Post on 28-Jan-2023
Journal of mathematics and computer science 7 (2013), 293-304
Prefunctions and System of Differential Equation via Laplace Transform
D. B. Dhaigude
Department of Mathematics, Dr. Babasaheb Ambedkar Marathawada UniversityAurangabad, Aurangabad, India
S. B. Dhaigude
SPMs Bhivarabai Savant Institute of Technology and Research (for women) Wagholi Pune-412207, India
Email: suresh.dhaigude@gmail.com Article history: Received April 2013 Accepted May 2013 Available online June 2013
Abstract The purpose of this paper is to study the system of Linear Differential Equations of first order, Second
order and Third order in two variables as well as in three variables and obtained the set of Pre-functions and extended prefunctions are the closed form of solutions via Laplace Transforms technique.
Keywords. : Pre-functions, Extended prefunctions, Initial Value Problems and Laplace Transforms 1.Introduction
It is well known that some basic functions such as polynomial functions, exponential functions, Logarithmic functions and trigonometric functions have played an important role in development of Mathematical, physical, biological as well as Engineering sciences .Deo and Howell in their paper [1] introduced an alternative method and studied trigonometric and trigonometric like functions. This new approach is simple and useful in the study of differential equations. Khandeparkar, Deo and Dhaigude [3] have defined new kind of Exponential, Trigonometric and hyperbolic functions. They are coined as Pre-functions and extended Pre-functions. It is shown that these pre-functions and extended pre-functions satisfy many interesting properties and relations. Some properties and Laplace Transforms of these functions are studied in the papers [2], [3], [4], [5], [6] and solution of initial value problems for
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non-homogeneous linear ordinary differential equations as well as solutions of voltera integral equations and volltera integro differential equations are obtained.
The aim of this paper is to study the system of Linear Differential Equations of first order, Second order and Third order in two variables and in three variables and obtained the set of Pre-functions and extended prefunctions are the closed form of solutions and Laplace Transforms technique is successfully applied.
2. Definitions of prefunctions and extended prefunctions
In this section, we take overview of pre-function and extended pre-functions and Laplace Transforms of these pre-functions [2] and [3] these pre-functions and Laplace Transforms are defined and their properties are also defined. We introduce definitions and Laplace Transforms of some Pre-functions and
extended pre-functions.
Definition: The pre-exponential functions is denoted by pexp (t, Ξ±) and is defined as,
ππππππππ(π‘π‘,Ξ±) = 1 + β tn +Ξ±
Ξ(n+1+Ξ±), t β¬ Rβ
n=1 and Ξ± β₯ 0
Also, ππππππππ(βπ‘π‘,Ξ±) = 1 + β (β1)n +Ξ± tn +Ξ±
Ξ(n+1+Ξ±), t β¬ Rβ
n=1 and Ξ± β₯ 0
The pre-cosine function is denoted by pcos (t, Ξ±) and is defined as,
ππππππππ(π‘π‘,Ξ±) = 1 + β (β1)n t2n +Ξ±
Ξ(2n+1+Ξ±), t β¬ Rβ
n=1 and Ξ± β₯ 0
The pre-sine function is denoted by ππππππππ(π‘π‘,Ξ±) and is defined as,
ππππππππ(π‘π‘,Ξ±) = β (β1)n t2n +Ξ±+1
Ξ(2n+2+Ξ±), t β¬ Rβ
n=0 and Ξ± β₯ 0
The extended pre function is denoted by pM32
ππππ32(π‘π‘,πΌπΌ) = β (β1)n t3n +2+Ξ±
Ξ(3n+3+Ξ±), t β¬ R β
n=0 and Ξ± β₯ 0
Laplace Transform of Prefunctions and extended Prefunctions:
We find the Laplace transforms of ππππππππ(π‘π‘,Ξ±): We know that,
ππππππππ(πππ‘π‘,Ξ±) = 1 + β (at )n +Ξ±
Ξ(n+1+Ξ±), t β¬ Rβ
n=1 and Ξ± β₯ 0
Now we calculate the Laplace Transforms of this prefunction as follows,
πΏπΏ[ππππππππ(πππ‘π‘,Ξ±)] = β« eβstππππππππ(πππ‘π‘,Ξ±)dtβ0
= β« eβst [1 + β (at )n +Ξ±
Ξ(n+1+Ξ±) βn=1 dtβ
0 ]
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= β« eβst 1dt + β 1Ξ(n+1+Ξ±)
βn=1
β0 β« eβstβ
0 (at)n+Ξ±dt
= 1ππ
+ β 1Ξ(n+1+Ξ±)
βn=1
(a)n +Ξ± Ξ(n+1+Ξ±)sn +1+Ξ±
= 1ππ
+ β (a)n +Ξ±
sn +1+Ξ±β1
= 1ππ
+ (a)1+Ξ±
(sβa)s1+Ξ±
In particular, for a=1,
πΏπΏ[ππππππππ(π‘π‘,Ξ±)] = 1ππ
+ 1(sβ1)s1+Ξ±
In particular Ξ±=0,
πΏπΏ[ππππππππ(π‘π‘, 0)] = πΏπΏ[ππππππππ(π‘π‘)] = 1(ππβ1)
Clearly, πΏπΏβ1 οΏ½1ππ
+ (a)1+Ξ±
(sβa)s1+Ξ± οΏ½ = ππππππππ(πππ‘π‘,Ξ±)
Laplace transforms of other prefunctions can be derived and listed as follows,
1. πΏπΏ[ππππππππ(βπππ‘π‘,Ξ±)] = 1ππβ (β1)Ξ± (a)1+Ξ±
(s+a)s1+Ξ±
2. πΏπΏ[ππππππππ(πππ‘π‘,Ξ±)] = (a)1+Ξ±
(s2 +a2)sΞ±
3. πΏπΏ[ππππππππ(πππ‘π‘,Ξ±)] = 1ππβ aΞ±+2
(s2 +a2)s1+Ξ±
4. πΏπΏ[ππππππππβ(πππ‘π‘,Ξ±)] = (a)1+Ξ±
(s2 βa2)sΞ±
5. πΏπΏ[ππππππππβ(πππ‘π‘,Ξ±)] = 1ππ
+ aΞ±+2
(s2 βa2)s1+Ξ±
6. πΏπΏ[ππππ30(πππ‘π‘,Ξ±)] = 1sβ (a)3+Ξ±
(s3 +a3)sΞ±+1
7. πΏπΏ[ππππ31(πππ‘π‘,Ξ±)] = (a)1+Ξ±
(s3 +a3)sΞ±β1
8. πΏπΏ[ππππ32(πππ‘π‘,Ξ±)] = (a)2+Ξ±
(s3 +a3)sΞ±
9. πΏπΏ[ππππ30(πππ‘π‘,Ξ±)] = 1s
+ (a)3+Ξ±
(s3 βa3)sΞ±+1
10. πΏπΏ[ππππ31(πππ‘π‘,Ξ±)] = (a)1+Ξ±
οΏ½s3 βa3οΏ½sΞ±β1
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11. πΏπΏ[ππππ32(πππ‘π‘,Ξ±)] = (a)2+Ξ±
(s3 βa3)sΞ±
Note that on putting Ξ±=0 and a=1 above we get Laplace Transforms of exponential functions, trigonometric functions and hyperbolic function.
3. Initial Value Problems for System of Differential Equations:
The solution of simultaneous differential equations with initial conditions can be obtained by applying Laplace Transform method. First we discuss the solution of first order system of ordinary differential equations with suitable initial conditions.
I. First Order System of Differential Equations:
IVP-I. Consider the system of first order differential equations,
ππβ²(π‘π‘,πΌπΌ) + π¦π¦(π‘π‘,πΌπΌ) = 0 (1.1)
π¦π¦β²(π‘π‘,πΌπΌ) β ππ(π‘π‘,πΌπΌ) = π‘π‘πΌπΌ
π€π€(πΌπΌ+1)β 1 (1.2)
With the initial conditions, ππ(0,πΌπΌ) = 1, π¦π¦(0,πΌπΌ) = 0 (1.3)
Apply Laplace transform to both the sides of differential equations (1.1) and (1.2), we have,
πΏπΏ{ππβ²(π‘π‘,πΌπΌ)} + πΏπΏ{π¦π¦(π‘π‘,πΌπΌ)} = 0
πΏπΏ{π¦π¦β²(π‘π‘,πΌπΌ)} β πΏπΏ{ππ(π‘π‘,πΌπΌ)} = πΏπΏ{ π‘π‘πΌπΌ
π€π€(πΌπΌ+1)} β πΏπΏ{1}
πππΏπΏ{ππ(π‘π‘,πΌπΌ)} β ππ(0,πΌπΌ) + πΏπΏ{π¦π¦(π‘π‘,πΌπΌ)} = 0
And πππΏπΏ{π¦π¦(π‘π‘,πΌπΌ)} β π¦π¦(0,πΌπΌ) β πΏπΏ{ππ(π‘π‘,πΌπΌ)} = 1π€π€(πΌπΌ+1)
π€π€(πΌπΌ+1)πππΌπΌ+1 β 1
ππ
Consider, πΏπΏ{ππ(π‘π‘,πΌπΌ) = ππ(ππ,πΌπΌ), πΏπΏ{π¦π¦(π‘π‘,πΌπΌ) = ππ(ππ,πΌπΌ). Using initial conditions (1.3), we obtain the System of equations in X and Y
ππππ(ππ,πΌπΌ) + ππ(ππ,πΌπΌ) = 1 (1.4)
βππ(ππ,πΌπΌ) + ππππ(ππ,πΌπΌ) = 1πππΌπΌ+1 β
1ππ (1.5)
Multiply equation (1.5) by s and add with equation (1.4), we get,
οΏ½ππ2 + 1οΏ½ππ(ππ,πΌπΌ) = 1πππΌπΌ
ππ(ππ,πΌπΌ) = 1(ππ2+1)πππΌπΌ
Taking inverse Laplace transform of both sides, we have,
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πΏπΏβ1{ππ(ππ,πΌπΌ)} = πΏπΏβ1{ 1(ππ2+1)πππΌπΌ
}
π¦π¦(π‘π‘,πΌπΌ) = ππππππππ(π‘π‘,πΌπΌ)
Now we obtain prefunction ππ(π‘π‘,πΌπΌ), multiply equation (1.4) by s and subtract equation (1.5) from it, we have,
οΏ½ππ2 + 1οΏ½ππ(ππ,πΌπΌ) = ππ2+1ππβ 1
πππΌπΌ+1
ππ(ππ,πΌπΌ) = 1ππβ 1
(ππ2+1)πππΌπΌ+1
Taking inverse Laplace transform of both sides, we have,
πΏπΏβ1{ππ(ππ,πΌπΌ)} = πΏπΏβ1{ 1ππβ 1
(ππ2+1)πππΌπΌ+1 }
ππ(π‘π‘,πΌπΌ) = ππππππππ(π‘π‘,πΌπΌ)
The solution set of the IVP (1.1)-(1.3) is {x(π‘π‘,πΌπΌ) = ππππππππ(π‘π‘,πΌπΌ), π¦π¦(π‘π‘,πΌπΌ) = ππππππππ(π‘π‘,πΌπΌ)}.
IVP-II. Consider the system of first order differential equations,
ππβ²(π‘π‘,πΌπΌ) + π§π§(π‘π‘,πΌπΌ) = 0 (2.1)
π¦π¦β²(π‘π‘,πΌπΌ) β ππ(π‘π‘,πΌπΌ) = π‘π‘πΌπΌ
π€π€(πΌπΌ+1)β 1 (2.2)
π§π§β²(π‘π‘,πΌπΌ) β π¦π¦(π‘π‘,πΌπΌ) = 0 (2.3)
With the initial conditions,
ππ(0,πΌπΌ) = 1, π¦π¦(0,πΌπΌ) = 0, π§π§(0,πΌπΌ) = 0 (2.4)
Apply Laplace transform to both the sides of differential equations (2.1), (2.2), and (2.3) we have,
πΏπΏ{ππβ²(π‘π‘,πΌπΌ)} + πΏπΏ{π§π§(π‘π‘,πΌπΌ)} = 0
πΏπΏ{π¦π¦β²(π‘π‘,πΌπΌ)} β πΏπΏ{ππ(π‘π‘,πΌπΌ)} = πΏπΏ{ π‘π‘πΌπΌ
π€π€(πΌπΌ+1)} β πΏπΏ{1}
πΏπΏ{π§π§β²(π‘π‘,πΌπΌ)} β πΏπΏ{π¦π¦(π‘π‘,πΌπΌ)} = 0
Simplification it gives,
πππΏπΏ{ππ(π‘π‘,πΌπΌ)} β ππ(0,πΌπΌ) + πΏπΏ{π§π§(π‘π‘,πΌπΌ)} = 0
πππΏπΏ{π¦π¦(π‘π‘,πΌπΌ)} β π¦π¦(0,πΌπΌ) β πΏπΏ{ππ(π‘π‘,πΌπΌ)} = 1π€π€(πΌπΌ+1)
π€π€(πΌπΌ+1)πππΌπΌ+1 β 1
ππ
And πππΏπΏ{π§π§(π‘π‘,πΌπΌ)} β π§π§(0,πΌπΌ) β πΏπΏ{π¦π¦(π‘π‘,πΌπΌ)} = 0
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Suppose,
πΏπΏ{ππ(π‘π‘,πΌπΌ)} = ππ(ππ,πΌπΌ), πΏπΏ{π¦π¦(π‘π‘,πΌπΌ)} = ππ(ππ,πΌπΌ), πΏπΏ{π§π§(π‘π‘,πΌπΌ)} = ππ(ππ,πΌπΌ)
Using initial condition (2.4), we get the system of equations in X, Y and Z
ππππ(ππ,πΌπΌ) + ππ(ππ,πΌπΌ) = 1
βππ(ππ,πΌπΌ) + ππππ(ππ,πΌπΌ) = 1πππΌπΌ+1 β
1ππ
βππ(ππ,πΌπΌ) + ππππ(ππ,πΌπΌ) = 0
We write this system of equations in matrix form as,
οΏ½ππ 0 1β1 ππ 00 β1 ππ
οΏ½ οΏ½ ππππππ
οΏ½ = οΏ½1
1πππΌπΌ+1 β
1ππ
0οΏ½
We know by matrix theory,
ππ = π·π·πππ·π·
, ππ = π·π·π¦π¦π·π·
, ππ = π·π·π§π§π·π·
(2.5)
Where,
π·π· = οΏ½ππ 0 1β1 ππ 0 0 β1 ππ
οΏ½ = ππ3 + 1
π·π·ππ = οΏ½ 1 0 β1
1πππΌπΌ+1 β
1ππ
ππ 00 β1 ππ
οΏ½ = ππ2 + 1ππβ 1
πππΌπΌ+1
π·π·π¦π¦ = οΏ½ ππ 1 1 β1 1
πππΌπΌ+1 β1ππ
00 0 ππ
οΏ½ = 1πππΌπΌβ1
π·π·π§π§ = οΏ½ππ 0 1β1 ππ 1
πππΌπΌ+1 β1ππ
0 β1 0οΏ½ = 1
πππΌπΌ
Substituting these values in equation (2.5), then we obtain
ππ(ππ,πΌπΌ) = 1ππβ 1
(ππ3+1)πππΌπΌ+1
ππ(ππ,πΌπΌ) = 1(ππ3+1)πππΌπΌβ1
ππ(ππ,πΌπΌ) = 1(ππ3+1)πππΌπΌ
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In each case, taking inverse Laplace transform of both sides, we have,
x(π‘π‘,πΌπΌ) = πΏπΏβ1 οΏ½ 1ππβ 1
(ππ3+1)πππΌπΌ+1 οΏ½ = ππππ30(π‘π‘,Ξ±)
y(π‘π‘,πΌπΌ) = πΏπΏβ1 οΏ½ 1(ππ3+1)πππΌπΌβ1οΏ½ = ππππ31(π‘π‘,Ξ±)
z(π‘π‘,πΌπΌ) = πΏπΏβ1 οΏ½ 1(ππ3+1)πππΌπΌ
οΏ½ = ππππ32(π‘π‘,Ξ±)
The solution set of the IVP (2.1)-(2.4) is,
{ x(π‘π‘,πΌπΌ) = ππππ30(π‘π‘,Ξ±), y(π‘π‘,πΌπΌ) = ππππ31(π‘π‘,Ξ±), z(π‘π‘,πΌπΌ) = ππππ32(π‘π‘,Ξ±) }
II. Second Order System of Differential Equation:
IVP-III: Consider the system of first order differential equations,
ππβ²β² (π‘π‘,πΌπΌ) β π¦π¦β²(π‘π‘,πΌπΌ) = 0 (3.1)
π¦π¦β²β² (π‘π‘,πΌπΌ) β ππβ²(π‘π‘,πΌπΌ) = π‘π‘πΌπΌβ1
π€π€(πΌπΌ) (3.2)
With initial conditions,
ππ(0,πΌπΌ) = 1, π¦π¦(0,πΌπΌ) = 0, ππβ²(0,πΌπΌ) = 0, π¦π¦β²(0,πΌπΌ) = 0 (3.3)
Apply Laplace transform to both the sides of differential equations (3.1) and (3.2),
We have,
πΏπΏ{ππβ²β² (π‘π‘,πΌπΌ)} β πΏπΏ{π¦π¦Μ(π‘π‘,πΌπΌ)} = 0
πΏπΏ{π¦π¦β²β² (π‘π‘,πΌπΌ)} β πΏπΏ{ππβ²(π‘π‘,πΌπΌ)} = πΏπΏ{ π‘π‘πΌπΌβ1
π€π€(πΌπΌ) }
ππ2πΏπΏ{ππ(π‘π‘,πΌπΌ)} β ππππ(0,πΌπΌ) β ππβ²(0,πΌπΌ) β πππΏπΏ{π¦π¦(π‘π‘,πΌπΌ)} + π¦π¦(π‘π‘,πΌπΌ) = 0
And ππ2πΏπΏ{π¦π¦(π‘π‘,πΌπΌ)} β πππ¦π¦(0,πΌπΌ) β π¦π¦β²(0,πΌπΌ) β πΏπΏ{ππ(π‘π‘,πΌπΌ)} + ππ(0,πΌπΌ) = 1π€π€(πΌπΌ)
π€π€(πΌπΌ)πππΌπΌ
Consider, πΏπΏ{ππ(π‘π‘,πΌπΌ)} = ππ(ππ,πΌπΌ), πΏπΏ{π¦π¦(π‘π‘,πΌπΌ)} = ππ(ππ,πΌπΌ) ,using initial condition (3.3)
We get the system of equations in X, Y
ππ2ππ(ππ,πΌπΌ) β ππππ(ππ,πΌπΌ) = ππ (3.4)
-ππππ(ππ,πΌπΌ) + ππ2ππ(ππ,πΌπΌ) = 1πππΌπΌβ 1 (3.5)
Multiply equation (3.5) by s and add with equation (3.4), we obtain
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οΏ½ππ3 β πποΏ½ππ(ππ,πΌπΌ) = 1πππΌπΌβ1
ππ(ππ,πΌπΌ) = 1(ππ2β1)πππΌπΌ
Taking inverse Laplace transform of both sides, we have,
y(π‘π‘,πΌπΌ) = πΏπΏβ1 οΏ½ 1(ππ2β1)πππΌπΌ
οΏ½ = ππππππππβ(π‘π‘,Ξ±)
Also multiply equation (3.4) by s and add with equation (3.5), we obtain
οΏ½ππ3 β πποΏ½ππ(ππ,πΌπΌ) = οΏ½ππ2 β 1οΏ½ + 1πππΌπΌ
X(ππ,πΌπΌ) = 1ππ
+ 1(ππ2β1)πππΌπΌ+1
Taking inverse Laplace transform of both sides, we have,
x(π‘π‘,πΌπΌ) = πΏπΏβ1 οΏ½ 1ππ
+ 1(ππ2β1)πππΌπΌ+1 οΏ½ = ππππππππβ(π‘π‘,Ξ±)
The solution set of the IVP (3.1)-(3.3) is {ππ(π‘π‘,πΌπΌ)} = ππππππππβ(π‘π‘,πΌπΌ), π¦π¦(π‘π‘,πΌπΌ) = ππππππππβ(π‘π‘,πΌπΌ)}
IVP-IV: Consider the system of second order differential equations
ππβ²β² (π‘π‘,πΌπΌ) β π§π§β²(π‘π‘,πΌπΌ) = 0 (4.1)
π¦π¦β²β² (π‘π‘,πΌπΌ) β ππβ²(π‘π‘,πΌπΌ) = π‘π‘πΌπΌβ1
π€π€(πΌπΌ) (4.2)
π§π§β²β² (π‘π‘,πΌπΌ) β π¦π¦β²(π‘π‘,πΌπΌ) = 0 (4.3)
With the initial conditions,
ππ(0,πΌπΌ) = 1, π¦π¦(0,πΌπΌ) = π§π§(0,πΌπΌ) = ππβ²(0,πΌπΌ) = π¦π¦β²(0,πΌπΌ) = π§π§β²(0,πΌπΌ) = 0 (4.4)
Apply Laplace transform to both the sides of differential equations (4.1), (4.2) and (4.3), we have,
πΏπΏ{ππβ²β² (π‘π‘,πΌπΌ)} β πΏπΏ{π§π§β²(π‘π‘,πΌπΌ)} = 0
πΏπΏ{π¦π¦β²β² (π‘π‘,πΌπΌ)} β πΏπΏ{ππβ²(π‘π‘,πΌπΌ)} = πΏπΏ{π‘π‘πΌπΌβ1
π€π€(πΌπΌ)}
πΏπΏ{π§π§β²β² (π‘π‘,πΌπΌ)} β πΏπΏ{π¦π¦β²(π‘π‘,πΌπΌ)} = 0
Suppose,
πΏπΏ{ππ(π‘π‘,πΌπΌ)} = ππ(ππ,πΌπΌ), πΏπΏ{π¦π¦(π‘π‘,πΌπΌ)} = ππ(ππ,πΌπΌ), πΏπΏ{π§π§(π‘π‘,πΌπΌ)} = ππ(ππ,πΌπΌ)
Using initial condition (4.4), we get the system of equations in X, Y and Z
ππ2ππ(ππ,πΌπΌ) β ππππ(ππ,πΌπΌ) = ππ
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βππππ(ππ,πΌπΌ) + ππ2ππ(ππ,πΌπΌ) = 1πππΌπΌβ 1
βππ(ππ,πΌπΌ) + ππ2ππ(ππ,πΌπΌ) = 0
We write this system of equations in matrix form as,
οΏ½ππ2 0 βππβππ ππ2 00 βππ ππ2
οΏ½ οΏ½ ππππππ
οΏ½ = οΏ½ππ
1πππΌπΌβ 10
οΏ½
We know by matrix theory, ππ = π·π·πππ·π·
, ππ = π·π·π¦π¦π·π·
, ππ = π·π·π§π§π·π·
(4.5)
Where,
π·π· = οΏ½ππ2 0 βππβππ ππ2 0 0 βππ ππ2
οΏ½ = ππ3(ππ3 β 1)
π·π·ππ = οΏ½ ππ 0 βππ
1πππΌπΌβ 1 ππ2 0
0 βππ ππ2
οΏ½ = ππ5 β ππ2 + ππ2
πππΌπΌ
π·π·π¦π¦ = οΏ½ ππ2 ππ βππ βππ 1
πππΌπΌβ 1 0
0 0 ππ2
οΏ½ = ππ4
πππΌπΌ
π·π·π§π§ = οΏ½ππ2 0 ππβππ ππ2 1
πππΌπΌβ 1
0 βππ 0οΏ½ = ππ3
πππΌπΌ
Substituting these values in equation (4.5), then we obtain
ππ(ππ,πΌπΌ) = 1ππ
+ 1(ππ3β1)πππΌπΌ+1
ππ(ππ,πΌπΌ) = 1(ππ3β1)πππΌπΌβ1
ππ(ππ,πΌπΌ) = 1(ππ3β1)πππΌπΌ
In each case, taking inverse Laplace transform of both sides, we have,
x(π‘π‘,πΌπΌ) = πΏπΏβ1 οΏ½ 1ππ
+ 1(ππ3β1)πππΌπΌ+1 οΏ½ = ππππ30(π‘π‘,Ξ±)
y(π‘π‘,πΌπΌ) = πΏπΏβ1 οΏ½ 1(ππ3β1)πππΌπΌβ1οΏ½ = ππππ31(π‘π‘,Ξ±)
z(π‘π‘,πΌπΌ) = πΏπΏβ1 οΏ½ 1(ππ3β1)πππΌπΌ
οΏ½ = ππππ32(π‘π‘,Ξ±)
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The solution set of the IVP (4.1)-(4.4) is,
{ x(π‘π‘,πΌπΌ) = ππππ30(π‘π‘,Ξ±), y(π‘π‘,πΌπΌ) = ππππ31(π‘π‘,Ξ±), z(π‘π‘,πΌπΌ) = ππππ32(π‘π‘,Ξ±) }
III. Third Order System of Differential Equation:
IVP-V: Consider the system of second order differential equations,
xβ²β²β² (π‘π‘,πΌπΌ) + π§π§β²β² (π‘π‘,πΌπΌ) = 0 (5.1)
π¦π¦β²β²β² (π‘π‘,πΌπΌ) β ππβ²β² (π‘π‘,πΌπΌ) = π‘π‘πΌπΌβ2
π€π€(πΌπΌβ1) (5.2)
π§π§β²β²β² (π‘π‘,πΌπΌ) β π¦π¦β²β² (π‘π‘,πΌπΌ) = 0 (5.3)
with the initial conditions,
ππ(0,πΌπΌ) = 1, π¦π¦(0,πΌπΌ) = π§π§(0,πΌπΌ) = ππβ²(0,πΌπΌ) = π¦π¦β²(0,πΌπΌ) = π§π§β²(0,πΌπΌ) = 0
ππβ²β² (0,πΌπΌ) = π¦π¦β²β² (0,πΌπΌ) = π§π§β²β² (0,πΌπΌ) = 0 (5.4)
Apply Laplace transform to both the sides of differential equations (5.1),(5.2) and (5.3), we have,
πΏπΏ{ππβ²β²β² (π‘π‘,πΌπΌ)} + πΏπΏ{π§π§β²β² (π‘π‘,πΌπΌ)} = 0
πΏπΏ{π¦π¦β²β²β² (π‘π‘,πΌπΌ)} β πΏπΏ{ππβ²β² (π‘π‘,πΌπΌ)} = πΏπΏ{ π‘π‘πΌπΌβ2
π€π€(πΌπΌβ1)}
πΏπΏ{π§π§β²β²β² (π‘π‘,πΌπΌ)} β πΏπΏ{π¦π¦β²β² (π‘π‘,πΌπΌ)} = 0
Suppose, πΏπΏ{ππ(π‘π‘,πΌπΌ)} = ππ(ππ,πΌπΌ), πΏπΏ{π¦π¦(π‘π‘,πΌπΌ)} = ππ(ππ,πΌπΌ), πΏπΏ{π§π§(π‘π‘,πΌπΌ)} = ππ(ππ,πΌπΌ)
Using initial condition (5.4),we get the system of equations in X, Y and Z
ππ3ππ(ππ,πΌπΌ) + ππ2ππ(ππ,πΌπΌ) = ππ2
βππ2ππ(ππ,πΌπΌ) + ππ3ππ(ππ,πΌπΌ) = 1πππΌπΌβ1 β ππ
ππ3ππ(ππ,πΌπΌ) β ππ2ππ(ππ,πΌπΌ) = 0
we write this system of equations in matrix form as,
οΏ½ππ3 0 ππ2
βππ2 ππ3 00 βππ2 ππ3
οΏ½ οΏ½ ππππππ
οΏ½ = οΏ½ππ2
1πππΌπΌβ1 β ππ
0οΏ½
we know by matrix theory,
ππ = π·π·πππ·π·
, ππ = π·π·π¦π¦π·π·
, ππ = π·π·π§π§π·π·
(5.5)
where,
D.B. Dhaigude, S.B. Dhaigude / J. Math. Computer Sci. 7 (2013), 293-304
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π·π· = οΏ½ππ3 0 ππ2
βππ2 ππ3 00 βππ2 ππ3
οΏ½ = ππ6(ππ3 + 1)
π·π·ππ = οΏ½ ππ2 0 ππ2
1πππΌπΌβ1 β ππ ππ3 0
0 βππ2 ππ3
οΏ½ = ππ8 + ππ5 β 1πππΌπΌβ5
π·π·π¦π¦ = οΏ½ ππ3 ππ2 ππ2 βππ2 1
πππΌπΌβ1 β ππ 0
0 0 ππ3
οΏ½ = ππ6
πππΌπΌβ1
π·π·π§π§ = οΏ½ππ3 0 ππ2
βππ2 ππ3 1πππΌπΌβ1 β ππ
0 βππ2 0
οΏ½ = ππ3
πππΌπΌβ1
Substituting these values in equation (5.5), then we obtain
ππ(ππ,πΌπΌ) = 1ππβ 1
(ππ3+1)πππΌπΌ+1
ππ(ππ,πΌπΌ) = 1(ππ3+1)πππΌπΌβ1
ππ(ππ,πΌπΌ) = 1(ππ3+1)πππΌπΌ
In each case, taking inverse Laplace transform of both sides, we have,
x(π‘π‘,πΌπΌ) = πΏπΏβ1 οΏ½ 1ππβ 1
(ππ3+1)πππΌπΌ+1 οΏ½ = ππππ30(π‘π‘,Ξ±)
y(π‘π‘,πΌπΌ) = πΏπΏβ1 οΏ½ 1(ππ3+1)πππΌπΌβ1οΏ½ = ππππ31(π‘π‘,Ξ±)
z(π‘π‘,πΌπΌ) = πΏπΏβ1 οΏ½ 1(ππ3+1)πππΌπΌ
οΏ½ = ππππ32(π‘π‘,Ξ±)
The solution set of the IVP (5.1)-(5.4) is,
{ x(π‘π‘,πΌπΌ) = ππππ30(π‘π‘,Ξ±), y(π‘π‘,πΌπΌ) = ππππ31(π‘π‘,Ξ±), z(π‘π‘,πΌπΌ) = ππππ32(π‘π‘,Ξ±) }
Conclusion: In this paper we have studied the system of Linear Differential Equations of first Order,
Second order and Third order in two variables as well as in three variables and the set of Prefunctions and extended Prefunctions are the closed form of solutions, Laplace Transforms method is successfully applied.
Acknowledgements: The second authors is thankful to Principal Dr.D.M.Yadav, JSPMs Bhivarabai
Savant Institute of Technology and Research (for women) Wagholi, Pune for the given the time and support to this work.
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