KindiakMath

Laplace Inversion

In this post, we apply the culmination of our discussions in complex analysis to sufficiently legitimise the inverse Laplace transform for polytechnic courses in differential equations, taking heavy inspiration from this thread. Unfortunately, Fourier analysis only gets us the desired intuition for the Laplace inversion formula—we’ll need complex analysis to prove it properly.

For any function f : [0, \infty) \to \mathbb C, recall the Laplace transform \mathcal L\{f\} \equiv F defined by

\displaystyle F(s) := \int_0^\infty f(t)e^{-st}\, \mathrm dt

whenever the integral converges. Recall also the Fourier transform \hat f of f defined by

\displaystyle \hat f(\xi) := \frac 1{2\pi} \int_{0}^{\infty} f(x)e^{-i \xi x}\,\mathrm dx.

whenever its integral converges. The Fourier inversion theorem tells us that if f is continuous and Lebesgue-integrable, then \hat f exists and

\displaystyle f(x) = \int_{-\infty}^{\infty} \hat f(\xi) e^{i \xi x}\, \mathrm d\xi.

Theorem 1. Suppose f is differentiable and there exists some g : [0, \infty) \to \mathbb C such that

\displaystyle f(u) = f(0) + \int_0^u g(v)\, \mathrm dv,

so that f' = g, and there exists \sigma \in \mathbb R such that f(t)e^{-\sigma t} and g(t)e^{-\sigma t} are Lebesgue-integrable. Then for any t > 0,

\displaystyle f(t) = \frac 1{2 \pi i} \cdot \lim_{R \to \infty} \oint_{\Gamma_R} F(s) e^{st}\, \mathrm ds,

where \Gamma_R = r( [-R, R] ), r(t) = \sigma + i t. In particular, if F = G, then f = g, yielding the required Laplace inversion.

Remark 1. The intuition for this formula is setting s = i\xi to recover the inverse Fourier transform. The technicality involves the Fourier’s limited effectiveness on only Lebesgue-integrable functions, and how even approximating techniques do not work since the powerful integration convergence theorems still require the baseline condition of integrability (or sadly, un-guaranteed non-negativity).

Proof. Firstly, we define the bounded function S : \mathbb R \to \mathbb C by

\displaystyle S(u) = \frac 12 + \frac 1{\pi} \int_0^u \frac{\sin (v)}{v}\, \mathrm dt \quad \Rightarrow \quad S'(u) = \frac 1{\pi} \cdot \frac{\sin(u)}{u}.

Using the Dirichlet integral,

\displaystyle \lim_{R \to \infty} S(Ru) = \begin{cases} 1, & u > 0, \\ 1/2, & u = 0, \\ 0, & u < 0. \end{cases}

For each u \neq 0, S(Ru) \to H(u) as R \to \infty, where H := \mathbb I_{[0, \infty)} denotes the unit step function. To simplify the contour integral, parameterise and expand the Laplace transform to obtain

\begin{aligned} \frac 1{2\pi i} \oint_{\Gamma_R} F(s) e^{ts}\, \mathrm ds &= \frac 1{2\pi i} \cdot \int_{-R}^R F(\sigma + i \xi) e^{t (\sigma + i \xi)}\cdot i \, \mathrm d\xi \\ &= \frac 1{2\pi} \cdot \int_{-R}^R \left( \int_0^\infty f(u) e^{-(\sigma + i \xi)u} \, \mathrm du \right) e^{t (\sigma + i \xi)} \, \mathrm d\xi \\ &= \frac 1{2\pi} \cdot \int_0^\infty f(u) \int_{-R}^R e^{-(\sigma + i \xi)u}e^{t (\sigma + i \xi)} \, \mathrm d\xi \, \mathrm du \\ &= \frac 1{2\pi} \cdot \int_0^\infty f(u) e^{-\sigma (u-t)} \int_{-R}^R e^{i \xi (t-u)} \, \mathrm d\xi \, \mathrm du \\ &= \frac 1{2\pi} \cdot e^{\sigma t} \cdot \int_0^\infty f(u) e^{-\sigma u}\cdot \frac{2 \cdot \sin (u-t)R}{(u-t)} \, \mathrm dt \\ &= \frac 1{2\pi} \cdot 2\pi e^{\sigma t} \cdot \int_0^\infty f(u) e^{-\sigma u}\cdot S'(R(u-t)) \cdot R \, \mathrm du \\ &= e^{\sigma t} \cdot \int_0^\infty f(u) e^{-\sigma u}\cdot S'(R(u-t)) \cdot R \, \mathrm du, \end{aligned}

where we exchanged the integrals using Fubini’s theorem. By hypothesis, f(u)e^{-\sigma u} and h(u):= (f(u)e^{-\sigma u})' = (f'(u) - \sigma f(u)) e^{-\sigma u} are Lebesgue-integrable, so that we can integrate by parts:

\begin{aligned} & \int_0^\infty f(u) e^{-\sigma u}\cdot S'(R(u-t)) \cdot R \, \mathrm du \\ &= \left[ f(u) e^{-\sigma u} \cdot S(R(u-t))\right]_0^\infty - \int_0^\infty h(u) \cdot S(R(u-t))\, \mathrm du \\ &= (0 - 0) - \int_0^\infty h(u) \cdot S(R(u-t))\, \mathrm du \\ &= - \int_0^\infty h(u) \cdot S(R(u-t))\, \mathrm du. \end{aligned}

Taking R \to \infty, apply the dominated convergence theorem to obtain

\begin{aligned} \lim_{R \to \infty} \frac 1{2\pi i} \oint_{\Gamma_R} F(s) e^{ts}\, \mathrm ds &= -e^{\sigma t} \cdot \lim_{R \to \infty} \int_0^\infty h(u) \cdot S(R(u-t))\, \mathrm du \\ &= -e^{\sigma t} \cdot \int_0^\infty h(u) \cdot \lim_{R \to \infty} S(R(u-t))\, \mathrm du \\ &= -e^{\sigma t} \cdot \int_0^\infty h(u) \cdot H(u-t)\, \mathrm du \\ &= -e^{\sigma t} \cdot \int_t^\infty h(u) \, \mathrm du \\ &= -e^{\sigma t} \cdot \left[ f(u) e^{-\sigma u} \right]_t^\infty \\ &= - e^{\sigma t} \cdot (0 - f(t) e^{-\sigma t}) = f(t),\end{aligned}

as required.

With that, we finally complete our discussion on complex analysis!

—Joel Kindiak, 29 Aug 25, 1158H

,

Published by

joelkindiak

Leave a comment