#21 (11/12/2025)

Laplace transforms

Formulae of the Fourier transform1 and the inverse Fourier transform are listed here for reference as

F(x)
=
1
2π
 ⌠
⌡ 		∞

−∞ 
F
 
 (ω) ei ωx dω,
F
 
 (ω)
=
⌠
⌡ 		∞

−∞ 
 F(x) e−i ωx dx.
(1)
The Laplace transform can be derived as a special case of the Fourier transforms by limiting the range of f(x) over (0, ∞). By combining Eq. (1) and Eq. (1), one obtains
F(x) = 1
2 π
 ⌠
⌡ 		∞

−∞ 
 ⎛
⎝ 		⌠
⌡ 		∞

−∞ 
 F(y) e−i ωy dy⎞
⎠ 		ei ωx dω.
(2)
laplace1.jpg
If we require that

F(x) = ⎧
⎪
⎨
⎪
⎩
e−γx f(x)
    x > 0
0
    x < 0,
(3)
where γ is a positive number (to ensure that F(x) → 0 as x→ ∞), then for x > 0,

e−γx f(x)
=
1
2 π
 ⌠
⌡ 		∞

−∞ 
 ⎛
⎝ 		⌠
⌡ 		∞

−∞ 
 F(y) e−i ωy dy⎞
⎠ 		eiωx dω
=
1
2π
 ⌠
⌡ 		∞

−∞ 
 ⎛
⎝ 		⌠
⌡ 		∞

0 
 f(y) e−(γ+ iω) y dy⎞
⎠ 		eiωx dω.
(4)
For x > 0, multiplying eγx on the both sides of Eq. (4) yields

f(x) = 1
2 π
 ⌠
⌡ 		∞

−∞ 
 ⎛
⎝ 		⌠
⌡ 		∞

0 
 f(y) e−(γ+iω) y dy⎞
⎠ 		e(γ+ i ω)x dω.
(5)
If we let
γ+ i ω ≡ s,
then
i dω = ds,
so that Eq. (5) becomes
f(x)
=
1
2π
 ⌠
⌡ 		γ+ i ∞

γ−i∞ 
 ⎛
⎝ 		⌠
⌡ 		∞

0 
 f(y) e−s y dy⎞
⎠ 		esx ds
i
=
1
2πi
 ⌠
⌡ 		γ+ i ∞

γ− i ∞ 
-
f
 
 (s) es x ds,
(6)
where
-
f
 
 (s) ≡ ⌠
⌡ 		∞

0 
 f(x)e− s x dx.
(7)
s-plane.jpg
Equation (7) is called the Laplace transform of f(x) and Eq. (6) is called the inverse Laplace transform.
As discussed in class, the Laplace transforms are special cases of the Fourier transforms by restricting F(t) 2 to exp(−γt) f(t) for t > 0, where γ is a positive number. For the convergence of the Laplace transform, f(t) must be a function such that e−γt f(t) → 0 as t→ ∞ for an appropriately chosen positive value of γ. Such functions include

f(t) = 1
Choose γ = 1,
(8)
f(t) = t
Choose γ = 1,
(9)
f(t) = t100
Choose γ = 1,
(10)
f(t) = e100t
Choose γ = 101,
(11)
but not
f(t) = exp (t2),
(12)
as it is not possible to choose γ such that e−γt exp(t2) vanishes as t→ ∞.

Properties of Laplace transforms


    (1) Linearity
    L (αf + βg) = αL (f) + βL (g).
    		(13)

    (2) Differentiation
    L f(n) = sL f(n−1) − f(n−1) (0).
    		(14)
    (Proof): For n=1,

    L(f′)
    =
    ⌠
    ⌡     		∞

    0 
    		 f′(t) e−st dt
    =
    [ f(t) e−st]0∞ − ⌠
    ⌡     		∞

    0 
    		 f(t) (−s) e−s t dt
    =
    fe−st|∞ − fe−st|0 +s ⌠
    ⌡     		∞

    0 
    		 f(t) e−st dt
    =
    s L(f) − f(0)
    =
    s
    -
    f
     
    		 (s) − f(0).
    		(15)
    Use mathematical induction to complete the proof.
    For n = 2,

    L(f  "(t))
    =
    s L(f  ′(t)) − f  ′(0)
    =
    s⎛
    ⎝     		s
    -
    f
     
    		 (s) − f(0) ⎞
    ⎠     		− f  ′(0)
    =
    s2
    -
    f
     
    		 (s) − s f(0) − f  ′(0).
    		(16)


    (3) Laplace convolution theorem
    The Laplace convolution is defined as

    f * g = ⌠
    ⌡     		t

    0 
    		 f(t − τ) g(τ) dτ.
    		(17)
    Note that the integral range is different from that in the Fourier convolution.
    Note also
    f * g = g * f.

    The Laplace convolution theorem is stated as
    L (f * g) =
    -
    f
     
    		 (s)
    -
    g
     
    		 (s).
    		(18)
(Proof):
L⎛
⎝ 		⌠
⌡ 		t

0 
 f(t − τ) g(τ) d τ⎞
⎠
=
⌠
⌡ 		∞

0 
 ⎛
⎝ 		⌠
⌡ 		t

0 
 f(t − τ) g(τ) d τ⎞
⎠ 		e−s t d t
=
⌠
⌡ 		∞

0 
 ⎛
⎝ 		⌠
⌡ 		∞

τ 
 f(t − τ) e−s t d t ⎞
⎠ 		g(τ) d τ
=
⌠
⌡ 		∞

0 
 ⎛
⎝ 		⌠
⌡ 		∞

0 
 f(p) e− s (τ+ p) d p ⎞
⎠ 		g(τ) dτ
=
⌠
⌡ 		∞

0 
 ⎛
⎝ 		⌠
⌡ 		∞

0 
 f(p) e− s p d p e−s τ ⎞
⎠ 		g(τ) dτ
=
⌠
⌡ 		∞

0 
 ⎛
⎝ 		⌠
⌡ 		∞

0 
 f(p) e− s p d p⎞
⎠ 		g(τ) e− s τ d τ
=
⌠
⌡ 		∞

0 
 f(p) e− s p d p  ⌠
⌡ 		∞

0 
 g(τ) e− s τ d τ
=
-
f
 
 (s)  
-
g
 
 (s),
(19)
where the order of integrations was changed.
convproof.jpg

τ
:
0 → t
t
:
0 → ∞
⇒

t
:
τ→ ∞
τ
:
0 → ∞

Some important Laplace transforms


t
s
ta
Γ(a + 1)
sa + 1
 3
ea t
1
s − a
sin at
a
s2 + a2
cos at
s
s2 + a2
H(t − a)
exp(−a s)
s
sinh at
a
s2 − a2
cosh at
s
s2 − a2
e−at f(t)
-
f
 
 (s+a)
(20)
You can verify the above by the following forms:
Laplace transform
Enter the function to be Laplace transformed.
with respect to .

Example:
t^4 (must be lowercase t) t

Inverse Laplace transform
Enter the function to be inverse Laplace transformed.
with respect to .

Example:
1/(s^2+4) (must be lowercase s)


Example 1 (initial value problem)

⋅⋅
x
 
 (t) + x(t) = f(t),
(21)
with x(0) and · x(0) are given as the initial conditions.
Equation (21) can be Laplace transformed to

s2
-
x
 
 − s x(0) −
⋅
x
 
 (0) +
-
x
 
 =
-
f
 
 (s),
(22)
which can be solved as

-
x
 
 =
-
f
 
 (s)
1 + s2
 +
⋅
x
 
 (0)
1 + s2
 + s
1 + s2
 x(0).
(23)
The inverse Laplace transform of eq.(23) is

x(t)
=
f(t) * sin t +
⋅
x
 
 (0) sin t + x(0) cos t
=
⌠
⌡ 		t

0 
 f(t−τ) sin τdτ+
⋅
x
 
 (0) sin t + x(0) cos t.
(24)
Note that the convolution theorem was used.
Example 2
Simultaneous differential equations:

⎧
⎨
⎩
dx
dt
 = x + y,     x(0)=1,
dy
dt
 = 4 x + y,     y(0)=0.
(25)
(Solution): Because of the way the initial condition is given at t=0 and the range of t is (0, ∞), the Laplace transform should be used. By transforming each equation into the s domain, one obtains

s
-
x
 
 (s) − x(0)
=
-
x
 
 (s) +
-
y
 
 (s)
(26)
s
-
y
 
 (s) − y(0)
=
4
-
x
 
 (s) +
-
y
 
 (s)
(27)
which can be solved as

-
x
 
 (s)
=
s−1
(s−1)2−4
(28)
-
y
 
 (s)
=
4
(s−1)2−4
(29)
From the table above, one can find
x(t)
=
et cosh 2t
(30)
y(t)
=
2 et sinh 2t.
(31)
Inverse Laplace transform
Enter the function to be inverse Laplace transformed.
with respect to .
Alternatively,
x(t)= s−1
(s+1)(s−3)
 = 1
2(s−3)
 + 1
2(s+1)
 ,
(32)

y(t)= 4
(s+1)(s−3)
 = 1
s−3
 − 1
s+1
 .
(33)
Therefore,
x(t) = 1
2
 e3t + 1
2
 e−t, .    y(t)=e3t−e−t.
(34)

Partial fraction


P(s)
(s−a1)m1 (s−a2)m2 …(s−an)mn
 = A0+A1 s + A2 s2 + …Am1−1sm1−1
(s−a1)m1
 +…
(35)

Example:
s+1
s3 (s+2)
 = a0+a1s + a2 s2
s3
 + b0
s+2
 .
(36)
In order to determine the unknowns, the denominators can be eliminated as

s + 1 = (a0+a1s + a2 s2)(s+2) + b0 s3.
(37)

s = −2
⇒
−1 = −8 b0,
(38)
s = 0
⇒
1 = 2 a0.
(39)
To obtain two more conditions (four unknowns !), eq.(37) can be differentiated as

1 = (a1 + 2a2 s) (s+2) + (a0 + a1 s + a2 s2) + 3 b0 s2.
(40)

s = 0
⇒
1 = 2 a1 + a0.
(41)
Differentiating eq.(40) again as

0 = 2a2 (s+2)+ (a1 + 2a2 s) + (a1 + 2a2 s)+ 6 b0 s,
(42)

s = 0
⇒
0 = 4 a2 + a1+a1,
(43)
so

b0= 1
8
 ,     a0= 1
2
 ,     a1= 1
4
 ,     a2=− 1
8
 .
(44)

s+1
s3 (s+2)
=
⎛
⎝ 		1
2
 + s
4
 − s2
8
 ⎞
⎠
s3
 +
⎛
⎝ 		1
8
 ⎞
⎠
s+2
=
1
2s3
 + 1
4s2
 − 1
8s
 + 1
8 (s+2)
 .
(45)

Footnotes:

1One of the three definitions. This is derived from Fourier series.
2 Because the variable used in the Laplace transform is often time, "t" is customarily used instead of "x."
3
Γ(s) ≡ ⌠
⌡ 		∞

0 
 ts − 1 e− t d t,     Γ(n) = (n − 1)!     n: integer


File translated from TEX by TTH, version 4.03.
On 11 Nov 2025, 19:52.