IBN AL- HAITHAM J. FOR PURE & APPL. SCI. VOL.24 (3) 2011 Using Semi-Analytic Method to Decreasing Dangers of Lead L. N. M. Tawfiq and M. N. M.Tawfiq Department of Mathematics, College of Education Ibn Al - Haitham , University of Baghdad Received in : 7 November 2010 Accepted in : 8 February 2011 Abstract The aim of this paper is to present a method for solving of system of first order initial value problems of ordinary differential equation by a semi-analytic technique with constructing polynomial solutions for decreasing dangers of lead. The original problem is concerned using two-point osculatory interpolation with the fit equals numbers of derivatives at the end points of an interval [0 , 1]. Key word : Osculator interpolation , Initial value problem Introduction Lead is a metal found naturally in the earth’s crust. In nature , it is found more often in chemical compounds than as a pure metal. When released into the air, it may travel long distances before settling to the ground , where it can contaminate water and soil . Lead has been used extensively throughout human history because it is easy both to extract and to work with. Where used in some Cosmetics and hair dyes, paint houses , Plumbing and solder , Gutters , window glazing , Leaded gasoline , crayons , Printers ( ink ) , batteries , plastics and used in the manufacturing of products others . [1] Where the deficiencies in some mineral nutrients, specifically calcium, iron and zinc, where involvement in building bones, may increase the amount of lead absorbed . As a result, Children, Pregnant Women, old Women and old man are more efficient at absorbing lead, than adult's because they need more calcium. Additionally, the kidneys and the intestine, less able to eliminate lead . Lead absorp tion rates vary ; the gastrointestinal tracts of adults typically absorb 10 - 15 percent of ingested lead , while those of pregnant women and children can absorb up to 50 percent . Lead poisoning occurs when there are adverse health effects due from lead in the body .Some of the more prominent symptoms of lead poisoning include headaches, irritability, abdominal pain , vomiting , anemia ( general weakness) , weight loss, poor attention span, noticeable learning difficulty, Nervous system and kidney damage. However, at very low exposure levels, lead may not produce specific symptoms, but can produce subtle adverse effects on children's development . There is a medical test to show whether I've been exposed to lead through a simple blood test, there are treatments to remove lead from the body although medication that used CAedta intravenously must have the attention into treatment because of probably the metal exciding ZINC from body. Therefore, suggesting treatment application CaNa2 EDTA from mouth way over a period of four weeks , if high levels of lead in blood between 90-100 mg / de , treatment application dimercaprol, also known as BAL the dose 500 and 1500 CAedta where inject in the time self ,but all cured types used at case ,lead poisoning where thrown metal into out of ,but there metals desirable in body may thrown therefore the safeguard best from the remedy IBN AL- HAITHAM J. FOR PURE & APPL. SCI. VOL.24 (3) 2011 Lead in the human body : A case study Rabinowitz , Wetherill and Kopple in [2] made a carefully controlled study of the lead intake and exertion of a healthy volunteer in an industrial urban setting . The data from this study were used to estimate the rate constants for the compartment model (1): ( Blood ) y'1 = – ( k01 + k21 + k31) y1 + k12 y2 + k13 y3 + I1 ( Tissue ) y '2 = k21 y1 - ( k02 + k12 ) y2 ..…. (1) ( Bone ) y'3 = k31y1 - k13 y3 Lead is measured in micrograms and time in days. For example, the rate term 49.3 in system (2) below is the ingestion rate I1 of lead in micrograms per day, while the coefficient 0.0361 (day)-1 in the first rate equation of (2) is the sum of the three compartment transfer coefficients k02 , k21 and k31 of lead from the blood into, respectively, the excretory system, tissue and bones. The full IVP is given by : y'1 = – 0.0361 y1 + 0.0124 y2 + 0.000035 y3 + 49.3 , y1(0) = 0 y'2 = 0.0111 y1 + 0.0286 y2 , y2(0) = 0 …(2) y'3 = 0.0039 y1 – 0.00035 y3 , y3(0) = 0 Assumed that initially there is no lead in the compartments . Now, we solve system (2) using semi-analytic method ,i.e., use two-point osculatory interpolation [3]. Essentially this is a generalization of interpolation using Taylor polynomials and for that reason osculatory interpolation is sometimes referred to as two-point Taylor interpolation. The idea is to approximate a function y(x) by a polynomial P(x) in which values of y(x) and any number of its derivatives at given points are fitted by the corresponding function values and derivatives of P(x) . In this paper, we are particularly concerned with fitting function values and derivatives at the two end points of a finite interval, say [0 , 1] ,wherein a useful and succinct way of writing osculatory interpolant P2n+1(x) of degree 2n + 1 was given for example by Phillips [4] as : P2n+1(x) =   n j 0 { y )( j (0) q j (x) + (-1) j y )( j (1) q j (1-x) } ………….(3) q j (x) = ( x j / j!)(1-x) 1n    jn s 0        s sn x s = Q j (x) / j! ...………..( 4 ) so that ( 3 ) with ( 4 ) satisfies : y )( j (0) = )( 12 j nP  (0) , y )( j (1) = )( 12 j nP  (1) , j = 0, 1, 2,…, n . implying that P2n+1(x) agrees with the appropriately truncated Taylor series for y(x) about x = 0 and x = 1. The error on [0, 1] is given by : R2n+1 = y(x)-P2n+1(x) = )!22( )()1()1( )22(1)1(1    n yxx nnnn  where ε  (0, 1) and y )22( n is assumed to be continuous. Finally we observe that ( 3 ) can be written directly in terms of the Taylor coefficients ai and bi about x = 0 and x = 1 respectively, as : P2n+1(x) =   n j 0 { a j Q j (x) + (-1) j b j Q j (1-x) } ….. (5) Now, the simple idea behind the use of two-point polynomials is to replace y(x) in problem by a P2n+1 ( equation(3) or (5) ) which enables any unknown derivatives of y (x) to be computed . The first step therefore is to construct the P2n+1 . To do this we need the Taylor coefficients of y1(x) and y2(x) respectively about x = 0 : IBN AL- HAITHAM J. FOR PURE & APPL. SCI. VOL.24 (3) 2011 y1 = a0 + a1 x +    2i a i x i ….…. ( 6a ) y2 = b 0 + b 1 x +    2i b i x i …….. ( 6b ) where y1(0) = a0 , y1 '(0) = a1 , … , y1 (i)(0) / i! = ai , i = 2, 3, …… and y2(0) = b0 , y2 '(0) = b1 , … , y2 (i)(0) / i! = bi , i = 2, 3, …… then insert the series forms (6a) and (6b) respectively into (2) and equate coefficients of powers of x . Also ,we need Taylor coefficients of y1(x) and y2(x) about x = 1, respectively y1 = c0 + c1 (x-1) +    2i ci (x-1)i …….. ( 7a ) y2 = d 0 + d 1 (x-1) +    2i di (x-1)i …….. ( 7b ) where y1(1) = c0 , y1 '(1) = c1 , … , y1 (i)(1) / i! = ci , i = 2, 3, …… and y2(1) = d0 , y2 '(1) = d1 , … , y2 (i)(1) / i! = di , i = 2, 3, …… then insert the series forms (7a) and (7b) respectively into (2) and equate coefficients of powers of ( x – 1 ) . The resulting system of equations can be solved using MATLAB version 7.9 to obtain ai , bi , ci and di for all i ≥ 2, we see that cisۥ and disۥ coefficients depend on indicated unknowns c0 and d0 . The algebraic manipulations needed for this process .We are now in a position to construct a P2n+1(x) and 12 ~  n (x) from (6) and (7) of the form ( 3 ) by the following : P2n+1(x) =   n i 0 { ai Qi(x) + (-1)i ci Qi(1-x) } ………… (8a) and 12 ~  n (x) =   n i 0 { bi Qi(x) + (-1) i di Qi(1-x) } ………… (8b) Where Qi(x) defined in (4) , We see that (8) have only two unknowns c0 and d0 . Now, integrate equation (2) to obtain : c0 – a0 =  1 0 f1( x, y1, y2 ) dx ………. (9a) d0 – b0 =  1 0 f2( x, y1, y2 ) dx ………. (9b) use P2n+1 and 12 ~   n as a replacement of y1 and y2 respectively in (9) . Since we have only the two unknowns c0 and d0 to compute for any n we only need to generate two equations from this p rocedure as two equations are already supplied by (9) and initial condition. Then solve this system of algebraic equations using MATLAB version 7.9 to obtain c0 and d0 ,so insert it into (8) thus (8) represent the solution of (2) . Extensive computations have shown that this generally provides a more accurate polynomial representation for a given n . Now from equations (3) and (4) we have : P9 = 0.00000000000000137t9– 0.00000000000028t8 + 0.00000000005t7 – 0.00000000802t6 + 0.00000109t5 –0.00012516007t4 + 0.0118401057t3 – 0.889865000005t2 + 49.3t IBN AL- HAITHAM J. FOR PURE & APPL. SCI. VOL.24 (3) 2011 P ~ 9=– 0.0000000000000009t9+0.00000000000019t8– 0.000000000035t7 + 0.000000005387t6– 0.0000007071t5 +0.0000750482t4 –0.0059009635t3 + 0.273615t2 T9=–0.00000000000000012t 9 +0.000000000000025t 8 –0.0000000000045t 7 + 0.000000000709t 6 – 0.00000009771t 5 +0.0000115542t 4 – 0.0011579461t 3 + 0.096135t 2 Table (1) give the results for n = 4 to different nodes in the domain . Figure (2) illustrate buildup of lead in the bones , tissue , blood using semi-analytic solution . Also , A comparison between different methods are given in table (2) to illustrate the accuracy of suggested method . With system (2) we can study the effect of changing the input rate I1 = 49.3 micrograms / day of lead into the bloodstream, or the effect of a medication that increases the diffusion coefficient k13 = 0.000035 (day)-1 of lead out of the bones. Figure (2) displays the buildup of lead in the body compartments over a period of 800 days. Lead levels in the bloodstream and the issue appear to have nearly reached steady state after the first 200 days, but the lead level in the bones is far from a steady state. The transfer coefficients k13 = 0.000035 (day)-1 of lead for bones back into the blood stream is so small that the skeleton acts like a storage reservoir for lead . Now, suppose that after 400 days the subject is placed in a completely lead-free environment (i.e., the term 49.3 in system (2) becomes 0 for t > 400 ). Figure (3) shows that the lead levels in the blood and tissue plunge dramatically, but the amount of lead in the bones, does not seem to drop very much, at least not in the next 400 days . Another way to remove lead from the bones is to administer and anti lead medication that increases the rate at which lead leaves the skeletal system In particular, suppose that the rate coefficient k13 = 0.000035 increases by an order of magnitude to 0.00035. Figure (4) shows the very small effect this change has if the medication is administered from the 400th day onward. However, if the subject moves to a lead free environment and takes the medication from the 400th day onward, then there is a noticeable drop in the skeletal lead levels (Figure 5). Let us see what happens when a massive dose is given, so large that k13 increases from 0.000035 to 0.035. Suppose from the 400th day on that the heavy dose is given and the subject is in a lead free environment. The lead exists the bones very quickly. Following a slight rise after the 400th day, lead exists the blood and tissue as well (Figure 6) of course that much medication is probably more harmful than the lead . Why might blood lead levels be higher in women after menopause ? After menopause, women often develop osteoporosis , which is a progressive and serious loss of bone mass . As a result of loss of bone mass, lead stored in bone may be released into the blood . What is lead "poisoning" compared to lead "exposure" ? Lead exposure refers to the entry of lead into the body, through ingestion, inhalation, the skin or the placenta . Lead poisoning occurs when there are adverse health effects due to lead in the body . Recommendations After above study ,we introduce some recommendations and advices which help to minimize harms of lead in society and we get some of this information from [5] . IBN AL- HAITHAM J. FOR PURE & APPL. SCI. VOL.24 (3) 2011 What role has the government played in reducing lead poisoning? 1. Using Benzene free of Lead which is called ( Green Benzene ) 2. Not renewing licenses for vehicles mechanically expired , with using transportation means that uses electricity , and the expansion in using collective transportation means such as ( Mitro) with improving them 3. Surveillance on batteries factories the private as well as the governmental , which conduct the melting of Lead in addition to newspapers and magazines pressing , with transferring them to deserted regions 4. Paying the attention for not to construct new schools near to highways that is crowded with vehicles 5. It was proved through scientific study in the college of agriculture / Ain Shams university , that the root part absorbs the Lead more than the green system , where there are many kinds of vegetables which the root part is eaten , that is planted beside the highways the fact that leaded to the Lead to go into the human as well as the animal's body , therefore the work must aim for vegetables and fruits not to be planted near highways 6. Surveillance on all kinds of paints factories also the imported products with their inspection to make sure that they are free of Lead substance What role has the family played in protecting children from lead poisoning ?[6] 1. Test your home for paint .if it content of lead or no 2. Hire a person with special training for correcting lead paint problems to remove lead paint from your home, someone who knows how to do this work safely. 3. Home Clean, window sills, and other surfaces weekly. 4. Don’t bring lead dust into your home from work . 5. Make sure your children wash their hands before meals . 6. A child who gets enough iron and calcium will absorb less lead . 7. Don’t store food or liquid in lead crystal glassware or imported or old pottery. 8. Household water will contain more lead if it has sat for a long time in the pipes 9. .therefore; let it run for 15–30 seconds before drinking it or cooking with it . References 1. Lewis , J. (1985), Lead poisoning : a historical perspective , EPA Journal. 2. Robert, L. B. and Courtney , S. C. (1996 ), Differential Equations A Modeling perspective" , United States of America. 3. Girault ,V. and Scott ,L. R. (2002),Hermite Interpolation of Non smooth Functions Preserving Boundary Conditions, Mathematics of Computation, 71( 239): 1043-1074 . 4. Phillips , G .M. (1973), Explicit forms for certain Hermite approximations, BIT 13 , 177- 180 . 5. New England Journal of Medicine, (2003), April, 17(348): 1517-1526 . 6. Agency for Toxic Substances and Disease Registry. Lead ToxFAQs sheet . August( 2007), www.atsdr.cdc.gov/tfacts13.html , viewed November 14 , (2007). IBN AL- HAITHAM J. FOR PURE & APPL. SCI. VOL.24 (3) 2011 Table (1) : The result of the method for n = 4 P9 P ~ 9 T9 b10 48.42185102755592 48.42185102755592 48.42185102755592 b20 0.2677883829038516 0.2677883829038516 0.2677883829038516 b30 0.09498851115891924 0.09498851115891924 0.09498851115891924 t P9 P ~ 9 T9 0 0 0 0 0.1 4.92111317760058 0.00273025653425228 0.000960193208372153 0.2 9.82449992093787 0.0108975121431540 0.00383615488695559 0.3 14.7102308217005 0.0244666301613681 0.00862097880837015 0.4 19.5783761737962 0.0434026523504553 0.0153077862420084 0.5 24.4290059746451 0.0676707980599557 0.0238897258380576 0.6 29.2621899264675 0.0972364633922961 0.0343599735120251 0.7 34.0779974375664 0.132065220371507 0.0467117323297653 0.8 38.8764976236032 0.172122816115731 0.0609382323930065 0.9 43.6577593088695 0.217375172013507 0.0770327307253737 1 48.4218510275513 0.267788382903812 0.0949885111589079 IBN AL- HAITHAM J. FOR PURE & APPL. SCI. VOL.24 (3) 2011 Table( 2) : A comparison between semi-analytic and other methods P9 by using Osculatory y1:ABM solution y1:RK solution t 0 0 0 0 4.92111317760058 4.9211131775747 7 4.92111317760100 0.1 9.82449992093787 9.8244999208060 5 9.82449992093843 0.2 14.7102308217005 14.710230821422 5 14.7102308217013 0.3 19.5783761737962 19.578376173370 2 19.5783761737973 0.4 24.4290059746451 24.429005974074 3 24.4290059746466 0.5 29.2621899264675 29.262189925928 3 29.2621899264696 0.6 34.0779974375664 34.077997437030 8 34.0779974375690 0.7 38.8764976236032 38.876497623070 6 38.8764976236066 0.8 43.6577593088695 43.657759308340 1 43.6577593088736 0.9 48.4218510275513 48.421851027025 1 48.4218510275559 1 T9 by using Osculatory y3:ABM solution y3:RK solution t 0 0 0 0 0.0009601932083 72153 0.0009601932106 90100 0.00096019320833 8816 0.1 0.0038361548869 5559 0.0038361548987 9100 0.00383615488692 262 0.2 0.0086209788083 7015 0.0086209788333 3617 0.00862097880833 768 0.3 0.0153077862420 084 0.0153077862802 708 0.01530778624197 67 0.4 0.0238897258380 576 0.0238897258893 462 0.02388972583802 68 0.5 0.0343599735120 251 0.0343599735605 295 0.03435997351199 56 0.6 0.0467117323297 653 0.0467117323780 034 0.04671173232973 76 0.7 0.0609382323930 065 0.0609382324410 489 0.06093823239298 09 0.8 0.0770327307253 737 0.0770327307732 064 0.07703273072535 09 0.9 0.0949885111589 079 0.0949885112065 369 0.09498851115892 05 1 P ~ 9 by using Osculatory y2:ABM solution y2:RK solution t 0 0 0 0 0.0027302565342 5228 0.0027302565510 2616 0.00273025653401 098 0.1 0.0108975121431 540 0.0108975122287 910 0.01089751214291 52 0.2 0.0244666301613 681 0.0244666303419 837 0.02446663016113 21 0.3 0.0434026523504 553 0.0434026526272 081 0.04340265235022 29 0.4 0.0676707980599 557 0.0676707984308 485 0.06767079805972 79 0.5 0.0972364633922 961 0.0972364637429 603 0.09723646339207 40 0.6 0.1320652203715 07 0.1320652207201 43 0.13206522037129 2 0.7 0.1721228161157 31 0.1721228164628 73 0.17212281611552 5 0.8 0.2173751720135 07 0.2173751723590 46 0.21737517201331 2 0.9 0.2677883829038 12 0.2677883832477 92 0.26778838290386 1 1 IBN AL- HAITHAM J. FOR PURE & APPL. SCI. VOL.24 (3) 2011 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 0 10 20 30 40 50 The solution at n=4 blood tissues bones Fig.( 1) : Buildup of lead in the bones , tissue , blood 0 100 200 300 400 500 600 700 800 0 500 1000 1500 2000 2500 3000 3500 4000 4500 5000 t(days) y1 ,y 2, y3 (m ic ro gr am s) blood tissues bones Fig.( 2) : Buildup of lead in the bones(circle), blood(line), tissue(plus) 0 100 200 300 400 500 600 700 800 0 500 1000 1500 2000 2500 t(days) y1 ,y 2,y 3( mi cro gr am s) blood tis sues bones Fig.( 3): lead intake stops on the 400 th day IBN AL- HAITHAM J. FOR PURE & APPL. SCI. VOL.24 (3) 2011 0 100 200 300 400 500 600 700 800 0 500 1000 1500 2000 2500 3000 3500 4000 4500 t(days ) y1 ,y2 ,y3 (m icr og ra ms ) blood ti ssues bones Fig.( 4): Anti lead medication taken from the 400 th day on does not help much 0 100 200 300 400 500 600 700 800 0 500 1000 1500 2000 2500 t(days) y1 ,y2 ,y 3( m icr og ram s) blood ti ssues bones Fig.( 5): lead-free environment and medication from the 400 th day on help a lot 0 100 200 300 400 500 600 700 800 0 500 1000 1500 2000 2500 t(days) y1 ,y 2, y3 (m ic ro gr am s) blood ti ssues bones Fig.( 6): lead-free environment and a heavy dose of antilead drug from the 400 th day on .Is the subject still alive? 2011) 3( 24مجلة ابن الهیثم للعلوم الصرفة والتطبیقیة المجلد شبة التحلیلیة لتقلیل أخطار الرصاص طریقةاستخدام ال محمد ناجي محمد توفیقلمى ناجي محمد توفیق و جامعة بغداد ،كلیة التربیة ابن الهیثم ،قسم الریاضیات 2010تشرین الثاني 7 :استلم البحث في 2011 شباط 8 :قبل البحث في خالصةال ـة األولـى لمســائل القـیم االبتدائیــة الهـدف مـن البحــث هـو أیجــاد طریقـة لحـل منظومــة معـادالت تفاضــلیة اعتیادیـة مـن الرتبـ .صاص التقنیة شبه التحلیلیة وذلك بإیجاد الحل بشكل متعددة حدود و ذلك للتقلیل من أخطار الر عمالباست النقطتین التي تتفق فیهـا الصـورة وعـدد متسـاو مـن المشـتقات المعرفـة ياالندراج التماسي ذ عمالالمسالة األصلیة تتعلق باست .مع البیانات المعطاة [1 , 0] مدةعند نقطتي نهایة ال االندراج التماسي، مسائل القیم االبتدائیة : الكلمات المفتاحیة