The following results let us evaluate some limits using the Laplace transform. This corresponds to:
Let’s get into the details!
1. Laplace Transform initial value
Proof:
Let us assume f is a bounded function. We may say there are a,K\in\mathbb{R} such that
|f(t)|\leq Ke^{at}
and consequently
e^{-st}|f(t)|\leq Ke^{-(s-a)t}
As F(s)=\mathscr{L}\{f(t)\}, applying the definition, we have
\displaystyle |F(s)|=\left|\int_{0}^{\infty}e^{-st}f(t)\right|dt\leq\int_{0}^{\infty}e^{-st}|f(t)|dt
And then
\displaystyle |F(s)|\leq\int_{0}^{\infty}e^{-(s-a)t}Kdt=K\int_{0}^{\infty}e^{-(s-a)t}dt
Since
\displaystyle\int_{0}^{\infty}e^{-(s-a)t}dt=\frac{1}{s-a}
we may conclude
\displaystyle |F(s)|\leq\frac{K}{s-a}
As
\displaystyle \lim_{s\to\infty}\frac{K}{s-a}=0
hence, from the squeeze theorem
\displaystyle \lim_{s\to\infty}F(s)=0
Now, let H(s) be Laplace Transform of f'(t):
H(s)=sF(s)-f(0)
If we assume f'(t) is bounded as the last result, then
\displaystyle\lim_{s\to\infty}H(s)=0
i.e.,
\begin{aligned} & \lim_{s\to0}(sF(s)-f(0))=0 \\ & \lim_{s\to0}sF(s)-f(0)=0 \\ & \lim_{s\to0}sF(s)=f(0) \end{aligned}
which concludes the proofing.
2. Illustrative example of initial value problem
Let us verify the theorem above for f(t)=\sin(t). As
\displaystyle F(s)=\frac{1}{s^{2}+1}
thus
\displaystyle sF(s)=\frac{s}{s^{2}+1}
Using the L’Hopital’s rule
\begin{aligned} \lim_{s\to\infty}sF(s) &=\lim_{s\to\infty}\frac{s}{s^{2}+1} \\ &=\lim_{s\to\infty}\frac{1}{2s} \\ &=0 \end{aligned}
Since
f(0)=\sin(0)=0
we may conclude
\displaystyle f(0)=\lim_{s\to\infty}sF(s)
3. Final value theorem Laplace
Proof:
We know from Laplace transform of derivatives:
\displaystyle\mathscr{L}\{f'(t)\}=\int_{0}^{\infty}e^{-st}f'(t)dt=sF(s)-f(0)
Taking the limit as s\to0
\begin{aligned} & \lim_{s\to0}\int_{0}^{\infty}e^{-st}f'(t)dt=\lim_{s\to0}[sF(s)-f(0)] \\ & \int_{0}^{\infty}\lim_{s\to0}e^{-st}f'(t)dt=\lim_{s\to0}sF(s)-f(0) \\ & \int_{0}^{\infty}1\cdot f'(t)dt=\lim_{s\to0}sF(s)-f(0) \\ & \lim_{t\to\infty}\int_{0}^{t}f'(t)dt=\lim_{s\to0}sF(s)-f(0) \end{aligned}
And from Fundamental Theorem of Calculus, the integral becomes
\begin{aligned} & \lim_{t\to\infty}\left.f(t)\right|_{t=0}^{t=t}=\lim_{s\to0}sF(s)-f(0) \\ & \lim_{t\to\infty}f(t)-f(0)=\lim_{s\to0}sF(s)-f(0) \end{aligned}
Hence
\displaystyle\lim_{t\to\infty}f(t)=\lim_{s\to0}sF(s)
Illustrative Example of final value problem
If f(t) is a function in the variable t such that
\displaystyle F(s)=\mathscr{L}\{f(t)\}=\frac{3}{s+9}
then
\displaystyle sF(s)=\frac{3s}{s+9}
To evaluate \displaystyle\lim_{t\to\infty}f(t) is equivalent to solve the following limit
\begin{aligned} \lim_{s\to0}sF(s) & =\lim_{s\to0}\frac{3s}{s+9} \\ & =\frac{3\cdot0}{0+9} \\ & =\frac{0}{9} \\ & =0 \end{aligned}
Thus
\displaystyle\lim_{t\to\infty}f(t)=0
We reached the end of this lesson about asymptotic values and how they can be calculated using the Laplace transform. For a full list of Laplace transform properties, check this post!