Lecture 15: The Laplace Transform, Part I
Definition and Basic Formulas
Winfried Just
Department of Mathematics, Ohio University
October 12, 2015
Winfried Just, Ohio University
MATH3400, Lecture 15: The Laplace transform I
Definition 7.1.1 of the Laplace transform
The Laplace transform is a method that allows us (among other
things) to reduce IVPs for linear DEs to solving algebraic equations.
Let f (t) be a function that is defined for t ≥ 0. Then the integral
R∞
L{f (t)} = 0 e −st f (t) dt
is said to be the Laplace transform of f , provided that the integral
converges.
L{f (t)} = F (s) is a function of another variable s.
For most reasonably nice functions f (t) the Laplace transform
exists for all s in an interval (a, ∞).
Ohio University – Since 1804
Winfried Just, Ohio University
Department of Mathematics
MATH3400, Lecture 15: The Laplace transform I
Example 1: L{1}
Definition of the Laplace transform: L{f (t)} =
L{1} =
R∞
0
R∞
0
e −st 1 dt (by the definition)
Making a substitution: u = −st, du = −sdt, dt =
L{1} =
−1
s
e −st f (t) dt.
R −∞
0
−du
s
e u du
Switching limits of integration and integrating:
L{1} =
1
s
R0
−∞ e
u
du =
1
s
e u ]0−∞
Expressing in terms of a limit:
L{1} = 1s (e 0 − limu→−∞ e u ) = 1s (1 − 0)
We finally obtain:
L{1} =
1
s
Ohio University – Since 1804
Winfried Just, Ohio University
Department of Mathematics
MATH3400, Lecture 15: The Laplace transform I
Basic Laplace transforms: The first item
L{1} =
1
s
Ohio University – Since 1804
Winfried Just, Ohio University
Department of Mathematics
MATH3400, Lecture 15: The Laplace transform I
Example 2: L{e at }
Definition of the Laplace transform: L{f (t)} =
R∞
R∞
L{e at } = 0 e −st e at dt = 0 e −(s−a)t dt
Making a substitution:
u = −(s − a)t, du = −(s − a)dt, dt =
L{e at } =
−1
s−a
R −∞
0
R∞
0
e −st f (t) dt.
−du
s−a
e u du
Switching limits of integration and integrating:
R0
1
1
u
u 0
L{e at } = s−a
−∞ e du = s−a e ]−∞
Expressing in terms of a limit:
L{e at } =
1
0
s−a (e
We finally obtain:
− limu→−∞ e u ) =
L{e at } =
1
s−a (1
− 0)
1
s−a
Ohio University – Since 1804
Winfried Just, Ohio University
Department of Mathematics
MATH3400, Lecture 15: The Laplace transform I
Basic Laplace transforms: The first two items
L{1} =
1
s
L{e at } =
1
s−a
Note that for a = 0 we get e 0t = e 0 = 1,
and the first item becomes a special case of the second.
Ohio University – Since 1804
Winfried Just, Ohio University
Department of Mathematics
MATH3400, Lecture 15: The Laplace transform I
Example 3: L{sin kt}
L{sin kt} =
R∞
0
e −st sin kt dt (by definition)
Integration by parts:
R∞
0
u dv = uv ]∞
0 −
R∞
0
v du
Let u = sin kt, du = k cos kt dt, dv = e −st dt, v =
L{sin kt} =
−1 −st
s e
sin kt
∞
0
+
k
s
R∞
0
−1 −st
.
s e
e −st cos kt dt
Doesn’t look like much progress, but lets try again:
R∞
0
e −st cos kt dt =
R∞
0
u1 dv by parts:
Let u1 = cos kt, du1 = −k sin kt dt, dv = e −st dt, v =
R∞
0
e −st cos kt dt =
−1 −st
s e
cos kt
∞
0
−
k
s
R∞
0
−1 −st
.
s e
e −st sin kt dt
Bummer! We are back to Square 1.
Ohio University – Since 1804
Winfried Just, Ohio University
Department of Mathematics
MATH3400, Lecture 15: The Laplace transform I
Example 3 continued: Just when we’d like to give up . . .
Instead of giving up, let’s look at what we have got and substitute:
R
−st sin kt ∞ + k ∞ e −st cos kt dt
L{sin kt} = −1
s e
s 0
0
∞
R ∞ −st
R∞
cos kt dt = 1s e −st cos kt 0 − ks 0 e −st sin kt dt
0 e
−st sin kt ∞ + −k e −st cos kt ∞ − k 2 L{sin kt}
L{sin kt} = −1
s e
s2
s2
0
0
Moving the last term to the right:
2
−st sin kt ∞ + −k e −st cos kt ∞
L{sin kt} + ks 2 L{sin kt} = −1
s e
s2
0
0
2
∞
∞
−st sin kt
1 + ks 2 L{sin kt} = −1
+ −k
e −st cos kt 0
s e
s2
0
∞ −k −st
∞ s2
−1 −st
L{sin kt} = s 2 +k 2 s e
sin kt 0 + s 2 e
cos kt 0
We got a result!
Ohio University – Since 1804
Winfried Just, Ohio University
Department of Mathematics
MATH3400, Lecture 15: The Laplace transform I
Example 3 completed: Simplifying our result
We have show that
2
−st sin kt ∞ +
L{sin kt} = s 2 s+k 2 −1
s e
0
k −st
e
s2
cos kt
∞ 0
This result doesn’t look particularly friendly; let’s simplify it.
−1 −st
s e
sin kt
∞
0
=
−1
s
limt→∞ e −st sin kt − e −s0 sin k0 = 0
Similarly,
−k −st
e
s2
cos kt
∞
0
=
−k
s2
limt→∞ e −st cos kt − e −s0 cos k0 =
k
s2
We conclude that
L{sin kt} =
s2
s 2 +k 2
0+
k
s2
=
k
s 2 +k 2
Ohio University – Since 1804
Winfried Just, Ohio University
Department of Mathematics
MATH3400, Lecture 15: The Laplace transform I
Basic Laplace transforms: Adding two items
L{1} =
1
s
L{e at } =
1
s−a
k
s 2 +k 2
s
= s 2 +k
2
L{sin kt} =
L{cos kt}
The result for L{cos kt} can be obtained by a similar calculation as
we performed for L{sin kt}, but we will shortly see a simpler
derivation.
Ohio University – Since 1804
Winfried Just, Ohio University
Department of Mathematics
MATH3400, Lecture 15: The Laplace transform I
Linearity of the Laplace transform
Suppose f (t), g (t) are functions that have Laplace transforms, and
let α, β be constants. Then
L{αf (t) + βg (t)} = αL{f (t)} + βL{g (t)}
Example 4: L{sinh kt} = L{ e
= 21 L{e kt } +
−1
−kt }
2 L{e
=
kt −e −kt
2
1
2(s−k)
+
} = L{ 21 e kt +
−1
2(s+k)
=
−1 −kt
}
2 e
k
s 2 −k 2
Example 5: L{sin(−kt)} = L{− sin kt} = −L{sin kt} =
Example 6: L{cos(−kt)} = L{cos kt} =
−k
s 2 +k 2
s
s 2 +k 2
The last two examples give a helpful hint for memorizing which
formula of the Laplace transform goes with sin kt and which with
cos kt: The former must change signs when k is replaced by −k,
the latter must stay the same.
Ohio University – Since 1804
Winfried Just, Ohio University
Department of Mathematics
MATH3400, Lecture 15: The Laplace transform I
Basic Laplace transforms: Adding two more items
L{1} =
1
s
L{e at } =
1
s−a
k
s 2 +k 2
s
L{cos kt} = s 2 +k
2
k
L{sinh kt} = s 2 −k 2
s
L{cosh kt} = s 2 −k
2
L{sin kt} =
The last formula follows from the definition
cosh kt =
e kt +e −kt
2
and linearity of the Laplace transform.
Ohio University – Since 1804
Winfried Just, Ohio University
Department of Mathematics
MATH3400, Lecture 15: The Laplace transform I
The Laplace transform of the first derivative
Assume f (t) is differentiable and has a Laplace transform. How is
the Laplace transform of a function f (t) related to the Laplace
transform of its derivative f 0 (t)?
Consider L{f 0 (t)} =
R∞
0
e −st f 0 (t) dt for fixed s.
For u = e −st dt and dv = f 0 (t) dt we get du = −se −st dt
and v = f (t). Integration by parts gives
R ∞ −st 0
R∞
−st f (t) dt
f (t) dt = e −st f (t)|∞
0 − 0 −se
0 e
R∞
= limt→∞ e −st f (t) − e −0s f (0) + s 0 e −st f (t) dt
= 0 − f (0) + sL{f (t)}
Note that limt→∞ e −st f (t) = 0 whenever
R∞
L{f (t)} = 0 e −st f (t) dt exists.
Ohio University – Since 1804
Winfried Just, Ohio University
Department of Mathematics
MATH3400, Lecture 15: The Laplace transform I
The Laplace transform of the first derivative: Examples
We have derived the equation
L{f 0 (t)} = sL{f (t)} − f (0).
Example 7: Let f (t) = e at .
Then L{f (t)} =
1
s−a
sL{f (t)} − f (0) =
and L{f 0 (t)} = L{ae at } =
s
s−a
−1=
s
s−a
−
s−a
s−a
=
a
s−a
a
s−a
by linearity.
= L{f 0 (t)}.
So this works as expected.
Example 8: Let f (t) = sin kt.
Then L{f (t)} =
k
s 2 +k 2
and
L{f 0 (t)} = L{k cos kt} = kL{cos kt} by linearity.
L{f 0 (t)} = sL{sin kt} − sin 0 =
sk
s 2 +k 2
s
− 0 = k s 2 +k
2 = kL{cos kt}.
This gives an easy derivation of the formula for L{cos kt}.
Ohio University – Since 1804
Winfried Just, Ohio University
Department of Mathematics
MATH3400, Lecture 15: The Laplace transform I
The Laplace transform of power functions
We have derived the equation
L{f 0 (t)} = sL{f (t)} − f (0).
Example 9: Let f (t) = t α+1 , where α > −1.
Then f 0 (t) = (α + 1)t α and f (0) = 0.
L{(α + 1)t α } = L{f 0 (t)} = sL{f (t)} − f (0) = sL{t α+1 }.
It follows from linearity that L{t α+1 } =
α+1
α
s L{t }.
We know L{t 0 } = L{1} = 1s .
1
1
It follows that L{t} = L{t 0+1 } = 0+1
s s = s2 .
1
2
It follows that L{t 2 } = L{t 1+1 } = 1+1
s s2 = s3 .
2+1
2
It follows that L{t 3 } = L{t 2+1 } = s s 3 = s64 .
6
24
It follows that L{t 4 } = L{t 3+1 } = 3+1
s s4 = s5 .
This is getting boring . . .
·n
n!
It follows that L{t n } == 1·2·3·...
= s n+1
.
s n+1
Ohio University – Since 1804
Winfried Just, Ohio University
Department of Mathematics
MATH3400, Lecture 15: The Laplace transform I
The Gamma-function
This works beautifully when α = n is a nonnegative integer.
L{t n+1 } =
(n+1)!
s n+2
=
n+1 n!
s s n+1
=
n+1
n
s L{t }.
For arbitrary α > 1 we have L{t α } = Γ(α+1)
, where
s α+1
R ∞ x−1 −t
Γ(x) = 0 t
e dt is called the Gamma function.
Properties of this function are outlined in Appendix I of your
textbook.
Most notably:
Γ(x + 1) = xΓ(x) for every x > 0.
Γ(n + 1) = n! for every nonnegative integer n.
Ohio University – Since 1804
Winfried Just, Ohio University
Department of Mathematics
MATH3400, Lecture 15: The Laplace transform I
Basic Laplace transforms: The complete list
L{1} =
1
s
n!
for n = 0, 1, 2, 3, . . .
s n+1
Γ(α+1)
L{t α } = s α+1 for arbitrary α > −1
1
L{e at } = s−a
k
L{sin kt} = s 2 +k
2
s
L{cos kt} = s 2 +k 2
k
L{sinh kt} = s 2 −k
2
s
L{cosh kt} = s 2 −k 2
L{t n } =
Ohio University – Since 1804
Winfried Just, Ohio University
Department of Mathematics
MATH3400, Lecture 15: The Laplace transform I
The Laplace transform of higher-order derivative
Now consider L{f 00 (t)} = L{(f 0 )0 (t)}.
From the equation L{f 0 (t)} = sL{f (t)} − f (0) we get
L{f 00 (t)} = sL{f 0 (t)} − f 0 (0) = s(sL{f (t)} − f (0)) − f 0 (0).
It follows that
L{f 00 (t)} = s 2 L{f (t)} − sf (0) − f 0 (0).
In general,
L{f (n) (t)} = s n L{f (t)} − s n−1 f (0) − s n−2 f 0 (0) − · · · − f (n−1) (0).
Note that the terms in the last expression are of the form
−s n−k f (k−1) (0) for k = 1, . . . n, where s 0 = 1 and f (0) = f .
Ohio University – Since 1804
Winfried Just, Ohio University
Department of Mathematics
MATH3400, Lecture 15: The Laplace transform I