IBN AL- HAITHAM J. FOR PURE & APPL. SCI. VOL.23 (2) 2010 Some Statistical Properties of Linear Volterra Integral Equation solutions A.A. Hussein Department of Mathematics, College of Education Ibn-Al-Haitham, University of Baghdad Abstract Our aim of this research is to find the results of numerical solution of Volterra linear integral equation of the second kind using numerical methods such that Trapezoidal and Simpson's rule. That is to derive some statistical properties expected value, the variance and the correlation coefficient between the numerical and exact solution□ 1. Introduction The linear integral equation in u(x) can be represented as[1]: dt)t(u )t,x(K)x(f)x(u)x(h )x(b a  ...………….………….(1) Equation (1) is called a Volterra integral equation when b(x)=x, dt)t(u )t,x(K)x(f)x(u)x(h x a  ……...…………………….(2) Equation (2) is called a Volterra equation of the first kind when h=0, and is called a Volterra equation of the second kind when h=1, K(x,t) is known function and is called the kernel of the integral equation, where a is constant[1]. If X be a continuous type of random variable having a p.d.f. f(x), and u(x) be a function of X such that   - dx f(x) )x(u , and if X is a discrete type of random variable[2] such that  x f(x) )x(u . Then the integral, or the sum called the expected value[2,3] of u(x). And the variance of X denoted by σ2 or V(x), defined [2,3] as: 222 )E(X )X(E)x(V   where X is a discrete or continuous of random variable and μ=E(X) □ 2. Solving Linear One Dimension Volterra Integral Equation of The Second Kind Using Trapezoidal Rule[1]: dttutxKxfxu x a )( ),()()(  ……………………………….……..(3) By dividing the interval [a,x] into n subintervals [xi,xi+1], i=0,1,..,n-1, such that xi=a+it, i=0,1,..,n, where t=(xn-a)/n, xn is the end point for x, then by setting x=xi, i=0,1,..,n, in equation (3) we can have: dttutxKxfxu ix a iii )( ),()()(  ….………….……………….(4) Then by replacing the integral term in the right hand side of equation (4) by the Trapezoidal rule, to get: IHJPAS IBN AL- HAITHAM J. FOR PURE & APPL. SCI. VOL.23 (2) 2010 ij n, 1,...,i ],),( 2 1 ),K(x ... ),(),( 2 1 [)()( )()( 11i 1100 00     jjijj iiii utxKut utxKutxKtxfxu xfxu ..…….(5) 3. Solving Linear One Dimension Volterra Integral Equation of The Second Kind Using Simpson Rule[1]. As above solution, we divide the interval [a,x] into n subintervals [xi,xi+1], i=0,1,..,n-1, such that xi=a+it, i=0,1,..,n, where n is restricted to be an even integer, and t=(xn-a)/n, xn is the end point for x. By replacing the integral term in the right hand side of equation (4) by Simpson rule, to get: ij n, 1,...,i , 3 h where ],),(),K(x4 ... ),(2),(4),([)()( )()( 11i 221100 00        t utxKut utxKutxKutxKhxfxu xfxu jjijj iiiii …(6) 4. The Formulation of Problem Consider the one-dimensional Volterra linear integral equation of the second kind: dt)t(u )t,x(K)x(f)x(u)x(h x a  …………………………(7) First step: Solve the integral equation in (7) using two numerical methods Trapezoidal rule, then by Simpson rule which illustrated in section two. Second step: Interpolate the data obtained in step one for each method respectively by using Newton Forward Formula [4]. Third step: Find the expected value and the variance [2,3] for the result obtained by Trapezoidal rule, then for the result obtained using Simpson rule. Fourth step: Compute the Correlation Coefficient [2,3] of the solutions obtained in above steps. Example 1x0 where dt )t(u )tx()x2exp() 3 4 xx2()x(u x 0 2    ………….….(8) We will solve this integral equation (8) by using two methods. First, by using trapezoidal rule. To do this, we divide the interval of integration [0,1] into 10 equal subintervals of width 10 1 10 01   t . Since we use trapezoidal rule, we apply equation (5) and we can written the equation (5) in the following compact form: IHJPAS IBN AL- HAITHAM J. FOR PURE & APPL. SCI. VOL.23 (2) 2010 ij )t,x(kk n1,2,...,i ]uk 2 1 uk ...ukuk 2 1 [tfu fu jiij jij1j1ij 11i00iii 00      ………………….(9) Here we have x tt,-xt)k(x, ),2exp() 3 4 2()( 22  xxxxf Then the equation (9) becomes: ij n1,2,...,i ]uk 2 1 uk...ukuk 2 1 [t )x2exp() 3 4 xx2(u fu jij1j1ij11i00i ii 2 ii 00      ……….….(10) By evaluating equation (10) at each i=1,2,…,10 one can get the following values: u0=1.3333333333, u1=1.1965553611, u2=1.1067485612, u3=1.0501770263, u4=1.0178222086, u5=1.0039651300, u6=1.0051677590, u7=1.0195489952, u8=1.0462767757, u9=1.0852176618, u10=1.1367003013. By using the following Newton Forward Formula[4]: n n n hn f xxxxxxxx h f xxxx h f xxxfxf ! )).....()()((. . . !2 ))(()()()( 0 1210 2 0 2 10 0 00         We can find the polynomial which is: f(x)=0.013214064 x 10 - 0.0766672228 x 9 + 0.2415794909 x 8 - 0.6073832681 x 7+ 1.3600009821 x6 - 2.5771430663 x5 + 3.8357067640 x4 – 4.0526880921 x3 + 3.3306605692 x2 – 1.6639099065 x + 1.3333333333. Now we can obtain the expected value[2,3] and the variance[2,3],     1 0 x 0 5260811871.0 x f(x) dx)x(E f(x) dx x)x(E IHJPAS IBN AL- HAITHAM J. FOR PURE & APPL. SCI. VOL.23 (2) 2010 330.07516707 ))x(E()x(E)x(V 88 0.45192848 dx f(x) x)x(E 22 1 0 22    Second, we will solve the integral equation (8) by using Simpson’s rule. To do this, we divide the interval of integration [0,1] into equal subintervals of width 10 1 10 0-1 t where 3 t h    Then to use Simpson’s rule, we apply equation(6) and we can write equation (6) in the following compact form: : ij )t,x(kk n1,2,...,i ]ukuk4 ...uk2uk4uk[hfu fu jiij jij1j1ij 22i11i00iii 00      ….......(11) Then the equation (11) becomes: ij n1,2,...,i ]ukuk4... uk2uk4uk[h )x2exp() 3 4 xx2(u fu jij1j1ij 22i11i00iii 2 ii 00     . ….…..(12) By evaluating equation (12) at each i=1,2,…,10 one can get the following values: u0=1.3333333333, u1=1.1943331389, u2=1.1062630050, u3=1.0477357372, u4=1.0168932262, u5=1.0011495203, u6=1.0037745311, u7=1.0162183017, u8=1.0443627039, u9=1.0812198620, u10=1.1341828617. By using Newton Forward Formula[4], we can find the polynomial which is: f(x)= 2387.0379995 x10 – 11940.289032 x9 + 25626.121937 x8 – 30845.226184x7 + 22833.549096 x 6 – 10715.042130 x 5 + 3164.5486860 x 4 – 563.76739209 x 3 + 56.57019699 x 2 – 3.7023280574 x +1.333333333. Now we can obtain the expected value[2,3] and the variance[2,3], 460.07569935 (E(x)) - )E(x )x(V 850.34985003 dx f(x) x )x(E 670.52359400 dx f(x) x)x(E 22 1 0 22 1 0      To find the exact solution we must evaluate integral equation (8) at each xi, i=0,1,…,10, then one can get following values: IHJPAS IBN AL- HAITHAM J. FOR PURE & APPL. SCI. VOL.23 (2) 2010 u0=1.3333333333, u1=1.1965553611, u2=1.1053807815, u3=1.0449205686, u4=1.0062157851, u5=0.9841623359, u6=0.9763178627, u7=0.9822767277, u8=1.0034686676, u9=1.0433565263, u10=1.1081311572. By using Newton Forward Formula[4], we can find the polynomial which is: f(x) = -0.0078939518 x 10 – 0.1017527499 x9 + 0.8462405483 x8 – 2.4760805677 x7 + 4.4612226593 x6 -5.7691603786 x5 + 6.0710939102 x4 – 5.0167100778 x3 + 3.4343687856 x2 – 1.6665858973 x + 1.3333333333. Now we can obtain the expected value[2,3] and the variance[2,3], 650.07918682 (E(x)) - )E(x )x(V 950.34063514 dx f(x) x )x(E 7650.5113201 dx f(x) x)x(E 22 1 0 22 1 0      We will denote the solutions of Trapezoidal, Simpson’s rule and exact by X,Y and Z, respectively and there mean by Z and Y,X respectively. To compute the correlation coefficient of these solutions by using[2,3]: 50389999411042.0 08081018352315.0 45781018292338.0 )YY()XX( )YY)(XX( r 10 0i 10 0i 2 i 2 i 10 0i ii XY           rXY represents the correlation coefficient of solution of Trapezoidal and Simpson’s rule. 28069912367115.0 36481110460085.0 30021100728803.0 )ZZ()XX( )ZZ)(XX( r 10 0i 10 0i 2 i 2 i 10 0i ii XZ           rxz represents the correlation coefficient of solution of Trapezoidal and exact. 7999919365744.0 66901116255890.0 93351107254991.0 )ZZ()YY( )ZZ)(YY( r 10 0i 10 0i 2 i 2 i 10 0i i YZ           rYZ represent the correlation coefficient of solution of Simpson’s rule and exact. References 1.Kythe, P.K.and Puri, P. (2002) ,Computational Methods of Linear Integral Equations, Springer-Verlag, New York. 2.Hogg Robert, V. and Craig Allen, T. (1978) ,Introduction to Mathematical Statistics,Macmillan Publishing Co., Inc. IHJPAS IBN AL- HAITHAM J. FOR PURE & APPL. SCI. VOL.23 (2) 2010 3.Neter, J. Kunter, M. H. Nachtsheim, C. J. and Wasserman, W. (1996) ,Applied Linear statistical Models, The McGraw-Hill Companies, Inc. 4.Burden, R.L. and Faires, J.D. (2001) ,Numerical Analysis" seventh edition, An International Publishing Company (ITP). Table :(1) represents the exact and the numerical solution of Volterra integral equation(3) i xi Ui(Trapezoidal) Ui(S impson) Exact solution 0 0 1.3333333333 1.3333333333 1.3333333333 1 0.1 1.1965553611 1.1943331389 1.1965553611 2 0.2 1.1067485612 1.1062630050 1.1053807815 3 0.3 1.0501770263 1.0477357372 1.0449205686 4 0.4 1.0178222086 1.0168932262 1.0062157851 5 0.5 1.0039651300 1.0011495203 0.9841623359 6 0.6 1.0051677590 1.0037745311 0.9763178627 7 0.7 1.0195489952 1.0162183017 0.9822767277 8 0.8 1.0462767757 1.0443627039 1.0034686676 9 0.9 1.0852176618 1.0812198620 1.0433565263 10 1 1.1367003013 1.1341828617 1.1081311572 IHJPAS 2010) 2( 23مجلة ابن الهیثم للعلوم الصرفة والتطبیقیة المجلد استخدام بعض االحصاءات لحلول معادلة فولتیرا الخطیة التكاملیة عادل عبد الكاظم حسین قسم الریاضیات، كلیة التربیة ابن الهیثم، جامعة بغداد الخالصة لمعادلة فولتیرا الخطیة التكاملیة من الدرجة الثانیة والحل العددي لهذه للبحث هو إیجاد الحل الصحیح الهدف الرئیس المتوقعة والتباین لكل الحلول ومعرفة العالقة بین الحل ثم إیجاد القیمة. المعادلة بطریقة شبه المنحرف وطریقة سیمبسون .المضبوط والحلول العددیة باستخدام معامل االرتباط IHJPAS