Laplace Transform of Derivative

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

Let $f$ be of exponential order $a$.

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

Let $\mathcal L$ be the Laplace transform.

Then $\mathcal L \left\{{f}\right\}$ exists for $\operatorname{Re} \left({s}\right) > a$, and:

$\mathcal L \left\{{f' \left({t}\right)}\right\} = s \mathcal L \left\{ {f \left({t}\right)}\right\} - f \left({0}\right)$


\(\displaystyle \mathcal L \left\{ {f' \left({t}\right)}\right\}\) \(=\) \(\displaystyle \int_0^{\mathop \to +\infty} e^{-s t} f' \left({t}\right) \rd t\) $\quad$ Definition of Laplace Transform $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \lim_{A \mathop \to +\infty} \int_0^A e^{-s t} f' \left({t}\right) \rd t\) $\quad$ Definition of Improper Integral on Closed Interval Unbounded Above $\quad$


$\displaystyle \int_0^A e^{-s t} f' \left({t}\right) \rd t$

By hypothesis, $f'$ is piecewise continuous with one-sided limits.

So by Piecewise Continuous Function with One-Sided Limits is Riemann Integrable, this integral exists.

This means that integration by parts can be invoked:

$\displaystyle \int h j \,' \rd t = hj - \int h' j \rd t$


\(\displaystyle h\) \(=\) \(\displaystyle e^{-s t}\) $\quad$ $\quad$
\(\displaystyle \leadsto \ \ \) \(\displaystyle h'\) \(=\) \(\displaystyle -s e^{-s t}\) $\quad$ $\quad$
\(\displaystyle j\,'\) \(=\) \(\displaystyle f' \left({t}\right)\) $\quad$ $\quad$
\(\displaystyle \leadsto \ \ \) \(\displaystyle j\) \(=\) \(\displaystyle f \left({t}\right)\) $\quad$ $\quad$


\(\displaystyle \int_0^A e^{-s t} f' \left({t}\right) \rd t\) \(=\) \(\displaystyle \left.{e^{-s t} f \left({t}\right)}\right\vert_{t \mathop = 0}^{t \mathop = A} + s \int_0^A e^{-s t} f \left({t}\right) \rd t\) $\quad$ $\quad$

Now, take the limit as $t = A \to +\infty$:

\(\displaystyle \mathcal L \left\{ {f' \left({t}\right)}\right\}\) \(=\) \(\displaystyle \lim_{A \mathop \to +\infty} e^{-s A} f \left({A}\right) - f \left({0}\right) + s \mathcal L \left\{ {f \left({t}\right)}\right\}\) $\quad$ $\quad$

Recall that $f$ is of exponential order $a$:

\(\displaystyle \left\vert {f \left({A}\right)} \right \vert\) \(<\) \(\displaystyle Ke^{at}\) $\quad$ $\quad$
\(\displaystyle \leadsto \ \ \) \(\displaystyle \left\vert {f \left({A}\right)} \right \vert \left \vert {e^{-s A} } \right \vert\) \(<\) \(\displaystyle Ke^{a t} \left\vert {e^{-sA} } \right \vert\) $\quad$ $\quad$
\(\displaystyle \leadsto \ \ \) \(\displaystyle \left\vert {e^{-s A} f \left({A}\right)} \right \vert\) \(<\) \(\displaystyle \left \vert {Ke^{a t} e^{-s A} } \right \vert\) $\quad$ Modulus of Product, Exponential Tends to Zero and Infinity $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \left \vert {K e^{\left({a - s}\right)t} }\right \vert\) $\quad$ Exponent Combination Laws, definition of $A$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \left \vert {K e^{\left({a- \operatorname{Re} \left({s}\right) - i \operatorname{Im} \left({s}\right)}\right)t} }\right \vert\) $\quad$ $\quad$
\(\displaystyle \) \(=\) \(\displaystyle \left \vert {K e^{\left({a- \operatorname{Re} \left({s}\right)}\right) t} }\right \vert\) $\quad$ Modulus of Exponential is Modulus of Real Part $\quad$
\(\displaystyle \) \(=\) \(\displaystyle K e^{\left({a- \operatorname{Re} \left({s}\right)}\right)t}\) $\quad$ Exponential Tends to Zero and Infinity $\quad$

This implies, from Complex Exponential Tends to Zero and the Squeeze Theorem:

$\displaystyle \lim_{A \mathop \to +\infty} e^{-s A} f \left({A}\right) = 0$

which produces:

$\mathcal L \left\{{f' \left({t}\right)}\right\} = s \mathcal L \left\{{f \left({t}\right)}\right\} - f \left({0}\right)$


Also see