The Laplace Transform

Definition 1
Let f(t) be a real-valued function defined for t > 0.
The Laplace transform of f(t) is defined by

for all s such that the improper integral converges.

Example 1 ***

Let f(t) = 1, t > 0. Find ℒ {f(t)}.

Solution
By Definition 1,

If s > 0, the limit converges to zero; if s < 0, the limit diverges; if s = 0, it yields indeterminate form 0/0. Thus,

❏

Example 2 ***

Let f(t) = e^at, t > 0. Find ℒ {f(t)}.

Solution

The limit converges to zero if s >a. Thus,

❏

Example 3 ***

Let f(t) = t, t > 0. Find ℒ {f(t)}.

Solution

The limits

converges to zero if s > 0. Thus,

❏

Example 4 ***

Show that, for t > 0,

Property — Linearity
Suppose ℒ {f} and ℒ {g} exist for s > a, then
ℒ {c₁f + c₂g} = c₁ℒ {f} + c₂ℒ {g},
i.e. the Laplace transform is a linear operation.

Example 5 *

Find ℒ {sin²bt}.

Solution
We note that

for any t, so that

which is, by Formulae (2) and (5),

❏

Example 6 **(Optional)

Show that the Laplace transform of power functions are

where Γ is the gamma function.

Definition 2 (piecewise continuity)
A function f is said to be piecewise continuous on a closed interval [a, b] if and only if the interval can be partitioned by a finite number of points a = t₀ < t₁ < … < t_n = b, such that
1. f is continuous for every interior points on each subinterval; in other words, f is continuous on t_i−1 < t < t_i, for i = 1, 2, …, n;
2. f approaches a finite limit as the endpoints of each subinterval are approached from within the subinterval (either leftward or rightward); in other words, f does not approach to infinity as t approaches to t_i, for i = 1, 2, …, n, from the interior points of subintervals.

Briefly, f is piecewise continuous on [a, b] if it is continuous there except for a finite number of jump discontinuities.

Remarks

1. If f is continuous on some closed interval [a, b], then it is piecewise continuous on [a, b].
2. If f is piecewise continuous on some closed interval [a, b], then f is integrable on [a, b].

Definition 3 (exponential order)
A function f is said to be of exponential order α for some α ∈ ℝ, if and only if there exists t₀ > 0 and M > 0, such that

for all t > t₀; in other words, the function e^−αt|f(t)| is bounded on t > t₀.

Remarks

1. If f is of exponential order α, then f is of exponential order β for any α < β.

2. If f is bounded on (0, ∞), then f is of exponential order α for any α ≥ 0. This is because if f is bounded for t > 0, then there exists M ∈ ℝ such that |f(t)| ≤ M for all t > 0, and thus |f(t)| ≤ Me^αt for all t > 0. For example, cos(bt) and sin(bt) are of exponential order α for any α ≥ 0.

3. The function f(t) = e^at is of exponential order α for any α ≥ a.

4. The function f(t) = tⁿ, n ∈ ℕ, is of exponential order α for any α ≥ 0. This is because, by L’Hôpital’s rule,

which implies that, given M > 0, there exist a number N > 0 such that if t > N, then |e^−αtf(t) − 0| < M, and thus e^−αt|f(t)| < M.

5. The function f₁(t) = e^t2 and f₂(t) = t^t = e^{t ln t} are not of exponential order α of any α because it can be proved that

and

Theorem 1 (existence of the Laplace transform)
Suppose that f is piecewise continuous on some close interval [0, A] for any A > 0, and that f is of exponential order α for some real number α. Then ℒ {f} exists for s > α.

Proof
Suppose that there exists t₀ > 0 and M > 0 such that |f(t)| ≤ Me^αt for all t > t. Then, by definition,

Here we deal with the improper integral (8) by dividing it into two terms, so we need to prove that both terms converges. For the latter term of Formula (8),

This limit converges to some finite value if s − α > 0. Since the integral

converges when s − α > 0, so does

So e^-stf(t) is integrable on [t₀, ∞) if s > a.

Now we examine the former term of Formula (8). Since f is piecewise continuous on [0, t₀], e^-stf(t) is piecewise continuous there as well. Thus, by Remark 2 of Definition 2, e^-stf(t) is integrable on [0, t₀].

As we have shown that both terms of Formula (8) is simultaneously integrable on specific intervals if s > α, the Laplace transform ℒ {f} exists for s > α. Note that we use the two conditions (piecewise continuity and exponential order) of Theorem 1 to prove that e^-stf(t) is integrable on the two intervals, respectively. ◪