Laplace transforms of derivatives

where

As shown in the parent entry (http://planetmath.org/LaplaceTransformOfDerivative), the Laplace transformDlmfMathworldPlanetmath of the first derivativeMathworldPlanetmath of a Laplace-transformable functionMathworldPlanetmath f(t) is got from the formula

{f(t)}=sF(s)-limt0+f(t). (1)

The rule can be applied also to the function f(t):

{f′′(t)}=s[sF(s)-limt0+f(t)]-limt0+f(t)=s2F(s)-sf(0+)-f(0+)

Here the short notation 0+ has been used for the right limits.

Further, one can use the rule to f′′(t), getting

{f′′′(t)}=s[s2F(s)-sf(0+)-f(0+)]-f′′(0+)=s3F(s)-s2f(0+)-sf(0+)-f′′(0+).

Continuing similarly, one comes to the general formula

{f(n)(t)}=snF(s)-sn-1f(0+)-sn-2f(0+)--f(n-1)(0+). (2)

Use of (2) requires that f(t), f(t), f′′(t), …, f(n)(t) are Laplace-transformable and that f(t), f(t), f′′(t), …, f(n-1)(t) are continuousMathworldPlanetmath when  t>0 (not only piecewise continuous).

Remark.  Suppose that f(t) and f(t) are Laplace-transformable and that f(t) is continuous for  t>0  except the point  t=a  where the function has a finite jump discontinuity.  Then

{f(t)}=sF(s)-f(0+)-e-as(limsa+f(s)-limsa-f(s)).

Application.  Derive the Laplace transform of sinat using the derivatives of sine (cf. Laplace transform of cosine and sine).

We have

f(t):=sinat,f(t)=acosat,f′′(t)=-a2sinat.

Using (2) with  n=2  we obtain

{-a2sinat}=s2{sinat}-ssin0-acos0,

i.e.

-a2{sinat}=s2{sinat}-a,

which implies

{sinat}=as2+a2.
Title Laplace transforms of derivatives
Canonical name LaplaceTransformsOfDerivatives
Date of creation 2014-04-06 8:24:33
Last modified on 2014-04-06 8:24:33
Owner pahio (2872)
Last modified by pahio (2872)
Numerical id 6
Author pahio (2872)
Entry type Topic
Classification msc 44A10