Mathematical Methods for Physicists: A concise introduction - Site Map

THE LAPLACE TRANSFORMATION
But f …u ‡ P† ˆf …u†; f …u ‡ 2P† ˆf …u†; etc: Also, let us replace the dummy
variable u by x, then the above equation becomes
Lf…x† ‰ Š ˆ
ˆ
Z P
0
Z P
0
e px f …x†dx ‡
Z P
0
e p…x‡P† f …x†dx ‡
Z P
0
e p…x‡2P† f …x†dx ‡
Z P
Z P
e px f …x†dx ‡ e pP e px f …x†dx ‡ e 2pP e px f …x†dx ‡
ˆ…1 ‡ e pP ‡ e 2pP ‡†
Z
1 P
ˆ
1 e pP e px f …x†dx:
0
0
Z P
0
e px f …x†dx
0
Laplace transforms of derivatives
If f …x† is a continuous for x 0, and f 0 …x† is piecewise continuous in every ®nite
interval 0 x k, and if j f …x†j Me bx (that is, f …x† is of exponential order),
then
L‰ f 0 …x†Š ˆ pL‰ f …x† Šf …0†; p > b:
We may employ integration by parts to prove this result:
Z
Z
udv ˆ uv vdu with u ˆ e px ; and dv ˆ f 0 …x†dx;
L‰ f 0 …x†Š ˆ
Z 1
0
Z 1
e px f 0 …x†dx ˆ‰e px f …x†Š 1 0 …p†e px f …x†dx:
Since j f …x†j Me bx for suciently large x, then j f …x†e px jMe …bp† for suciently
large x. If p > b, then Me …bp† ! 0 as x !1; and e px f …x† !0 as
x !1. Next, f …x† is continuous at x ˆ 0, and so e px f …x† !f …0† as x ! 0.
Thus, the desired result follows:
L‰ f 0 …x†Š ˆ pL‰ f …x†Š f …0†; p > b:
This result can be extended as follows:
If f …x† is such that f …n1† …x† is continuous and f …n† …x† piecewise continuous in
every interval 0 x k and furthermore, if f …x†; f 0 …x†; ...; f …n† …x† are of
exponential order for 0 > k, then
L‰ f …n† …x†Š ˆ p n L‰ f …x†Š p n1 f …0†p n2 f 0 …0†f …n1† …0†:
0
Example 9.6
Solve the initial value problem:
y 00 ‡ y ˆ 0; y…0† ˆy 0 …0† ˆ0; and f …t† ˆ0 for t < 0 but f …t† ˆ1 for t 0:
382