Existence and Properties of the Laplace Transform

The Laplace transform exists for a broad class of functions and possesses algebraic properties that make it a powerful tool for solving differential equations.

Existence Theorem

Theorem4.7Existence of the Laplace transform

If $f: [0, \infty) \to \mathbb{R}$ is piecewise continuous on every finite interval $[0, T]$ and of exponential order $\alpha$ (i.e., there exist constants $M, T_0$ such that $|f(t)| \leq Me^{\alpha t}$ for all $t \geq T_0$ ), then $\mathcal{L}\{f\}(s)$ exists for all $s > \alpha$ and

$|\mathcal{L}\{f\}(s)| \leq \frac{M}{s - \alpha}$

for $s > \alpha$ . Moreover, $\lim_{s \to \infty}F(s) = 0$ .

Proof

For $s > \alpha$ : $|F(s)| \leq \int_0^{T_0}|e^{-st}f(t)|\,dt + \int_{T_0}^\infty Me^{-st}e^{\alpha t}\,dt$ . The first integral is finite (piecewise continuous on a finite interval). The second integral equals $M e^{-(s-\alpha)T_0}/(s-\alpha) < \infty$ for $s > \alpha$ .

For $\lim_{s\to\infty}F(s) = 0$ : the bound $|F(s)| \leq C/(s-\alpha)$ gives $F(s) \to 0$ . $\blacksquare$

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Differentiation and Integration Properties

Theorem4.8Transforms of derivatives and integrals

Suppose $f, f', \ldots, f^{(n-1)}$ are continuous on $[0, \infty)$ and of exponential order, and $f^{(n)}$ is piecewise continuous. Then:

$\mathcal{L}\{f^{(n)}\}(s) = s^n F(s) - s^{n-1}f(0) - s^{n-2}f'(0) - \cdots - f^{(n-1)}(0).$

For integration: $\mathcal{L}\left\{\int_0^t f(\tau)\,d\tau\right\}(s) = \frac{F(s)}{s}$ .

Also: $\mathcal{L}\{tf(t)\}(s) = -F'(s)$ and more generally $\mathcal{L}\{t^n f(t)\}(s) = (-1)^n F^{(n)}(s)$ .

ExampleUsing the derivative property

Find $\mathcal{L}\{t\sin(at)\}$ . Since $\mathcal{L}\{\sin(at)\} = a/(s^2+a^2)$ :

$\mathcal{L}\{t\sin(at)\} = -\frac{d}{ds}\frac{a}{s^2+a^2} = \frac{2as}{(s^2+a^2)^2}.$

Initial and Final Value Theorems

Theorem4.9Initial value theorem

If $\lim_{s\to\infty}sF(s)$ exists, then $f(0^+) = \lim_{s\to\infty}sF(s)$ .

Theorem4.10Final value theorem

If $\lim_{t\to\infty}f(t)$ exists and is finite, and all poles of $sF(s)$ have negative real part (except possibly a simple pole at $s = 0$ ), then $\lim_{t\to\infty}f(t) = \lim_{s\to 0}sF(s)$ .

ExampleFinal value computation

For $F(s) = \frac{5}{s(s^2+3s+2)}$ : $\lim_{t\to\infty}f(t) = \lim_{s\to 0}sF(s) = \lim_{s\to 0}\frac{5}{s^2+3s+2} = \frac{5}{2}$ .

The poles of $sF(s)$ are $s = -1, -2$ (both with negative real part), so the theorem applies.

Uniqueness and Inversion

RemarkThe Laplace transform is injective

On the class of piecewise continuous functions of exponential order, the Laplace transform is one-to-one (Lerch's theorem). This ensures that the solution obtained by inverting the transform is the unique solution to the original IVP.