IBN AL- HAITHAM J. FOR PURE & APPL. S CI. VOL.23 (2) 2010 Some Statistical Properties of Linear Volterra Integral Equation so lutions A.A. Hussein Departme nt of Mathematics, College of Education Ibn-Al-Haitham, Unive rsity 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 Trap ezoidal and Simp son's rule. That is to derive some st atist ical p rop erties exp ected value, the variance and the correlation coeff icient 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 integr al 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 Vo lterra 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 const ant[1]. If X be a continuous typ e of random variable h avin g a p .d.f. f( x), and u( x) b e a function of X such that   - dx f (x) )x(u , and if X is a discrete typ e of random variable[2] such that  x f (x) )x(u . Then the integral, or the sum called the exp ected value[2,3] of u(x). And t he variance of X denoted by σ 2 or V( x), defin ed [2,3] as: 222 )E(X )X(E)x(V   where X is a d iscrete or continuous of random variable and μ= E(X) □ 2. Solving Linear One Dimension Volterra Inte gral 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 p oint for x, then by setting x=xi, i=0,1,..,n, in equation (3) we can hav e: dttutxKxfxu ix a iii )( ),()()(  ….………….……………….(4) Then by rep lacing the integral term in the right hand side of equation (4) by the Trap ezoidal rule, to get: IHJPAS IBN AL- HAITHAM J. FOR PURE & APPL. S CI. 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 Inte gral Equation of The Second Kind Using Simpson Rule[1]. As above solution, we d ivide 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 r estricted to be an ev en integer, and t=(xn- a)/n, xn is the end p oint for x. By replacing the integr al term in the right hand side of equation (4) by Simp son 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-dimension al Volterra linear integr al equation of the second kind: dt)t(u )t,x(K)x(f)x(u)x(h x a  …………………………(7) First step : Solve the integr al equation in (7) usin g two numerical methods Trapezoidal rule, then by Simp son rule which illustrated in section two. Second st ep: Interp olate the data obtained in st ep one for each method resp ectively by using Newt on Forward Formula [4]. Third step: Find the exp ected value and the variance [2,3] for the result obtained by Trapezoidal rule, then for the result obtained using Simpson rule. Fourt h step : Compute the Correlation Coefficient [2,3] of the solutions obtained in above st ep s. Exam ple 1x0 where dt )t(u )tx()x2exp() 3 4 xx2()x(u x 0 2    ………….….(8) We will solv e 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 trap ezoidal rule, we apply equation (5) and we can written the equation (5) in the followin g comp act form: IHJPAS IBN AL- HAITHAM J. FOR PURE & APPL. S CI. 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 t he 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 Newt on 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 f ind the p olynomial which is: f(x)=0.013214064 x 10 - 0.0766672228 x 9 + 0.2415794909 x 8 - 0.6073832681 x 7+ 1.3600009821 x 6 - 2.5771430663 x 5 + 3.8357067640 x 4 – 4.0526880921 x 3 + 3.3306605692 x 2 – 1.6639099065 x + 1.3333333333. Now we can obtain the exp ected 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. S CI. 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    Se cond, we will solve the inte gral equation (8) by using Simp son’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 Simp son’s rule, we app ly equation(6) and we can write equation (6) in the followin g comp act 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 t he 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 Newt on Forward Formula[4], we can find the p olynomial which is: f(x)= 2387.0379995 x 10 – 11940.289032 x 9 + 25626.121937 x 8 – 30845.226184 x 7 + 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 exp ected 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 inte gral equation (8) at each xi, i=0,1,…,10, then one can get following values: IHJPAS IBN AL- HAITHAM J. FOR PURE & APPL. S CI. 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 Newt on Forward Formula[4], we can find the p olynomial which is: f(x) = -0.0078939518 x 10 – 0.1017527499 x 9 + 0.8462405483 x 8 – 2.4760805677 x 7 + 4.4612226593 x 6 -5.7691603786 x 5 + 6.0710939102 x 4 – 5.0167100778 x 3 + 3.4343687856 x 2 – 1.6665858973 x + 1.3333333333. Now we can obtain the exp ected 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 Trap ezoidal, Simp son’s rule and exact by X,Y and Z, resp ectively and there mean by Z and Y,X resp ectively. To comp ute the correlation coefficient of t hese 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 X Y           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 Trap ezoidal 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. Re ferences 1.Kyt he, P.K.and Puri, P. (2002) ,Computational M ethods of Linear Integral Equations, Sp ringer-Verlag, New York. 2.Hogg Robert, V. and Craig Allen, T. (1978) ,Introduction to M athematical Statist ics,M acmillan Publishing Co., Inc. IHJPAS IBN AL- HAITHAM J. FOR PURE & APPL. S CI. VOL.23 (2) 2010 3.Neter, J. Kunt er, M . H. Nachtsheim, C. J. and Wasserman, W. (1996) ,Ap p lied Linear st atistical M odels, T he M cGraw-Hill Companies, Inc. 4.Burden, R.L. and Faires, J.D. (2001) ,Numerical Analy sis" seventh edition, An International Publishing Co mpany (ITP). Table :(1) represents the exact and the numerical soluti on of Volterra integral equati on(3) i xi Ui(Trapezoidal) Ui(S impson) Exact sol uti on 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