Matrices and Tensors - Continuum Mechanics of Anisotropic Materials - page 410

ð

t

f ð t Þ g ð t Þ¼ f ð t x Þ g ð x Þ

d x

:

(A.217)

0

It can be shown that the convolution is commutative ( f g ¼ g f ), associative

ðð h ). The

Laplace transform of a convolution is the product of the Laplace transforms, thus

f g Þ h ¼ f ð g h Þ ), and distributive ( f ð g þ h Þ¼ f g þ f

Þg ¼ f L f f ð t Þ g ð t ð s Þ~ g ð s Þ: (A.218)

Example A.16.2

Problem : Solve the differential equation d d t 0 in the general

case where q ( t ) is not specified, other than it has a Laplace transform.

5 y ¼ q ð t Þ for y (0) ¼

Solution: Noting that the Laplace transform of e at

1

sþa and that the Laplace
transform of the derivative of a function is equal to s times the Laplace transform of
the function minus the value of the function at t

is

¼ 0, the Laplace transform of

Þ¼ q ð s Þ
ðs 5 Þ

d y

d t

5 y ¼ q ð t Þ is sy ð s Þ 5 y ð s Þ¼q ð s Þ , thus y ð s .

is the product of two transformed functions like those that
appear on the right-hand side of (A.218) above, thus from (A.218) one can see by
partial fractions that

Observe that y ~ ð s Þ

< =

; ¼

ð ð

;

t t

q ~ ð s Þ q ~ ð s Þ

e 5 ðtxÞ q e 5 ðtxÞ q L 1 L ð x Þ d x or ð x Þ d x ¼

:

ð s 5 Þ ð s 5 Þ

0 0

Þ¼ Ð

t

thus y

ð t

e 5 ðtxÞ q

ð x Þ

d x . This is a general integral to the differential equation for

0

any function q ( t ).
In the material above it was assumed for simplicity that the functions were
continuous in the interval 0 . However, one of the most attractive features

of using Laplace transforms is their simplicity in dealing with functions that contain
discontinuities. The unit or Heaviside step function h ( t

t 1

t o ) is defined as 0 for t < t o

and as 1 for t > t o . Since the function jumps from the value 0 to the value 1 on

passage through t ¼ t o , the approach of h ( t t o ) to the value t o is therefore different

from below than it is from above,

t o Þ¼ t o Þ¼

h ð lim h ð t Þ¼ 0 ; h ð lim

t>t o

h ð t Þ¼ 1 : (A.219)

t > t o

The function h ( t t o ) is multiple valued at t ¼ t o and its value there depends on

how the point t t o is approached. The Laplace transform of the discontinuous

function, because it integrates the function, removes the discontinuity, thus

¼