Laplace transform - Essential Mathematics for Chemical Engineers

Definition and properties¶

The Laplace transform is used in signals/controls. It is also a way to solve differential equations using algebra. It is defined as:

F(s) = L[f(t)] = \int_0^\infty e^{-st} f(t) \d{t}

(1)

The inverse Laplace transform of F is f, i.e., $f(t) = L^{-1}[F(s)]$ . The Laplace transform of many functions can be computed using integration by parts. For example, the Laplace transform of $f(t) = t$ is:

\begin{align} L[t] &= \int_0^\infty e^{-st} t \d{t} \\ &= \left.-\frac{te^{-st}}{s}\right|_0^\infty + \int_0^\infty \frac{e^{-st}}{s} \d{t} \\ &= -\left.\frac{e^{-st}}{s^2}\right|_0^\infty \\ &= \frac{1}{s^2} \end{align}

(2)

Importantly, the Laplace transform of the first derivative of an unknown function $f(t) = y'(t)$ is:

\begin{align} L[y'(t)] &= \int_0^\infty e^{-st} y'(t) \d{t} \\ &= \left.e^{-st} y(t)\right|_0^\infty - \int_0^\infty y(t) \left(-s e^{-st}\right) \d{t} \\ &= [0 - y (0)] + s \int_0^\infty e^{-st} y \d{t} \\ &= -y(0) + s L[y(t)] \\ &= s Y(s) - y(0) \end{align}

(3)

using $u = e^{-st}$ and $\d{v} = y'(t) \d{t}$ . Some common Laplace transforms are:

$f(t)$	$F(s) = L[f(t)]$
$t^n$ , $n=0, 1, 2, \cdots$	$\dfrac{n!}{s^{n+1}}$
$e^{at}$	$\dfrac{1}{s - a}$
$\sin(at)$	$\dfrac{a}{s^2 + a^2}$
$\cos(at)$	$\dfrac{s}{s^2 + a^2}$
$e^{at} \sin(bt)$	$\dfrac{b}{(s - a)^2 + b^2}$
$e^{at} \cos(bt)$	$\dfrac{s - a}{(s - a)^2 + b^2}$

The Laplace transform is a linear operator, so

\begin{align} L[kf] &= kL[f] \\ L[f + g] &= L[f] + L[g] \end{align}

(4)

Solving first-order ODEs¶

Nonhomogeneous, first-order ODEs with constant coefficients are nice to solve with Laplace transforms:

y' + by = r(t), \quad y(0) = y_0

(5)

Apply the Laplace transform to both sides of the equation:

\begin{align} L[y' + by] &= L[r] \\ L[y'] + bL[y] &= L[r]\\ \left(sY - y_0\right) + bY &= R\\ Y &= \frac{y_0 + R}{s+b} \end{align}

(6)

where $Y = L[y(t)]$ and $R = L[r(t)]$ . If we can invert $y = L^{-1}[Y]$ using tables, we have a solution!

Example 1: Laplace transform¶

Solve the initial value problem

y' =-ky, \quad y(0) =2

(7)

Rearrange, apply Laplace transform $Y = L[y]$ , and solve for $Y$ :

\begin{align} y'+ ky &= 0 \\ L[y'+ky] &= L[0] \\ L[y']+kL[y] &= L[0] \\ sY - y(0) + k Y &= 0 \\ (s+k) Y - 2 &= 0 \\ Y &= \frac{2}{s+k} \end{align}

(8)

Invert Laplace transform:

y = L^{-1}\left[\frac{2}{s+k}\right] = 2L^{-1}\left[\frac{1}{s+k}\right] = 2 e^{-kt}

(9)

Example 2: Laplace transform with partial fractions¶

Solve the initial value problem:

y'-y = t, \quad y(0) = 1

(10)

Rearrange, apply Laplace transform $Y = L[y]$ , and solve for $Y$ :

\begin{align} L[y'-y] &= L[t]\\ L[y']-L[y] &= L[t]\\ sY - y(0) - Y &= \frac{1}{s^2}\\ (s-1) Y - 1 &= \frac{1}{s^2}\\ Y &= \frac{1}{s-1} + \frac{1}{s^2(s-1)} \end{align}

(11)

These terms do not immediately correspond to the Laplace transform table, but we can separate them using partial fraction decomposition:

\frac{1}{(s-1) s^2} = \frac{A_1}{s-1} + \frac{A_2}{s} + \frac{A_3}{s^2}

(12)

$A_1$ and $A_3$ can be found using the coverup method:

\begin{align} A_1 &= \frac{1}{1^2} = 1 \\ A_3 &= \frac{1}{0-1} = -1 \end{align}

(13)

To find $A_2$ , substitute and cross multiply:

1 = s^2 + A_2(s-1)s - (s-1)

(14)

then compare the coefficient of the $s^2$ terms on either side:

1 + A_2 = 0 \to A_2 = -1

(15)

So, all together:

Y = \frac{2}{s-1} - \frac{1}{s} - \frac{1}{s^2}

(16)

Solve for $y$ by applying the inverse Laplace transform

\begin{align} y &=L^{-1}[Y]\\ &=2L^{-1}\left[\frac{1}{s-1}\right]-L^{-1}\left[\frac{1}{s}\right]-L^{-1}\left[\frac{1}{s^2}\right]\\ &=2e^t-1-t \end{align}

(17)

Example: Hormone level¶

The concentration of a hormone in the blood c varies due to sinusoidal production by the thyroid and continuous removal according to:

c' = A + B\cos\left(\frac{\pi t}{12}\right) - kc

(18)

The concentration is $c_0$ at 6 AM ( $t = 0$ ). What is the average concentration between 6 PM and 6 AM the same day?

To solve, rearrange and use the Laplace transform:

\begin{align} c' + kc &= A + B\cos\left(\frac{\pi t}{12}\right) \\ [sC(s) - c_0] + kC(s) &= L\left[A + B\cos\left(\frac{\pi t}{12}\right)\right] \\ (s+k) C - c_0 &= \frac{A}{s} + \frac{Bs}{s^2 + (\pi/12)^2} \\ C(s) &= \left(\frac{1}{s + k}\right)\left[c_0 + \frac{A}{s} + \frac{Bs}{s^2 + (\pi/12)^2}\right] \\ &= \frac{c_0}{s + k} + \frac{A}{s(s + k)} + \frac{Bs}{[s^2 + (\pi/12)^2](s + k)} \end{align}

(19)

Use partial fraction decomposition on both fractions. The first one is:

\frac{1}{s(s + k)} = \frac{c_1}{s} + \frac{c_2}{s + k}

(20)

and the cover-up method gives $c_1 = 1/k$ and $c_2 = -1/k$ . For the next one:

\frac{s}{[s^2 + (\pi/12)^2](s + k)} = \frac{c_1 s + c_2}{s^2 + (\pi/12)^2} + \frac{c_3}{s + k}

(21)

The cover-up method gives

c_3 = \frac{-k}{k^2 + (\pi/12)^2}

(22)

while cross-multiplying and matching coefficients gives

\begin{align} s &= (c_1 s + c_2)(s + k) + c_3\left[s^2 + (\pi/12)^2\right] \\ s &= (c_1 + c_3)s^2 + (c_1 k + c_2)s + c_2 k + c_3(\pi/12)^2 \end{align}

(23)

\begin{align} c_1 + c_3 &= 0 \\ c_1 k + c_2 &= 1 \\ c_2 k + c_3 (\pi/12)^2 &= 0 \end{align}

(24)

Using our solution for $c_3$ in the first equation gives $c_1$ and in the third equation gives $c_2$ :

c_1 = \frac{k}{k^2 + \left(\frac{\pi}{12}\right)^2}, \quad c_2 = \frac{(\pi/12)^2}{k^2 + (\pi/12)^2}

(25)

All together,

\begin{align} C(s) &= \frac{c_0}{s + k} + \frac{A}{k} \left[\frac{1}{s} - \frac{1}{s+k} \right] \\ &+ \frac{B}{k^2 + (\pi/12)^2} \left[\frac{ks}{s^2 + (\pi/12)^2} + \frac{(\pi/12)^2}{s^2 + (\pi/12)^2} - \frac{k}{s+k} \right] \end{align}

(26)

Invert the Laplace transforms term by term:

\begin{align} c(t) &= c_0 e^{-kt} + \frac{A}{k} \left(1 - e^{-kt}\right) \\ &+ \frac{B}{k^2 + (\pi/12)^2} \left[k \cos\left(\frac{\pi t}{12}\right) + \frac{\pi}{12} \sin\left(\frac{\pi t}{12}\right) - k e^{-kt}\right] \\ &= \frac{A}{k} + \frac{B}{k^2 + (\pi/12)^2} \left[k \cos\left(\frac{\pi t}{12}\right) + \frac{\pi}{12} \sin\left(\frac{\pi t}{12}\right)\right] \\ &+ \left[c_0 - \frac{A}{k} - \frac{Bk}{k^2 + \left(\frac{\pi}{12}\right)^2}\right] e^{-kt} \end{align}

(27)

The average concentration is:

\begin{align} \langle c \rangle &= \frac{1}{t_1-t_0} \int_{t_0}^{t_1} c(t) \d{t} \\ &= \frac{1}{24 - 12} \int_{12}^{24} \Biggl( \frac{A}{k} + \frac{B}{k^2 + (\pi/12)^2} \left[k \cos\left(\frac{\pi t}{12}\right) + \frac{\pi}{12} \sin\left(\frac{\pi t}{12}\right)\right] \\ &+ \left[c_0 - \frac{A}{k} - \frac{Bk}{k^2 + (\pi/12)^2}\right] e^{-kt} \Biggr) \d{t} \\ &= \frac{1}{12} \Biggl[ \frac{A}{k} t + \frac{B}{k^2 + (\pi/12)^2} \left( \frac{12k}{\pi} \sin\left( \frac{\pi t}{12} \right) - \cos\left( \frac{\pi t}{12} \right) \right) \\ &+ \frac{1}{k} \left( \frac{A}{k} + \frac{Bk}{k^2 + (\pi/12)^2} \right) e^{-kt} \Biggr]_{12}^{24} \\ &= \frac{A}{k} - \frac{1}{6} \frac{B}{k^2 + (\pi/12)^2} - \frac{1}{12}\left[\frac{A}{k^2} + \frac{B}{k^2 + (\pi/12)^2}\right] (e^{-12k} - e^{-24k}) \end{align}

(28)

Skill builder problems¶

Solve the following IVPs using Laplace transforms:

Solution to Exercise 1

\begin{align} L[y' + 4y] &= L[e^{4x}] \\ sY - y(0) + 4Y &= \frac{1}{s-4} \\ (s+4) Y &= \frac{1}{s-4} \\ Y &= \frac{1}{(s+4)(s-4)} \end{align}

(29)

Use partial fractions

\frac{1}{(s+4)(s-4)} = \frac{A_1}{s+4} + \frac{A_2}{s-4}

(30)

and cover up to find $A_1 = -1/8$ and $A_2 = 1/8$ .

Solve by inverting the Laplace transforms:

\begin{align} Y &= \frac{1}{8} \left(\frac{1}{s-4} - \frac{1}{s+4} \right) \\ y = L^{-1}[Y] &= \frac{1}{8}\left( L^{-1}\left[\frac{1}{s-4}\right] - L^{-1}\left[\frac{1}{s+4}\right] \right) \\ &= \frac{1}{8}(e^{4x} - e^{-4x}) \end{align}

(31)

Solution to Exercise 2

\begin{align} L[y' + 2y] &= L[8] \\ sY - y(0) + 2Y &= \frac{8}{s} \\ (s+2) Y - 1 &= \frac{8}{s} \\ Y &= \frac{1}{s+2} + \frac{8}{s(s+2)} \end{align}

(32)

Use partial fractions and the cover-up method for the second fraction:

\frac{8}{s(s+2)} = \frac{A_1}{s} + \frac{A_2}{s+2}

(33)

to find $A_1 = 4$ and $A_2 = -4$ . Solve by simplifying and inverting the Laplace transforms:

\begin{align} Y &= \frac{4}{s} - \frac{3}{s+2} \\ y = L^{-1}[Y] &= 4L^{-1}\left[\frac{1}{s}\right] - 3L^{-1}\left[\frac{1}{s+2}\right] \\ &= 4 - 3e^{-2x} \end{align}

(34)

Solution to Exercise 3

\begin{align} L[y' - y] &= L[1 - 2x + \sin 3x ] \\ sY - y(0) - Y &= \frac{1}{s} - \frac{2}{s^2} + \frac{3}{s^2 + 9} \\ (s-1)Y &= -1 + \frac{1}{s} - \frac{2}{s^2} + \frac{3}{s^2 + 9} \\ Y &= -\frac{1}{s - 1} + \frac{1}{s(s - 1)} - \frac{2}{s^2(s - 1)} + \frac{3}{(s^2 + 9)(s - 1)} \end{align}

(35)

Use partial fractions:

\begin{align} \frac{1}{s(s-1)} &= \frac{A_1}{s} + \frac{A_2}{s-1} \\ \frac{-2}{s^2(s - 1)} &= \frac{A_3}{s} + \frac{A_4}{s^2} + \frac{A_5}{s - 1} \\ \frac{3}{(s^2 + 9)(s - 1)} &= \frac{A_6s + B_6}{s^2 + 9} + \frac{A_7}{s - 1} \end{align}

(36)

Cover up to find:

\begin{align} A_1 &=\frac{1}{0-1} = -1 & A_4 &= \frac{-2}{0 - 1} = 2 & A_7 &= \frac{3}{1^2 + 9} = \frac{3}{10} \\ A_2 &=\frac{1}{1} = 1 & A_5 &= \frac{-2}{1} = -2 \end{align}

(37)

Finish the rest by cross-multiplying or pluggin in values. For the first fraction, plug in $s=-1$ and known coefficients:

1 = -A_3 + 2 + 1

(38)

then solve $A_3 = 2$ . For the third fraction, cross-multiply

3 = (A_6 s + B_6)(s-1) + \frac{3}{10}(s^2+9)

(39)

Then compare like powers of $s$ . Here, I use $s^2$ and $s^0$ :

\begin{align} A_6 + \frac{3}{10} &= 0 \\ -B_6 + \frac{27}{10} &= 3 \end{align}

(40)

so $A_6 = -3/10$ and $B_6 = -3/10$ . Simplify Y

Y = \frac{1}{s} + \frac{2}{s^2} - \frac{17}{10} \frac{1}{s-1} - \frac{3}{10} \frac{1}{s^2+9} - \frac{3}{10} \frac{s}{s^2+9}

(41)

Apply inverse Laplace transform across the equation:

y = L^{-1}\left[\frac{1}{s}\right] + 2 L^{-1}\left[\frac{1}{s^2}\right] - \frac{17}{10} L^{-1}\left[\frac{1}{s - 1}\right] - \frac{1}{10} L^{-1}\left[\frac{3}{s^2 + 9}\right] - \frac{3}{10} L^{-1}\left[\frac{s}{s^2 + 9}\right]

(42)

So, the solution is

y = 1 + 2x - \frac{17}{10} e^x - \frac{1}{10} \sin 3x - \frac{3}{10} \cos 3x

(43)