The Method of Laplace Transforms

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In this section we describe the basic properties of Laplace transforms and show how these properties lead to a method for solving forced equations. We also discuss the kind of information that we will need about Laplace transforms in order to solve a general second order forced equation.

The Basic Properties

Suppose that there is an operation or transform that transforms functions $x$ defined on a variable $t$ to functions ${\cal L}[x]$ defined on a variable $s$ in such a way that:

$\begin{equation} \label{Tprops} \begin{array}{cl} (a) & {\cal L}\; \mbox{is linear},\\ (b) & {\cal L}\; \mbox{is invertible},\; \mbox{and}\\ (c) & {\cal L}[\dot{x}](s) = s{\cal L}[x](s) - x(0). \end{array} \end{equation}$ Note that (??)(c) relates the transform of the derivative of a function to the transform of the function itself. We call a transform that satisfies (??) the Laplace transform, as there is only one transform that satisfies these three properties.

The Laplace Transform of $e^{at}$

We can use properties (??) to compute the Laplace transform of the function $x(t) = e^{at}$ . To perform this calculation we need only recall that $e^{at}$ is the unique solution to the differential equation $\dot {x}=ax$ with initial value $x(0)=1$ . In essence, we can compute the Laplace transform of $e^{at}$ without actually having defined the Laplace transform.

It follows from the differential equation $\dot {x}=ax$ and (??)(a) that ${\cal L}[\dot {x}](s) = {\cal L}[ae^{at}](s) = a{\cal L}[e^{at}](s)$ and from (??)(c) that ${\cal L}[\dot {x}](s) = s{\cal L}[e^{at}](s)-1.$ Equating these two expressions for ${\cal L}[\dot {x}]$ leads to $s{\cal L}[e^{at}](s)-1 = a{\cal L}[e^{at}](s),$ and hence to

$\begin{equation} \label{eq:Lexp} {\cal L}[e^{at}](s) = \frac{1}{s-a}. \end{equation}$

Solving an Inhomogeneous Equation by Laplace Transforms

Properties (??) and formula (??) allow us to solve the initial value problem

$\begin{equation} \label{eq:expODE} \begin{array}{crcl} (a) & \dot{x} - 3x & = & e^{2t}\\ (b) & x(0) & = & 1. \end{array} \end{equation}$

Before proceeding, note that (??) can be solved directly using the method of undetermined coefficients. Indeed, undetermined coefficients show that $x_p(t) = 2e^{2t}$ is a particular solution to (??)(a). Since the general solution to the homogeneous equation is $\alpha e^{3t}$ , the general solution to the inhomogeneous equation (??)(a) is: $x(t) = 2e^{2t} + \alpha e^{3t}.$ Setting $\alpha =-1$ solves the initial value problem (??)(b).

We illustrate the method of Laplace transforms by solving (??) using Laplace transforms. Apply the transform $\cal L$ to both sides of (??), obtaining ${\cal L}[\dot {x}-3x](s) = {\cal L}[e^{2t}](s).$ From (??) with $a=2$ , we see that ${\cal L}[e^{2t}](s) = \frac {1}{s-2}$ Using (??)(c,a), we see that ${\cal L}[\dot {x}-3x](s) = (s-3){\cal L}[x](s) - 1.$ Hence $(s-3){\cal L}[x](s) = 1 + \frac {1}{s-2}.$

Next, we solve for ${\cal L}[x](s)$ , obtaining ${\cal L}[x](s) = \frac {1}{s-3} + \frac {1}{(s-2)(s-3)}.$ Using partial fractions, we find ${\cal L}[x](s) = \frac {2}{s-3} - \frac {1}{s-2}.$ We describe the method of partial fractions in more detail in Section ??. The particular uses of partial fractions in this section all follow from Exercise ??.

Finally, from (??) we see that ${\cal L}[2e^{3t}-e^{2t}](s) = \frac {2}{s-3} - \frac {1}{s-2},$ and using (??)(b) we conclude that $x(t) = 2e^{3t}-e^{2t}$ is the unique solution to the initial value problem (??).

Three Steps in Solving Equations by Laplace Transforms

We can summarize the method for solving ordinary differential equations by Laplace transforms in three steps. In this summary it will be useful to have defined the inverse Laplace transform.

The inverse Laplace transform of a function $Y(s)$ is the function $y(t)$ satisfying ${\cal L}[y(t)](s) = Y(s)$ , and is denoted by ${\cal L}\inv [Y(s)]$ .

The summary of the Laplace transform method is:

(a): Compute the Laplace transforms of both sides of the differential equation in (??) using the linearity of Laplace transforms, the formulas for Laplace transforms of derivatives (??)(c), and specific Laplace transforms such as in (??).
(b): Explicitly solve the transformed equation for ${\cal L}[x(t)](s)=Y(s)$ .
(c): Find a function $x(t)$ whose Laplace transform is $Y(s)$ ; that is, find ${\cal L}\inv [Y(s)]$ .

An Example of a First Order Forced Equation

As an example, we use this three step method to solve the initial value problem:

$\begin{equation} \label{eq:dxx2} \begin{array}{rcl} \dot x - x & = & 2\\ x(0) & = & 4. \end{array} \end{equation}$

The first step is to apply the Laplace transform to both sides of the differential equation. Using formula (??)(c) and the linearity of $\cal L$ , we obtain $\frac {2}{s} = {\cal L}[2] = {\cal L}[\dot x]-{\cal L}[x] = (s-1){\cal L}[x]-4.$ Note that $e^{0t}=1$ so that ${\cal L}[1] = \frac {1}{s-0}=\frac {1}{s}$ .

The second step requires solving this equation explicitly for ${\cal L}[x]$ obtaining

$\begin{equation} \label{eq:Lapdxx2} {\cal L}[x] = \frac{\frac{2}{s}+4}{s-1} = \frac{2+4s}{s(s-1)} = \frac{6}{s-1} - \frac{2}{s}. \end{equation}$ Note that we have simplified the right hand side by use of partial fractions.

The third step requires using (??) to see that $x(t) = {\cal L}\inv \left [\frac {6}{s-1} - \frac {2}{s}\right ](t) = 6{\cal L}\inv \left [\frac {1}{s-1}\right ](t) - 2{\cal L}\inv \left [\frac {1}{s}\right ](t) = 6e^t-2.$

Laplace Transforms of Second Order Equations

We end this section with a discussion of the type of information that we will need to solve initial value problems of second order inhomogeneous linear equations by the method of Laplace transforms. Consider the differential equation

$\begin{equation} \label{e:2ndforced} \begin{array}{rcl} \ddot{x} + a\dot{x} + bx & = & g(t)\\ x(0) & = & x_0\\ \dot{x}(0) & = & \dot{x}_0, \end{array} \end{equation}$ where $a,b\in \R$ are constants. The function $g(t)$ is called the forcing term.

A Second Order Homogeneous Example

We begin our discussion by considering a simple example of an unforced equation:

$\begin{equation} \label{Lexam2} \begin{array}{rcl} \ddot{x} + 3\dot{x} +2x & = & 0 \\ x(0) & = & 1 \\ \dot{x}(0) & = & 2. \end{array} \end{equation}$ We begin by applying (??)(c) twice to obtain $\begin {array}{rcl} {\cal L}[\ddot {x}](s) & = & s{\cal L}[\dot {x}](s) - \dot {x}(0)\\ & = & s(s{\cal L}[x](s) - x(0)) - \dot {x}(0)\\ & = & s^2{\cal L}[x](s) - sx(0) - \dot {x}(0). \end {array}$ Where there is no ambiguity we will drop the argument $(s)$ in ${\cal L}[x](s)$ ; this change should make subsequent formulas easier to read. For instance, the previous equation is: $\begin{equation} \label{eq:dderLap} {\cal L}[\ddot{x}] = s^2{\cal L}[x] - sx(0) - \dot{x}(0). \end{equation}$ Now apply (??) and (??) to the differential equation (??) to obtain

$\begin{eqnarray*} 0 & = & {\cal L}[\ddot{x} + 3\dot{x} +2x] \\ & = & (s^2{\cal L}[x] - sx(0) - \dot{x}(0)) + 3(s{\cal L}[x] - x(0)) + 2{\cal L}[x]\\ & = & (s^2+3s+2){\cal L}[x] -(sx(0)+\dot{x}(0)+3x(0)) \\ & = & (s^2+3s+2){\cal L}[x] - (s+5). \end{eqnarray*}$

Next, use this equation and partial fractions to solve for ${\cal L}[x]$ as $\dps {\cal L}[x] = \frac {s+5}{(s+1)(s+2)} = 4\frac {1}{s+1} - 3\frac {1}{s+2}.$ Using linearity and (??), we see that ${\cal L}[4e^{-t}-3e^{-2t}] = 4\frac {1}{s+1} - 3\frac {1}{s+2}.$ Hence $x(t) = 4e^{-t}-3e^{-2t}$ is the solution to (??).

Information Needed to Solve Second Order Equations

What information about Laplace transforms will we need to solve the initial value problem (??) in general? We can answer this question just by taking the Laplace transform of (??) and solving for ${\cal L}[x]$ . Using (??) and (??)(c), compute

$\begin{eqnarray*} {\cal L}[g(t)] & = & {\cal L}[\ddot{x}+a\dot{x} + bx] \\ & = & {\cal L}[\ddot{x}] + a{\cal L}[\dot{x}]+b {\cal L}[x] \\ & = & (s^2{\cal L}[x]-sx_0-\dot{x}_0) +a(s{\cal L}[x]-x_0)+b{\cal L}[x]\\ & = & (s^2+as+b){\cal L}[x]-(x_0s+ax_0+\dot{x}_0). \end{eqnarray*}$

Then solve for ${\cal L}[x]$ as $\begin{equation} \label{e:laplace2nd} {\cal L}[x] = \frac{x_0s + ax_0+\dot{x}_0}{s^2+as+b} + \frac{G(s)}{s^2+as+b} \end{equation}$ where $G = {\cal L}[g(t)]$ . The third step in solving (??) by Laplace transforms is to compute the inverse Laplace transform of the right hand side of (??).

To find the inverse Laplace transform of the first term, we use partial fractions. For example, suppose that the polynomial $s^2+as+b$ has real distinct roots $r_1$ and $r_2$ , then we can rewrite the first term as $\frac {c_1}{s-r_1} + \frac {c_2}{s-r_2}$ for some real constants $c_1$ and $c_2$ . We can now use (??) to find the inverse Laplace transform of this first term.

If this polynomial has a double real root or a complex conjugate pair of roots, then we need to find inverse Laplace transforms of functions like

$\begin{equation} \label{e:sampleLT} \frac{1}{(s-r_1)^2} \AND \frac{1}{(s-B)^2+C^2} \AND \frac{s}{(s-B)^2+C^2}. \end{equation}$

Finding the inverse Laplace transform for the second term is in general more difficult, since it depends on the hitherto unspecified function $g(t)$ . As it happens this inverse Laplace transform can be computed for a number of important functions, as we discuss in the next section.

To summarize: in order to use the method of Laplace transforms successfully to solve forced second order linear equations, we must be able to

compute partial fraction expansions,
compute the inverse Laplace transforms for functions in (??), and
compute the inverse Laplace transforms of the second term in (??), which involves the forcing function $g(t)$ .

Exercises

Find the Laplace transform of $x(t)=e^{3t}-2e^{-4t}$ .

Find the Laplace transform of $x(t)=14e^{-6t}+3e^{7t}$ .

Given distinct roots $r_1$ and $r_2$ , the method of partial fractions states that $\begin{equation} \label{E:pfquad} \frac{1}{(s-r_1)(s-r_2)} = \frac{a_1}{s-r_1} + \frac{a_2}{s-r_2} \end{equation}$ for some scalars $a_1$ and $a_2$ . By putting the right side of (??) over a common denominator, verify that $a_1$ and $a_2$ are found by solving the system of linear equations

$\begin{eqnarray*} a_1 + a_2 & = & 0\\ r_2a_1 + r_1a_2 & = & -1. \end{eqnarray*}$

In Exercises ?? – ?? use partial fractions to find a function $x(t)$ whose Laplace transform is the given function $Y(s)$ . Hint: In your calculations, you may use the result of Exercise ??.

$Y(s) = \dps \frac {1}{s^2-4s+3}$ .

$Y(s) = \dps \frac {s-1}{s^2+5s+6}$ .

In Exercises ?? – ??, use Laplace transforms to compute the solution to the given initial value problem.

$\ddot {x} + 3\dot {x} + 2x = 0, \quad x(0) = 1, \quad \dot {x}(0) = -1$ .

$\ddot {x} - 3\dot {x} - 4x = 0, \quad x(0) = 2, \quad \dot {x}(0) = -1$ .

Let $\alpha ,\beta ,a,b,c$ be real numbers. Show that the solution of the initial value problem $\ddot x +a\dot x +b x=c,\quad x(0)=\alpha ,\quad \dot x(0)=\beta ,$ has the Laplace transform ${\cal L}[x] = \frac {\alpha s^2 +(\beta +a\alpha )s+c}{s(s^2+as+b)}.$

Identity (??) may be generalized to: $\begin{equation} \label{LapDerk} {\cal L}\left[\frac{d^k y}{dt^k}\right] = s^k{\cal L}[y] - s^{k-1} y(0) - \cdots - \frac{d^{k-1} y}{dt^{k-1}}(0). \end{equation}$ Use induction to verify (??).