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| < αΆ―), G1=(x0 ≤ s ≤ x≤ X,|y| ≤ αΆ―)
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 πΎ(π₯,π ,π¦)=Ξ£(ππ(x)ππ(s,y) )ππ=1
Then equation can be written as
y'(x) = f(x,y) + Ξ£ππ(x)yππ=0∫ππ(s,y(s))ds,yπ₯π₯0 y(x) = yo
Using the notation
Vi(x)=∫ππ(s,y(s))π₯π₯0ds, (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)+ Ξ£(ππ(x)π£π(x))ππ=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
Ξ£πΌππ=0i yn+i = hΞ£π½ππ=0ifn+i + hΞ£π½ππ=0ian+ivn+i (n=0,1,2,⋯ ⋯,π−π)
Ξ£πΌππ=0ivn+i = hΞ£π½ππ=0ibn+i (n=0,1,2,⋯⋯,π−π)
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
αΊn+k = Ξ£(πΌππ=0iyn+i + hΞ²ifn+i + hΞ²ian+i)
yn+k = Ξ£(π−1π=0Ξ±'iyn+i + hΞ²'ifn+i + hΞ²'ian+ivn+i) + hΞ²' k f (f ( xn+k+yn+k) +an+kvn+k)
Where the coefficients of prediction are method are denoted Ξ±i, Ξ²j (l = 0,1,2, k-i) and by a="/a
Bi - B/a (i = 0,1,2....k-1).
Ξ²' k=Ξ²'k /Ξ±k, 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 = Σαπ−1π=0'ivn+i + h Σβπ−1π=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
αΊ(xn+k) = Ξ£[πΌπ−1π=0iy(xn+i) + hΞ²if(xn+i,y(xn+i) + hΞ²ia(xn+i)v(xn+i)] +Εn
y(xn+k) = Ξ£[πΌ′π−1π=0iy(xn+i) + hΞ²'if(xn+i,y(xn+i) + hΞ²'ia(xn+i)v(xn+i)] + hΞ²'kf(xn+k,y(xn+k) +
hΞ²'ka(xn+k)v(xn+i) + Rn
V(xn+k) = Σαπ−1π=0'iv(xn+i) + h Σβππ=0'ib(xn+I,y(xn+i)) + Rn
Here the error of corresponding methods is denoted by Ε,Rn
Note that αΊ(xm) = y(xm) ( m = 0,1,2,⋯⋯)
Definition
Equation is stable if the roots of the polynomial
π(π) = Ξ£πΌππ=0iπi
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
Ξ£[πΌππ=0iy(x+i) - hΞ²iy'(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≤π≤π(β°m,Γm) ≤ A exp (BX)(πmax0≤π₯≤1| β°π|+ππ)
Here A, B and d are some bounded values and Γπ=π£(π₯π)−π£π,β°π=π¦(π₯π)−π¦π(π=0,1,2,⋯⋯,)
Proof:
β°π+π=Ξ£(πΌπβ°π+π+βπ½πβπ+πβπ+π+βπ½πππ+πππ+π )+Εππ−1π=0
β°π+π=Ξ£(πΌ′πβ°π+π+βπ½πβπ+πβπ+π+βπ½′πππ+πππ+π )+βπ½′πβπ+πβπ+π+π−1π=0
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βπ½′πππ+πβπ+π+Επ
Here βπ=π′π¦(π₯π,ππ),βπ=π′π¦(π₯π,ππ),ππ=π(π₯π),Γπ=π¦(π₯π)−αΊπ
,( m = 0,1,2,.....)
Where ππ is found between y(xm) and ym and ππ between y(xm) and αΊm.
We have ππ+π=Ξ£ππππ+π+βΞ£ππππ+π+βΞ£πππ−1π=0ππ+π+βπ½′πππ+πππ+π+βπ½′πβπ+πΕπ+βππ−1π=0π−1π=0
Where, ππ=πΌ′π+βπ½′πβπ+ππΌπ, ππ=π½′πβπ+π+βπ½′πβπ+ππ½πβπ+1, ππ=π½′πππ+π+βπ½′πβπ+πππ+π
we will obtain, ππ+π=Ξ£(π′πππ+π+βπ½′πβπ+ππ−1π=0ππ+π)+βπ½′πβπ+πππ+π+βπ
Where βπ=π′π¦(π₯π,π¦π) (m= 0,1,2,⋯⋯⋯,)ππ are between y(xm) and ym for error estimation of the method applied to the solution
We will obtain, ππ+π=Ξ£ππππ+ππ−1π=0+βΞ£πππ−1π=0ππ+π+βπ
ππ+π=Ξ£(π−1π=0π′πππ+π+βπ½′πβπ+πππ+π+π½′πβπ+πΞ£(π−1π=0ππππ+π+ β2ππππ+π)+βπ½′πβπ+πβπ+βπ
Where, ππ=ππ+βππ1−β2ππ+π ππ=ππ+ππ+ππ½′ππΌ′π1−β2ππ+π ππ=ππ+βππ+ππ½′ππ½′πβπ+π ππ+π=ππ+π(π½′π)2βπ+π
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βπ=(βπ½′πβπ+πβπ+βπ+βππ+πβπ)/1−β2ππ+π
Consider the following relation
ππ+βππ/1−β2ππ+π=ππ+βππ
Where ππ=(ππ+βππππ+π)/1−β2ππ+π
For smallness of h we can assume |1−β2ππ+π|≥12
Then we obtain that |ππ|≤π(i= 0,1,2, ⋯⋯-k-1)
It is easy to show that ππ=(ππ+ππ+ππ½′ππΌπ)/(1−β2ππ+π) (π= 0,1,2,⋯ ⋯,π−1)
are also bounded, that is,|ππ| ≤ b.
we can write ππ+π=Ξ£π′πππ+ππ−1π=0+βΞ£π£ππ−1π=0ππ+π+βΞ£πππ−1π=0ππ+π+βπ
ππ+π=Ξ£π′πππ+ππ−1π=0+βΞ£π£ππ−1π=0ππ+π+β2Ξ£πππ½′πβπ+ππ−1π=0ππ+π+Επ
Where, Επ=βπ+βπ½′πβπ+π+βπ π£π=π½′πβπ+ππΌπ+ππ π£π=π½′πβπ+π+πΌππ½′πβπ+π (π=0,1,2,…,π−1)
Here we can also write that lvi| ≤v and |αΏ‘i|≤αΏ‘
Consider the following vectors,
Yn+k-1 (π n+k-1, π n+k-2, … … … , π n)
αΊn+k-1 (π n+k-1, π n+k-2, … … … , π n)
Then adding to equation the following identities respectively,
ππ+π=π΄ππ+π−1+βππ+πππ+π−1+βπ΅π+παΊπ+π−1+Ε΄π,π
αΊπ+π=π΄αΊπ+π−1+βαΉΌπ+πππ+π−1+β2π΅π+παΊπ+π−1+ππ,π
Where the matrices A, Vn+kαΏ‘n+ki En+k. Γn+k and the vectors Wnk Ε΄nk are
Determined in the following form π΄=(πΌ′π−1πΌ′π−2.10.00..πΌ′1πΌ′0 ....10) π£π+π=(π£π−1π£π−2.00.00..π£0...0)
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π©π+π=(π΅π−1π΅π−2.00.00..π0.0.0) π©π+π=(ππ−1ππ−2.00.00..π0.000)(π½′ππΏπ+π)−1 π£π+π=(π£π−1π£π−2.00.00. .π£0 .0 .0) ππ,π=(π π00) ππ,π=(Επ00)
Using the vector Zn+k =(π¦π+π,αΊπ+π)
We can rewrite equations in the following form ππ+π=π΄ππ+π−1βππ+πππ+π−1+Ε΄π,π
Where, π΄=(π΄00π΄) ππ+π=(ππ+ππ΅π+πππ+πβπ΅π+π) Ε΄π,π=(Ε΄π,πΕ΄π,π)
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 π¦π+π=π·π¦π+π−1βπΈπ+ππ¦π+π−1+Ε΄π,π
determined in the following form π·=πΆ−1αΊ¬πΆ πΈπ+π=πΆ−1αΉΌπ+ππΆ Ε΄π,π=πΆ−1Ε΄π,π
Passing to the norm in equation
we have, ‖π¦π+π‖≤‖π·‖‖π¦π+π‖+β‖πΈπ+π‖‖π¦π+π−1‖+‖Ε΄π,π‖
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+kiαΏ‘n+k have the bounded norms.
Then we can write
||En+k|| ≤ B
Subject to the above metrices Vn+k,En+k,αΏ‘n+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=||Ε΄n,k||
we will obtain that ‖π¦π‖≤(1+βπ΅)π−π‖π¦π‖+Ξ£(1+βπ΅)ππ−π−1π=0π
It is known that (1+hB) m exp (mhB)
For h ≤ X-xo≤ X, We can rewrite in the following form ‖π¦π‖≤exp(π΅π)(‖π¦π‖+πβ−1) ‖π¦π§‖≤π΄exp(π΅π)(‖π‖‖π§π‖+πβ−1)
Where, π΄=‖πΆ‖ π=‖πΆ−1‖
From the last it follows that maxπ≤π≤π(ππ,ππ)≤πexp(π΅π)(πmax0≤π≤π−1|ππ|+ππ)
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≤π≤π−1ππ=π(βπ)
Then from the assertions of the theorem it follows that ππ=π(βπ),β→0
As we see from the proof of the theorem by using equation (1.7) it was assumed that
π¦π+π is known. However, for finding π¦π+π is to be known π£π+π
Note that in real calculations for finding the valuesπ£π+π - it is used approximate values of π¦π+π calculated by the prediction method.
In this case, the proposed method for the solution of equation (1.5) is written in the
following form αΊπ+π=Ξ£(πππ¦π+πππ=0+βπ½πππ+π+βπ½πππ+π,π£π+π) π£π+π=Ξ£π′ππ£π+ππ−1π=0+βΞ£π½′πππ+π+π−1π=0βπ½′ππ(π₯π+π,αΊπ+π) π¦π+π=Ξ£π′ππ¦π+ππ−1π=0+βΞ£π½′π(ππ+π+π−1π=0ππ+ππ£π+π)+βπ½′π(π₯π+π,αΊπ+π)+ππ+ππ£π+π
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