# Laplace Transform of Second Derivative

## Theorem

Let $f: \R \to \R$ or $\R \to \C$ be a continuous function on any interval of the form $0 \le t \le a$.

Let $f$ be twice differentiable.

Let $f'$ be continuous and $f''$ piecewise continuous with one-sided limits on said intervals.

Let $f$ and $f'$ be of exponential order.

Let $\laptrans f$ denote the Laplace transform of $f$.

Then $\laptrans {f''}$ exists for $\map \Re s > a$, and:

$\laptrans {\map {f''} t} = s^2 \laptrans {\map f t} - s \, \map f 0 - \map {f'} 0$

## Proof

 $\displaystyle \laptrans {\map {f''} t}$ $=$ $\displaystyle s \laptrans {\map {f'} t} - \map {f'} 0$ Laplace Transform of Derivative $\displaystyle$ $=$ $\displaystyle s \paren {s \laptrans {\map f t} - \map f 0} - \map {f'} 0$ Laplace Transform of Derivative $\displaystyle$ $=$ $\displaystyle s^2 \laptrans {\map f t} - s \, \map f 0 - \map {f'} 0$

$\blacksquare$