Laplace Transform of Second Derivative

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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

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

$\blacksquare$


Also see


Sources

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