6.4 The Laplace transform
Definition 6.4.1. A function \(f\) on \([0, \infty )\) is of exponential order \(c\) if it is piecewise continuous and \(|f(t)| \leq Ke^{ct}\) for large \(t\). Its Laplace transform is \[ F(s) = \mathcal L[f](s) = \int _0^\infty e^{-st}f(t)\,dt, \qquad \Real s > c . \]
The integral converges absolutely for \(\Real s > c\), by comparison with \(\int Ke^{-(\Real s - c)t}\,dt\).
Theorem 6.4.2. For functions of exponential order, and \(\Real s\) large enough:
- (i)
- \(\mathcal L[e^{at}f](s) = F(s - a)\) (shift in \(s\));
- (ii)
- \(\mathcal L[f(t - a)H(t - a)](s) = e^{-as}F(s)\) for \(a \geq 0\), where \(H\) is the Heaviside step function (shift in \(t\));
- (iii)
- \(\mathcal L[f'](s) = sF(s) - f(0)\), and \(\mathcal L[f^{(n)}] = s^nF - s^{n-1}f(0) - \dots - f^{(n-1)}(0)\), if \(f, \dots , f^{(n-1)}\) are continuous and \(f^{(n)}\) piecewise continuous;
- (iv)
- \(\mathcal L[f * g] = FG\), where \((f * g)(t) = \int _0^tf(t - s)g(s)\,ds\) (convolution theorem);
- (v)
- \(\mathcal L[t^n] = n!/s^{n+1}\), \(\mathcal L[e^{at}] = 1/(s - a)\), \(\mathcal L[\cos bt] = s/(s^2 + b^2)\), \(\mathcal L[\sin bt] = b/(s^2 + b^2)\).
Proof. (i) \(\int e^{-st}e^{at}f\,dt = \int e^{-(s - a)t}f\,dt\). (ii) Substitute \(u = t - a\) in \(\int _a^\infty e^{-st}f(t - a)\,dt\). (iii) Integrate by parts: \(\int _0^\infty e^{-st}f'\,dt = \bigl [e^{-st}f\bigr ]_0^\infty + s\int _0^\infty e^{-st}f\,dt\), and the boundary term at infinity vanishes for \(\Real s > c\); repeat for higher derivatives. (iv) By Fubini’s theorem, justified by absolute convergence, \[ \int _0^\infty e^{-st}\int _0^tf(t - u)g(u)\,du\,dt = \int _0^\infty g(u)e^{-su}\int _u^\infty e^{-s(t - u)}f(t - u)\,dt\,du = G(s)F(s) . \] (v) \(\mathcal L[1] = 1/s\); then (i) gives \(e^{at}\); \(\cos bt\) and \(\sin bt\) are the real and imaginary parts of \(e^{ibt}\), with transform \(1/(s - ib) = (s + ib)/(s^2 + b^2)\); and \(\mathcal L[t^n]\) follows from (iii) by induction, or from the Gamma integral. □
A function of exponential order is determined by its transform, up to its values at points of discontinuity (Lerch’s theorem), so a transform can be inverted by recognising it, usually after a partial fraction decomposition. Applied to \(p(D)y = g\) with initial data, rule (iii) turns the differential equation into an algebraic one: \begin {gather*} p(s)Y(s) - \bigl (\text {polynomial in } s \text { built from the initial values}\bigr ) = G_g(s),\\ Y(s) = \frac {G_g(s) + \text {initial terms}}{p(s)} , \end {gather*} where \(G_g = \mathcal L[g]\). With zero initial data, \(Y = H(s)G_g(s)\) with the transfer function \(H(s) = 1/p(s)\). By the convolution theorem, \(H\) is the transform of the impulse response of Theorem 6.3.1: \(\mathcal L[G] = 1/p(s)\), since \(G\) has zero initial data except \(G^{(n-1)}(0) = 1\). The poles of \(H\) are the roots of \(p\), so the system is asymptotically stable if and only if every pole of the transfer function lies in the left half-plane \(\Real s < 0\), the criterion that control engineers use.
Discontinuous and impulsive forcing
The transform handles switched forcing without piecewise solving: a force that is switched on at \(t = a\) and off at \(t = b\) is \(H(t - a) - H(t - b)\), with transform \((e^{-as} - e^{-bs})/s\). An idealised blow at time \(a\), a force of unit total impulse concentrated at one instant, is represented by the Dirac delta \(\delta (t - a)\), with \(\mathcal L[\delta (t - a)] = e^{-as}\); the response of a system at rest to \(\delta (t)\) is the impulse response \(G\), which is why it has that name. A rigorous treatment of \(\delta \) requires distributions, but the transform rule can be justified as the limit of rectangular pulses of width \(\varepsilon \) and height \(1/\varepsilon \) as \(\varepsilon \to 0\) (Problem 6.6.5).
Example 6.4.3. Solve \(y'' + 3y' + 2y = 1\), \(y(0) = y'(0) = 0\). Transforming, \((s^2 + 3s + 2)Y = 1/s\), so \[ Y = \frac {1}{s(s + 1)(s + 2)} = \frac {1}{2s} - \frac {1}{s + 1} + \frac {1}{2(s + 2)}, \qquad y = \tfrac 12 - e^{-t} + \tfrac 12e^{-2t} . \] The constant \(\frac 12\) is the new equilibrium, and the exponentials are the transient, decaying at the rates given by the poles \(-1\) and \(-2\).
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