3.1. Mixed Formulation and Projections
In order to get the mixed scheme, we first split (1) into the following coupled system of two lower-order equations by introducing an auxiliary variable
=
∇
u
:
(10)
∂
α
u
(
x
,
t
)
∂
t
α
-
∇
·
σ
=
f
(
x
,
t
)
,
σ
-
∇
u
=
0
,
Based on the new mixed method in [26, 27], using Green's formula, the new mixed weak formulation of (10) is to determine
{
u
,
σ
}
:
[
0
,
T
]
↦
H
0
1
×
(
L
2
(
Ω
)
)
2
such that
(11a)
(
∂
α
u
(
x
,
t
)
∂
t
α
,
v
)
+
(
σ
,
∇
v
)
=
(
f
,
v
)
,
∀
v
∈
H
0
1
,
(11b)
(
σ
,
w
)
-
(
∇
u
,
w
)
=
0
,
∀
w
∈
(
L
2
(
Ω
)
)
2
.
In order to formulate a new mixed finite element scheme, we first define the mixed finite element spaces. As shown in the literatures [26, 27], we choose the mixed space
(
V
h
,
W
h
)
with finite element pair
P
1
-
P
0
2
as
(12)
V
h
=
{
v
h
∈
C
0
(
Ω
)
∩
H
0
1
∣
v
h
∈
P
1
(
K
)
,
∀
K
∈
𝒦
h
}
,
W
h
=
{
w
h
=
(
w
1
h
,
w
2
h
)
∈
(
L
2
(
Ω
)
)
2
∣
w
2
h
∈
P
0
(
K
)
,
∀
K
∈
𝒦
h
(
L
2
(
Ω
)
)
2
}
.
As discussed in [26, 27], we know that
(
V
h
,
W
h
)
satisfies the so-called discrete Ladyzhenskaya-Babuska-Brezzi condition.
In view of the definition of the above mixed space, the corresponding semidiscrete mixed scheme of (11a) and (11b) is to find
{
u
h
,
σ
h
}
:
[
0
,
T
]
↦
V
h
×
W
h
such that
(13a)
(
∂
α
u
h
(
x
,
t
)
∂
t
α
,
v
h
)
+
(
σ
h
,
∇
v
h
)
=
(
f
,
v
h
)
,
∀
v
h
∈
V
h
,
(13b)
(
σ
h
,
w
h
)
-
(
∇
u
h
,
w
h
)
=
0
,
∀
w
h
∈
W
h
.
Remark 4.
(i) If the standard mixed method is considered, the mixed weak formulation for problem (1) is to find
{
u
,
σ
}
:
[
0
,
T
]
↦
X
×
H
such that
(14a)
(
∂
α
u
(
x
,
t
)
∂
t
α
,
v
)
-
(
∇
·
σ
,
v
)
=
(
f
,
v
)
,
∀
v
∈
X
,
(14b)
(
σ
,
w
)
+
(
u
,
∇
·
w
)
=
0
,
∀
w
∈
H
,
where
X
=
{
w
∈
L
2
(
Ω
)
∣
w
|
∂
Ω
=
0
}
,
H
=
H
(
div
;
Ω
)
=
{
v
∈
(
L
2
(
Ω
)
)
2
∣
∇
·
v
∈
L
2
(
Ω
)
}
.
(ii) Compared with the classical mixed weak formulation (14a) and (14b), the gradient in our scheme (11a) and (11b) belongs to the simple square integrable
(
L
2
(
Ω
)
)
2
space avoiding the use of the complex
H
(
div
;
Ω
)
space. Obviously, the regularity requirements on the solution
σ
=
∇
u
is reduced.
(iii) So far, we have not seen any related reports on the study of mixed finite element methods for solving Fractional PDEs. Here, we will give some detailed theoretical analysis on a kind of new mixed element method for solving the fractional PDE (1).
In order to analyze the convergence of the method, we first introduce two mixed elliptic projection associated with our equations.
Lemma 5.
There exists a linear operator
Π
h
:
(
L
2
(
Ω
)
)
2
→
W
h
such that
(15)
(
σ
-
Π
h
σ
,
∇
v
h
)
=
0
,
∀
v
h
∈
V
h
,
(16)
∥
σ
-
Π
h
σ
∥
L
2
(
Ω
)
≤
C
h
m
∥
σ
∥
m
.
Lemma 6.
There exists a linear operator
P
h
:
H
0
1
(
Ω
)
→
V
h
such that
(17)
(
∇
(
u
-
P
h
u
)
,
w
h
)
=
0
,
∀
w
h
∈
W
h
,
(18)
∥
u
-
P
h
u
∥
+
h
∥
u
-
P
h
u
∥
1
≤
C
h
m
+
1
∥
u
∥
m
+
1
.
From [26–28], we can obtain the proof for Lemmas 5 and 6.
3.2. A Priori Error Estimates for Fully Discrete Scheme
In the following discussion, we will analyze some a priori error estimates for fully discrete schemes based on the case
0
<
α
<
1
. For the convenience of theoretical analysis, we now denote
(19)
B
n
-
k
α
=
(
n
-
k
+
1
)
1
-
α
-
(
n
-
k
)
1
-
α
,
D
t
u
k
+
1
=
3
u
k
+
1
-
4
u
k
+
u
k
-
1
2
Δ
t
.
By the discrete formula (3) for time-fractional derivative, (11a) and (11b) have the following equivalent formulation:
(20a)
Δ
t
1
-
α
Γ
(
2
-
α
)
∑
k
=
0
n
B
n
-
k
α
(
D
t
u
k
+
1
,
v
)
+
(
σ
n
+
1
,
∇
v
)
=
(
f
n
+
1
+
E
0
n
+
1
,
v
)
,
∀
v
∈
H
0
1
,
(20b)
(
σ
n
+
1
,
w
)
-
(
∇
u
n
+
1
,
w
)
=
0
,
∀
w
∈
(
L
2
(
Ω
)
)
2
.
Now, we formulate a completely discrete procedure: find
(
u
h
n
+
1
,
σ
h
n
+
1
)
∈
V
h
×
W
h
,
(
n
=
0,1
,
…
,
M
-
1
)
such that
(21a)
Δ
t
1
-
α
Γ
(
2
-
α
)
∑
k
=
0
n
B
n
-
k
α
(
D
t
u
h
k
+
1
,
v
h
)
+
(
σ
h
n
+
1
,
∇
v
h
)
=
(
f
n
+
1
,
v
h
)
,
∀
v
h
∈
V
h
,
(21b)
(
σ
h
n
+
1
,
w
h
)
-
(
∇
u
h
n
+
1
,
w
h
)
=
0
,
∀
w
h
∈
W
h
.
For the convenience of the analysis, we now decompose the errors as
(22)
u
(
t
n
)
-
u
h
n
=
(
u
(
t
n
)
-
P
h
u
n
)
+
(
P
h
u
n
-
u
h
n
)
=
η
n
+
ς
n
;
σ
(
t
n
)
-
σ
h
n
=
(
σ
(
t
n
)
-
Π
h
σ
n
)
+
(
Π
h
σ
n
-
σ
h
n
)
=
ρ
n
+
ξ
n
.
Subtracting (21a) and (21b) from (20a) and (20b) and using two projections (15) and (17), we get the error equations
(23a)
Δ
t
1
-
α
Γ
(
2
-
α
)
∑
k
=
0
n
B
n
-
k
α
(
D
t
ς
k
+
1
,
v
h
)
+
(
ξ
n
+
1
,
∇
v
h
)
=
-
Δ
t
1
-
α
Γ
(
2
-
α
)
∑
k
=
0
n
B
n
-
k
α
(
D
t
η
k
+
1
,
v
h
)
wwwii
+
(
E
0
n
+
1
,
v
h
)
,
∀
v
h
∈
V
h
,
(23b)
(
ξ
n
+
1
,
w
h
)
-
(
∇
ς
n
+
1
,
w
h
)
=
0
,
∀
w
h
∈
W
h
.
In the following discussion, we will derive the detailed process of proof for the fully discrete a priori error estimates.
Theorem 7.
Supposing that
u
h
0
=
P
h
u
(
0
)
,
σ
h
0
=
Π
h
σ
(
0
)
; then the error estimates hold with a parameter
0
<
α
<
1
:
(24a)
∥
u
(
t
n
)
-
u
h
n
∥
≤
C
(
u
,
α
,
T
)
T
α
1
-
α
(
Δ
t
+
Δ
t
-
α
h
m
+
1
)
,
(24b)
Δ
t
α
Γ
(
2
-
α
)
∥
σ
n
-
σ
h
n
∥
≤
C
(
u
,
α
,
T
)
T
α
1
-
α
(
Δ
t
+
Δ
t
-
α
h
m
+
1
+
h
m
)
.
Proof.
Noting that
(25)
∑
k
=
0
n
B
n
-
k
α
(
D
t
ς
k
+
1
,
v
h
)
=
∑
k
=
0
n
B
k
α
(
D
t
ς
n
-
k
+
1
,
v
h
)
.
Then (23a) may be rewritten as
(26)
Δ
t
1
-
α
Γ
(
2
-
α
)
∑
k
=
0
n
B
k
α
(
D
t
ς
n
-
k
+
1
,
v
h
)
+
(
ξ
n
+
1
,
∇
v
h
)
=
-
Δ
t
1
-
α
Γ
(
2
-
α
)
∑
k
=
0
n
B
k
α
(
D
t
η
n
-
k
+
1
,
v
h
)
wwwii
+
(
E
0
n
+
1
,
v
h
)
,
∀
v
h
∈
V
h
.
We add (26) to (23b), take
(
v
h
,
w
h
)
=
(
ς
n
+
1
,
ξ
n
+
1
)
, and multiply by
2
Δ
t
α
Γ
(
2
-
α
)
to arrive at
(27)
∑
k
=
0
n
B
k
α
(
3
ς
n
-
k
+
1
-
4
ς
n
-
k
+
ς
n
-
k
-
1
,
ς
n
+
1
)
+
2
Δ
t
α
Γ
(
2
-
α
)
∥
ξ
n
+
1
∥
2
=
-
2
Δ
t
∑
k
=
0
n
B
k
α
(
D
t
η
n
-
k
+
1
,
ς
n
+
1
)
wwwwii
+
2
Δ
t
α
Γ
(
2
-
α
)
(
E
0
n
+
1
,
ς
n
+
1
)
.
Now we consider the first term on the left-hand side of (27). Noting that
B
0
α
=
1
and denoting
B
n
+
1
α
=
B
-
1
α
=
0
, we have
(28)
∑
k
=
0
n
B
k
α
(
3
ς
n
-
k
+
1
-
4
ς
n
-
k
+
ς
n
-
k
-
1
,
ς
n
+
1
)
=
(
3
∑
k
=
0
n
B
k
α
ς
n
-
k
+
1
-
4
∑
k
=
0
n
B
k
α
ς
n
-
k
w
w
w
w
i
+
∑
k
=
0
n
B
k
α
ς
n
-
k
-
1
,
ς
n
+
1
)
=
(
3
∑
k
=
-
1
n
-
1
B
k
+
1
α
ς
n
-
k
-
4
∑
k
=
0
n
B
k
α
ς
n
-
k
+
∑
k
=
1
n
+
1
B
k
-
1
α
ς
n
-
k
,
ς
n
+
1
)
=
3
∥
ς
n
+
1
∥
2
+
(
∑
k
=
0
n
(
3
B
k
+
1
α
-
4
B
k
α
+
B
k
-
1
α
)
ς
n
-
k
,
ς
n
+
1
)
+
B
n
α
(
ς
-
1
,
ς
n
+
1
)
.
Substitute (28) into (27) to arrive at
(29)
3
∥
ς
n
+
1
∥
2
+
2
Δ
t
α
Γ
(
2
-
α
)
∥
ξ
n
+
1
∥
2
wiiii
=
-
2
Δ
t
∑
k
=
0
n
B
k
α
(
D
t
η
n
-
k
+
1
,
ς
n
+
1
)
wwww
+
2
Δ
t
α
Γ
(
2
-
α
)
(
E
0
n
+
1
,
ς
n
+
1
)
wwww
-
(
∑
k
=
0
n
(
3
B
k
+
1
α
-
4
B
k
α
+
B
k
-
1
α
)
ς
n
-
k
,
ς
n
+
1
)
wwww
-
B
n
α
(
ς
-
1
,
ς
n
+
1
)
.
For (29), we use Cauchy-Schwarz inequality to get
(30)
3
∥
ς
n
+
1
∥
2
+
2
Δ
t
α
Γ
(
2
-
α
)
∥
ξ
n
+
1
∥
2
≤
(
∑
k
=
0
n
|
3
B
k
+
1
α
-
4
B
k
α
+
B
k
-
1
α
|
∥
ς
n
-
k
∥
+
B
n
α
∥
ς
-
1
∥
wwww
+
∥
Δ
t
∑
k
=
0
n
-
1
B
k
α
D
t
η
n
-
k
-
1
∥
wwww
+
2
Δ
t
α
Γ
(
2
-
α
)
∥
E
0
n
+
1
∥
∑
k
=
0
n
)
∥
ς
n
+
1
∥
.
By (30), we can arrive at
(31)
3
∥
ς
n
+
1
∥
2
+
2
Δ
t
α
Γ
(
2
-
α
)
∥
ξ
n
+
1
∥
2
≤
C
(
u
,
α
,
T
)
B
n
α
(
Δ
t
1
+
α
+
Δ
t
2
+
α
+
h
m
+
1
)
∥
ς
n
+
1
∥
.
Now, we use induction to prove the conclusion (31).
Step 1. Setting
n
=
0
in (30) and noting that
B
0
α
>
B
1
α
and
B
-
1
α
=
0
, we can arrive easily at
(32)
3
∥
ς
1
∥
2
+
2
Δ
t
α
Γ
(
2
-
α
)
∥
ς
1
∥
2
≤
(
∑
k
=
0
-
1
|
3
B
1
α
-
4
B
0
α
+
B
-
1
α
|
∥
ς
0
∥
+
B
0
α
∥
ς
-
1
∥
wwww
+
∥
Δ
t
∑
k
=
0
-
1
B
k
α
D
t
η
-
1
∥
wwww
+
2
Δ
t
α
Γ
(
2
-
α
)
∥
E
0
1
∥
∑
k
=
0
-
1
)
∥
ς
1
∥
=
B
0
α
(
3
[
1
-
B
1
α
B
0
α
]
∥
ς
0
∥
+
(
∥
ς
0
∥
+
∥
ς
-
1
∥
)
wwwwww
+
1
B
0
α
∥
Δ
t
∑
k
=
0
-
1
B
k
α
D
t
η
-
1
∥
wwwwww
+
2
Δ
t
α
Γ
(
2
-
α
)
B
0
α
∥
E
0
1
∥
)
∥
ς
1
∥
≤
C
(
u
,
α
,
T
)
1
B
0
α
(
Δ
t
1
+
α
+
Δ
t
2
+
α
+
h
m
+
1
)
∥
ς
1
∥
.
So, when
n
=
0
, (31) holds.
Step 2. Supposing that (31) holds, for
n
≤
j
,
(33)
3
∥
ς
j
+
1
∥
2
+
2
Δ
t
α
Γ
(
2
-
α
)
∥
ξ
j
+
1
∥
2
≤
C
(
u
,
α
,
T
)
B
j
α
(
Δ
t
1
+
α
+
Δ
t
2
+
α
+
h
m
+
1
)
∥
ς
j
+
1
∥
.
Now, we consider the case for
n
=
j
+
1
. By (30) and the supposition (33), we have
(34)
3
∥
ς
(
j
+
1
)
+
1
∥
2
+
2
Δ
t
α
Γ
(
2
-
α
)
∥
ξ
(
j
+
1
)
+
1
∥
2
≤
(
∑
k
=
0
j
+
1
|
3
B
k
+
1
α
-
4
B
k
α
+
B
k
-
1
α
|
∥
ς
j
+
1
-
k
∥
+
B
j
+
1
α
∥
ς
-
1
∥
wwww
+
∥
Δ
t
∑
k
=
0
j
B
k
α
D
t
η
j
-
k
∥
wwww
+
2
Δ
t
α
Γ
(
2
-
α
)
∥
E
0
(
j
+
1
)
+
1
∥
∑
k
=
0
j
+
1
)
∥
ς
(
j
+
1
)
+
1
∥
≤
(
C
(
u
,
α
,
T
)
B
j
-
k
α
(
Δ
t
1
+
α
+
Δ
t
2
+
α
+
h
m
+
1
)
w
w
w
w
×
∑
k
=
0
j
+
1
|
3
B
k
+
1
α
-
4
B
k
α
+
B
k
-
1
α
|
+
B
j
+
1
α
∥
ς
-
1
∥
w
w
w
w
+
∥
Δ
t
∑
k
=
0
j
B
k
α
D
t
η
j
-
k
∥
w
w
w
w
+
2
Δ
t
α
Γ
(
2
-
α
)
∥
E
0
(
j
+
1
)
+
1
∥
C
(
u
,
α
,
T
)
B
j
-
k
α
)
∥
ς
(
j
+
1
)
+
1
∥
.
Noting that
1
/
B
j
-
k
α
<
1
/
B
j
+
1
α
in inequality (34); then we have
(35)
3
∥
ς
(
j
+
1
)
+
1
∥
2
+
2
Δ
t
α
Γ
(
2
-
α
)
∥
ξ
(
j
+
1
)
+
1
∥
2
≤
C
(
u
,
α
,
T
)
B
j
+
1
α
(
Δ
t
1
+
α
+
Δ
t
2
+
α
+
h
m
+
1
)
w
w
w
i
×
∑
k
=
0
j
+
1
|
3
B
k
+
1
α
-
4
B
k
α
+
B
k
-
1
α
|
∥
ς
(
j
+
1
)
+
1
∥
.
In order to obtain the estimate for (35), we have to discuss the boundedness for
∑
k
=
0
j
+
1
|
3
B
k
+
1
α
-
4
B
k
α
+
B
k
-
1
α
|
. Noting that
B
k
α
>
B
k
+
1
α
, we have
(36)
∑
k
=
0
j
+
1
|
3
B
k
+
1
α
-
4
B
k
α
+
B
k
-
1
α
|
=
∑
k
=
0
j
+
1
|
3
(
B
k
α
-
B
k
+
1
α
)
+
(
B
k
-
1
α
-
B
k
α
)
|
≤
3
∑
k
=
0
j
+
1
(
B
k
α
-
B
k
+
1
α
)
+
∑
k
=
1
j
+
1
(
B
k
-
1
α
-
B
k
α
)
+
B
0
α
=
5
B
0
α
-
3
B
j
+
2
α
-
B
j
+
1
α
.
Combining (36) with (35), we arrive at
(37)
3
∥
ς
(
j
+
1
)
+
1
∥
2
+
2
Δ
t
α
Γ
(
2
-
α
)
∥
ξ
(
j
+
1
)
+
1
∥
2
≤
C
(
u
,
α
,
T
)
B
j
+
1
α
(
Δ
t
1
+
α
+
Δ
t
2
+
α
+
h
m
+
1
)
∥
ς
(
j
+
1
)
+
1
∥
.
Making use of induction based on (32) and (37), we claim that (31) holds.
Note that the relationship [15]
(
n
+
1
)
-
α
/
B
n
α
→
1
/
(
1
-
α
)
holds; then we have
(38)
C
(
u
,
α
,
T
)
B
n
α
(
Δ
t
1
+
α
+
Δ
t
2
+
α
+
h
m
+
1
)
=
C
(
u
,
α
,
T
)
(
n
+
1
)
-
α
B
n
α
(
n
+
1
)
α
Δ
t
α
×
(
Δ
t
+
Δ
t
2
+
Δ
t
-
α
h
m
+
1
)
≤
C
(
u
,
α
,
T
)
T
α
1
-
α
(
Δ
t
+
Δ
t
-
α
h
m
+
1
)
.
By a substitution (38) into (31), we get
(39)
∥
ς
n
∥
+
Δ
t
α
Γ
(
2
-
α
)
∥
ξ
n
∥
≤
C
(
u
,
α
,
T
)
T
α
1
-
α
(
Δ
t
+
Δ
t
-
α
h
m
+
1
)
.
By a combination of (16) and (18) with triangle inequality, we get the error results of theorem.
Theorem 8.
With the same condition, one has the following a priori error estimates:
(40)
∥
u
(
t
n
)
-
u
h
n
∥
1
+
∥
σ
(
t
n
)
-
σ
h
n
∥
≤
C
(
u
,
α
,
T
)
T
α
1
-
α
(
Δ
t
+
Δ
t
-
α
h
m
+
1
+
h
m
)
.
Proof.
By (21a) and (21b), we easily get
(41)
Δ
t
1
-
α
Γ
(
2
-
α
)
∑
k
=
0
n
B
n
-
k
α
(
D
t
ξ
k
+
1
,
w
h
)
-
Δ
t
1
-
α
Γ
(
2
-
α
)
∑
k
=
0
n
B
n
-
k
α
(
∇
D
t
ς
k
+
1
,
w
h
)
=
0
.
We take
w
h
=
ξ
n
+
1
in (41) and
v
h
=
(
Δ
t
1
-
α
/
Γ
(
2
-
α
)
)
∑
k
=
0
n
B
n
-
k
α
∇
D
t
ς
k
+
1
in (41) and use Cauchy-Schwarz inequality and Young’s inequality to get
(42)
∥
Δ
t
1
-
α
Γ
(
2
-
α
)
∑
k
=
0
n
B
n
-
k
α
∇
D
t
ς
k
+
1
∥
2
+
Δ
t
1
-
α
Γ
(
2
-
α
)
∑
k
=
0
n
B
n
-
k
α
(
D
t
ξ
k
+
1
,
ξ
n
+
1
)
=
(
-
Δ
t
1
-
α
Γ
(
2
-
α
)
∑
k
=
0
n
B
n
-
k
α
D
t
η
k
+
1
+
E
0
n
+
1
,
Δ
t
1
-
α
Γ
(
2
-
α
)
∑
k
=
0
n
B
n
-
k
α
∇
D
t
ς
k
+
1
)
≤
(
∥
-
Δ
t
1
-
α
Γ
(
2
-
α
)
∑
k
=
0
n
B
n
-
k
α
D
t
η
k
+
1
∥
+
∥
E
0
n
+
1
∥
)
w
w
w
i
×
∥
Δ
t
1
-
α
Γ
(
2
-
α
)
∑
k
=
0
n
B
n
-
k
α
∇
D
t
ς
k
+
1
∥
≤
C
(
Δ
t
2
+
h
2
m
+
2
)
+
1
2
∥
Δ
t
1
-
α
Γ
(
2
-
α
)
∑
k
=
0
n
B
n
-
k
α
∇
D
t
ς
k
+
1
∥
2
.
By (42), we have
(43)
Δ
t
1
-
α
Γ
(
2
-
α
)
∑
k
=
0
n
B
n
-
k
α
(
D
t
ξ
k
+
1
,
ξ
n
+
1
)
≤
C
(
Δ
t
2
+
h
2
m
+
2
)
.
By a similar discussion to Theorem 7, we get
(44)
∥
ξ
n
∥
≤
C
(
u
,
α
,
T
)
T
α
1
-
α
(
Δ
t
+
Δ
t
-
α
h
m
+
1
)
.
Taking
w
h
=
∇
ς
n
in (23b) and using (31), we arrive at
(45)
∥
∇
ς
n
+
1
∥
≤
∥
ξ
n
+
1
∥
≤
C
(
u
,
α
,
T
)
T
α
1
-
α
(
Δ
t
+
Δ
t
-
α
h
m
+
1
)
.
Combining (44), (45), (16), and (18) with triangle inequality, we complete the proof.
Remark 9.
It is easy to find that a priori error estimate in
H
1
-norm for the variable
u
, which cannot be derived based on the classical mixed scheme (14a) and (14b), is gotten.