1Introduction A.Tari OntheExistence,UniquenessandSolutionoftheNonlinearVolterraPartialIntegro-DifferentialEquations 2ExistenceandUniquenessofSolution

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International Journal of Nonlinear Science Vol.15(2013) No.3,pp.1-12

On the Existence, Uniqueness and Solution of the Nonlinear Volterra Partial Integro-Differential Equations

A. Tari

^∗

Department of Mathematics, Shahed University, P. O. Box 18151-159, Tehran, Iran (Received August 13 2012 , accepted December 31 2012)

Abstract: In this paper, we study a form of the nonlinear Volterra partial integro-differential equations (NVPIDEs). First, we prove a theorem and some corollaries about existence and uniqueness of solution of the NVPIDEs. Next, we develop the two-dimensional reduced differential transform method (RDTM) for solving a class of the NVPIDEs. For this purpose, based on some preliminary results of the reduced differential transform, we prove some theorems to develop RDTM for the aforementioned equations. The paper concludes with some numerical examples to demonstrate the accuracy of the presented method.

Keywords:Nonlinear Volterra partial integro-differential equations; Differential transform; Reduced differential transform.

1 Introduction

The Volterra partial integro-differential equations enjoy many applications in Physics, Mechanics and other applied sciences [1], however, to solve these equations, not many numerical methods of high accuracy exist (see, for example [2-3]).

Here, we investigate such types of equations and develop RDTM for solving them.

The concept of differential transform was first introduced by Zhou [4] for solving linear and nonlinear initial value prob- lems in electrical analysis. Recently, a reduced form of the differential transform was introduced and developed for some equations. Some instances follow. In [5], this method was used in solving partial differential equations, while in [6], it was used to solve the fractional partial differential equations. In [7], the RDTM was developed for solving nonlinear PDEs with proportional delay. In [8], the RDTM was applied for solving gas dynamics equation. In [9], it was applied to solve the linear and nonlinear wave equations and in [10] to solve generalized KDV equation. And quite recently, the authors of [11] developed the RDTM for Wu-Zhang equation and the authors of [12] for fractional Benney-Lin equation.

As mentioned above, the subject of this paper is applying RDTM for solving the linear and nonlinear Volterra partial integro-differential equations. For this purpose, we consider the Volterra partial integro-differential equations of the form

Du(x, t)− Z t

0

Z x 0

K(x, t, y, z, u(y, z))dydz=f(x, t), x∈[0, b], t∈[0, d] (1) with some supplementary conditions, whereDu(x, t)denotes the differential part of equation and the functionsKandf are assumed to be continuous on their domains. Note should be made that throughout the rest of this paper,banddare taken to be finite.

2 Existence and Uniqueness of Solution

In this section, we investigate the existence and uniqueness of solution of the equation (1). To this end, we first consider the equations of the form

u(x, t)− Z t

0

Z x 0

K(x, t, y, z, u(y, z))dydz=f(x, t), x∈[0, b], t∈[0, d] (2)

∗ E-mail address: [email protected], [email protected]

Copyright c°World Academic Press, World Academic Union IJNS.20xx.xx.xx/xx

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and we extend the theorem 4.1 from [13], to two-dimensional case in the following theorem.

Theorem 2.1 Let the functionsf andKbe continuous for allx, y∈[0, b],z, t∈[0, d]and−∞< u <∞. Let also the kernelKsatisfies the Lipschitz condition, i.e.

|K(x, t, y, z, u1)−K(x, t, y, z, u2)| ≤L|u1−u₂| (3) whereLis independent ofx, t, y, z, u1andu2.

Then the equation (2) has a unique continuous solution.

Proof.We define

u0(x, t) =f(x, t) u_n(x, t) =f(x, t) +

Z t 0

Z x 0

K(x, t, y, z, un−1(y, z))dydz, n= 1,2, ... (4) then we have

u_n(x, t)−un−1(x, t) = Z t

0

Z x 0

h

K(x, t, y, z, un−1(y, z))−K(x, t, y, z, un−2(y, z))i

dydz. (5)

We also introduce the functionφ_n(x, t)as

φ0(x, t) =u0(x, t)

φ_n(x, t) =u_n(x, t)−un−1(x, t), n= 1,2, ... (6) so

u_n(x, t) = Xn

i=0

φ_i(x, t) and

|φn(x, t)| ≤L Z t

0

Z x 0

|φn−1(y, z)|dydz which implies

|φn(x, t)| ≤ M(Lxt)ⁿ

(n!)² , ∀x∈[0, b], t∈[0, d]

where

M = max

0≤x≤b,0≤t≤d|f(x, t)|

so the sequence{φn(x, t)}is convergence by Weierstrass M-test.

Therefore

u(x, t) = X∞

n=0

φn(x, t) (7)

exists and is a continuous function.

Now we prove that the continuous functionu(x, t)in (7) satisfies the equation (2). To this end, we set

u(x, t) =un(x, t) + ∆n(x, t). (8)

Then from (4) we have

u(x, t)−∆n(x, t) =f(x, t) + Z t

0

Z x 0

K(x, t, y, z, u(y, z)−∆n−1(y, z))dydz

or equivalently

u(x, t)−f(x, t)− Z t

0

Z x 0

K(x, t, y, z, u(y, z))dydz= ∆n(x, t)+

(3)

Z t 0

Z x 0

hK(x, t, y, z, u(y, z)−∆n−1(y, z))−K(x, t, y, z, u(y, z))i dydz and by using the Lipschitz condition (3) we obtain

¯¯

¯¯u(x, t)−f(x, t)− Z t

0

Z x 0

K(x, t, y, z, u(y, z))dydz

¯¯

¯¯≤ |∆n(x, t)|+Lbdk∆n−1k (9) where

k∆n−1k= max

0≤x≤b,0≤t≤d|∆n−1(x, t)|.

But from (8) we have

n→∞lim |∆n(x, t)|= 0 therefore whenn→ ∞, (9) yields

u(x, t)− Z t

0

Z x 0

K(x, t, y, z, u(y, z))dydz=f(x, t) and thereforeu(x, t)is a solution of the equation (2).

Finally, we prove uniqueness. To this end, leteu(x, t)be another continuous solution of (2), then we have u(x, t)−u(x, t) =e

Z t 0

Z x 0

hK(x, t, y, z, u(y, z))−K(x, t, y, z,eu(y, z))i dydz

and using (3) implies

|u(x, t)−u(x, t)| ≤e L Z t

0

Z x 0

|u(y, z)−eu(y, z)|dydz. (10) But|u(y, z)−u(y, z)|e is bounded, so we set

B= max

0≤x≤b,0≤t≤d|u(x, t)−u(x, t)|e then from (10) we obtain

|u(x, t)−u(x, t)| ≤e BLxt and using this relation repeatedly yields

|u(x, t)−eu(x, t)| ≤B(Lxt)ⁿ

(n!)² ≤B(Lbd)ⁿ

(n!)² (11)

and passing to the limitn→ ∞in (11) yieldsu(x, t) =eu(x, t).¤

As the results of the above theorem, we consider some cases of the Volterra partial integro-differential equations and investigate the existence and uniqueness of solution of them.

Corollary 2.1 If the hypothesis of theorem 2.1 hold, then the equation

∂u(x, t)

∂t +u(x, t)− Z t

0

Z x 0

K(x, t, y, z, u(y, z))dydz =f(x, t), x∈[0, b], t∈[0, d] (12) with initial conditionu(x,0) =α1(x), whereα1(x)is a continuous function, has a unique continuous solution.

Proof.By integrating from two sides of the relation (12) with respect totwe obtain u(x, t) =α1(x)+

Z t 0

½

−u(x, t) + Z t

0

Z x 0

K(x, t, y, z, u(y, z))dydz+f(x, t)

¾

dt, x∈[0, b], t∈[0, d] (13) so if we set

K1(x, t, y, z, u) =−u(x, t) + Z t

0

Z x 0

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then we have

|K1(x, t, y, z, u1)−K₁(x, t, y, z, u2)|

≤ |u1−u2|+ Z t

0

Z x 0

|K(x, t, y, z, u1)−K(x, t, y, z, u2)|dydz

≤(1 +bdL)|u1−u₂|.

Hence the kernel of the equation (13) satisfies in Lipschitz condition and therefore by theorem 2.1 it has a unique continuous solution.¤

Corollary 2.2 If the hypothesis of theorem 2.1 hold for the equation

∂²u(x, t)

∂t² +a(x, t)u(x, t)− Z t

0

Z x 0

K(x, t, y, z, u(y, z))dydz=f(x, t), x∈[0, b], t∈[0, d] (14) with initial conditions

u(x,0) =α1(x), ∂

∂tu(x,0) =α2(x)

and ifa, α1andα2are continuous functions forx∈[0, b], t∈[0, d], then the problem has a unique continuous solution.

Proof.By integrating twice from both sides of the relation (14) with respect totwe obtain u(x, t) =α1(x) +α2(x)t+

Z t 0

½

−a(x, t)u(x, t) + Z t

0

Z x 0

¾

dtdt, x∈[0, b], t∈[0, d] (15) then by setting

K1(x, t, y, z, u) =−a(x, t)u(x, t) + Z t

0

Z x 0

K(x, t, y, z, u(y, z))dydz+f(x, t) we have

|K1(x, t, y, z, u1)−K1(x, t, y, z, u2)|

≤ |a(x, t)(u1−u2)|+ Z t

0

Z x 0

|K(x, t, y, z, u1)−K(x, t, y, z, u2)|dydz

≤(M +bdL)|u1−u2| where

M = max

0≤x≤b,0≤t≤d|a(x, t)|

so by theorem 2.1 the equation (15) has a unique continuous solution.¤ Corollary 2.3 Let the assumptions of theorem 2.1 satisfy for the equation

∂²u(x, t)

∂x² +a(x, t)u(x, t)− Z t

0

Z x 0

K(x, t, y, z, u(y, z))dydz=f(x, t), x∈[0, b], t∈[0, d] (16) with conditions

u(0, t) =β1(t), ∂

∂xu(0, t) =β2(t).

Furthermore assume that the functionsa, β1andβ2are continuous forx∈[0, b], t∈[0, d].

Then this problem has a unique continuous solution.

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Proof.Integrating twice from both sides of (16) with respect toximplies that u(x, t) =β1(t) +β2(t)x+

Z x 0

½

−a(x, t)u(x, t) + Z t

0

Z x 0

¾

dxdx, x∈[0, b], t∈[0, d] (17) and let

K1(x, t, y, z, u) =−a(x, t)u(x, t) + Z t

0

Z x 0

K(x, t, y, z, u(y, z))dydz+f(x, t) we have

|K1(x, t, y, z, u1)−K1(x, t, y, z, u2)| ≤(M+bdL)|u1−u2| where

M = max

0≤x≤b,0≤t≤d|a(x, t)|

so the proof is completed.¤

3 Two-dimensional RDT

In this section, we remind some preliminary results of the two-dimensional RDT. Then we prove some theorems to develop the two-dimensional RDT for solving the NVPIDEs.

Definition 3.1 The two-dimensional RDT of the analytic bivariate functionf(x, t)att0is defined by (see [5], [9-11])

F_n(x) = 1 n!

·∂ⁿf(x, t)

∂tⁿ

¸

t=t0

(18) and its inverse transform is defined as

f(x, t) = X∞

n=0

F_n(x)(t−t0)ⁿ. ¤ (19)

Note that the relations (18) and (19) imply f(x, t) =

X∞

n=0

1 n!

·∂ⁿf(x, t)

∂tⁿ

¸

t=t0

(t−t0)ⁿ. (20)

In the following theorem, we summarize some fundamental properties of the two dimensional reduced differential transform (see [9-11]).

Theorem 3.1 Let Fn(x), Un(x) and Vn(x) be the two-dimensional reduced differential transforms of the functions f(x, t),u(x, t)andv(x, t)att0= 0respectively, then we have

a.Iff(x, t) =u(x, t)±v(x, t), then

F_n(x) =U_n(x)±V_n(x).

b.Iff(x, t) =au(x, t), then

Fn(x) =aUn(x).

c.Iff(x, t) =u(x, t)v(x, t), then

F_n(x) = Xn

k=0

U_n(x)Vn−k(x).

d.Iff(x, t) =x^kt^l, then

Fn(x) =x^kδn,l. e.Iff(x, t) =sin(ax+bt), then

F_n(x) =bⁿ

n!sin(ax+nπ 2 ).

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f.Iff(x, t) =cos(ax+bt), then

F_n(x) =bⁿ

n!cos(ax+nπ 2 ).

g.Iff(x, t) =e^ax+bt, then

Fn(x) = bⁿ

n!e^ax. ¤

For applying the reduced differential transform to the differential part of the equation (1), we state the following theorem(See [9-10]).

Theorem 3.2 Let Fn(x), Un(x) and Vn(x) be the two-dimensional reduced differential transforms of the functions f(x, t),u(x, t)andv(x, t)att0= 0respectively, then we have

a.Iff(x, t) = ^∂

ru(x,t)

∂x^r , r∈N, then

F_n(x) = d^r dx^rU_n(x).

b.Iff(x, t) =^∂

su(x,t)

∂t^s , s∈N, then

Fn(x) = (n+ 1)(n+ 2)· · ·(n+s)Un+s(x).

c.Iff(x, t) =^∂^r+s_∂^r_x∂^u(x,t)^s_t , r, s∈N, then

Fn(x) = (n+ 1)(n+ 2)· · ·(n+s) d^r

dx^rUn+s(x). ¤

Now for applying the reduced differential transform to the integral part of (1), we give the following theorem.

Theorem 3.3 Assume thatUn(x),Vn(x),Hn(x)andGn(x)be the differential transforms of the functionsu(x, t),v(x, t), h(x, t)andg(x, t)att0= 0respectively, then we have

a.Ifg(x, t) =Rt 0

Rx

0 u(y, z)dydz, then

G0(x) = 0, x∈[0, b]

G_n(x) = 1 n

Z x 0

Un−1(y)dy, n= 1,2, . . . . (21)

b.Ifg(x, t) =Rt 0

Rx

0 u(y, z)v(y, z)dydz, then G0(x) = 0, x∈[0, b]

Gn(x) = 1 n

n−1X

k=0

Z x 0

Uk(y)Vn−k−1(y)dy, n= 1,2, . . . . (22) c.Ifg(x, t) =h(x, t)Rt

0

Rx

0 u(y, z)dydz, then G0(x) = 0, x∈[0, b]

G_n(x) =

n−1X

k=0

1

n−kH_k(x) Z x

0

Un−k−1(y)dy, n= 1,2, . . . . (23)

Proof :It is obvious thatG0(x) = 0in all three parts.

a.SinceUn(x)is the reduced differential transform ofu(x, t)att0= 0, we haveu(x, t) =P∞

n=0Un(x)tⁿthus g(x, t) =

Z t 0

Z x 0

Ã_∞ X

n=0

U_n(y)zⁿ

! dydz =

X∞

n=0

1 n+ 1tⁿ⁺¹

Z x 0

U_n(y)dy

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therefore

Gn+1(x) = 1 n+ 1

Z x 0

Un(y)dy, n= 0,1, . . . or

G_n(x) = 1 n

Z x 0

U_n−1(y)dy, n= 1,2, . . . which yields

n d

dxGn(x) =Un−1(x), n= 1,2, . . . . b.By differentiating fromg(x, t)with respect totandx, we obtain

∂²g(x, t)

∂t∂x =u(x, t)v(x, t) and by using partcof the theorems 3.2 and 3.1, we get

(n+ 1) d

dxGn+1(x) = Xn

k=0

U_k(x)Vn−k(x), n= 0,1, . . . . Replacingn+ 1byngives the result.

c.First we setf(x, t) =Rt 0

Rx

0 u(y, z)dydzthen by parta, we have Fn(x) = 1

n Z x

0

Un−1(y)dy, n= 1,2, . . . . Since

g(x, t) =h(x, t)f(x, t) we have

G_n(x) = Xn

k=0

H_k(x)Fn−k(x) =

n−1X

k=0

1

n−kH_k(x) Z x

0

Un−k−1(y)dy sinceFn−k(x) = 0fork=n.¤

Theorem 3.4 LetU_n(x),V_n(x)and G_n(x)be the reduced differential transforms of the functionsu(x, t),v(x, t)and g(x, t)att₀= 0respectively, andv(x, t)6= 0forx∈[0, b], t∈[0, d]. Then we have

a.Ifg(x, t) =Rt 0

Rx 0

u(y,z)

v(y,z)dydz, then

G0(x) = 0, x∈[0, b]

Un(x) = Xn

k=0

(k+ 1) d

dxGk+1(x)Vn−k(x), n= 0,1, . . . . (24) b.Ifg(x, t) = _v(x,t)¹ Rt

0

Rx

0 u(y, z)dydz, then

G0(x) = 0, x∈[0, b]

Xn

k=0

Gk(x)Vn−k(x) = 1 n

Z x 0

Un−1(y)dy, n= 1,2, . . . . (25) Proof : a.We have

∂²g(x, t)

∂t∂x =u(x, t) v(x, t) by settingeg(x, t) =^∂²_∂t∂x^g(x,t), it can be written

eg(x, t)v(x, t) =u(x, t)

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and by using partcof the theorem 3.1, we obtain Xn

k=0

Gek(x)Vn−k(x) =Un(x)

whereG(x)e is the reduced differential transform of the functioneg(x, t). Therefore Xn

k=0

(k+ 1) d

dxGk+1(x)Vn−k(x) =Un(x).

b.We have

g(x, t)v(x, t) = Z t

0

Z x 0

u(y, z)dydz

and by using partaof the theorem 3.3 Xn

k=0

Gk(x)Vn−k(x) = 1 n

Z x 0

Un−1(y)dy

so the proof is completed.¤

4 Method and Examples

In this section, we describe the RDTM and illustrate it by some numerical examples. Here, we consider the equations of the form

Du(x, t)− Z t

0

Z x 0

K(x, t, y, z)u^q(y, z)dydz=f(x, t), x∈[0, b], t∈[0, d] (26) with some supplementary conditions, whereqis a positive integer. Obviously, in caseq= 1, the equation (26) is reduced to a linear equation.

We also assume thatKhas the following degenerate form K(x, t, y, z) =

Xp

i=0

vi(x, t)wi(y, z).

By substituting thisK(x, t, y, z)into (26), we obtain Du(x, t)−

Xp

i=1

v_i(x, t) Z t

0

Z x 0

w_i(y, z)u^q(y, z)dydz=f(x, t) (27) which is solvable by using RDTM.

By using the theorems 3.1, 3.2, 3.3 and 3.4 in equation (27), we obtain a recursive relation forU_n(x)(reduced differential transform ofu(x, t)) which is solved by programming in MAPLE environment to obtainU_n(x), n= 0,1, ..., N. At last, we use the relation (19) foruN(x, t)(approximate ofu(x, t)) in truncated form

u_N(x, t) = XN

n=0

U_n(x)(t−t₀)ⁿ

to obtainu_N(x, t).

Note should be made that, we also need some starting values ofU_n(x)for solving the recursive relation that can be obtained from integro-differential equation and supplementary conditions.

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Example 4.1 Consider the integro-differential equation

∂²u(x, t)

∂t² −u(x, t)− Z t

0

Z x 0

xe^y−zu(y, z)dydz=1

2xt(1−e^2x), x, tǫ[0,1] (28) with supplementary conditions

u(x,0) =e^x

∂u

∂t(x,0) =e^x

which has the exact solutionu(x, t) =e^x+t.

By applying the RDTM on the equation (28), we obtain

Un+2(x) = 1 (n+ 1)(n+ 2)

"

Un(x) + 1 nx

n−1X

k=0

(−1)^k k!

Z x 0

e^yUn−k−1(y)dy+1

2xδn,1(1−e^2x)

#

(29) forn= 1,2, . . . , N−2.

Also from supplementary conditions we have

U0(x) =e^x, U1(x) =e^x and by settingt= 0in the equation (28) we obtain

U2(x) = 1 2e^x finally the equation (29) leads to

Un(x) = 1

n!e^x, n= 3, ..., N.

Note should be added that, the obtained solution is the same as the firstN + 1terms of the Poisson series of the exact solution. Table 1 shows the absolute errors at some points.

Table 1

(x, t) Error(N = 10) Error(N = 12) Error(N = 14) (0.25,0.25) 0.7832e−14 0.3000e−17 0.2000e−18 (0.50,0.75) 0.1860e−8 0.6645e−11 0.1767e−13 (0.50,1.00) 0.4503e−7 0.2850e−9 0.1344e−11 (0.75,0.75) 0.2388e−8 0.8532e−11 0.2270e−13 (0.75,1.00) 0.5782e−7 0.3660e−9 0.1726e−11 (1.00,1.00) 0.7424e−7 0.4700e−9 0.2217e−11

Example 4.2 We consider a nonlinear equation with variable coefficients of the form

∂²u(x, t)

∂t² +t²u(x, t)− Z t

0

Z x 0

ye^3zu³(y, z)dydz= (t³+t−2)e^x−t

−1

36(3x−1)t⁴e^3x− 1

36t⁴, x, tǫ[0,1] (30)

with supplementary conditions u(x,0) = 0

∂u

∂t(x,0) =e^x

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and the exact solutionu(x, t) =te^x−t. By the same way of the example 1. we obtain

U_n+2(x) = 1 (n+ 1)(n+ 2)

"

1 n

n−1X

k=0 n−k−1X

r=0

n−k−r−1X

l=0

3^k k!

Z x 0

yU_r(y)Ul(y)Un−k−r−l−1(y)dy

− Xn

k=0

δ_k,2U_n−k(x) +e^x Xn

k=0

(δk,3+δ_k,1−2δk,0)(−1)^n−k (n−k)!

−1

36(3x−1)e^3xδn,4− 1 36δn,4

¸

, n= 1,2, . . . . (31)

And from supplementary conditions we have

U0(x) = 0, U1(x) =e^x. Settingt= 0in the both sides of (30) yields

U2(x) =−e^x

And similar to the previous examples the obtained solution is the same as the firstN+ 1terms of the Poisson series of the exact solution. Table 2 shows the absolute errors at some points.

Table 2

Example 4.3 We consider in this example a linear integro-differential equation as

∂²u(x, t)

∂x² +∂²u(x, t)

∂t² + 3u(x, t)− Z t

0

Z x 0

1

coszu(y, z)dydz=−2sintcos(x+t)

−sinx−sint+sin(x+t), x, tǫ[0,1]. (32) with supplementary conditions

u(x,0) =sinx

∂u

∂t(x,0) =cosx

and the exact solutionu(x, t) =costsin(x+t).

Similar to the previous examples, we obtain

(n+ 1)(n+ 2)(n+ 3) d

dxUn+3(x) =Un(x)−

n−1X

k=0

(k+ 1)

· d dx

µd²Uk+1

dx² + 3Uk+1(x)

+(k+ 2)(k+ 3)Uk+3(x) + 2

k+1X

r=0

1

r!(k+ 1−r)!sin(rπ

2 )sin(x+(k+ 1−r)π

2 ) +sinxδ_k+1,0

+ 1

(k+ 1)!(sin((k+ 1)π

2 )−sin(x+(k+ 1)π 2 ))

¶¸ · 1

(n−k)!cos((n−k)π

2 )

¸

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(11)

And from the given conditions, we have

U0(x) =sinx, U1(x) =cosx and by settingt= 0in the equation (32) we have

U2(x) =−sinx.

Finally we obtainUn(x)forn= 3, ..., Nfrom the recurrence relation (33).

U3(x) =−2

3cos(x) U4(x) =1

3sin(x) U5(x) = 2

15cos(x) U6(x) =− 2

45sin(x) U7(x) =− 4

315cos(x) U8(x) = 1

315sin(x) U9(x) = 2

2835cos(x) U10(x) =− 2

14175sin(x) U11(x) =− 4

155925cos(x) U12(x) = 4

467775sin(x) U13(x) = 4

6081075cos(x) U14(x) =− 4

42567525sin(x) And thus, Table 3 shows the absolute errors for this example.

Table 3

5 Conclusion

In this study, we applied the reduced form of the two-dimensional differential transform method for solving the linear and nonlinear Volterra partial integro-differential equations. For illustration purposes, we solved some examples. The results of the examples showed that this method enjoys high accuracy. Moreover, this method can be applied to many similar equations without linearization, discretization and perturbation. Also, since this is a simple method, it can be used by researchers in applied sciences and engineering.

To close the discussion, it seems to me that RDTM can be developed for solving Fredholm partial integro-differential equations, as well as systems of Volterra and Fredholm partial integro-differential equations.

6 Acknowledgment

The author would like to thank Dr. Sedaghat Shahmorad from University of Tabriz for his careful reading of the preliminary forms of this paper and providing me with useful comments which greatly improved the quality of this paper.

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