DOI : 10.17577/IJERTV2IS120767
- Open Access
- Total Downloads : 440
- Authors : Ranjit R. Dhunde, G. L. Waghmare
- Paper ID : IJERTV2IS120767
- Volume & Issue : Volume 02, Issue 12 (December 2013)
- Published (First Online): 18-12-2013
- ISSN (Online) : 2278-0181
- Publisher Name : IJERT
- License:
This work is licensed under a Creative Commons Attribution 4.0 International License
Double Laplace Transform & It’s Applications
Ranjit R. Dhunde1 And G. L. Waghmare2
1Department of Mathematics, Datta Meghe Institute of Engineering Technology & Research, Wardha (MH- India)
2Department of Mathematics, Institute of Science, Nagpur (MH- India)
Abstract: In this paper, we applied the method of Double Laplace Transform for solving the Partial Differential Equations, that is, one dimensional Wave & Heat equation.
Through this methodology we tried to prove that this method is very effective & convenient for solving Partial Differential Equation. The scheme is tested through some examples & the results demonstrate reliability.
Mathematics Subject Classifications: 44A30
Keywords: Double Laplace Transform, Inverse Laplace Transform, Heat equation, Wave equation, Partial derivatives.
-
Introduction
Integral Transform [1, 2] is one of the most known methods to solve partial differential equations. The Wave equation & Heat equation as the fundamental equations in mathematical Physics & occur in many branches of Physics, in Applied mathematics as well as in Engineering. Eltayeb and Kilicman [3] have applied the double Laplace transform to solve general linear telegraph and partial integro-differential equations.
In 2011 [4], Aghilli & Moghaddam proved certain Theorems on Two Dimensional Laplace Transform & applied on Non-Homogeneous parabolic Partial differential equations.
Recently in 2013 [5], R. R. Dhunde has discussed & proved different properties of Double Laplace Transform.
In this study, we use the Double Laplace Transform to solve a first & second order Partial Differential equation, specially one dimensional Wave & Heat equation. Through
this method the Partial Differential equation is solved without converting it into Ordinary Differential equation, therefore no need to find complete solution of Ordinary Differential equation. This is the biggest advantage of this method. Therefore Double Laplace Transform technique is very convenient & effective.
The scheme is tested through three different examples which are being referred from [6, 7].
Definition of double Laplace transform:
First of all, we recall the following definitions given by Estrin & Higgins [2].
Let f(x, t) be a function of two variables x and t, where x, t > 0. The double Laplace transform of f(x, t) is defined as
Lt Lx
f (x, t)
f ( p, s )
0
est
0
e p x
f (x, t) dx dt
(1)
whenever the improper integral converges. Here p, s are complex numbers.
Existence of double Laplace transforms:
Let f(x, t) be a continuous function on the interval [0, ) which is of exponential order, that is, for some a, b .
Consider Sup
(,) <
>0,>0 +
In this case, the double Laplace transform of f(x, t) that is
Lt Lx
f (x, t)
f ( p, s )
0
est
0
e p x
f (x, t) dx dt
exists for all > & > & is in fact infinitely differentiable with respect to >
& > .
All functions in this study are assumed to be of exponential order.
-
Double Laplace Transforms of Partial Derivatives:
Double Laplace Transform for first partial derivatives with respect to x is defined as follows:
L t L x (, ) = p (p, s) (0, s) (2)
Similarly, Double Laplace Transform for first partial derivatives with respect to t is given by
L t L x (, ) = s (p, s) (p, 0) (3)
Double Laplace Transform for second partial derivatives with respect to x is defined by
L t L x (, ) = = 2 p, s 0, s 0, (4)
In a similar manner, Double Laplace Transform for second partial derivatives with respect to t can be deduced from a single Laplace Transform
L t L x (, ) = 2 p, s p, 0 , 0 (5)
-
Applications of Double Laplace Transform:
-
Double Laplace Transform & First order Partial Differential Equation Example: Find the bounded solution of ux = 2ut + u, u(x, o) = 3 for x> 0, t > 0. Solution: Taking the Double Laplace Transform, we obtain
L t L x = L t L x 2 +
p (p, s) 0, = 2 p, s p, 0 + (p, s)
(p 2s 1) (p, s) = 0, 2 (p, 0) But u(x, 0) = 3 (p, 0) = 1
+3
(p, s) = 1
(p 2s 1)
0, 2
p 2s 1 (p+3)
(p, s) = 1
(p 2s 1)
-1
-1
Using Lx , we get
0, 1
+2
1
21
1
+3
(x, s) = 2+1 0, 1
+2
2+1 + 1
+2
3
(x, s) = 2+1 0, 1
+2
+ 1
+2
3 (6)
Now u(x, t) is bounded as x & hence (x, s) is bounded as x .
Hence from (6), 0, 1 = 0
+2
0, = 1
+2
Therefore, (x, s) = 1
+2
3
By Inverse Laplace Transform, we get bounded solution u(x, t) = 3 2 .
-
Double Laplace Transform & One Dimensional Heat Equation
Example: Solve = k , u(x, 0) = sin , u(0, t) = 0 , u(1, t) = 0, 0< < 1, > 0.
Solution: L t L x = k L t L x
s (p, s) , 0 = k 2 p, s p 0, s (0, s)
2 2
2 2
But u(0, t) = 0 0, s = 0 & u(x, 0) = sin p, 0 =
p +
2 2
2 2
s (p, s)
p +
= k 2 p, s (0, s)
p2+2
p2+2
(k 2 s) (p, s) = 0, s
2
2
(p, s) =
(k s)
0, s
0, s
k 2 s p2+2
=
=
1
)
)
k
k
( 2 s
0, s
k 2 s p2+2
k 2 s p2+2
k
= 1 s
0, s
1 1
k
k
k
k
( 2 k)
k s + 2
2 s
p2 +2
=
=
1
)
)
k
k
( 2 s
0, s
s +k 2 p2s
s +k 2 p2s
k
+
+
s +k 2 p2+2
= 1
s +k 2
s +k 2
+
0, s
+
+
s +k 2 p2+2
= 1
1
+
2
0, s
+
+
s +k 2 p2+2
s +k 2
s +k 2
-1
-1
Applying Lx , we get
(x, s)=
2
0, s
s +k 2
+ 1
2
2
s +k
sinx (7)
Taking limit as x 1
(1, s)=
2
0, s
s +k 2
+ 1
2
2
s +k
sin
But u(1, t) = 0 (1, s) = 0
0 =
2
0, s
s +k 2
s +k 2
s +k 2
0, s =
1(7) (x, s)=
(x, s)= 1
sinx
2
s +k 2
s +k 2
+ s +k 2
sinx
s +k 2
Applying Lt-1 , we get
(x, t)= e k2t sinx (8)
The problem can be interpreted physically. The given equation is the heat equation, where u(x, t) gives the temperature at a point x at time t. Consider a section bounded by planes x=0 & x=1 & the boundary conditions u(0, t)=0=u(1, t) give the temperature zero at the planes. The condition u(x, 0) = sinx indicates the initial temperature in 0< < 1. The u in (8) represents the temperature at time t> 0.
-
Double Laplace Transform & One Dimensional Wave Equation
Example: Semi-infinite String
Find the displacement w(x,t) of an elastic string subject to the following conditions:
-
The string is initially at rest on the x-axis from x=0 to .
-
For time t>0 the left end of the string is moved in a given fashion, namely,
w(0,t) = = sint if 0 t 2
0
-
Furthermore lim (, ) = 0 for t0.
-
-
Of course, there is no infinite string, but our model describes a long string or rope (of negligible weight) with its right end fixed far out on the x-axis.
Solution:
We have to solve the wave equation 2 = c2 2 where c2 =
2 2
For positive x & t, subject to the boundary conditions w(0,t) = f(t) , lim (, ) = 0 (t0)
with f as given above, & the initial conditions w(x,0) = 0,
t=0 = 0
we use the Double Laplace Transform on wave equation
L t L
{2} = c2L
2
2
x
x
t
t
L {2}
2
2
x
x
s2 (p, s) s (p, 0) t(p, 0) = c2 s2 w p, s p w 0, s x (0, s)
Given: w(x, 0) = 0 (p, 0) = 0 &
t=0
= 0
(p, 0) = 0
s2 (p, s) = c2 p2 w p, s p w 0, s wx (0, s)
(c2 p2 s2) (p, s) = c2p w 0, s + c2 x (0, s)
2 2 2
2 2 2
(p, s) = c2p
(c p s )
w 0, s + c2
2 2 2
2 2 2
(c p s )
x
(0, s) (9)
Now, w 0, s = w(0, t) = f(t) = (
) &
w x (0, s) = lim 0 w x p, s = lim
0
wx(x, t) =
lim
w
(x, t)dx
0 0
0 x
= w
(x, t)dx = lim
(, ) (0, )
0 0 x
0
0
0
= (0, ) = (
)
2 2 2
2 2 2
From (9), (p, s) = c2p
(c p s )
c2
(c2 p2 s2)
(
)
(p, s) = 1 1 + 1 2 1 1 (
)
2 +
+
By applying 1, we get
1
2
(x, s) =
2
+
()
(x, s) =
+
( ) (10)
where A(s) = 1 1 & B(s) = 1 1 +
2 2
Again since lim (, ) = 0 for t0 Therefore,
since lim
, = lim (, ) = lim
(, ) = 0
0
0
This implies A(s) = 0 in (10) because c>0, so that for every fixed positive s the function
increases as x increases.
Therefore, (10) (x, s) = ()
Since A(s) = 0 c = s Therefore B(s) = 1
2
1 + = 1
(x, s) = ()
Taking Lt-1, From second shifting theorem, we obtain w(x, t) = f(t – ) u(t – )
w(x, t) = sin(t – ) if < t < + 2 or ct> > 2
and zero otherwise.
This is a single sine wave travelling to the right with speed c.
Note that a point x remains at rest until t = , the time needed to reach that x if one starts
at t = 0 (starts of the motion of the left end) & travels with speed c.
References:
-
D. G. Duff, Transform Methods for solving Partial Differential Equations, Chapman and Hall/CRC, Boca Raton, F. L. 2004.
-
A. Estrin & T. J. Higgins, The Solution of Boundary Value Problems by Multiple Laplace Transformation, Journal of the Franklin Institute, 252 (2), 153 167, 1951.
-
H. Eltayeb & A. Kilicman, A Note on Double Laplace Transform and Telegraphic Equations, Abstract & Applied Analysis, Volume 2013.
-
A. Aghili, B. Parsa Moghaddam, Certain Theorems on Two Dimensional Laplace Transform and Non-Homogeneous Parabolic Partial Differential Equations, Surveys in Mathematics & Its Applications, Volume 6, 2011, 165-174.
-
R. R. Dhunde, N. M. Bhondge & P. R. Dhongle, Some Remarks on the Properties of Double Laplace Transforms, International Journal of Applied Physics & Mathematics, Vol. 3, No. 4, July 2013, 293-295.
-
Erwin Kreyszig, Advanced Engineering Mathematics, eight Edition, John Wiley & Sons (ASIA) Pte Ltd, 643-645, 2001.
-
Murray R. Spiegel, Theory & Problems of Laplace Transforms, Schaums Outlines series, McGraw Hill, 1965.