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]

Let

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

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

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

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

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

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

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

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

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

${\displaystyle \lim _{s\to \infty }sF(s)=\lim _{s\to \infty }(s-c)F(s)=\lim _{s\to \infty }sF(s+c)=\lim _{s\to \infty }sG(s),}${\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 ${\displaystyle \lim _{s\to \infty }F(s)=0}${\displaystyle \lim _{s\to \infty }F(s)=0}.

• Final value theorem

• Theorems in mathematical analysis
• Mathematical analysis stubs

• CS1 maint: others
• Articles with short description
• Short description matches Wikidata
• All stub articles