(I) Definition of the Laplace Transform

Similar to the differential operator \( D \), where \( D\{f(t)\} = f'(t) \), the Laplace transform is an integral transform that converts a function of time \( t \) into a function of a complex variable \( s \).

The Laplace Transform

Given a function \( f(t) \) defined for all \( t \ge 0 \), the Laplace transform of \( f \) is the function \( F(s) \) defined by the following improper integral:

\[ F(s) = \mathcal{L}\{f(t)\} = \int_{0}^{\infty} e^{-st} f(t) \, dt = \lim_{b \to \infty} \int_{0}^{b} e^{-st} f(t) \, dt \]

The transform exists for all values of \( s \) for which the improper integral converges. If the limit exists, the integral converges; if the limit does not exist, the integral diverges.

Example 1

Find the Laplace transform of \( f(t) = e^{at} \) for \( t \ge 0 \).

\[ F(s) = \mathcal{L}\{e^{at}\} = \int_{0}^{\infty} e^{-st} e^{at} \, dt = \int_{0}^{\infty} e^{(a-s)t} \, dt \] \[ = \lim_{b \to \infty} \left[ \frac{e^{(a-s)t}}{a-s} \right]_{0}^{b} \]

This integral converges if \( a-s < 0 \) (or \( s > a \)):

\[ F(s) = \lim_{b \to \infty} \left( \frac{e^{(a-s)b}}{a-s} - \frac{1}{a-s} \right) = 0 - \frac{1}{a-s} = \frac{1}{s-a} \]

Result:

\[ \mathcal{L}\{e^{at}\} = \frac{1}{s-a} \ qquad (s > a) \].

Example 2: Power Functions

Find the Laplace transform of \( f(t) = t^a \).

Using the substitution \( u = st \), then \( t = u/s \) and \( dt = (1/s) du \):

\[ \mathcal{L}\{t^a\} = \int_{0}^{\infty} e^{-st} t^a \, dt = \int_{0}^{\infty} e^{-u} \left( \frac{u}{s} \right)^a \frac{1}{s} \, du = \frac{1}{s^{a+1}} \int_{0}^{\infty} e^{-u} u^a \, du \]

The integral is the definition of the Gamma Function \( \Gamma(x) \):

\[ \Gamma(x) = \int_{0}^{\infty} e^{-t} t^{x-1} \, dt, \qquad x > 0 \]

Key properties include \( \Gamma(x+1) = x\Gamma(x) \), \( \Gamma(n+1) = n! \) for positive integers \( n \), and \( \Gamma(1/2) = \sqrt{\pi} \).

Result: If n is a positive integer, then

\[ \mathcal{L}\{t^n\} = \frac{n!}{s^{n+1}} \ \ (s>0)\]

(II) Linearity of Transforms

Theorem 1: Linearity of the Laplace Transform

The Laplace transform is a linear operation. If \( a \) and \( b \) are constants:

\[ \mathcal{L}\{a f(t) + b g(t)\} = a \mathcal{L}\{f(t)\} + b \mathcal{L}\{g(t)\} \]

This holds for all \( s \) such that \( F(s) \) and \( G(s) \) both exist.

Example 3: Trigonometric and Hyperbolic Functions

Recall the definitions of hyperbolic functions:

\[ \cosh(kt) = \frac{e^{kt} + e^{-kt}}{2}, \quad \sinh(kt) = \frac{e^{kt} - e^{-kt}}{2} \]

Using linearity and Example 1:

\[ \mathcal{L}\{\cosh(kt)\} = \frac{1}{2} \left( \frac{1}{s-k} + \frac{1}{s+k} \right) = \frac{s}{s^2 - k^2} \qquad (s>|k|)\] \[ \mathcal{L}\{\sinh(kt)\} = \frac{1}{2} \left( \frac{1}{s-k} - \frac{1}{s+k} \right) = \frac{k}{s^2 - k^2}\qquad (s>|k|)\]

Similarly for trigonometric functions using Euler's formula: \( e^{ikt} = \cos(kt) + i \sin(kt) \), we have

\[ \cos(kt) = \frac{e^{ikt} + e^{-ikt}}{2}, \qquad \sin(kt) = \frac{e^{ikt} - e^{-ikt}}{2i} \]
\( \begin{aligned} \mathcal{L}\{\cos(kt)\} &= \frac{1}{2} \left( \mathcal{L}\{e^{ikt}\} + \mathcal{L}\{e^{-ikt}\} \right) = \frac{1}{2} \left( \frac{1}{s-ik} + \frac{1}{s+ik} \right) \\ &= \frac{1}{2} \left( \frac{s+ik + s-ik}{(s-ik)(s+ik)} \right) = \frac{1}{2} \left( \frac{2s}{s^2 + k^2} \right) = \frac{s}{s^2 + k^2} \end{aligned} \)
\( \begin{aligned} \mathcal{L}\{\sin(kt)\} &= \frac{1}{2i} \left( \frac{1}{s-ik} - \frac{1}{s+ik} \right) = \frac{1}{2i} \left( \frac{s+ik - (s-ik)}{s^2 + k^2} \right) \\ &= \frac{1}{2i} \left( \frac{2ik}{s^2 + k^2} \right) = \frac{k}{s^2 + k^2} \end{aligned} \)

Conclusion:

\[ \mathcal{L}\{\cos(kt)\} = \frac{s}{s^2 + k^2}, \quad \mathcal{L}\{\sin(kt)\} = \frac{k}{s^2 + k^2} \qquad (s>0) \]

Example 4: Combined Terms

Find \( \mathcal{L}\{3e^{2t} + 2\sin^2(3t)\} \).

Using the identity \( \sin^2(3t) = \frac{1 - \cos(6t)}{2} \):

\( \mathcal{L}\{3e^{2t} + 1 - \cos(6t)\} = 3\mathcal{L}\{e^{2t}\} + \mathcal{L}\{1\} - \mathcal{L}\{\cos(6t)\} = \frac{3}{s-2} + \frac{1}{s} - \frac{s}{s^2 + 36} \quad (s > 2)\)

(III) Inverse Transforms

If \( F(s) \) is the transform of a continuous function \( f(t) \), then \( f(t) \) is uniquely determined. This allows us to define the inverse Laplace transform.

If \( F(s) = \mathcal{L}\{f(t)\} \), then \( f(t) = \mathcal{L}^{-1}\{F(s)\} \).

Example 5
  • \( \mathcal{L}^{-1}\{s^{-3}\} = \frac{1}{2} t^2 \)
  • \( \mathcal{L}^{-1}\{\frac{1}{s+5}\} = e^{-5t} \)
  • \( \mathcal{L}^{-1}\{\frac{3s+1}{s^2+4}\} = 3\mathcal{L}^{-1}\{\frac{s}{s^2+4}\} + \frac{1}{2}\mathcal{L}^{-1}\{\frac{2}{s^2+4}\} = 3\cos(2t) + \frac{1}{2}\sin(2t) \)

(IV) Piecewise Continuous Functions

A function \( f(t) \) is piecewise continuous on an interval \([a, b]\) if the interval can be subdivided into finitely many subintervals such that:

  1. \( f \) is continuous in the interior of each subinterval.
  2. \( f \) has finite limits as \( t \) approaches each endpoint of each subinterval from the interior.

The Unit Step Function

\[ u(t) = \begin{cases} 0 & \text{for } t < 0 \\ 1 & \text{for } t \ge 0 \end{cases} \implies \mathcal{L}\{u(t)\} = \frac{1}{s} \qquad (s>0) \]

The shifted unit step function is defined as:

\[ u_a(t) = u(t-a) = \begin{cases} 0 & \text{for } t < a \\ 1 & \text{for } t \ge a \end{cases} \]
Example 6

Find \( \mathcal{L}\{u_a(t)\} \) for \( a > 0 \).

\( \begin{aligned} \mathcal{L}\{u_a(t)\} &= \int_{0}^{\infty} e^{-st} u_a(t) \, dt = \int_{a}^{\infty} e^{-st} \, dt \\ &= \lim_{b \to \infty} \left[ -\frac{e^{-st}}{s} \right]_{a}^{b} = \lim_{b \to \infty} \left( \frac{e^{-as}}{s} - \frac{e^{-bs}}{s} \right) \\ &= \frac{e^{-as}}{s} \quad (s > 0, a > 0) \end{aligned} \)

(V) Existence of Laplace Transforms

Theorem 2: Existence

If a function \( f(t) \) is piecewise continuous for \( t \ge 0 \) and is of exponential order as \( t \to \infty \) (meaning there exist constants \( M, c, \) and \( T \) such that \( |f(t)| \le Me^{ct} \) for all \( t \ge T \)), then its Laplace transform \( F(s) \) exists.