Applied Mathematics Letters 45 (2015) 93–97
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An improved error estimate for Maxwell’s equations with a power-law nonlinear conductivity✩ Tong Kang a,∗ , Yanfang Wang a , Liyun Wu a , Kwang Ik Kim b a
Department of Applied Mathematics, School of Sciences, Communication University of China, Beijing, 100024, China
b
Department of Mathematics, Pohang University of Science and Technology, Pohang, 790-784, Republic of Korea
article
info
Article history: Received 11 December 2014 Received in revised form 11 January 2015 Accepted 11 January 2015 Available online 7 February 2015
abstract In this work we investigate a fully discrete finite element method for Maxwell’s equations with a power-law nonlinear conductivity and prove an improved error estimate with
O τ + hmin{1,α} (α > 0) for both time and space discretization, which is a more accurate approximation than the suboptimal order estimate in [1]. © 2015 Elsevier Ltd. All rights reserved.
Keywords: Nonlinear Maxwell’s equations Backward Euler Edge elements Error estimate
1. Introduction A power-law for the conductivity occurs in various physical models such as the constitutive law for type-II superconductors (see [2]) and modeling of the nonlinear conductivity of the charge-density wave state of NbSe3 (see [3]). In this paper, we will study Maxwell’s equations with a power-law nonlinear conductivity:
ε∂t E + σ (|E |)E = ∇ × H + J , ∂t H + ∇ × E = 0,
(1)
where E and H are the electric and the magnetic field and J denotes the current density. Let Ω ⊂ R3 be a polyhedral domain with the boundary ∂ Ω . This domain is occupied by a nonlinear conducting material with the electric conductivity σ (|E |), which is assumed to be a monotone function of the form |E |α−1 with α > 0. Let a(E ) = |E |α−1 E. For simplification, put the electric permittivity ε = 1. We eliminate the magnetic field H for the system (1) and obtain the following initial and boundary value problem:
∂tt E + ∂t a(E ) + ∇ × ∇ × E = F , E (x, 0) = E0 , ∂t E (x, 0) = E0′ , n × E (x, t ) = 0,
(x, t ) ∈ Ω × (0, T ], x ∈ Ω, (x, t ) ∈ ∂ Ω × (0, T ],
(2)
where we define a new source term F .
✩ This work was supported by National Basic Research Program of China under grant number 2014CB845906, R&D of Key Instruments and Technologies for Deep Resources Prospecting (the National R&D Projects for Key Scientific Instruments) under grant number ZDYZ2012-1-02-04, NSFC under grant 91130015 and BSRP through POSTECH BK21 Plus Mathematics. ∗ Corresponding author. E-mail address:
[email protected] (T. Kang).
http://dx.doi.org/10.1016/j.aml.2015.01.017 0893-9659/© 2015 Elsevier Ltd. All rights reserved.
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T. Kang et al. / Applied Mathematics Letters 45 (2015) 93–97
In [1], Durand and Slodička presented a fully discrete finite element method for the problem (2) based on backward Euler discretization in time and curl-conforming finite elements in space. The obtained error estimate is suboptimal, namely, of order τ 1/2 +hmin{1/2,α} . According to the authors, the optimality of the backward Euler method cannot be expected a priori due to the combination of the hyperbolicity and the nonlinearity of the problem. The suboptimality of space discretization is due to the estimation of S3 (see Page 15, [1]). If the nonlinear function a(E ) is replaced by a Lipschitz continuous approximation, then we could use Young’s inequality to estimate S3 and obtain the optimality with respect to h under proper conditions. Some similar suboptimal estimates for eddy current problems and type-II superconductor models are found in [4–6]. The purpose of this work is to prove an improved error estimate with O(τ + hmin{1,α} ) for the problem (2) with α > 0. The theoretical result can be extended to other similar problems. We state the fully discrete scheme and the main result in Section 2 and the proof of the result is given in Section 3. 2. Fully discrete scheme and the main result We set p = max{2, 1 + α} and define the function space
V = u ∈ L p (Ω )∇ × u ∈ L 2 (Ω ), n × u = 0 on ∂ Ω .
This is a reflexive Banach space (see [1]). Before we give the fully discrete scheme, we define Th as a regular tetrahedral mesh partition of Ω with a mesh size h. We construct a finite element space based on Nédélec’s curl-conforming edge elements:
Wh0 = ϕ ∈ V : ∀K ∈ Th , ∃ a, b ∈ R3 , ϕ|K = a + b × x . We introduce the interpolation operator Π h to have the following estimate (see [7,8])
u − Π h u
(3) ≤ ChuW 1,p (curl ,Ω ) , for any u ∈ W 1,p (curl , Ω ) = u ∈ W 1,p (Ω )∇ × u ∈ W 1,p (Ω ) . Let ∥ · ∥ denote the L 2 (Ω )-norm. We divide the time interval (0, T ) into M equal subintervals by using nodal points 0 = t0 < t1 < t2 < · · · < tM = T with tn = nτ , and denote the nth subinterval by I n = (tn−1 , tn ]. Set δ ui − δ ui−1 ui − ui−1 , δ 2 ui = . ui = u(ti ), δ ui = τ τ L p (Ω )
With the finite element setting we can formulate the fully discrete scheme as follows: Find ehi ∈ Wh0 such that
δ 2 eh , ϕ h + δ aeh , ϕ h + ∇ × eh , ∇ × ϕ h = F h , ϕ h , i
i
eh = Π h E , 0 0
i
δ eh0 = Π h E0′ ,
i
∀ϕ h ∈ Wh0 ,
(4)
x ∈ Ω.
Now, we have the main result of this paper. Theorem 2.1. Let E and ehi (1 ≤ i ≤ M) be the solutions of the problem (2) and the scheme (4) respectively. For the source function F and the initial data E0 and E0′ , we assume F (x, t ) ∈ H 1 (0, T ; L 2 (Ω )),
E0 , E0′ ∈ W 1,2 (curl , Ω ).
Suppose that E ∈ H 2 (0, T ; W 1,2 (curl , Ω )). Then we have
2
max ehi − E (ti ) +
1≤i≤M
M M 2 2 τ ∇ × ehi − E (ti ) ≤ C (τ 2 + h2·min{1,α} ), τ ∇ × ehi − E (ti ) + i=1
(5)
i=1
where C is a positive constant independent of both the time step length τ and the mesh size h. 3. Proof of Theorem 2.1 Proof. First, multiplying the equation of the problem (3) by any ϕ h ∈ Wh0 and integrating the results over Ω , then we obtain
∂tt E , ϕ h + ∂t (a(E )), ϕ h + ∇ × E , ∇ × ϕ h = F , ϕ h .
(6)
Integrating (6) over [ti−1 , ti ] in time yields
h h ∂t Ei − ∂t Ei−1 , ϕ + a(Ei ) − a(Ei−1 ), ϕ +
ti ti−1
∇ × E, ∇ × ϕ
h
=
ti ti−1
F , ϕh .
(7)
T. Kang et al. / Applied Mathematics Letters 45 (2015) 93–97
95
For i = 1, . . . , M, setting (1)
Ri
(2)
Ri
ti
1
(t − ti−1 )∂tt E dt , = ∂t Ei − δ Ei = τ ti ti ti−1 (t − ti−1 )∂t (∇ × E ) dt , ∇ × E dt = = τ ∇ × Ei − ti−1
ti−1
we have
δ Ei − δ Ei−1 , ϕ h + a(Ei ) − a(Ei−1 ), ϕ h + τ ∇ × Ei , ∇ × ϕ h ti (1) (1) (2) h h F , ϕh . = −Ri + Ri−1 , ϕ + Ri , ∇ × ϕ +
(8)
ti−1
It is clear that
ti 2 (1) 2 Ri ≤ C τ ∂tt E dt , ti−1 ti 2 (2) 2 3 ∂t (∇ × E ) dt . Ri ≤ C τ
(9) (10)
ti−1
(1)
Define E−1 = E0 − τ E0′ . Then R0
= 0. Making summation for i = 1, . . . , j leads to
j δ Ej , ϕ h + a(Ej ), ϕ h + τ ∇ × Ei , ∇ × ϕ h i=1 j j (2) Ri , ∇ × ϕ h + = −Rj(1) , ϕ h + i=1
i =1
ti
F , ϕ h + δ E0 , ϕ h + a(E0 ), ϕ h .
(11)
ti−1
Next, we sum (4) up for i = 1, . . . , j and then subtract (11) to obtain
j δ(ehj − Ej ), ϕ h + a(ehj ) − a(Ej ), ϕ h + τ ∇ × (ehj − Ej ), ∇ × ϕ h i =1
= Rj(1) , ϕ h
j j (2) Ri , ∇ × ϕ h + + − i=1
i=1
ti
(Fi − F ), ϕ h + δ eh0 − δ E0 , ϕ h + a(eh0 ) − a(E0 ), ϕ h . (12)
ti−1
For brevity, set 2hj := ehj − Π h Ej . Thus,
j δ 2hj , 2hj + a(ehj ) − a(Π h Ej ), 2hj + τ ∇ × 2hi , ∇ × 2hj i=1 j ti j (2) (Fi − F ), 2hj Ri , ∇ × 2hj + = Rj(1) , 2hj + − i=1 ti−1 i =1 h ′ ′ h h + Π E0 − E0 , 2j + a(Π E0 ) − a(E0 ), 2hj + δ(Ej − Π h Ej ), 2hj
j + a(Ej ) − a(Π h Ej ), 2hj + τ ∇ × (Ej − Π h Ej ), ∇ × 2hj .
(13)
i=1
Further, we multiply both sides of the equation (13) by τ and sum up for j = 1, . . . , n to have j n n n τ δ 2hj , 2hj + τ a(ehj ) − a(Π h Ej ), 2hj + τ ∇ × 2hi , τ ∇ × 2hj j=1
j =1
j =1
i =1
j j n n n (2) = τ Rj(1) , 2hj + − Ri , τ ∇ × 2hj + τ j =1
j =1
i=1
j =1
i=1
ti
(Fi − F ), 2hj
ti−1
n n n + τ Π h E0′ − E0′ , 2hj + τ a(Π h E0 ) − a(E0 ), 2hj + τ δ(Ej − Π h Ej ), 2hj j =1
j=1
j =1
j n + τ a(Ej ) − a(Π h Ej ), 2hj + τ ∇ × (Ej − Π h Ej ), τ ∇ × 2hj . n
j =1
j =1
i=1
(14)
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T. Kang et al. / Applied Mathematics Letters 45 (2015) 93–97
Now, we deal with each term on both sides of the equation (14). Using the inequality a(a − b) ≥ a2 /2 − b2 /2 yields n n 2 1 2 1 2 1 h 2 1 τ δ 2hj , 2hj ≥ 2j − 2hj−1 = 2hn − 2h0 . j =1
2
j =1
2
2
2
(15)
We use the monotonicity of the function a to conclude that n τ a(ehj ) − a(Π h Ej ), 2hj ≥ 0.
(16)
j =1
Applying Abel’s summation rule and defining j n j =1
τ ∇ × 2hi , τ ∇ × 2hj
=
i=1
0
j =1
τ ∇ × 2hj = 0, we have
n n 2 1 2 1 τ ∇ × 2hj . τ ∇ × 2hj + 2 j =1 2 j =1
(17)
Using Young’s inequality, (3) and (9)–(10), we have the following estimates. j n j =1
(2)
Ri , τ ∇ × 2hj
i=1 n
≤C
j n n 2 (2) 2 1 Ri + τ ∇ × 2hj
≤C
τ3
T
t0
j =1
n 2 2 1 ∂t (∇ × E ) dt + τ ∇ × 2hj
8 j =1
n 2 2 1 τ ∇ × 2hj ∂t (∇ × E ) dt +
T
8 j =1
t0
2 ≤ C τ 2 E 1
H (0,T ;W 1,2 (curl ,Ω ))
+
n 2 1 τ ∇ × 2hj , 8 j =1
n 2 τ Rj(1) , 2hj ≤ C τ 2 E 2
H (0,T ;L 2 (Ω ))
j =1
j n τ j =1
i=1 n
≤
8 j =1
j=1 i=1
≤ Cτ 2
ti ti−1
3
τ2
0
j =1
n 1 h 2 τ 2j , 16 j=1
j n (Fi − F )dt , 2hj ≤ τ j =1
T
+
(18)
i =1
ti ti−1
(19)
(Fi − F )dt 2hj
2 1 ∂t F dt · τ 2 2hj ≤ C τ 2 F 1
H (0,T ;L 2 (Ω ))
+
n 1 h 2 τ 2j , 16 j=1
(20)
n n 2 1 h 2 τ Π h E0′ − E0′ , 2hj ≤ C Π h E0′ − E0′ + τ 2j
16 j=1
j =1
2 ≤ Ch2 E0′
W 1,2 (curl ,Ω )
+
n 1 h 2 τ 2j , 16 j=1
(21)
n n τ a(E0 ) − a(Π h E0 ), 2hj ≤ τ a(E0 ) − a(Π h E0 ) 2hj j =1
j =1
n n 2·min{1,α} 1 h 2 ≤C τ E0 − Π h E0 + τ 2j
16 j=1 (the α -Hölder continuity of the function a with 0 < α ≤ 1 in Lemma 6.4, [6]) n 2·min{1,α} 1 h 2 τ 2j , ≤ Ch2·min{1,α} E0 1,2 + W (curl ,Ω ) 16 j=1 j=1
n n τ δ(Ej − Π h Ej ), 2hj ≤ τ δ(Ej − Π h Ej ) 2hj j =1
≤C
j =1
n j =1
2 τ h2 δ Ej
W 1,2 (curl ,Ω )
+
n 1 h 2 τ 2j 16 j=1
(22)
T. Kang et al. / Applied Mathematics Letters 45 (2015) 93–97
2 ≤ Ch2 ∂t E ∞ L
(0,T ;W 1,2 (curl ,Ω ))
2 ≤ Ch2 E 2
+
+
H (0,T ;W 1,2 (curl ,Ω ))
97
n 1 h 2 τ 2j 16 j=1
n 1 h 2 τ 2 j , 16 j=1
(23)
n n τ a(Ej ) − a(Π h Ej ), 2hj ≤ τ a(Ej ) − a(Π h Ej ) 2hj j=1
j =1
n n 2·min{1,α} 1 h 2 τ 2j ≤C τ Ej − Π h Ej +
16 j=1
j =1
(the α -Hölder continuity of the function a with 0 < α ≤ 1 in Lemma 6.4, [6]) n 2 1 h 2 ≤ Ch2·min{1,α} E ∞ + τ 2j 1,2 L
(0,T ;W
(curl ,Ω ))
2 ≤ Ch2·min{1,α} E 1
H (0,T ;W 1,2 (curl ,Ω ))
j n j=1
(24)
16 j=1
+
τ ∇ × (Ej − Π h Ej ), τ ∇ × 2hj
n 1 h 2 τ 2j , 16 j=1
i=1
≤C
j n n 2 1 2 τ ∇ × (Ej − Π h Ej ) + τ ∇ × 2hj j
≤C
8 j =1
i =1
j =1
n j=1 i=1
2 τ 2 Ej − Π h Ej
2 ≤ Ch2 E ∞ L
2 ≤ Ch2 E 1
W 1,2 (curl ,Ω )
(0,T ;W 1,2 (curl ,Ω ))
H (0,T ;W 1,2 (curl ,Ω ))
+
n 2 1 τ ∇ × 2hj 8 j =1
+
n 2 1 τ ∇ × 2hj 8 j =1
+
n 2 1 τ ∇ × 2hj . 8 j=1
(25)
Thus, making use of the previous estimates (18)–(25) to (14) and applying Grönwall’s inequality lead to n n 2 2 2 h τ ∇ × 2hj ≤ C (τ 2 + h2·min{1,α} ). 2n + τ ∇ × 2hj + j=1
(26)
j =1
Finally, we use the triangle inequality to arrive at the error estimate for the fully discrete scheme (4), which completes the proof.
2
Remark 3.1. We use the term max1≤i≤M ehi − E (ti ) as the error E in Section 8, [1] approximately. By comparing the result of Theorem 2.1 with the numerical experiments in [1], Experiment 1 shows the same convergence rate as the estimated O(τ 2 ) rate. Experiment 2 and 3 shows the less convergence rates than the theoretical result in time, but we obtain the almost same convergence rates with respective to the mesh size h as the predicted O(h2·min{1,α} ) rate with α = 0.6 and α = 3.0.
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