Session 8a
Laplace Transform Method
Ernesto Gutierrez-Miravete
March 9, 2000
1 Introduction
Consider a real function F(t). The Laplace transform of F(t),
L[F(t)] = [`F](s) is defined as
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| L[F(t)] = |
_
F
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(s) = |
ó
õ
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¥
t¢ = 0
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e-s t¢ F(t¢) dt¢ |
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The inverse transform (to recover F(t) from L[F(t)]) is
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| F(t) = |
1
2 pi
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ó
õ
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g+ i¥
s = g- i ¥
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e s t |
_
F
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(s) ds |
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where g is large enough that all the singularities of
[`F](s) lie to the left of the imaginary axis.
For the transform and inverse to exist, F(t) must be at least
piecewise continuous, of exponential order and tn F(t) must be bounded
as t ® 0+.
Laplace transforms have the following properties:
Linearity.
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| L [c1 F(t) + c2 G(t)] = c1 |
_
F
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(s) + c2 |
_
G
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(s) |
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Transforms of Derivatives.
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| L [F¢(t)] = s |
_
F
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(s) - F(0) |
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| L [F¢¢(t)] = s2 |
_
F
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(s) - s F(0) - F¢(0) |
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| L [F¢¢¢(t)] = s3 |
_
F
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(s) - s2 F(0) - s F¢(0) - F¢¢(0) |
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| L [F(n)(t)] = sn |
_
F
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(s) - sn-1 F(0) - sn-2 F(1)(0) - sn-3 F(2)(0) - ... - F¢(n-1)(0) |
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Transforms of Integrals.
Let g(t) = ò0t F(t) d t (i.e. g¢(t) = F(t)).
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| L[ |
ó
õ
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t
0
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F(t)] = |
1
s
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_
F
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(s) |
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| L[ |
ó
õ
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t
0
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ó
õ
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t2
0
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F(t1) dt1 dt2] = |
1
s2
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_
F
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(s) |
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| L[ |
ó
õ
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t
0
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... |
ó
õ
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tn
0
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F(t1) dt1 ... dtn] = |
1
sn
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_
F
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(s) |
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Scale change.
Let a > 0 be a real number.
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| L [F( |
1
a
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t)] = a |
_
F
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(a s) |
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Shift.
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| L [e±at F(t)] = |
_
F
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(s -±a) |
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Transform of translated functions.
The translated function is
where
The Laplace transform of the translated function is
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| L [U(t-a) F(t-a)] = e-a s |
_
F
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(s) |
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Also
Transform of the d function.
Transform of convolution.
The convolution of two functions f(t) and g(t) is
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| f*g = |
ó
õ
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t
0
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f(t-t) g(t) dt = g*f |
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The Laplace transform is
Derivatives.
If [`F](s) is the Laplace transform of F(t) then
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ds
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= |
_
F
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¢(s) = L[(-t) F(t)] |
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dsn
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= |
_
F
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¢(s) = L[(-t)n F(t)] |
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Integrals.
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ó
õ
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¥
s
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_
F
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(s¢) ds¢ = L[ |
F(t)
t
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] |
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2 Inversion of Laplace Transforms
The inverse Laplace transform of L[F(t)] is
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| F(t) = |
1
2 pi
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|
ó
õ
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g+ i¥
s = g- i ¥
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e s t |
_
F
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(s) ds |
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The integration must be carried out in the complex domain and requires
a basic understanding of analytic functions of a complex variable and
residue calculus. Fortunately, inversion formulae for many transforms
have been computed and are readily available in tabular form.
See Table 7.1, pp. 268-271.
3 Applications
Consider the problem of finding T(x,t) in 0 £ x < ¥
satisfying
and subject to
and
Taking Laplace transforms one gets
The solution of this problem is
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_
T
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(x,s) = |
_
f
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(s) e-x Ö{s/a} = |
_
f
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(s) |
_
g
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(x,s) = L[f(t)*g(x,t)] |
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Inversion then produces
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| T(x,t) = f(t)*g(x,t) = |
ó
õ
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t
0
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f(t) g(x,t-t) dt |
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Inversion of [`g](x,s) to get g(x,t) finally gives
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| T(x,t) = |
x
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ó
õ
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t
t = 0
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f(t)
(t-t)3/2
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exp[ - |
x2
4 a(t-t)
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] dt |
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If f(t) = T0 = constant, the solution is
Consider now the transient problem of quenching a solid sphere (radius b)
initially at T0 by maintaining its surface at zero. The Laplace transform
method can be used to find and expression for T(r,t).
The transformed problem is
the solution of which is
To invert this the trigonometric functions are
expressed as asymptotic series and then inverted term by term. The final
result is
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| T(r,t) = T0 - |
b T0
r
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¥
å
n = 0
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{ erfc( |
b(2n+1) - r
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) - erfc( |
b(2n+1) + r
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)} |
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File translated from TEX by TTH, version 2.34.
On 9 Mar 2000, 21:22.