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

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

be the (one-sided) Laplace transform of ƒ(t). If ${\displaystyle f}$ is bounded on ${\displaystyle (0,\infty )}$ (or if just ${\displaystyle f(t)=O(e^{ct})}$) and ${\displaystyle \lim _{t\to 0^{+))f(t)}$ exists then the initial value theorem says^[2]

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

Proofs

Proof using dominated convergence theorem and assuming that function is bounded

Suppose first that ${\displaystyle f}$ is bounded, i.e. ${\displaystyle \lim _{t\to 0^{+))f(t)=\alpha }$. A change of variable in the integral ${\displaystyle \int _{0}^{\infty }f(t)e^{-st}\,dt}$ shows that

${\displaystyle sF(s)=\int _{0}^{\infty }f\left({\frac {t}{s))\right)e^{-t}\,dt}$.

Since ${\displaystyle f}$ is bounded, the Dominated Convergence Theorem implies that

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

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

Generalizing to non-bounded functions that have exponential order

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

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

${\displaystyle \lim _{s\to \infty }sF(s)=\lim _{s\to \infty }(s-c)F(s)=\lim _{s\to \infty }sF(s+c)=\lim _{s\to \infty }sG(s),}$

since ${\displaystyle \lim _{s\to \infty }F(s)=0}$.