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

ℒ⁢{f′⁢(t)}=s⁢F⁢(s)-limt→0+⁡f⁢(t). (1)

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

ℒ⁢{f′′⁢(t)}=s⁢[s⁢F⁢(s)-limt→0+⁡f⁢(t)]-limt→0+⁡f′⁢(t)=s2⁢F⁢(s)-s⁢f⁢(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

ℒ⁢{f′′′⁢(t)}=s⁢[s2⁢F⁢(s)-s⁢f⁢(0+)-f′⁢(0+)]-f′′⁢(0+)=s3⁢F⁢(s)-s2⁢f⁢(0+)-s⁢f′⁢(0+)-f′′⁢(0+).

Continuing similarly, one comes to the general formula

ℒ⁢{f(n)⁢(t)}=sn⁢F⁢(s)-sn-1⁢f⁢(0+)-sn-2⁢f′⁢(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

ℒ⁢{f′⁢(t)}=s⁢F⁢(s)-f⁢(0+)-e-a⁢s⁢(lims→a+⁡f⁢(s)-lims→a-⁡f⁢(s)).

Application.  Derive the Laplace transform of sin⁡a⁢t using the derivatives of sine (cf. Laplace transform of cosine and sine).

We have

f⁢(t):=sin⁡a⁢t,f′⁢(t)=a⁢cos⁡a⁢t,f′′⁢(t)=-a2⁢sin⁡a⁢t.

Using (2) with  n=2  we obtain

ℒ⁢{-a2⁢sin⁡a⁢t}=s2⁢ℒ⁢{sin⁡a⁢t}-s⁢sin⁡0-a⁢cos⁡0,

i.e.

-a2⁢ℒ⁢{sin⁡a⁢t}=s2⁢ℒ⁢{sin⁡a⁢t}-a,

which implies

ℒ⁢{sin⁡a⁢t}=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