As regular readers know, time evolution of classical and quantum systems is a popular topic on this blog, forming a major thread of the columns to date. But the majority of column space dealing with the initial value problem has either been devoted to time domain analysis or, less frequently, with frequency domain techniques that fall under the Fourier Transform heading. This week starts what will be a lazy meandering through the Laplace Transform.
The Laplace Transform is a favorite of engineers, specifically ones who have to work with feedback control, and it would be worth exploring simply for that reason alone. However, it is also curious that its existence is almost wholely absent from the standard training for physicists. Exactly why the latter tend to use the Fourier transform at the expense of all else can be confidently guessed at (the Fourier Transform seems to be built into Quantum Mechanics) but a deeper understanding will result simply by comparing and contrasting the two methods.
Well before that last goal can be reached, the Laplace Transform must be examined on its own merits.
The Laplace Transform is a functional that maps input functions to output functions via
\[ {\mathcal L}[f(t)] = \int_0^{\infty} dt \, e^{-st} f(t) \; .\]
It is often written in a shorthand way as
\[ {\mathcal L}[f(t)] \equiv F(s) \; , \]
since this notation makes for convenient manipulation.
The properties of the Laplace Transform that make it so useful can be summarized as:
- it is a linear transformation,
- it turns operations of differentiation and integration into algebraic operations,
- incorporates the initial conditions seamlessly, and
- is general, thus affording a single approach applicable to an extraordinary wide range of situations.
One caveat is needed on point 4. All Laplace Transforms can only be applied when the system obeys a set of linear evolution equations. In this regard, it is unable to tackle the range and scope of problems that the Lie Series can deal with but its convenience for linear (or linearized) systems makes it a favorite of the controls engineer.
Let’s look at the first 3 points in more detail. For an operation $${\mathcal O}$$ to be linear, the following relationship must hold
\[ {\mathcal O}[a F + b G] = a {\mathcal O}[F] + b {\mathcal O}[G] \; ,\]
where $$a,b$$ are constants and $$F,G$$ are functions. The properties of the integral insure that the Laplace Transform is linear as follows
\[ {\mathcal L}[a f(t) + b g(t) ] = \int_0^{\infty} dt \, e^{-st} \left( a f(t) + b g(t) \right) \\ = a \int_0^{\infty} dt \, e^{-st} f(t) + b \int_0^{\infty} dt \, e^{-st} g(t) \\ = a {\mathcal L}[f(t)] + b {\mathcal L}[g(t)] \; .\]
The Laplace Transform is a machine which takes an input function and maps it to an output function. Differentiation is also a machine that maps input functions to output functions. It’s logical to ask how do these two operations interact.
To figure this out, feed the derivative of a function $$f'(t)$$ into the Laplace Transform.
\[ {\mathcal L}[f'(t)] = \int_0^{\infty} dt \, e^{-st} f'(t) \]
and integrate by parts to move the derivative off of $$f(t)$$ and onto the exponential. This gives
\[ {\mathcal L}[f'(t)] = \left. e^{-st} f(t) \right|_0^{\infty} – \int_0^{\infty} dt \, \left( \frac{d}{dt} e^{-st} \right) f(t) \; , \]
which simplifies to
\[ {\mathcal L}[f'(t)] = -f(0) + \int_0^{\infty} dt \, s e^{-st} f(t) = s F(s) – f(0) \; .\]
The same steps can be followed for higher order derivatives but a more useful way is to simply use an induction argument. Define a second derivative to be the derivative of first derivative and apply the theorem twice.
\[{\mathcal L}[f^{\prime\prime}(t)] = {\mathcal L}[(f'(t))’] = s {\mathcal L}[f'(t)] – f'(0) = s^2 F(s) – s f(0) – f'(0) \; \]
This process can be repeated as often as desired yielding the general form
\[ {\mathcal L}\left[ \frac{d^n}{dt^n} f(t) \right] = s^n F(s) – s^{n-1} f(0) – s^{n-2} f^{(1)}(0) – s^{(n-3)} f^{(2)}(0) – \ldots \; ,\]
where
\[ f^{(n)}(0) = \left. \frac{d^n}{dt^n} f(t) \right|_0 \; .\]
Likewise, the Laplace Transform takes integrals of functions to algebraic relations as well. As in the derivative case, start with the most basic form of
\[{\mathcal L}\left[ \int dt’ \, f(t’) \right] = \int_0^{\infty} dt \; e^{-st} \int dt’ \, f(t’) \; . \]
An application of an integration-by-parts is next employed with
\[ dV = e^{-st} \implies V = – \frac{1}{s} e^{-st} \]
and
\[ U = \int dt’ \, f(t’) \implies \frac{dU}{dt} f(t) \; , \]
yielding
\[ \int_0^{\infty} dt \, e^{-st} \int dt’ \, f(t’) = \left.-\frac{1}{s} e^{-st} \int dt’ \, f(t’) \right|_0^{\infty} – \int_0^{\infty} \left(-\frac{1}{s} \right) e^{-st} f(t) \; \]
This expression simplifies to
\[ {\mathcal L}\left[ \int dt’ \, f(t’) \right] = \frac{f^{(-1)}(0)}{s} + \frac{1}{s} F(s) \; ,\]
where $$f^{(-1)}(0) = \int dt’ \, f(t’) \left.\right|_0$$ is a convenient short-hand.
Repeated applications yields the generalization
\[ {\mathcal L}\left[ \int dt’ \, \int dt^{\prime\prime} \, \ldots \int dt^{(n)} \, f(t^{(n)})\right] = \frac{f^{(-n)}(0)}{(s)} + \frac{f^{(-n+1)}(0)}{(s^2)} \\ + \ldots + \frac{F(s)}{s^n} \; .\]
In contrast, the Laplace Transform of products and divisions, results in derivatives and integrations in the $$s$$-space.
First consider multiplication of $$f(t)$$ by some power of $$t$$:
\[ {\mathcal L}[t f(t)] = \int_0^{\infty} dt \, e^{-st} t f(t) \; .\]
Recognize that
\[ \frac{d}{dt} \left( e^{-st} \right) = -t e^{-st} \;.\]
Thus
\[ {\mathcal L}[t f(t)] = -\int_0^{\infty} dt \, \frac{d}{ds} \left( e^{-st} \right) f(t) = -\frac{d}{ds} \int_0^{\infty} dt \, e^{-st} f(t) = -\frac{d}{ds} F(s) \; .\]
This is easily generalized to any arbitrary power to yield
\[ {\mathcal L}[t^n f(t) ] = (-1)^n \frac{d^n}{ds^n} {\mathcal L}[f(t)] = (-1)^n \frac{d^n}{ds^n} F(s) \; .\]
As the last piece, let’s look at division by a power of $$t$$, starting with the simplest case
\[ {\mathcal L}[f(t)/t] = \int_0^{\infty} dt \, e^{-st} f(t)/t \; .\]
Recognizing that
\[ \int_s^{\infty} ds’ \, e^{-s’ t} = e^{-st}/t \]
allows for the Laplace Transform to become
\[{\mathcal L}[f(t)/t] = \int_0^{\infty} dt \int_s^{\infty} ds’ \, e^{-s’ t} = \int_s^{\infty} ds’ \, F(s’) \; .\]
This is also easily generalized to arbitrary powers to yield
\[{\mathcal L}[f(t)/t^n] = \int_s^{\infty} ds’ \, \int_{s’}^{\infty} ds^{\prime\prime} \ldots \int_{s^{(n-1)}}^{\infty} ds^{(n)} F(s(n)) \;.\]
Next week, I’ll take a look at what functions possess a Laplace Transform and how the Laplace Transform relates to the Fourier Transform.