Limiting Theorems

The following two theorems give us the behaviour of the function $f(t)$ when $t \longrightarrow 0^+$ and when $t \longrightarrow \infty$ .

THEOREM 10.4.1 (First Limit Theorem)   Suppose ${\mathcal L}(f(t))$ exists. Then

$$\lim_{t \longrightarrow 0^+} f(t) = \lim_{s \longrightarrow \infty} s F(s).$$

Proof. We know $s F(s) - f(0) = {\mathcal L} \left( f^\prime(t) \right).$ Therefore

$$\lim_{s \longrightarrow \infty} s F(s) = f(0) + \lim_{s \longrightarrow \infty} \int_0^\infty e^{-st} f^\prime(t) dt$$

$$= f(0) + \int_0^\infty \lim_{s \longrightarrow \infty} e^{-st} f^\prime(t) dt = f(0).$$

as $\lim\limits_{s \longrightarrow \infty} e^{-st} = 0.$ height6pt width 6pt depth 0pt

EXAMPLE 10.4.2  

1. For $t \geq 0,$ let $Y(s) = {\mathcal L}(y(t)) = a (1 + s^2)^{-1/2}.$ Determine $a$ such that $y(0) = 1.$
   Solution: Theorem 10.4.1 implies
   $1 = \displaystyle \lim\limits_{s \longrightarrow \infty} s Y(s) = \lim\limits... ... \lim\limits_{s \longrightarrow \infty} \frac{a}{ (\frac{1}{s^2} + 1)^{1/2} }.$ Thus, $a = 1.$

2. If $F(s) = \displaystyle\frac{(s+1)(s+3)}{s(s+2)(s+8)}$ find $f(0^+).$
   Solution: Theorem 10.4.1 implies
   $$f(0^+) = \lim_{s \longrightarrow \infty} s F(s) = \lim_{s \longrightarrow \infty} s \cdot \frac{(s+1)(s+3)}{s(s+2)(s+8)} = 1.$$

On similar lines, one has the following theorem. But this theorem is valid only when $f(t)$ is bounded as $t$ approaches infinity.

THEOREM 10.4.3 (Second Limit Theorem)   Suppose ${\mathcal L}(f(t))$ exists. Then

$$\lim_{t \longrightarrow \infty} f(t) = \lim_{s \longrightarrow 0} s F(s)$$

provided that $s F(s)$ converges to a finite limit as $s$ tends to 0 .

Proof.

$$\lim_{s \longrightarrow 0} s F(s) = f(0) + \lim_{s \longrightarrow 0} \int_0^\infty e^{-st} f^\prime(t) dt$$

$$= f(0) + \lim_{s \longrightarrow 0} \lim_{t \longrightarrow \infty} \int_0^t e^{-s \tau} f^\prime(\tau) d \tau$$

$$= f(0) + \lim_{t \longrightarrow \infty} \int_0^t \lim_{s \longrightarrow 0} e^{-s \tau} f^\prime(\tau) d\tau = \lim_{t \longrightarrow \infty} f(t).$$

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EXAMPLE 10.4.4   If $F(s) = \displaystyle\frac{2(s+3)}{s(s+2)(s+8)}$ find $$\lim_{t \longrightarrow \infty} f(t).$$
Solution: From Theorem 10.4.3, we have

$$\lim_{t \longrightarrow \infty} f(t) = \lim_{s \longrightarrow 0} s F(s) = \lim_{s \longrightarrow 0} s \cdot \frac{2(s+3)}{s(s+2)(s+8)} = \frac{6}{16} = \frac{3}{8}.$$

We now generalise the lemma on Laplace transform of an integral as convolution theorem.

DEFINITION 10.4.5 (Convolution of Functions)   Let $f(t)$ and $g(t)$ be two smooth functions. The convolution, $f \star g,$ is a function defined by

$$(f\star g) (t) = \int_0^t f(\tau) g( t - \tau) d \tau.$$

Check that

1. $(f\star g)(t) = g\star f(t).$
2. If $f(t) = \cos (t)$ then $(f\star f)(t) = \displaystyle\frac{t \cos (t) + \sin (t)}{2}.$

THEOREM 10.4.6 (Convolution Theorem)   If $F(s) = {\mathcal L}(f(t))$ and $G(s) = {\mathcal L}(g(t))$ then

$${\mathcal L}\left[\int_0^t f(\tau) g(t - \tau) d\tau \right] = F(s) \cdot G(s).$$

Remark 10.4.7   Let $g(t) = 1$ for all $t \geq 0.$ Then we know that ${\mathcal L}(g(t)) = G(s) = \displaystyle\frac{1}{s}.$ Thus, the Convolution Theorem 10.4.6 reduces to the Integral Lemma 10.3.12.