error estimates for semidiscrete galerkin approximation to a time dependent parabolic...

CALCOLO 37, 181 – 205 (2000) CALCOLO© Springer-Verlag 2000

Error estimates for semidiscrete Galerkinapproximation to a time dependent parabolicintegro-differential equation with nonsmooth data

A. K. Pani1, R. K. Sinha2

1 Department of Mathematics, Indian Institute of Technology, Bombay, Powai,Mumbai-400 076, Indiae-mail: [email protected]

2 Department of Mathematics, Indian Institute of Technology, Guwahati Institute ofEngineers Building, Panbazar, Guwahati-781 001, Indiae-mail: [email protected]

Received: December 1998 / Revised version: January 2000

Abstract. In this paper, an attempt has been made to carry over knownresults for the finite element Galerkin method for a time dependent parabolicequation with nonsmooth initial data to an integro-differential equation ofparabolic type. More precisely, for the homogeneous problem a standardenergy technique in conjunction with a duality argument is used to obtain

anL2-error estimate of orderO(h2

t

)for the semidiscrete solution when

the given initial function is only inL2. Further, for the nonhomogeneouscase with zero initial condition, an error estimate of orderO

(h2 log

(1h

))uniformly in time is proved, provided that the nonhomogeneous term is inL∞(L2). The present paper provides a complete answer to an open problemposed on p. 106 of the bookFinite Element Methods for Integro-differentialEquations by Chen and Shih.

1 Introduction

We consider a semidiscrete Galerkin method for a parabolic integro-differ-ential equation of the form

ut + A(t)u =∫ t

0B(t, s)u(s)ds + f in � × J,

u = 0 on∂� × J, (1.1)

u(·,0) = u0 in �.

182 A. K. Pani, R. K. Sinha

Here,u = u(x, t) andf = f (x, t) are real-valued functions in� × J ,where� is a bounded domain inRd with smooth boundary∂�, J = (0, T ],T < ∞ andut = ∂u

∂t. Further,A(t) is a selfadjoint, uniformly positive

definite second-order elliptic partial differential operator of the form

A(t) = −d∑

i,j=1

∂

∂xj

(aij (x, t)

∂

∂xi

)+ a0(x, t)I,

andB(t, s) is a general second-order partial differential operator

B(t, s) = −d∑

i,j=1

∂

∂xj

(bij (x; t, s) ∂

∂xi

)+

d∑j=1

bj (x; t, s) ∂

∂xj+b0(x; t, s)I.

The nonhomogeneous termf and the coefficients ofA(t) andB(t, s) areassumed to be smooth.

Parabolic integro-differential equations (1.1) occur in many applications,such as heat conduction in materials with memory, compression of poro-viscoelastic media, nuclear reactor dynamics, etc.

For the purpose of the spatially semidiscrete Galerkin procedure, we de-fine H 1

0 = {φ ∈ H 1(�) | φ = 0 on∂�

}. Further, let A(t; ·, ·) and

B(t, s; ·, ·) be bilinear forms onH 10 × H 1

0 corresponding to operatorsA(t)andB(t, s), i.e.,

A(t; u, φ) =∫�

d∑

i,j=1

aij (x, t)∂u

∂xi

∂φ

∂xj+ a0(x, t)uφ

dx,

and

B(t, s; u(s), φ) =∫�

d∑

i,j=1

bij (x; t, s)∂u(s)∂xi

∂φ

∂xj

+d∑

j=1

bj (x; t, s)∂u(s)∂xj

φ + b0(x; t, s)u(s)φ dx.

The weak formulation of the problem (1.1) may be stated as: findu : J →H 1

0 such that

(ut , φ) + A(t; u, φ) =∫ t

0B(t, s; u(s), φ)d s + (f, φ)

∀φ ∈ H 10 , t ∈ J,

u(0) = u0.

(1.2)

Error estimates for semidiscrete Galerkin approximation 183

Here and below, we denote by(·, ·) and‖ · ‖ theL2 inner product and theinduced norm onL2(�). Let Hm = Hm(�), m ∈ Z, denote the standardSobolev space with norm denoted by‖ · ‖m. We shall assume thatA(t)satisfies the following elliptic regularity result and coercive property: thereexist positive constantsα andc such that

‖A(t)φ‖ ≥ α‖φ‖2, φ ∈ H 2 ∩ H 10 ,

and

A(t;φ, φ) ≥ c‖φ‖21, φ ∈ H 1

0 .

For the finite element Galerkin approximation, we assume that we aregiven a family{Sh}, 0< h < 1, of finite dimensional subspaces ofH 1

0 suchthat

infχ∈Sh

{‖φ − χ‖ + h‖φ − χ‖1

} ≤ Chr‖φ‖r ,φ ∈ Hr ∩ H 1

0 , r = 1,2.(1.3)

The standard semidiscrete finite element approximation is then defined as afunctionuh : J → Sh such that

(uht , χ) + A(t; uh, χ) =∫ t

0B(t, s; uh(s), χ) ds + (f, χ),

∀χ ∈ Sh, t ∈ J,

uh(0) =Phu0,

(1.4)

wherePhu0 is theL2-projection ofu0 ontoSh.Numerical solution by means of finite element methods has been inves-

tigated by several authors. Earlier, Greenwell Yanik and Fairweather [18]derived optimal error estimates for smooth solutions in the case of a non-linear problem with first-order partial differential operatorB. For problem(1.1) and its nonlinear variant a priori error estimates have also been provedin Cannon and Lin [2, 3], Lin et al. [7] and Lin and Zhang [8] for smoothinitial data by using the Ritz–Volterra projection as an intermediate solution.WhenA is independent of time, i.e.,A(t) = A in (1.1), special attentionhas been paid to time stepping for the computation of the memory term byeconomical quadrature in Sloan and Thomée [14] and Zhang [19]. For arelated result, see Le Roux and Thomée [6]. In all these papers, a spectralargument is used in a crucial way. Subsequently, Pani et al. [10] extendedthese results to the time dependent problem (1.1) by using energy argu-ments. For both smooth and nonsmooth initial data, Thomée and Zhang[16] have obtained optimalL2-error estimates whenA(t) ≡ A. It was alsoshown that, in the case of a homogeneous equation with nonsmooth data, the


parabolic integro-differential equation has a limited smoothing property andthe Galerkin solutionuh approximates the exact solutionu to orderO(h

2

t)

in theL2-norm. Their analysis is based on a semigroup-theoretic approachand the proof is quite complicated.Can it be simplified by using the Ritz–Volterra projection? This is an open problem proposed in the book by Chenand Shih [4, p. 106]. In the present paper, a complete answer to this problemis provided by using energy arguments and the Ritz–Volterra projection.

For discretization in time with advantageous storage requirements,Thomée and Zhang [17] considered backward Euler methods and obtainedrelated error estimates for nonsmooth initial data. More recently, Pani andPeterson [11] studied the finite element method with quadrature and ob-tained convergence of orderO

(h2

t

)for theL2-norm andO

(h2

tlog

(1h

))for

theL∞-norm via the energy method for the homogeneous equation, whenthe initial function is inH 1

0 ∩ H 2.In the absence of a memory term, i.e., whenB(t, s) ≡ 0, error analyses

for the cases of both smooth and nonsmooth data can be found in Brambleet al. [1], Luskin and Rannacher [9], Huang and Thomeé [5], Sammon [12],Thomeé [15] and its references.

In the present paper, for the time dependent problem (1.1) we use energytype arguments and the duality technique to obtain anL2-error estimate oforderO

(h2

t

)when the given initial functionu0 is only inL2. In particular,

an attempt has been made to carry over the energy technique in Luskinand Rannacher [9] for purely parabolic cases to time dependent parabolicintegro-differential equations. However, such an extension is not trivial dueto the presence of a memory term in the equation. Throughout this paperC denotes a generic positive constant independent ofh and of any functioninvolved; it is not necessarily the same at each occurrence.

The plan of the paper is as follows. In Sect. 2, some a priori estimatesfor the exact solution are derived using standard energy methods. Further,the Ritz–Volterra projection is introduced and related estimates are carriedout in this section. Section 3 is devoted to error estimates for smooth data.Finally, Sect. 4 deals withL2-error estimates for nonsmooth initial data.Subsequently, some results are obtained for nonhomogeneous equationswith zero initial data.

2 A priori estimates and Ritz–Volterra projection

In this section, we derive some a priori bounds which are needed in oursubsequent error analysis. Further, some error estimates related to the Ritz–Volterra projection are also discussed.


Lemma 2.1 Let u be the solution of (1.1). If u0 ∈ L2 and f ∈ H−1, then

‖u(t)‖2 +∫ t

0‖u(s)‖2

1ds ≤ C

(‖u0‖2 +

∫ t

0‖f (s)‖2

−1ds

).

Moreover, when u0 ∈ H 10 and f ∈ L2, then

‖u(t)‖21 +

∫ t

0{‖us(s)‖2 + ‖u(s)‖2

2} ds ≤ C

(‖u0‖2

1 +∫ t

0‖f ‖2 ds

).

Proof Settingφ = u in (1.2) and using standard energy arguments, it iseasy to prove the first part. For the second estimate, chooseφ = ut in (1.2)and then integrate from 0 tot to obtain∫ t

0‖us(s)‖2 ds + 1

2

∫ t

0

d

ds{A(s; u(s), u(s))}ds

= 1

2

∫ t

0As(s; u(s), u(s)) ds +

∫ t

0

∫ s

0B(s, τ ; u(τ), us(s)) dτ ds

+∫ t

0(f (s), us(s)) ds.

For the second term on the right-hand side, integration by parts yields∫ t

0

∫ s

0B(s, τ ; u(τ), us(s)) dτ ds =

∫ t

0B(t, τ ; u(τ), u(t)) dτ

−∫ t

0B(τ, τ ; u(τ), u(τ )) dτ −

∫ t

0

∫ s

0Bs(s, τ ; u(τ), u(s)) dτ ds.

Hence,

‖u(t)‖21 +

∫ t

0‖us(s)‖2 ds

≤ C

[‖u0‖2

1 +∫ t

0‖u(s)‖2

1 ds +∫ t

0‖f (s)‖2 ds

]. (2.1)

To estimate‖u‖2, we have from (1.1) and elliptic regularity forA(t) that

‖u(t)‖2 ≤ 1

α‖A(t)u(t)‖ ≤ C(α)

[‖ut‖ + ‖f ‖ +

∫ t

0‖u(s)‖2ds

].

An application of Gronwall’s Lemma yields

‖u(t)‖2 ≤ C

(‖ut‖ + ‖f ‖ +

∫ t

0‖us‖ds +

∫ t

0‖f ‖ds

), (2.2)

and the desired estimate now follows from (2.1) and (2.2).��


Lemma 2.2 Let u satisfy (1.1). If u0 ∈ L2 and f ≡ 0, then

(a) t‖u(t)‖21 +

∫ t

0s‖us(s)‖2ds ≤ C‖u0‖2,

(b) t2‖ut(t)‖2 +∫ t

0s2‖us(s)‖2

1ds ≤ C‖u0‖2 and

(c) t‖u(t)‖2 ≤ C‖u0‖, t ∈ J .

Proof Settingφ = tut in (1.2) and integrating from 0 tot we obtain∫ t

0s‖us(s)‖2 ds +

∫ t

0sA(s; u(s), us(s)) ds

=∫ t

0

∫ s

0sB(s, τ ; u(τ), us(s))dτ ds.

For the term on the right-hand side we now integrate by parts to obtain∫ t

0

∫ s

0sB(s, τ ; u(τ), us(s)) dτ ds

=∫ t

0tB(t, τ ; u(τ), u(t)) dτ

−∫ t

0τB(τ, τ ; u(τ), u(τ )) dτ

−∫ t

0

∫ s

0sBs(s, τ ; u(τ), u(s)) dτ ds

−∫ t

0

∫ s

0B(s, τ ; u(τ), u(s))dτds.

Since1

2

d

dt{tA(t; u, u)} = tA(t; u, ut ) + t

2At(t; u, u) + 1

2A(t; u, u),

we obtain

tA(t; u(t), u(t)) +∫ t

0s‖us(s)‖2 ds

≤ C

∫ t

0‖u(s)‖2

1ds + C

∫ t

0s‖u(s)‖2

1 ds.

We use Lemma 2.1 and Gronwall’s Lemma to complete the proof of (a).For (b), differentiating (1.2) with respect to time gives

(utt , φ) + A(t; ut , φ) = − At(t; u, φ)+ B(t, t; u(t), φ) +

∫ t

0Bt(t, s; u(s), φ) ds.

(2.3)


Settingφ = t2ut in the above equation, then yields

1

2

d

dt{t2‖ut‖2} + t2A(t; ut , ut ) ≤ t‖ut‖2 + Ct2‖u‖1‖ut‖1

+ C

∫ t

0t2‖u(s)‖1‖ut(t)‖1 ds.

We next integrate from 0 tot and apply Young’s inequality to obtain

t2‖ut(t)‖2 +∫ t

0s2‖us(s)‖2

1ds

≤ C

(∫ t

0s‖us(s)‖2ds +

∫ t

0‖u(s)‖2

1 ds

).

The estimate (b) now follows from Lemma 2.1 and estimate (a).Finally, to prove (c), we defineu(t) = ∫ t

0 u(s)ds and, hence,ut(t) =u(t). Now, by integrating by parts with respect tot , equation (1.1) yields

A(t)ut (t) = A(t)u(t) = −ut +∫ t

0B(t, s)us(s)ds

= −ut + B(t, t)u(t) −∫ t

0Bs(t, s)u(s)ds.

(2.4)

By elliptic regularity,

t‖u(t)‖2 ≤ t

α‖A(t)u(t)‖

≤ C(α)

[t‖ut(t)‖ + t‖u(t)‖2 + t

∫ t

0‖u(s)‖2ds

].

(2.5)

To obtain an estimate foru, integration of (2.4) from 0 tot leads to

A(t)u(t) = − u(t) + u0 +∫ t

0B(s, s)u(s)ds

−∫ t

0

(∫ s

0Bτ(s, τ )u(τ )dτ

)ds +

∫ t

0As(s)u(s)ds.

Again, use of elliptic regularity yields

‖u(t)‖2 ≤ C(α)

(‖u(t)‖ + ‖u0‖ +

∫ t

0‖u(s)‖2ds

).

We next apply Lemma 2.1 and Gronwall’s Lemma to obtain

‖u(t)‖2 ≤ C‖u0‖. (2.6)

Now, (2.5), (2.6) and estimate (b) imply (c). This completes the proof.��


Remark 2.1 The last property, i.e., estimate (c), states the smoothing prop-erty for the time dependent problem (1.1) fort ∈ J .

Lemma 2.3 For u0 ∈ H 2∩H 10 and f ≡ 0, we have the following estimates:

(a) ‖ut(t)‖2 +∫ t

0‖us‖2

1ds ≤ C[‖ut(0)‖2 + ‖u0‖2

],

(b) t‖ut(t)‖21 +

∫ t

0s‖uss‖2ds ≤ C‖u0‖2

2 and

(c) ‖u(t)‖2 ≤ C‖u0‖2.

Proof With φ = ut in (2.3), integrate the resulting equation from 0 tot andapply Young’s inequality to obtain

1

2‖ut(t)‖2 + (c − 3ε)

∫ t

0‖us(s)‖2

1ds ≤ 1

2‖ut(0)‖2 + C(ε)

∫ t

0‖u(t)‖2

1ds.

Chooseε appropriately and then use Lemma 2.1 withf = 0 to completethe proof of (a).

Setφ = tutt in (2.3) and proceed as in Lemma 2.2. Then use estimate(a) and Lemma 2.1 withf = 0 to obtain the required estimate (b).

Finally, since‖ut(0)‖ ≤ C‖u(0)‖2, estimate (c) is an immediate conse-quence of (2.2) and estimate (a). This completes the proof.��

In a similar way, one can easily derive the following a priori boundsfor the continuous time Galerkin solutionuh satisfying (1.4). The proof is,therefore, omitted here.

Lemma 2.4 Let uh satisfy (1.4). If f ≡ 0, then

(a) ‖uh(t)‖2 +∫ t

0‖uh(s)‖2

1ds ≤ C‖uh(0)‖2,

(b) ‖uht (t)‖2 +∫ t

0‖uhs(s)‖2

1ds ≤ C[‖uht (0)‖2 + ‖uh(0)‖2] and

(c) t2‖uht (t)‖2 +∫ t

0s2‖uhs(s)‖2

1ds ≤ C‖uh(0)‖2.

Now consider the following backward problems. For fixedt > 0 andgiven anyf ∈ L2 , let v : [0, t) → H 1

0 andvh : [0, t) → Sh, respectively,be solutions of

(φ, vs) − A(s;φ, v) = −∫ t

s

B(τ, s;φ, v(τ))dτ + (φ, f ),

∀φ ∈ H 10 , s ≤ t,

v(t) = g,

(2.7)


and

(χ, vhs) − A(s;χ, vh) = −∫ t

s

B(τ, s;χ, vh(τ ))dτ + (χ, f ),

∀χ ∈ Sh, s ≤ t,

vh(t) = gh.

(2.8)

With a simple change of variables in the proofs of Lemmas 2.1–2.3 andby using the backward Gronwall’s Lemma, it is easy to obtain a priori boundsfor the backward solutionsv andvh. We remark that the smoothing propertyfor the backward problem (2.7) can be proved in a similar way to the proofof (c) in Lemma 2.2. In this case we first definev(s) = − ∫ t

sv(τ )dτ and,

using similar to estimate (2.6), we obtain an estimate withf ≡ 0,

‖v(s)‖2 ≤ C‖g‖. (2.9)

Now the rest of the proof follows as in the case of Lemma 2.2. In fact wehave

(t − s)‖v(s)‖2 ≤ C‖g‖ for 0 ≤ s < t.

Following Lin et al. [7] (see [2, 3]), define the Ritz–Volterra projectionWh : [0, T ] → Sh by

A(t; (Whu − u)(t), χ) =∫ t

0B(t, s; (Whu − u)(s), χ)ds,

∀ χ ∈ Sh, t ∈ J .

(2.10)

To see that the Ritz–Volterra projection is well-defined, we refer to Cannonand Lin [3], Chen and Shih [4] and Lin et al. [7]. We also use the RitzprojectionRh = Rh(t) : H 1

0 → Sh defined by

A(t; (Rhu − u)(t), χ) = 0, ∀ χ ∈ Sh, t ∈ J . (2.11)

It is now quite standard to prove that

‖Rhu − u‖ + h‖Rhu − u‖1 ≤ Chj‖u‖j , u ∈ H 10 ∩ Hj, j = 1,2.

(2.12)

Let ρ = Whu − u. By using Lemma 2.1 in Pani et al. [10] the followingestimates ofρ and its temporal derivatives can be easily proved.

Theorem 2.1 Let ρ satisfy (2.10). Then, for r = {1,2}, we have

‖ρ(t)‖ + h‖ρ(t)‖1 ≤ Chr[‖u(t)‖r +

∫ t

0‖u(s)‖rds

],

and

‖ρt(t)‖ + h‖ρt(t)‖1 ≤ Chr[‖u(t)‖r + ‖ut(t)‖r +

∫ t

0‖u(s)‖rds

].


3 Error estimates for smooth initial data

In this section, we deriveL2 estimates ofe = u−uh for smooth initial data.For this purpose, the following lemmas will prove convenient.

Lemma 3.1 Let u and uh, respectively, be the solutions of (1.1) and (1.4).Then, for u0 ∈ H 1

0 and uh(0) = Whu0, it follows that∫ t

0‖e(s)‖2

1ds ≤ Ch2

(‖u0‖2

1 +∫ t

0‖f ‖2ds

).

Proof Setχ = Rhe in the error equation

(et , χ) + A(t; e, χ) =∫ t

0B(t, s; e(s), χ)ds (3.1)

to obtain

1

2

d

dt‖e‖2 + A(t; e, e) = (et , u − Rhu) + A(t; e, u − Rhu)

−∫ t

0B(t, s; e(s), u − Rhu) ds +

∫ t

0B(t, s; e(s), e) ds

≤ ‖et‖‖u − Rhu‖ + C

(‖e‖1 +

∫ t

0‖e(s)‖1 ds

)‖u − Rhu‖1

+ C

(∫ t

0‖e(s)‖1 ds

)‖e‖1.

Using (2.12) and integrating over 0 tot , we obtain with an application ofGronwall’s Lemma∫ t

0‖e(s)‖2

1 ds ≤ Ch2

(‖e(0)‖2 +

∫ t

0{‖u‖2

2 + ‖es‖2} ds).

Again, we use Lemma 2.1 with its discrete analogue to get∫ t

0‖e(s)‖2

1 ds ≤ Ch2

(‖u0‖2

1 + ‖uh(0)‖21 +

∫ t

0‖f ‖2 ds

).

Since‖uh(0)‖1 = ‖Whu0‖1 ≤ C‖u0‖1, the desired estimate follows andthis completes the proof.��Lemma 3.2 Let u and uh be the solutions of homogeneous problems (1.1)and (1.4), respectively. Then, for u0 ∈ H 1

0 ∩ Hr−j with r = {1,2} andj = {0,1} (r − j �= 0), we have∫ t

0‖e(s)‖2 ds ≤ Ct1−jh2r‖u0‖2

r−j .


Further, for u0 ∈ L2, we have∫ t

0‖e(s)‖2 ds ≤ Ch2‖u0‖2.

Proof Here we use the following duality argument. Letz(s) ∈ H 10 and

zh(s) ∈ Sh, respectively, be the solutions of the following backward prob-lems

(φ, zs) − A(s;φ, z) = −∫ t

s

B(τ, s;φ, z(τ ))dτ + (φ, e),

∀φ ∈ H 10 , s ≤ t,

z(t) = 0,

(3.2)

and

(χ, zhs) − A(s;χ, zh) = −∫ t

s

B(τ, s;χ, zh(τ )) dτ + (χ, e),

∀χ ∈ Sh, s ≤ t,

zh(t) = 0.

(3.3)

From Lemmas 2.1 and 3.1 (with time reversed), it follows easily that∫ t

0[‖zs − zhs‖2 + h−2‖z − zh‖2

1] ds ≤ C

∫ t

0‖e(s)‖2 ds. (3.4)

Takingφ = e in (3.2), we use error equation (3.1) withχ = zh to get

‖e(s)‖2 = d

ds(e, zh) + (e, zs − zhs) − A(s; e, z − zh)

+∫ t

s

B(τ, s; e(s), (z − zh)(τ )) dτ

−∫ s

0B(s, τ ; e(τ ), zh(s))dτ +

∫ t

s

B(τ, s; e(s), zh(τ )) dτ.Integrating the above equation from 0 tot we obtain∫ t

0‖e(s)‖2 ds

=∫ t

0(u − Rhu, zs − zhs)ds −

∫ t

0A(s; u − Rhu, z − zh) ds

+∫ t

0

∫ t

s

B(τ, s; (u − Rhu)(s), (z − zh)(τ )) dτ ds

≤∫ t

0‖zs − zhs‖‖u − Rhu‖ds + C

∫ t

0‖z − zh‖1‖u − Rhu‖1 ds

+ C

∫ t

0

∫ τ

0‖(u − Rhu)(s)‖1‖(z − zh)(τ )‖1 ds dτ.


From (2.12) and Young’s inequality, it follows that∫ t

0‖e(s)‖2ds ≤ ε

∫ t

0[‖zs − zhs‖2 + h−2‖z − zh‖2

1] ds

+ C(ε)h2r∫ t

0‖u‖2

r ds.

Finally, estimate (3.4) and a priori estimates in Lemmas 2.1, 2.3, 2.4 and anappropriate choice ofε complete the proof. ��

Setθ(t) = Whu(t) − uh(t). From (1.2), (1.4) and (2.10), we write anerror equation inθ as

(θt , χ) + A(t; θ, χ) =∫ t

0B(t, s; θ(s), χ) ds − (ρt , χ) ∀χ ∈ Sh, t ∈ J.

(3.5)

Define θ (t) = ∫ t

0 θ(s) ds. Thenθt (t) = θ(t). In order to derive our mainresult, we require the following estimates.

Lemma 3.3 Let θ (t) be defined as above. Then there is a positive constantC such that

‖θ (t)‖2 +∫ t

0‖θ (t)‖2

1 ds ≤ C

[∫ t

0‖θ(s)‖2 ds +

∫ t

0‖ρ(s)‖2 ds

].

Proof Setχ = θ in (3.5) to obtain

(θt , θ ) + A(t; θ, θ) =∫ t

0B(t, s; θs(s), θ (t))ds − (ρt , θ ).

Note thatd

dt(θ, θ) = (θt , θ ) + (θ, θ),

d

dt(ρ, θ) = (ρt , θ ) + (ρ, θ)

and1

2

d

dtA(t; θ , θ ) = A(t; θ, θ) + 1

2At(t; θ , θ ).

Further, integration by parts with respect tos yields∫ t

0B(t, s; θs(s), θ (t)) ds = B(t, t; θ (t), θ (t))−

∫ t

0Bs(t, s; θ (s), θ (t)) ds.

Again, integrating with respect tot and using(θ, θ) = (θt , θ ) with thecoercive property ofA(t; ·, ·), we obtain

1

2

d

dt‖θ (t)‖2 + c‖θ (t)‖2

1 ≤ C

[∫ t

0‖θ(s)‖2 ds+

∫ t

0‖θ (s)‖2

1 ds+‖θ (t)‖2

+ ‖ρ(t)‖2 +∫ t

0‖ρ(s)‖2 ds

].


Now, integrating both the sides from 0 tot and noting that

‖θ (t)‖ ≤∫ t

0‖θ(s)‖ ds,

we have

‖θ (t)‖2 +∫ t

0‖θ (s)‖2

1 ds ≤ C

[∫ t

0‖θ(s)‖2 ds +

∫ t

0‖ρ(s)‖2 ds

]

+ C

∫ t

0

∫ s

0‖θ (τ )‖2

1dτ ds.

Finally, apply Gronwall’s Lemma to complete the proof.��Lemma 3.4 Assume that u0 ∈ L2. Then there is a positive constant Cindependent of h such that

t‖θ(t)‖2 +∫ t

0s‖θ(s)‖2

1 ds ≤ Ch2‖u0‖2.

Proof Settingχ = tθ in (3.5) and integrating by parts we obtain

1

2

d

dt{t‖θ‖2} + tA(t; θ, θ) = 1

2‖θ‖2 + tB(t, t; θ (t), θ(t))

−∫ t

0tBs(t, s; θ (s), θ(t)) ds − t (ρt , θ).

Integrate from 0 tot and apply Young’s inequality to get

t‖θ(t)‖2 + 2(c − ε)

∫ t

0s‖θ(s)‖2

1 ds

≤ C(ε)

∫ t

0‖θ (s)‖2

1 ds + C(ε)

∫ t

0

∫ s

0‖θ (τ )‖2

1 dτ ds

+ C

∫ t

0{‖θ‖2 + s2‖ρs‖2} ds.

By choosingε appropriately, an application of Lemma 3.3 yields

t‖θ(t)‖2 +∫ t

0s‖θ(s)‖2

1 ds ≤ C

∫ t

0{‖ρ(s)‖2 + s2‖ρs(s)‖2} ds

+ C

∫ t

0‖e(s)‖2ds.

(3.6)

The desired estimate now follows from Theorem 2.1 and Lemmas 3.2, 2.1and 2.2. ��


Theorem 3.1 Let u and uh, respectively, be the solutions of (1.1) and (1.4)with f ≡ 0. Then, for u0 ∈ Hj ∩ H 1

0 , j = {1,2}, we have

‖e(t)‖ ≤ Chj‖u0‖j .Proof We write the errore(t)ase(t) = θ(t)−ρ(t). Then, usingTheorem 2.1and (3.6), we obtain

t‖e‖2 ≤ Ch2j

(t‖u‖2

j +∫ t

0[‖u‖2

j + s2‖us‖2j ] ds

)+ C

∫ t

0‖e‖2 ds.

The proof of the theorem follows by using Lemma 3.2 and the a prioriestimates of Lemmas 2.3 and 2.4.��

4 Error estimates for nonsmooth initial data

In this section, we prove a sequence of lemmas which will be used to deriveerror estimates for nonsmooth initial data.

Lemma 4.1 Let u and uh, respectively, satisfy (1.1) and (1.4) with f ≡ 0.Then, for u0 ∈ L2, there is a positive constant C independent of h such that

‖e(t)‖ ≤ Ct−12h‖u0‖.

Proof Note thatt12 ‖e(t)‖ ≤ t

12 ‖θ‖ + t

12 ‖ρ‖. By using Lemma 3.4, Theo-

rem 2.1 and Lemma 2.2, the desired result now follows.��Remark 4.1 Let e = v − vh be the error associated with the backwardproblem. Then, forg ∈ L2 andf ≡ 0, we can apply Lemma 4.1 toe withappropriate modifications to obtain

‖e(s)‖ ≤ C(t − s)−12h‖g‖, 0 ≤ s < t. (4.1)

For a detailed analysis, we refer to [13].

From (2.7) and (2.8) withf ≡ 0, we now have

d

ds{(u, v) − (uh, vh)} =

∫ s

0B(s, τ ; u(τ), v(s)) dτ

−∫ t

s

B(τ, s; u(s), v(τ )) dτ

−∫ s

0B(s, τ ; uh(τ), vh(s)) dτ

+∫ t

s

B(τ, s; uh(s), vh(τ )) dτ + (f, v − vh).

(4.2)


Integration from 0 tot leads to

(u(t), v(t)) − (uh(t), vh(t)) = (u0, v(0))

− (uh(0), vh(0)) +∫ t

0(f, v − vh) ds.

TakingPhu0 = uh(0) andPhg = gh we obtain

(e(t), g) = (u0, e(0)) +∫ t

0(f, e) ds. (4.3)

Definee(t) = ∫ t

0 e(s) ds. Thenet = e. To improve theL2-error estimatein Lemma 4.1, it is convenient to prove the following negative norm estimatesfor e ande.

Lemma 4.2 Let u0 ∈ L2 and f ≡ 0. Then there is a generic constant Cindependent of h such that

‖e(t)‖−j ≤ Chj‖u0‖ (4.4)

and

‖e(t)‖−j ≤ Cthj‖u0‖ (4.5)

hold for j = 1,2.

Proof We note from (4.3) that

|(e(t), g)| = |(u0, e(0))| ≤ ‖u0‖ ‖e(0)‖.ApplyingTheorem 3.1 to the backward errorewithg ∈ H 1

0 ∩Hj ,j = {1,2},we obtain

|(e(t), g)| ≤ Chj‖u0‖‖g‖j ,and this yields (4.4). To obtain (4.5), integrate (4.3) from 0 tot to get

|(e(t), g)| =∣∣∣∣∫ t

0(u0, e(0)) ds

∣∣∣∣ ≤ t‖u0‖‖e(0)‖.

An argument similar to the above completes the proof.��Lemma 4.3 If u0 ∈ L2 and f ≡ 0, then

‖e(t)‖1 ≤ Ch‖u0‖. (4.6)


Proof From the coercive property ofA(t), we have

c‖e‖21 ≤ A(t; e, e) = [A(t; e, e − Rhe) + A(t; e, Rhe)]. (4.7)

Using the definition ofe, we integrate the error equation (3.1) from 0 totand obtain, withuh(0) = Phu0,

(e, χ) + A(t; e, χ) =∫ t

0As(s; e, χ) ds +

∫ t

0B(s, s; e(s), χ) ds

−∫ t

0

∫ s

0Bs(s, τ ; e(τ ), χ)dτ ds.

(4.8)

Taking χ = Rhe in the above equation and substituting for the termA(t; e, Rhe) in (4.7) we get

c‖e‖21 ≤A(t; e, e − Rhe) − (e, e) + (e, e − Rhe)

+∫ t

0As(s; e(s), e) ds −

∫ t

0As(s; e(s), e − Rhe) ds

−∫ t

0B(s, s; e(s), e − Rhe) ds +

∫ t

0B(s, s; e(s), e) ds

+∫ t

0

∫ s

0Bs(s, τ ; e(τ ), (e − Rhe)(s)) dτ ds

−∫ t

0

∫ s

0Bs(s, τ ; e(τ ), e(s)) dτ ds,

and hence we obtain

‖e‖21 ≤ C

[‖e‖1‖u − Rhu‖1

+ ‖e‖−1‖e‖1 + ‖e‖‖u − Rhu‖ +∫ t

0‖e(s)‖1‖e‖1 ds

+∫ t

0‖e‖1‖u − Rhu‖1 ds

+∫ t

0

∫ s

0‖e(τ )‖1‖(u − Rhu)(s)‖1d τ ds

+∫ t

0

∫ s

0‖e(τ )‖1‖e(s)‖1 dτ ds

].

Applying the inequalityab ≤ ε2a

2 + b2

2ε , a, b ≥ 0, ε > 0, and (2.12), we get

(1 − 2ε)‖e‖21 ≤ C(ε)

(h2(‖u‖2

2 +∫ t

0‖u‖2

2 ds + ‖e‖2) + ‖e‖2−1

)

+ C

∫ t

0‖e‖2

1 ds.


With ε = 14, use (2.6), Lemmas 4.2, 2.1, and Gronwall’s Lemma to complete

the proof. ��Lemma 4.4 Let u0 ∈ L2 and f ≡ 0. Then there is a generic constant Cindependent of h such that

∫ t

0‖e‖2ds ≤ Cth4‖u0‖2 (4.9)

and

‖e(t)‖2 ≤ Ch4‖u0‖2. (4.10)

Proof Letw ∈ H 2 ∩H 10 be the solution of the following backward problem

ws − A(s)w = −∫ t

s

B∗(τ, s)w(τ) dτ + e, s ≤ t,

w(t) = 0,(4.11)

whereB∗(τ, s) is the adjoint ofB(τ, s). By reversing time and using changeof variables, it is an easy exercise to check that the solutionw(s) and itssemidiscrete solutionwh(s), which may be stated in a manner similar to(3.2)-(3.3), satisfy the following estimate

∫ t

0[‖ws − whs‖2 + h−2‖w − wh‖2

1 + ‖w‖22] ds ≤ C

∫ t

0‖e‖2 ds. (4.12)

Next we multiply (4.11) bye and integrate by parts with respect tox. Thenwe use (4.8) withχ = wh to get

‖e(s)‖2 = d

ds(e, w) − (e, w − wh) − A(s; e, w − wh)

−∫ s

0Aτ(τ ; e(τ ), wh(s)) dτ

−∫ s

0B(τ, τ ; e(τ ), wh(s)) dτ

+∫ s

0

∫ τ

0Bτ ′(τ, τ ′; e(τ ′), wh(s)) dτ

′ dτ

+∫ t

s

B(τ, s; e(s), (w − wh)(τ )) dτ

+∫ t

s

B(τ, s; e(s), wh(τ )) dτ.


Rewriting the above identity, we obtain

‖e(s)‖2 = d

ds(e, wh) + (u−Rhu, ws−whs)−A(s; u−Rhu, w−wh)

+∫ t

s

B(τ, s; (u − Rhu)(s), (w − wh)(τ )) dτ

+∫ s

0Aτ(τ ; e(τ ), (w − wh)(s)) dτ

+∫ s

0B(τ, τ ; e(τ ), (w − wh)(s)) dτ

−∫ s

0

∫ τ

0Bτ ′(τ, τ ′; e(τ ′), (w − wh)(s)) dτ

′ dτ

−∫ s

0Aτ(τ ; e(τ ), w(s)) dτ −

∫ s

0B(τ, τ ; e(τ ), w(s)) dτ

+∫ s

0

∫ τ

0Bτ ′(τ, τ ′; e(τ ′), w(s)) dτ ′ dτ

−∫ t

s

B(τ, s; e(s), (w − wh)(τ )) dτ

+∫ t

s

B(τ, s; e(s), w(τ)) dτ.

Integrating with respect tos from 0 tot and changing the order of integrationfor the fourth and the last two terms, we obtain

∫ t

0‖e(s)‖2ds ≤ C(ε)

[∫ t

0

{‖u − Rhu‖2

+ h2∫ s

0‖u − Rhu‖2

1dτ + h2∫ s

0‖e(τ )‖2

1dτ

}ds

]

+ ε

∫ t

0

{‖ws − whs‖2 + h−2‖w − wh‖21 + ‖w‖2

2

}ds

+ C(ε)

∫ t

0

∫ s

0‖e(τ )‖2 dτ ds.

By using (2.12), (2.6), (4.12) and Lemma 4.3, an appropriate choice ofε

leads to

∫ t

0‖e(s)‖2ds ≤ Ch4t‖u0‖2 + C

∫ t

0

∫ s

0‖e(τ )‖dτds.


An application of Gronwall’s Lemma yields (4.9). To prove (4.10), we formtheL2-inner product between (4.11) ande, and use (3.1) to obtain

1

2

d

ds‖e(s)‖2 = d

ds(e,wh) + (e, ws − whs) − A(s; e,w − wh)

+∫ t

s

B(τ, s; e(s), (w − wh)(τ )) dτ

−∫ s

0B(s, τ ; e(τ ), wh(s)) dτ

+∫ t

s

B(τ, s; e(s), wh(τ )) dτ.

Multiplying both sides bys and integrating from 0 tot , we get

1

2t‖e(t)‖2 = 1

2

∫ t

0‖e(s)‖2 ds +

∫ t

0[(e, w − wh) − (e, w)] ds

+∫ t

0s(u − Rhu,ws − whs) ds

−∫ t

0sA(s; u − Rhu,w − wh) ds

+∫ t

0

∫ t

s

sB(τ, s; (u − Rhu)(s), (w − wh)(τ ))dτ ds

+∫ t

0sB(s, s; e(s), (w − wh)(s)) ds

−∫ t

0sB(s, s; e(s), w(s)) ds

−∫ t

0

∫ s

0sBτ (s, τ ; e(τ ), (w − wh)(s)) dτ ds

+∫ t

0

∫ s

0sBτ (s, τ ; e(τ ), w(s)) dτ ds

+∫ t

0

∫ t

s

sB(τ, s; e(s), (wh − w)(τ)) dτ ds

+∫ t

0

∫ t

s

sB(τ, s; e(s), w(τ)) dτ ds.


A simple change of variables along with integration by parts in the temporalvariable yields∫ t

0

∫ t

s

sB(τ, s; e(s), (wh − w)(τ)) dτ ds

=∫ t

0

∫ s

0τB(s, τ ; eτ (τ ), (wh − w)(s)) dτ ds

=∫ t

0sB(s, s; e(s), (wh − w)(s))ds

−∫ t

0

∫ s

0τBτ (s, τ ; e(τ ), (wh − w)(s))dτds

−∫ t

0

∫ s

0B(s, τ ; e(τ ), (wh − w)(s))dτds.

Similarly, we rewrite the term∫ t

0

∫ t

ssB(τ, s; e(s), w(τ))dτds. Thus,

t‖e(t)‖2 ≤ C

[ ∫ t

0{h2(‖e‖2

−1 + ‖e‖21) + ‖e‖2

−2} ds

+∫ t

0s2{‖u − Rhu‖2 + h2‖u − Rhu‖2

1} ds]

+ C

∫ t

0{‖ws − whs‖2 + h−2‖w − wh‖2

1 + ‖w‖22} ds

+ C

∫ t

0‖e(s)‖2 ds.

Now we use (2.12), (4.12), Lemmas 4.2 and 4.3 and the a priori estimatesin Lemma 2.2 to obtain

t‖e(t)‖2 ≤ Cth4‖u0‖2 + C

∫ t

0‖e(s)‖2ds.

Finally, we apply (4.9) to complete the proof.��Remark 4.2 For the errore = v−vh associated with the backward problem,set ˜e(s) = − ∫ t

se(τ )dτ , s ≤ t . With a change of variables and the estimates

of the forward problem, it is easy to derive the following estimates for˜e‖˜e(s)‖j ≤ Ch2−j‖g‖, j = 0,1, (4.13)

and ∫ t

s

‖˜e‖2dτ ≤ C(t − s)h4‖g‖2. (4.14)


We are now in a position to prove our main result in the semidiscretecase.

Theorem 4.1 Let u and uh be solutions of (1.1) and (1.4), respectively. Thenfor f ≡ 0 and u0 ∈ L2, the following error estimate holds:

‖e(t)‖ ≤ Ct−1h2‖u0‖, t ∈ J.

Proof Integrate (4.2) fromt2 to t to obtain

(e(t), g) = (e(t/2), v(t/2)) − (e(t/2), e(t/2)) + (u(t/2), e(t/2))

+∫ t

2

0

∫ t

t2

[B(τ, s; u(s), e(τ )) + B(τ, s; e(s), v(τ ))

− B(τ, s; e(s), e(τ ))] dτ ds= I1 + I2 + I3 + I4.

For I1, apply Lemma 4.2 and a priori estimates for the backward problemto get

|I1| ≤ ‖e(t/2)‖−2‖v(t/2)‖2 ≤ Ch2

t‖u0‖‖g‖.

By using (4.1) and Lemma 4.1,I2 can be estimated as

|I2| ≤ ‖e(t/2)‖‖e(t/2)‖ ≤ Ch2

t‖u0‖‖g‖.

For I3, an appropriate modification of Lemma 4.2 yields an estimate fore.Then, using Lemma 2.2, we obtain

|I3| ≤ ‖u(t/2)‖2‖e(t/2)‖−2 ≤ Ch2

t‖u0‖‖g‖.

To estimate the first termI4,1 in I4, we integrate by parts with respect toτto get

I4,1 =∫ t

2

0

∫ t

t2

B(τ, s; u(s), e(τ )) dτ ds

=∫ t

2

0

∫ t

t2

B(τ, s; us(s), ˜eτ (τ )) dτ ds

−∫ t

2

0B(t/2, s; us(s), ˜e(t/2)) ds

−∫ t

2

0

∫ t

t2

Bτ(τ, s; us(s), ˜e(τ )) dτ ds.


Again integrating by parts with respect tos, we rewriteI4,1 as

I4,1 = − B(t/2, t/2, u(t/2), ˜e(t/2)) +∫ t

2

0Bs(t/2, s; u(s), ˜e(t/2)) ds

−∫ t

t2

Bτ(τ, t/2; u(t/2), ˜e(τ )) dτ

+∫ t

2

0

∫ t

t2

Bτs(τ, s; u(s), ˜e(τ )) dτ ds.

Hence, using (2.6) and (4.13) withj = 0, we obtain

|I4,1| ≤ C

[‖u(t/2)‖2‖˜e(t/2)‖ +

∫ t2

0‖u(s)‖2‖˜e(t/2)‖ ds

+∫ t

t2

‖u(t/2)‖2‖˜e(τ )‖ dτ

+∫ t

2

0

∫ t

t2

‖u(s)‖2‖˜e(τ )‖ dτ ds]

≤ Ch2‖u0‖‖g‖.

Similarly, integrating by parts first byτ and then bys, we rewrite the secondtermI4,2 and third termI4,3 in I4 as we did withI4,1. Then, we use estimates(4.10) and (2.9) to obtain

|I4,2| ≤ C

[‖e(t/2)‖‖v(t/2)‖2 +

∫ t2

0‖e(s)‖‖v(t/2)‖2 ds

+∫ t

t2

‖e(t/2)‖‖v(τ )‖2 dτ

+∫ t

2

0

∫ t

t2

‖e(s)‖‖v(τ )‖2 dτ ds

]≤ Ch2‖u0‖‖g‖,

and finally, forI4,3, we apply (4.6) and (4.13) withj = 1 to get

|I4,3| ≤ C

[‖e(t/2)‖1‖˜e(t/2)‖1 +

∫ t2

0‖e(s)‖1‖˜e(t/2)‖1 ds

+∫ t

t2

‖e(t/2)‖1‖˜e(τ )‖1 dτ

+∫ t

2

0

∫ t

t2

‖e(s)‖1‖˜e(τ )‖1d τ ds

]≤ Ch2‖u0‖‖g‖.


Combining all the above estimates we now obtain

|(e(t), g)| ≤ C

(h2

t‖u0‖ + h2‖u0‖

)‖g‖,

and this completes the proof.��For a nonhomogeneous source term i.e.,f �= 0 with u0 = 0, we obtain

the following error estimates.

Theorem 4.2 For f (t) ∈ L2 and u0 = 0, we have

‖e(t)‖ ≤ Ch2

(log

1

h

)maxs∈[0,t] ‖f (s)‖, t ∈ [0, T ].

Proof From (4.3) we note that

(e(t), g) =∫ t

0(f, e) ds. (4.15)

Consider the case when 0≤ t ≤ h2. Using the a priori estimates max[0,t] ‖e‖ ≤C‖g‖ (with time reversed), we obtain∫ t

0(f, e) ds ≤ C max

s∈[0,t] ‖f (s)‖∫ t

0‖e(s)‖ ds ≤ Ch2‖g‖ max

s∈[0,t] ‖f (s)‖.

Whent ≥ h2, we rewrite the term on the right-hand side of (4.15) as∫ t

0(f, e)ds =

∫ t−h2

0(f, e) ds +

∫ t

t−h2(f, e) ds

= I1 + I2.

To estimatee, first we use change of variables to obtain a forward problemand apply Theorem 4.1 to obtain

‖e(s)‖ ≤ Ch2

(t − s)‖g‖, 0 ≤ s < t,

and therefore

|I1| ≤∫ t−h2

0‖f (s)‖‖e(s)‖ ds ≤ Ch2

(log

1

h

)‖g‖ max

s∈[0,t] ‖f (s)‖.

For I2, we use the a priori estimates to obtain

|I2| ≤∫ t

t−h2‖f (s)‖‖e(s)‖ ds ≤ Ch2‖g‖ max

s∈[0,t] ‖f (s)‖.

By combining these two estimates, the result now follows fort ≥ h2 andthis completes the proof.��


We note that it is possible to avoid the logarithmic term in the previoustheorem, provided thatft ∈ L1(L2).

Theorem 4.3 For u0 = 0, f (t) ∈ L2 and ft ∈ L1(L2), we have

‖e(t)‖ ≤ Ch2

[‖f (0)‖ +

∫ t

0‖fs(s)‖ds

].

Proof From (4.15), it follows that

(e(t), g) =∫ t

0(f, e) ds =

∫ t

0(f, ˜es) ds.

where˜e is defined earlier. Integrate the right-hand side of the above equationby parts to get

(e(t), g) = −(f (0), ˜e(0)) −∫ t

0(fs, ˜e) ds

≤ C‖f (0)‖‖˜e(0)‖ +∫ t

0‖fs‖‖˜e‖ ds.

Use (4.13) withj = 0 to complete the proof. ��

Remark 4.3 Note that the linear problem (1.1) may be split asu = u1 +u2,whereu1 satisfies equation (1.1) withf ≡ 0 andu2 satisfies equation (1.1)with u0 = 0. We now denote the corresponding Galerkin approximations asu1,h andu2,h, respectively. Therefore the total error

e = u − uh = (u1 − u1,h) + (u2 − u2,h)

= e1 + e2.

Using Theorems 4.1 and 4.2, we now obtain for the nonhomogeneousproblem

‖e(t)‖ ≤ Ch2

[t−1‖u0‖ +

(log

1

h

)maxs∈[0,t] ‖f (s)‖

], t ∈ J.

Remark 4.4 Following the techniques of proof of Luskin and Rannacher [9]or Huang and Thomée [5], it is possible to obtain the above error estimatesfor the problem (1.1) when the operatorA(t) is not selfadjoint.


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