6.2 Operational Properties
Theorem 6.2.1 (Linearity, shifting and scaling). For constants \(\alpha ,\beta \) and \(a>0\):
- (i)
- \(\mathcal {L}\{\alpha f+\beta g\} = \alpha F(s)+\beta G(s)\);
- (ii)
- first shifting: \(\mathcal {L}\{e^{at}f(t)\} = F(s-a)\);
- (iii)
- second shifting: \(\mathcal {L}\{f(t-a)H(t-a)\} = e^{-as}F(s)\);
- (iv)
- scaling: \(\mathcal {L}\{f(at)\} = \frac {1}{a}F(s/a)\);
- (v)
- multiplication by \(t\): \(\mathcal {L}\{t^{n}f(t)\} = (-1)^{n}F^{(n)}(s)\).
Note 6.2.2. The two shifting rules are duals and are easily confused. Multiplying by an exponential in \(t\) shifts the transform in \(s\); shifting in \(t\) multiplies the transform by an exponential in \(s\). The Heaviside factor in (iii) is not decorative — without it the formula is false, because the Laplace integral knows nothing of \(f\) before \(t=0\) and a shifted function must be told to remain zero until \(t=a\).
Theorem 6.2.3 (Transform of derivatives). If \(f\) is continuous with piecewise continuous derivative and of exponential order, \[\mathcal {L}\{f'\}(s) = s\,F(s) - f(0),\] and by iteration \[\mathcal {L}\{f^{(n)}\}(s) = s^{n}F(s) - s^{n-1}f(0) - s^{n-2}f'(0) - \cdots - f^{(n-1)}(0).\]
Proof. Integrate by parts: \[\int ^{\infty }_{0}f'(t)e^{-st}dt = \left [f(t)e^{-st}\right ]^{\infty }_{0} + s\int ^{\infty }_{0}f(t)e^{-st}dt = -f(0) + sF(s),\] the boundary term vanishing at infinity by exponential order and \(\operatorname {Re}s\) large. □
Note 6.2.4. Theorem 6.2.3 is the whole reason the transform exists. It converts differentiation into multiplication by \(s\) and absorbs the initial conditions into the algebra, so a linear differential equation with constant coefficients becomes a linear algebraic equation in \(F(s)\) that already knows \(f(0)\) and \(f'(0)\). Compare the Fourier property \(\widehat {f'}=i\omega \widehat {f}\) of Section 5.9, which converts differentiation just as neatly but carries no initial data — which is why Fourier suits boundary-value problems on the line and Laplace suits initial-value problems in time.
Theorem 6.2.5 (Convolution). With \((f*g)(t) = \displaystyle \int ^{t}_{0}f(\tau )g(t-\tau )\,d\tau \), \[\mathcal {L}\{f*g\} = F(s)\,G(s).\]
Note 6.2.6. The convolution here runs from \(0\) to \(t\), not over the whole line as in Section 5.9, because both functions are taken to vanish for negative argument. The theorem is what makes the inverse transform of a product tractable: rather than inverting \(F(s)G(s)\) directly one inverts each factor and convolves.
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