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

[math]\displaystyle{ F(s) = \int_0^\infty f(t) e^{-st}\,dt }[/math]

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

[math]\displaystyle{ \lim_{t\,\to\, 0}f(t)=\lim_{s\to\infty}{sF(s)}. }[/math]

Proofs

Proof using dominated convergence theorem and assuming that function is bounded

Suppose first that [math]\displaystyle{ f }[/math] is bounded, i.e. [math]\displaystyle{ \lim_{t\to 0^+}f(t)=\alpha }[/math]. A change of variable in the integral [math]\displaystyle{ \int_0^\infty f(t)e^{-st}\,dt }[/math] shows that

[math]\displaystyle{ sF(s)=\int_0^\infty f\left(\frac ts\right)e^{-t}\,dt }[/math].

Since [math]\displaystyle{ f }[/math] is bounded, the Dominated Convergence Theorem implies that

[math]\displaystyle{ \lim_{s\to\infty}sF(s)=\int_0^\infty\alpha e^{-t}\,dt=\alpha. }[/math]

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

Generalizing to non-bounded functions that have exponential order

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

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

[math]\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), }[/math]

since [math]\displaystyle{ \lim_{s\to\infty}F(s)=0 }[/math].

See also

• Final value theorem