Math 2.0

Integral transforms, operational calculus Laplace transform: operational rules Proposition

Scaling rule for the Laplace transform

Builds on Laplace transform

Statement

Let f ⁣:[0,∞)→ℂf \colon [0, \infty) \to \C be integrable on each interval [0,T][0, T], and let FF be its Laplace transform. Let c>0c > 0 be real, and let ss be a complex number such that the Laplace integral of ff converges at s/cs / c. Then the Laplace integral of t↦f(ct)t \mapsto f(c t) converges at ss, and ℒ{f(ct)}(s)=1cF ⁣(sc).\mathcal{L}\{f(c t)\}(s) = \frac{1}{c} F\!\left(\frac{s}{c}\right).

Remarks

The substitution τ=ct\tau = c t turns ∫0∞f(ct)e−st dt\int_0^\infty f(c t) e^{-s t} \, dt into 1c∫0∞f(τ)e−(s/c)τ dτ\frac{1}{c} \int_0^\infty f(\tau) e^{-(s / c) \tau} \, d\tau. Example: from ℒ{sin⁡t}(s)=1/(s2+1)\mathcal{L}\{\sin t\}(s) = 1 / (s^2 + 1) one gets ℒ{sin⁡ωt}(s)=1ω⋅1(s/ω)2+1=ωs2+ω2\mathcal{L}\{\sin \omega t\}(s) = \frac{1}{\omega} \cdot \frac{1}{(s / \omega)^2 + 1} = \frac{\omega}{s^2 + \omega^2} for ω>0\omega > 0. The condition c>0c > 0 is needed: for c<0c < 0 the function f(ct)f(c t) uses values of ff at negative times.

Going deeper

For a student

The rule lets one derive a whole family of transforms from one transform, for example all sin⁡ωt\sin \omega t from sin⁡t\sin t. It also explains how the abscissa of convergence changes: it is multiplied by cc.

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  1. Function concept
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Further reading

  1. Web https://en.wikipedia.org/wiki/Laplace_transform
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AboutLaplace transform
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