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.^[1]

It is also known under the abbreviation IVT.

Let

F(s)=\int _{0}^{\infty }f(t)e^{-st}\,dt

be the (one-sided) Laplace transform of ƒ(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^[2]

\lim _{t\,\to \,0}f(t)=\lim _{s\to \infty }{sF(s)}.

Proof[]

Suppose first that $f$ is bounded. Say $\lim _{t\to 0^{+}}f(t)=\alpha$ . 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 {t}{s}}\right)e^{-t}\,dt

.

Since $f$ is bounded, the Dominated Convergence Theorem shows that

\lim _{s\to \infty }sF(s)=\int _{0}^{\infty }\alpha e^{-t}\,dt=\alpha .

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 $\lim _{s\to \infty }f\left({\frac {t}{s}}\right)=\alpha$ uniformly for $t\in (0,A]$ .

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