Differential Equations 2301312: ISE Program, Chulalongkorn University:
First Semester 2007
The Laplace Transform
Definition: Let f(t) be given for t≥0, and suppose thatf(t) is integrable on [0,∞). The Laplace transform of f, which we will denote by L[f(t)] or by F(s), is defined by the equation
L[f(t)] =F(s) = Z ∞
0
e−stf(t)dt, whenever this improper integral converges.
Examples:
1. Let f(t) = 1, t≥0. Then L[1] =
Z ∞
0
e−st·1dt = lim
A→∞
Z A 0
e−stdt
=− lim
A→∞
e−st s |A0
= 1
s, s >0 2.
L[eat] = Z ∞
0
e−st·eatdt = Z ∞
0
e−(s−a)tdt= 1
s−a, s > a.
3.
L[sin(at)] = F(s) = Z ∞
0
e−st ·sin(at)dt, s > 0
= lim
A→∞[−e−stcos(at) a |A0 − s
a Z A
0
e−stcos(at)dt]
= 1 a − s
a Z ∞
0
e−stcos(at)dt
= 1 a − s2
a2 Z ∞
0
e−stsin(at)dt So F(s) = 1a− sa22F(s),or F(s) = s2+aa 2, s >0. Hence
L[sin(at)] = a
s2+a2, s >0. Similarly, we can prove that
L[cos(at)] = s
s2+a2, s >0.
Suppose f1, f2 are two functions whose Laplace transform exist for s > a1 and s > a2, respectively. Then fors greater than the maximum ofa1 and a2,
L[c1f1(t) +c2f2(t)] = Z ∞
0
e−st[c1f1(t) +c2f2(t)]dt
=c1 Z ∞
0
e−stf1(t)dt+c2 Z ∞
0
e−stf2(t)dt hence
L[c1f1(t) +c2f2(t)] =c1L[f1(t)] +c2L[f2(t)].
This equation states that the Laplace transform is a linear operator. The sum in this equation can be extended to an arbitrary number of terms.
Example: Find the Laplace transform of f(t) = 5e−3t−4 sin(3t), t≥0.
Solution:
L[5e−3t−4 sin(3t)] = 5L[e−3t]−4L[sin(3t)]
= 5
s−(−3)−4( 3
s2+ 9) = 5
s+ 3 − 12
s2+ 9, s >0.
Example: Find the Laplace transform of f(t) = cos2(2t), t≥0.
Solution:
L[cos2(2t)] =L[1
2 +cos(4t) 2 ]
= 1
2s + s
2(s2+ 16), s >0.
Gamma function Definition:
Gamma function is the function Γ(x) =R∞
0 e−ttx−1dt, f orx >0.
Γ(1) =R∞
0 e−tdt = 1.
Γ(x+ 1) =xΓ(x), since Γ(x+1) =
Z ∞
0
e−ttxdt=−[txe−t|∞0 − Z ∞
0
e−ttx−1dt] =x Z ∞
0
e−ttx−1dt =xΓ(x), x >0.
By using this propery, we get that
Γ(2) = Γ(1 + 1) = 1Γ(1) = 1 = 1!.
Γ(3) = Γ(2 + 1) = 2Γ(2) = 2(1!) = 2!. Γ(n) = (n−1)!.
Γ(n+ 1) =nΓ(n) = n(n−1)! =n!.
So sometimes the Gamma function is called the generalized factorial function.
We can prove that Γ(12) =√ π.
We then have the formula L[tp] =
Z ∞
0
e−sttpdt = 1 sp+1
Z ∞
0
e−uupdu= Γ(p+ 1)
sp+1 , p > −1.
Example:
L[t−12] = Γ(−12 + 1)
s12 = Γ(12) s12 =
√π
√s = rπ
s, s >0.
Piecewise continuous functions
Definition: A function f(t) is called a piecewise continuous function on the interval [a, b] if there exists points a =t0 < t1 < ... < tn = b, such that f is continuous on subintervals (ti−1, ti), i= 1,2, ..., n, and the left limit and the right limit of f at ti exist for all i.
Example: Let
f(t) =
e−t, if 0≤t ≤1 2, if 1< t <3 0, if t≥3,
then f(t) is piecewise continuous on any closed interval [0, T] for any T >0. Piecewise continuous functions are very important in Laplace transform the- ory.
Note that every continuous functions on an interval [a, b] are piecewise con- tinuous on the interval.
Exponential order
Definition: A function f(t) is said to have exponential order property if there is a point a ∈ R, M > 0 and t0 in [0,∞) such that |f(t)| ≤ M eat for t ≥t0. We denote by f(t) =O(eat) for f is of exponential order.
If
tlim→∞
e−at|f(t)| exists and finite then f(t) =O(eat).
If
tlim→∞
e−at|f(t)|=∞ then f(t) is not of exponential order.
Example:
1. f(t) = sin(t) is of exponential order because |sin(t)| ≤1 for all t ≥0.
2. f(t) = et2 is not of exponential order because
tlim→∞
e−atet2 = lim
t→∞
et2−at = lim
t→∞
et(t−at)=∞. Theorem:
If f(t) is piecewise continuous on any closed subinterval [0, T], T > 0 and f(t) has an exponential order property then the Laplace transform of f(t) exists for s > a for some constanta.
Inverse Laplace Transform: Consider the following two functions:
f1(t) = e−t and
f2(t) =
½ e−t, if t6= 1 1, if t= 1.
We see that
L[f1(t)] =L[f2(t)],
but f1 6= f2, this means that the Laplace transform is not a one-to-one function on the space of piecewise continuous functions, but the Laplace transform is a one-to-one function on the space of continuous functions, that is if we have
L[f1(t)] =L[f2(t)],
and f1(t), f2(t) are continuous on [0,∞), we can conclude that f1(t) =f2(t) for allt. So we definef(t) is the inverse Laplace transform of a functionF(s) if
L[f(t)] = F(s),
and f(t) is unique only if f is continuous, we use L−1[F(s)] for the inverse Laplace transform of F(s), that is f(t) = L−1[F(s)].
Example: We know that L[sin(at)] = s2+aa 2 then we have L−1[ a
s2+a2] = sin(at).
So
L−1[ 1
s2+ 16] =L−1[1 4( 4
s2+ 16)] = 1
4L−1[ 4
s2+ 16] = 1
4sin(4t). The inverse Laplace transform has linear property, i.e.,
L−1[c1F1(s) +c2F2(s)] = c1L−1[F1(s)] +c2L−1[F2(s)]. Examples:
1. Find the Laplace transform of f(t) =t52 − 1
2sin(t
2) + 2 cos(2t) +e−3t. 2. Find the inverse Laplace transform of
F(s) = 2s2−3s+ 2 (s+ 2)(s2 + 4). Properties of Laplace transform:
1. If f(t), f0(t), ..., f(n−1)(t) are continuous on the interval [0,∞) and are of exponential order for some constant a and if f(n)(t) is piecewise continuous on any closed interval [0, T] then the Laplace transform of f(n)(t) exists for all n and
L[f(n)(t)] =snF(s)−sn−1f(0)−sn−2f0(0)−...−f(n−1)(0), where F(s) =L[f(t)].
For examples:
L[f0(t)] =sF(s)−f(0).
L[f00(t)] = s2F(s)−sf(0)−f0(0).
L[f000(t)] =s3F(s)−s2f(0)−sf0(0)−f00(0).
Example: Solve the following initial value problem by using the Laplace transform
y0+ 3y = 3, y(0) = 0.
Solution: Let Y(s) = L[y(t)]. Take the Laplace transform at each term in the equation, we obtain
{sY(s)−y(0)}+ 3Y(s) = 3 s.
Since y(0) = 0, we write
Y(s) = 3
s(s+ 3) = 1
s − 1
s+ 3. Take the inverse Laplace transform, we get
L−1[Y(s)] =L−1[1
s]− L−1[ 1 s+ 3], thus
y(t) = 1−e−3t.
