Laplace Transform of the First Order Derivative 📂Odinary Differential Equations

Laplace Transform of the First Order Derivative

Theorem¹

Let’s assume the following two conditions.

Let function $f(t)$ be continuous on interval $0 \le t \le A$ , and let its first derivative $f^{\prime}(t)$ be piecewise continuous.
There exists real numbers $a$ and positive numbers $K$ , $M$ such that when $t \ge M$ , it satisfies $|f(t)| \le Ke^{at}$ .

Then, the first derivative of $f$ ’s Laplace transform $\mathcal{L} \left\{ f^{\prime}(t) \right\}$ exists when $s>a$ and its value is as follows.

$\mathcal {L} \left\{ f^{\prime}(t) \right\} = s\mathcal {L} \left\{ f(t) \right\} -f(0)$

Explanation

To put the conditions simply, ‘The Laplace transform of $f(t)$ exists and $f^{\prime}(t)$ is piecewise continuous’.

Proof

If $\displaystyle \lim_{A \to \infty} \int _{0} ^A e^{-st}f^{\prime}(t)dt$ converges, then the proof is complete. Since it is stated that $f^{\prime}$ is piecewise continuous within the interval, let’s say there are $k$ points of discontinuity. And let’s name these points of discontinuity $t_{1}$ , $t_2$ , $\cdots$ , $t_{k}$ . If we divide the integration by these points of discontinuity, it would look as follows.

$\int _{0} ^A e^{-st}f^{\prime}(t)dt = \int _{0} ^{t_{1}} e^{-st}f^{\prime}(t)dt + \int _{t_{1}} ^{t_2} e^{-st}f^{\prime}(t)dt + \cdots + \int _{t_{k}} ^A e^{-st}f^{\prime}(t)dt$

Now, let’s perform integration by parts for each term. Separating the definite and indefinite integrals, we organize them as follows.

$\begin{align*} \int_{0}^A e^{-st}f^{\prime}(t)dt &= \left[ e^{-st}f(t)\right]_{0}^{t_{1}} + \left[ e^{-st}f(t)\right]_{t_{1}}^{t_2} + \cdots + \left[ e^{-st}f(t)\right]_{t_{k}}^{A} \\ &\quad +s \left[ \int _{0} ^{t_{1}} e^{-st}f(t)dt + \int _{t_{1}} ^{t_2} e^{-st}f(t)dt + \cdots + \int _{t_{k}} ^A e^{-st}f(t)dt \right] \end{align*}$

Calculating the definite integral part and combining the integral parts, we get the following.

$\begin{align*} &\left[ e^{-st_{1}}f(t_{1}) - e^{-s0}f(0) \right] + \left[ e^{-st_2}f(t_2) - e^{-st_{1}}f(t_{1}) \right] + \cdots + \left[ e^{-sA}f(A) - e^{-st_{k}}f(t_{k}) \right] + s \int _{0} ^{A} e^{-st}f(t)dt \\ &= e^{-sA}f(A) - e^{-s0}f(0) + s \int _{0} ^{A} e^{-st}f(t)dt \end{align*}$

As assumed, since the Laplace transform of $f(t)$ exists,

$\int _{0} ^A e^{-st}f^{\prime}(t)dt = e^{-sA}f(A) - f(0) + s \mathcal{L} \left\{ f(t) \right\}$

Now, to check if $\displaystyle \lim_{A \to \infty} e^{-sA}f(A)$ converges, by demonstrating that the Laplace transform of $f^{\prime}(t)$ exists. By Condition 2., it follows that $|f(A)| \le Ke^{aA}$ . Multiplying both sides by $e^{-sA}$ ,

$|e^{-sA}f(A)| \le Ke^{-(s-a)A}$

Then, taking the limit $\lim \limits_{A \to \infty}$ on both sides, when $s>a$ , the right-hand side converges to $0$ . Therefore, the left-hand side also converges to $0$ . Thus,

$\begin{align*} \int_{0}^\infty e^{-st}f^{\prime}(t)dt &= \lim_{A \to \infty} \int _{0} ^A e^{-st}f^{\prime}(t)dt \\ &= \lim_{A \to \infty} e^{-sA}f(A) - f(0) + s \mathcal{L} \left\{ f(t) \right\} \\ &= s \mathcal{L} \left\{ f(t) \right\} -f(0) \\ &= s F(s) -f(0) \end{align*}$

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