Hindawi Publishing Corporation
Mathematical Problems in Engineering
Volume 2010, Article ID 279038, 26 pages
doi:10.1155/2010/279038
Review Article
A Fully Discrete Discontinuous Galerkin Method
for Nonlinear Fractional Fokker-Planck Equation
Yunying Zheng,1, 2 Changpin Li,1 and Zhengang Zhao1
1
2
Department of Mathematics, Shanghai University, Shanghai 200444, China
Department of Mathematics, Huainan Normal University, Huainan 232038, China
Correspondence should be addressed to Changpin Li, lcp@shu.edu.cn
Received 13 July 2010; Revised 16 September 2010; Accepted 19 October 2010
Academic Editor: J. Jiang
Copyright q 2010 Yunying Zheng et al. This is an open access article distributed under the
Creative Commons Attribution License, which permits unrestricted use, distribution, and
reproduction in any medium, provided the original work is properly cited.
The fractional Fokker-Planck equation is often used to characterize anomalous diffusion. In this
paper, a fully discrete approximation for the nonlinear spatial fractional Fokker-Planck equation is
given, where the discontinuous Galerkin finite element approach is utilized in time domain and the
Galerkin finite element approach is utilized in spatial domain. The priori error estimate is derived
in detail. Numerical examples are presented which are inline with the theoretical convergence rate.
1. Introduction
Many models in physics, chemistry are successfully described by the Langevin equation,
which has been introduced almost 100 years before. And for some particular cases, say
diffusion, the original Langevin equation can be transformed into the Fokker-Planck
equation. Hänggi and Thomas 1 associated a Gaussian distribution of the increments of the
noise generating process with the classical Fokker-Planck equation. Sun et al. 2 discussed
the fractional model for anomalous diffusion. Metzler et al. 3 and Dubkov and Spagnolo
4 derived the fractional Fokker-Planck equation from different anomalous diffusion
procedures. Metzler and Klafter 5 discussed fractional kinetic equation and its relation to
the fractional Fokker-Planck equation. Dubkov et al. 6 introduced Fokker-Planck equation
for Lévy flights. Now the Fokker-Planck equation is one of the best tools for characterizing
anomalous diffusion, especially sub-/super-diffusion. Meanwhile the fractional FokkerPlanck equation has been found to be used in relatively wide field of applied sciences, such
as plasma physics, population dynamics, biophysics, engineering, neuroscience, nonlinear
hydrodynamics, and marking; see 7–13.
The Fokker-Planck equation describes the changes of a random function in space
and in time. So different assumptions on probability density function lead to a variety of
2
Mathematical Problems in Engineering
space-time equations. In this paper, we mainly study the model described by the following
fractional Fokker-Planck equation, which is a special case in 12:
∂u β
2β
C D0,x bx, t, u C D0,x ax, t, u u,
∂t
0 < β < 1, x, t ∈ Ω × 0, T ,
1.1
β
where C D0,x denotes left fractional derivative with order β in the sense of Caputo.
There are some numerical methods to find the approximate solutions of the fractional
differential equations 14–21. But the discontinuous Galerkin finite element method is a very
attractive method for partial differential equations because of its flexibility and efficiency in
terms of mesh and shape functions. And the higher order of convergence can be achieved
without over many iterations. Such a method was first proposed and analyzed in the
early 1970s as a technique to seek numerical solutions of partial differential equations. The
discontinuous Galerkin finite element method becomes a very attractive tool for the initial
problems of the ODEs and the initial-boundary problems of PDEs; see 22–27.
The rest of this paper is constructed as follows. The fractional derivative space is
introduced in Section 2. In Section 3, the discontinuous Galerkin finite element scheme is
introduced. The existence and uniqueness of numerical solution are proved in Section 3. And
the error estimate of the discontinuous Galerkin finite element approximation is studied in
Section 4. Finally in Section 5, numerical examples are also taken to show the efficiency of the
theoretical results.
2. Fractional Derivative Space
In this section, we firstly introduce the fractional integral, fractional Caputo derivative, and
their properties.
Definition 2.1. The αth order left and right Riemann-Liouville integrals of function ux are
defined as follows:
−α
ux
Da,x
1
Γα
1
−α
Dx,b
ux Γα
x
a
b
x
us
x − s1−α
us
s − x1−α
ds,
2.1
ds,
where α > 0 and a < x < b.
Definition 2.2. The αth order left and right Caputo derivatives of function ux are defined in
α
C D a,x ux
α
C D x,b ux
where n − 1 < α < n ∈ Z and a < x < b.
α−n n
Da,x
D ux,
α−n
Dx,b
−Dn ux,
2.2
Mathematical Problems in Engineering
3
α,p
Definition 2.3. Let α > 0; define a fractional derivative space Js ΩΩ a, b as
α,p
Js Ω ux ∈ H n Ω :
α
α
C D a,x ux, C D x,b ux
∈ Lp Ω, n − 1 ≤ α < n ,
2.3
endowed with the seminorm
1/2 |u|Jsα,p C Dαa,x ux, C Dαx,b ux
Lp Ω
2.4
and the norm
⎞1/p
p
p
β
⎝
D ux |ux|J α,p ⎠ ,
⎛
uxJsα,p
s
k≤α
β ∈ Z, 0 ≤ β ≤ α − 1.
2.5
With the
Transform,
we can conclude that
the following three
help of Fourier
α
ux|dΩ, Ω C Dαa,x ux2 dΩ, and Ω C Dαx,b ux2 dΩ, are
expressions: Ω | C Dαa,x ux·C Dx,b
1/2
1/2
equivalent 20. So Ω C Dαa,x ux2 dΩ
and Ω C Dαx,b ux2 dΩ
can also be recogα,p
nized as the seminorms of the fractional space Js , and when we use the seminorm of
α,p
fractional space Js , there is no difference among them. Now we restrict our discussion in
case p 2, and the following notations are used, Jsα,2 Ω being rewritten as Jsα Ω with norm
α
Ω as the closure of C0∞ Ω under its
· Jsα or · α and seminorm | · |Jsα or | · |α . We denote Js,0
norm.
The following lemmas are useful for our discussions later on.
α
a, b, then
Lemma 2.4 see 19. Let n − 1 ≤ α < n; if u ∈ Js,0
−α
α
Da,x
C D a,x u x ux,
−α
α
Dx,b
C D x,b u x ux.
2.6
α
a, b, 0 < β < α, then
Lemma 2.5 see 20. For u ∈ Js,0
α
C D a,x ux
α−β
β
C D a,x C D a,x ux.
2.7
Lemma 2.6. For u ∈ C02α a, b, v ∈ C0α a, b, n − 1 ≤ α < n, then
2α
C D a,x u, v
−1n
α
α
C D a,x u, C D x,b v
.
2.8
α
Lemma 2.7 see 21. For u ∈ Js,0
Ω, 0 < β < α, then
|u|β ≤ |u|α ,
uβ ≤ uα .
2.9
4
Mathematical Problems in Engineering
For u ∈ H0α Ω, then
uL2 Ω ≤ c|u|H0α .
2.10
uH β Ω ≤ C|u|H0α .
2.11
For 0 < β < α, then
0
Lemma 2.8. Let μ > 0. The following mapping properties hold:
μ
α
C D a,x
: Js Ω −→ L2 Ω is a bounded linear operator,
α
C D x,b
: Js Ω −→ L2 Ω is a bounded linear operator.
μ
2.12
The proof is similar to that in [19].
3. The Space-Time Discontinuous Galerkin Finite
Element Approximation
In this section, we formulate a fully discrete discontinuous Galerkin finite element method
for a type of nonlinear spacial fractional Fokker-Planck equation.
Problem 1 nonlinear fractional Fokker-Planck equation. For 0 < β < 1,
β
2β
ut − C D0,x bx, t, u C D0,x ax, t, u u,
ux, t φx, t,
x, t ∈ Ω × 0, T ,
x ∈ ∂Ω × 0, T ,
ux, 0 gx,
3.1
x ∈ Ω,
where the Ω is a bounded domain. For positive constants m and M, the coefficients a and b
satisfy
0 < m < au < M,
0 < m < bu < M.
3.2
Throughout the paper, we always assume that the following mild Lipschitz continuity
conditions on a, b are satisfied: there exists a positive constant L such that for x ∈ Ω, t ∈ 0, T ,
and s, r ∈ R, there have
|ax, t, s − ax, t, r| ≤ L|s − r|,
|bx, t, s − bx, t, r| ≤ L|s − r|.
3.3
Mathematical Problems in Engineering
5
Let Kh {K} be a partition of spatial domain Ω. Define hk as the diameter of the
element K and h maxK∈Kh hK . And let Sh be a finite element space
β
Sh v ∈ H0 Ω : v|K ∈ Pr−1 K, K ∈ Kh ,
3.4
where Pr−1 K is a set of polynomials with degree r − 1 on a given domain K. And the
functions in Sh are continuous on Ω.
Let us consider a partition of time domain I 0, T , 0 t0 < t1 < · · · < tN T , and
In tn , tn1 , kn tn1 − tn , n 0, 1, . . . N − 1, k max kn . On each time slab In , we define a
discrete function space Ik as
Ik v : In −→ R; v|t∈In ∈ Pq In .
3.5
The functions in the space Ik can be discontinuous at the time node tn , but is at least left
continuous and right continuous. And the functions in the space Ik are polynomials, whose
degree is no more than q − 1.
Define a discrete function space Tkh on Ω × I as
Tkh {v : Ω × I −→ R | ∀t ∈ I, v·, t ∈ Ik , ∀x ∈ Ω, vx, · ∈ Sh }.
3.6
Moreover, the space Tkh can be verified as follows:
⎧
⎨
Tkh
⎫
q−1
⎬
v | vx, t tj χj x, χj ∈ Sh ,
⎩
⎭
j0
3.7
where q is a positive integer. In other words, for each t ∈ In the functions in Tkh are the
elements of Sh , and for each x ∈ Ω piecewise polynomial functions in t of degree q − 1 with
n
{v|Ω×In : v ∈ Tkh }.
possible discontinuities at the nodes tn , n 0, 1, . . . , N − 1. Set Tkh
We introduce some norms on different spaces which will be used later on. And
β
L2 I, Js Ω is equipped with the norms
v0,β I
v2β dt
1/2
,
v∞,β maxvβ max vβ .
0≤t≤tN
I
3.8
β
And L2 In , Js Ω is equipped with the norms
v0,β,n 1/2
In
v2β dt
,
v∞,β,n maxvβ max vβ .
In
tn <t≤tn1
3.9
6
Mathematical Problems in Engineering
In order to derive a variational form of Problem 1, we assume that u is a sufficiently
smooth solution of Problem 1, then multiply an arbitrary v ∈ Tkh to obtain the integration
formulation
I
∂u
2β
β
, v dt C D 0,x auu, v dt −
C D 0,x buu, v dt,
∂t
I
I
3.10
where ·, · denotes the inner product on L2 Ω. Integrating by parts in the right, noting that
β
for all t ∈ I, v ∈ J0,s Ω, and the discontinuous property at the time node tn , n 0, . . . , N − 1,
one has
N−1
n0
In
∂u
β
C D0,x buu, v
∂t
β
β
C D 0,x auu, C D x,b v
dt un , vn 0.
3.11
The notation un denotes the jump of the function u at the time node tn ; that is,
un lim u·, tn |ε| − u·, tn − |ε|.
3.12
ε→0
Using the superscripts “–” and “” for left and right limits, respectively, the jump is
described by
un u·, tn − u ·, t−n .
3.13
It shows the discontinuity of the scheme. The inner product un , vn also denotes the
transport process during different time-space slabs.
Thus, we define
!
Au, v In
∂u
,v
∂t
β
β
C D 0,x auu, C Dx,b v
β
C D 0,x buu, v
"
dt un , vn . 3.14
Definition 3.1. For all v ∈ Tkh , the function u ∈ Tkh is a variational solution of Problem 1
provided that
N−1
Au, v 0,
∀v ∈ Tkh .
3.15
n0
Now we are ready to describe a fully discrete space-time finite element method to
solve the nonlinear problem 1, where the Galerkin finite element method is used in spacial
domain and discontinuous finite element approximation is used in time domain.
Applying the same notations with Definition 3.1, the space-time discontinuous
n
, satisfying,
Galerkin scheme for Problem 1 can be now formulated as follows. Find U ∈ Tkh
n
for all v ∈ Tkh , n 0, 1, . . . , N − 1,
AU, v 0,
∀v ∈ Tkh .
3.16
Mathematical Problems in Engineering
7
The value of U on t−0 is replaced by the initial condition gx. The term including U0
can be moved to the right-hand side of 3.16. Once it is computed, the value of U on t0 is
different from that on u−0 , and it is the error introduced by the discretization in the numerical
scheme.
Noting the discontinuity at each node tn , n 0, 1, . . . , N − 1 in Thk , the computation of
uh can be decoupled in each time slab. Once uh is known on t−n , this value is taken as an initial
condition for the time slab In and the following equation needs to be solved:
−
−
, vn1
Un1
! "
∂v
β
β
β
− U,
C D0,x aUU, C Dx,b v C D0,x bUU, v dt Un− , vn .
∂t
In
3.17
Next, we investigate the uniqueness and existence about the numerical scheme. First,
q
we give the scheme in detail. For a fixed integer q ≥ 1, let li τi1 be the Lagrange polynomials
associated with the abscissa 0 < τ1 < · · · < τq 1; that is,
#
li τ j 1, . . . , q
i
/j
τ − τj
.
τi − τj
3.18
For the quadrature in time slab tn , tn1 , we use the Radau quadrature rule. For a given
function gτ, τ ∈ 0, 1, the following approximation holds:
1
gτdτ ≈
0
q
wj g τj ,
3.19
j1
1
where wj 0 lj τdτ. This quadrature rule is exact for all polynomials of degree ≤ 2q − 2.
Using the linear transformation t tn τkn , which maps 0, 1 into I n , one gets
tn,j tn τj kn ,
wn,j
j 1, . . . , q, tn,q tn1 ,
ln,j t lj τ, t tn τkn ,
1
tn1
ln,j dt kn lj τdτ kn wj , j 1, . . . , q.
3.20
0
tn
Then the Radau rule in I n can be got by
gtdt In
q 1
j1
0
lj τdτg tn τj kn
q
wn,j g tn,j .
j1
3.21
8
Mathematical Problems in Engineering
q
We choose {ln,i t}j1 as the basis functions for the piecewise polynomial function
space Pq−1 In . Then U|In is uniquely determined by the functions Un,j Un,j x, Ux ∈ Sh ,
such that
Ux, t q
ln,j tUn,j x,
3.22
x, t ∈ Ω × In .
j1
n
Taking v ln,i tψx ∈ Vhk
into 3.17, where ψ ∈ Snh , we have
ln,q tn,q Un,q , ln,i tn,q ψx −
In
⎛
In
⎝ C Dβ a
0,x
In
j1
⎞
q
q
β ln,k Un,k x
ln,j Un,j x, C Dx,b ln,i tψx ⎠dt
j1
k1
⎛
⎞
q
⎝ ln,j Un,j x, l
tψx⎠dt
n,i
⎛
⎝ C Dβ b
0,x
q
ln,k Un,k x
q
3.23
⎞
ln,j Un,j x, ln,i tψx⎠dt
j1
k1
ln,1 tn,1 Un,1 , ln,i tn,1 ψx .
On time slab In tn , tn1 , we define a Lagrange interpolation operator I$n : CIn →
Pq−1 In , such that
I$n y tn,j y tn,j ,
j 1, 2, . . . , q,
3.24
where the interpolation points are Radau points. It is easy to see that for given x ∈ Ω,
I$n ux, t ∈ Pq−1 Tn and I$n ux, t ux, tn1 are available. On time slab In tn , tn1 ,
tn,1 tn , tn,q tn1 , with the definition of interpolation operator, the discontinuous Galerkin
scheme can be rewritten in detail as follows:
δqi Un,q , ψx −
In
In
In
⎞
q
⎝ ln,j Un,j x, l
tψx⎠dt
n,i
⎛
⎛
j1
⎝ C Dβ a
0,x
q
k1
⎛
⎝ C Dβ b
0,x
⎞
q
β ln,k Un,k x
ln,j Un,j x,C Dx,b ln,i tψx ⎠dt
j1
3.25
⎞
q
q
ln,k Un,k x
ln,j Un,j x, ln,i tψx⎠dt
k1
j1
ln,i tn Un , ψx .
Next we introduce a lemma which is useful to prove the existence and estimate the
error.
Mathematical Problems in Engineering
9
Consider a q × q matrix W,
Wij wn,j ln,i
tn,j wj li
τj ,
M eq eqT − N,
3.26
where eq 0, 0, . . . , 1.
It is clear that W, M are independent of Kn . And if Y T yn,1 , . . . , yn,q ∈ Rq , then
Y T MY q
q
δqi yn,q yn,i − wn,j ln,i
tn,j yn,j yn,i .
j1
3.27
j1
% D be the matrixes
Lemma 3.2 see 27. Let M,
% D−1/2 MD1/2 ,
M
D diag τ1 , . . . , τq .
3.28
If λ α0 /2 min{w1 /τ1 , w2 /τ2 , . . . , wq−1 /τq−1 , 1 wq } > 0, then there holds
% ≥ λ|X|2 ,
X T MX
where |x| &q
j1
∀X ∈ Rq ,
3.29
Xi2 .
In order to prove the existence and uniqueness, we need to define a new exhibition for
U by
U
q
τj1/2 ln,j T Un,j x,
3.30
x, t ∈ Ω × In ,
j1
' τ −1/2 Un,j ∈ Sn . Then U|I is uniquely determined by the function U
' Ux
'
where U
∈ Snh .
j
n
h
We choose ψ τi−1/2 φ, where φ ∈ Snh , and use the new expression U to obtain the
following results:
' φ δqi U,
⎞
q
q
⎝− τ 1/2 ln,j U
' n,j , τ −1/2 ln,k l
tn,k φ⎠dt
n,i
j
i
⎛
In
j1
k1
⎞
q
q
β
β
1/2
−1/2
1/2
⎝ τ CD a
' n,m ln,j U
' n,j , C D τ
τm ln,m U
ln,i φ⎠dt
0,x
j
x,b i
⎛
In
j1
⎛
⎝
In
m1
⎞
q
q
β
1/2
1/2
'
' n,j , τ −1/2 ln,i φ⎠dt
τj C D0,x b
τm ln,m Un,m ln,j U
i
j1
' n, φ .
ln,i tn τ −1/2 U
m1
3.31
10
Mathematical Problems in Engineering
1/2
' a&q τm
' n,m and Wn,j ln,m U
Set that aU
m1
rewritten as
−
l2 tdt,
In n,j
the above equation can be
q
q
' n,j , φ Wn,j τ 1/2 τ −1/2 C Dβ a U
' U
' n,j , C Dβ φ
Wn,j τj1/2 τi−1/2 ln,i
tn,j U
0,x
j
i
x,b
j1
j1
q
β
' φ
' U
' n,j , φ δqi U,
Wn,j τj1/2 τi−1/2 C D0,x b U
3.32
j1
' n, φ .
ln,i tn τ −1/2 U
' n,j }q ∈ Sn
Theorem 3.3. Let Un be given in Snh , then for the sufficiently small kn , there exists {U
j1
h
satisfying 3.32; hence, 3.17 has a unique solution U ∈ Thk .
Proof of Theorem 3.3. The vector space Snh q is a Hilbert space with finite dimension. For all
x x1 , x2 , . . . , xq T , φ φ1 , φ2 , . . . , φq T ∈ Snh , equipped with the norm
⎞1/2
q |x| ⎝ xi2 ⎠ ,
⎛
⎞1/2
q |x|β ⎝ xi2 ⎠ .
j1
j1
⎛
β
3.33
Define a map F from Snh q to itself by
q
Fvi , φ vi , φ δqi vq , φ − Wn,j τj1/2 τi−1/2 ln,i
vi , φ
j1
Wn,i
β
β
C D 0,x avvi , C D x,b φ
Wn,i
β
C D 0,x bvvi , φ
− Un , τi−1/2 ln,i tn φ .
3.34
Since
q
vi , φ − Un , τi−1/2 ln,i tn φ
vi , φ δqi vq , φ − Wn,j τj1/2 τi−1/2 ln,i
3.35
j1
is a linear system, ax, t, u, bx, t, u are all continuous maps, and the map F is a continuous
map from Snh q into itself. So according to the Brouwer fixed point theorem, there exists
at least one fixed point vi such that Fvi , φ vi , φ. So 3.32 has at least one solution,
denoted by v. Next we investigate the uniqueness of the solution.
Mathematical Problems in Engineering
11
Let v and v∗ be two solutions of 3.32, setting φ vi − vi∗ and summing i from 1 to q,
then we have
q
Fv − v∗i , vi − vi∗
i1
q q
1/2 −1/2 tn,j τj τi
vj − vj∗ , vi − vi∗
δqi vq − vq∗ , vi − vi∗ −
Wn,j ln,i
i1 j1
q q
(
β
β Wn,j τj1/2 τi−1/2 C D0,x av − v∗ vj − vj∗ , C Dx,b vi − vi∗
3.36
i1 j1
q q
β
Wn,j τj1/2 τi−1/2 C D0,x bv − v∗ vj − vj∗ , vi − vi∗
i1 j1
q
vi − vi∗ , vi − vi∗ .
i1
According to Lemma 3.2, we can see that for the first two terms of the right-hand side of
3.36, there exists a constant α, such that
q q
1/2 −1/2 vj − vj∗ , vi − vi∗
Wn,j ln,i
tn,j τj τi
δqi vq − vq∗ , vi − vi∗ −
i1 i1
% − v∗ , v − v∗ ≥ α|v − v∗ |2 ≥ Cα|v − v∗ |2β .
Mv
3.37
With the help of bounded assumption of a and b, we have
q q
(
β
β )
Wn,j τj1/2 τi−1/2 C D0,x av − v∗ vj − vj∗ , C Dx,b vi − vi∗
i1 j1
⎛
⎞
q q
β
β
1/2 −1/2
∗
∗
∗ ⎠
⎝
≤
dt
C D0,x av − v τj τi ln,j t vj − vj , ln,i t C Dx,b vi − vi
i1 In j1
q
β
β ≤
M τj1/2 τi−1/2 ln,i tln,j t C D0,x vj − vj∗ , C Dx,b vi − vi∗ dt
i,j1
In
2
≤ Ckn vk − v∗ ,
k
β
q q
β
Wn,j τj1/2 τi−1/2 C D0,x bv − v∗ vj − vj∗ , vi − vi∗
i1 j1
⎛
⎞
q q
β
1/2
−1/2
∗
∗
∗
dt
⎝ τ τ
⎠
v
,
l
≤
l
bv
−
v
−
v
−
v
v
j
n,j t C D 0,x
n,i t
i
j
i
j
i
i1 In j1
q
β
≤
M τj1/2 τi−1/2 ln,i tln,j t C D0,x vj − vj∗ , vi − vi∗ dt.
i,j1
In
3.38
12
Mathematical Problems in Engineering
According to the boundedness of fractional operator
Cvi − vi∗ holds. Furthermore, we have
q
β
C D 0,x ,
β
inequality C D0,x vi − vi∗ ≤
β
M τj1/2 τi−1/2 ln,i tln,j t C D0,x vj − vj∗ , vi − vi∗ dt
In
i,j1
≤
q
τj1/2 τj−1/2 Wn,k vk − vk∗ , vk − vk∗ j1
3.39
q
2
M Wn,k vk − vk∗ k1
2
≤ Ckn vk − vk∗ β .
The fifth term of the right-hand side of 3.36 is estimated in
q
vi − vi∗ , vi − vi∗ |||v − v∗ |||2 ≤ |||v − v∗ |||2β .
3.40
j1
And by the Brouwer fixed point theorem, one gets
Fv − v∗ i , vi − vi∗ vi − vi∗ , vi − vi∗ ,
3.41
furthermore,
q
q
2
Fv − v∗ i , vi − vi∗ vi − vi∗ , vi − vi∗ vi − vi∗ .
i1
3.42
i1
By using inequalities 3.37–3.40 and 3.42, one has
2
Cα − kn vi − vi∗ β ≤ 0.
3.43
vi − v∗ 2 ≤ 0.
i
β
3.44
Choosing α ≥ kn , we have
That is impossible. So the uniqueness is proved.
4. Error Estimation
Now we turn to analyze the error estimate of the D-G scheme.
Mathematical Problems in Engineering
13
Let unh be the approximate solution on time slab In . Denote Ph un as a Ritz-Galerkin
projection operator defined as follows:
β
− Ph un , C Dx,b v 0,
β
β
n
n
C D 0,x au0 u − Ph u , C D x,b v 0.
β
n
C D 0,x auu
4.1
And ρn Ph un − un , θn unh − W n , then one has
unh − un unh − W n W n − un θn W n − un ,
4.2
β
where W : I → H0 Ω is defined by
q
Wx, t I$n Ph ux, t,
x, t ∈ Ω × In .
4.3
n
.
It can be seen that W is an element of Thk
Lemma 4.1. Let au, bu be smooth functions on Ω, and 0 < m ≤ au, bu ≤ M, 0 < β < 1,
and Ph un is defined as above, then
1β
Ph un − un ≤ c2 hn u1β .
4.4
Proof of Lemma 4.1. For all φ ∈ L2 Ω, w is the solution of the following equation:
2β
−aw C D0,x w φ,
w 0,
w ∈ Ω,
w ∈ ∂Ω.
4.5
So the next equation holds
w2β ≤ c2 φ.
4.6
And for all χ ∈ Sh , by the aid of approximation properties of Sh and the weak form,
we can derive
2β
β
β
Ph un − un , φ − Ph un − un , aw C D0,x w C Dx,b awPh un − un , C D0,x w
β
β C Dx,b awPh un − un , C D0,x w − χ ≤ MPh un − un β w − χβ
≤ MPh un − un β inf w − χβ
χ∈Sh
≤ c2 hr−β ur hβ wβ c2 hr ur φ.
4.7
14
Mathematical Problems in Engineering
So
P h u n − un , φ
≤ c2 hr ur .
φ
Ph u − u n
n
sup
0/
φ∈L2 Ω
4.8
β
Lemma 4.2. Let Ph un be defined as 4.1, then DPh un and C D0,x Ph un are bounded, where
0 ≤ β ≤ 1.
Proof of Lemma 4.2. Using the properties of the norm, then the following inequality is valid:
β
C D0,x Ph un L∞
β
≤ C D0,x Ph un − Ih un L∞
β
C D0,x Ih un L∞
4.9
.
According to the estimate of interpolation, we get
β
C D0,x Ph un − Ih un L∞
β
≤ c C D0,x ρn L∞
β
C D0,x Ih un − un L∞
.
4.10
β
From Lemmas 2.8 and 4.1, and the inverse estimate, we know that C D0,x ρn L∞ ≤ cun .
Therefore, it is easy to see that
β
C D0,x Ih un β
≤ c C D0,x un L∞
L∞
β
C D0,x Ih un − un L∞
,
β
≤ c C D0,x un 4.11
L∞
.
So
β
C D0,x Ph un L∞
β
≤ c C D0,x un L∞
cun .
4.12
β
The boundedness C D0,x Ph un can be derived from above. The proof of DPh un ≤ cu can
be similarly given.
Let I$n : CIn → Pq In be the usual Lagrange interpolation operator at the Radau
points on In tn , tn1 ; that is,
q
q
I$n y
tn,j y tn,j ,
j 1, 2, . . . , q,
4.13
where tn,j , j 1, 2, . . . , q are the Radau points. Therefore, we can see that
q
I$n u x, · ∈ Pq−1 In ,
q
I$n u x, tn1 ux, tn1 ,
4.14
x ∈ Ω.
Mathematical Problems in Engineering
15
q
Lemma 4.3. The interpolation operation I$n on Ph In has the following property:
$q
In u − u ≤ chs us ,
β
∀u ∈ H s ∩ H0 .
4.15
Lemma 4.4. For W defined above, we have the following error estimates:
q
maxu − Wβ ≤ Ckn maxuq chs−β maxus ,
In
β
In
u − W0,β,n ≤
q
Ckn uq 0,0,n
In
2 ≤ s ≤ r 1,
4.16
s−β
Chn u0,s,n ,
2 ≤ s ≤ r.
Proof of Lemma 4.4. Since
u − Wβ u − Ph u Ph u − Wβ ≤ u − Ph uβ Ph u − Wβ ,
4.17
according to Lemma 4.1, we have
u − Ph uβ Ph u − Wβ ≤ hs−β us Ph u − Wβ
q
q
≤ hs−β us Ckn Ph uq ≤ hs−β us Ckn uq q
≤ hs−β us Ckn uq .
β
β
4.18
β
The first inequality can be similarly proved.
We are now ready to prove convergence result. Putting W into D-G finite element
formula 3.17, and setting θ U − W, we can get the basic error equation as follows:
−
−
, vn1
θn1
−
−
θn− , vn
−
−
Wn1
, vn1
θ, vt dt −
In
Wn− , vn
In
β
C D 0,x aUθ
W, vt dt −
In
In
β
− bUU − W, C Dx,b v dt
β
C D 0,x aUW
β
bUW, C Dx,b v dt,
4.19
where W0 Ph u0 and θ0 U0 − W0 u0 − Ph u0 .
Theorem 4.5. Let u be the solution of Problem 1. If U is a solution of the discontinuous Galerkin
scheme 3.17, then the following error estimation holds:
u − U∞,β C max hs max ut s us h−β us
0≤n≤N−1
In
!
"
q
2β
β
C max kn max uq1 uq Dx uq Dx uq ,
0≤n≤N−1
where 2 ≤ s ≤ r.
In
4.20
16
Mathematical Problems in Engineering
Proof of Theorem 4.5. It is necessary to rewrite W and discrete error θ as follows:
W
q
ln,j tPh un,j ,
j1
4.21
q
ln,j t Un,j − Ph un,j ,
θ U−W j1
&q
&q
where un,j ux, tn,j . We also denote a j1 ln,j tUn,j , b j1 ln,j tUn,j as aU and bU.
2
tdt and
Noting that tn,q tn1 , tn,1 tn , so ln,q tn,q 1 and ln,1 tn,1 1. Setting Wn,i In ln,i
v ln,i tφx, φ ∈ Snh , then the basic error equation can be rearranged as follows:
− δqi
⎞
q
Ph un,q , φ ln,i tn Ph un , φ ⎝ Wn,j ln,i
tn,j Ph un,j , φ⎠ − u−n , ln,i tn φ
⎛
j1
⎛
⎞
q
q
β
β
β
− ⎝ Wn,j C D0,x aUPh un,j , C Dx,b φ⎠ Wn,j bUtPh un,j , C Dx,b φ
j1
j1
−
I−n
u, ln,i
tφ dt In
β
C D 0,x auu
β
buu, ln,i t C Dx,b φ
4.22
dt
u−n1 , ln,i tn1 φ
β
β
R1 , φ R2 , φ R3 , C Dx,b φ R4 , C Dx,b φ R5 , φ ,
where
R1 q
Wn,j ln,i
tn,j Ph un,j − un,j − δqi Ph un,q − un,q − ln,i tn un − Ph un ,
j1
q
R2 Wn,j ln,i tn,j un,j −
j1
R3 −
I−n
ln,i
tu dt,
q
β
β
Wn,j C D0,x aUPh un,j ln,i t C Dx,b auu dt,
In
j1
R4 −
q
Wn,j bU Ph un,j ln,i tbuu dt,
j1
In
R5 ln,i un − Ph un ln,i tn un − Ph un − u−n − Ph u−n .
4.23
Mathematical Problems in Engineering
17
Let φ θn,i and sum i from 1 to q, then the error equation can be derived as
−
−
, θn1
θn1
−
θn− , θn
θ, θt dt −
In
In
β
β
C D 0,x aUθ, C D x,b θ
dt In
β
bUθ, C Dx,b θ dt
q
β
β
R1 , θn,i R2 , θn,i R3 , C Dx,b θn,i R4 , C Dx,b θn,i R5 , θn,i .
i1
4.24
Note that
−
θ, θt dt −
In
1
2
In
d
1 − 2 1 θn 2 .
θ2 − θn1
dt
2
2
4.25
Consider the bounded assumption of au and bu; the following inequality can be
got:
In
β
β
C D 0,x aUθ, C D x,b θ
dt ≥ m
In
β
β
C D 0,x θ, C D x,b θ
2
β
dt m
C D0,x θ dt.
Since bu is bounded, and from Lemma 2.8, where
the following inequality is established:
In
β
C D x,b
4.26
is a linear bounded map,
β
β
bUθ, C Dx,b θ dt ≥ m
θ, C Dx,b θ dt
In
In
≥ Cm
In
β
θ C Dx,b θdt ≥ Cm
θ2 dt.
4.27
In
The first and second terms of the left side of 4.24 can be seen as
−
− 2
−
θn1 , θn1
θn1 ,
2 θ 2
θn− , θn ≤ θn− · θn ≤ θn− n .
4
4.28
Taking 4.24–4.27 into the left side of 4.22 yields
1
θ− 2 1 θn 2 m
n1
2
2
2
β
θ2 dt
C D0,x θ dt Cm
In
In
2 θ 2
≤ θn− n
4
q β
β
R1 , θn,i R2 , θn,i R3 , C Dx,b θn,i R4 , C Dx,b θn,i R5 , θn,i .
i1
4.29
18
Mathematical Problems in Engineering
As to R1 , for i 1, 2, . . . , q, there exists
δqi −
q
Wn,j ln,i
tn,i −
ln,i tn j1
δqi −
In
ln,i
tdt − ln,i tn 0.
4.30
This means that there exists constants Cij and C, which are independent of n, such that
q
R1 Cij ρn,j − ρn,j−1 ≤ Cρt τdτ In
j1
≤C
In
4.31
hs ut τs dτ ≤ Chs kn1/2 ut 0,s,n .
In order to estimate R2 , we apply the interpolation operator. Let I$qn : CI$n → Pq I$n be the interpolation operator on the time slab In , whose order is less than q. The interpolation
q
points involve not only Radau points, but also tn , and these points satisfy that I$n ytn,j q
$q
ytn,j , I$n ytn ytn , j 1, 2, . . . , q. Then, for every x ∈ Ω, such that ln,i
In u is a polynomial
whose order is 2q − 2,
q
W
l
−
l
dt
t
u
tu
R2 n,j n,i n,j
n,j
n,i
j1
In
q
q1/2 q1 ln,i
.
t I$n u − u dt ≤ Ckn
u
In
0,0,n
4.32
As to R3 , when kn is sufficiently small, with the aid of Hölder inequality, we obtain
β
β
R3 , C Dx,b θn,i − C D0,x R3 , θn,i
q
2β
≤ Wn,j C D0,x aU − auun,j · θn,i j1
q
2β
2β
ln,i t C D0,x auu dt
Wn,j C D0,x auun,j −
· θn,i j1
In
R31 R32 θn,i ,
4.33
Mathematical Problems in Engineering
where
19
q
2β
W
−
auu
aU
R31 D
n,j C 0,x
n,j j1
2β 2
2β
2
2
≤
ln,i t C D0,x θ dt LC
ln,i
t C D0,x W − udt
In
In
≤ LC
In
2β 2
q1/2 2β
2
ln,i
t C D0,x θ dt Ckn
C D0,x uq 0,0,n
2β
C D0,x uq q1/2 ≤ LC
In
2
ln,i
tθ2 dt Ckn
0,0,n
,
q
2β
2β
ln,i t C D0,x auu dt
R32 Wn,j C D0,x auun,j −
j1
In
q
2β
q
2β
$
≤ M Wn,j C D0,x In un,j −
ln,i t C D0,x u dt
j1
In
q
2β
2
≤ M ln,l
t I$n − I C D0,x u dt
In
q1/2 2β
≤ CMkn
.
C D0,x uq 4.34
0,0,n
Almost similar to the estimation of R3 , the analysis of R4 can be got by the help of
boundedness of bu, u, and Ph un
β
R4 , C Dx,b θn,i
q
β
θn,i ≤
W
u
−
buP
bU
D
n,j
C
h
n,j
0,x
j1
4.35
q
β
β
Wn,j C D0,x buPh un,j −
ln,i t C D0,x buu dtθn,i j1
In
≤C
In
q1/2 β
β ln,j C D0,x θdt kn
C D0,x uq 0,0,n
M Lhs−β kn1/2 u0,s,n θn,i .
The last term contains R5 which is estimated as follows:
2 1
2
2
R5 , θn ≤ Cun − Ph un Cu−n − Ph u−n θn 4
≤
2
Ch2s
n us
2
Ch2s
n−1 us
where hn is the space partition in the time slab In .
1
2
θn ,
4
4.36
20
Mathematical Problems in Engineering
Taking the estimations from R1 to R5 into 4.29, and setting Wn,j kn Wj , since
&q
&q
1/2
1/2
θ0,0,n j1 Wn,j θn,j 2 kn1/2 j1 Wj θn,j 2 , the following inequality holds:
1
θ− 2 m
n1
2
2
β
θ2 dtθ0,2,n
C D0,x θ dt Cm
2
≤ θn− In
q !
In
2 2 "
β
β
R1 2 R2 2 C D0,x R3 C D0,x R4 i1
C
4.37
q
2
2
2s
θn,i 2 Ch2s
n us Chn−1 us .
i1
Taking the interpolation representation of θ on time slab In into the error equation
4.22, and multiplying τi−1/2 in both side of the error equation, we have
−
q
1/2 −1/2 β
β
Wn,j ln,i
tn,j τj τi
θ'n,j , φ Wn,j τj1/2 τi−1/2 C D0,x aUθ'n,j , C Dx,b φ
j1
β
Wn,j τj1/2 τi−1/2 bUθ'n,j , C Dx,b φ δqi θ'n,j , φ
β
β
τi−1/2 θn− , ln,i tn φ R1 , φ R2 , φ R3 , C Dx,b φ R4 , C Dx,b φ R5 , φ .
4.38
Let φ θ'n,i , and sum from i 1 to q, with Lemma 3.2, we know that
q
q
q 2
%θ'n , θ'n ≥ α δqi θ'n,q , θ'n,i −
τi−1/2 τj1/2 θ'n,j , θ'n,i M
θ'n,j ,
i1
i,j1
4.39
j1
where θ'n θ'n,1 , θ'n,2 , . . . , θ'n,q .
The second term of the left side of 4.42 is estimated as
q
q
β
β
β
β
Wn,i C D0,x aUθ'n,i , C Dx,b θ'n,i ≥ m Wn,j C D0,x θ'n,i , C Dx,b θ'n,i
i1
i1
≥ mC
q 2
β
C D0,x θ'n,i ,
i1
q
i1
β
Wn,i bUθ'n,i , C Dx,b θ'n,i ≥ m
β
Wn,i θ'n,i , C Dx,b θ'n,i
q
i1
≥ mC
q
β
Wn,i θ'n,i · C Dx,b θ'n,i i1
≥ mCkn1/2
q 2
β
C D0,x θ'n,i .
i1
4.40
Mathematical Problems in Engineering
21
From 4.39-4.40, the following estimation can be derived:
q q q 2
β ' 2
β
' 2
1/2
α θn,i mc C D0,x θn,i mCkn
C D0,x θ'n,i i1
i1
q ' 2
≤C
θn,i i1
1/2 q 2
− 2
β
θn R1 2 R2 2 C D0,x R3 i1
4.41
i1
2 β
C D0,x R4 ⎫
⎬
1/2
un − Ph un .
⎭
Introduce the estimations from R1 to R4 and the inequality containing un − Ph un into the above expression, by the definition and characteristics of the norm of the fractional
derivative space and the relation of θ0,β,n , there holds
θ0,β,n
−
s−β
q q1 2
s
≤ Ckn { θn β hn ut 0,s,n hn u0,s,n kn u
0,0,n
uq 0,s,n
2β
C D0,x uq
0,0,n
β
C D0,x uq
0,0,n
hsn
hsn−1
us .
4.42
Introducing the estimations from R1 to R5 into 4.37, based on the boundedness of the
fractional operation, one has
2
− 2
q1 θ ≤ Ckn2q u
n1
0,0,n
2β q 2
D
u
C
0,x
0,0,n 2
uq 0,0,n
2
2
2
2s
2s
−2s
h2s
n hn−1 us hn ut 0,s,n hn ut 0,s,n
2
β
C D0,x uq 0,0,n
4.43
2
Cθ20,β,n θn− .
From inequalities 4.42 and 4.43, there holds
2
− 2
q1 θ ≤ Ckn2q 1 kn u
n1
0,0,n
1 kn h2s
n
h2s
n−1
2
2β
C D0,x uq u2s
0,0,n
1 2
β
C D0,x uq 2
kn h2s
n ut 0,s,n
0,0,n
2
1 kn θn− .
β
4.44
22
Mathematical Problems in Engineering
According to the above inequality, we can estimate θn− . With the θj , 0 ≤ j ≤ n,
n
− 2 #
θ ≤
C 1 kj θ0 2β
n1 β
j0
C
!
2q q1 2
1 kj Ckm u
n
n #
0,0,m
m0jm1
2
2β
C D0,x uq 0,0,m
2
β
C D0,x uq 0,0,m
2
2s
2s
h2s
m hm−1 us h ut 0,s,n
2s−2β
1 kn hn
2
2q
u0,s,n kn 1 kn uq 0,0,n
"
.
4.45
For a fixed n, if kj → 0, j 1, 2, . . . , n, we choose k max0≤j≤n kj , then there holds
n
#
n
#
1 kj ≤
1 k 1 kn ≤ eCtn1 .
j0
4.46
j0
Setting C1 eCtn1 , since θ0− u0 − Ph u0 , we get
n !
− 2
2q q1 2
θ ≤ Cu0 − Ph u0 2 C1
k
m u
β
n1
m0
0,0,m
2
2β
C D0,x uq 0,0,m
2
β
C D0,x uq 0,0,m
2
2s
2s
2s−2β
h2s
u2s
m hm−1 us hm ut 0,s,m h
2s−2β
1 kn hn
2
2q
u0,s,n kn 1 kn uq 0,0,n
"
.
4.47
Taking it into 4.43, we have
θ0,β,n ≤ Ckn1/2 h0 u0 s C1 kn1/2
n !
m0
2
2q km uq1 0,0,m
2
2β
C D0,x uq 0,0,m
2
β
C D0,x uq 0,0,m
2s−2β
2
2s
2s
h2s
u0,s,n
m hm−1 us hm ut 0,s,m 1 kn hn
2
2q
kn 1 kn uq 0,0,n
"
.
4.48
Note that U − u∞,β,n ≤ θ∞,β,n W − u∞,β,n . By Lemma 2.8, the estimate of W − u is
proved. Therefore, the proof of Theorem 4.5 is ended.
Mathematical Problems in Engineering
23
Table 1: Numerical error result for Example 5.1.
u − uh ∞, 0
0.1220e − 003
0.7433e − 004
0.3244e − 004
0.1642e − 004
0.7880e − 005
0.3952e − 005
h
1/10
1/20
1/40
1/80
1/160
1/320
Cvge. rate
—
0.7147
1.1962
0.9828
1.0588
0.9955
5. Numerical Examples
In this section, we present numerical results for the Galerkin approximations which support
the theoretical analysis derived in Section 4.
Let Sh denote a uniform partition on spacial domain 0, a and Xh the space of
continuous piecewise linear functions on Sh . In order to implement the discontinuous
Galerkin finite element approximation, we adapt the finite elements scheme in space domain
and the discontinuous finite element scheme along time domain. We associate shape function
of space Xh with the standard basis of hat functions on the uniform grid of size h 1/n and
adapt the same shape functions along time axis. In our scheme, the finite element trial and
test spaces for Problem 1 are chosen to be the same.
Example 5.1. Consider the following equation:
∂u
∂t
− C D1.6
0,x
x1.6
x0.8
− C D0.8
0,x
Γ4.6
Γ3.8
ux, 0 x2 ,
u0, t 0,
u,
0 ≤ x ≤ 1, 0 ≤ t ≤ 1,
5.1
0 ≤ x ≤ 1,
u1, t et ,
0 ≤ t ≤ 1.
The exact solution of the equation is ux, t e−t x2 .
If we select kn ch and note ux, 0 is smooth enough, then we have the following
numerical results presented in Table 1.
Table 1 includes numerical calculations over a regular partition of 0, 1. We can see
that, when the size of grid becomes smaller, the finite element approximation becomes better.
We can also observe that the experimental rates of convergence agree well with the theoretical
rates for the numerical solution.
Example 5.2. We consider another space fractional nonlinear Fokker-Planck differential
equation. The ux, t e−t x3 is the exact solution of the following equation:
∂u
∂t
− C D1.8
0,x
12et u
6et u
0.9
ux, t,
D
C
0,x
Γ5.8x1.2
Γ4.9x2.1
ux, 0 x3 ,
u0, t 0,
x ∈ 0, 2, t ∈ 0, 1,
0 ≤ x ≤ 2,
u1, t et ,
0 ≤ t ≤ 1.
5.2
24
Mathematical Problems in Engineering
Table 2: Numerical error result for Example 5.2.
h
u − uh ∞, 0
Cvge. rate
1/10
0.0231
—
1/20
0.0127
0.8610
1/40
0.0067
0.9315
1/80
0.0034
0.9660
1/160
0.0017
0.9830
1/320
0.0009
0.9915
Table 3: Numerical error result for Example 5.3.
h
u − uh ∞, 0
Cvge. rate
1/10
0.6125e − 003
—
1/20
0.4199e − 003
0.5447
1/40
0.1765e − 003
1.2500
1/80
0.0985e − 003
0.8425
1/160
0.0527e − 003
0.9003
We can also select kn ch. Table 2 shows the error results at different sizes of space
grid. We can still observe that the experimental rates of convergence are inline with the
theoretical convergence rates.
Example 5.3. Equation ux, t e−t x2 − x1.9 is the exact solution of the following problem:
−0.4
x u x1.5 u
x0.9 Γ3.5
x2
∂u
1.6
− C D0,x
− e−t ,
∂t
Γ4.6
Γ4.6Γ1.9
2
ux, 0 x2 − x1.9 ,
u0, t 0,
0 ≤ x ≤ 1,
u1, t 0,
0 ≤ x ≤ 1, 0 ≤ t ≤ 1,
5.3
0 ≤ t ≤ 1.
Table 3 includes numerical calculations over a regular partition of 0, 1. We can see that the
experimental numerical is valid and inline with the theoretical result. We can also see that
this method is valid for more generalized fractional diffusion-type equation, and the rate of
convergence depends on the highest order of the equation and the order of the shape function.
Acknowledgments
This work was partially supported by the National Natural Science Foundation of China
under Grant no. 10872119, Shanghai Leading Academic Discipline Project under Grant no.
S30104, and the Natural Science Foundation of Anhui province Grant no. KJ2010B442.
Mathematical Problems in Engineering
25
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