Applied Mathematics and Computation 236 (2014) 161–168

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Applied Mathematics and Computation

journal homepage: www.elsevier .com/ locate/amc

Rothe’s method for solving some fractional integral diffusionequation

http://dx.doi.org/10.1016/j.amc.2014.03.0250096-3003/� 2014 Elsevier Inc. All rights reserved.

⇑ Corresponding author.E-mail addresses: araheem.iitk3239@gmail.com (A. Raheem), dhiren@iitk.ac.in (D. Bahuguna).

Abdur Raheem ⇑, Dhirendra BahugunaDepartment of Mathematics and Statistics, Indian Institute of Technology Kanpur, Kanpur 208016, India

a r t i c l e i n f o

Keywords:Fractional integral equationDiffusion equationStrong solutionSemigroup of bounded linear operatorsMethod of semidiscretization

a b s t r a c t

In this paper, we apply the Rothe’s method to a fractional integral diffusion equation andestablish the existence and uniqueness of a strong solution. As an application, we includean example to illustrate the main result.

� 2014 Elsevier Inc. All rights reserved.

1. Introduction

In this paper we apply the Rothe’s method to the following fractional integral diffusion equation in a Banach space X

@uðtÞ@tþ AuðtÞ ¼ 1

CðaÞ

Z t

0

uðsÞðt � sÞ1�a dsþ f ðtÞ; t 2 ð0; T�; ð1Þ

uð0Þ ¼ u0; ð2Þ

where 0 < a < 1; �A is the infinitesimal generator of a C0-semigroup of contractions, f is a given map from ½0; T� intoX; u0 2 DðAÞ � A, the domain of A.

The problem considered in this paper is a particular case of the fractional integral diffusion problem

DbuðtÞ þ AuðtÞ ¼ 1CðaÞ

Z t

0

uðsÞðt � sÞ1�a dsþ f ðtÞ; uð0Þ ¼ u0;

where 0 < a 6 1; 0 < b 6 1. If we take b ¼ 1 and 0 < a < 1, then above problem reduces to the problem (1) and (2).In 1930, E. Rothe [8] has introduced a method to solve the following scalar parabolic initial boundary value problem of

second order

Rðt; xÞ @u@t� @

2u@x2 ¼ Sðt; x;uÞ; 0 < x < 1; t > 0;

uð0; xÞ ¼ u0ðxÞ;

uðt;0Þ ¼ uðt;1Þ ¼ 0; t P 0;

162 A. Raheem, D. Bahuguna / Applied Mathematics and Computation 236 (2014) 161–168

where R and S are sufficiently smooth functions of t and x in ½0; T� � ð0;1Þ satisfying certain additional conditions. Here Tmeans an arbitrary finite positive number. His method consist in dividing ½0; T� into n number of subintervals½tn

j�1; tnj �; tn

j ¼ jh; j ¼ 1;2; . . . ; n with tn0 ¼ 0 of equal lengths hðh ¼ T

nÞ and replacing the partial derivative @u@t of the unknown

function u by the difference quotientsun

j�un

j�1h . After defining a sequence of polygonal functions

Unðx; tÞ ¼ unj�1ðxÞ þ

1hðt � tn

j�1Þðunj ðxÞ � un

j�1ðxÞÞ; t 2 ½tnj�1; t

nj �:

Rothe has proved the convergence of the sequence fUng to the unique solution of the problem as n!1 using some a prioriestimates on fUng. The problem treated by Rothe is a simple one but the method introduced by him turns out to be a verypowerful theoretical tools for proving the existence and uniqueness of solutions of linear as well as nonlinear parabolic andhyperbolic problems of higher orders. This method is known as ‘‘Rothe’s method’’. It is also known as the method of semi-discretization or the method of lines. After that many authors have applied this method to various classical types of initialboundary value problem; for instance [9–12,14–18] and references therein.

Dubey [4], has established the existence and uniqueness of a strong solution for the following nonlinear nonlocal func-tional differential equation in a Banach X, using the method of semidiscretization:

u0ðtÞ þ AuðtÞ ¼ f ðt;uðtÞ;utÞ; t 2 ð0; T�;hðu0Þ ¼ / on ½�s;0�;

where 0 < T <1, / 2 C0 :¼ Cð½�s;0�; XÞ; s > 0, the nonlinear operator A is singlevalued and m-accretive defined from thedomain DðAÞ � X into X, the nonlinear map f is defined from ½0; T� � X � C0 :¼ Cð½�s;0�; XÞ into X, the map h is defined fromC0 into C0. For u 2 CT :¼ Cð½�s; T�; XÞ, function ut 2 C0 is given by utðsÞ ¼ uðt þ sÞ for s 2 ½�s;0�. Here Ct :¼ Cð½�s; t�; XÞ fort 2 ½0; T� is the Banach space of all continuous functions from ½�s; t� into X endowed with the supremum norm

k/kt ¼ sup�s6g6t

k/ðgÞk; / 2 Ct :

For the application of Rothe’s method to delay differential equation, delayed cooperation diffusion system, integrodifferen-tial equation, parabolic and hyperbolic problems, we refer the readers to [1–3,5,9–22].

By literature, it is clear that Rothe’s method or the method of semidiscretization is applicable in many physical, mathe-matical, biological problems modeled by partial differential equations.

In the present paper our aim is to apply the Rothe’s method to a fractional integral diffusion problem and to establish theexistence and uniqueness of a strong solution. This work is motivated by the work of Lin and Xu [23]. In which authors usedmethod based on time discretization to the following time fractional diffusion problem

@auðx; tÞ@ta

� @2uðx; tÞ@x2 ¼ f ðx; tÞ; x 2 ^; 0 < t 6 T:

Subject to the following initial and boundary conditions

uðx;0Þ ¼ gðxÞ; x 2 ^;

uð0; tÞ ¼ uðL; tÞ ¼ 0; 0 6 t 6 T;

where 0 < a < 1 is the order of the time fractional derivative. @auðx;tÞ@ta is defined as Caputo fractional derivative of order a, given

by

@auðx; tÞ@ta

¼ 1Cð1� aÞ

Z t

0

@uðx; sÞ@s

dsðt � sÞa

:

In [24], authors develop the Crank–Nicolson finite difference method to solve the following linear time-fractional diffusionequation with Dirichlet boundary conditions

@auðx; tÞ@ta

¼ @2uðx; tÞ@x2 ;

uðx;0Þ ¼ f ðxÞ;

uð0; tÞ ¼ uð1; tÞ ¼ 0;

where 0 < x < 1; 0 6 t 6 T and the parameter 0 < a < 1 refers to the fractional order of the time derivative. For the timediscretization in fractional differential equations, we refer the readers to [25].

The plan of the rest paper is as follows. In Section 2, we state some preliminaries and the main result. In Section 3, westate and prove all the lemmas that are required to prove the main result and at the end of this section, we prove the mainresult. In the last section, as an application, we include an example to illustrate the main result.

A. Raheem, D. Bahuguna / Applied Mathematics and Computation 236 (2014) 161–168 163

2. Preliminaries and main result

Definition 2.1. Let X be a Banach space and let X� be its dual. For every x 2 X we define the duality map J as:

JðxÞ ¼ fx� : x� 2 X� and ðx�; xÞ ¼ kxk2 ¼ kx�k2g;

where ðx�; xÞ denotes the value of x� at x.

Definition 2.2. A nonlinear operator A : DðAÞ � X�!X is called m-accretive if

ðAx� Ay; Jðx� yÞÞP 0 8x; y 2 DðAÞ; and RðI þ AÞ ¼ X;

where Rð�Þ is the range of an operator.

Lemma 2.3 (Theorem 1.4.3, [6]). If �A is the infinitesimal generator of a C0-semigroup of contractions then A is m-accretive, i.e.

ðAu� Av ; Jðu� vÞÞP 0 8u; v 2 DðAÞ;

where J is the duality mapping and RðI þ kAÞ ¼ X for k > 0; I is the identity operator on X and Rð�Þ is the range of an operator.

Lemma 2.4 (Lemma 2.5, [7]). If �A is the infinitesimal generator of a C0-semigroup of contractions. If Xn 2 DðAÞ; n ¼ 1;2;3; . . . ;

Xn ! u 2 H and if kAXnk are bounded, then u 2 DðAÞ and AXn* Au.

Lemma 2.5 [9]. If a1; . . . ;aj be nonnegative numbers satisfying

a1 6 A;

ai 6 Aþ Bhða1 þ � � � þ ai�1Þ; i ¼ 2; . . . ; j;

where A; B; h are positive constants. Then

ai 6 AeBði�1Þh; i ¼ 1; . . . ; j:

Definition 2.6. By a strong solution u of (1) and (2) on ½0; T�, we mean a function, u 2 Cð½0; T�;XÞ such that uðtÞ 2 DðAÞ for a.e.t 2 ½0; T�; u is differentiable a.e. on ½0; T� and

@uðtÞ@tþ AuðtÞ ¼ 1

CðaÞ

Z t

0

uðsÞðt � sÞ1�a dsþ f ðtÞ a:e: t 2 ð0; T�;

uð0Þ ¼ u0:

We assume the following assumptions:

(H1) There exists a constant k > 0 s.t.

kf ðtÞ � f ðsÞk 6 kjt � sj 8t; s 2 ½0; T�:

(H2) Suppose that T and a satisfy the following relation

T1þa

Cð1þ aÞ < 1;

where C is the Gamma function.

Theorem 2.7. Suppose that conditions (H1) and (H2) are satisfied. Then for every u0 2 DðAÞ, the initial value problem (1) and (2)has a unique strong solution on the interval ½0; T�.

3. Approximation

To apply the method of semidiscretization, we divide the interval ½0; T� into the subintervals of length hn ¼ Tn. We replace

(1) and (2) by the following approximate equations

un0 ¼ u0;

un1 � un

0

hnþ Au1 ¼ f0;

164 A. Raheem, D. Bahuguna / Applied Mathematics and Computation 236 (2014) 161–168

and for j ¼ 2;3; . . . ;n, we have

unj � un

j�1

hnþ Aun

j ¼1

CðaÞXj�1

i¼1

Z tni

tni�1

uni

ðtnj � sÞ1�a dsþ f ðtn

j�1Þ ¼1

CðaÞXj�1

i¼1

uni �ðtj � sÞa

a

� �ti

ti�1

þ f nj�1

¼ 1Cð1þ aÞ

Xj�1

i¼1

ui½ðtnj � tn

i�1Þa � ðtn

j � tni Þ

a� þ f nj�1;

where f nj ¼ f ðtn

j Þ and f0 ¼ f nð0Þ.Next we establish successively existence and uniqueness of solution of the following approximate equations

un0 ¼ u0;

un1 � un

0

hnþ Au1 ¼ f0; ð3Þ

unj � un

j�1

hnþ Aun

j ¼1

Cð1þ aÞXj�1

i¼1

ui½ðtnj � tn

i�1Þa � ðtn

j � tni Þ

a� þ f nj�1; ð4Þ

j ¼ 2; . . . ;n:

Existence and uniqueness of unj 2 DðAÞ satisfying Eq. (3) and (4) is a consequence of Lemma 2.3.

We define the Rothe’s sequence fUnj g as:

UnðtÞ ¼u0 if t ¼ 0;un

j�1 þ 1hnðt � tn

j�1Þðunj � un

j�1Þ if t 2 ðtnj�1; t

nj �:

(ð5Þ

Next we prove some a priori estimates which are required to prove the main result. Throughout the paper C denotes genericconstant, this constant may have different value in the same discussion.

Lemma 3.1. For each n 2 N; j ¼ 1;2;3; . . . ;n,

kunj k 6 C;

where C is a generic constant independent of n; j; and hn.

Proof. From (3), we have

un1 þ hnAun

1 ¼ u0 þ hnf0:

Applying Jðun1Þ on both sides and using the definition of accretivity of the operator A, we obtain

kun1k 6 ku0k þ hnkf0k 6 ku0k þ Tkf0k � C ðsayÞ:

From (4), for j ¼ 2; . . . ;n, we have

unj þ hnAun

j ¼ unj�1 þ

h1þan

Cð1þ aÞXj�1

i¼1

uni ½ðj� iþ 1Þa � ðj� iÞa� þ hnf n

j�1:

Applying Jðunj Þ on both sides and using the definition of accretivity of the operator A, we obtain

kunj k 6 kun

j�1k þh1þa

n

Cð1þ aÞXj�1

i¼1

kuni k½ðj� iþ 1Þa � ðj� iÞa� þ hnkf n

j�1k:

By using (H1), we obtain

kunj k 6 kun

j�1k þh1þa

n

Cð1þ aÞXj�1

i¼1

kuni k½ðj� iþ 1Þa � ðj� iÞa� þ hnðkjtn

j�1j þ kf0kÞ: ð6Þ

We have

ðj� iþ 1Þa � ðj� iÞa ¼ ðj� iÞa�1 1þ 1j� i

� �a�1

ðj� iþ 1Þ � ðj� iÞ" #

6 ðj� iÞa�1 1þ 11

� �a�1

ðj� iþ 1Þ � ðj� iÞ" #

6 ja�1½2a�1ðj� iþ 1Þ � ðj� iÞ� ¼ ja�1 �ðj� iÞ 1� 121�a

� �þ 1

21�a

� �6 ja�1 � 1� 1

21�a

� �þ 1

21�a

� �

¼ ja�1½2a � 1� 6 ja�1:

A. Raheem, D. Bahuguna / Applied Mathematics and Computation 236 (2014) 161–168 165

Using above inequality in (6), we obtain

kunj k 6 kun

j�1k þja�1h1þa

n

Cð1þ aÞXj�1

i¼1

kuik þ hnðkjtnj�1j þ kf0kÞ:

Repeating above inequality, we obtain

kunj k 6 ku0k þ

jah1þan

Cð1þ aÞXj�1

i¼1

kuik þ jhn kjtnj�1j þ kf0k

� �6 ku0k þ

Tahn

Cð1þ aÞXj�1

i¼1

kuik þ TðkT þ kf0kÞ:

Applying Lemma 2.5, we obtain

kunj k 6 ðku0k þ kT2 þ Tkf0kÞ exp

ðj� 1ÞhnTa

Cð1þ aÞ

� �6 ðku0k þ kT2 þ Tkf0kÞ exp

T1þa

Cð1þ aÞ

!� C ðsayÞ:

This completes the proof. h

Lemma 3.2. For each n 2 N, and j ¼ 1; . . . ;n,

unj � un

j�1

hn

�������� 6 C;

where C is a generic constant independent of n; j; and hn.

Proof. From (3), we obtain

un1 � un

0

hnþ Aun

1 � Aun0 ¼ f0 � Aun

0:

Applying Jðunj � u0Þ on both sides and using the definition of accretivity of the operator A, we obtain

un1 � un

0

hn

�������� 6 kf0k þ kAu0k � C ðsayÞ: ð7Þ

Subtracting from (4), the same equation written for j� 1, we obtain

unj � un

j�1

hnþ Aun

j � Aunj�1 ¼

unj�1 � un

j�2

hnþ ha

n

Cð1þ aÞunj�1½2

a � 1a� þ han

Cð1þ aÞXj�2

i¼1

uni ½ðj� iþ 1Þa � 2ðj� iÞa þ ðj� i� 1Þa�

þ f nj�1 � f n

j�2:

Applying Jðunj � un

j�1Þ on both sides and using the definition of accretivity of the operator A, we obtain

unj � un

j�1

hn

�������� 6 un

j�1 � unj�2

hn

��������þ ha

n

Cð1þ aÞ kunj�1k½2

a � 1a� þ han

Cð1þ aÞXj�2

i¼1

kuni k½ðj� iþ 1Þa � 2ðj� iÞa þ ðj� i� 1Þa� þ kf n

j�1

� f nj�2k:

By using Lemma 3.1, we obtain

unj � un

j�1

hn

�������� 6 un

j�1 � unj�2

hn

��������þ Cha

n

Cð1þ aÞ ½2a � 1a� þ Cha

n

Cð1þ aÞXj�2

i¼1

½ðj� iþ 1Þa � 2ðj� iÞa þ ðj� i� 1Þa� þ khn:

We have

Xj�2

i¼1

½ðj� iþ 1Þa � 2ðj� iÞa þ ðj� i� 1Þa� þ ð2a � 1aÞ ¼ ja � ðj� 1Þa:

Using above equality, we obtain

unj � un

j�1

hn

�������� 6 un

j�1 � unj�2

hn

��������þ Cha

n

Cð1þ aÞ ½ja � ðj� 1Þa� þ khn 6

unj�1 � un

j�2

hn

��������þ Cha

n

Cð1þ aÞ ja�1 þ khn:

Repeating above inequality, we obtain

unj � un

j�1

hn

�������� 6 un

1 � un0

hn

��������þ Cha

n

Cð1þ aÞ ja þ kjhn:

166 A. Raheem, D. Bahuguna / Applied Mathematics and Computation 236 (2014) 161–168

As jhn 6 T and using (7), we obtain

unj � un

j�1

hn

�������� 6 C þ CTa

Cð1þ aÞ þ kT � C ðsayÞ:

This completes the proof. h

We define a sequence of step functions

XnðtÞ ¼u0 if t ¼ 0;un

j if t 2 ðtnj�1; t

nj �:

(ð8Þ

Remark 3.3. From Lemma 3.2, it is clear that UnðtÞ is uniformly Lipschitz continuous and UnðtÞ � XnðtÞ ! 0; n!1.If we suppose that

f nðtÞ ¼ f ðtnj Þ;

then (3) and (4) can be written as:

d�

dtUnðtÞ þ AXnðtÞ ¼ 1

CðaÞ

Z t

0

XnðsÞðt � sÞ1�a dsþ f nðtÞ; t 2 ð0; T�; ð9Þ

where d�

dt denotes the left derivative in ð0; T�.Also, for t 2 ð0; T�, we have

Z t

0AXnðsÞds ¼ u0 � UnðtÞ þ 1

CðaÞ

Z t

0

Z s

0

Xnðs1Þðt � s1Þ1�a ds1dsþ

Z t

0f nðsÞds: ð10Þ

Next we prove the convergence of Un to u in Cð½0; T�;XÞ.

Lemma 3.4. There exists u 2 Cð½0; T�;XÞ, such that Un ! u in Cð½0; T�;XÞ as n!1. Moreover, u is Lipschitz continuous on ½0; T�.

Proof. From (9), we see that

d�

dtUnðtÞ � d�

dtUmðtÞ þ AXnðtÞ � AXmðtÞ ¼ 1

CðaÞ

Z t

0

XnðsÞ � XmðsÞðt � sÞ1�a dsþ f nðtÞ � f mðtÞ:

Applying JðXnðtÞ � XmðtÞÞ on both sides and using the definition of accretivity of the operator A, we obtain

d�

dtUnðtÞ � d�

dtUmðtÞ; JðXnðtÞ � XmðtÞÞ

� �6

1CðaÞ

Z t

0

kXnðsÞ � XmðsÞk2

ðt � sÞ1�a dsþ kf nðtÞ � f mðtÞkkXnðtÞ � XmðtÞk:

Using above inequality, we obtain

ddtkUnðtÞ � UmðtÞk6 1

CðaÞ

Z t

0

kXnðsÞ � XmðsÞk2

ðt � sÞ1�a dsþ e1nmðtÞ;

where

e1nmðtÞ ¼ kjtn

j � tmk jkX

nðtÞ � XmðtÞk þ ddtðUnðtÞ � UmðtÞÞ

�������� kUnðtÞ � XnðtÞk þ kUmðtÞ � XmðtÞk½ �;

and e1nmðtÞ ! 0 as n;m!1.

ddtkUnðtÞ � UmðtÞk2

61

CðaÞ

Z t

0

kUnðsÞ � UmðsÞk2

ðt � sÞ1�a dsþ e2nmðtÞ;

where

e2nmðtÞ ¼ e1

nmðtÞ þ1

CðaÞ

Z t

0

kXnðsÞ � UnðsÞk2 þ kXmðsÞ � UmðsÞk2

ðt � sÞ1�a ds;

and e2nmðtÞ ! 0 as n;m!1.

ddtkUnðtÞ � UmðtÞk2

6

supt2½0;T�kUnðtÞ � UmðtÞk2

CðaÞ

Z t

0

1

ðt � sÞ1�a dsþ e2nmðtÞ 6

Ta

Cðaþ 1Þ supt2½0;T�kUnðtÞ � UmðtÞk2 þ e2

nmðtÞ:

A. Raheem, D. Bahuguna / Applied Mathematics and Computation 236 (2014) 161–168 167

This implies that

kUnðtÞ � UmðtÞk26

Taþ1

Cðaþ 1Þ supt2½0;T�kUnðtÞ � UmðtÞk2 þ e3

nmðtÞ;

where

e3nmðtÞ ¼

Z t

0e2

nmðsÞds;

and e3nmðtÞ ! 0 as n;m!1. Taking supremum, we obtain

supt2½0;T�kUnðtÞ � UmðtÞk2

6Taþ1

Cðaþ 1Þ supt2½0;T�kUnðtÞ � UmðtÞk2 þ e3

nmðtÞ ) supt2½0;T�kUnðtÞ � UmðtÞk2

6 1� T1þa

Cð1þ aÞ

!�1

e3nmðtÞ:

Using above inequality, we get that Un ! u in Cð½0; T�;XÞ. As each Un is uniformly Lipschitz continuous, u is Lipschitzcontinuous.

Next we show that this u is a unique strong solution of the problem (1) and (2). h

Remark 3.5. Clearly XnðtÞ 2 DðAÞ, for each n. As UnðtÞ � XnðtÞ ! 0 as n!1;XnðtÞ ! uðtÞ 2 H. Also kAXnk are bounded there-fore by Lemma 2.4, it is clear that AXn

* Au.So for every x� 2 X� and t 2 ð0; T�, we have

Z t

t0

ðAXnðsÞ; x�Þds ¼ ðu0; x�Þ � ðUnðtÞ; x�Þ þ 1CðaÞ

Z t

0

Z s

0

Xnðs1Þðt � s1Þ1�a ds1; x�

!dsþ

Z t

0ðf nðsÞ; x�Þds:

Using Lemma 3.4, Remark 3.5 and the bounded convergence theorem, we obtain as n!1,

Z t

0ðAuðsÞ; x�Þds ¼ ðu0; x�Þ � ðuðtÞ; x�Þ þ

1CðaÞ

Z t

0

Z s

0

uðs1Þðt � s1Þ1�a ds1; x�

!dsþ

Z t

0ðf ðsÞ; x�Þds:

As AuðtÞ is Bochner integrable on ½0; T�, from above equation, we have

ddt

uðtÞ þ AuðtÞ ¼ 1CðaÞ

Z t

0

uðsÞðt � sÞ1�a dsþ f ðtÞ a:e: t 2 ð0; T�: ð11Þ

Clearly u 2 Cð½0; T�; XÞ and differentiable a.e. on ð0; T� with uðtÞ 2 DðAÞ a.e. on ð0; T� and uð0Þ ¼ u0 satisfying (11). Hence it willbe a strong solution of the problem (1) and (2) on ½0; T�.

Next we will prove the uniqueness. For this suppose that u1; u2 are two strong solutions of the problem (1) and (2). Letu ¼ u1 � u2, from (11) we have

duðtÞdt

; JðuðtÞÞ� �

þ ðAu1ðtÞ � Au2ðtÞ; Jðu1ðtÞ � u2ðtÞÞÞ ¼1

CðaÞ

Z t

0

uðsÞðt � sÞ1�a ds; JðuðtÞÞ

!:

By using the definition of accretivity of the operator A, we obtain

ddtkuðtÞk2

61

CðaÞ

Z t

0

kuðsÞk2

ðt � sÞ1�a ds 61

CðaÞ supt2½0;T�kuðsÞk2

Z t

0

1

ðt � sÞ1�a ds 6Ta

Cð1þ aÞ supt2½0;T�kuðsÞk2

:

Integrating, we obtain

kuðtÞk26

T1þa

Cð1þ aÞ supt2½0;T�kuðsÞk2 ) sup

t2½0;T�kuðtÞk2

6T1þa

Cð1þ aÞ supt2½0;T�kuðsÞk2

:

If T1þa

Cð1þaÞ < 1, then supt2½0;T�

kuðtÞk2 ¼ 0, i.e. kuðtÞk ¼ 0; t 2 ½0; T�. This shows the uniqueness of the strong solution.

4. Application

We can apply our result to the following problem.Consider the following example

@uðt; xÞ@t

� @2uðt; xÞ@x2 ¼ 1

Cð12Þ

Z t

0

uðs; xÞðt � sÞ

12

ds on ½0;1� � ½0;p�;

uð0; xÞ ¼ u0ðxÞ;uðt;0Þ ¼ uðt;pÞ ¼ 0 8t 2 ½0;1�;

168 A. Raheem, D. Bahuguna / Applied Mathematics and Computation 236 (2014) 161–168

where u : ½0;1� � ½0;p� ! R is an unknown function and u0 : ½0;p� ! R is a given initial value function.If we identify u : ½0;1� ! L2ð½0;p� by uðtÞðxÞ ¼ uðt; xÞ, and define

Au ¼ � @2u@x2 ; DðAÞ ¼ fu 2 L2½0;p�ju00 2 L2½0;p�g;

then above problem reduces to

@uðtÞ@tþ AuðtÞ ¼ 1

C 12

Z t

0

uðsÞðt � sÞ

12

ds; t 2 ½0; T�;

uð0Þ ¼ u0;

which is same as the problem (1) and (2).Here f ðtÞ ¼ 0; T ¼ 1 and a ¼ 1. So

T1þa

Cð1þ aÞ ¼1

C 32

< 1:

Thus conditions (H1) and (H2) are satisfied, so by applying the Theorem 2.7, we obtain a unique strong solution of the givenproblem.

Acknowledgements

The authors thanks the referee for his valuable suggestions. The second author acknowledges the financial help from theDepartment of Science and Technology, New Delhi, under its research project SR/S4/MS: 796/12.

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