The Laplace Transform

The expression “It is easy to see that…” was used many and many times by the scientist Pierre-Simon Laplace when he didn’t want to come into the details of its ideas. One of the main contributions of Laplace, the Laplace Transform, will be explained here, hopefully in an easy to see fashion.

Definition

Let f:\mathcal{R}\rightarrow\mathcal{R}. Its Laplace transform is defined as:

\displaystyle \mathcal{L}\{f(t)\} = F(s) = \int_{0}^{\infty}{e^{-st}f(t)dt}.

If f is piecewise continuous and there exist real positive numbers a,K,M such that |f(t)| \leq Ke^{at} \quad \forall t \geq M, then the integral above is well defined for all values s > a.

Two important properties make the Laplace Transform interesting. The first one is its linearity, which follows simply because the integral is a linear operator.

\mathcal{L}\{a\cdot f(t) + b \cdot g(t)\} = a\cdot \mathcal{L}\{f\} + b\cdot \mathcal{L}\{g\}.

The second one tells us how to compute the Laplace transform of the derivative of a function, and it follows from integration by parts.

\begin{array}{ll} \mathcal{L}\{f^\prime(t)\} &= \displaystyle \int_{0}^{\infty}{e^{-st}f^\prime(t)dt}\\[0.25in] &= \displaystyle e^{-st}f(t)|^{\infty}_{0} - (-s)\int_{0}^{\infty}{e^{-st}f(t)dt}\\[0.25in] &= \displaystyle -f(0) + s\mathcal{L}\{f(t)\}. \end{array}

Generally:

\displaystyle \mathcal{L}\{f^{(n)}\} = s^n\mathcal{L}\{f(t)\} - s^{(n-1)}f(0) - s^{(n-2)}f^{(1)}(0) - \cdots - sf^{(n-2)}(0) - f^{(n-1)}(0).

For the last property, one needs some assumptions on f, namely, f^{(1)}, f^{(2)} \cdots f^{n-1} must be continuous, and f^{(n)} piecewise continuous.

Laplace transform of basic functions.

The Laplace Transform between any two distinct continuous functions will always be different, but it could be the case that two different piecewise continuous functions f,g that differs only in a finite set of points have the same Laplace Transform. In applications, however, the inverse transform can be computed even for piecewise continuous functions and gives the right answer to the problem. In order to compute inverse transforms it will be useful to have a table listing the Laplace Transform of some basic functions.

Constant function: f(t) = C.

\begin{array}{ll} \mathcal{L}\{f(t)\} &= \displaystyle C \cdot \int_{0}^{\infty}{e^{-st}dt}\\[0.25in] &= \displaystyle \lim_{x\rightarrow \infty } -\frac{C}{s} \left( e^{-st}|_{t=x} - e^{-st}|_{s=0} \right) = \frac{C}{s}. \end{array}

Exponential function: f(t) = C \cdot e^{at}.

\begin{array}{ll} \mathcal{L}\{f(t)\} &= \displaystyle  C \cdot \int_{0}^{\infty}{e^{-st}e^{at}dt}\\[0.25in] &= \displaystyle C \cdot \int_{0}^{\infty}{e^{(s-a)t}dt} = \frac{C}{(s-a)}\\[0.25in] \end{array}

Sine function: f(t) = \sin(at).

\begin{array}{rl} \mathcal{L}\{f(t)\} &= \displaystyle \int_{0}^{\infty}{e^{-st}\sin(at)dt}\\[0.25in] &= \displaystyle \frac{-1}{s}e^{-st}\sin(at)|_0^{\infty} - \frac{-a}{s}\int_0^{\infty}{e^{-st}\cos(at)dt}\\[0.25in] &= \displaystyle \frac{a}{s}\int_0^{\infty}{e^{-st}\cos(at)dt}\\[0.25in] &= \displaystyle \frac{a}{s}\left( \frac{1}{s}e^{-st}cos(at)|_0^{\infty} + \frac{-a}{s}\int_{0}^{\infty}{e^{-st}\sin(at)dt}\right)\\[0.25in] &= \displaystyle \frac{a}{s^2} -\frac{a^2}{s^2}\mathcal{L}\{f(t)\}\\[0.25in] \displaystyle \mathcal{L}\{f(t)\}\left( 1+\frac{a^2}{s^2}\right) &= \displaystyle \frac{a}{s^2}\\[0.25in] \displaystyle \mathcal{L}\{f(t)\} &= \displaystyle \frac{a}{s^2} \cdot \frac{1}{1+a^2/s^2} = \frac{a}{s^2+a^2}. \end{array}

Polynomial function: f(t) = t^n.

\begin{array}{ll} \mathcal{L}\{f(t)\} &=\displaystyle \int_{0}^{\infty}{e^{-st}t^ndt}\\[0.25in] &= \displaystyle \frac{-1}{s}e^{-st}t^n|_0^{\infty} + \frac{n}{s}\int_{0}^{\infty}{e^{-st}t^{n-1}dt}\\[0.25in] &= \displaystyle \frac{n}{s}\int_{0}^{\infty}{e^{-st}t^{n-1}dt}\\[0.25in] &= \displaystyle \frac{n!}{s^{n+1}}. \end{array}

Applications

One can use the Laplace Transform to solve initial value problems. For example, solve:

y'' - 2y' + 2y = \sin(t) \quad y(0)=1, \; y'(0) = 0.

To solve it we apply the Laplace transform in both sides and use its two main properties listed in the beginning of the post,.

\begin{array}{rl} \displaystyle \mathcal{L}\{y''\} - 2\mathcal{L}\{y'\} + 2\mathcal{L}\{y\} &=\displaystyle \mathcal{L}\{\sin(t)\}\\[0.25in] \displaystyle \mathcal{L}\{y\}\left( s^2 - 2s +2 \right) - y'(0) - sy(0) + 2y(0) &=\displaystyle \frac{1}{s^2+1}\\[0.25in] \displaystyle \mathcal{L}\{y\} &=\displaystyle \frac{1+s^3 - 2s^2 + s - 1}{(s^2+1)(s^2-2s+2)}. \end{array}

Using the technique of partial fractions:

\begin{array}{rl} \displaystyle \mathcal{L}\{y\} &= \displaystyle \frac{-7/5 + 3/5s}{s^2 -2s +2} + \frac{1/5 + 2/5s}{s^2+1}\\[0.25in] &= \displaystyle \frac{-7}{5}\left( \frac{1}{(s-1)^2+1}\right) + \frac{3}{5}\left(\frac{(s-1)}{(s-1)^2+1}\right)\\[0.25in] &+ \displaystyle \frac{3}{5}\left( \frac{1}{(s-1)^2+1}\right) + \frac{1}{5}\left(\frac{1}{s^2+1}\right) + \frac{2}{5}\left(\frac{s}{s^2+1}\right) \end{array}

Applying the inverse transform on both sides (with the help of the Laplace transform’s table):

\begin{array}{rl} y &= \displaystyle \frac{-7}{5}e^t\sin(t) + \frac{3}{5}e^t\cos(t) + \frac{3}{5}e^t\sin(t) + \frac{1}{5}\sin(t) + \frac{2}{5}\cos(t)\\[0.25in] &= \displaystyle \frac{1}{5} \left( \sin(t) - 4e^t\sin(t) + 2\cos(t) + 3e^t\cos(t) \right) . \end{array}

The last problem could be easily solved without recurring to the Laplace transform. In fact, the real power of Laplace transforms shows up when considering problems modelled using what is called step functions u_c(t).

u_c(t) = \left\{ \begin{array}{ll} 0 & t< c\\ 1 & t \geq c \end{array}\right. \quad c\geq 0.

Such functions are recurring in physics, in particular in the modelling of electric circuits. For example, the differential equation

2y'' + y' + 2y = g(t), \quad y(0) = 0, \; y'(0) = 0,

where g(t) = u_5(t) - u_{20}(t)  = \left\{ \begin{array}{ll} 1, & 5 \leq t < 20\\ 0, & 0 \leq t < 5 \text{ and } t \geq 20 \end{array} \right.

,governs the charge on the capacitor in a electric circuit with a unit voltage pulse 5 \leq t < 20. In order to see why the Laplace transform is useful in those cases, consider the following properties.

\begin{array}{rl} \mathcal{L}\{u_c(t)f(t-c)\} &= e^{-sc}\mathcal{L}\{f(t)\} \\[0.25in] u_c(t)f(t-c) &= \mathcal{L}^{-1}\{ e^{-sc}\mathcal{L}\{f(t)\} \} \\[0.25in] \mathcal{L}\{e^{ct}f(t)\} &= F(s-c) \\[0.25in] e^{ct}f(t) &= \mathcal{L}^{-1}\{F(s-c)\}. \end{array}

Let’s see how it works in practice by solving the model above for a simple electric circuit:

\begin{array}{ll} \mathcal{L}\{2y''\} + \mathcal{L}\{y'\} + \mathcal{L}\{2y\} & = \mathcal{L}\{u_5(t) - u_{20}(t)\} \\[0.25in] \mathcal{L}\{y\} &=  \mathcal{L}\{u_5(t)\}H(s) - \mathcal{L}\{u_{20}(t)\}H(s), \end{array}

where \displaystyle H(s) = \frac{1}{2s^3 + s^2 + 2s} = \mathcal{L}\{h(t)\}, and h(t) can be computed with the help of the Laplace tranform’s table of basic functions. Therefore,

y(t) = u_5(t)h(t-5) - u_{20}(t)h(t-20).

Convolutions.

Unfortunately, the Laplacian of the product is not equal to the product of the Laplacians. However, by defining an operation similar to the product of functions (in fact this operation is also known as the generalized product) called convolution, we can derive another interesting property of the Laplace transform. The convolution of function f with function g is defined as:

\displaystyle (f*g)(t) = \int_0^t{f(t-\tau)g(\tau)d\tau}.

Then

\displaystyle \mathcal{L}\{(f*g)\}(s) = \mathcal{L}\{f\} \cdot \mathcal{L}\{g\}.

 

In order to see that, work out the expression on the right-hand side and apply a substitution of variables.

Convolutions are seen very often in variational problems, and the Laplace transform can be extremely useful for solving them. See for example the post about the tautochrone curve.

Further Reading

[1] – Boyce, W.E. and Diprima R.C – Elementary Differential Equations and Boundary Value Problems (Chapter 6).

 

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