5  Laplace Transforms

These notes are far from complete. However, you may find the examples helpful. Your main resource for this material should be Chapter 1 of Tupper’s Notes.

5.1 Definition and basic properties

We will see two integral transforms in this course: the Fourier Transform and the Laplace Transform. In general, an integral transform takes a function, \(f\), of one variable defined on an interval \([a,b]\), multiplies it by a kernel, \(k\), depending on two variables, and integrates the product over the domain \([a,b]\). The result is a function of the second variable appearing in the kernel: \[F(s) = \int_a^b k(t,s) f(t)\, dt\] The Laplace Transform takes a function \(f\) defined on \((0,\infty)\) and transforms it using the kernel \(k(t,s) = e^{-st}\). \[F(s) = \int_0^\infty e^{-st} f(t)\, dt\] We can and will think of the transform as a function whose inputs are functions of time, and whose outputs are functions of the second variable \(s\). We’ll use a fancy L to denote this function. Hence, the Laplace Transform of a function \(f\) is the function \(F\) defined by \[F(s) = {\mathcal L}\left\{f\right\}(s) = \int_0^\infty e^{-st}f(t)\, dt \tag{5.1}\]

If we just want to refer to the transformed function we’ll use \({\mathcal L}\left\{f\right\}\). Sometimes we’ll use the notation \({\mathcal L}\left\{f(t)\right\}\), which is inconsistent with the convention that \(f\) refers to the function and \(f(t)\) refers to the function evaluated at a given time \(t\). However, the abuse is a convenient one, since it allows us to use \(f(t-c)\) to denote the time-shift of \(f\). Hence, depending on the context, \(f(t-c)\) might refer to both the shifted function or the value of the function.

Example 5.1 Find the Laplace Transform of \(e^{-at}\) for any constant \(a\) \[\begin{align} {\mathcal L}\left\{e^{-at}\right\}(s) &= \int_0^\infty e^{-st}e^{-at}\, dt \\ &= \int_0^\infty e^{-(s+a)t}\, dt \\ &= \left. -\frac{1}{s+a}e^{-(s+a)t} \right|_{t=0}^\infty \\ &= -\frac{1}{s+a} \quad s+a>0 \end{align}\] Note we treat \(s\) and \(a\) as constants for the integration step. Think of this as finding the Transform at a given value of \(s\). For this example, the integral diverges if \(s+a\le0\), hence, the domain of the transformed function is \((a,\infty)\).

A special case of the previous example is \({\mathcal L}\left\{1\right\}(s) = \frac{1}{s}\).

Example 5.2 Find the Laplace Transform of \(t\) \[\begin{align} {\mathcal L}\left\{t\right\}(s) &= \int_0^\infty te^{-st}\, dt \\ &= \left. t\left(-\frac{1}{s}e^{-st}\right)\right|_{t=0}^\infty - \int_0^\infty \left(-\frac{1}{s}e^{-st}\right)\, dt \\ &= 0 - \left. \frac{1}{s^2}e^{-st}\right|_{t=0}^\infty \\ &= \frac{1}{s^2}, \quad s>0 \end{align}\] Note we’ve used integration by parts with \(u = t\) and \(dv = e^{-st}dt\). And again note that the integral diverges if \(s\le0\).

Example 5.3 Find the Laplace Transform of \(t^n\) \[\begin{align} {\mathcal L}\left\{t^n\right\}(s) &= \int_0^\infty t^ne^{-st}\, dt \\ &= \left. t^n\left(-\frac{1}{s}e^{-st}\right)\right|_{t=0}^\infty - \int_0^\infty \left(nt^{n-1}\right)\left(-\frac{1}{s}e^{-st}\right)\, dt \\ &= 0 + \frac{n}{s}\int_0^\infty t^{n-1}e^{-st}\, dt , \quad s > 0\\ &= 0 + \frac{n}{s}{\mathcal L}\left\{t^{n-1}\right\} \end{align}\] Repeating this step another \(n-1\) times, we should arrive at \[{\mathcal L}\left\{t^n\right\} = \frac{n!}{s^n}{\mathcal L}\left\{1\right\} = \frac{n!}{s^{n+1}}\] Indeed, using this formula with \(n=0\) and \(n=1\) gives the same results as before. To complete the proof by induction, simply note that replacing \(n\) by \(n-1\) gives \({\mathcal L}\left\{t^{n-1}\right\} = \frac{(n-1)!}{s^n}\) and \[\frac{n}{s}{\mathcal L}\left\{t^{n-1}\right\} =\frac{n}{s} \frac{(n-1)!}{s^n} = \frac{n!}{s^{n+1}}\] as expected.

Example 5.4 The transforms of sine and cosine can be viewed as special cases of our first example. Simply set \(a\) to \(i\) and note that \[\begin{gather} e^{-it} = \cos t) - i\sin t \\ {\mathcal L}\left\{e^{-it}\right\} = {\mathcal L}\left\{\cos t - i\sin t \right\} \\ {\mathcal L}\left\{e^{-it}\right\} = {\mathcal L}\left\{\cos t \right\} - i{\mathcal L}\left\{\sin t \right\} \\ \frac{1}{s+i} = {\mathcal L}\left\{\cos t \right\} - i{\mathcal L}\left\{\sin t \right\} \\ \frac{s-i}{(s+i)(s-i)} = {\mathcal L}\left\{\cos t \right\} - i{\mathcal L}\left\{\sin t \right\} \\ \frac{s-i}{s^2-1} = {\mathcal L}\left\{\cos t \right\} - i{\mathcal L}\left\{\sin t \right\} \\ \frac{s}{s^2-1} - i\frac{1}{s^2-1} = {\mathcal L}\left\{\cos t \right\} - i{\mathcal L}\left\{\sin t \right\} \end{gather}\] Matching the real and imaginary parts of each side implies \[\begin{align} {\mathcal L}\left\{\cos t \right\} &= \frac{s}{s^2+1} \\ {\mathcal L}\left\{\sin t \right\} &= \frac{1}{s^2+1} \end{align}\]

Theorem 5.1 (First Shifting Theorem) If \({\mathcal L}\left\{f(t)\right\} = F(s)\) for \(s>c\), then \({\mathcal L}\left\{e^{at}f(t)\right\} = F(s-a)\) for \(s>c+a\).

Proof. The result follows immediately from \[{\mathcal L}\left\{e^{at}f(t)\right\} = \int_0^\infty e^{at}f(t) e^{-st} \, ds = \int_0^\infty f(t) e^{-(s-a)t} \, ds\]

Exercise 5.1 Use partial fractions (if necessary) and the First Shifting Theorem (if necessary) to find the inverse Laplace transform of \[\displaystyle F(s) = \frac{s+7}{s^2 + 4s + 13}.\]

\[\begin{align*} F(s) &= \frac{s+7}{s^2 + 4s + 13}, \\ &= \frac{s+7}{(s+2)^2 + 3^2}. \\[5pt] f(t) &= {\mathcal L}^{-1}\left\{\frac{s+7}{(s+2)^2 + 3^2}\right\}, \\ &= e^{-2t}{\mathcal L}^{-1}\left\{\frac{s+5}{s^2 + 3^2}\right\}, \qquad \text{ Shifting $(s+2)\mapsto s$}\\ &= e^{-2t}\left(\cos{3t} + \frac{5}{3}\sin{3t}\right). \\ \end{align*}\]

Exercise 5.2 Use partial fractions (if necessary) and the First Shifting Theorem (if necessary) to find the inverse Laplace transform of \[\displaystyle F(s) = \frac{s+7}{s^2 + 2s + 5}.\]

\[\begin{align*} F(s) &= \frac{s+7}{s^2 + 2s + 5}, \\ &= \frac{s+7}{(s+1)^2 + 2^2}. \\[5pt] f(t) &= {\mathcal L}^{-1}\left\{\frac{s+7}{(s+1)^2 + 2^2}\right\}, \\ &= {\mathcal L}^{-1}\left\{\frac{(s+1)+6}{(s+1)^2 + 2^2}\right\}, \\ &= e^{-t}{\mathcal L}^{-1}\left\{\frac{s+6}{s^2 + 2^2}\right\}, \qquad \text{ Shifting $(s+1)\mapsto s$}\\ &= e^{-t}\left(\cos{2t} + 3\sin{2t}\right). \\ \end{align*}\]

5.2 Laplace Transforms and Differential Equations

Our interest in Laplace Transforms for this course is that they are (1) linear and (2) transform linear ordinary differential equations into algebraic equations.

Exercise 5.3 Suppose \({\mathcal L}\left\{f\right\} = F\) and \({\mathcal L}\left\{g\right\} = G\). Show that \[{\mathcal L}\left\{af + bg\right\} = a{\mathcal L}\left\{f\right\}+b{\mathcal L}\left\{g\right\}\] for any constants \(a\) and \(b\).

Exercise 5.4 Show that \({\mathcal L}\left\{y'\right\} = s{\mathcal L}\left\{y\right\} - y(0)\).

Exercise 5.5 Show that \({\mathcal L}\left\{y''\right\} = s^2{\mathcal L}\left\{y\right\} - sy(0) - y'(0)\).

Consider our generic second order constant coefficient ordinary differential equation: \[\begin{equation} ay'' + by' + cy = g, \end{equation}\] with \(y(0)\) and \(y'(0)\) specified. Taking the Laplace transform of both sides of the equation leads to \[a{\mathcal L}\left\{y''\right\} + b{\mathcal L}\left\{y'\right\} + c{\mathcal L}\left\{y\right\} = {\mathcal L}\left\{g\right\}.\] Using \(Y\) for \({\mathcal L}\left\{y\right\}\) and \(G\) for \({\mathcal L}\left\{g\right\}\) and our previous result for the Laplace transform of derivatives results in the equation \[a\left( s^2Y(s) - sy(0) - y'(0) \right) \ +\ b\left(sY(s) - y(0) \right) \ +\ cY(s) = G(s).\] Thus, the Laplace transform transforms the differential equation for \(y\) to an algebraic equation for \(Y\). Solving for \(Y\), we find \[\left( as^2 + bs + c \right) Y(s) = G(s) + (as+b)y(0) + ay'(0),\] or \[Y(s) = H(s)G(s) + H(s) I(s),\] where \[H(s) = \frac{1}{as^2 + bs + c} \qquad \text{and} \qquad I(s) = (as+b)y(0) + ay'(0).\] Finally, taking the inverse transforms gives the solution to the original problem: \[y(t) = {\mathcal L}^{-1}\left\{H(s)G(s)\right\} \ +\ {\mathcal L}^{-1}\left\{H(s)I(s)\right\}. \tag{5.2}\]

Equation 5.2 hides the details of the computations of the transforms and nicely shows the structure of the solution. The first part of the solution is the response to the forcing, and the second is the echo of the initial conditions.

Example 5.5 Solve the IVP \(y''(t) - 4y(t) = 8t^2 - 4\), with \(y(0) = 5\) and \(y'(0) = 10\).

Taking the Laplace transforms of both sides of the equations we find \[{\mathcal L}\left\{y''\right\} - 4{\mathcal L}\left\{y\right\} = 8{\mathcal L}\left\{t^2\right\} - {\mathcal L}\left\{4\right\}.\]

\[\left(s^2Y(s) - 5s - 10\right) \ -\ 4Y(s) = \frac{16}{s^3} - \frac{4}{s},\] which when solved for \(Y\) gives \[\begin{align*} (s^2 - 4 ) Y(s) &= \frac{16}{s^3} - \frac{4}{s} + 5s + 10\\ Y(s) &= \frac{16}{s^3(s^2 - 4 )} - \frac{4}{s(s^2 - 4 )} + \frac{5s + 10}{s^2 - 4 }\\ &= \frac{16 - 4s^2}{s^3(s^2 - 4 )} \ +\ \frac{5s + 10}{s^2 - 4 }\\ &= -\frac{4}{s^3} \ +\ \frac{5}{s - 2 }, \end{align*}\] where the last step results from cancelling common factors in the numerators and denominators. Inverting leads to \[y(t) = -{\mathcal L}^{-1}\left\{\frac{4}{s^3}\right\} \ +\ 5{\mathcal L}^{-1}\left\{\frac{1}{s - 2 }\right\} = -2t^2 \ +\ 5e^{2t}.\]

5.3 Step functions

We often want to include sudden changes in an applied force into our differential equation models of systems. For example, an electric circuit where a voltage is repeatedly switched on and off. A convenient building block for such functions in the unit step function, \(u_a\), defined by \[u_a(t) = \begin{cases} 0 & t < a \\ 1 & t\ge a \end{cases}\] To model an applied force, \(f\), turned-on at a time \(a>0\) we can use \(f(t-a)u_a(t)\). Notice we’ve shifted \(f\) to the right.

If \(F(s) = {\mathcal L}\left\{f(t)\right\}\), then \[\begin{align} {\mathcal L}\left\{f(t-a)u_a(t)\right\} &= \int_0^\infty e^{-st}f(t-a)u_a(t)\, dt \\ &= \int_a^\infty e^{-st}f(t-a)\, dt \\ &= \int_0^\infty e^{-s(t+a)}f(t)\, dt \\ &= e^{-sa}\int_0^\infty e^{-st}f(t)\, dt \\ &= e^{-sa}F(s) \end{align}\]

Exercise 5.6 Using the table of Laplace transforms and the Second Shifting Theorem, find the Laplace transform of \[g(t) = (t-3)^3 u_3(t)\]

\[\begin{align*} {\mathcal L}\left\{(t-3)^3 u_3(t)\right\} &= e^{-3s}{\mathcal L}\left\{t^3 \right\} \qquad \text{ shifting $(t-3)\mapsto t$,}\\ &= \frac{3!}{s^4}e^{-3s}\\ &= \frac{6e^{-3s}}{s^4}\\ \end{align*}\]

Exercise 5.7 Express the function \(f\) defined below using step functions and find its Laplace transform using the Second Shifting Theorem. \[f(t) = \begin{cases} t, & 0 \le t < 4, \\ 0, & 4 \le t < 5, \\ 1, & t \ge 5. \end{cases}\]

First, express \(f\) using step functions, then use the second shifting theorem. \[\begin{align*} f(t) &= t(1 - u_4(t)) + u_5(t) \\ f(t) &= t - t u_4(t) + u_5(t) \\ F(s) &= \frac{1}{s^2} - \frac{1}{s^2}e^{-4s} + \frac{1}{s}e^{-5s} \end{align*}\]

Exercise 5.8 Use Laplace transforms to solve the initial value problem \[y'' + 5y' + 6y = f(t), \quad y(0)=0, \quad y'(0)=0,\] where \[f(t) =\begin{cases} 0, & 0 \le t < 1, \\ 1, & t \ge 1 \end{cases}\]

\[\begin{align*} y'' + 5y' + 6y &= f(t)\\ y'' + 5y' + 6y &= u_1(t)\\ {\mathcal L}\left\{y''\right\} + 5{\mathcal L}\left\{y'\right\} + 6{\mathcal L}\left\{y\right\} &= {\mathcal L}\left\{u_1(t)\right\}\\ (s^2Y(s) - sy(0) - y'(0)) + 5(sY(s) - y(0)) + 6Y(s) &= \frac{e^{-s}}{s} \\ (s^2+5s+6)Y(s) &= (s + 5)y(0) + \frac{e^{-s}}{s} \\ (s^2+5s+6)Y(s) &= s+5 + \frac{e^{-s}}{s} \\ Y(s) &= \frac{s+5}{(s^2+5s+6)} + \frac{e^{-s}}{s(s^2+5s+6)} \\ Y(s) &= \frac{s+5}{(s+3)(s+2)} + \frac{e^{-s}}{s(s+3)(s+2)} \\ Y(s) &= -\frac{2}{s+3} + \frac{3}{s+2} + \frac{e^{-s}}{6s} + \frac{e^{-s}}{3(s+3)} - \frac{e^{-s}}{2(s+2)} \\ y(t) &= -2e^{-3t} + 3e^{-2} + \tfrac{1}{6} u_1(t) + \frac{1}{3}u_1(t)e^{-(t-1)} - \frac{1}{2}u_1(t)e^{-2(t-1)} \end{align*}\]

Exercise 5.9 Use Laplace transforms to solve the initial value problem \[y'' + 2y' + 6y = f(t), \quad y(0)=0, \quad y'(0)=0,\] where \[f(t) =\begin{cases} 2t, & 0 \le t < 1, \\ 0, & t \ge 1 \end{cases}\]

Step 1: express \(f\) using step functions and find \(F(s) = {\mathcal L}\left\{f(t)\right\}\): \[\begin{align*} f(t) &= 2t(1-u_1(t)) \\ F(s) &= {\mathcal L}\left\{2t(1-u_1(t))\right\} \\ &= 2{\mathcal L}\left\{t\right\} - 2{\mathcal L}\left\{t u_1(t))\right\}\\ &= 2{\mathcal L}\left\{t\right\} - 2{\mathcal L}\left\{((t-1) + 1)u_1(t)\right\}\\ &= \frac{2}{s^2} - 2e^{-s}{\mathcal L}\left\{t + 1\right\}\\ &= \frac{2}{s^2} - 2e^{-s}\big(\frac{1+s}{s^2} \big) \end{align*}\] Step 2: transform the IVP and solve for \(Y(s) = {\mathcal L}\left\{y(t)\right\}\) \[\begin{gather*} y'' + 2y' + 6y = f(t) \\ {\mathcal L}\left\{y''\right\} + 2{\mathcal L}\left\{y'\right\} + 6{\mathcal L}\left\{y\right\} = F(s)\\ (s^2Y(s) - sy'(0) - y(0)) + 2(sY(s) - y(0)) + 6Y(s) = F(s)\\ (s^2+2s+6)Y(s) = F(s)\\ Y(s) = G(s) - e^{-s}H(s) \end{gather*}\] with \(G(s) = \dfrac{2}{s^2(s^2+2s+6)}\) and \(H(s) = \dfrac{2+2s}{s^2(s^2+2s+6)}\)

Step 3: perform a partial fraction expansions on \(G\) and \(H\). \[\begin{gather*} \frac{2}{s^2(s^2+2s+6)} = \frac{A}{s} + \frac{B}{s^2} + \frac{C + Ds}{s^2+2s+6} \\ 2 = As(s^2+2s+6) + B(s^2+2s+6) + Cs^2 + Ds^3 \\ 2 = (A+D)s^3 + (2A+B+C)s^2 + (6A+2B)s + (6B) \end{gather*}\] Matching the coefficients of each power of \(s\) leads to \(B = 1/3\), \(A = -1/9\), \(C = -1/9\), \(D = 1/9\). Hence \[G(s) = \frac{1}{9}\left(-\frac{1}{s} + \frac{3}{s^2} + \frac{-1 + s}{s^2+2s+6} \right) \] Similarly, for \(H\) we obtain \[2+2s = (A+D)s^3 + (2A+B+C)s^2 + (6A+2B)s + (6B)\] Matching the coefficients of each power of \(s\) leads to \(B = 1/3\), \(A = 2/9\), \(C = -7/9\), \(D = -2/9\). \[H(s) = \frac{1}{9}\left(\frac{2}{s} + \frac{3}{s^2} + \frac{-7 -2s}{s^2+2s+6} \right) \] Step 4: invert \(G\) and \(H\)
The tricky term is \(\dfrac{-1 + s}{s^2+2s+6}\).
To invert this, first note the denominator is \((s+1)^2 + 4\).
So we write \(\dfrac{-1 + s}{s^2+2s+6} = \dfrac{-2 + (s+1)}{(s+1)^2+5}\),
which has an inverse Laplace transform of \(-\frac{2}{\sqrt{5}}e^{-t}\sin(\sqrt{5}t) + e^{-t}\cos(\sqrt{5}t)\) (lines 10 and 11 in Tupper’s table).
Hence, \[g(t) = {\mathcal L}^{-1}\left\{G(s)\right\} = \frac{1}{9}\left(-1 + 3t -e^{-t}\sin(\sqrt{5}t) + e^{-t}\cos(\sqrt{5}t)\right)\] \[h(t) = {\mathcal L}^{-1}\left\{H(s)\right\} = \frac{1}{9}\left(2 + 3t -\sqrt{5}e^{-t}\sin(\sqrt{5}t) - 2e^{-t}\cos(\sqrt{5}t)\right)\] Step 5: combine \(g\) and \(h\) to make \(y\) \[y(t) = {\mathcal L}^{-1}\left\{G(s)\right\} - {\mathcal L}^{-1}\left\{e^{-s}H(s)\right\} = g(t) - u_1(t)h(t-1)\]

5.4 Impulsive forcing

We are often interested in forcing the system with an impulse, a large force applied over a short time period. Define a unit box function, \[\delta_a(t) = \begin{cases}\frac{1}{2a}, & -a < t < a, \\ 0, & \text{otherwise}. \end{cases}\] Note that \(\int_{0}^\infty \delta_a(t-t_o) \, dt = 1\) provided \(t_o > a\). Since we are interested in impulses over very short times (small \(a\)), it is natural to think of a limit as \(a\rightarrow0\).
The limit of such a sequence of unit boxes is called the Dirac delta function. Conceptually, at least for this course, we can view it as being defined as \[\delta(t-t_o) = \lim_{a\rightarrow0} \delta_a(t-t_0).\] However, this limit is not a function in the sense we are used to and it is certainly not Riemann integrable. The mathematical machinery needed to make sense of this limit is outside the scope of this course. Nevertheless, the concept is interesting and turns out to be useful in practice.

Note that \[\lim_{\rightarrow0^+} \int_{0}^\infty \delta_a(t-t_o) \, dt = 1\] since the integral is one for every positive \(a\). Moreover, for any continuous function \(f\) defined on \([0,\infty)\) \[\begin{align} \lim_{a\rightarrow0^+} \int_{0}^\infty f(t) \delta_a(t-t_o) \, dt &= \lim_{a\rightarrow0^+} \frac{1}{2a}\int_{t_o-a}^{t_o+a} f(t) \, dt &= f(t_o) \end{align}\] The unit impulses behave nicely in the limit as long as they stay inside integrals! In fact, another way to define the delta-function is by this property: \(\int_0^\infty \delta(t-t_o) f(t) \, dt = f(t_o)\) for every continuous function \(f\) and time \(t_o \ge 0\).

In particular, \[{\mathcal L}\left\{\delta(t-t_o)\right\} = \int_0^\infty e^{-st} \delta(t-t_o)\, dt = e^{-st_o}.\]

Theoretically, we’ve wandered off the edge of the map. \(\delta\) is not a function in any sense we are familiar with, it’s not Riemann ingegrable, and all the previous lines are nothing more than wishful symbol manipulation. Despite this, the symbol manipulation usually turns out nicely and proves useful.

Example 5.6 Solve the initial value problem \[y'' + 9y = \delta(t-\frac{\pi}{2}), \qquad y(0) = 0, \ y'(0) = 0.\] Note this is the classic model for a simple pendulum starting at rest at time zero, and given a unit of force concentrated at time \(\pi/2\). Think of tapping the pendulum with a hammer.

Taking the Laplace transform of both sides yields
\[(s^2 + 9) Y(s) = e^{-\pi s/2}\] \[Y(s) = \frac{e^{-\pi s/2}}{s^2 + 9}\] and the inverse transform gives the solution \[\begin{align*} y(t) &= \frac{1}{3}\sin\left(3(t-\pi/2)\right) u_{\pi/2}(t) \\ &= \begin{cases} 0 & 0 \le t < \pi/2 \\ \frac{1}{3}\sin\left(3(t-\pi/2)\right), & t \ge \pi/2. \end{cases} \end{align*}\]