Final Value Theorem of Laplace Transform

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Theorem

Let $\laptrans {\map f t} = \map F s$ denote the Laplace transform of the real function $f$.

Then:

$\displaystyle \lim_{t \mathop \to \infty} \map f t = \lim_{s \mathop \to 0} s \, \map F s$

if those limits exist.


General Result

Let $\displaystyle \lim_{t \mathop \to \infty} \dfrac {\map f t} {\map g t} = 1$.


Then:

$\displaystyle \lim_{s \mathop \to 0} \dfrac {\map F s} {\map G s} = 1$

if those limits exist.


Proof

From Laplace Transform of Derivative:

$(1): \quad \laptrans {\map {f'} t} = s \, \map F s - \map f 0$


We have that:

\(\displaystyle \lim_{s \mathop \to 0} \laptrans {\map {f'} t}\) \(=\) \(\displaystyle \lim_{s \mathop \to 0} \int_0^\infty e^{-s t} \map {f'} t \rd t\) Definition of Laplace Transform
\(\displaystyle \) \(=\) \(\displaystyle \int_0^\infty \map {f'} t \rd t\)
\(\displaystyle \) \(=\) \(\displaystyle \lim_{L \mathop \to \infty} \int_0^L \map {f'} t \rd t\)
\(\displaystyle \) \(=\) \(\displaystyle \lim_{L \mathop \to \infty} \paren {\map f L - \map f 0}\) Fundamental Theorem of Calculus
\(\text {(2)}: \quad\) \(\displaystyle \) \(=\) \(\displaystyle \lim_{t \mathop \to \infty} \map f t - \map f 0\)


Hence:

\(\displaystyle \lim_{s \mathop \to 0} \laptrans {\map {f'} t}\) \(=\) \(\displaystyle \lim_{s \mathop \to 0} s \, \map F s - \map f 0\) from $(1)$
\(\displaystyle \leadsto \ \ \) \(\displaystyle \lim_{t \mathop \to \infty} \map f t - \map f 0\) \(=\) \(\displaystyle \lim_{s \mathop \to 0} s \, \map F s - \map f 0\) from $(2)$
\(\displaystyle \leadsto \ \ \) \(\displaystyle \lim_{t \mathop \to \infty} \map f t\) \(=\) \(\displaystyle \lim_{s \mathop \to 0} s \, \map F s\)

$\Box$


Suppose that $f$ is not continuous at $t = 0$.

From Laplace Transform of Derivative with Discontinuity at Zero:

$\laptrans {\map {f'} t} = s \, \map F s - \map f {0^+}$

which means:

$(3): \quad \laptrans {\map {f'} t} = s \, \map F s - \displaystyle \lim_{u \mathop \to 0} \map f u$


We have that:

\(\displaystyle \lim_{s \mathop \to 0} \laptrans {\map {f'} t}\) \(=\) \(\displaystyle \lim_{u \mathop \to 0} \paren {\lim_{s \mathop \to 0} \int_u^\infty e^{-s t} \map {f'} t \rd t}\) Definition of Laplace Transform
\(\displaystyle \) \(=\) \(\displaystyle \lim_{u \mathop \to 0} \paren {\int_u^\infty \map {f'} t \rd t}\)
\(\displaystyle \) \(=\) \(\displaystyle \lim_{u \mathop \to 0} \paren {\lim_{L \mathop \to \infty} \int_u^L \map {f'} t \rd t}\)
\(\displaystyle \) \(=\) \(\displaystyle \lim_{u \mathop \to 0} \paren {\lim_{L \mathop \to \infty} \paren {\map f L - \map f u} }\) Fundamental Theorem of Calculus
\(\displaystyle \) \(=\) \(\displaystyle \lim_{u \mathop \to 0} \paren {\lim_{t \mathop \to \infty} \map f t - \map f u}\)
\(\text {(4)}: \quad\) \(\displaystyle \) \(=\) \(\displaystyle \lim_{t \mathop \to \infty} \map f t - \lim_{u \mathop \to 0} \map f u\)


Hence:

\(\displaystyle \lim_{s \mathop \to 0} \laptrans {\map {f'} t}\) \(=\) \(\displaystyle \lim_{s \mathop \to 0} s \, \map F s - \lim_{u \mathop \to 0} \map f u\) from $(3)$
\(\displaystyle \leadsto \ \ \) \(\displaystyle \lim_{t \mathop \to \infty} \map f t - \lim_{u \mathop \to 0} \map f u\) \(=\) \(\displaystyle \lim_{s \mathop \to 0} s \, \map F s - \lim_{u \mathop \to 0} \map f u\) from $(4)$
\(\displaystyle \leadsto \ \ \) \(\displaystyle \lim_{t \mathop \to \infty} \map f t\) \(=\) \(\displaystyle \lim_{s \mathop \to 0} s \, \map F s\)

$\blacksquare$


Examples

Example 1

Consider the real function $f: \R \to \R$ defined as:

$\map f t = 3 e^{-2 t}$

From Laplace Transform of Exponential:

$\laptrans {\map f t} = \dfrac 3 {s + 2}$


Then by the Initial Value Theorem of Laplace Transform:

\(\displaystyle \lim_{t \mathop \to 0} 3 e^{-2 t}\) \(=\) \(\displaystyle \lim_{s \mathop \to \infty} \dfrac 3 {s + 2}\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle 3\) \(=\) \(\displaystyle 3\)


Sources