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 ) = ∫ 0 ∞ f ( t ) e − s t d t {\displaystyle F(s)=\int _{0}^{\infty }f(t)e^{-st}\,dt}be the (one-sided) Laplace transform of ƒ(t). If f {\displaystyle f} is bounded on ( 0 , ∞ ) {\displaystyle (0,\infty )} (or if just f ( t ) = O ( e c t ) {\displaystyle f(t)=O(e^{ct})} ) and lim t → 0 + f ( t ) {\displaystyle \lim _{t\to 0^{+}}f(t)} exists then the initial value theorem says
lim t → 0 f ( t ) = lim s → ∞ s F ( s ) . {\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 f {\displaystyle f} is bounded, i.e. lim t → 0 + f ( t ) = α {\displaystyle \lim _{t\to 0^{+}}f(t)=\alpha } . A change of variable in the integral ∫ 0 ∞ f ( t ) e − s t d t {\displaystyle \int _{0}^{\infty }f(t)e^{-st}\,dt} shows that
s F ( s ) = ∫ 0 ∞ f ( t s ) e − t d t {\displaystyle sF(s)=\int _{0}^{\infty }f\left({\frac {t}{s}}\right)e^{-t}\,dt} .Since f {\displaystyle f} is bounded, the Dominated Convergence Theorem implies that
lim s → ∞ s F ( s ) = ∫ 0 ∞ α e − t d t = α . {\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 A {\displaystyle A} so that ∫ A ∞ e − t d t < ϵ {\displaystyle \int _{A}^{\infty }e^{-t}\,dt<\epsilon } , and then note that lim s → ∞ f ( t s ) = α {\displaystyle \lim _{s\to \infty }f\left({\frac {t}{s}}\right)=\alpha } uniformly for t ∈ ( 0 , A ] {\displaystyle t\in (0,A]} .
Generalizing to non-bounded functions that have exponential order
The theorem assuming just that f ( t ) = O ( e c t ) {\displaystyle f(t)=O(e^{ct})} follows from the theorem for bounded f {\displaystyle f} :
Define g ( t ) = e − c t f ( t ) {\displaystyle g(t)=e^{-ct}f(t)} . Then g {\displaystyle g} is bounded, so we've shown that g ( 0 + ) = lim s → ∞ s G ( s ) {\displaystyle g(0^{+})=\lim _{s\to \infty }sG(s)} . But f ( 0 + ) = g ( 0 + ) {\displaystyle f(0^{+})=g(0^{+})} and G ( s ) = F ( s + c ) {\displaystyle G(s)=F(s+c)} , so
lim s → ∞ s F ( s ) = lim s → ∞ ( s − c ) F ( s ) = lim s → ∞ s F ( s + c ) = lim s → ∞ s G ( 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 lim s → ∞ F ( s ) = 0 {\displaystyle \lim _{s\to \infty }F(s)=0} .
See also
Notes
References
Fourier and Laplace transforms. R. J. Beerends. Cambridge: Cambridge University Press. 2003. ISBN 978-0-511-67510-2. OCLC 593333940.{{cite book}}: CS1 maint: others (link) 978-0-511-67510-2 ↩
Robert H. Cannon, Dynamics of Physical Systems, Courier Dover Publications, 2003, page 567. /wiki/Courier_Dover_Publications ↩