8.1: Introduction to the Laplace Transform

Last updated
Save as PDF

<h2 class="lt-math-9432">Definition of the Laplace Transform</h2> <p class="lt-math-9432">This is Paul Seeburger's version of this Section.</p> <p class="lt-math-9432">To define the Laplace transform, we first recall the definition of an <a title="7.7: Improper Integrals" href="/Bookshelves/Calculus/Calculus_(OpenStax)/07:_Techniques_of_Integration/7.07:_Improper_Integrals" mt-page-link-identifier="5822369bac8f43aba6815d3a6c197ad5">improper integral</a>. If \(g\) is integrable over the interval \([a,T]\) for every \(T>a\), then the <span><em>improper integral of \(g\)</em></span><em> over</em> \([a,\infty)\) is defined as</p> <p class="lt-math-9432">\[\label{eq:8.1.1} \int^\infty_a g(t)\,dt=\lim_{T\to\infty}\int^T_a g(t)\,dt.\]</p> <p class="lt-math-9432">We say that the improper integral <span><em>converges</em></span> if the limit in Equation \ref{eq:8.1.1} exists; otherwise, we say that the improper integral <span> <em>diverges</em></span> or <span><em>does not exist</em></span>. Here’s the definition of the Laplace transform of a function \(f\).</p> <section class="box-definition"> <header> <h5><span class="lt-icon-default">Definition <span class="script">template.Index(ID:1)</span>: Laplace Transform</span></h5> </header> <p class="lt-math-9432">Let \(f\) be defined for \(t\ge0\) and let \(s\) be a real number. Then the <em>Laplace transform</em> of \(f\) is the function \(F\) defined by</p> <p class="lt-math-9432">\[\label{eq:8.1.2} F(s)=\int_0^\infty e^{-st} f(t)\,dt,\]</p> <p class="lt-math-9432">for those values of \(s\) for which the improper integral converges.</p> </section> <p class="lt-math-9432">It is important to keep in mind that the variable of integration in Equation \ref{eq:8.1.2} is \(t\), while \(s\) is a parameter independent of \(t\). We use \(t\) as the independent variable for \(f\) because in applications the Laplace transform is usually applied to functions of time.</p> <p class="lt-math-9432">The Laplace transform can be viewed as an operator \({\mathscr L}\) that transforms the function \(f=f(t)\) into the function \(F=F(s)\). Thus, Equation \ref{eq:8.1.2} can be expressed as</p> <p class="lt-math-9432">\[F={\mathscr L}(f).\nonumber \]</p> <p class="lt-math-9432">The functions \(f\) and \(F\) form a <span><em>transform pair</em></span>, which we’ll sometimes denote by</p> <p class="lt-math-9432">\[f(t)\leftrightarrow F(s).\nonumber\]</p> <p class="lt-math-9432">It can be shown that if \(F(s)\) is defined for \(s=s_0\) then it is defined for all \(s>s_0\) (<em>Exercise 8.1.14b</em>).</p> <h2 class="lt-math-9432">Computation of Some Simple Laplace Transforms</h2> <section class="box-example"> <header> <h5><span class="lt-icon-default">Example <span class="script">template.Index(ID:1)</span></span></h5> </header> <p class="lt-math-9432">Find the Laplace transform of \(f(t)=1\).</p> <h6>Solution</h6> <p class="lt-math-9432">From Equation \ref{eq:8.1.2} with \(f(t)=1\),</p> <p class="lt-math-9432">\[F(s)=\int_0^\infty e^{-st}\,dt=\lim_{T\to\infty}\int_0^T e^{-st}\, dt.\nonumber\]</p> <p class="lt-math-9432">If \(s\ne 0\) then</p> <p class="lt-math-9432">\[\label{eq:8.1.3} \int_0^T e^{-st}dt=-{1\over s}e^{-st}\Big|_0^T={1-e^{-sT}\over s}.\]</p> <p class="lt-math-9432">Therefore</p> <p class="lt-math-9432">\[\label{eq:8.1.4} \lim_{T\to\infty}\int_0^T e^{-st}dt=\lim_{T\to\infty} {1-e^{-sT}\over s}=\left\{\begin{array}{rr} {1\over s}, & \text{if }s>0,\\ \infty, & \text{if }s<0. \end{array}\right.\]</p> <p class="lt-math-9432">If \(s=0\) the integrand reduces to the constant \(1\), and</p> <p class="lt-math-9432">\[\lim_{T\to\infty}\int_0^T 1\,dt=\lim_{T\to\infty}\int_0^T 1\,dt= \lim_{T\to\infty}T=\infty.\nonumber\]</p> <p class="lt-math-9432">Therefore \(F(0)\) is undefined, and</p> <p class="lt-math-9432">\[F(s)=\int_0^\infty e^{-st}dt={1\over s},\quad s>0.\nonumber\]</p> <p class="lt-math-9432">This result can be written in operator notation as</p> <p class="lt-math-9432">\[{\mathscr L}(1)={1\over s},\quad s>0,\nonumber\]</p> <p class="lt-math-9432">or as the transform pair</p> <p class="lt-math-9432">\[1\leftrightarrow{1\over s},\quad s>0.\nonumber\]</p> </section> <p> </p> <section class="box-example"> <header> <h5><span class="lt-icon-default">Example <span class="script">template.Index(ID:2)</span></span></h5> </header> <p class="lt-math-9432">Find the Laplace transform of \(f(t)=t\).</p> <p class="lt-math-9432">From Equation \ref{eq:8.1.2} with \(f(t)=t\),</p> <p class="lt-math-9432">\[\label{eq:8.1.5} F(s)=\int_0^\infty e^{-st}t\,dt = \lim_{T\to\infty}\int_0^T e^{-st}t\, dt.\]</p> <p class="lt-math-9432">If \(s\ne0\), integrating by parts with</p> <p class="lt-math-9432">\[\begin{array}{ll} u = t & dv = e^{-st}\,dt \\ du = dt & v = {-1 \over s} e^{-st}\end{array} \nonumber\]</p> <p class="lt-math-9432">yields</p> <p class="lt-math-9432">\[\begin{align*} \int_0^T e^{-st}t\,dt&=-{te^{-st}\over s}\bigg|_0^T +{1\over s}\int_0^T e^{-st}\,dt \\[5pt]<br /> &=-\left[{t\over s}+{1\over s^2}\right]e^{-st}\bigg|_0^T \\[5pt]<br /> &=-\left[{T\over s}+{1\over s^2}\right]e^{-sT} + \left[{0\over s}+{1\over s^2}\right]e^{0} \\[5pt]<br /> &={1\over s^2} - \left[{T\over s}+{1\over s^2}\right]e^{-sT} \end{align*}\nonumber\]</p> <p class="lt-math-9432">Then if \(s>0,\) we have</p> <p class="lt-math-9432">\[\begin{align*} F(s)&=\int_0^\infty e^{-st}t\,dt = \lim_{T\to\infty}\int_0^T e^{-st}t\, dt \\[4pt]<br /> &= \lim_{T\to\infty} \bigg( {1\over s^2} - \left[{T\over s}+{1\over s^2}\right]e^{-sT} \bigg) \\[4pt]<br /> &= \lim_{T\to\infty} \bigg( {1\over s^2} - \frac{T}{s e^{sT}}-{1\over s^2}e^{-sT} \bigg) \\[4pt]<br /> &= \lim_{T\to\infty}{1\over s^2} - \lim_{T\to\infty}\frac{T}{s e^{sT}}-\lim_{T\to\infty}\frac{1}{s^2}e^{-sT} \\[4pt]<br /> &= {1\over s^2} - \lim_{T\to\infty}\cancelto{\infty/\infty}{\frac{T}{s e^{sT}}}-\quad\lim_{T\to\infty}{\cancelto{0}{\frac{1}{s^2e^{sT}}}} \end{align*}\]</p> <p class="lt-math-9432">Since the middle term is in the indeterminate form \(\infty/\infty,\) we use L'Hôpital's rule to determine this limit:</p> <p class="lt-math-9432">\[\begin{align*} F(s) &= {1\over s^2} - \lim_{T\to\infty}\frac{T}{s e^{sT}} \\[4pt]<br /> &= {1\over s^2} - \lim_{T\to\infty}\cancelto{0}{\frac{1}{s^2 e^{sT}}} & & \text{We apply L'Hôpital's rule in this step.}\\[4pt]<br /> &= {1\over s^2} \end{align*}\]</p> <p class="lt-math-9432">If instead we assume that \(s<0,\) we have</p> <p class="lt-math-9432">\[\begin{align*} F(s)&=\int_0^\infty e^{-st}t\,dt = \lim_{T\to\infty}\int_0^T e^{-st}t\, dt \\[4pt]<br /> &= \lim_{T\to\infty} \bigg( {1\over s^2} - \left[{T\over s}+{1\over s^2}\right]e^{-sT} \bigg) \\[4pt]<br /> &= \lim_{T\to\infty} {1\over s^2} - \lim_{T\to\infty} \left[\frac{Ts+1}{s^2}\right]e^{-sT} \\[4pt]<br /> &= {1\over s^2} - \lim_{T\to\infty} \cancelto{-\infty\cdot\infty}{\left[\frac{Ts+1}{s^2}\right]e^{-sT}} & & \text{Note that } e^{-sT}\to\infty\text{ as }T\to\infty,\text{ since }s<0.\\[4pt]<br /> &= {1\over s^2} - (-\infty) \; = \; \infty \end{align*}\]</p> <p class="lt-math-9432">Thus we can conclude that \[ F(s)=\left\{\begin{array}{rr} \dfrac{1}{s^2},& \text{if }s>0,\\ \infty,&\text{if }s<0.\end{array}\right.\nonumber\]</p> <p class="lt-math-9432">If \(s=0\), Equation \ref{eq:8.1.5} becomes</p> <p class="lt-math-9432">\[\begin{align*} F(0)&=\int_0^\infty e^{-st}t\,dt = \int_0^\infty t\,dt \\[4pt]<br /> &= \lim_{T\to\infty}\int_0^T t\, dt= \lim_{T\to\infty} \left(\,{t^2\over2}\bigg|_0^T \,\right)\\[4pt]<br /> &= \frac{1}{2}\lim_{T\to\infty} \left(T^2 - 0\right) = \infty. \end{align*}\]</p> <p class="lt-math-9432">Therefore \(F(0)\) is undefined and</p> <p class="lt-math-9432">\[F(s)={1\over s^2},\quad s>0.\nonumber\]</p> <p class="lt-math-9432">This result can also be written as</p> <p class="lt-math-9432">\[{\mathscr L}(t)={1\over s^2},\quad s>0,\nonumber\]</p> <p class="lt-math-9432">or as the transform pair</p> <p class="lt-math-9432">\[t\leftrightarrow{1\over s^2},\quad s>0.\nonumber\]</p> </section> <section class="box-example"> <header> <h5><span class="lt-icon-default">Example <span class="script">template.Index(ID:3)</span></span></h5> </header> <p class="lt-math-9432">Find the Laplace transform of \(f(t)=e^{at}\), where \(a\) is a constant.</p> <p class="lt-math-9432">From Equation \ref{eq:8.1.2} with \(f(t)=e^{at}\),</p> <p class="lt-math-9432">\[F(s)=\int_0^\infty e^{-st}e^{at}\,dt.\nonumber\]</p> <p class="lt-math-9432">Combining the exponentials yields</p> <p class="lt-math-9432">\[F(s)=\int_0^\infty e^{-(s-a)t}\,dt.\nonumber\]</p> <p class="lt-math-9432">However, we know from Example <span class="script">template.Index(ID:1)</span> that</p> <p class="lt-math-9432">\[\int_0^\infty e^{-st}\,dt={1\over s},\quad s>0.\nonumber\]</p> <p class="lt-math-9432">Replacing \(s\) by \(s-a\) here shows that</p> <p class="lt-math-9432">\[F(s)={1\over s-a},\quad s>a.\nonumber\]</p> <p class="lt-math-9432">This can also be written as</p> <p class="lt-math-9432">\[{\mathscr L}(e^{at})={1\over s-a},\quad s>a, \text{ or } e^{at}\leftrightarrow{1\over s-a},\quad s>a.\nonumber\]</p> </section> <section class="box-example"> <header> <h5><span class="lt-icon-default">Example <span class="script">template.Index(ID:4)</span></span></h5> </header> <p class="lt-math-9432">Find the Laplace transforms of \(f(t)=\sin\omega t\) and \(g(t)=\cos\omega t\), where \(\omega\) is a constant.</p> <p class="lt-math-9432">Define</p> <p class="lt-math-9432">\[\label{eq:8.1.6} F(s)=\int_0^\infty e^{-st}\sin\omega t\,dt\]</p> <p class="lt-math-9432">and</p> <p class="lt-math-9432">\[\label{eq:8.1.7} G(s)=\int_0^\infty e^{-st}\cos\omega t\,dt.\]</p> <p class="lt-math-9432">If \(s>0\), integrating Equation \ref{eq:8.1.6} by parts with</p> <p class="lt-math-9432">\[\begin{array}{ll} u = \sin\omega t & dv = e^{-st}dt \\ du = \omega\cos\omega t & v = -\dfrac{e^{-st}}{s} \end{array} \nonumber\]</p> <p class="lt-math-9432">yields</p> <p class="lt-math-9432">\[\begin{align*} F(s) &= \int_0^\infty e^{-st}\sin\omega t\,dt = \lim_{T\to\infty}\int_0^T e^{-st}\sin\omega t\,dt \\[4pt]<br /> &=\lim_{T\to\infty} \bigg[ -{e^{-st}\over s}\sin\omega t\Big|_0^T+{\omega\over s} \int_0^T e^{-st}\cos\omega t\,dt \bigg] \\[4pt]<br /> &= \lim_{T\to\infty} \bigg[ -\cancelto{0}{{e^{-sT}\over s}\sin\omega T} + 0 \bigg] + {\omega\over s} \int_0^\infty e^{-st}\cos\omega t\,dt & & \text{ The limit of the first term is }0\text{ since } |\sin\omega T| \le 1.\\[4pt]<br /> & = {\omega\over s}G(s), \end{align*}\]</p> <p class="lt-math-9432">so</p> <p class="lt-math-9432">\[\label{eq:8.1.8} F(s)={\omega\over s}G(s).\]</p> <p class="lt-math-9432">Similarly, if \(s>0\), integrating Equation \ref{eq:8.1.7} by parts yields</p> <p class="lt-math-9432">\[\begin{align*} G(s) &= \int_0^\infty e^{-st}\cos\omega t\,dt = \lim_{T\to\infty}\int_0^T e^{-st}\cos\omega t\,dt \\[4pt]<br /> &=\lim_{T\to\infty} \bigg[ -{e^{-st}\over s}\cos\omega t\Big|_0^T-{\omega\over s} \int_0^T e^{-st}\sin\omega t\,dt \bigg] \\[4pt]<br /> &= \lim_{T\to\infty} \bigg[ -\cancelto{0}{{e^{-sT}\over s}\cos\omega T} + {1 \over s} \bigg] - {\omega\over s} \int_0^\infty e^{-st}\sin\omega t\,dt & & \text{ The limit of the first term is }0\text{ since } |\cos\omega T| \le 1.\\[4pt]<br /> & = {1 \over s} - {\omega\over s}F(s), \end{align*}\]</p> <p class="lt-math-9432">so</p> <p class="lt-math-9432">\[G(s)={1\over s} - {\omega\over s} F(s).\nonumber\]</p> <p class="lt-math-9432">Now substitute from Equation \ref{eq:8.1.8} into this to obtain</p> <p class="lt-math-9432">\[G(s)={1\over s} - {\omega^2\over s^2} G(s).\nonumber\]</p> <p class="lt-math-9432">Solving this for \(G(s)\) yields</p> <p class="lt-math-9432">\[G(s)={s\over s^2+\omega^2},\quad s>0.\nonumber\]</p> <p class="lt-math-9432">This and Equation \ref{eq:8.1.8} imply that</p> <p class="lt-math-9432">\[F(s)={\omega\over s^2+\omega^2},\quad s>0.\nonumber\]</p> </section> <h2 class="lt-math-9432">Tables of Laplace Transforms</h2> <p class="lt-math-9432">Extensive tables of Laplace transforms have been compiled and are commonly used in applications. The brief table of Laplace transforms in the Appendix will be adequate for our purposes.</p> <section class="box-example"> <header> <h5><span class="lt-icon-default">Example <span class="script">template.Index(ID:5)</span></span></h5> </header> <p class="lt-math-9432">Use the table of Laplace transforms to find \({\mathscr L}(t^3e^{4t})\).</p> <p class="lt-math-9432">The table includes the transform pair</p> <p class="lt-math-9432">\[t^ne^{at}\leftrightarrow {n!\over(s-a)^{n+1}}.\nonumber\]</p> <p class="lt-math-9432">Setting \(n=3\) and \(a=4\) here yields</p> <p class="lt-math-9432">\[{\mathscr L} (t^3e^{4t})={3!\over(s-4)^4}={6\over(s-4)^4}.\nonumber\]</p> </section> <p class="lt-math-9432">We’ll sometimes write Laplace transforms of specific functions without explicitly stating how they are obtained. In such cases you should refer to the table of Laplace transforms.</p> <h2 class="lt-math-9432">Linearity of the Laplace Transform</h2> <p class="lt-math-9432">The next theorem presents an important property of the Laplace transform.</p> <section class="box-theorem"> <header> <h5><span class="lt-icon-default">Theorem <span class="script">template.Index(ID:2)</span> Linearity Property</span></h5> </header> <p class="lt-math-9432">Suppose \({\mathscr L}(f_i)\) is defined for \(s>s_i,\) \(1\le i\le n).\) Let \(s_0\) be the largest of the numbers \(s_1\), \(s_{2},\) …,\(s_n,\) and let \(c_1\), \(c_2\),…, \(c_n\) be constants. Then</p> <p class="lt-math-9432">\[{\mathscr L}(c_1f_1+c_2f_2+\cdots+c_nf_n)=c_1{\mathscr L}(f_1)+c_2{\mathscr L}(f_2) +\cdots+c_n{\mathscr L}(f_n)\mbox{ for } s>s_0.\nonumber\]</p> <dl> <dt><strong class="emphasis bold">Proof</strong></dt> <dd> <p class="lt-math-9432">We give the proof for the case where \(n=2\). If \(s>s_0\) then</p> <p class="lt-math-9432">\[\begin{aligned} {\mathscr L}(c_1f_1+c_2f_2)&= \int_0^\infty e^{-st}\left(c_1f_1(t)+c_2f_2(t))\right)\,dt\\ &= c_1\int_0^\infty e^{-st}f_1(t)\,dt+c_2\int_0^\infty e^{-st}f_2(t)\,dt\\ &= c_1{\mathscr L}(f_1)+c_2{\mathscr L}(f_2).\end{aligned}\nonumber\]</p> </dd> </dl> </section> <section class="box-example"> <header> <h5><span class="lt-icon-default">Example <span class="script">template.Index(ID:6)</span></span></h5> </header> <p class="lt-math-9432">Use Theorem <span class="script">template.Index(ID:2)</span> and the known Laplace transform</p> <p class="lt-math-9432">\[{\mathscr L}(e^{at})={1\over s-a} \nonumber\]</p> <p class="lt-math-9432">to find \({\mathscr L}(\cosh bt)\,(b\ne0)\).</p> <h6>Solution</h6> <p class="lt-math-9432">By definition,</p> <p class="lt-math-9432">\[\cosh bt={e^{bt}+e^{-bt}\over 2}. \nonumber\]</p> <p class="lt-math-9432">Therefore</p> <p class="lt-math-9432">\[\label{eq:8.1.9} \begin{array}{ccl} {\mathscr L}(\cosh bt)&=& {\mathscr L}\left( {1\over 2} e^{bt}+ {1\over 2}e^{-bt}\right)\\[4pt] &=& {1\over 2} {\mathscr L}(e^{bt}) + {1\over 2} {\mathscr L}(e^{-bt}) \qquad \hbox{(by the Linearity Property)}\\[4pt] &=& \dfrac{1}{2}\cdot \dfrac{1}{s-b} + \dfrac{1}{2}\cdot \dfrac{1}{s+b}, \end{array}\]</p> <p class="lt-math-9432">where the first transform on the right is defined for \(s>b\) and the second for \(s>-b\); hence, both are defined for \(s>|b|\). Simplifying the last expression in Equation \ref{eq:8.1.9} yields</p> <p class="lt-math-9432">\[{\mathscr L}(\cosh bt)={s\over s^2-b^2},\quad s>|b|.\nonumber\]</p> </section> <p class="lt-math-9432">The next theorem enables us to start with known transform pairs and derive others. (For other results of this kind, see <em>Exercises 8.1.6</em> and <em>8.1.13</em>.)</p> <section class="box-theorem"> <header> <h5><span class="lt-icon-default">Theorem <span class="script">template.Index(ID:3)</span> First Shifting Theorem</span></h5> </header> <p class="lt-math-9432">If \(\displaystyle F(s)=\int_0^\infty e^{-st} f(t)\,dt\) is the Laplace transform of \(f(t)\) for \(s>s_0\), then</p> <p class="lt-math-9432">\[{\mathscr L}\big(e^{at}f(t)\big) = F(s - a), \quad \text{for }s >s_0+a. \nonumber\]</p> <p class="lt-math-9432">That is, \(F(s-a)\) is the Laplace transform of \(e^{at}f(t)\) for \(s >s_0+a\).</p> <dl> <dt><strong class="emphasis bold">Proof</strong></dt> <dd> <p class="lt-math-9432">Replacing \(s\) by \(s-a\) in Equation \ref{eq:8.1.10} yields</p> <p class="lt-math-9432">\[\label{eq:8.1.11} F(s-a)=\int_0^\infty e^{-(s-a)t}f(t)\,dt\]</p> <p class="lt-math-9432">if \(s-a>s_0\); that is, if \(s>s_0+a\). However, Equation \ref{eq:8.1.11} can be rewritten as</p> <p class="lt-math-9432">\[F(s-a)=\int_0^\infty e^{-st}\left(e^{at}f(t)\right)\,dt,\nonumber\]</p> <p class="lt-math-9432">which implies the conclusion.</p> </dd> </dl> </section> <section class="box-example"> <header> <h5><span class="lt-icon-default">Example <span class="script">template.Index(ID:7)</span></span></h5> </header> <p class="lt-math-9432">Use Theorem <span class="script">template.Index(ID:3)</span> and the known Laplace transforms of \(1\), \(t\), \(\cos\omega t\), and \(\sin\omega t\) to find</p> <p class="lt-math-9432">\[{\mathscr L}(e^{at}),\quad {\mathscr L}(te^{at}),\quad {\mathscr L}(e^{\lambda t}\sin \omega t),\mbox{and } {\mathscr L}(e^{\lambda t}\cos\omega t).\nonumber\]</p> <h6>Solution</h6> <p class="lt-math-9432">In the following table the known transform pairs are listed on the left and the required transform pairs listed on the right are obtained by applying Theorem <span class="script">template.Index(ID:3)</span>.</p> <table class="mt-responsive-table"> <caption>Table <span class="script">template.Index(ID:1)</span></caption> <thead> <tr class="header"> <th style="text-align:center;vertical-align:middle;">\(f(t)\leftrightarrow F(s)\)</th> <th style="text-align:center;vertical-align:middle;">\(e^{at}f(t)\leftrightarrow F(s-a)\)</th> </tr> </thead> <tbody> <tr class="even"> <td class="lt-math-9432" data-th="\(f(t)\leftrightarrow F(s)\)" style="text-align:center;vertical-align:middle;">\(1\leftrightarrow \dfrac{1}{s},\quad s>0\)</td> <td class="lt-math-9432" data-th="\(e^{at}f(t)\leftrightarrow F(s-a)\)" style="text-align:center;vertical-align:middle;">\(e^{at}\leftrightarrow \dfrac{1}{s-a},\quad s>a\)</td> </tr> <tr class="odd"> <td class="lt-math-9432" data-th="\(f(t)\leftrightarrow F(s)\)" style="text-align:center;vertical-align:middle;">\(t\leftrightarrow \dfrac{1}{s^{2}},\quad s>0\)</td> <td class="lt-math-9432" data-th="\(e^{at}f(t)\leftrightarrow F(s-a)\)" style="text-align:center;vertical-align:middle;">\(te^{at}\leftrightarrow \dfrac{1}{(s-a)^{2}},\quad s>a\)</td> </tr> <tr class="even"> <td class="lt-math-9432" data-th="\(f(t)\leftrightarrow F(s)\)" style="text-align:center;vertical-align:middle;">\(\sin\omega t\leftrightarrow \dfrac{\omega }{s^{2}+\omega ^{2}},\quad s>0\)</td> <td class="lt-math-9432" data-th="\(e^{at}f(t)\leftrightarrow F(s-a)\)" style="text-align:center;vertical-align:middle;">\(e^{\lambda t}\sin\omega t\leftrightarrow \dfrac{\omega}{(s-\lambda)^{2}+\omega ^{2}},\quad s>\lambda\)</td> </tr> <tr class="odd"> <td class="lt-math-9432" data-th="\(f(t)\leftrightarrow F(s)\)" style="text-align:center;vertical-align:middle;">\(\cos\omega t\leftrightarrow \dfrac{s}{s^{2}+\omega ^{2}},\quad s>0\)</td> <td class="lt-math-9432" data-th="\(e^{at}f(t)\leftrightarrow F(s-a)\)" style="text-align:center;vertical-align:middle;">\(e^{\lambda t}\sin\omega t\leftrightarrow \dfrac{s-\lambda }{(s-\lambda )^{2}+\omega ^{2}},\quad s>\lambda\)</td> </tr> </tbody> </table> </section> <h2 class="lt-math-9432">Existence of Laplace Transforms</h2> <p class="lt-math-9432">Not every function has a Laplace transform. For example, it can be shown (<em>Exercise 8.1.3</em>) that</p> <p class="lt-math-9432">\[\int_0^\infty e^{-st}e^{t^2} dt=\infty\nonumber\]</p> <p class="lt-math-9432">for every real number \(s\). Hence, the function \(f(t)=e^{t^2}\) does not have a Laplace transform.</p> <p class="lt-math-9432">Our next objective is to establish conditions that ensure the existence of the Laplace transform of a function. We first review some relevant definitions from calculus.</p> <p class="lt-math-9432">Recall that a limit</p> <p class="lt-math-9432">\[\lim_{t\to t_0} f(t)\nonumber\]</p> <p class="lt-math-9432">exists if and only if the one-sided limits</p> <p class="lt-math-9432">\[\lim_{t\to t_0-}f(t)\quad \text{and} \quad \lim_{t\to t_0+}f(t)\nonumber\]</p> <p class="lt-math-9432">both exist and are equal; in this case,</p> <p class="lt-math-9432">\[\lim_{t\to t_0}f(t)=\lim_{t\to t_0-}f(t)=\lim_{t\to t_0+}f(t) .\nonumber\]</p> <p class="lt-math-9432">Recall also that \(f\) is continuous at a point \(t_0\) in an open interval \((a,b)\) if and only if</p> <p class="lt-math-9432">\[\lim_{t\to t_0}f(t)=f(t_0),\nonumber\]</p> <p class="lt-math-9432">which is equivalent to</p> <p class="lt-math-9432">\[\label{eq:8.1.12} \lim_{t\to t_0+}f(t)=\lim_{t\to t_0-}f(t)=f(t_0).\]</p> <p class="lt-math-9432">For simplicity, we define</p> <p class="lt-math-9432">\[f(t_0+)=\lim_{t\to t_0+}f(t)\quad\hbox{and }\quad f(t_0-)=\lim_{t\to t_0-}f(t),\nonumber\]</p> <p class="lt-math-9432">so Equation \ref{eq:8.1.12} can be expressed as</p> <p class="lt-math-9432">\[f(t_0+)=f(t_0-)=f(t_0).\nonumber\]</p> <p class="lt-math-9432">If \(f(t_0+)\) and \(f(t_0-)\) have finite but distinct values, we say that \(f\) has a <span><em>jump discontinuity</em></span> at \(t_0\), and</p> <p class="lt-math-9432">\[f(t_0+)-f(t_0-)\nonumber\]</p> <p class="lt-math-9432">is called the <span><em>jump</em></span> in \(f\) at \(t_0\) (Figure <span class="script">template.Index(ID:1)</span>).</p> <figure><img alt="clipboard_e78f1a38a6c5d8fae002e0ac2e97ee81b.png" src="/@api/deki/files/30344/clipboard_e78f1a38a6c5d8fae002e0ac2e97ee81b.png" /> <figcaption>Figure <span class="script">template.Index(ID:1)</span>: A jump discontinuity.</figcaption> </figure> <p class="lt-math-9432">If \(f(t_0+)\) and \(f(t_0-)\) are finite and equal, but either \(f\) isn’t defined at \(t_0\) or it is defined but</p> <p class="lt-math-9432">\[f(t_0)\ne f(t_0+)=f(t_0-),\nonumber\]</p> <p class="lt-math-9432">we say that \(f\) has a <span><em>removable discontinuity</em></span> at \(t_0\) (Figure <span class="script">template.Index(ID:2)</span>). This terminolgy is appropriate since a function \(f\) with a removable discontinuity at \(t_0\) can be made continuous at \(t_0\) by defining (or redefining)</p> <p class="lt-math-9432">\[f(t_0)=f(t_0+)=f(t_0-).\nonumber\]</p> <section class="box-note"> <header> <h5><span class="lt-icon-default">Note</span></h5> </header> <p class="lt-math-9432">We know from calculus that a definite integral isn’t affected by changing the values of its integrand at isolated points. Therefore, redefining a function f to make it continuous at removable discontinuities does not change \(\cal{L}(f)\).</p> </section> <section class="box-definition"> <header> <h5><span class="lt-icon-default">Definition <span class="script">template.Index(ID:4)</span>: Piecewise Continuous</span></h5> </header> <ul> <li class="lt-math-9432">A function \(f\) is said to be <span><em>piecewise continuous</em></span> on a finite closed interval \([0,T]\) if \(f(0+)\) and \(f(T-)\) are finite and \(f\) is continuous on the open interval \((0,T)\) except possibly at finitely many points, where \(f\) may have jump discontinuities or removable discontinuities.</li> <li class="lt-math-9432">A function \(f\) is said to be <span><em>piecewise continuous</em></span> on the infinite interval \([0,\infty)\) if it is piecewise continuous on \([0,T]\) for every \(T>0\).</li> </ul> </section> <p class="lt-math-9432">Figure <span class="script">template.Index(ID:3)</span> shows the graph of a typical piecewise continuous function.</p> <p class="lt-math-9432">It is shown in calculus that if a function is piecewise continuous on a finite closed interval then it is integrable on that interval. But if \(f\) is piecewise continuous on \([0,\infty)\), then so is \(e^{-st}f (t)\), and therefore</p> <p class="lt-math-9432">\[\int_0^T e^{-st}f(t)\,dt \nonumber\]</p> <figure><img alt="clipboard_e8a13bcff17c498b55d624b6011453b79.png" src="/@api/deki/files/30345/clipboard_e8a13bcff17c498b55d624b6011453b79.png" /> <figcaption>Figure <span class="script">template.Index(ID:2)</span></figcaption> </figure> <figure><img alt="clipboard_ef2ae35576ae06c2ad537e4572cc6fc43.png" src="/@api/deki/files/30346/clipboard_ef2ae35576ae06c2ad537e4572cc6fc43.png" /> <figcaption>Figure <span class="script">template.Index(ID:3)</span></figcaption> </figure> <p class="lt-math-9432">exists for every \(T>0\). However, piecewise continuity alone does not guarantee that the improper integral</p> <p class="lt-math-9432">\[\label{eq:8.1.13} \int_0^\infty e^{-st}f(t)\,dt=\lim_{T\to\infty}\int_0^T e^{-st}f(t)\, dt\]</p> <p class="lt-math-9432">converges for \(s\) in some interval \((s_0,\infty)\). For example, we noted earlier that Equation \ref{eq:8.1.13} diverges for all \(s\) if \(f(t)=e^{t^2}\). Stated informally, this occurs because \(e^{t^2}\) increases too rapidly as \(t\to\infty\). The next definition provides a constraint on the growth of a function that guarantees convergence of its Laplace transform for \(s\) in some interval \((s_0,\infty)\).</p> <section class="box-definition"> <header> <h5><span class="lt-icon-default">Definition <span class="script">template.Index(ID:5)</span>: of exponential order</span></h5> </header> <p class="lt-math-9432">A function \(f\) is said to be <span><em>of exponential order</em></span> \(s_0\) if there are constants \(M\) and \(t_0\) such that</p> <p class="lt-math-9432">\[\label{eq:8.1.14} |f(t)|\le Me^{s_0t},\quad t\ge t_0.\]</p> <p class="lt-math-9432">In situations where the specific value of \(s_0\) is irrelevant we say simply that \(f\) is <span><em>of exponential order</em></span>.</p> </section> <p class="lt-math-9432">The next theorem gives useful sufficient conditions for a function \(f\) to have a Laplace transform. The proof is sketched in <em>Exercise 8.1.10</em>.</p> <section class="box-theorem"> <header> <h5><span class="lt-icon-default">Theorem <span class="script">template.Index(ID:6)</span></span></h5> </header> <p class="lt-math-9432">If \(f\) is piecewise continuous on \([0,\infty)\) and of exponential order \(s_0,\) then \({\mathscr L}(f)\) is defined for \(s>s_0\).</p> </section> <section class="box-note"> <header> <h5><span class="lt-icon-default">Note</span></h5> </header> <p class="lt-math-9432">We emphasize that the conditions of Theorem <span class="script">template.Index(ID:6)</span> are sufficient, but <em>not necessary</em>, for \(f\) to have a Laplace transform. For example, <em>Exercise 8.1.14(c)</em> shows that \(f\) may have a Laplace transform even though \(f\) isn’t of exponential order</p> </section> <section class="box-example"> <header> <h5><span class="lt-icon-default">Example <span class="script">template.Index(ID:8)</span></span></h5> </header> <p class="lt-math-9432">If \(f\) is bounded on some interval \([t_0,\infty)\), say</p> <p class="lt-math-9432">\[|f(t)|\le M,\quad t\ge t_0,\nonumber\]</p> <p class="lt-math-9432">then Equation \ref{eq:8.1.14} holds with \(s_0=0\), so \(f\) is of exponential order zero. Thus, for example, \(\sin\omega t\) and \(\cos \omega t\) are of exponential order zero, and Theorem <span class="script">template.Index(ID:6)</span> implies that \({\mathscr L}(\sin\omega t)\) and \({\mathscr L}(\cos \omega t)\) exist for \(s>0\). This is consistent with the conclusion of Example <span class="script">template.Index(ID:4)</span>.</p> </section> <section class="box-example"> <header> <h5><span class="lt-icon-default">Example <span class="script">template.Index(ID:9)</span></span></h5> </header> <p class="lt-math-9432">It can be shown that if \(\lim_{t\to\infty}e^{-s_0t}f(t)\) exists and is finite then \(f\) is of exponential order \(s_0\) (<em>Exercise 8.1.9</em>). If \(\alpha\) is any real number and \(s_0>0\) then \(f(t)=t^\alpha\) is of exponential order \(s_0\), since</p> <p class="lt-math-9432">\[\lim_{t\to\infty}e^{-s_0t}t^\alpha=0,\nonumber\]</p> <p class="lt-math-9432">by L’Hôpital’s rule. If \(\alpha\ge 0\), \(f\) is also continuous on \([0,\infty)\). Therefore <em>Exercise 8.1.9</em> and Theorem <span class="script">template.Index(ID:6)</span> imply that \({\mathscr L}(t^\alpha)\) exists for \(s\ge s_0\). However, since \(s_0\) is an arbitrary positive number, this really implies that \({\mathscr L}(t^\alpha)\) exists for all \(s>0\). This is consistent with the results of Example <span class="script">template.Index(ID:2)</span> and <em>Exercises 8.1.6</em> and<em> 8.1.8</em>.</p> </section> <section class="box-example"> <header> <h5><span class="lt-icon-default">Example <span class="script">template.Index(ID:10)</span></span></h5> </header> <p class="lt-math-9432">Find the Laplace transform of the piecewise continuous function</p> <p class="lt-math-9432">\[f(t)=\left\{\begin{array}{cl} 1,& \text{if }0\le t<1,\\ -3e^{-t},&\text{if }t\ge 1.\end{array}\right.\nonumber\]</p> <h6>Solution</h6> <p class="lt-math-9432">Since \(f\) is defined by different formulas on \([0,1)\) and \([1,\infty)\), we write</p> <p class="lt-math-9432">\[F(s)=\int_0^\infty e^{-st} f(t)\,dt =\int_0^1e^{-st}(1)\,dt+\int_1^\infty e^{-st}(-3e^{-t})\,dt.\nonumber\]</p> <p class="lt-math-9432">Since</p> <p class="lt-math-9432">\[\int_{0}^{1}e^{-st}dt = \left\{\begin{array}{cl} {\dfrac{1-e^{-s}}{s},}&{\text{if }s\neq 0} \\ {1,}&{\text{if }s=0} \end{array} \right. \nonumber \]</p> <p class="lt-math-9432">and</p> <p class="lt-math-9432">\[\int_1^\infty e^{-st}(-3e^{-t})\,dt=-3\int_1^\infty e^{-(s+1)t}\,dt\; =\; -{3e^{-(s+1)}\over s+1},\quad s>-1,\nonumber\]</p> <p class="lt-math-9432">it follows that</p> <p class="lt-math-9432">\[F(s) = \left\{\begin{array}{cl}{\dfrac{1-e^{-s}}{s}-3\dfrac{e^{-(s+1)}}{s+1},}&{\text{if }s>-1, s\neq 0} \\ {1-\dfrac{3}{e},}&{\text{if }s=0} \end{array} \right. \nonumber \]</p> <p class="lt-math-9432">This is consistent with Theorem <span class="script">template.Index(ID:6)</span>, since</p> <p class="lt-math-9432">\[|f(t)|\le 3e^{-t},\quad t\ge 1,\nonumber\]</p> <p class="lt-math-9432">and therefore \(f\) is of exponential order \(s_0=-1\).</p> </section> <section class="box-note"> <header> <h5><span class="lt-icon-default">Note</span></h5> </header> <p class="lt-math-9432">In Section 8.4 we’ll develop a more efficient method for finding Laplace transforms of piecewise continuous functions.</p> </section> <section class="box-example"> <header> <h5><span class="lt-icon-default">Example <span class="script">template.Index(ID:11)</span></span></h5> </header> <p class="lt-math-9432">We stated earlier that</p> <p class="lt-math-9432">\[\int_0^\infty e^{-st} e^{t^2} dt=\infty \nonumber\]</p> <p class="lt-math-9432">for all \(s\), so Theorem <span class="script">template.Index(ID:6)</span> implies that \(f(t)=e^{t^2}\) is not of exponential order, since</p> <p class="lt-math-9432">\[\lim_{t\to\infty} {e^{t^2}\over Me^{s_0t}}=\lim_{t\to\infty} {1\over M} e^{t^2-s_0t}=\infty, \nonumber\]</p> <p class="lt-math-9432">so</p> <p class="lt-math-9432">\[e^{t^2}>Me^{s_0t} \nonumber\]</p> <p class="lt-math-9432">for sufficiently large values of \(t\), for any choice of \(M\) and \(s_{0}\) (<em>Exercise 8.1.3</em>).</p> </section> <div class="comment"> <div class="mt-comment-content"> <p><a href="/Bookshelves/Differential_Equations/Elementary_Differential_Equations_with_Boundary_Value_Problems_(Trench)/08:_Laplace_Transforms/8.01:_Introduction_to_the_Laplace_Transform" mt-page-link-identifier="e468555b3f9544bcb07620aadc8d6e2e">Content Reuse Link: Bookshelves/Differential_Equations/Book%3A_Elementary_Differential_Equations_with_Boundary_Values_Problems_(Trench)/08%3A_Laplace_Transforms/8.01%3A_Introduction_to_the_Laplace_Transform</a></p> </div> </div> <h2>Contributors:</h2> <ul> <li>Paul Seeburger edited this section extensively to improve all examples using the limit definition of the Laplace transform. In particular, he added significant content to Examples \(\PageIndex{2}\) and \(\PageIndex{4}\).</li> </ul>