ANZIAM J. 50 (CTAC2008) pp.C610–C624, 2008 C610
A Petrov–Galerkin method for
integro-differential equations with a memory term
K. Mustapha
1(Received 29 July 2008; revised 16 December 2008)
Abstract
We investigate the numerical solution of an integro-differential equation with a memory term. For the time discretization we ap- ply the continuous Petrov–Galerkin method considered by Lin et al.
[SIAM J. Numer. Anal., 38, 2000]. We combined the Petrov–Galerkin scheme with respect to time with continuous finite elements for the space discretization and obtained a fully discrete scheme. We show op- timal error bounds of the numerical solutions for both schemes, and compare our theoretical error bounds with the results of numerical computations.
http://anziamj.austms.org.au/ojs/index.php/ANZIAMJ/article/view/1382 gives this article, c Austral. Mathematical Soc. 2008. Published December 22, 2008.
issn 1446-8735. (Print two pages per sheet of paper.)
Contents C611
Contents
1 Introduction C611
2 Stability results C615
3 Error estimates C618
4 Numerical experiments C622
References C624
1 Introduction
We study a class of numerical solutions for a linear integro-differential equa- tion of the form
u
0(t) + Au(t) + Z
t0
b(t, s)Au(s) ds = f(t) , t ∈ (0, T ] with u(0) = u
0, (1) where u
0(t) :=
∂u∂t(t), u(t) = u(x, t) and f(t) = f(x, t) with x ∈ Ω (Ω is a bounded domain subset of R
m, m ≥ 1). Here A is a linear, positive definite, self-adjoint operator with domain D(A) in a real Hilbert space H.
The solution u and the source term f take values in H, and the initial data u
0is an element of H. We set kvk
p= kA
p/2vk = hA
pv, vi
1/2, where k.k is the norm and h·, ·i is the inner product in H.
For 0 ≤ i + j ≤ 1 and for some 0 < α < 1 , we assume that
∂
i+j∂t
i∂s
jb(t, s)
≤ C(t − s)
α−1. (2)
1 Introduction C612
Many have considered numerical methods of the problem (1). Typically, time discretization is performed by either using finite difference or continuous and discontinuous finite elements with a quadrature. Finite difference and dis- continuous Galerkin in time, and finite elements in space have been discussed in the case of a smooth and non-smooth memory term (Larsson et al. [1] and Zhang [4]). Another approach using Laplace transform has been studied, for example, by McLean et al. [3].
In this article we solve the problem (1) using the Petrov–Galerkin method with respect to time (pgmt). Then we combine the pgmt with the finite el- ements discretization in space, which defines a fully discrete Petrov–Galerkin (pg) finite element method (pgfem) for ( 1). We derive the error estimates from both schemes. In addition, we present numerical evidence that our error bounds are optimal.
Unlike the standard Galerkin method, which is a special case of the pg
method, the pg method allows the trial and test spaces to be different. The
considered pgmt in this work is similar to the Petrov–Galerkin method stud-
ied by Lin et al. [2] for linear Volterra integro-differential equations. The
pgmt is a hybrid of the continuous and discontinuous Galerkin methods
with respect to time. The trial space is the same as of that the continu-
ous Galerkin method and the test space is a duplication of the test space in
the discontinuous Galerkin method. Because of the discontinuity of the test
space, the present method is highly stable. Theoretically, if the approximate
solution is a piecewise linear polynomial in time, optimal order convergence
results using L
∞-norm in time and the norm k · k
`in space (for ` = 0, 1) have
been shown globally of the pgfem. Proving the convergence in the L
∞-norm
in time and the norm k · k
1in space is a marked theoretical advantage over
the other existing numerical methods for solving (1). Practically, the number
of unknowns in the pgmt method is fewer than the number of unknowns in
the discontinuous Galerkin method with respect to time.
1 Introduction C613
By setting ˜ u(t) = u(t) − u(0), equation (1) becomes
˜
u
t+ A˜ u(t) + Z
t0
b(t, s)A˜ u(s) ds = f(t) − Au(0) − Z
t0
b(t, s)Au(0) ds for t ∈ (0, T ] with ˜ u(0) = 0 . Therefore, without loss of generality, instead of (1), we consider the problem
u
t+ Au(t) + Z
t0
b(t, s)Au(s) ds = f(t) for t ∈ (0, T ] with u(0) = 0 . (3) To define our time pgmt, we introduce the time partitions: 0 = t
0< t
1<
· · · < t
N= T of [0, T ] and denote the step sizes by k
n= t
n−t
n−1. We set the time intervals I
n= (t
n−1, t
n], and the maximum step size k = max
nk
n. For m ≥ 1 , let P
m(D(A
1/2)) denote the space of polynomials in the variable t of degree strictly less than m with coefficients in D(A
1/2). Let
W
k(D(A
1/2)) := {X ∈ C([0, T], D(A
1/2)) : X(0) = 0, X |
In∈ P
2(D(A
1/2)), 1 ≤ n ≤ N },
T
k(D(A
1/2)) := {X ∈ L
2([0, T ], D(A
1/2)) : X |
In∈ P
1(D(A
1/2)), 1 ≤ n ≤ N }, be the trial and test spaces respectively.
The pgmt of problem ( 3) is: compute U : [0, T ] × Ω → W
k(D(A
1/2)) such that
G
N(U, X) = Z
T0
hf(t), X(t)i dt for all X ∈ T
k(D(A
1/2)) (4) where
G
N(U, X) = Z
T0
[hU
0(t), X(t)i + A(U(t), X(t))] dt +
Z
T0
Z
t0
b(t, s)A(U(s), X(t)) ds dt , (5)
1 Introduction C614
and A(u, w) denote the bilinear forms on D(A
1/2) generated by hAu, wi.
The solution u of (3) satisfies G
N(u, X) = R
tN0
hf(t), X(t)i dt for all X ∈ T
k(D(A
1/2)). So U − u satisfies the orthogonality condition
G
N(U − u, X) = 0 for all X ∈ T
k(D(A
1/2)) . (6) Because the function X in T
k(D(A
1/2)) is not required to be continuous at the nodes t
nfor 1 ≤ n ≤ N , we choose its values on the different time intervals independently. By choosing X to vanish outside I
n, our numerical scheme (4) reduces to one equation for each I
nfor 1 ≤ n ≤ N . Therefore, the scheme (4) requires us to determine U ∈ W
k(D(A
1/2)) such that, for n = 1, . . . , N ,
Z
tntn−1
hU
0(t), X(t)i + A(U(t), X(t))
dt +
Z
tntn−1
Z
t 0b(t, s)A(U(s), X(t)) ds dt
= Z
tntn−1
hf(t), X(t)i dt for all X ∈ P
1(D(A
1/2)) . (7)
For the fully discrete scheme we assume that H = L
2(Ω) for a bounded, convex domain Ω, and that A is a strongly elliptic, second order partial differential operator subject to homogeneous Dirichlet boundary conditions.
We also assume that H
r(Ω) = H
r(Ω) and H
10(Ω) = H
10(Ω) denotes the space of all functions φ ∈ H
1(Ω) with φ = 0 on ∂Ω.
To describe our fully discrete scheme, we triangulate and define the usual continuous, piecewise linear finite element space S
h⊆ H
10(Ω), where h de- notes the maximum diameter of the elements. Let the mesh be quasi-uniform so that the Ritz projector R
h: H
10(Ω) → S
hfor the positive definite bilinear form A(u, v) which is defined by A(R
hv − v, χ) = 0 for all χ ∈ S
h, has the approximation property
kv − R
hvk
`≤ Ch
r−`kvk
rfor v ∈ H
10(Ω) ∩ H
r(Ω) with r ≥ 2 , (8)
2 Stability results C615
where kvk
m:= kvk
Hm(Ω)= kvk
Hm(Ω). Let P
2(S
h) and P
1(S
h) denote the spaces of polynomials in the variable t of degree strictly less than 2 and 1 respectively with coefficients in S
h. Define the corresponding trial and test spaces of piecewise polynomials W
k(S
h) and T
k(S
h) respectively.
By combining the pgmt introduced above with the use of finite elements for the approximation in the spatial variables, we arrive at a fully discrete pgfem of problem ( 3) which is: compute U
h: [0, T ] × Ω → W
k(S
h) such that
G
N(U
h, X) = Z
tN0
hf(t), X(t)i dt for all X ∈ T
k(S
h) . (9) The next section, §2, proves the stability of the proposed numerical schemes.
Optimal order convergence using L
∞-norm in time and H
`-norm in space is shown in Section 3 for ` = 0, 1 . Some numerical experiments demonstrating our theoretical results are given in Section 4
2 Stability results
We study next the stability of the approximate solution U defined by (4) or equivalently by (7). Since (4) and (9) have the same form, the stability of the fully discrete solution U
hdefined by (9) can be obtained in a similar manner.
Throughout the rest of the article, we assume that b(t, t) = 0 ; however, our approach to prove the stability and the convergence can be extended to cover the case that b(t, t) is not zero. In the next theorem we show the stability of the approximate solution U. For brevity, we introduce the following notations
U
n= U(t
n) , kUk
Jn= sup
t∈Jn
kU(t)k and
2 Stability results C616
kUk
p,Jn= sup
t∈Jn
kU(t)k
pwhere J
n=
n
[
j=1
I
j.
Theorem 1 Given U
0∈ H and f ∈ L
2(0, T ); H , then the approximate solution U defined by (4) is stable and
kUk
21,Jn
≤ C Z
tn0
kf(t)k
2dt for n = 1, 2, . . . , N .
Proof: Choose X = U
0in (4) and noting that hU
0(t), U
0(t)i =
12kU
0(t)k
2≥ 0 and A(U(t), U
0(t)) =
12dtdA(U(t), U(t)), we have
Z
tjtj−1
kU
0(t)k
2dt + A(U
j, U
j) − A(U
j−1, U
j−1)
+ 2 Z
tjtj−1
Z
t 0b(t, s)A(U(s), U
0(t)) ds dt = 2 Z
tjtj−1
hf(t), U
0(t)i dt . Hence, summing from j = 1 to j = n and using the Cauchy–Schwarz inequal- ity, we find that
Z
tn0
kU
0(t)k
2dt + kU
nk
21= 2 Z
tn0
hf(t), U
0(t)i dt − 2I
n≤ 2 Z
tn0
kf(t)kkU
0(t)k dt − 2I
n≤ Z
tn0
kf(t)k
2dt + Z
tn0
kU
0(t))k
2dt − 2I
n, (10) where
I
n= Z
tn0
Z
t 0b(t, s)A(U(s), U
0(t)) ds dt.
Reversing the order of integrals, integrating by parts, and using b(t, t) = 0 and U
0= 0 , we obtain
I
n= Z
tn0
Z
tns
hb(t, s)AU(s), U
0(t)i dt ds
2 Stability results C617
= Z
tn0
hb(t
n, s)AU(s), U(t
n)i − Z
tns
hb
t(t, s)AU(s), U(t)i dt
ds
= Z
tn0
b(t
n, s) d ds
Z
s 0AU(q) dq, U(t
n)
ds
− Z
tn0
Z
t 0hb
t(t, s)AU(s), U(t)i ds dt
= − Z
tn0
Z
s0
hb
s(t
n, s)AU(q), U(t
n)i dq ds
− Z
tn0
Z
t 0hb
t(t, s)AU(s), U(t)ids dt . Hence, using (2), we find that
|I
n| ≤ C Z
tn0
(t
n− t)
α−1Z
t0
kU(s)k
1kU(t
n)k
1ds dt + C
Z
tn0
kU(t)k
1Z
t0
(t − s)
α−1kU(s)k
1ds dt and thus, the inequality and Lemma 6.3 of Larsson et al. [1] yield
|I
n| ≤
1t
α+1nkU(t
n)k
21+ C
1Z
tn0
(t
n− t)
α−1Z
t0
kU(s)k
21ds dt
+
2Z
tn0
kU(t)k
21dt + C
2Z
tn0
Z
t 0(t − s)
α−1kU(s)k
1ds
2dt
≤
1t
α+1nkU(t
n)k
21+
2Z
tn0
kU(t)k
21dt + C
1,2Z
tn0
(t
n− t)
α−1Z
t0
kU(s)k
21ds dt .
Inserting this bound in (10) and choosing
1sufficiently small, we obtain kU
nk
21≤ C
Z
tn0
kf(t)k
2dt + 2
2Z
tn0
kU(t)k
21dt
3 Error estimates C618
+ C
2Z
tn0
(t
n− t)
α−1Z
t0
kU(s)k
21ds dt . (11) Since kUk
1,Jn= kU
n0k
1for some 1 ≤ n
0≤ n , and so by using (11) for n = n
0, we achieve
Z
tn0 0kU(t)k
21dt ≤ t
n0kU
n0k
21≤ Ct
n0Z
tn00
kf(t)k
2dt + 2t
n2Z
tn00
kU(t)k
21dt + C
2t
n0Z
tn00
(t
n0− t)
α−1Z
t0
kU(s)k
21ds dt . Thus, for
2sufficiently small, we have
Z
tn00
kU(t)k
21dt ≤ C Z
tn00
kf(t)k
2dt + C Z
tn00
(t
n0− t)
α−1dt Z
tn00
kU(s)k
21ds .
Now, an application of Lemma 6.4 of Larsson et al. [1] yields R
tn00
kU(t)k
21dt ≤ C R
tn00
kf(t)k
2dt . Inserting this into (11), we find that kUk
21,Jn= kU
n0k
21,Jn≤ C Z
tn00
kf(t)k
2dt + C Z
tn0
(t
n0− t)
α−1Z
tn00
kf(s)k
2ds dt
≤ C Z
tn00
kf(s)k
2ds ≤ C Z
tn0
kf(s)k
2ds .
♠
3 Error estimates
We estimate the errors U−u and U
h−u where U and U
hare the approximate
solutions obtained using the pgmt and pgfem respectively. In the next two
3 Error estimates C619
theorems we assume that there exists a constant C
0such that ku
00k
`,Jn+
Z
tn0
kAu
00(t)k
2dt ≤ C
0for all 1 ≤ n ≤ N .
In the following theorem we derive the error U − u when U is given by (4) and u is the continuous solution of (3).
Theorem 2 Let U and u be the solutions of (4) and (3) respectively. Then kU − uk
`,Jn≤ Ck
2for ` = 0, 1 and for all 1 ≤ n ≤ N .
Proof: Let Πu ∈ W
k(D(A
1/2)) be a continuous piecewise linear interpolant of u defined by
Πu(t
n) = u(t
n) for n = 0, 1, . . . , N .
Put θ = U − Πu and η = Πu − u . We decompose the error into two terms,
U − u = θ + η , (12)
and estimate each term separately. For ` = 0, 1 , kηk
`,Jn≤ C max
1≤j≤n
k
nZ
tjtj−1
ku
00(t)k
`dt
!
≤ Ck
2ku
00k
`,Jn≤ Ck
2. (13)
To estimate the term θ in (12), we use the orthogonality relation (6) and get G
N(θ, X) = −G
N(η, X) for X ∈ T
k(D(A
1/2)) with 1 ≤ n ≤ N . Using the defining properties of Πu and X is constant on I
n, we conclude
G
N(θ, X) = − Z
tN0
Aη(t) + Z
t0
b(t, s)Aη(s) ds, X(t)
dt (14)
3 Error estimates C620
for all X ∈ T
k(D(A
1/2)). Equation (14) has the same form as the equation (4) which is satisfied by U, so we apply the stability result of Theorem 1 and obtain
kθk
21,Jn≤ C Z
tn0
kAη(t)k + Z
t0
|b(t, s)| kAη(s)k ds
2for 1 ≤ n ≤ N . The Cauchy–Schwarz inequality, (2), Lemma 6.3 of Larsson et el. [1], and (13), yield
kθk
21,Jn≤ 2 Z
tn0
kAη(t)k
2dt + C Z
tn0
Z
t 0(t − s)
α−1kAη(s)k ds
2dt
≤ 2 Z
tn0
kAη(t)k
2dt + C Z
tn0
(t
n− t)
α−1Z
t0
kAη(s)k
2ds dt
≤ C X
nj=1
Z
tjtj−1
kAη(t)k
2dt
≤ C X
nj=1
k
jk
jZ
tjtj−1
kAu
00(s)k ds
!
2dt
≤ C X
nj=1
k
4jZ
tjtj−1
kAu
00(s)k
2ds
≤ Ck
4. (15)
Therefore, using (12), (13), and (15), we obtain the desired result. ♠
Next we estimate the error U
h− u where U
his the approximate solution obtained using the pgfem and u is the continuous solution of ( 3).
Theorem 3 If u is the solution of the problem (3) and if U
his the approx- imate solution defined by (9), then for r ≥ 2 and ` = 0, 1 ,
kU
h− uk
`,Jn≤ Ck
2+ Ch
r−`Z
tn0
ku
0(t)k
2rdt
1/2for all 1 ≤ n ≤ N .
3 Error estimates C621
Proof: We use the Ritz projector R
hand decompose the error as
U
h− u = (U
h− ΠR
hu) + (ΠR
hu − u) . (16) So,
kU
h− uk
`,Jn≤ kΠR
hu − uk
`,Jn+ kU
h− ΠR
huk
`,Jnfor ` = 0, 1 . (17) To estimate the first term ΠR
hu − u , we write ξ = R
hu − u , using (13), ξ(0) = 0 , and (8) with u
0in place of v, we observe that for ` = 0, 1 ,
kΠR
hu − uk
`,Jn= kΠu − u + Π(R
hu − u)k
`,Jn≤ kΠu − uk
`,Jn+ kξk
`,Jn≤ kΠu − uk
`,Jn+ kξ(0)k
`+ Z
tn0
kξ
0k
`dt
≤ Ck
2+ Ch
r−`Z
tn0
ku
0(t)k
rdt . (18) The Galerkin orthogonality property (6) now takes the form
G
N(U
h− u, X) = 0 for all X ∈ T
k(S
h). (19) Adapting the proof of Theorem 2, we see from (19) that
G
N(U
h− V, X) = −G
N(V − u, X) for all X ∈ T
k(S
h), (20) where V := ΠR
hu , and because
Z
tntn−1
hV
0, Xi dt = hΠR
hu(t
n) − ΠR
hu(t
n−1), X(t
n−1/2)i
= hR
hu(t
n) − R
hu(t
n−1), X(t
n−1/2)i
= Z
tntn−1
h(R
hu)
0, Xi dt = Z
tntn−1
hR
hu
0, Xi dt ,
4 Numerical experiments C622
the formula (5) gives
G
N(V − u, X) = X
N n=1Z
tntn−1
ξ
0+ A(V − u) + Z
t0
b(t, s)A(V − u)(s) ds, X
dt . The definition of the Ritz projector gives
A V − u, X = A R
hΠu − u, X = A Πu − u, X = hAη, Xi , so
G
N(V − u, X) = Z
tN0
ξ
0+ Aη + Z
t0
b(t, s)Aη(s) ds, X
dt . Thus, from (20),
G
N(U
h− V, X) = − Z
tN0
ξ
0+ Aη + Z
t0
b(t, s)Aη(s) ds, X
dt .
for all X ∈ W
k(S
h) . Stability of the pgmt (Theorem 1 with H = S
h) and the already estimated term R
tn0
kAη + R
t0
b(t, s)Aη(s) dsk
2dt in Theorem 2 now yield the estimate
kU
h− Vk
21,Jn
≤ C Z
tn0
kξ
0k
2+ Aη +
Z
t0
b(t, s)Aη(s) ds
2
dt
≤ Ck
4+ Ch
2rZ
tn0