Initial value theorem

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

: $F(s) = \int_0^\infty f(t) e^\,dt$

be the (one-sided) Laplace transform of ''ƒ''(''t''). If $f$ is bounded on $(0,\infty)$ (or if just $f(t)=O(e^)$) and $\lim_f(t)$ exists then the initial value theorem saysRobert H. Cannon, ''Dynamics of Physical Systems'', Courier Dover Publications, 2003, page 567.

: $\lim_f(t)=\lim_.$

Proofs

Proof using dominated convergence theorem and assuming that function is bounded

Suppose first that $f$ is bounded, i.e. $\lim_f(t)=\alpha$. A change of variable in the integral
$\int_0^\infty f(t)e^\,dt$ shows that
:$sF(s)=\int_0^\infty f\left(\frac ts\right)e^\,dt$.
Since $f$ is bounded, the Dominated Convergence Theorem implies that
:$\lim_sF(s)=\int_0^\infty\alpha e^\,dt=\alpha.$

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 $A$ so that $\int_A^\infty e^\,dt<\epsilon$, and then
note that $\lim_f\left(\frac ts\right)=\alpha$ ''uniformly'' for $t\in(0,A]$.

Generalizing to non-bounded functions that have exponential order

The theorem assuming just that $f(t)=O(e^)$ follows from the theorem for bounded $f$:

Define $g(t)=e^f(t)$. Then $g$ is bounded, so we've shown that $g(0^+)=\lim_sG(s)$.
But $f(0^+)=g(0^+)$ and $G(s)=F(s+c)$, so
:$\lim_sF(s)=\lim_(s-c)F(s)=\lim_sF(s+c) =\lim_sG(s),$
since $\lim_F(s)=0$.

See also

* Final value theorem

Notes

Theorems in analysis

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