Laplace Transform of Higher Order Derivatives

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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 $n$ times differentiable on said intervals.

Let $f, f', \ldots, f^{\paren {n - 1} }$ be continuous and $f^{\paren n}$ piecewise continuous with one-sided limits on said intervals.

Let $f, f', \ldots, f^{\paren {n - 1} }$ be of exponential order.

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


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

\(\ds \laptrans {\map {f^{\paren n} } t}\) \(=\) \(\ds s^n \laptrans {\map f t} - \sum_{j \mathop = 1}^n s^{j - 1} \map {f^{\paren {n - j} } } 0\)
\(\ds \) \(=\) \(\ds s^n \map F s - s^{n - 1} \, \map f 0 - s^{n - 2} \, \map {f'} 0 - s^{n - 3} \, \map {f''} 0 - \ldots - s \, \map {f^{\paren {n - 2} } } 0 - \map {f^{\paren {n - 1} } } 0\)


Proof

The proof proceeds by induction on $n$, the order of the derivative of $f$.


Basis for the Induction

By hypothesis $f$ is of exponential order.

Thus Laplace Transform of Derivative can be invoked:

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


This is the basis for the induction.


Induction Hypothesis

Fix $n - 1 \in \N$ with $n \ge 2$.

Assume:

$\ds \laptrans {\map {f^{\paren {n - 1} } } t} = s^{n - 1} \laptrans {\map f t}- \sum_{j \mathop = 1}^{n - 1} s^{j - 1} \map {f^{\paren {n - 1 - j} } } 0$

This is our induction hypothesis.


Induction Step

This is our induction step:

By hypothesis $f^\paren {n - 1}$ is of exponential order.

Thus Laplace Transform of Derivative can be invoked:

\(\ds \laptrans {\map {f^{\paren n} } t}\) \(=\) \(\ds s \laptrans {\map {f^{\paren {n - 1} } } t} - \map f 0\) Laplace Transform of Derivative
\(\ds \) \(=\) \(\ds s \paren {s^{n - 1} \laptrans {\map f t} - \sum_{j \mathop = 1}^{n - 1} s^{j - 1} \map {f^{\paren {n - 1 - j} } } 0} - \map f 0\) Induction Hypothesis
\(\ds \) \(=\) \(\ds s^n \laptrans {\map f t} - \sum_{j \mathop = 1}^{n - 1} s^{j - 1 + 1} \map {f^{\paren {n - j} } } 0 - s^0 \map {f^{\paren 0} } 0\) Definition of Zeroth Derivative
\(\ds \) \(=\) \(\ds s^n \laptrans {\map f t} - \sum_{j \mathop = 0}^{n - 1} s^j \map {f^{\paren {n - j} } } 0\)
\(\ds \) \(=\) \(\ds s^n \laptrans {\map f t} - \sum_{j \mathop = 1}^n s^{j - 1} \map {f^{\paren {n - j} } } 0\) replace $j$ with $j - 1$


The result follows by the Principle of Mathematical Induction.

$\blacksquare$


Also see


Sources

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