+r2
8 Z
Ω
(uN(t))4dx ≤ c +γ
2kPNDu0k20+r2
2 Z
Ω
(PNu0(t))4dx
≤ c +γ
2kDu0k20+2πr2
2 kPNu0k4∞
≤ c +γ
2kDu0k20+ πr2c3kPNu0k41 (By H1(Ω) ,→ L∞(Ω))
≤ c +γ
2kDu0k20+ πr2c3ku0k41= c4
By Poincar´e inequality and Lemma2.2,
kuN(t)k1≤ c(kDuN(t)k0+ |(uN(t), 1)|
≤ c(c4+ |(uN(t), 1)|) = c5
Consequently, (2.12) can be obtained from (2.11) by Sobolev imbedding theorem.
Using the same proof of Theorem 2.1 of [5], we have the following theorem:
Theorem 2.4: For any T > 0, if u0∈ Hp2(Ω), then (2.6)(2.7) has a unique solution uN(t) ∈ L2(0, T ; H4(Ω)), uNt (t) ∈ L2(0, T ; L2(Ω))
Now we estimate the error ku(t) − uN(t)k0. Denote eN(t) = PNu(t) − uN(t), we have from (1.1)(2.6) (eNt (t), vN) + γ(D2eN(t), D2vN) = (φ(u(t)) − φ(uN(t)), D2vN) ∀vN ∈ SN (2.15) choose vN = eN(t) in (2.15), we obtain
1 2
d
dtkeN(t)k20+ γkD2eN(t)k20= (φ(u(t)) − φ(uN(t)), D2eN(t)) (2.16)
≤ kφ(u(t)) − φ(uN(t))k0kD2eN(t)k0
≤ 1
2γkφ(u(t)) − φ(uN(t))k20+γ
2kD2eN(t)k20 But
φ(u(t)) − φ(uN(t)) = φ0(uN(t) − θ(u(t) − uN(t)))(u(t) − uN(t)), 0 < θ < 1
From (1.9)(2.12), we have
kφ0(uN(t) − θ(u(t) − uN(t)))k∞≤ c Thus
kφ(u(t)) − φ(uN(t))k0 ≤ cku(t) − uN(t))k0 (2.17)
≤ c(ku(t) − PNu(t)k0+ keN(t)k0) From (2.16)(2.17), we have
d
dtkeN(t)k20+ γkD2eN(t)k20≤ 2c2
γ keN(t)k20+2c2
γ ku(t) − PNu(t)k20) By Gronwall Lemma, we have
keN(t)k20 ≤ 2c2 γ e2c2Tγ
Z t
0
e−2c2τγ ku(τ ) − PNu(τ )k20dτ ∀t ≤ T
≤ cT
Z t
0
ku(τ ) − PNu(τ )k20dτ (2.18)
Thus we obtain the following theorem:
Theorem 2.5 Let u0 ∈ Hp1(Ω), u(t) is the solution for periodic initial value problem (1.1) (1.2) (1.3) (1.5) satisfying u ∈ L∞(0, T ; Hpm(Ω)) for any T > 0, uN(t) is the solution of semi-discrete approximation (2.6)(2.7), then there exists a constant c, independent of N , such that
ku(t) − uN(t)k0≤ cN−m(ku(t)km+ ( Z t
0
ku(τ )k2mdτ )12) (2.19)
Proof: It can be easily obtain from (2.3) (2.18) and the triangle inequality ku(t) − uN(t)k0≤ ku(t) − PNu(t)k0+ keN(t)k0.
3 A Fully Discrete Scheme
In this section we give a fully discrete scheme which keeps the Lyapunov functional (1.6) in discrete form and implies the pointwise bounded of the solution.
Let ∆t = T /M , for a positive integer M . Then we consider the following fully discrete problem:
Find uNn+1∈ SN, n = 0, 1, 2, · · · , M − 1, such that uNn+1− uNn
∆t + γD4uNn+1+ uNn
2 = D2φ(u˜ Nn+1, uNn) (3.1)
uN0 = PNu0 (3.2)
where
φ(u, v) =˜ r2
4(u3+ u2v + uv2+ v3) +r1
3(u2+ uv + v2) +r0
2(u + v)
Obviously, (3.1) is equivalent to the following equation : (uNn+1− uNn
∆t , vN) + γ(D2uNn+1+ uNn
2 , D2vN) = ( ˜φ(uNn+1, uNn), D2vN), ∀vN ∈ SN (3.3) Lemma 3.1 If {uNn} solves (3.1) (3.2), then
1
∆t(F(uNn+1) − F(uNn)) ≤ 0, n = 0, 1, · · · , M − 1 (3.4) i.e. (3.1) (3.2) keeps the Lyapunov functional (1.6) in its discrete form
Proof: By using (3.1) (3.2) and the definition of F(·), we obtain 1
∆t(F(uNn+1) − F(uNn)
= γ(DuNn+1+ uNn
2 , DuNn+1− uNn
∆t ) + ( ˜φ(uNn+1, uNn),uNn+1− uNn
∆t )
= (−γD2uNn+1+ uNn
2 + ˜φ(uNn+1, uNn),uNn+1− uNn
∆t )
= (−γD2uNn+1+ uNn
2 + ˜φ(uNn+1, uNn), D2(−γD2uNn+1+ uNn
2 + ˜φ(uNn+1, uNn)))
= −k − γD3uNn+1+ uNn
2 + D ˜φ(uNn+1, uNn)k20. Lemma 3.2
(uNn, 1) = (u0, 1), n = 0, 1, · · · , M. (3.5)
Proof: It can be easily obtained by choosing vN = 1 in (3.3).
Lemma 3.3 Let u0∈ Hp1(Ω), If {uNn} solves (3.1) (3.2), then
kuNnk1 ≤ c 0 < n ≤ M (3.6)
kuNnk∞ ≤ c 0 < n ≤ M (3.7)
Proof: It can be proved as the same as Lemma 2.3 by using Lemma 3.1, Lemma 3.2.
About the existence of the solution uNn of (3.1) (3.2) and the error estimates for this fully discrete scheme, we have the following results(Theorem 3.4,Theorem 3.5). They can be proofed similarly as in [8].
Theorem 3.4 Let u0∈ Hp1(Ω), then there exists a constant β > 0, depending only on u0 and constants in the inverse inequalities, such that for ∆tN2< β, (3.1) (3.2) has a unique solution uNn for all n ≥ 0.
Theorem 3.5: Assume ∆t is sufficiently small, the solution u(x, t) of (1.1) (1.2) (1.3) (1.5) satisfies u ∈ L∞(0, T ; Hpm(Ω)), ut∈ L∞(0, T ; Hm(Ω)) ∩ L2(0, T ; L4(Ω))
utt∈ L2(0, T ; H2(Ω)), uttt∈ L2(0, T ; L2(Ω))
uNn is the solution of (3.1) (3.2). Then there exists a constant c, independent of ∆t, N , such that, for n = 0, 1, · · · , M ,
kuNn − u(tn)k0≤ c(K1N−m+ K2(∆t)2) (3.8)
where
K1= (ku(tn)k2m+ Xn i=0
∆tku(tn)k2m)12
K2= ( Z tn
0
(kutk4L4(Ω)+ kuttk22+ kutttk20)dt)12
4 Numerical Results
Let uNn =PN/2−1
k=−N/2uˆk,neikx. From (3.1) (3.2), we have ˆ
uNk,n+1− ˆuNk,n
∆t + γk4uˆNk,n+1+ ˆuNk,n
2 = (D2φ(u˜ dNn+1, uNn))k, k = −N
2 , · · · ,N
2 − 1 (4.1)
ˆ
uNk,0= d(u0)k, k = −N
2 , · · · ,N
2 − 1 (4.2)
The implementation of (4.1) (4.2) requires the solution of a nonlinear system of equations at each time step. We use the following predictor-corrector algorithm:
ˆ
uN,[0]k,n+1= 1 − 12γk4∆t
1 + 12γk4∆tuˆNk,n+ ∆t
1 + 12γk4∆t(D2dφ(uNn))k, (4.3) k = −N
2, · · · ,N 2 − 1 ˆ
uN,[m+1]k,n+1 =1 − 12γk4∆t
1 + 12γk4∆tuˆNk,n+ ∆t 1 +12γk4∆t
1
2[(D2dφ(uNn))k (4.4)
+( d
D2φ(u˜ N,[m]n+1 , uNn))k], k = −N 2, · · · ,N
2 − 1, m = 0, 1, 2, · · ·
The iterations to solve the nonlinear system on each time step were judged to have converged when the difference between successive iterates was less than approximately 1.0 × 10−8; generally 2-3 iterations were used on each time step. According to [19],with sufficient resolution, aliased calculations are quite acceptable, we can compute the Fourier coefficients (D2dφ(uNn))k, ( d
D2φ(u˜ N,[m]n+1 , uNn))kby FFT. The procedure is as follows:
Let xj= 2πjN , j = 0, 1, · · · , N − 1, F (x) = D2φ(u(x)), then F (x)dk= 1
N
N −1X
j=0
F (xj)eikxj, −N/2 ≤ k ≤ N/2 − 1 (4.5)
Simply computing, we have
F (x) = (6r2u + 2r1)(Du)2+ (3r2u2+ 2r1u + r0)D2u (4.6) Note
u(xj) =
N 2−1
X
k=−N2
ˆ
ukeikxj (4.7)
Du(xj) = i
N2X−1 k=−N2
kˆukeikxj (4.8)
D2u(xj) = −
N2X−1 k=−N2
k2uˆkeikxj (4.9)
we know that u(xj), Du(xj), D2u(xj) can be computed by u’s Fourier coefficients ˆuk. But (4.7) (4.8) (4.9)(
so F (xj)) and (4.5) can be implemented by FFT, IFFT, therefore (D2dφ(uNn))k can be quickly calculated, so
( d
D2φ(u˜ N,[m]n+1 , uNn))k .
Choose u0= cos(x)/2, γ = 0.03, r2 = 1, r1 = 0, r0= −1, N = 64, ∆t = 0.001, The results are shown in Figure 4.1 and Figure 4.2.
0 1 2 3 4 5 6 7
−1
−0.8
−0.6
−0.4
−0.2 0 0.2 0.4 0.6 0.8 1
t=0 t=0.5 t=1 t=2 t=3
0 1 2 3 4 5 6 7
−1
−0.8
−0.6
−0.4
−0.2 0 0.2 0.4 0.6 0.8
1 t=5
t=0
(a) (b)
Figure 4.1
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
−1.3
−1.2
−1.1
−1
−0.9
−0.8
−0.7
−0.6
−0.5
−0.4
−0.3
t
Energy
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
−0.1
−0.08
−0.06
−0.04
−0.02 0 0.02 0.04 0.06 0.08 0.1
t
Mass
(a) (b)
Figure 4.2
Figure 4.1(a), (b) show the evolution from t=0 to t=5. From these plots, we can easily see that the total mass remains indeed constant since the initial data has null mean value (M = 0). In numerical experiments, we observed that the solution u(t) tends to ,and remains at for a long time , patterns that nearly piecewise constant, the steady-state solution is very close to a piecewise constant function. In [11], Elliott and French also observed the same results. In this experiment, the patterns hardly changed after t=3.
Figure 4.2(a) shows the time dependency of energy functional (1.6) of numerical solutions in Figure 4.1.As shown in the energy functional dissipation property, the energy functional F(uNn) of numerical solutions theoretically decreases as time pass(see Lemma2.1,Lemma3.1). Figure 4.2(a) indicates that the energy functional F(uNn) of numerical solutions decreases and agree with the dissipation property.
Figure 4.2(b) shows the time dependency of mass of numerical solutions in Figure 4.1. The mass of numerical solutions is theoretically independent of time and this is shown in Lemma2.2 and Lemma3.2.
Looking at Figure 4.2(b), the mass of numerical solution is conservation quite well and it agrees with the conservation property also.
About the restriction of the time step ∆t, we discovered that it isn’t severe as stated in Theorem 3.4.
When u0= cos(x)/2, γ = 0.03, r2 = 1, r1 = 0, r0= −1, N = 64, we chose ∆t = 0.1, 0.05, 0.01, 0.001, 0.0002 respectively to solve (4.1) (4.2), Table 4.1 show the results at t = 1, where
err(t, ∆t) = (1 N
N −1X
j=0
|U (xj, t, 0.0002) − U (xj, t, ∆t)|2)12
U (xj, t, ∆t) express the solution of (4.1) (4.2) at (xj, t) by using time step ∆t. As no exact solution to (1.1)(1.2)(1.4)(1.5) is known, a comparison between the solution of (4.1) (4.2) on a coarse mesh with that on a fine mesh was made.
Table 4.1
∆t err(1,∆t) 0.1 0.0010 0.05 4.1313×10−4 0.01 7.5728×10−5 0.001 6.1699×10−6
All above computing procedures are stable, and the different time step affects only the exactness of the solution, which is consistent with the result of Theorem 3.5 that the error is O((∆t)2).
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