Laplace Transforms of Derivatives and Integrals

Section 11.6 Laplace Transforms of Derivatives and Integrals

Theorem 11.6.1. Differentiation of $f(t)$.

Let $f(t)$ and $f\,'(t)$ be continuous for $t \ge 0\text{,}$ and of exponential order. Then,

\begin{equation*} \mathcal{L}(\,f\,'(t)) = sF(s) -f(0), \text{ where } F(s) = \mathcal{L}\big(\,f(t)\big) \end{equation*}

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Proof.

Let $K$ be chosen large enough so that both $f(t)$ and $f\,'(t)$ are of exponential order $K\text{.}$ If $\mathrm{Re}(s)>K\text{,}$ then $\mathcal{L}\big(\,f\,'(t)\big)$ is given by

\begin{equation*} \mathcal{L}(\,f\,'(t)) = \int_0^{\infty}f\,'(t)e^{-st}\,dt\text{.} \end{equation*}

Next, using integration by parts, we can write the equation this as

\begin{equation*} \mathcal{L}\big(f\,'(t)\big) = \lim_{R \to +\infty}\left[f(t) e^{-st}\right]\Big|_{t=0}^{t=R}+s\int_0^{\infty}f(t)e^{st}\,dt\text{.} \end{equation*}

Now, $f(t)$ is of exponential order $K\text{,}$ and $\mathrm{Re}(s)>K\text{.}$ It follows that $\lim\limits_{R \to +\infty}f(R)e^{-sR}=0\text{,}$ so

\begin{equation*} \mathcal{L}\big(f\,'(t)\big) = -f(0) + s\int_0^{\infty}f(t)e^{-st}\,dt = sF(s) - f(0) \end{equation*}

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An easy consequence of Theorem 11.6.1 is the following, which we state without proof.

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Corollary 11.6.2.

If $f(t), \, f\,'(t)\text{,}$ and $f\,''(t)$ are of exponential order, then

\begin{equation*} \mathcal{L}\big(f\,''(t)\big) = s^2f(s)-sf(0)-f\,'(0) \end{equation*}

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Example 11.6.3.

Show that $\mathcal{L}(\cos^2t)=\frac{s^2+2}{s(s^2+4)}\text{.}$

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Solution.

Let $f(t) =\cos^2t\text{,}$ then $f(0) =1$ and $f\,'(t) =-2\sin t\cos t=-\sin 2t\text{.}$ Using the fact that $\mathcal{L}(-\sin 2t) = -\frac{2}{s^2+4}\text{,}$ Theorem 11.6.1 implies that

\begin{equation*} -\frac{2}{s^2+4}=\mathcal{L}\big(\,f\,'(t)\big) = s\mathcal{L}(\cos^2t) -1\text{.} \end{equation*}

from which it follows that $\mathcal{L}(\cos^2t) = -\frac{2}{s(s^2+4)}+\frac{1}{s} = \frac{s^2+2}{s(s^2+4)}\text{.}$

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Theorem 11.6.4.

(Integration of $f(t)$] Let $f(t)$ be continuous for $t \ge 0\text{,}$ and of exponential order. Further, let $F(s)$ be its Laplace transform, then

\begin{equation*} \mathcal{L}\left(\int_0^tf(\tau)\,d\tau\right) =\frac{F(s)}{s} \end{equation*}

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Proof.

Let $g(t) =\int_0^tf(\tau ) d\tau\text{,}$ then $g'(t) =f(t)$ and $g(0) =0\text{.}$ If we can show that $g$ is of exponential order, then Theorem 11.6.1 implies that

\begin{equation*} \mathcal{L}\big(f(t)\big) = \mathcal{L}\big(g'(t)\big) = s\mathcal{L}\big(g(t)\big) - 0 = s \mathcal{L}\left(\int_0^tf(\tau)\,d\tau\right)\text{.} \end{equation*}

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Since $f(t)$ is of exponential order, there are positive values $M$ and $K$ so that

\begin{equation*} |g(t)| \le \int_0^tf(\tau)\,d\tau \le M\int_0^t e^{K\tau}\,d\tau = \frac{M}{K}(e^{Kt}-1) \le e^{Kt}\text{.} \end{equation*}

Therefore, $g$ is of exponential order.

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Example 11.6.5.

Show that $\mathcal{L}(t^2) =\frac{2}{s^3}$ and $\mathcal{L}(t^3) = \frac{6}{s^4}\text{.}$

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Solution.

Using Theorem 11.6.4 and the fact that $\mathcal{L}(2t) =\frac{2}{s^2}$ we obtain

\begin{equation*} \mathcal{L}(t^2) = \mathcal{L}\left(\int_0^t 2\tau \,d\tau\right) = \frac{1}{s}\mathcal{L}(2t) =\frac{1}{s}\frac{2}{s^2}=\frac{2}{s^3}\text{.} \end{equation*}

Now we can use the first result $\mathcal{L}(t^2) =\frac{2}{s^3}$ to establish the second one:

\begin{equation*} \mathcal{L}(t^3) =\mathcal{L}\left(\int_0^t3\tau^2\,d\tau\right) = \frac{1}{s}\mathcal{L}(3t^2) =\frac{1}{s}\frac{6}{s^3}=\frac{6}{s^4} \end{equation*}

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One of the main uses of the Laplace transform is its role in the solution of differential equations. The utility of the Laplace transform lies in the fact that the transform of the derivative $f\,'(t)$ corresponds to multiplication of the transform $F(s)$ by $s$ and then the subtraction of $f(0)\text{.}$ This permits us to replace the calculus operation of differentiation with simple algebraic operations on transforms.

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This idea is used to develop a method for solving linear differential equations with constant coefficients. Consider the initial value problem

\begin{equation*} y\,''(t) + ay\,'(t) + by(t) = f(t) \end{equation*}

with initial conditions $y(0) =y_0$ and $y\,'(0)= d_0\text{.}$ The linearity property of the Laplace transform can be used to obtain the equation

\begin{equation*} \mathcal{L}\big(y\,''(t)\big) + a\mathcal{L}\big(y\,'(t)\big) + b\mathcal{L}\big(y(t)\big) = \mathcal{L}\big(f(t)\big)\text{.} \end{equation*}

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Let $Y(s) =\mathcal{L}\big(y(t)\big)$ and $F(s) = \mathcal{L}\big(f(t)\big)\text{.}$ Now apply Theorem 11.6.1 and Corollary 11.6.2 to get

\begin{equation*} \mathcal{L}\big(y\,'(t)\big) = sY(s) - y(0) \text{ and } \mathcal{L}\big(y\,''(t)\big) = s^2Y(s) - sy(0) - y\,'(0)\text{.} \end{equation*}

We can rewrite the last equation in the form

\begin{equation} s^2Y(s) + asY(s) + bY(s) = F(s) + sy(0) + y\,'(0) + ay(0)\text{.}\tag{11.6.1} \end{equation}

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The Laplace transform $Y(s)$ of the solution $y(t)$ can be shown to be

\begin{equation} Y(s) = \frac{F(s) +sy(0) + y\,'(0) + ay(0)}{s^2+as+b}\text{.}\tag{11.6.2} \end{equation}

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For many physical problems involving mechanical systems and electric circuits, the transform $F(s)$ is known, and the inverse of $Y(s)$ can easily be computed. This process is referred to as the operational calculus and has the advantage of changing problems in differential equations into problems in algebra. Then the solution obtained will satisfy the specific initial conditions.

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Example 11.6.6.

Solve the initial value problem

\begin{equation*} y\,''(t) + y(t) = 0 \text{ with } y(0) =2 \text{ and } y\,'(0) = 3\text{.} \end{equation*}

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Solution.

Since the right-hand side of the differential equation is $f(t) \equiv 0$ we have $F(s) \equiv 0\text{.}$ The initial conditions yield $\mathcal{L}\big(y\,''(t)\big) =s^2Y(s) - 2s- 3$ and Equation (11.6.1) becomes $s^2Y(s) + Y(s) = 2s + 3\text{.}$ Solving we get $Y(s) =\frac{2s+3}{s^2+1}$ and the solution $y(t)$ is assisted by using Table 11.5.10 and the computation

\begin{equation*} y(t) = \mathcal{L}^{-1}\left(\frac{2s+3}{s^2+1}\right) = 2\mathcal{L}^{-1}\left(\frac{s}{s^2+1}\right) + 3\mathcal{L}^{-1}\left(\frac{1}{s^2+1}\right) =2\cos t+3\sin t\text{.} \end{equation*}

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Example 11.6.7.

Solve the initial value problem

\begin{equation*} y\,''(t) +y\,'(t) - 2y(t) = 0 \text{ with } y(0) =1 \text{ and } y\,'(0) = 4\text{.} \end{equation*}

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Solution.

In the spirit of Example 11.6.6, we use the initial conditions and Equation (11.6.2) becomes

\begin{equation*} Y(s) = \frac{s+4+1}{s^2+s-2}=\frac{s+5}{(s-1)(s+2)}\text{.} \end{equation*}

Using partial fraction expansion $Y(s) =\frac{2}{s-1}-\frac{1}{s+2}$ and the solution $y(t)$ is

\begin{equation*} y(t) = \mathcal{L}^{-1}\big(Y(s)\big) = 2\mathcal{L}^{-1}\left(\frac{1}{s-1}\right) - \mathcal{L}^{-1}\left(\frac{1}{s+2}\right) = 2e^t-e^{-2t}\text{.} \end{equation*}

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