Английская Википедия:Initial value theorem

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In mathematical analysis, the initial value theorem is a theorem used to relate frequency domain expressions to the time domain behavior as time approaches zero.

Let

<math> F(s) = \int_0^\infty f(t) e^{-st}\,dt </math>

be the (one-sided) Laplace transform of ƒ(t). If <math>f</math> is bounded on <math>(0,\infty)</math> (or if just <math>f(t)=O(e^{ct})</math>) and <math>\lim_{t\to 0^+}f(t)</math> exists then the initial value theorem says

<math>\lim_{t\,\to\, 0}f(t)=\lim_{s\to\infty}{sF(s)}. </math>

Proofs

Proof using dominated convergence theorem and assuming that function is bounded

Suppose first that <math> f</math> is bounded, i.e. <math>\lim_{t\to 0^+}f(t)=\alpha</math>. A change of variable in the integral <math>\int_0^\infty f(t)e^{-st}\,dt</math> shows that

<math>sF(s)=\int_0^\infty f\left(\frac ts\right)e^{-t}\,dt</math>.

Since <math>f</math> is bounded, the Dominated Convergence Theorem implies that

<math>\lim_{s\to\infty}sF(s)=\int_0^\infty\alpha e^{-t}\,dt=\alpha.</math>

Proof using elementary calculus and assuming that function is bounded

Of course we don't really need DCT here, one can give a very simple proof using only elementary calculus:

Start by choosing <math>A</math> so that <math>\int_A^\infty e^{-t}\,dt<\epsilon</math>, and then note that <math>\lim_{s\to\infty}f\left(\frac ts\right)=\alpha</math> uniformly for <math>t\in(0,A]</math>.

Generalizing to non-bounded functions that have exponential order

The theorem assuming just that <math>f(t)=O(e^{ct})</math> follows from the theorem for bounded <math>f</math>:

Define <math>g(t)=e^{-ct}f(t)</math>. Then <math>g</math> is bounded, so we've shown that <math>g(0^+)=\lim_{s\to\infty}sG(s)</math>. But <math>f(0^+)=g(0^+)</math> and <math>G(s)=F(s+c)</math>, so

<math>\lim_{s\to\infty}sF(s)=\lim_{s\to\infty}(s-c)F(s)=\lim_{s\to\infty}sF(s+c)

=\lim_{s\to\infty}sG(s),</math> since <math>\lim_{s\to\infty}F(s)=0</math>.

See also

• Final value theorem

Notes

Шаблон:Mathanalysis-stub