Motivation and physical interpretation of the Laplace transform

Question

Concerning the one-sided Laplace transform,

$$\mathcal{L}\{f\}(s) = \int_0^\infty f(t)e^{-st} dt$$

what is a motivation to come up with that formula? I am particularly interested in "physical" interpretations of the Laplace transform.

By comparison, the Fourier transform,

$$ \mathcal{F}\{f\}(\xi) = \int_{-\infty}^{\infty} f(x) e^{-i 2 \pi \xi x} dx $$

can be motivated as a continuous decomposition into frequency modes, which is a physics-oriented interpretation: $\mathcal{F}\{f\}(\xi)$ tells you the amplitude of the Fourier mode $e^{-i 2 \pi \xi x}$ in the signal $f$.

I have not yet encountered any good explanation of how the Laplace transform formula arises. If the function $f$ vanishes on negative values, then $\mathcal{L}\{f\}(s) = \mathcal{F}\{f\}(s/(2\pi i))$, which is a nice relationship between the Laplace and Fourier transform, but not particularly insightful.

Answer 1

The physical motivation for the Laplace transform is causality.

Consider the linear input-output relation $$f_{\text{output}}(t)=\int_{0}^\infty R(t-t')f_{\text{input}}(t')\,dt'.$$ Causality dictates that the response function $R(t)$ is zero for $t<0$, to ensure that the output only depends on the input at earlier times.

Then if we know the Laplace transform $\hat{R}(s)$ of the response function (or transfer function), the output is related to the input by $$\hat{f}_{\text{output}}(s)=\hat{R}(s)\hat{f}_{\text{input}}(s).$$ One could of course work instead with the Fourier transform, with the complication that the Fourier transform of a constant input only exists as a delta-function distribution (while the Laplace transform is simply $1/s$).

Answer 2

Besides the important physical motivation pointed out by Carlo Beenakker, there is another one, purely mathematical. Laplace transform is a generalization of a power series (and Dirichlet series). In $f(x)=\sum_{n=0}^\infty c_n x^n$, set $x=e^{-t}$, then $$f(e^{-t})=\sum_{n=0}^\infty c_ne^{-nt}=\int_{0-}^\infty e^{-st}d\mu(s),$$ where $\mu$ is a discrete measure with atoms at non-negative integers, each atom of size $c_n$. If instead of a discrete measure you take a continuous one with density $F$, that is $d\mu(s)=F(s)ds$, you obtain the classical Laplace integral. If $d\mu$ is any discrete measure, you obtain a Dirichlet series $$\sum_{n=0}^\infty c_n e^{-\lambda_n x}.$$

Another motivation is that Laplace transform is analytic, under much weaker assumptions on a function then those of analyticity of Fourier transform. This allows to generalize Fourier transform to a broader class of objects than Schwarz distributions.

For example, let $f$ be a locally integrable function satisfying only $$f(x)=O(e^{\epsilon x})\quad\mbox{for every}\quad\epsilon>0.$$ Then the usual Fourier transform is not defined. However both integrals $$F_1(z)=\int_0^\infty f(x)e^{-ixz}dx, \quad\mbox{and}\quad F_2(z)=-\int_{-\infty}^0 f(x)e^{-ixz}dx$$ are well defined and analytic: the first one for $z$ in the lower half-plane and the second one in the upper half-plane. The pair of analytic functions $(F_1,F_2)$ is called the Fourier transform in the sense of Carleman, and in the case when $f\in L^1$ that is usual Fourier transform exists, we have $\hat{f}=F_1-F_2$, where boundary values of $F_1,F_2$ should be used. When $f$ is a tempered distribution, the boundary values of $F_1,F_2$ still exist, as tempered distributions, and $\hat{f}=F_1-F_2$.

This idea is the basis of the theory of hyperfunctions. $F_1$ and $F_2$ are essentially Laplace transforms, up to the substitution $z\mapsto iz$.

Answer 3

Interest is continuously compounded at rate $r,$ so that if you deposit $\\\$1$ now it will be worth $\\\$e^{rx}$ at time $x.$ How much do you need to deposit now in order to withdraw at rate $\\\$f(x)$ per unit of time $x$ forever? The answer is $$ (\mathcal Lf)(r) = \int_0^\infty f(x) e^{-rx}\, dx. $$ In financial mathematics, this quantity is called the present value of a revenue stream.

Answer 4

The Laplace transform is the fundamental operation encoding the canonical ensemble in statistical mechanics. It converts the density of states $d(\varepsilon )$ (a non-statistical concept) into the canonical partition function $Z(\beta)$, supplying the requisite Boltzmann factor $e^{-\beta \varepsilon } $: $$ Z(\beta ) = \int_0^\infty d\varepsilon \ d(\varepsilon) \, e^{-\beta \varepsilon } $$

Answer 5

I will not discuss the uses of the Laplace transform.

Myself I think of the analogy with sequences. A sequence $(a_n)_{n\geq 0}$ is determined by its generating function (convergence issues aside) $$ A(s)=\sum_{n\geq 0} a_n s^n. $$ By making the change in variables $s=e^\lambda$ we obtain its Laplace transform (again set aside convergence issues) $$ L(\lambda) =\sum_{n\geq 0} a_n e^{n\lambda}. $$ Now think of the sequence as a signed measure on $[0,\infty)$ $$ \mu=\sum_{n\geq 0} a_n\delta_n, $$ where $\delta_n$ is the Dirac delta concentrated at $n$. Then $$ L_\mu(\lambda)=\int_{\mathbb{R}_{\geq 0}} e^{\lambda t}\mu[dt]. $$ This expression makes sense for any signed measure on $[0,\infty)$. If you now choose the measure $\mu$ to be of the form $\mu[dt]=f(t) dt$ you get the Laplace transform of $f$.

Why would the Laplace transform determine a signed measure $\mu$ uniquely? Here's an argument when $\mu$ is finite.

Denote by $C_\infty([0,\infty))$ the space of continuous functions $[0,\infty\to\mathbb{R}$ that have a finite limit at $\infty$. This is a Banach space with respect to the sup-norm. A finite measure on $[0,\infty)$ is uniquely determined by what it does to the functions in $C_\infty([0,\infty))$, i.e., the integrals $$ \mu[f]=\int_{[0,\infty)} f(t)\mu[dt],\;\;f\in C_\infty([0,\infty)). $$ The family of functions $E_\lambda(t):=e^{\lambda t}$, $\lambda\leq 0$ spans a dense subspace of $C_\infty([0,\infty)$. Thus $\mu$ is uniquely determined by the integrals $$ L_\mu(\lambda) =\mu[E_\lambda],\;\;\lambda \leq 0. $$

Answer 6

From R. N. Bracewell, "The Fourier Transform and Applications", McGraw Hill 3rd ed., pp.381:

"Advantages of the Laplace transform over the Fourier transform for handling electrical transient and other problems are often quoted in books. The essential advantage of the Fourier transform is its physical interpretability -- as a spectrum, a diffraction pattern, and so on. Laplace transforms are not so interpretable, and once we take the Laplace transform of an equation we retain only a mathematical, not a physical, grasp of its meaning."

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EDIT. I cited this excerpt, because this is where I personally found my answer to the question "is there a physical interpretation of the Laplace transform?". Bracewell gives a fairly clear and definite answer in the negative. As far as I know, this book is still a standard reference for science and engineering, so I stick to this.

Answer 7

Re "I have not yet encountered any good explanation of how the Laplace transform formula arises." I think it's useful to look at early origins of the LPT, relations to other transforms, and some specific uses in operator calculus, analysis, physics, and combinatorics.

I)

Consider these delta function resolutions for the Fourier, Mellin, and Laplace transforms:

$$\int^{\infty}_{-\infty}{\exp(2 \pi ifx)\exp(-2 \pi ify)df} = \delta(x-y),$$

$$\frac{1}{2\pi i} \int_{\sigma - i \infty}^{\sigma + i \infty} x^{-s} y^{s} ds= \delta(\ln(x)-\ln(y))= y \delta(x-y),$$

$$\frac{1}{2\pi i} \int_{\sigma - i \infty}^{\sigma + i \infty} e^{-xp} e^{yp} dp=\delta(x-y).$$

For me, the delta fct. results seem intuitive (also analogous to the orthogonality relationship for characters of character groups), and the eqns. encapsulate the properties of the transforms (try deriving the transform pairs and other relations from them, e.g., Plancherel, convolution, Poisson summation) and illustrate the transformations from one transform to another.

II}

Before the Fourier transform and before the LPT's use in probability theory by Laplace (characteristic functions), in projection for the orthogonal associated Laguerre polynomials, and in electrical circuit theory and associated operational calculus beginning with Heaviside and Bromwich (see Operational Calculus by van der Pol and Bremmer and The Theory of Linear Operators by Harold Davis), Euler used the fundamental Laplace transform the Euler integral for the gamma function, a hybrid Mellin-Laplace transform, to develop his version of a fractional calculus. This can be couched the following way:

(From my notes in my response to the MO-Q "What does Mellin inversion "really mean"?", see also this MSE-A);

The fundamental Laplace transform the iconic Euler integral for the gamma function for $Real(s) > 0$ (easily extended to all complex $s$ with appeal to the inverse Mellin transform) is

$$ \int_0^\infty \frac{t^{s-1}}{(s-1)!} \; e^{-t\;p} \; dt = p^{-s}.$$

Note that when the integral is analytically continued, this provides an interpolation of the sequence $p^n$, the coefficients of the e.g.f. $e^{-tp}$, to $p^{-s}$, a Mellin transform interpolation, or example of Ramanujan's master formula/heuristic.

The Euler integral is easily established by considering it as a modifed Mellin transform and then using the inverse modified Mellin transform to determine for $\sigma > 0$

$$ \frac{1}{2\pi i} \int_{\sigma -i \infty}^{\sigma + i \infty} \frac{\pi}{\sin(\pi s)} p^{-s} \frac{t^{-s}}{(-s)!}ds = e^{-pt}.$$

Generalizing from a variable $p$ to an operator, this suggests that a natural interpolation of the derivative as the fractional integro-derivative of a fractional calculus (Euler's) can be obtained by using the Euler-Laplace-Mellin transform to interpolate / extrapolate the sequence of derivatives $D_x^n = (\frac{\partial}{\partial x})^n$, the operator coefficients of the shift operator $\displaystyle e^{tD_x}$, to $D_x^{-s}$. Specifically, incorporating the Heaviside step function $H(t)$,

$$ \int_0^\infty \frac{t^{s-1}}{(s-1)!} \; e^{-tD_x} \; dt \; H(x) g(x) = D_x^{-s} H(x) g(x)$$

$$ = \int_0^\infty \frac{t^{s-1}}{(s-1)!} \; e^{-tD_x} \; H(x) g(x)\; dt$$

$$= \int_0^\infty \frac{t^{s-1}}{(s-1)!} \; H(x-t) \; g(x-t) dt \; = H(x) \int_0^x \frac{t^{s-1}}{(s-1)!}\; g(x-t) dt ,$$

a convolution integral, such as Carlo presents.

Then acting on the power function for $\displaystyle \alpha > -1$, i.e., with $g(x) = x^{\alpha}$,

$$ \int_0^\infty \frac{t^{s-1}}{(s-1)!} \; H(x-t) \; (x-t)^\alpha dt =H(x) \int_0^x \frac{t^{s-1}}{(s-1)!} \; (x-t)^\alpha \; dt $$

$$ = \int_0^x \frac{t^{s-1}}{(s-1)!} \; \sum_{k \ge 0} (-1)^k \; x^{\alpha-k} \frac{\alpha!}{(\alpha-k)} \; \frac{t^k}{k!} \; dt$$

$$ = \frac{1}{(s-1)!} \sum_{k \ge 0} (-1)^k \; x^{\alpha-k} \binom{\alpha}{k} \; \frac{t^{s+k}}{s+k} \; |_{t=0}^{x}$$

$$ = x^{\alpha + s} \; (-s)! \; \sum_{k \ge 0} \; \binom{\alpha}{k} \; \frac{sin(\pi (s+k))}{\pi (s+k)} = x^{\alpha +s} \frac{\alpha!}{(\alpha+s)!} $$

$$ = D_x^{-s} x^\alpha \; ,$$

where the Heaviside step function is suppressed in the last two lines. (Another way to interpret this convolution is as a Mellin transform interpolation of the falling factorials, equivalent to the Euler integral for the Euler beta function--another instance of Ramanujan's master heuristic.)

The last summation converges for any complex $s$. So, we see that the Euler-Laplace-Mellin transform does give a continuation of the coefficients $ D_x^k$ of the shift op $e^{tD_x}$ to $ D_x^{-s}$ consistent with one version of fractional calculus, the Heaviside operational calculus, in agreement with criteria proposed by Pincherle, who was one of the pioneers of Mellin transform theory (as Mellin explicitly acknowledged).

(This is also related to the Fourier transform over a circle in the MO-Q "Geometric interpretation of the half-derivative?". In this way, the Fourier, Mellin, and Laplace transform are related. This can also be related to Newton series interpolation.)

The ultimate results can be extricated from the transform and convolution integrals by analytic continuation and extended to all complex $\alpha$ and $\beta$ with $D_x^{\beta} H(x)\frac{x^\alpha}{\alpha!} = H(x)\frac{x^{\alpha-\beta}}{(\alpha-\beta)!}$ and the operators saisfying the law of exponents $D_x^{a}D_x^{b} = D_x^{a+b}$, realizing an extension of the Laplace convolution theorem (see, e.g., Generalized Functions by Gelfand and Shilov (Vol. I) or Kanwal).

The mathemage Euler took the initial step. Much later a cottage industry developed around the fractional calculus and Laplace transform after Heaviside's successes in the theory and applications of electrodynamics, Bromwich's parallel development of the Laplace transform pair, and Dirac's adoption of the delta function in quantum mechanics, leading to the the $i\epsilon$ cabbala of Feynman, the distribution theory of Schwartz, the algebraic operator calculus of Mikusinski, and the hyperfunctions of Yosida.

III)

A good deal of classical and quantum physics involving vibrational, or oscillatory, modes of confined systems can be characterized with the confluent hypergeometric functions (CHGF), which can be defined in terms of Laplace transforms:

the Kummer CHGF $M(a,b,z)$ is given by

$$\frac{M(a+1,b+1,-z)}{b!} = \int_{0}^{\infty} e^{-zt}\frac{t^a}{a!} H(1-t)\frac{(1-t)^{b-a-1}}{(b-a-1)!}dt$$

$$ = D_{x=1}^{a-b}e^{-zx}\frac{x^a}{a!}$$

and the Tricomi CHGF $U(a,b,z))$, by

$$\frac{U(a+1,b+1,z)}{(b-a-1)!} = \int_{0}^{\infty} e^{-zt}\frac{t^a}{a!}\frac{(1+t)^{b-a-1}}{(b-a-1)!}dt.$$

(For the Kummer CHGF, see also this MO-Q, "The Kummer confluent hypergeometric function and some of its applications ..." by Georgiev et al., "The influence of elasticity on analysis: The classic heritage" by Truesdell, and the more recent "A physicist's guide to the solution of Kummer's equation and confluent hypergeometric functions" by Mathews Jr., Esrick, Teoh, and Freericks.)

The confluent hypergeometric functions include the families of Laguerre polynomials from which the Hermite polynomials can be constructed. These appear not only in quantum physics, probability theory, and analysis involving the heat, harmonic oscillator, and Schrodinger equations but also combinatorics. They also have associated ladder operators which form instances of a Graves/Lie/Heisenberg/Weyl algebra. Operators associated with the Laguerre polynomials can be related to a Witt-Lie algebra as well. OEIS A131758 has associations among the confluent hypergeometric functions and basic functions and number sequences arising in number theory and topology/characteristic classes. Special functions of the hypergeometric type are rife in group/representation theory and operational calculus, explored by Miller, Gilmore, Vilenkin, Carlitz, Al-Salam, Rota, Roman, Askey, Wilson, among others. Peruse also the book Hypergeometric Functions, My Love by Yoshida.

IV)

The Laplace transform also arises naturally in one method of deriving the Lagrange inversion formula for compositional inversion based on arguments surrounding the change of variables (for a suitable class of functions or Weierstrass approximations)

$$\int_{0}^\infty e^{\frac{-f(xt)}{x}} dt = \int_{0}^\infty \frac{1}{x}e^{-\frac{t}{x}}df^{(-1)}(t)$$

with $f(x)$ and $f^{(-1)}(x)$ a compositional inverse pair about the origin.

See, e.g., my blog post with pdf "The Laplace transform and compositional inversion" with links to earlier notes of mine on the same topic with different approaches and with explicit connections to partition polynomials of significance in combinatorics. This is also related to Hirzebruch genera.

Added 2/26/2024:

V)

I'll sketch some of Heaviside's legacy in using a formalism related to the LPT in characterizing the dynamics of electric circuits.

In the analysis of linear, time-ivariant (LTI), causal sytems, the action of a system on an input, or changes of the input induced by the system, can be modeled by differential equations as an intial value problem. The Laplace transforms of the derivatives of suitable functions are given by integration by parts as

$$ \int_{0}^{\infty} e^{-pt} f'(t) dt = p\tilde{f}(p)-f(0),$$

$$ \int_{0}^{\infty} e^{-pt} f''(t) dt = p^2\tilde{f}(p)-pf(0)-f'(0),$$

and so on with

$$\tilde{f}(p) = \int_{0}^{\infty} e^{-pt} f(t) dt,$$

so the differential formulation of the action is changed into an algebraic formulation with initial conditions included.

The ratio of the LPT of the output to the LPT of the input is called the transfer function, equivalent to the LPT of the output for a unit impulse input, i.e., a Heaviside-Dirac delta function input. The output for a unit impulse input is naturally called the impulse response function (IRF).

This website gives an example of a computation of the transfer function for a specific mechanical system.

This pdf from Linear Dynamic Systems and Signals by Zoran Gajic gives an example of computations for a specific electric circuit and further discussion.

Since the ratio of the LPTs for the input and output are sufficient to characterize such systems, the method can be applied to any 'blackbox' that represents the system but for which an explicit model of the system is not available. If the inverse LPT is available, the Laplace convolution theorem allows the system to be modelled as a Laplace convolution of the IRF with the input.

Of course, these considerations for systems analysis don't provide the fundamental LPT--the Euler integral for the gamma function--and I would also be at a loss to discern the inverse LPT from this perspective although I believe Heaviside could give the IRF in his own singular way.

Answer 8

The following simple motivation for the formula was given in my undergrad class, which I hope I'm not misremembering:

Typically the way the Laplace transform arises in applications is when solving initial-value problems by transforming differential equations to polynomial equations. This technique was used extensively in the early theory of electrodynamics (as the Heaviside calculus), and it was only eventually justified using Laplace transforms.

Suppose one were looking for an integral transform $\mathcal{L}[f] = \int_a^bK(s,x)f(x)dx$ with the property that for all $s,$ $$\mathcal{L}[Df](s) = s\cdot \mathcal{L}[f](s).$$ The reason for this specific functional equation is hopefully clear: such a transform would take a differential equation to the 'obvious' associated polynomial. As to why an integral transform in the first place, one can motivate that in much the same way as the Fourier transform: we are decomposing $f$ into a sum of projections onto a one-parameter system of functions (see e.g. chapter 18 of Zorich's Mathematical Analysis for how this is done.)

A natural domain to work with for an IVP would be the positive real axis, and we then have by the above equation: $$\mathcal{L}[Df]:=\int_0^\infty K(s,x)f'(x) dx = s\cdot \int_0^\infty K(s,x)f(x) dx = s\cdot \mathcal{L}[f](s)$$ Integrating by parts and solving will yield the kernel $K(s,x) = e^{-sx}$ (and a boundary term).