Applied Mathematical Sciences, Vol. 7, 2013, no. 90, 4455 - 4460 HIKARI Ltd, www.m-hikari.com http://dx.doi.org/10.12988/ams.2013.36301
The Laplace Transform of Derivative Expressed by Heaviside Function Ingoo Cho and Hwajoon Kim* Incheon National University 119 Academy-ro, Songdo-dong, Incheon, Korea
[email protected] *
[email protected] c 2013 Ingoo Cho and Hwajoon Kim. This is an open access article disCopyright tributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
Abstract We have showed that the Laplace transform of derivative can be expressed by an infinite series or Heaviside function. Related to this topic, the proposed idea can be also applied to other transforms(Sumudu/Elzaki).
Mathematics Subject Classification: 44A10, 34A12 Keywords: Laplace transform of derivative, integral transform, Heaviside function
1
Introduction
Integral transform methods have been researched to solve many problems in the differential equations with initial or boundary conditions[1-12]. Laplace, Sumudu and Elzaki transforms are such typical things[3-10, 14-15]. Among these, the Laplace transform method is a powerful and primary tool, and this plays a role to solve directly initial value problems without first determining a general solution, and nonhomogeneous ordinary differential equations(ODE) without first solving the corresponding homogeneous ODE[12]. The Laplace transforms of derivatives have been researched in many ways to solve differential equations. The principal contents are £(f ) = s£(f ) − f (0)
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Ingoo Cho and Hwajoon Kim
£(f ) = s2 £(f ) − sf (0) − f (0) for the Laplace transform of the first and second derivatives of f (t). In this article, we would like to propose the new approach of £(f ) by changing the choice of function of differential form in integration by parts. The obtained result is £(f ) can be represented by an infinite series or Heaviside function.
2
Preliminary Notes
Definition 2.1 If f (t) is a function defined for all t ≥ 0, its Laplace transform is the integral of f (t) times e−st form t = 0 to ∞. It is a function of s, and is defined by £(f ); thus F (s) = £(f ) =
∞ 0
e−st f (t)dt
provided the integral of f (t) exists[12]. In the above equation, if the kernel be changed to 1 −t/u e /ue−t/u , u we call Sumudu/Elzaki transform, respectively. The detailed contents in Sumudu/Elzaki transform can be find in [15]/[6], respectively. Definition 2.2 A function f (t) has a Laplace transform if it satisfies the growth restriction f (t) ≤ Mekt for all t ≥ 0 and some constant M and k.
3
Main Results
We would like to propose £(y ) can be represented as an infinite series of sk by changing the choice of function of differential form in integration by parts, and deal with the expression by Heaviside function of it. Theorem 3.1 The Laplace transform of the first derivative of y(t) satisfies £(y ) =
n
1 1 (k) y (0) + n £(y (n+1) ) k s k=1 s
4457
Laplace transform of derivative
for y (k) is the k-th derivative of a given function y(t). As n → ∞, we have £(y ) =
n
1 (k) y (0). k k=1 s
The above formula holds if y(t) and y (t) are continuous for all t ≥ 0 and satisfies the growth restriction. Proof. We would like to establish the validity of the statement by the mathematical induction. For k = 1, by integration by parts, £(y ) =
∞ 0
e−st y (t)dt
1 ∞ −st 1 e y (t)dt = [− y (t)e−st ]∞ 0 + s s 0 1 1 = y (0) + £(y ) s s holds. Next, we suppose that £(y ) =
m
1 (k) 1 y (0) + m £(y (m+1) ), k s s k=1
(∗)
and show that £(y ) can be expressed by
£(y ) =
m+1 k=1
1 (k) 1 y (0) + £(y (m+2) ). k m+1 s s
In (*), £(y (m+1) ) =
∞ 0
e−st y (m+1)(t) dt
1 1 ∞ −st (m+2) = − e−st y (m+1) (t)]∞ e y (t)dt 0 + s s 0 1 1 = y (m+1) (0) + £(y (m+2)). s s Hence, from (*), £(y ) =
m
1 (k) 1 1 1 y (0) + m [ y (m+1)(0) + £(y (m+2) )] k s s s k=1 s =
m+1 k=1
1 (k) 1 y (0) + m+1 £(y (m+2) ). k s s
Thus, if the equality holds for k, it holds for k + 1. Therefore, by mathematical induction, the equality is true for all natural number n.
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Ingoo Cho and Hwajoon Kim
Theorem 3.2
−ns
£(y ) = e
n
y(n) − y(0) + s
0
e−st y(t)dt + £[y (t)u(t − n)]
for all n and for u is the unit step function. Proof. We would like to verify by mathematical induction. If n = 1, ∞
£(y ) = −st
= [e
y(t)]10
+s
1
−s
−st
e
0
= e y(1) − y(0) + s
0
1 0
e−st y (t)dt
y(t)dt +
∞ 1
e−st y (t)dt
e−st y(t)dt + £[y (t)u(t − 1)].
Next, we assume that the equality holds for n = k i.e.,
−ks
£(y ) = e
y(k) − y(0) + s
k 0
e−st y(t)dt + £[y (t)u(t − k)].
(∗∗)
Let us show that £(y ) = e−(k+1)s y(k + 1) − y(0) + s
k+1 0
e−st y(t)dt + £[y (t)u(t − k − 1)].
From (**),
−ks
£(y ) = e
y(k) − y(0) + s
Here,
k 0
∞
=
k+1 k
k
−st
e
+s = [e−st y(t)]k+1 k
y (t)dt +
k+1 k
= e−s(k+1) y(k +1)−e−sk y(k)+s
k
e−st y (t)dt.
∞ k+1
e−st y (t)dt
e−st y(t)dt + £[y (t)u(t − k − 1)]
k+1 k
e−st y(t)dt+£[y (t)u(t−k −1)]. (∗ ∗ ∗∗)
Substituting (****) to (***), £(y ) = e−ks y(k) − y(0) + s −e−sk y(k) + s
k+1 k
(∗ ∗ ∗)
e−st y (t)dt
−st
e
y(t)dt +
∞
k 0
e−st y(t)dt + e−s(k+1) y(k + 1)
e−st y(t)dt + £[y (t)u(t − k − 1)]
4459
Laplace transform of derivative −(k+1)s
=e
y(k + 1) − y(0) + s
k+1 0
e−st y(t)dt + £[y (t)u(t − k − 1)].
The validity of the equality for all natural number n follows by mathematical induction. It is clear that Theorem 3.2 be rewritten by £(y ) = e−ns y(n) − y(0) + s
n 0
e−st y(t)dt
for t < n. With the proposed idea, other integral transform, such as Sumudu/Elzaki, can be modified by a similar form.
ACKNOWLEDGEMENTS. This work was supported by the University of Incheon Reserch Grant in 2013.
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