Initial value theorem

Template:Short description 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

$F(s)={\int }_0^{\mathrm{\infty }}f(t)e^{-st}dt$

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

$\underset{t\to 0}{lim}f(t)=\underset{s\to \mathrm{\infty }}{lim}sF(s).$

Proofs

Proof using dominated convergence theorem and assuming that function is bounded

Suppose first that $f$ is bounded, i.e. $\underset{t\to 0^{+}}{lim}f(t)=\alpha$. A change of variable in the integral ${\int }_0^{\mathrm{\infty }}f(t)e^{-st}dt$ shows that

$sF(s)={\int }_0^{\mathrm{\infty }}f\left(\frac{t}{s}\right)e^{-t}dt$.

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

$\underset{s\to \mathrm{\infty }}{lim}sF(s)={\int }_0^{\mathrm{\infty }}\alpha e^{-t}dt=\alpha .$

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 ${\int }_A^{\mathrm{\infty }}{e}^{-t}dt<ϵ$, and then note that $\underset{s\to \mathrm{\infty }}{lim}f\left(\frac{t}{s}\right)=\alpha$ 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 $g(0^{+})=\underset{s\to \mathrm{\infty }}{lim}sG(s)$. But $f(0^{+})=g(0^{+})$ and $G(s)=F(s+c)$, so

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

since $\underset{s\to \mathrm{\infty }}{lim}F(s)=0$.

See also

• Final value theorem

Notes

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