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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.[1]
Let
![{\displaystyle F(s)=\int _{0}^{\infty }f(t)e^{-st}\,dt}](https://wikimedia.org/api/rest_v1/media/math/render/svg/b49caba366b94ee1ecb91eee31d5a709f8b0beaa)
be the (one-sided) Laplace transform of ƒ(t). If
is bounded on
(or if just
) and
exists then the initial value theorem says[2]
![{\displaystyle \lim _{t\,\to \,0}f(t)=\lim _{s\to \infty }{sF(s)}.}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d1a9de2bca39353ab674c8ebbdc706b1cda9647a)
Proof using dominated convergence theorem and assuming that function is bounded
[edit]
Suppose first that
is bounded, i.e.
. A change of variable in the integral
shows that
.
Since
is bounded, the Dominated Convergence Theorem implies that
![{\displaystyle \lim _{s\to \infty }sF(s)=\int _{0}^{\infty }\alpha e^{-t}\,dt=\alpha .}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7dfcdbee624ab25e36954ea35485ae7c031d03de)
Proof using elementary calculus and assuming that function is bounded
[edit]
Of course we don't really need DCT here, one can give a very simple proof using only elementary calculus:
Start by choosing
so that
, and then
note that
uniformly for
.
Generalizing to non-bounded functions that have exponential order
[edit]
The theorem assuming just that
follows from the theorem for bounded
:
Define
. Then
is bounded, so we've shown that
.
But
and
, so
![{\displaystyle \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),}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ab6736bfe909a883472e53637b178e1da28d0d76)
since
.