{{Short description|Mathematical theorem using Laplace transform}} 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.<ref>{{Cite book |title=Fourier and Laplace transforms |date=2003 |publisher=Cambridge University Press |others=R. J. Beerends |isbn=978-0-511-67510-2 |location=Cambridge |oclc=593333940}}</ref>
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<ref>Robert H. Cannon, ''Dynamics of Physical Systems'', Courier Dover Publications, 2003, page 567.</ref>
: <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== <references/>
Category:Theorems in mathematical analysis
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