# Initial Value Theorem of Laplace Transform

## Theorem

Let $\laptrans {\map f t} = \map F s$ denote the Laplace transform of the real function $f$.

Then:

$\ds \lim_{t \mathop \to 0} \map f t = \lim_{s \mathop \to \infty} s \, \map F s$

if those limits exist.

### General Result

Let $\ds \lim_{t \mathop \to 0} \dfrac {\map f t} {\map g t} = 1$.

Then:

$\ds \lim_{s \mathop \to \infty} \dfrac {\map F s} {\map G s} = 1$

if those limits exist.

## Proof

We have that $\map {f'} t$ is piecewise continuous with one-sided limits and of exponential order.

Hence:

$\ds \lim_{s \mathop \to \infty} \int_0^\infty e^{-s t} \map {f'} t \rd t = 0$

Suppose that $f$ is continuous at $t = 0$.

$(1): \quad \laptrans {\map {f'} t} = s \map F s - \map f 0$

Taking the limit as $s \to \infty$ in $(1)$, where it is assumed that $\map f t$ is continuous at $t = 0$:

$0 = \ds \lim_{s \mathop \to \infty} s \map F s - \map f 0$

or:

$\ds \lim_{s \mathop \to \infty} s \map F s = \map f 0 = \lim_{t \mathop \to 0} \map f t$

$\Box$

Suppose that $f$ is not continuous at $t = 0$.

$\laptrans {\map {f'} t} = s \map F s - \map f {0^+}$

which means:

$(2): \quad \laptrans {\map {f'} t} = s \map F s - \ds \lim_{u \mathop \to 0} \map f u$

Similarly taking the limit as $s \to \infty$ in $(2)$, where it is assumed that $\map f t$ is continuous at $t = 0$:

$0 = \ds \lim_{s \mathop \to \infty} s \map F s - \lim_{u \mathop \to 0} \map f u$

and so:

$\ds \lim_{s \mathop \to \infty} s \map F s = \lim_{u \mathop \to 0} \map f u = \lim_{t \mathop \to 0} \map f t$

$\blacksquare$

## Examples

### Example 1

Consider the real function $f: \R \to \R$ defined as:

$\map f t = 3 e^{-2 t}$
$\laptrans {\map f t} = \dfrac 3 {s + 2}$

Then by the Initial Value Theorem of Laplace Transform:

 $\ds \lim_{t \mathop \to 0} 3 e^{-2 t}$ $=$ $\ds \lim_{s \mathop \to \infty} \dfrac 3 {s + 2}$ $\ds \leadsto \ \$ $\ds 3$ $=$ $\ds 3$