6 The Laplace transform

In this chapter we will discuss the Laplace transform . The Laplace transform turns out to be a very efficient method to solve certain ODE problems. In particular, the transform can take a differential equation and turn it into an algebraic equation. If the algebraic equation can be solved, applying the inverse transform gives us our desired solution. The Laplace transform is also useful in the analysis of certain systems such as electrical circuits, NMR spectroscopy, signal processing and others. Finally, understanding the Laplace transform will also help with understanding the related Fourier transform, which, however, requires more understanding of complex numbers. We will not cover the Fourier transform.

The Laplace transform also gives a lot of insight into the nature of the equations we are dealing with. It can be seen as converting between the time and the frequency domain. For example, take the standard equation

We can think of $t$ as time and $f(t)$ as incoming signal. The Laplace transform will convert the equation from a differential equation in time to an algebraic (no derivatives) equation, where the new independent variable $s$ is the frequency.

We can think of the Laplace transform as a black box. It eats functions and spits out functions in a new variable. We write $L{f(t)} = F (s)$ . It is common to write lower case letters for functions in the time domain and upper case letters for functions in the frequency domain. We will use the same letter to denote that one function is the Laplace transform of the other, for example $F (s)$ is the Laplace transform of $f (t)$ . Let us define the transform.

We note that we are only considering $t ≥ 0$ in the transform. Of course, if we think of $t$ as time there is no problem, we are generally interested in finding out what will happen in the future (Laplace transform is one place where it is safe to ignore the past). Let us compute the simplest transforms.

Example 6.1.1: Suppose $f (t) = 1$ , then

∫ ∞ [e-st]∞ 1 L {1} = e-st dt = ---- = -. 0 -s t=0 s

Of course, the limit only exists if $s > 0$ . So $L {1}$ is only defined for $s > 0$ .

Example 6.1.2: Suppose $-at f (t) = e$ , then

∫ ∫ [ ]∞ -at ∞ -st -at ∞ -(s+a)t -e-(s+a)t- -1--- L {e } = e e dt = e dt = -(s + a ) = s + a. 0 0 t=0

Of course, the limit only exists if $s + a > 0$ . So $-at L{e }$ is only defined for $s + a > 0$ .

Example 6.1.3: Suppose $f (t) = t$ , then using integration by parts

∫ ∞ L {t} = e-stt dt [0 ] -te-st∞ 1 ∫ ∞ -st = ------ + -- e dt s [ t=0]∞s 0 1-e-st = 0 + s -s t=0 = 1-. s2

Of course, again, the limit only exists if $s > 0$ .

Example 6.1.4: A common function is the unit step function, which is sometimes called the Heaviside function . This function is generally given as

( ||{0 if t < 0, u(t) = || (1 if t ≥ 0.

Let us find the Laplace transform of $u(t - a)$ , where $a ≥ 0$ is some constant. That is, the function which is 0 for $t < a$ and 1 for $t ≥ a$ .

∫ ∞ ∫ ∞ [ -st]∞ -as L {u(t - a)} = e-stu(t - a) dt = e-st dt = e-- = e---, 0 a -s t=a s

where of course $s > 0$ (and $a ≥ 0$ as we said before).

By applying similar procedures we can compute the transforms of many elementary functions. Many basic transforms are listed in Table 6.1.

$f (t)$	$L{f(t)} = F (s)$






$C$	$C s-$

$t$	$1s2-$

$t2$	$23- s$

$t3$	$6- s4$

$n t$	$-n!- sn+1$

$-at e$	$1 s+a$

$sinωt$	$s2ω+ω2$

$cos ωt$	$s2s+ω2$

$sinh ωt$	$s2ω-ω2$

$coshωt$	$s2s-ω2$

$u(t - a )$	$e-as s$

Table 6.1: Some Laplace transforms ( $C$ , $ω$ , and $a$ are constants).

Exercise 6.1.1: Verify Table 6.1.

Since the transform is defined by an integral. We can use the linearity properties of the integral. For example, suppose $C$ is a constant, then

So we can “pull out” a constant out of the transform. Similarly we have linearity. Since linearity is very important we state it as a theorem.

Theorem 6.1.1 (Linearity of Laplace transform). Suppose that $A$ , $B$ , and $C$ are constants, then

|----------------------------------------| | L{Af (t) + Bg(t)} = AL {f(t)} + BL {g (t)}, | ------------------------------------------

and in particular

L{Cf (t)} = CL {f(t)}.

Exercise 6.1.2: Verify the theorem. That is, show that $L {Af(t) + Bg (t)} = AL {f(t)} + BL {g(t)}$ .

These rules together with Table 6.1 make it easy to already find the Laplace transform of a whole lot of functions already. It is a common mistake to think that Laplace transform of a product is the product of the transforms. But in general

It must also be noted that not all functions have Laplace transform. For example, the function $1 t$ does not have a Laplace transform as the integral diverges. Similarly $tant$ or $t2 e$ do not have Laplace transforms.

6.1.2 Existence and uniqueness

Let us consider in more detail when does the Laplace transform exist. First let us consider functions of exponential order. $f (t)$ is of exponential order as $t$ goes to infinity if

for some constants $M$ and $c$ , for sufficiently large $t$ (say for all $t > t0$ for some $t0$ ). The simplest way to check this condition is to try and compute

If the limit exists and is finite (usually zero), then $f(t)$ is of exponential order.

Exercise 6.1.3: Use L’Hopital’s rule from calculus to show that a polynomial is of exponential order. Hint: Note that a sum of two exponential order functions is also of exponential order. Then show that $tn$ is of exponential order for any $n$ .

For an exponential order function we have existence and uniqueness of the Laplace transform.

Theorem 6.1.2 (Existence). Let $f(t)$ be continuous and of exponential order for a certain constant $c$ . Then $F (s) = L {f (t)}$ is defined for all $s > c$ .

You may have existence of the transform for other functions, that are not of exponential order, but that will not relevant to us. Before dealing with uniqueness, let us also note that for exponential order functions you also obtain that their Laplace transform decays at infinity:

Theorem 6.1.3 (Uniqueness). Let $f(t)$ and $g (t)$ be continuous and of exponential order. Suppose that there exists a constant $C$ , such that $F(s) = G(s)$ for all $s > C$ . Then $f(t) = g(t)$ for all $t ≥ 0$ .

Both theorems hold for piecewise continuous functions as well. Recall that piecewise continuous means that the function is continuous except perhaps at a discrete set of points where it has jump discontinuities like the Heaviside function. Uniqueness however does not “see” values at the discontinuities. So you can only conclude that $f (t) = g(t)$ outside of discontinuities. For example, the unit step function is sometimes defined using $u(0 ) = 1 2$ . This new step function, however, we defined has the exact same Laplace transform as the one we defined earlier where $u(0) = 1$ .

6.1.3 The inverse transform

As we said, the Laplace transform will allow us to convert a differential equation into an algebraic equation which we can solve. Once we do solve the algebraic equation in the frequency domain we will want to get back to the time domain, as that is what we are really interested in. We, therefore, need to also be able to get back. If we have a function $F (s)$ , to be able to find $f(t)$ such that $L{f(t)} = F (s)$ , we need to first know if such a function is unique. It turns out we are in luck by Theorem 6.1.3. So we can without fear make the following definition.

If $F(s) = L{f(t)}$ for some function $f(t)$ . We define the inverse Laplace transform as

There is an integral formula for the inverse, but it is not as simple as the transform itself (requires complex numbers). The best way to compute the inverse is to use the Table 6.1.

Example 6.1.5: Take $1 F(s) = s+1$ . Find the inverse Laplace transform.

We look at the table and we find

{ 1 } L -1 ----- = e-t. s + 1

We note that because the Laplace transform is linear, the inverse Laplace transform is also linear. That is,

We can of course also just pull out constants. Let us demonstrate how linearity is used by the following example.

Example 6.1.6: Take $s2+s+1 F(s) = -s3+s-$ . Find the inverse Laplace transform.

First we use the method of partial fractions to write $F$ in a form where we can use Table 6.1. We factor the denominator as $s(s2 + 1)$ and write

s2 + s + 1 A Bs + C --------- = --+ -------. s3 + s s s2 + 1

Hence $A(s2 - 1) + s(Bs + C ) = s2 + s + 1$ . Therefore, $A + B = 1$ , $C = 1$ , $A = 1$ . In other words,

s2 + s + 1 1 1 F(s) = ---3----- = --+ -2----. s + s s s + 1

By linearity of Laplace transform (and thus of its inverse) we get that

{ 2 } { } { } L -1 s-+-s-+-1 = L -1 1- + L-1 --1--- = 1 + sint. s3 + s s s2 + 1

A useful property is the so-called shifting property or the first shifting property

Exercise 6.1.4: Derive this property from the definition.

The shifting property can be used when the denominator is a more complicated quadratic that may come up in the method of partial fractions. You always want to write such quadratics as $(s + a)2 + b$ by completing the square and then using the shifting property.

Example 6.1.7: Find ${ } L-1 s2+14s+8-$ .

First we complete the square to make the denominator $(s + 2)2 + 4$ . Next we find

{ } -1 --1--- 1- L s2 + 4 = 2 sin2t.

Putting it all together with the shifting property we find

{ } { } -1 ----1------ -1 -----1----- 1- -2t L s2 + 4s + 8 = L (s + 2)2 + 4 = 2 e sin 2t.

In general, we will want to be able to apply the Laplace transform to rational functions, that is functions of the form

where $F (s)$ and $G (s)$ are polynomials. Since normally (for functions that we are considering) the Laplace transform goes to zero as $s → ∞$ , it is not hard to see that the degree of $F (s)$ will always be smaller than that of $G (s)$ . Such rational functions are called proper rational functions and we will always be able to apply the method of partial fractions. Of course this means we will need to be able to factor the denominator into linear and quadratic terms, which involves finding the roots of the denominator.

6.1.4 Exercises

Exercise 6.1.5: Find the Laplace transform of $5 3 + t + sinπt$ .

Exercise 6.1.6: Find the Laplace transform of $a + bt + ct2$ for some constants $a$ , $b$ , and $c$ .

Exercise 6.1.7: Find the Laplace transform of $A cos ωt + B sinωt$ .

Exercise 6.1.8: Find the Laplace transform of $2 cos ωt$ .

Exercise 6.1.9: Find the inverse Laplace transform of $s24-9-$ .

Exercise 6.1.10: Find the inverse Laplace transform of $22s-- s-1$ .

Exercise 6.1.11: Find the inverse Laplace transform of $---1---- (s-1)2(s+1)$ .

6.2 Transforms of derivatives and ODEs

6.2.1 Transforms of derivatives

Let us see how the Laplace transform is used for differential equations. First let us try to find the Laplace transform of a function that is a derivative. That is, suppose $g(t)$ is a continuous differentiable function of exponential order.

We can keep doing this procedure for higher derivatives. The results are listed in Table 6.2. The procedure also works for piecewise smooth functions, that is functions which are piecewise continuous with a piecewise continuous derivative. The fact that the function is of exponential order is used to show that the limits appearing above exist. We will not worry much about this fact.

Exercise 6.2.1: Verify Table 6.2.

6.2.2 Solving ODEs with the Laplace transform

If you notice, the Laplace transform turns differentiation essentially into multiplication by $s$ . Let us see how to apply this to differential equations.

Example 6.2.1: Take the equation

′′ ′ x (t) + x(t) = cos2t, x (0) = 0, x (0) = 1.

We will take the Laplace transform of both sides. By $X (s)$ we will, as usual, denote the Laplace transform of $x(t)$ .

′′ L{x (t) + x(t)} = L{cos 2t}, 2 ′ --s--- s X (s) - sx(0) - x (0) + X(s) = s2 + 4 .

We can plug in the initial conditions now (this will make computations more streamlined) to obtain

2 --s--- s X(s) - 1 + X(s) = s2 + 4 .

We now solve for $X(s)$ ,

X (s) =-------s------ + ---1--. (s2 + 1)(s2 + 4) s2 + 1

We use partial fractions (exercise) to write

X(s) = 1---s---- 1---s---+ --1---. 3 s2 + 1 3 s2 + 4 s2 + 1

Now take the inverse Laplace transform to obtain

x(t) = 1cos t - 1-cos2t + sin t. 3 3

The procedure is as follows. You take an ordinary differential equation in the time variable $t$ . You apply the Laplace transform to transform the equation into an algebraic (non differential) equation in the frequency domain. All the $x(t)$ , $′ x (t)$ , $′′ x (t)$ , and so on, will be converted to $X (s)$ , $sX (s) - x(0)$ , $s2X(s) - sx(0) - x′(0)$ , and so on. If the differential equation we started with was constant coefficient linear equation, it is generally pretty easy to solve for $X (s)$ and we will obtain some expression for $X(s)$ . Then taking the inverse transform if possible, we find $x(t)$ .

It should be noted that since not every function has a Laplace transform, not every equation can be solved in this manner.

6.2.3 Using the Heaviside function

Before we move on to more general functions than those we could solve before, we want to consider the Heaviside function. See Figure 6.1 for the graph.

This function is useful for putting together functions, or cutting functions off. Most commonly it is used as $u (t - a )$ for some constant $a$ . This just shifts the graph to the right by $a$ . That is, it is a function which is zero when $t < a$ and 1 when $t ≥ a$ . Suppose for example that $f(t)$ is a “signal” and you started receiving the signal $sin t$ at time $t$ . The function $f(t)$ should then be defined as

Similarly the step function which is 1 on the interval $[1,2)$ and zero everywhere else can be written as

The Heaviside function is useful to define functions defined piecewise. If you want the function $t$ on when $t$ is in $[0,1 ]$ and the function $- t + 2$ when $t$ is in $[1,2]$ and zero otherwise, you can use the expression

Hence it is useful to know how the Heaviside function interacts with the Laplace transform. We have already seen that

Example 6.2.2: Suppose that the forcing function is not periodic. For example, suppose that we had a mass spring system

x′′(t) + x(t) = f(t), x (0) = 0, x′(0) = 0,

where $f (t) = 1$ if $1 ≤ t < 3$ and zero otherwise. We could imagine a mass and spring system where a rocket was fired for 2 seconds starting at $t = 1$ . Or perhaps an RLC circuit, where the voltage was being raised at a constant rate for 2 seconds starting at $t = 1$ and then held steady again starting at $t = 3$ .

We can write $f (t) = u (t - 1 ) - u(t - 3)$ . We transform the equation and we plug in the initial conditions as before to obtain

2 e-s e-3s s X(s) + X (s) = s - s .

We solve for $X (s)$ to obtain

e-s e-3s X (s ) =---2---- - ---2----. s(s + 1) s(s + 1)

We leave it as an exercise to the reader to show that

{ } L-1 ---1---- = 1 - cost. s(s2 + 1)

In other words $L {1 - cos t} =--1-- s(s2+1)$ . So using (6.1) we find

{ } -1 e-s -s ( ) L ---2---- = e L {1 - cos t} = 1 - cos(t - 1) u(t - 1). s(s + 1)

Similarly

{ } -1 --e-2s-- -2s ( ) L s(s2 + 1) = e L{1 - cost} = 1 - cos(t - 3)u(t - 3).

Hence, the solution is

( ) ( ) x(t) = 1 - co s(t - 1 )u(t - 1) - 1 - cos(t - 2) u(t - 2).

The plot of this solution is given in Figure 6.2.

Figure 6.2: Plot of $x(t)$ .

6.2.4 Transforms of integrals

A feature of Laplace transforms is that it is also able to easily deal with integral equations. That is, equations in which integrals rather than derivatives of functions appear. The basic property, which can be proved by applying the definition and again doing integration by parts, is the following.

Example 6.2.3: To compute the inverse transform of $1 s(s2+1)$ we could proceed by applying this integration rule.

{ } ∫ { } ∫ -1 1---1--- t -1 --1--- t L s s2 + 1 = L s2 + 1 dτ = sin τ dτ = 1 - cost. 0 0

If an equation contains an integral of the unknown function the equation is called an integral equation. For example, take the equation

More complicated integral equations can also be solved using the convolution that we will learn next.

6.2.5 Exercises

Exercise 6.2.2: Using the Heaviside function write down the piecewise function that is 0 for $t < 0$ , $2 t$ for $t$ in $[0,1]$ and $t$ for $t > 1$ .

Exercise 6.2.3: Using the Laplace transform solve

mx ′′ + cx ′ + kx = 0, x(0) = 0, x′(0 ) = 0,

where $m > 0$ , $c > 0$ , $k > 0$ , and $2 c - 4km > 0$ (system is overdamped).

Exercise 6.2.4: Using the Laplace transform solve

mx ′′ + cx ′ + kx = 0, x(0) = 0, x′(0 ) = 0,

where $m > 0$ , $c > 0$ , $k > 0$ , and $2 c - 4km < 0$ (system is underdamped).

Exercise 6.2.5: Using the Laplace transform solve

mx ′′ + cx ′ + kx = 0, x(0) = 0, x′(0 ) = 0,

where $m > 0$ , $c > 0$ , $k > 0$ , and $c 2 = 4km$ (system is critically damped).

Exercise 6.2.6: Solve $x′′ + x = u(t - 1)$ for initial conditions $x(0 ) = 0$ and $x′(0 ) = 0$ .

Exercise 6.2.7: Show the differentiation of the transform property. Suppose $L {f(t)} = F (s)$ , then show

L {-tf(t)} = F′(s).

Hint: differentiate under the integral sign.

6.3 Convolution

6.3.1 The convolution

We said that the Laplace transformation of a product is not the product of the transforms. All hope is not lost however. We simply have to use a different type of a “product.” Take two functions $f(t)$ and $g(t)$ defined for $t ≥ 0$ . Define the convolution of $f(t)$ and $g(t)$ as

As you can see, the convolution of two functions of $t$ is another function of $t$ .

Example 6.3.1: Take $f(t) = et$ and $g(t) = t$ for $t ≥ 0$ . Then

∫ t (f * g )(t) = eτ(t - τ ) dτ = et - t - 1. 0

To solve the integral we did one integration by parts.

Example 6.3.2: Take $f (t) = sin ωt$ and $g(t) = cosωt$ for $t ≥ 0$ . Then

∫ t( )( ) (f * g)(t) = sin ωτ cosω 0(t - τ) dτ. 0

We will apply the identity

1( ) cos θsinψ = --sin(θ + ψ) - sin(θ - ψ ). 2

Hence,

∫ t (f * g)(t) = 1-(sin(ω t) - sin(ω t - 2ω τ)) dτ 0 2 0 0 0 [ ]t = 1τsinω 0t +-1--co s(2ω 0τ - ω 0t) 2 4ω 0 τ=0 1 = -tsinω 0t. 2

Of course the formula only holds for $t ≥ 0$ . We did assume that $f$ and $g$ are zero (or just not defined) for negative $t$ .

The convolution has many properties that make it behave like a product. Let $c$ be a constant and $f$ , $g$ , and $h$ be functions then

Theorem 6.3.1. Let $f (t)$ and $g (t)$ be of exponential type, then

|------------------{∫--t-------------}-------------------| | | | L {(f * g)(t)} = L 0 f (τ)g (t - τ ) dτ = L {f(t)}L {g(t)}.| ---------------------------------------------------------

In other words, the Laplace transform of a convolution is the product of the Laplace transforms. The simplest way to use this result is in reverse.

Example 6.3.3: Suppose we have the function of $s$ defined by

----1--- = --1---1 . (s + 1 )s2 s + 1s2

We recognize the two entries of Table 6.2. That is

{ } { } -1 --1-- -t -1 -1 L s + 1 = e and L s2 = t.

Therefore,

{ } ∫ t L-1 --1---1 = τe-(t-τ) dτ = e-t + t - 1. s + 1s2 0

The calculation of the integral involved an integration by parts.

6.3.2 Solving ODEs

The next example will demonstrate the full power of the convolution and Laplace transform. We will be able to give a solution to the forced oscillation problem for any forcing function as a definite integral.

Example 6.3.4: Find the solution to

x′′ + ω 20x = f(t), x (0) = 0, x′(0) = 0,

for an arbitrary function $f(t)$ .

We first apply the Laplace transform to the equation. Denote the transform of $x(t)$ by $X (s)$ and the transform of $f(t)$ by $F (s)$ as usual.

s2X (s) + ω2X (s) = F(s), 0

or in other words

1 X (s) = F (s)-----2. s2 + ω0

We know

{ 1 } sinω 0t L-1 -2----2 = -------. s + ω0 ω 0

Therefore,

∫ t sin ω0(t - τ) x(t) = f(τ)------------dτ, 0 ω 0

or if we reverse the order

∫ t x(t) = sinω-0tf (t - τ) dτ. 0 ω0

Let us notice one more thing with this example. We can now also notice how Laplace transform handles resonance. Suppose that $f (t) = cosω 0t$ . Then

We have already computed the convolution of sine and cosine in Example 6.3.2. Hence

Note the $t$ in front of the sine. This solution will, therefore, grow without bound as $t$ gets large, meaning we get resonance.

Using convolution you can also find a solution as a definite integral for arbitrary forcing function $f (t)$ for any constant coefficient equation. A definite integral is usually enough for most practical purposes. It is usually not hard to numerically evaluate a definite integral.

6.3.3 Volterra integral equation

where $f (t)$ and $g (t)$ are known functions and $x (t)$ is an unknown. To solve this equation we apply the Laplace transform to get

where $X (s)$ , $F(s)$ , and $G (s)$ are the Laplace transforms of $x(t)$ , $f(t)$ , and $g(t)$ respectively. We find

Example 6.3.5: Solve

∫ t x(t) = e-t + sinh(t - τ)x(τ) d τ. 0

We apply Laplace transform to obtain

X(s) = --1--+ ---1--X (s ), s + 1 s2 - 1

1-- X (s) =---s+1---= s-- 1-= --s---- --1--. 1 - -12-- s2 - 2 s2 - 2 s2 - 2 s-1

It is not hard to apply Table 6.1 to find

√-- -1-- √ -- x(t) = cosh 2t - √ 2-sinh 2t.

$f (t)$	$L{f(t)} = F (s)$






$′ g (t)$	$sG(s) - g(0)$

$′′ g (t)$	$2 ′ s G (s) - sg(0) - g (0 )$

$′′′ g (t)$	$3 2 ′ ′′ sG (s) - s g(0) - sg (0) - g (0)$

Chapter 6
The Laplace transform

6.1 The Laplace transform

6.1.1 The transform

6.1.2 Existence and uniqueness

6.1.3 The inverse transform

6.1.4 Exercises

6.2 Transforms of derivatives and ODEs

6.2.1 Transforms of derivatives

6.2.2 Solving ODEs with the Laplace transform

6.2.3 Using the Heaviside function

6.2.4 Transforms of integrals

6.2.5 Exercises

6.3 Convolution

6.3.1 The convolution

6.3.2 Solving ODEs

6.3.3 Volterra integral equation

6.3.4 Exercises

Chapter 6The Laplace transform

6.1 The Laplace transform

6.1.1 The transform

6.1.2 Existence and uniqueness

6.1.3 The inverse transform

6.1.4 Exercises

6.2 Transforms of derivatives and ODEs

6.2.1 Transforms of derivatives

6.2.2 Solving ODEs with the Laplace transform

6.2.3 Using the Heaviside function

6.2.4 Transforms of integrals

6.2.5 Exercises

6.3 Convolution

6.3.1 The convolution

6.3.2 Solving ODEs

6.3.3 Volterra integral equation

6.3.4 Exercises

Chapter 6
The Laplace transform