Math 2.0

Integral transforms, operational calculus Laplace transform: operational rules Proposition

Laplace transform of a derivative

Builds on Laplace transform

Statement

Let f:[0,∞)→ℂf : [0, \infty) \to \mathbb{C} be continuously differentiable, and let M,aM, a be real constants with ∣f(t)∣≤Meat|f(t)| \le M e^{a t} for all t≥0t \ge 0. Let FF be the Laplace transform of ff. Then for each s∈ℂs \in \mathbb{C} with Re⁡s>a\operatorname{Re} s > a, the Laplace transform of f′f' converges at ss and ℒ{f′}(s)=sF(s)−f(0).\mathcal{L}\{f'\}(s) = s F(s) - f(0).

Remarks

Proof: integration by parts gives ∫0Tf′(t)e−st dt=f(T)e−sT−f(0)+s∫0Tf(t)e−st dt\int_0^T f'(t) e^{-st}\, dt = f(T) e^{-sT} - f(0) + s \int_0^T f(t) e^{-st}\, dt. The term f(T)e−sTf(T) e^{-sT} tends to 00 for Re⁡s>a\operatorname{Re} s > a. Applied twice, under the same hypotheses on ff and f′f': ℒ{f′′}(s)=s2F(s)−sf(0)−f′(0)\mathcal{L}\{f''\}(s) = s^2 F(s) - s f(0) - f'(0). This rule is the basis of solving initial value problems with the Laplace transform. The initial values enter the transformed equation directly.

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Further reading

  1. Web https://en.wikipedia.org/wiki/Laplace_transform, Transform of a function's derivative
  2. Web https://en.wikipedia.org/wiki/Laplace_transform, Properties and theorems
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