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


 * $$ F(s) = \int_0^\infty f(t) e^{-st}\,dt $$

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


 * $$\lim_{t\,\to\, 0}f(t)=\lim_{s\to\infty}{sF(s)}. $$

Proof using dominated convergence theorem and assuming that function is bounded
Suppose first that $$ f$$ is bounded, i.e. $$\lim_{t\to 0^+}f(t)=\alpha$$. A change of variable in the integral $$\int_0^\infty f(t)e^{-st}\,dt$$ shows that
 * $$sF(s)=\int_0^\infty f\left(\frac ts\right)e^{-t}\,dt$$.

Since $$f$$ is bounded, the Dominated Convergence Theorem implies that
 * $$\lim_{s\to\infty}sF(s)=\int_0^\infty\alpha e^{-t}\,dt=\alpha.$$

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

Generalizing to non-bounded functions that have exponential order
The theorem assuming just that $$f(t)=O(e^{ct})$$ follows from the theorem for bounded $$f$$:

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