Numerical Solution Of High Dimensional functional ...

The Second Khansar Conference on Mathematical Sciences 4th Conference on Mathematical Analysis and its Applications Faculty of Khansar, Khansar, Iran, May 7-8, 2013

Numerical Solution Of High Dimensional functional Hammerestein Integral Equation by Using Mean Value Theorem Method Yaghoub Mahmoodia Farhad dastmalchi saeib Samad Noeiaghdamc a,b,c Department

of Mathematics, Islamic Azad University, Tabriz Branch, Tabriz, Iran.

Abstract In this work, we present the numerical method for solving functional Hammerstein integral equations. The presented method can be used for solving functional integral equations in high dimensions. In this paper, the mean value theorem for integrals has been used to solve the high dimensional functional integral equations. This method transforms any integral equation into a system of equations. Some example are presented to show the ability, effective and simple of the method. AMS subject Classification 2010: Primary: 45B05; Secondary: 45B99 Keyword and phrases: Hammerstein functional integral equations, Integral mean value theorem

1

Introduction

It is well known that functional integral equations of various types create an important branch of nonlinear analysis and find numerous applications in describing of miscellaneous real world problems. Let us mention that those problems arise frequently in mechanics, physics, engineering, biology, economics and so on. Apart from that, integral equations are often investigated in research papers and monographs [2, 3, 4, 5, 6, 7, 9, 10, 12]. In this paper we presented a new algorithm for solving high dimensional functional Hammerstein integral equations. The mean value theorem method has proved to be one of the useful techniques in solving numerous Fredholm integral equations.[2, 8] This paper organized as follow : section 2 introduces the main idea of method for solving two dimensional Hammerstein integral equations and 1

2 generalized the method for solving high dimensional functional integral equations. this section includes the extended formulae to clarify the generalization process. section 3 contains numerical example, graph of error functions and comparison between exact and approximate solution to describe the method, and section 4 is discussion and conclusion.

2 2.1

Main Results Solving of two dimensional Hammerstein functional integral equations by mean value theorem method

Following functional Hammerstein integral equation is given by Z b u(x) = f (x) + K(x, t) G(t, u(t), u(h(t)) dt, x ∈ [a, b]

(1)

a

where a, b ∈ R, a < b, K : [a, b] × [a, b] → R, h, f : [a, b] → R are known continuous functions. In this paper we consider following two dimensional Hammerstein functional integral equation Z bZ d u(x, y) = f (x, y)+ K(x, y, s, t)G(s, t, u(s, t), u(h(s, t), h0 (s, t))) ds dt a

c

(2) where a, b, c, d ∈ R, a < b, c < d, K, h, f are known continuous functions and u(x, y) is unknown function to be determined. Theorem 2.1 (mean value theorem for integrals [11]). If w(x) is continuous in [a, b], then there is a point c ∈ [a, b], such that Z b w(x)dx = (b − a)w(c) (3) a

Corollary(mean value theorem for integrals). If w(x, y) is continuous in [a, b] × [c, d], then there are points c1 ∈ [a, b] and c2 ∈ [c, d], such that Z

b

w(s, t)ds = (b − a)w(c1 , t)

(4)

w(s, t)dt = (d − c)w(s, c2 )

(5)

a

and Z c

d

3 Proof :It is clear using(3)

Theorem 2.2 (mean value theorem for double integrals). If w(x, y) is continuous in [a, b] × [c, d], then there are points c1 ∈ [a, b] and c2 ∈ [c, d], such that Z bZ

d

w(s, t) ds dt = (b − a)(d − c)w(c1 , c2 ). a

(6)

c

Proof:It is clear using Theorem 1. Algorithm

1. Apply(5) in (2) to get Z u(x, y) = f (x, y)+(d−c)

b

  K(x, y, s, c2 ) G s, c2 , u(s, c2 ), u(h(s, c2 ), h0 (s, c2 )) ds

a

(7) 2. Apply(4) in (7)to get   u(x, y) = f (x, y)+(b−a)(d−c)K(x, y, c1 , c2 ) G c1 , c2 , u(c1 , c2 ), u(h(c1 , c2 ), h0 (c1 , c2 )) (8) 3. Let (x, y) = (c1 , c2 ) in the (8). It is obtained as u(c1 , c2 ) = f (c1 , c2 ) + (b − a)(d − c)K(c1 , c2 , c1 , c2 ) (9) × G[c1 , c2 , u(c1 , c2 ), u(h(c1 , c2 ), h0 (c1 , c2 ))] 4. Let (x, y) = (h(c1 , c2 ), h0 (c1 , c2 )) in the (8). It is obtained as u (h(c1 , c2 ), h0 (c1 , c2 )) = f (h(c1 , c2 ), h0 (c1 , c2 )) + (b − a)(d − c)   × K(h(c1 , c2 ), h0 (c1 , c2 ), c1 , c2 ) G c1 , c2 , u(c1 , c2 ), u(h(c1 , c2 ), h0 (c1 , c2 )) (10)

4 5. Replace (8) into (2)

Z bZ

d

K(x, y, s, t)G[s, t, f (s, t) + (b − a)(d − c)

u(x, y) = f (x, y) + a

c

× K(s, t, c1 , c2 ) G[c1 , c2 , u(c1 , c2 ), u(h(c1 , c2 ), h0 (c1 , c2 ))], f (h(s, t), h0 (s, t)) + (b − a)(d − c)K(h(s, t), h0 (s, t), c1 , c2 ) G(c1 , c2 , u(c1 , c2 ), u(h(c1 , c2 ) , h0 (c1 , c2 )))] ds dt (11)

6. Let (x, y) = (c1 , c2 ) in the (11). It is obtained as

Z bZ

d

K(c1 , c2 , s, t)G[s, t, f (s, t) + (b − a)(d − c)

u(c1 , c2 ) = f (c1 , c2 ) + a

c

× K(s, t, c1 , c2 )G[c1 , c2 , u(c1 , c2 ), u(h(c1 , c2 ), h0 (c1 , c2 ))], f (h(s, t), h0 (s, t)) + (b − a)(d − c)K(h(s, t), h0 (s, t), c1 , c2 ) G(c1 , c2 , u(c1 , c2 ), u(h(c1 , c2 ) , h0 (c1 , c2 )))] ds dt (12)

7. Replace (8) into (7)

Z u(x, y) = f (x, y) + (d − c)

b

K(x, y, s, c2 ) G[s, c2 , f (s, c2 ) + (b − a)(d − c) a

× K(s, c2 , c1 , c2 ) G[c1 , c2 , u(c1 , c2 ), u(h(c1 , c2 ), h0 (c1 , c2 ))], × f (h(s, c2 ), h0 (s, c2 )) + (b − a)(d − c)K(h(s, c2 ), h0 (s, c2 ), c1 , c2 ) × G[c1 , c2 , u(c1 , c2 ), u(h(c1 , c2 ), h0 (c1 , c2 ))]] ds (13)

5 8. Let (x, y) = (c1 , c2 ) in the (13). It is obtained as b

Z u(c1 , c2 ) = f (c1 , c2 ) + (d − c)

K(c1 , c2 , s, c2 ) G[s, c2 , f (s, c2 ) a

+ (b − a)(d − c)K(s, c2 , c1 , c2 ) G[c1 , c2 , u(c1 , c2 ), u(h(c1 , c2 ) , h0 (c1 , c2 ))], f (h(s, c2 ), h0 (s, c2 )) + (b − a)(d − c)K(h(s, c2 ), h0 (s, c2 ), c1 , c2 ) G[c1 , c2 , u(c1 , c2 ), u(h(c1 , c2 ), h0 (c1 , c2 ))]] ds (14) 9. Solve the obtained equations (9), (10), (12) and (14) simultaneously as a system including four equations of for unknowns c1 , c2 , u(c1 , c2 ) and u(h(c1 , c2 ), h0 (c1 , c2 )). 10. Substitute the obtained c1 , c2 , u(c1 , c2 ) and u(h(c1 , c2 ), h0 (c1 , c2 )) into (8) to get the approximate solution of the fractional integral equation.

2.2

Solving of high dimensional functional Hammerstein integral equations via IMVM

Consider the following functional Hammerstein integral equation Z

b1

Z

b2

u(x1 , x2 , ..., xn ) = f (x1 , x2 , ..., xn ) +

Z

bn

K(x1 , x2 , ..., xn , t1 , t2 , ..., tn )

... a1

a2

an

G[t1 , t2 , ..., tn , u(t1 , t2 , ..., tn ), u(h1 (t1 , t2 , ..., tn ), ..., hn (t1 , t2 , ..., tn ) dtn dtn−1 ... dt1 (15) which can be considered in the other useful type as follows Z

b1

Z

b2

u(x) = f (x) +

Z

bn

... a1

a2

K(x, t)G(t, u(t), u(h1 (t), ..., hn (t))) an

dtn dtn−1 ... dt1 (16) where x = (x1 , x2 , ..., xn )and t = (t1 , t2 , ..., tn ) . These notations are valid in the following lines. Similar to the previous section, instead of using mean value theorem for multiple integral, we apply one dimensional integral mean value theorem directly to provide needful linearly

6 independent equations.

Theorem 2.3 (mean value theorem for multiple integrals). If s(x) is continuous in [ai , bi ]n , i = 1, 2, ..., n, then there are points ci ∈ [ai , bi ], i = 1, 2, ..., n such that Z

b1

b2

Z

Z

bn

... a1

=

a2

(s(x)) dxn dxn−1 ... dx1 an

(17)

n−1 Y

(bn−j − an−j )s(c1 , c2 , ..., cn )

j=0

Now, to find u(x1 , x2 , ..., xn ), we have to obtain c1 , c2 , ..., cn and u(c1 , c2 , ..., cn ).

Algorithm 1. Apply the integral mean value theorem for interval[an , bn ]: Z

bn

K(x, t)G[t, u(t), u(h1 (t), ..., hn (t)) dtn = an

(18) (bn − an )K(x, ξn )G[ξn , u(ξn ), u(h1 (ξn ), ..., hn (ξn ))]

where ξn = (t1 , t2 , ..., tn−1 , cn ). 2. Substitute 18 into 16 to obtain Z

b1

Z

b2

u(x) = f (x) + (bn − an )

Z

bn−1

... a1

a2

an−1

(19)

K(x, ξn )G[ξn , u(ξn ), u(h1 (ξn ), ..., hn (ξn ))]dtn−1 ... dt1

3. Again, we use the integral mean value theorem with i = n − 1,as

7 follows Z bn−1 K(x, t1 , ..., cn )]G(t1 , ..., cn , u(t1 , ..., cn ), an−1

u(h1 (t1 , ..., cn ), ..., hn (t1 , ..., cn )) dtn−1 = (bn−1 − an−1 )K(x, ξn−1 )G[ξn−1 , u(ξn−1 ), u(h1 (ξn−1 ), ..., hn (ξn−1 ))] (20) where ξn−1 = (t1 , t2 , ..., cn−1 , cn ). 4. Replace the above equation into (19). So, we have Z b1 Z bn−2 ... K(x, ξn−1 ) u(x) = f (x) + (bn − an )(bn−1 − an−1 ) a1

an−2

G[ξn−1 , u(ξn−1 ), u(h1 (ξn−1 ), ..., hn (ξn−1 ))]dtn−2 ... dt1 (21) 5. Repeat the process for i = 3, ..., n − 1, as Z

b1

u(x) = f (x) + (bn − an )...(bn−i+1 − an−i+1 )

Z

bn−i

K(x, ξn−i+1 )

... a1

an−i

G[ξn−i+1 , u(ξn−i+1 ), u(h1 (ξn−i+1 ), ..., hn (ξn−i+1 ))] dtn−i ... dt1 (22) where ξn−i+1 = (t1 , t2 , ..., tn−i , cn−i+1 , cn−i+2 , ..., cn ). Also, the n-th step of the above process leads to u(x) = f (x) + (bn − an )...(b1 − a1 )K(x, ξ1 )]G[ξ1 , u(ξ1 ), u(h1 (ξ1 ), ..., hn (ξ1 ))] (23) where ξ1 = (c1 , c2 , ..., cn ). 6. Let x = ξ1 into 23 to get u(ξ1 ) = f (ξ1 ) +

n−1 Y

(bn−j − an−j )K(ξ1 , ξ1 )

j=0

G[ξ1 , u(ξ1 ), u(h1 (ξ1 ), ..., hn (ξ1 ))]

(24)

8 Now the first equation of the proposed system including (n + 1)unknowns and (n + 1) equations is constructed.

7. To make the other equations, the demonstrated process must be repeated such that the i-th achieved equation is as follows (substitute 23 in 22)

u(x) = f (x) +

i−1 Y

Z (bn−j − an−j )

n−1 Y

bn−i

Z ...

K(x, ξn−i+1 )G[ξn−i+1 , (f (ξn−i+1 ) an−i

a1

j=0

+

b1

(bn−j − an−j )K(ξn−i+1 , ξ1 )G(ξ1 , u(ξ1 ), u(H1 ))), f (H1 )+

j=0 i−1 Y

(bn−j − an−j )K(H1 , ξ1 )G(ξ1 , u(ξ1 ), u(H1 ))]dtn−i ... dt1

j=0

(25) where i = 1, ..., n − 1 and H1 = (h1 (ξ1 ), ..., hn (ξ1 )) , and gives

u(ξ1 ) = f (ξ1 ) +

i−1 Y

Z

b1

(bn−j − an−j )

j=0

Z

bn−i

... an−i

a1

K(ξ1 , ξn−i+1 )G[ξn−i+1 , (f (ξn−i+1 ) +

n−1 Y

(bn−j − an−j )

j=0

K(ξn−i+1 , ξ1 )G(ξ1 , u(ξ1 ), u(H1 , ))), f (H1 ) +

Qi−1

j=0 (bn−j

− an−j )

K(H1 , ξ1 )G(ξ1 , u(ξ1 ), u(H1 ))]dtn−i ... dt1 (26) as the i-th equation. Therefore, we obtain the (n − 1), new equations when i = 1, ..., n − 1, for final mentioned system including (n + 1)unknowns and (n + 1) equations. Also, 24 can be provided 26 with i = n.

8. Implement the previous step to obtain the n-th equation by substitute 23 in 16 as

9

Z

b1

u(x) = f (x) +

Z

K(x, t)G[t, f (t) an

a1

+

bn

...

n−1 Y

(bn−j − an−j )K(t, ξ1 )G(ξ1 , u(ξ1 ), u(H1 )), f (Ht )

j=0

+

n−1 Y

(bn−j − an−j )K(Ht , ξ1 )G(ξ1 , u(ξ1 ), u(H1 ))] dtn ... dt1

j=0

(27) which lead to Z

b1

u(ξ1 ) = f (ξ1 ) +

Z

a1

+

bn

...

K(ξ1 , t)G[t, f (t) an

n−1 Y

(bn−j − an−j )K(t, ξ1 )G(ξ1 , u(ξ1 ), u(H1 )), f (Ht )

j=0

+

n−1 Y

(bn−j − an−j )K(Ht , ξ1 )G(ξ1 , u(ξ1 ), u(H1 ))] dtn ... dt1

j=0

(28) 9. Let(H1 ) in 23 which lead to u(H1 ) = f (H1 ) +

n−1 Y

(bn−j − an−j )K(H1 , ξ1 )

j=0

(29)

G[ξ1 , u(ξ1 ), u(h1 (ξ1 ), ..., hn (ξ1 ))] 10. Solve the nonlinear system including (24), (26), , (28) and (29) with the Newtons method or other efficient methods.

Remark: The equations of the final system include the numerous integrals containing long terms. It is recommended to use the numerical integration rules such as Gauss quadrature rule or trapezoidal integration method.

10

3

Numerical results

In this section we presented different dimensional functional Hammerstein integral equations and graph of error function and compression between exact and approximate solution to show this method for solving high dimensional problems is simple and precise. Example1: Consider the integral equations : Z 1Z 1 x2 y(1+st2 )(st+u(s, t)+u(h(s, t), h0 (s, t))) ds dt u(x, y) = f (x, y)+ 0

0

where 1 1 133x2 y h(s, t) = st, h0 (s, t) = st, f (x, y) = x + y − 2 3 72 for which the exact solution is u(x, y) = x + y. Using the presented method leads to the following system of equations:  (133c21 c2)    u(c1 , c2 ) = c1 + c2 − + c21 c2 (1 + c1 c22 )(c1 c2 + u(c1 , c2 )   72    + u(h(c1 , c2 ), h0 (c1 , c2 )))         1 1 (133(1/2c1 c2 )2 (1/2c1 c2 ))  0  u(h(c , c ), h (c , c )) = c c + c c −  1 2 1 2 1 2 1 2   2 3 72  2 2 0  + ((1/2c c ) (1/3c c )(1 + c c2 )(c c + u(c  1 2 1 2 1 1 2 1 , c2 ) + u(h(c1 , c2 ), h (c1 , c2 ))))        1 (133c21 c2 ) + c21 c2 (8640 + c2 (80(281 + 72u(c1 , c2 ) u(c1 , c2 ) = c1 + c2 − 72 17280    + 72u(h(c1 , c2 ), h0 (c1 , c2 ))) + c2 (5760 + 72c12 c22 (80 + 65c22 + 4c42 )+     4c32 (−133 + 72u(c1 , c2 ) + 72u(h(c1 , c2 ), h0 (c1 , c2 ))) + 5c2 (2111 + 936u(c1 , c2 )+     936u(h(c1 , c2 ), h0 (c1 , c2 ))) + 72c1 (80 + 65c22 + 4c42 )(1 + c2 (u(c1 , c2 )+      u(h(c1 , c2 ), h0 (c1 , c2 )))))))         u(c1 , c2 ) = c1 + c2 − (133c21 c2 )/72 + (c21 c2 (292201+     49176u(c1 , c2 ) + 49176u(h(c1 , c2 ), h0 (c1 , c2 )) + 49176c1 c2    (1 + c2 (c1 c2 + u(c1 , c2 ) + u(h(c1 , c2 ), h0 (c1 , c2 ))))))/207360 By mentioned method, values of c1 , c2 , u(c1 , c2 )andu(h(c1 , c2 ), h0 (c1 , c2 )) are found as follows: c1 = 0.5189727638344717 c2 = 0.5522330010440364 u(c1 , c2 ) = 1.0712057648785083 u(h(c1 , c2 ), h0 (c1 , c2 )) = 0.2370164456396298

11 Hence, we have u(x, y) = x + y − 4.44089 × 10−15 . Fig. 1.show the comparison between exact solution and approximate solution and error function is demonstrated in Fig. 2.

Insert Fig.1.

Insert Fig.2.

Example2: Consider the integral equations : Z 1Z 1 ex+y+s+t [stu(s, t) + u(h(s, t), h0 (s, t))]ds dt u(x, y) = f (x, y) + 0

0

where 1 1 1 h(s, t) = st, h0 (s, t) = st, f (x, y) = − e(x+y) (−19+12e)+x+y 2 3 6 for which the exact solution is u(x, y) = x + y. By mentioned method, values of c1 , c2 , u(c1 , c2 )andu(h(c1 , c2 ), h0 (c1 , c2 )) are found as follows: c1 = 0.5898248126729583 c2 = 0.5886747070573066 u(c1 , c2 ) = 1.178499519730266 u(h(c1 , c2 ), h0 (c1 , c2 )) = 0.2893457906794877 Hence, we have u(x, y) = x + y − 3.55271 × 10−15 . Fig. 3.show the comparison between exact solution and approximate solution and error function is demonstrated in Fig. 4.

Insert Fig.3.

Insert Fig.4.

Example3: Consider the integral equations : Z 1Z 1 u(x, y) = f (x, y) + (s t exy [s + t + u(s, t) + u(h(s, t), h0 (s, t))]ds dt 0

0

where h(s, t) =

1 s t, 5

h0 (s, t) =

1 s t, 3

f (x, y) = −

74exy + x2 − y 2 225

12 for which the exact solution is u(x, y) = x2 − y 2 . By mentioned method, values of c1 , c2 , u(c1 , c2 )andu(h(c1 , c2 ), h0 (c1 , c2 )) are found as follows: c1 = 0.5927422175624245 c2 = 0.4672278611153644 u(c1 , c2 ) = 0.13304146227838248 u(h(c1 , c2 ), h0 (c1 , c2 )) = −0.005454144628785653 Hence, we have u(x, y) = x2 − y 2 − 7.54475 × 10−16 . Fig. 5.show the comparison between exact solution and approximate solution and error function is demonstrated in Fig. 6.

Insert Fig.5.

4

Insert Fig.6.

Conclusions

In this paper, we used the new efficient method for solving high dimensional Hammerstein functional integral equations. We exhibit the applied main idea that it just is the famous integral mean value theorem.Exampels that were presented show the ability of the model.The results confirm that the method is applied and simple.

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13 [4] T.A. Burton, Volterra Integral and Differential Equations, Academic Press, New York, 1983. [5] K. Deimling, Nonlinear Functional Analysis, Springer Verlag, Berlin, 1985. [6] N. Dunford, J.T. Schwartz, Linear Operators, Int. Publ., Leyden, 1963. [7] G. Emmanuele, Integrable solutions of a functional-integral equation, J. Integral Equations 4 (1992) 89-94. [8] M. Heydari, Z. Avazzadeh, H. Navabpour, G.B. Loghmani, Numerical solution of Fredholm integral equations of the second kind by using integral mean value theorem II. High dimensional problems, Appl. Math. Model. (2012)In press . [9] M.A. Krasnoselskii, P.P. Zabrejko, J.I. Pustylnik, P.J. Sobolevskii, Integral Operators in Spaces of Summable Functions, Noordhoff, Leyden, 1976. [10] D. ORegan, M. Meehan, Existence Theory for Nonlinear Integral and Integradifferential Equations, Kluwer Academic Publishers, Dordrecht, 1998. [11] R.A. Silverman, Essential Calculus with Applications, Dover, 1989. [12] P.P. Zabrejko, A.I. Koshelev, M.A. Krasnoselskii, S.G. Mikhlin, L.S. Rakovshchik, V.J. Stecenko, Integral Equations, Noordhoff, Leyden, 1975.