Sunday 13 November 2022
Wednesday 26 October 2022
Developing Self Esteem
Friday 23 September 2022
Kanakkum Inikkum
Tuesday 16 August 2022
Character Encodings- Pi
Character Encodings -Pi
The various forms of pi are present in Unicode as:
- U+03A0 Π GREEK CAPITAL LETTER PI (Π)
- U+03C0 π GREEK SMALL LETTER PI (π)
- U+03D6 ϖ GREEK PI SYMBOL (ϖ, ϖ)
- U+220F ∏ N-ARY PRODUCT (∏, ∏)
- U+1D28 ᴨ GREEK LETTER SMALL CAPITAL PI
- U+1D70B 𝜋 MATHEMATICAL ITALIC SMALL PI
- U+1D6D1 𝛑 MATHEMATICAL BOLD SMALL PI.
Character Encodings Table
- Greek / Coptic Pi
Preview | Π | π | ϖ | Ⲡ | ⲡ | |||||
---|---|---|---|---|---|---|---|---|---|---|
Unicode name | GREEK CAPITAL LETTER PI | GREEK SMALL LETTER PI | GREEK PI SYMBOL | COPTIC CAPITAL LETTER PI | COPTIC SMALL LETTER PI | |||||
Encodings | decimal | hex | dec | hex | dec | hex | dec | hex | dec | hex |
Unicode | 928 | U+03A0 | 960 | U+03C0 | 982 | U+03D6 | 11424 | U+2CA0 | 11425 | U+2CA1 |
UTF-8 | 206 160 | CE A0 | 207 128 | CF 80 | 207 150 | CF 96 | 226 178 160 | E2 B2 A0 | 226 178 161 | E2 B2 A1 |
Numeric character reference | Π | Π | π | π | ϖ | ϖ | Ⲡ | Ⲡ | ⲡ | ⲡ |
Named character reference | Π | π | ϖ, ϖ | |||||||
DOS Greek | 143 | 8F | 167 | A7 | ||||||
DOS Greek-2 | 198 | C6 | 234 | EA | ||||||
Windows 1253 | 208 | D0 | 240 | F0 | ||||||
TeX | \Pi | \pi | \varpi |
MARUDHAR KESARI JAIN COLLEGE FOR WOMEN
PG and Research Department of Mathematics
Activities for the Month of July
2021-2022
Academic
· ISO external audit held on 20.07.2022.
· Semester exam started from 04.07.2022 to 01.08.2022.
· Subject allotment completed for the staffs.
· Prepared workload and timetable for the staffs.
Department activity
· Staff Meeting was conducted on 19.07.2022 & 21.07.2022
· Website upload completed.
· Induction program for the first year students on 25.07.2022
· Notes prepared for the next academic year.
· Lesson plan prepared by the staffs.
Co-Curricular (Students)
· I M. Sc. Students had training classes from 20.07.2022 to 30.07.2022.
· Bridge course stared for the first year students from 26.07.2022.
STAFF ACHIEVEMENT
· One staff participated in One week FDP on ELT in Digital World: Developing Global Citizenship from 27.06.2022 to 02.07.2022.
· One staff got Asiriyar- Perunthalaivarin Sirpy Sathanai Viruthu on 15.07.2022.
· Four staff member got an Kalvi Mamani Award.
· Two staffs got Dr. APJ Abdul Kalam Education Excellence Awards 2022.
Friday 24 June 2022
SOME APPLICATIONS OF VOLTERRA TYPE INTEGRO DIFFERENTIAL EQUATIONS
SOME APPLICATIONS OF VOLTERRA TYPE INTEGRO DIFFERENTIAL EQUATIONS
Dr,K.R.Salini[1] , M. Deivanai[2], S.Madhavan[3] and Dr.S.Bamini[4]
[1] Guest Lecturer, PG & Department of Mathematics, Government Arts & Science College, Kariyampatti, Tirupattur, Tamilnadu, India.
Email: shalinimanoj1985@gmail.com
[2],[3] II M.Sc, Mathematics, PG & Department of Mathematics Government Arts & Science College, Kariyampatti, Tirupattur, Tamilnadu, India.
[4] Research Coordinator, Assistant Professor, PG & Research Department of Mathematics,
Marudhar Kesari Jain College for Women, Vaniyambadi, Tamilnadu, India.
Email: saransham07@gmail.com
Abstract
We determine the numerical solution of the specific nonlinear Voltera-Fredholm Integro-Differential Equation is proposed. The method is based on new vector forms for representation of triangular functions and its operational matrix.
Introduction
The nonlinear Volterra type integro-differential equation of the first order
y'(x)=f(x,y)+K(x. s. y(s))ds, y(x)= yox € [x.X]
Here the functions f(x,y). K(x.s, y) are determined in the domains
G=(x0 < x < x , |y| < ᶯ), G1=(x0 ≤ s ≤ x≤ X,|y| ≤ ᶯ)
and respectively and equation (1.1) has a unique solution.
In case K(x.s, y) ≡ 0 equation (1.1) turns into the Cauchy problem for ordinar
differential equations of the first order.
Non-Linear Volterra Type Integro Differentia Equation Applying Multi Step Method
The aim is to apply the multistep method with constant coefficients to the
solution of equation
Assume that the kernel K(x.s, y) is degenerate that is 𝐾(𝑥,𝑠,𝑦)=Σ(𝑎𝑖(x)𝑏𝑖(s,y) )𝑚𝑖=1
Then equation can be written as
y'(x) = f(x,y) + Σ𝑎𝑖(x)y𝑚𝑖=0∫𝑏𝑖(s,y(s))ds,y𝑥𝑥0 y(x) = yo
Using the notation
Vi(x)=∫𝑏𝑖(s,y(s))𝑥𝑥0ds, (i=1.2,...m)
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We can rewrite equation in the form of a system of differential equations
y' = f(x,y)+ Σ(𝑎𝑖(x)𝑣𝑖(x))𝑚𝑖=1
v'i(x)=bi (x,y(x)), ( i=1,2,....,m )
The initial conditions for the equation (1.3) of differential equations have the
following form
y(x0) = y0, vi(x0) = 0 ( i =1.2 ,……., m)
Consider the simple case and assume m = 1 then we have
y' ≡ f(x,y)+a(x)v(x),y(x 0 ) ≡ y0
v' (x) = b(x,y(x)),v(x0 ) = 0
Let's divide the section [x0 , X] by means of the constant step h > 0 into N equal parts
We apply k - th step method with constant coefficients to the numerical solution of
equation Then we can write
Σ𝛼𝑘𝑖=0i yn+i = hΣ𝛽𝑘𝑖=0ifn+i + hΣ𝛽𝑘𝑖=0ian+ivn+i (n=0,1,2,⋯ ⋯,𝑁−𝑘)
Σ𝛼𝑘𝑖=0ivn+i = hΣ𝛽𝑘𝑖=0ibn+i (n=0,1,2,⋯⋯,𝑁−𝑘)
Where, am= a(xm)
vm = v(Xm)
bm = b(xm, ym)
fm = f(xm,ym)
xm = x0 + mh ( m = 0,1,2,..........)
If we assume that ak ≠ 0, then from equation (1.6) and equation (1.7) we can find yn+k and Vn+k respectively. However, relation (1.7) is the implicit nonlinear finite difference equation. Usually in these cases the different varients of prediction correction method is used
Non-Linear Volterra Type Integro Differential Equation Applying Prediction-Correction Method
Applying the prediction-correction method to equation (1.6) ,we have
ẏn+k = Σ(𝛼𝑘𝑖=0iyn+i + hβifn+i + hβian+i)
yn+k = Σ(𝑘−1𝑖=0α'iyn+i + hβ'ifn+i + hβ'ian+ivn+i) + hβ' k f (f ( xn+k+yn+k) +an+kvn+k)
Where the coefficients of prediction are method are denoted αi, βj (l = 0,1,2, k-i) and by a="/a
Bi - B/a (i = 0,1,2....k-1).
β' k=β'k /αk, we denote the coefficients of the correction method.
Then the numerical method for solution of equation is obtained from methods
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vn+k = Σα𝑘−1𝑖=0'ivn+i + h Σβ𝑘−1𝑖=0'ibn+i (n=0,1,2, ⋯⋯)
t is obvious that the found value ym vm (m ≥ k) by methods aren't the exact values of equation at the points xm (m ≥0).
Therefore, if in methods instead of the approximate values ym,vm we put their exact values y(xm), v(xm), these methods will take the following form
ẏ(xn+k) = Σ[𝛼𝑘−1𝑖=0iy(xn+i) + hβif(xn+i,y(xn+i) + hβia(xn+i)v(xn+i)] +Řn
y(xn+k) = Σ[𝛼′𝑘−1𝑖=0iy(xn+i) + hβ'if(xn+i,y(xn+i) + hβ'ia(xn+i)v(xn+i)] + hβ'kf(xn+k,y(xn+k) +
hβ'ka(xn+k)v(xn+i) + Rn
V(xn+k) = Σα𝑘−1𝑖=0'iv(xn+i) + h Σβ𝑘𝑖=0'ib(xn+I,y(xn+i)) + Rn
Here the error of corresponding methods is denoted by Ř,Rn
Note that ẏ(xm) = y(xm) ( m = 0,1,2,⋯⋯)
Definition
Equation is stable if the roots of the polynomial
𝜌(𝜆) = Σ𝛼𝑘𝑖=0i𝜆i
lie inside a unique circle on whose boundary there are no multi roots.
Definition:
The integer-valued p is called of the degree of method (1.6),if the following
Σ[𝛼𝑘𝑖=0iy(x+i) - hβiy'(x+ih)) = o(hp),h→0
holds
Determination of Convergence of The Proposed Method
Let us prove the following theorem for determination of convergence of the
proposed method.
Theorem
Let method (1.6) be stable, satisfy the conditions A, B, C. The first derivatives
of the function b (x, y) and f(x, y) by y be bounded. Then it holds
max0≤𝑚≤𝑁(ℰm,Êm) ≤ A exp (BX)(𝜇max0≤𝑥≤1| ℰ𝑖|+𝑚𝑑)
Here A, B and d are some bounded values and Ê𝑚=𝑣(𝑥𝑚)−𝑣𝑚,ℰ𝑚=𝑦(𝑥𝑚)−𝑦𝑚(𝑚=0,1,2,⋯⋯,)
Proof:
ℰ𝑛+𝑘=Σ(𝛼𝑖ℰ𝑛+𝑖+ℎ𝛽𝑖ℒ𝑛+𝑘ℇ𝑛+𝑖+ℎ𝛽𝑖𝑎𝑛+𝑖𝜀𝑛+𝑖 )+Ř𝑛𝑘−1𝑖=0
ℰ𝑛+𝑘=Σ(𝛼′𝑖ℰ𝑛+𝑖+ℎ𝛽𝑖ℒ𝑛+𝑘ℇ𝑛+𝑖+ℎ𝛽′𝑖𝑎𝑛+𝑖𝜀𝑛+𝑖 )+ℎ𝛽′𝑘ℒ𝑛+𝑘ℇ𝑛+𝑘+𝑘−1𝑖=0
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ℎ𝛽′𝑘𝑎𝑛+𝑘ℇ𝑛+𝑘+Ř𝑛
Here ℒ𝑚=𝑓′𝑦(𝑥𝑚,𝜉𝑚),ℒ𝑚=𝑓′𝑦(𝑥𝑚,𝜉𝑚),𝑎𝑚=𝑎(𝑥𝑚),Ê𝑚=𝑦(𝑥𝑚)−ẏ𝑚
,( m = 0,1,2,.....)
Where 𝜉𝑚 is found between y(xm) and ym and 𝜉𝑚 between y(xm) and ẏm.
We have 𝜀𝑛+𝑘=Σ𝑎𝑖𝑒𝑛+𝑖+ℎΣ𝑏𝑖𝑒𝑛+𝑖+ℎΣ𝑐𝑖𝑘−1𝑖=0𝜀𝑛+𝑖+ℎ𝛽′𝑘𝑎𝑛+𝑘𝜀𝑛+𝑘+ℎ𝛽′𝑘ℒ𝑛+𝑘Ř𝑛+ℛ𝑛𝑘−1𝑖=0𝑘−1𝑖=0
Where, 𝑎𝑖=𝛼′𝑖+ℎ𝛽′𝑖ℒ𝑛+𝑖𝛼𝑖, 𝑏𝑖=𝛽′𝑖ℒ𝑛+𝑖+ℎ𝛽′𝑘ℒ𝑛+𝑘𝛽𝑖ℒ𝑛+1, 𝑐𝑖=𝛽′𝑖𝑎𝑛+𝑖+ℎ𝛽′𝑖ℒ𝑛+𝑘𝑎𝑛+𝑖
we will obtain, 𝜀𝑛+𝑘=Σ(𝑎′𝑖𝑒𝑛+𝑖+ℎ𝛽′𝑖ℒ𝑛+𝑖𝑘−1𝑖=0𝜀𝑛+𝑖)+ℎ𝛽′𝑘ℒ𝑛+𝑘𝜀𝑛+𝑘+ℛ𝑛
Where ℒ𝑚=𝑏′𝑦(𝑥𝑚,𝑦𝑚) (m= 0,1,2,⋯⋯⋯,)𝜂𝑚 are between y(xm) and ym for error estimation of the method applied to the solution
We will obtain, 𝜀𝑛+𝑘=Σ𝑎𝑖𝜀𝑛+𝑖𝑘−1𝑖=0+ℎΣ𝑏𝑖𝑘−1𝑖=0𝜀𝑛+𝑖+ℛ𝑛
𝜀𝑛+𝑘=Σ(𝑘−1𝑖=0𝑎′𝑖𝜀𝑛+𝑖+ℎ𝛽′𝑖ℒ𝑛+𝑘𝜀𝑛+𝑖+𝛽′𝑘ℒ𝑛+𝑘Σ(𝑘−1𝑖=0𝑎𝑖𝜀𝑛+𝑖+ ℎ2𝑏𝑖𝜀𝑛+𝑖)+ℎ𝛽′𝑘ℒ𝑛+𝑘ℛ𝑛+ℛ𝑛
Where, 𝑎𝑖=𝑎𝑖+ℎ𝑏𝑖1−ℎ2𝑑𝑛+𝑘 𝑏𝑖=𝑐𝑖+𝑎𝑛+𝑘𝛽′𝑘𝛼′𝑖1−ℎ2𝑑𝑛+𝑘 𝑏𝑖=𝑏𝑖+ℎ𝑎𝑛+𝑘𝛽′𝑘𝛽′𝑖ℒ𝑛+𝑖 𝑑𝑛+𝑘=𝑎𝑛+𝑘(𝛽′𝑘)2ℒ𝑛+𝑘
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ℛ𝑛=(ℎ𝛽′𝑘ℒ𝑛+𝑘ℛ𝑛+ℛ𝑛+ℎ𝑎𝑛+𝑖ℛ𝑛)/1−ℎ2𝑑𝑛+𝑘
Consider the following relation
𝑎𝑖+ℎ𝑏𝑖/1−ℎ2𝑑𝑛+𝑘=𝑎𝑖+ℎ𝑙𝑖
Where 𝑙𝑖=(𝑏𝑖+ℎ𝑎𝑖𝑑𝑛+𝑘)/1−ℎ2𝑑𝑛+𝑘
For smallness of h we can assume |1−ℎ2𝑑𝑛+𝑘|≥12
Then we obtain that |𝑙𝑖|≤𝑙(i= 0,1,2, ⋯⋯-k-1)
It is easy to show that 𝑏𝑖=(𝑐𝑖+𝑎𝑛+𝑘𝛽′𝑖𝛼𝑖)/(1−ℎ2𝑑𝑛+𝑘) (𝑖= 0,1,2,⋯ ⋯,𝑘−1)
are also bounded, that is,|𝑏𝑖| ≤ b.
we can write 𝜀𝑛+𝑘=Σ𝑎′𝑖𝜀𝑛+𝑖𝑘−1𝑖=0+ℎΣ𝑣𝑖𝑘−1𝑖=0𝜀𝑛+𝑖+ℎΣ𝑏𝑖𝑘−1𝑖=0𝜀𝑛+𝑖+ℛ𝑛
𝜀𝑛+𝑘=Σ𝑎′𝑖𝜀𝑛+𝑖𝑘−1𝑖=0+ℎΣ𝑣𝑖𝑘−1𝑖=0𝜀𝑛+𝑖+ℎ2Σ𝑏𝑖𝛽′𝑘ℒ𝑛+𝑘𝑘−1𝑖=0𝜀𝑛+𝑖+Ř𝑛
Where, Ř𝑛=ℛ𝑛+ℎ𝛽′𝑘ℒ𝑛+𝑘+ℛ𝑛 𝑣𝑖=𝛽′𝑘ℒ𝑛+𝑘𝛼𝑘+𝑙𝑖 𝑣𝑖=𝛽′𝑖ℒ𝑛+𝑖+𝛼𝑖𝛽′𝑘ℒ𝑛+𝑘 (𝑖=0,1,2,…,𝑘−1)
Here we can also write that lvi| ≤v and |ῡi|≤ῡ
Consider the following vectors,
Yn+k-1 (𝜀 n+k-1, 𝜀 n+k-2, … … … , 𝜀 n)
Ẏn+k-1 (𝜀 n+k-1, 𝜀 n+k-2, … … … , 𝜀 n)
Then adding to equation the following identities respectively,
𝑌𝑛+𝑘=𝐴𝑌𝑛+𝑘−1+ℎ𝑉𝑛+𝑘𝑌𝑛+𝑘−1+ℎ𝐵𝑛+𝑘Ẏ𝑛+𝑘−1+Ŵ𝑛,𝑘
Ẏ𝑛+𝑘=𝐴Ẏ𝑛+𝑘−1+ℎṼ𝑛+𝑘𝑌𝑛+𝑘−1+ℎ2𝐵𝑛+𝑘Ẏ𝑛+𝑘−1+𝑊𝑛,𝑘
Where the matrices A, Vn+kῡn+ki En+k. Ên+k and the vectors Wnk Ŵnk are
Determined in the following form 𝐴=(𝛼′𝑘−1𝛼′𝑘−2.10.00..𝛼′1𝛼′0 ....10) 𝑣𝑛+𝑘=(𝑣𝑘−1𝑣𝑘−2.00.00..𝑣0...0)
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𝑩𝑛+𝑘=(𝐵𝑘−1𝐵𝑘−2.00.00..𝑏0.0.0) 𝑩𝑛+𝑘=(𝑏𝑘−1𝑏𝑘−2.00.00..𝑏0.000)(𝛽′𝑘𝐿𝑛+𝑘)−1 𝑣𝑛+𝑘=(𝑣𝑘−1𝑣𝑘−2.00.00. .𝑣0 .0 .0) 𝑊𝑛,𝑘=(𝑅𝑛00) 𝑊𝑛,𝑘=(Ř𝑛00)
Using the vector Zn+k =(𝑦𝑛+𝑘,Ẏ𝑛+𝑘)
We can rewrite equations in the following form 𝑍𝑛+𝑘=𝐴𝑍𝑛+𝑘−1ℎ𝑉𝑛+𝑘𝑍𝑛+𝑘−1+Ŵ𝑛,𝑘
Where, 𝐴=(𝐴00𝐴) 𝑉𝑛+𝑘=(𝑉𝑛+𝑘𝐵𝑛+𝑘𝑉𝑛+𝑘ℎ𝐵𝑛+𝑘) Ŵ𝑛,𝑘=(Ŵ𝑛,𝑘Ŵ𝑛,𝑘)
Note that Zn+k = Cyn+k
Then after the multiplication of equations (1.24) on the left hand side by C-1 we will obtain 𝑦𝑛+𝑘=𝐷𝑦𝑛+𝑘−1ℎ𝐸𝑛+𝑘𝑦𝑛+𝑘−1+Ŵ𝑛,𝑘
determined in the following form 𝐷=𝐶−1Ậ𝐶 𝐸𝑛+𝑘=𝐶−1Ṽ𝑛+𝑘𝐶 Ŵ𝑛,𝑘=𝐶−1Ŵ𝑛,𝑘
Passing to the norm in equation
we have, ‖𝑦𝑛+𝑘‖≤‖𝐷‖‖𝑦𝑛+𝑘‖+ℎ‖𝐸𝑛+𝑘‖‖𝑦𝑛+𝑘−1‖+‖Ŵ𝑛,𝑘‖
For the stability of equation (1.6) we can assert that the characteristic numbers of the matrix A satisfy the following condition: all characteristic numbers of the matrix by the modulus are less than unit, and the roots equal by the modulus of a unit are simple
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Since the matrix A is a Frobenius matrix smallness of h and 15,| bi |≤b,I = 0,1,2,...k-1 Then subject to the sufficient we will obtain that all characteristic numbers of the matrix A are situated in a unit circle on whose boundary that are no multi roots, to roots by the module equal to unit correspond the Jordan cells of dimension one.
Consequently,
||D||≤1
We can show that the matrices Vn+k. En+kiῡn+k have the bounded norms.
Then we can write
||En+k|| ≤ B
Subject to the above metrices Vn+k,En+k,ῡn+k have the bounded norms
There we can write
||En+k||≤ B
Subject to the above mentioned in (1.26) we will obtain
||ym|| ≤ (1+hB)||ym-1||+d
Where.
d=||Ŵn,k||
we will obtain that ‖𝑦𝑚‖≤(1+ℎ𝐵)𝑚−𝑘‖𝑦𝑘‖+Σ(1+ℎ𝐵)𝑖𝑚−𝑘−1𝑖=0𝑑
It is known that (1+hB) m exp (mhB)
For h ≤ X-xo≤ X, We can rewrite in the following form ‖𝑦𝑚‖≤exp(𝐵𝑋)(‖𝑦𝑘‖+𝑑ℎ−1) ‖𝑦𝑧‖≤𝐴exp(𝐵𝑋)(‖𝜇‖‖𝑧𝑘‖+𝑑ℎ−1)
Where, 𝐴=‖𝐶‖ 𝜇=‖𝐶−1‖
From the last it follows that max𝑘≤𝑚≤𝑁(𝜀𝑚,𝜀𝑚)≤𝑎exp(𝐵𝑋)(𝜇max0≤𝑖≤𝑘−1|𝜀𝑚|+𝑑𝑛)
If we assume that equation (1.6) is of degree p the initial values are calculated with
accuracy p.
That is,
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max0≤𝑖≤𝑘−1𝜀𝑖=𝑂(ℎ𝑝)
Then from the assertions of the theorem it follows that 𝜀𝑚=𝑂(ℎ𝑝),ℎ→0
As we see from the proof of the theorem by using equation (1.7) it was assumed that
𝑦𝑛+𝑘 is known. However, for finding 𝑦𝑛+𝑘 is to be known 𝑣𝑛+𝑘
Note that in real calculations for finding the values𝑣𝑛+𝑘 - it is used approximate values of 𝑦𝑛+𝑘 calculated by the prediction method.
In this case, the proposed method for the solution of equation (1.5) is written in the
following form Ẏ𝑛+𝑘=Σ(𝑎𝑖𝑦𝑛+𝑖𝑘𝑖=0+ℎ𝛽𝑖𝑓𝑛+𝑖+ℎ𝛽𝑖𝑎𝑛+𝑖,𝑣𝑛+𝑖) 𝑣𝑛+𝑘=Σ𝑎′𝑖𝑣𝑛+𝑖𝑘−1𝑖=0+ℎΣ𝛽′𝑖𝑏𝑛+𝑖+𝑘−1𝑖=0ℎ𝛽′𝑖𝑏(𝑥𝑛+𝑘,Ẏ𝑛+𝑘) 𝑦𝑛+𝑘=Σ𝑎′𝑖𝑦𝑛+𝑖𝑘−1𝑖=0+ℎΣ𝛽′𝑖(𝑓𝑛+𝑖+𝑘−1𝑖=0𝑎𝑛+𝑖𝑣𝑛+𝑖)+ℎ𝛽′𝑘(𝑥𝑛+𝑘,Ẏ𝑛+𝑘)+𝑎𝑛+𝑘𝑣𝑛+𝑘
Reference
1. G.Yu.Mekhtieva, V.R.Ibrahimov “On one numerical method for solution of integro-differential equations”, Vestnik BSU, ser.phys.-math. Scien., 2003, No3, pp.21-27.(Russian).
2. H.Brunner, “Collection method for Volterra integral and Relation Functional Equations”, Cambridge University Press, Cambridge,2004.
3. P.Darania, E.Abadian, “A method for the numerical solution of the integro-differential equations”, Appl.Math. and Comput. 188(2007) 657-668.
4. B. Ahmad, S. Siva Sundaram, Existence Results For Nonlinear Impulsive Hybrid Boundary Value Problems involving Fractional Differential Equations, Non linear Analysis, hybrid System, 3(2009).
5. Z.B. Bai, H.S. Lii, Positive Solutions Of Boundary Valuve problems Of Nonlinear Fractional Differential Equations, J. Math Anal.Appl.311(2005)\
6. v. Lakshmikantham, D.D. Bainov And P.S. Simeonov, Theory Of Impulsive Differential Equations, modern Applied Mathematics (1989).
7. K.S. Miller, B.Ross, An Introduction To The Fractional Caculus And Fractional Differential Equations, Wiley, New York, 1993.
Wednesday 11 May 2022
NUMERICAL SOLUTION OF NONLINEAR VOLTERRA-FREDHOLM INTEGRO-DIFFERENTIAL EQUATIONS VIA DIRECT METHOD USING TRIANGULAR FUNCTIONS
Dr.K.R.Salini[1] and Dr.S.Bamini[2]
[1] Guest Lecturer, PG & Department of Mathematics, Government Arts & Science College, Kariyampatti, Tirupattur, Tamilnadu, India.
Email: shalinimanoj1985@gmail.com
[2] Research Coordinator, Assistant Professor, PG & Research Department of Mathematics, Marudhar Kesari Jain College For Women, Vaniyambadi, Tamilnadu, India.
Email: saransham07@gmail.com
Abstract
We determine the numerical solution of the specific nonlinear Voltera-Fredholm Integro-Differential Equation is proposed. The method is based on new vector forms for representation of triangular functions and its operational matrix.
Introduction
A set of basic functions and an appropriate projection method or a direct method. These methods often transform an integro-differential equation to a linear or nonlinear system of algebraic equations which can be solved by direct or iterative methods.
Consider a Volterra-Fredhlom Integro-differential equations of the form 𝑥′(𝑠)+𝑞(𝑠)𝑥(𝑠)+𝜆1∫𝑘1(𝑠,𝑡)𝐹(𝑥(𝑡))𝑑𝑡10+𝜆2∫𝑘2(𝑠,𝑡)𝐺(𝑥(𝑡))𝑑𝑡10=𝑦(𝑠)
𝑥(0)=𝑥0
Where the functions 𝐹(𝑥(𝑡)) and 𝐺(𝑥(𝑡)) are polynomials of 𝑥(𝑡) with constant coefficients.
For convenience, we put 𝐹(𝑥(𝑡))=[𝑥(𝑡)]𝑛1 and 𝐺(𝑥(𝑡))=[𝑥(𝑡)]𝑛2, where 𝑛1 & 𝑛2 are positive integers.
For 𝑛1,𝑛2=1, equation (1.1) is a linear integro- differential equation.
We present new vector forms of triangular functions (TFs), operational matrix of integration, expansion of functions of one and two variables with respect to TFs and other TFs properties.
By using new representations, a nonlinear integro-differential equation can be easily reduced to a nonlinear system of algebraic equations.
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Triangular Functions
Definition
Two 𝑚-sets of triangular functions are defined over the intercal [0,𝑡) as 𝑇1𝑖(𝑡)={1−𝑡−𝑖ℎℎ𝑖ℎ≤𝑡<(𝑖+1)ℎ0𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒 𝑇2𝑖(𝑡)={𝑡−𝑖ℎℎ𝑖ℎ≤𝑡<(𝑖+1)ℎ0𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
Where 𝑖=0,1,2…𝑚−1 with a positive integer value for m.
Also consider ℎ=𝑇/𝑚, and 𝑇1𝑖 as the 𝑖𝑡ℎ left-handed triangular function and 𝑇2𝑖 as
the 𝑖𝑡ℎ right-handed triangular function.
It is assumed that 𝑇=1, so TFs are defined over [0,1) and ℎ=1/𝑚.
From the definition of TFs, it is clear that triangular functions are disjoint, orthogonal
and complete.
We can write ∫𝑇1𝑖10(𝑡)𝑇1𝑗(𝑡)𝑑𝑡=∫𝑇2𝑖10(𝑡)𝑇2𝑗(𝑡)𝑑𝑡={ℎ3𝑖=𝑗0𝑖≠𝑗
Also, ∅𝑖(𝑡)=𝑇1𝑖(𝑡)+𝑇2𝑖,𝑖=0,1,…𝑚−1
Where ∅𝑖(𝑡) is the 𝑖th block-pulse function defined as ∅𝑖(𝑡)={1𝑖ℎ≤𝑡<(𝑖+1)ℎ0𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒
Where 𝑖=0,1,…𝑚−1.
Vector Forms
Consider the first 𝑚 terms of the left-handed triangular functions and the first 𝑚 terms of the right- handed triangular functions and write them concisely as 𝑚-vectors. 𝑇1(𝑡)=[𝑇10(𝑡),𝑇11(𝑡),…𝑇1𝑚−1(𝑡)]𝑇
𝑇2(𝑡)=[𝑇20(𝑡),𝑇21(𝑡),…𝑇2𝑚−1(𝑡)]𝑇
Where 𝑇1(𝑡) and 𝑇2(𝑡) are called left-handed triangular functions (LHTF) and right-handed triangular functions (RHTF) vector, respectively.
The product of two TFs vectors are
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𝑇1(𝑡)𝑇1𝑇(𝑡)≅( 𝑇10(𝑡)0⋯⋯⋯00𝑇11(𝑡)⋯⋯⋯0⋮⋮0⋮⋮0⋮⋮⋯⋮⋮⋯⋮⋮⋯⋮⋮𝑇1𝑚−1(𝑡))
𝑇2(𝑡)𝑇2𝑇(𝑡)≅( 𝑇20(𝑡)0⋯⋯⋯00𝑇21(𝑡)⋯⋯⋯0⋮⋮0⋮⋮0⋮⋮⋯⋮⋮⋯⋮⋮⋯⋮⋮𝑇2𝑚−1(𝑡)) 𝑇1(𝑡)𝑇2𝑇(𝑡)=0
𝑇2(𝑡)𝑇1𝑇(𝑡)=0
Where 0 is the 𝑧𝑒𝑟𝑜 𝑚×𝑚 matrix.
Also ∫𝑇110(𝑡)𝑇1𝑇(𝑡)𝑑𝑡=∫𝑇2(𝑡)𝑇1𝑇(𝑡)𝑑𝑡10≅ℎ3𝐼
∫𝑇110(𝑡)𝑇2𝑇(𝑡)𝑑𝑡=∫𝑇2(𝑡)𝑇1𝑇(𝑡)𝑑𝑡10≅ℎ6𝐼
In which 𝐼 is an 𝑚×𝑚 identity matrix.
TFs Expansion
The expansion of a function 𝑓(𝑡) over [0,1) with respect to TFs written as 𝑓(𝑡)≅Σ𝑐𝑖𝑚−1𝑖=0𝑇1𝑖(𝑡)+Σ𝑑𝑖𝑚−1𝑖=0𝑇2𝑖(𝑡) 𝑓(𝑡)=𝑐𝑇𝑇1(𝑡)+𝑑𝑇𝑇2(𝑡)
Where we may put 𝑐𝑖=𝑓(𝑖ℎ) and 𝑑𝑖=𝑓((𝑖+𝑖)ℎ) for 𝑖=0,1,2,..𝑚−1.
Operational Matrix of Integration
Expressing ∫𝑇1(𝜏)𝑑𝜏≅𝑃1𝑇1(𝑠)+𝑃2𝑇2(𝑠)𝑠0
∫𝑇2(𝜏)𝑑𝜏≅𝑃1𝑇1(𝑠)+𝑃2𝑇2(𝑠)𝑠0
Where 𝑃1𝑚×𝑚 and 𝑃2𝑚×𝑚 are called operational matrices of integration in TFs domain and represented as follows 𝑃1=ℎ2( 011⋯1001⋯10⋮00⋮00⋮0⋯⋱⋯1⋮0)
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𝑃2=ℎ2( 111⋯1011⋯10⋮00⋮00⋮0⋯⋱⋯1⋮1)
So the integral of any function 𝑓(𝑡) can be approximated as ∫𝑓(𝜏)𝑑𝜏≅𝑐𝑇𝑇1(𝑠)+𝑑𝑇𝑇2(𝑠)𝑑𝜏𝑠0 ∫𝑓(𝜏)𝑑𝜏≅(𝑐+𝑑)𝑇𝑃1𝑇1(𝑠)+(𝑐+𝑑)𝑇𝑃2𝑇2(𝑠)𝑠0
New Representation of TFs Vector Forms and Other Properties
We define a new representation of TFs vector forms. Then, some characteristics of TFs are presented using the new definition.
Definition and Expansion
Let 𝑇(𝑡) be a 2𝑚-vector defined as 𝑇(𝑡)=(𝑇1(𝑡)𝑇2(𝑡)),0<𝑡<1
Where 𝑇1(𝑡) and 𝑇2(𝑡) have been defined in (1.2).
Now, the expansion of 𝑓(𝑡) with respect to TFs can be written as 𝑓(𝑡)≅𝐹1𝑇𝑇1(𝑡)+𝐹2𝑇𝑇2(𝑡)
≅𝐹𝑇𝑇(𝑡)
Where 𝐹1 &𝐹2 are TFs co-efficients with
𝐹1𝑖=𝑓(𝑖ℎ)& 𝐹2𝑖=𝑓((𝑖+1)ℎ),
For 𝑖=0,1,…𝑚−1.
Also, 2𝑚-vector 𝐹 is defined as 𝐹=(𝐹1𝐹2)
Now, assume that 𝑘(𝑠,𝑡) is a function of two variables. It can be expanded with respect to TFs as follows
𝑘(𝑠,𝑡)≅𝑇𝑇(𝑠)𝐾𝑇(𝑡)
Where 𝑇(𝑠)& 𝑇(𝑡) are 2𝑚1&2𝑚2 dimensional triangular functions and 𝐾 is a
2𝑚1×2𝑚2 TFs coefficient matrix.
For convenience, we put 𝑚1=𝑚2=𝑚. So matrix 𝐾 can be written as
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𝐾=((𝐾11)𝑚×𝑚(𝐾12)𝑚×𝑚(𝐾21)𝑚×𝑚(𝐾22)𝑚×𝑚)
Where 𝐾11,𝐾12,𝐾21,𝐾22 can be compted by sampling the function 𝑘(𝑠,𝑡) at points 𝑠𝑖 & 𝑡𝑗 such that 𝑠𝑖=𝑖ℎ & 𝑡𝑗=𝑗ℎ, for 𝑖,𝑗=0,1,…𝑚. (𝐾11)𝑖𝑗=𝑘(𝑠𝑖,𝑡𝑗),𝑖=0,1,…𝑚−1,𝑗=0,1,…𝑚−1 (𝐾12)𝑖𝑗=𝑘(𝑠𝑖,𝑡𝑗),𝑖=0,1,…𝑚−1,𝑗=0,1,…𝑚 (𝐾21)𝑖𝑗=𝑘(𝑠𝑖,𝑡𝑗),𝑖=0,1,…𝑚,𝑗=0,1,…𝑚−1 (𝐾22)𝑖𝑗=𝑘(𝑠𝑖,𝑡𝑗),𝑖=0,1,…𝑚,𝑗=0,1,…𝑚
Product Properties
Let 𝑋 be an 2𝑚-vertor which can be written as 𝑋𝑇=(𝑋1𝑇 𝑋2𝑇) such that 𝑋1 & 𝑋2 are 𝑚-vectors.
It can be concluded from (1.3) & (1.4) that 𝑇(𝑡)𝑇𝑇(𝑡)𝑋=(𝑇1(𝑡)𝑇2(𝑡))(𝑇𝑇(𝑡)𝑇2𝑇(𝑡))(𝑋1𝑋2)
≅(𝑑𝑖𝑔(𝑇1(𝑡))𝑜𝑚×𝑚𝑜𝑚×𝑚𝑑𝑖𝑔(𝑇2(𝑡)))(𝑋1𝑋2)
=𝑑𝑖𝑔 (𝑇(𝑡))𝑋
=𝑑𝑖𝑔(𝑋)𝑇(𝑡)
Therefore,
𝑇(𝑡)𝑇𝑇(𝑡)𝑋≅𝑋̅ 𝑇(𝑡)
Where 𝑋̅=𝑑𝑖𝑔(𝑋) is an 2𝑚×2𝑚 diagonal matrix.
Let 𝐵 be a 2𝑚×2𝑚 matrix as 𝐵=((𝐵11)𝑚×𝑚(𝐵12)𝑚×𝑚(𝐵21)𝑚×𝑚(𝐵22)𝑚×𝑚)
So, it can be similarly concluded from equations (1.3) and (1.4) that 𝑇(𝑡)𝐵𝑇(𝑡)=𝑇1𝑇(𝑡)𝑇2𝑇(𝑡)(𝐵11𝐵12𝐵21𝐵22)(𝑇1(𝑡)𝑇2(𝑡)) ≅𝑇1𝑇(𝑡)𝐵11𝑇1(𝑡)+𝑇2𝑇(𝑡)𝐵22𝑇2(𝑡) ≅𝐵̂11𝑇𝑇1(𝑡)+𝐵̂22𝑇𝑇2(𝑡)
Where 𝐵̂11 and 𝐵̂22 are 𝑚−vectors with elements equal to the diagonal entries of matrices 𝐵11 and 𝐵22 respectively.
Therefore,
𝑇𝑇(𝑡)𝐵𝑇(𝑡)≅𝐵̂ 𝑇𝑇(𝑡)
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In which 𝐵̂ is a 2𝑚- vector with elements equal to the diagonal entries of matrix 𝐵.
It is immediately concluded from equation (1.5) that ∫𝑇10(𝑡)𝑇𝑇(𝑡)𝑑𝑡=∫(𝑇1(𝑡)𝑇2(𝑡))𝑇1𝑇(𝑡)𝑇2𝑇(𝑡)𝑑𝑡10
=∫(𝑇1(𝑡)𝑇1𝑇(𝑡)𝑇1(𝑡)𝑇2𝑇(𝑡)𝑇2(𝑡)𝑇1𝑇(𝑡)𝑇2(𝑡)𝑇2𝑇(𝑡))𝑑𝑡10
≅(ℎ3𝐼𝑚×𝑚ℎ6𝐼𝑚×𝑚ℎ6𝐼𝑚×𝑚ℎ3𝐼𝑚×𝑚)
Therefore,
∫𝑇10(𝑡)𝑇𝑇(𝑡)𝑑𝑡≅𝐷
Where 𝐷 is the following 2𝑚×2𝑚 matrix. 𝐷=ℎ3( 10⋯012⁄0⋯001⋯0012⁄⋯0⋮012⁄0⋮0⋮0012⁄⋮0⋱⋯⋯⋯⋱⋯⋮100⋮12⁄⋮010⋮0⋮001⋮0⋱…⋯⋯⋱⋯⋮12⁄00⋮1)
Operational Matrix
Expressing ∫𝑇(𝜏)𝑑𝜏𝑠0 in terms of 𝑇(𝑠), we can write ∫𝑇(𝜏)𝑑𝜏𝑠0=∫(𝑇1(𝜏)𝑇2(𝜏))𝑑𝜏𝑠0
≅(𝑃1𝑇1(𝑠)+𝑃2𝑇2(𝑠)𝑃1𝑇1(𝑠)+𝑃2𝑇2(𝑠))
=(𝑃1𝑃2𝑃1𝑃2)(𝑇1(𝑠)𝑇2(𝑠))
So, ∫𝑇(𝜏)𝑑𝜏𝑠0≅𝑃𝑇(𝑠)
Where 𝑃2𝑚×2𝑚, operational matrix of 𝑇(𝑠) is
𝑃=(𝑃1𝑃2𝑃1𝑃2) (
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Where 𝑃1 & 𝑃2 are given by
The integral of any function 𝑓(𝑡) can be approximated as ∫𝑓(𝜏)𝑑𝜏𝑠0≅∫𝐹𝑇𝑇(𝜏)𝑑𝜏𝑠0
≅𝐹𝑇𝑃𝑇(𝑠)
Solving Nonlinear Integro-Differential Equation
Consider the following nonlinear Volterra-Fredhlom integro-differential equation
{𝑥′(𝑠)+𝑞(𝑠)𝑥(𝑠)+𝜆1∫𝑘1(𝑠,𝑡)[𝑥(𝑡)]𝑛1𝑑𝑡+𝑠0𝜆2∫𝑘2(𝑠,𝑡)[𝑥(𝑡)]𝑛2𝑑𝑡=𝑦(𝑠)𝑠0𝑥(0)=𝑥0, 0≤𝑠<1,𝑛1,𝑛2≥1
Where the parameters 𝜆1 and 𝜆2 and ℒ2 functions 𝑞(𝑠),𝑦(𝑠),𝑘1(𝑠,𝑡),𝑘2(𝑠,𝑡) are known but 𝑥(𝑠) is not.
The appearance of initial condition equation. This is necessary to ensure the existence of a solution.
Approximating functions
𝑥(𝑠),𝑥′(𝑠),𝑞(𝑠),𝑦(𝑠),[𝑥(𝑡)]𝑛1,[𝑥(𝑡)]𝑛2,𝑘1(𝑠,𝑡),𝑘2(𝑠,𝑡)
with respect to TFs, 𝑥(𝑠)≅𝑥𝑇(𝑠)𝑋 𝑥′(𝑠)≅𝑋′𝑇𝑇(𝑠)=𝑇𝑇𝑋′
𝑞(𝑠)≅ 𝑄𝑇𝑇(𝑠)=𝑇𝑇(𝑠)𝑄
𝑦(𝑠)≅ 𝑌𝑇𝑇(𝑠)=𝑇𝑇(𝑠)𝑌 [𝑥(𝑡)]𝑛1≅𝑋𝑛1𝑇𝑇(𝑠)=𝑇𝑇(𝑠)𝑋𝑛1 [𝑥(𝑡)]𝑛2≅𝑋𝑛2𝑇𝑇(𝑠)=𝑇𝑇(𝑠)𝑋𝑛2 𝑘1(𝑠,𝑡)≅𝑇𝑇𝐾1𝑇(𝑡)
𝑘2(𝑠,𝑡)≅𝑇𝑇𝐾2𝑇(𝑡)
Where 2𝑚-vectors 𝑋,𝑋′,𝑄,𝑌,𝑋𝑛1,𝑋𝑛2 and 2𝑚×2𝑚 matrices 𝐾1 & 𝐾2 are TFs coefficients of
𝑥(𝑠),𝑥′(𝑠),𝑞(𝑠),𝑦(𝑠),[𝑥(𝑡)]𝑛1,[𝑥(𝑡)]𝑛2,𝑘1(𝑠,𝑡),𝑘2(𝑠,𝑡) respectively.
Lemma
Let 2𝑚-vectors 𝑋 and 𝑋𝑛 be TFs coefficients of 𝑥(𝑠) and [𝑥(𝑠)]𝑛 respectively. If
𝑋=(𝑋1𝑇𝑋2𝑇)𝑇=(𝑋10,𝑋11,…,𝑋1𝑚−1,𝑋20,𝑋21…𝑋2𝑚−1)𝑇, then
𝑋𝑛=(𝑋10,𝑋11,…,𝑋1𝑚−1,𝑋20,𝑋21…𝑋2𝑚−1)𝑇
Where 𝑛≥1 is a positive integer.
Proof:
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When 𝑛=1, , follows at once from [𝑥(𝑠)]𝑛=𝑥(𝑠).
Suppose that (4.16), holds for 𝑛, we shall deduce it for 𝑛+1.
Since [𝑥(𝑠)]𝑛+1≅(𝑋𝑇𝑇(𝑠)).(𝑋1𝑇𝑇(𝑠))
=𝑋𝑇𝑇(𝑠)𝑇𝑇(𝑠)𝑋𝑛 =𝑋𝑇𝑋𝑛̃𝑇(𝑠)
we obtain 𝑋𝑇𝑋𝑛̃=(𝑋10𝑛+1,𝑋11𝑛+1,……,𝑋1𝑚−1𝑛+1,𝑋20𝑛+1,𝑋21𝑛+1,……,𝑋2𝑚−1𝑛+1)𝑇
Equation holds for 𝑛+1.
Components of 𝑿𝒏 can be Computed in terms of Components of Unknown Vector 𝑿
We substitute (1.15) into (1.14), we have 𝑌𝑇𝑇(𝑠)≅𝑋′𝑇𝑇(𝑠)+𝑄𝑇𝑇(𝑠)𝑇𝑇(𝑠)𝑋+𝜆1𝑇𝑇(𝑠)𝐾1∫𝑇(𝑡)𝑇𝑇(𝑡)𝑋𝑛1𝑑𝑡+𝑠0𝜆2𝑇𝑇(𝑠)𝐾2∫𝑇(𝑡)𝑇𝑇(𝑡)𝑋𝑛2𝑑𝑡𝑠0
it follows that 𝑌𝑇𝑇(𝑠)≅𝑋′𝑇𝑇(𝑠)+(𝑄̃ 𝑇(𝑠))𝑇𝑋+𝜆1𝑇𝑇(𝑠)𝐾1𝑋̃𝑛1∫𝑇(𝑡)𝑑𝑡+𝜆2𝑇𝑇(𝑠)𝐾2𝐷𝑋𝑛2𝑠0
Using Operational matrix 𝑃,
𝑌𝑇𝑇(𝑠)≅𝑋′𝑇𝑇(𝑠)+𝑋𝑇𝑄̃ 𝑇(𝑠)+𝜆1𝑇𝑇(𝑠)𝐾1𝑋̃𝑛1𝑃𝑇(𝑠)+𝜆2(𝐾2𝐷𝑋𝑛2)𝑇𝑇(𝑠)
In which 𝜆1𝐾1𝑋̃𝑛1𝑃 is a 2𝑚×2𝑚 matrix
𝑇𝑇(𝑠)𝜆1𝐾1𝑋̃𝑛1𝑃𝑇(𝑠)≅𝑋𝑛1𝑇̂𝑇(𝑠)
Where 𝑋̃𝑛1 is an 2𝑚-vector with components equal to the diagonal entries of the matrix 𝜆1𝐾1𝑋̃𝑛1𝑃.
𝑌𝑇𝑇(𝑠)≅𝑋′𝑇𝑇(𝑠)+𝑋𝑇𝑄̃ 𝑇(𝑠)+𝑋𝑛1𝑇̂𝑇(𝑠)+𝜆2(𝐾2𝐷𝑋𝑛2)𝑇𝑇(+𝑄̃ 𝑋+𝑋𝑛1̂+ 𝜆2𝐾2𝐷𝑋𝑛2≅𝑌
Where 𝑄̃ is a diagonal matrix, so 𝑄̃ 𝑇=𝑄̃.
𝑋′ must be computed in terms of 𝑋. 𝑥(𝑠)−𝑥(0)=∫𝑥′(𝜏)𝑑𝜏𝑠0
≅∫𝑋′𝑇𝑇(𝜏)𝑑𝜏𝑠0 ≅𝑋′𝑇𝑃𝑇(𝑠) 𝑥(𝑠)≅𝑋′𝑇𝑃𝑇(𝑠)+𝑋0𝑇𝑇(𝑠)
Where 𝑋0 is the 2𝑚- vector of the form
Journal of the Maharaja Sayajirao University of Baroda ISSN: 0025-0422
75
Volume-56, No.1 (IV) 2022
𝑋0=[𝑥0,𝑥0,…𝑥0]𝑇
Consequently,
𝑋≅𝑃𝑇𝑋′+𝑋0
Now, combining (1.19) and (1.20) and replacing ≅ with =, it follows that
(𝐼+𝑃𝑇𝑄̃)𝑋+𝑃𝑇𝑋𝑛1̂+𝜆2𝑃𝑇𝐾2𝐷𝑋𝑛2≅𝑃𝑇𝑌+𝑋0
Equation is a nonlinear system of 2𝑚 algebraic equations for the 2𝑚 unknowns 𝑋10,𝑋11,…𝑋1𝑚−1,𝑋20,𝑋21,…𝑋2𝑚−1.
Components of 𝑋𝑇=(𝑋1𝑇 𝑋2𝑇) can be obtainbed by an iterative method.
Hence, an approximate solution 𝑥(𝑠)≅𝑋𝑇𝑇(𝑠)
𝑥(𝑠)≅𝑋1𝑇𝑇1(𝑠)+𝑋2𝑇𝑇2(𝑠)
Can be computed for equation (1.14) without using any projection method.
Conclusion
We investigated the numerical solution of nonlinear and singularly perturbed for Vulture integro-differential equations. Also, numerical solution of high-order and fractional nonlinear Volterra-Fredholm integro-differential equations and its error analysis are discussed. Finally, some numerical results of integro-differential equations are presented to illustrated the efficiency and accuracy of the proposed methods.
Reference
1. G.Yu.Mekhtieva, V.R.Ibrahimov “On one numerical method for solution of integro-differential equations”, Vestnik BSU, ser.phys.-math. Scien., 2003, No3, pp.21-27.(Russian).
2. H.Brunner, “Collection method for Volterra integral and Relation Functional Equations”, Cambridge University Press, Cambridge,2004.
3. P.Darania, E.Abadian, “A method for the numerical solution of the integro-differential equations”, Appl.Math. and Comput. 188(2007) 657-668.
4. B. Ahmad, S. Siva Sundaram, Existence Results For Nonlinear Impulsive Hybrid Boundary Value Problems involving Fractional Differential Equations, Non linear Analysis, hybrid System, 3(2009).
5. Z.B. Bai, H.S. Lii, Positive Solutions Of Boundary Valuve problems Of Nonlinear Fractional Differential Equations, J. Math Anal.Appl.311(2005)\
6. v. Lakshmikantham, D.D. Bainov And P.S. Simeonov, Theory Of Impulsive Differential Equations, modern Applied Mathematics (1989).
7. K.S. Miller, B.Ross, An Introduction To The Fractional Caculus And Fractional Differential Equations, Wiley, New york, 1993
Friday 29 April 2022
CRIME DETECTION MODELLING BY USING NUMERICAL METHODS
CRIME DETECTION MODELLING BY USING NUMERICAL METHODS
1 CRIME:
In most countries the detection of crime is the responsibility of the police though special law enforcement agencies may be responsible for the discovery of the particular type of crime.
Example: Customs departments may be charged with computing smuggling and related offenses.
Crime detection falls in to three distinguishable phases the discovery that a crime has been committed the identification of a suspect and the collection of sufficient evidence to indict the suspect before a court.
Example: Electronic eavesdropping surveillance interception of communication and infiltration of gangs.
Fig: 1 FLOWCHAT OF CRIME DEFELATION
1.3. STEPS OF CRIMINAL CASE:
The 5 basic steps of a criminal proceeding are
·
Arrest
·
Preliminary hearing
·
Grand jury investigation
·
Arraignment in criminal court
·
Trial by jury
1.4. CAUSES OF CRIME:
·
Poverty this is perhaps one of the most concrete
reasons why people
·
commit crime
·
Peer pressure new form
of concern in the modern world
·
Drugs always been highly criticized by critics
·
Politics
·
Religion ,family conditions, the society ,unemployment.
1.5. THE CRIME SCENE SKETCH:
·
Measure and outline
area
·
North should be at the top
of the paper
·
Determine the scale
·
Take the longest
measurement at the scene and divide it by the longest measurement of the paper used for the Sketching.
Ø 1/2″ = 1 small rooms.
Ø
1/8″ = 1 very large rooms.
Ø
1/8″ = 10 large land area.
2. CRIME PREVENTION AND TECHNIQUES
2.1. INVOLVES:
Recognition.
↓
Identification
↓
Individualization
↓
Reconstruction
2.1.1 RECOGNITION:
• Scene survey documentation collection
2.1.2 IDENTIFICATION:
• Classification of evidence
2.1.3 INDIVIDUATION:
• Comparison testing
evaluation and interpretation
2.1.4 RECONSTRUCTION:
• Sequencing events
reporting and presenting
2.2. THE CRIME SCENE SKETCH:
•
Measure and outline
area
•
It doesn’t stretch
•
Use conventional units of measurements
•
Inches
•
Feet
•
Centimeters
•
Meters
2.3. CRIME SCENE RECONSTRUCTION:
•
Data collection
•
Conjecture
•
Hypothesis formulation
•
Testing
•
Theory
3.FLOWCHART AND CRIME NOTES
The location of possible targets homes people as well as general environmental goes like levels of disorder housing density and surveillance researches also include knowledge offenders might have about target vulnerability for instance tricks learned from burglarizing nearby and what they found was two kinds of crime hotspots.
Supercritical which formed from a rapid chain reaction of law less ness and subcritical a large spike in a crime in an otherwise stable area in a supercritical situation small spikes in crime grow and spread when increased police force is applied new hotspots.
But off existing once they reform around small spikes in a crime to continue away from areas of police presence the crime is displaced police suppression is a cat and mouse game something very different happens when police are added to a sub critical hots
The surrounding are relatively stable so one suppression is applied the hotpots is reduced or eradicated completely and even when suppression efforts are removed the hotpots doesn’t really. Since strategy won’t work for both kinds of hotpots.
This key difference may help law enforcement tailor their crime strategies to different neighborliness ultimately thought hotpots maybe similar sizes and distribution
The research shows that looking at the dynamic of crime is important in gauging the best police response and that could be a fundamental step towards better prediction and crime prevention strategies and an even safer future for cities.
4. NOTES OF CRIME:
Imagine someone is committing crime throughout local area on a regular basis may be they have committed 5 crimes in the last month at these locations how could we go about catching this person.
One thing we can do is try finding a patterns in crime location in order to make a prediction about the next one that can be tricky through since there likely a huge elements of randomness but decades ago Kim Ross mo. a PhD criminologist had another idea he tried to find a formula instead could find.
Where the criminal likely lives based on the past data he knows that criminals often don’t commit crimes right by their own home but also they don’t go too far away so from the data you can determine a quote hot zone fig★
Fig ★ HOT ZONE
Which is not too close or too far from the crime scene it has a high probability of the person living there this is his equation for determining those probabilities. I know it looks complex but its actually not as bad as you think like take this Pi, j parts.
To see the crime scene in map and put a gird over it any given squares will be in some row we will label i and column will label j, Pi, j is probability that the criminal lives in the square. How you calculate that value for any square is with this right side of the equation.
Formula for crime detection:
(| Xi-xn|+|Yj -yn|)f → 1 𝑠𝑡 𝑡𝑒𝑟m
(2𝐵| Xi-xn|+|Yj -yn|) g
→ 2 𝑛𝑑 𝑡𝑒𝑟𝑚
Now take this denominator equation, (|Xi-xn |+|Yj -yn|) f this 1stpart just means take the arbitrary grid you are analysing and one of the crime scene and subtract these x corps donets which gives you this distance.
The absolute value just ensure that its positive then |Yj -yn| parts just says do the same thing with the y axis which gives us this length add those up and we get the distance between the gird and the crime scene no it’s not the straight – line distance.
If you cannot move to agonal and as we saw this term is in the denominator which means as that distance goes up the entire fraction and thus the probability go down this is expected because like .I said these criminals usually don’t go super far to commit a crime so a larger distance between our square and the crime scene means a smaller Probability the criminal lies.
There at least for this laughter but remember criminals don’t commit these crime close to this homes either and that where this side of the equation comes.
You have that same distance down here subtracted from something known as a buffer zone which is just a 2B constant determined by what works best with known data or past crime so as
distance goes up the entire denominator actually decreasing making the whole fraction.
Probability go up so physically if you are too close to the crime scene probability is low that the criminal lives there but as you get further the probability increase that is of course until you pass a certain point which is where the left side of the formula comes in again after the distance and these 2 fractions change in opposite ways which is essentially.
The balancing act that creeps the hot zone of high probability that isn’t too close and isn’t too far from the crime scene for is sort of a constant that I am not going to go in to ∅ i j and that g, f and B are constant. That just make certain parts of this equation matter more than other or they add more weights to certain parts then lastly this part.
We calculated these 2 fractions for every single crime committed and add the results do this for square on our gird and we create a heat map of probability. Actually is the equation output based on real crimes of a serial killer from the 70s named Richard chase you can see the crime locations in green and the formula predicts his residence to be somewhere in this dark region his actual residence is plotted here in purple exactly as expected so that least in the case Rasco’s formula works.
5. CRIME SCENE CONTROL
The actions which the first arriving officer at the crime scene takes to make sure that the integrity of the scene maintained. Control also includes preventing people at the scene from becoming combatants and separating witnesses.
5.1 DRAWBACKS:
·
No bobby on the beat
·
Local police officer
in the local community
·
Large amount of data but is there a significant impact on detecting and reducing crime
overal
5.2 BENEFITS:
·
Administrative duties completed
more quickly
·
Tools available to help track down criminals
·
Communication improved
·
Evidence collection tracking
analysis and availability improved
·
Amount of data available.
5.3 SCALING THE INVESTIGATION TO THE EVENT:
·
The crime scene must be secured preserved and recorded until evidence is collected
·
Existing contamination must be
considered and recorded
·
Cross contamination must be prevented
·
Exhibits must be identified preserved
collected and secured
to pressure the chain of continuity
5.4 CRIME SCENE ANALYSIS PROCESS:
·
Crime analysts study
crime reports arrests
reports and police cells for service calls for service
to identify emerging
patterns series and trends
as quickly as possible.
·
The analyse these phenomena for all relevant factors
sometimes predict or forecast future occurrence and issue bulletins reports and alert to their
agencies.
5.5 CRIME SCENE CONTROL:
·
The actions
which the first arriving officer
at the crime scene takes to make sure that the integrity of the scene is maintained.
·
Control also includes
preventing people at the scene from becoming
combatants and separating witnesses.
5.6.1 CRIME SCENE MAPPING:
·
Triangulation- uses to two points
at the crime scene to map each piece of evidence
·
Coordinate or grid – divides the crime scene into squares
for mapping
·
Suspended polar coordinate-for use in mapping evidence in a hole
·
Baseline- set a north or south line and measure each piece of evidence
from this line.
5.6.2 MICROSCOPIC CRIME SCENE:
·
Based on the size
·
Trace evidence
·
Gunshot residue
·
Tire treads
·
Hair samples
·
Mites found in clothes
5.6.3 CRIME SCENE DOCUMENTATION:
·
Take notes at the crime scene
·
Videotape the crime scene
·
Photograph the crime
scene