Properties of Laplace Transform

LEMMA 10.3.1 (Linearity of Laplace Transform)  

1. Let $a, b \in {\mathbb{R}}$ . Then
   

   $${\mathcal L}\bigl(a f(t) + b g(t)\bigr) = \int_0^\infty \bigl(a f(t) + b g(t)\bigr) e^{-s t} dt$$

   $$= a {\mathcal L}(f(t)) + b {\mathcal L}(g(t)).$$

   
   
2. If $F(s) = {\mathcal L}(f(t)), \;$ and $G(s) = {\mathcal L}(g(t))$ , then
   $${\mathcal L}^{-1} \bigl( a F(s) + b G(s) \bigr) = a f(t) + b g(t).$$

The above lemma is immediate from the definition of Laplace transform and the linearity of the definite integral.

EXAMPLE 10.3.2  

1. Find the Laplace transform of $\cosh (at).$
   Solution: $\cosh (at) = \displaystyle\frac{e^{at} + e^{-at}}{2}.$ Thus
   $${\mathcal L}(\cosh(at) ) = \displaystyle\frac{1}{2} \left( \frac... ... + \frac{1}{s+a} \right) = \frac{s}{s^2 - a^2}, \hspace{1in} s > \vert a\vert.$$

2. Similarly,
   $${\mathcal L}(\sinh(at) ) = \displaystyle\frac{1}{2} \left( \frac... ... - \frac{1}{s+a} \right) = \frac{a}{s^2 - a^2}, \hspace{1in} s > \vert a\vert.$$

3. Find the inverse Laplace transform of $$\frac{1}{s(s+1)}$$ .
   Solution:
   

   $${\mathcal L}^{-1} \bigl(\frac{1}{s(s+1)}\bigr) = {\mathcal L}^{-1} \bigl(\frac{1}{s} - \frac{1}{s+1}\bigr)$$

   $$= {\mathcal L}^{-1} \bigl(\frac{1}{s}\bigr) - {\mathcal L}^{-1}\bigl(\frac{1}{s+1}\bigr) = 1 - e^{-t}.$$

   
   Thus, the inverse Laplace transform of $$\frac{1}{s(s+1)}$$ is $f(t) = 1 - e^{-t}.$

THEOREM 10.3.3 (Scaling by $a$ )   Let $f(t)$ be a piecewise continuous function with Laplace transform $F(s).$ Then for $\; a > 0,\;\; {\mathcal L}(f(at)) = \displaystyle\frac{1}{a} F(\frac{s}{a}).$

Proof. By definition and the substitution $z= at,$ we get

$${\mathcal L}(f(at)) = \int_0^\infty e^{-st} f(at) d t =\frac{1}{a} \int_0^\infty e^{-s \frac{z}{a}} f(z) dz$$

$$= \frac{1}{a} \int_0^\infty e^{-\frac{s}{a} z} f(z) dz = \frac{1}{a} F(\frac{s}{a}).$$

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EXERCISE 10.3.4  

1. Find the Laplace transform of
   $$t^2 + a t + b, \;\; \cos (w t + \theta), \;\; \cos^2 t, \;\; \sinh^2 t;$$
   where $a, b, w$ and $\theta$ are arbitrary constants.

   Figure 10.2: $f(t)$

   
2. Find the Laplace transform of the function $f(\cdot)$ given by the graphs in Figure 10.2.
3. If ${\mathcal L}(f(t)) = \displaystyle\frac{1}{s^2 + 1} + \frac{1}{2s+1}$ , find $f(t)$ .

The next theorem relates the Laplace transform of the function $f^\prime(t)$ with that of $f(t)$ .

THEOREM 10.3.5 (Laplace Transform of Differentiable Functions)   Let $f(t),$ for $t > 0,$ be a differentiable function with the derivative, $f^\prime(t),$ being continuous. Suppose that there exist constants $M$ and $T$ such that $\vert f(t) \vert \leq M e^{{\alpha}t}$ for all $t \geq T.$ If ${\mathcal L}(f(t)) = F(s)$ then

$${\mathcal L}\left( f^\prime(t) \right) = s F(s) - f(0) \;\; {\mbox{ for }} \;\; s > {\alpha}.$$ (10.3.1)

Proof. Note that the condition $\vert f(t) \vert \leq M e^{{\alpha}t}$ for all $t \geq T$ implies that

$$\lim\limits_{b \longrightarrow \infty} f(b) e^{-sb} = 0 \;\; {\mbox{ for }} \;\; s > {\alpha}.$$

So, by definition,

$${\mathcal L}\bigl( f^\prime(t) \bigr) = \int_0^\infty e^{-s t} f^\prime(t) dt = \lim_{b \longrightarrow \infty} \int_0^b e^{-s t} f^\prime(t) dt$$

$$= \lim_{b \longrightarrow \infty} \biggl.f(t) e^{-st} \biggr\vert _0^b - \lim_{b \longrightarrow \infty}\int_0^b f(t) (-s) e^{-st} dt$$

$$= - f(0) + s F(s).$$

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We can extend the above result for $n^{\mbox{th}}$ derivative of a function $f(t)$ , if $f^\prime(t), \ldots, f^{(n-1)}(t), f^{(n)}(t)$ exist and $f^{(n)}(t)$ is continuous for $t \geq 0$ . In this case, a repeated use of Theorem 10.3.5, gives the following corollary.

COROLLARY 10.3.6   Let $f(t)$ be a function with ${\mathcal L}(f(t)) = F(s).$ If $f^\prime(t), \ldots, f^{(n-1)}(t), f^{(n)}(t)$ exist and $f^{(n)}(t)$ is continuous for $t \geq 0,$ then

$${\mathcal L}\bigl( f^{(n)}(t)\bigr) = s^n F(s) - s^{n-1} f(0) - s^{n-2} f^\prime(0) - \cdots - f^{(n-1)}(0).$$ (10.3.2)

In particular, for $n = 2$ , we have

$${\mathcal L}\bigl( f^{\prime\prime}(t) \bigr) = s^2 F(s) - s f(0) - f^\prime(0).$$ (10.3.3)

COROLLARY 10.3.7   Let $f^\prime(t)$ be a piecewise continuous function for $t \geq 0$ . Also, let $f(0) = 0$ . Then

$${\mathcal L}(f^\prime(t))= s F(s) \;\; {\mbox{ or equivalently }} \;\; {\mathcal L}^{-1}(s F(s)) = f^\prime(t).$$

EXAMPLE 10.3.8  

1. Find the inverse Laplace transform of $$\frac{s}{s^2 + 1}.$$
   Solution: We know that ${\mathcal L}^{-1} ( \displaystyle\frac{1}{s^2 + 1} ) = \sin t.$ Then $\sin (0) = 0$ and therefore, ${\mathcal L}^{-1} ( \displaystyle\frac{s}{s^2 + 1} ) = \cos t.$
2. Find the Laplace transform of $f(t) = \cos^2 (t).$
   Solution: Note that $f(0) = 1$ and $f^\prime(t) = - 2 \cos t \; \sin t = - \sin (2t).$ Also,
   $${\mathcal L}(- \sin (2t) ) = \frac{-2}{s^2 + 4}.$$
   Now, using Theorem 10.3.5, we get
   $${\mathcal L}(f(t)) = \frac{1}{s} \left(- \frac{2}{s^2 + 4} + 1 \right) = \frac{s^2 + 2}{s(s^2 + 4)}.$$

LEMMA 10.3.9 (Laplace Transform of $t f(t)$ )   Let $f(t)$ be a piecewise continuous function with ${\mathcal L}(f(t)) = F(s).$ If the function $F(s)$ is differentiable, then

$${\mathcal L}(t f(t)) = -\displaystyle\frac{d}{ds}F(s).$$

$${\mbox{ Equivalently,}} \;\;\; {\mathcal L}^{-1}(-\frac{d}{ds}F(s)) = t f(t).$$

Proof. By definition, $F(s) = \displaystyle\int_0^\infty e^{-st} f(t) dt.$ The result is obtained by differentiating both sides with respect to $s$ . height6pt width 6pt depth 0pt

Suppose we know the Laplace transform of a $f(t)$ and we wish to find the Laplace transform of the function $g(t) = \displaystyle\frac{f(t)}{t}.$ Suppose that $G(s) = {\mathcal L}(g(t))$ exists. Then writing $f(t) = t g(t)$ gives

$$F(s) = {\mathcal L}(f(t)) = {\mathcal L}(t g(t)) = - \frac{d}{ds} G(s).$$

Thus, $G(s) = -\int\limits_a^s F(p) dp$ for some real number $a$ . As $\lim\limits_{s {\longrightarrow}\infty} G(s) = 0$ , we get $G(s) = \int\limits_s^\infty F(p) dp.$

Hence,we have the following corollary.

COROLLARY 10.3.10   Let ${\mathcal L}(f(t)) = F(s)$ and $g(t) = \displaystyle\frac{f(t)}{t}.$ Then

$${\mathcal L}(g(t)) = G(s) = \int\limits_s^\infty F(p) dp.$$

EXAMPLE 10.3.11  

1. Find ${\mathcal L} (t \sin (at)).$
   Solution: We know ${\mathcal L} (\sin (at)) = \displaystyle\frac{a}{s^2 + a^2}.$ Hence ${\mathcal L} (t \sin (at)) = \displaystyle\frac{2 a s}{(s^2 + a^2)^2}.$
2. Find the function $f(t)$ such that $F(s) = \displaystyle\frac{4}{(s-1)^3}.$
   Solution: We know ${\mathcal L} (e^t) = \displaystyle\frac{1}{s-1}$ and
   $${\frac{4}{(s-1)^3} = 2 \frac{d}{ds} \left( - \frac{1}{(s-1)^2} \right) = 2 \frac{d^2}{ds^2} \left( \frac{1}{s-1} \right).}$$

   By lemma 10.3.9, we know that ${\mathcal L}(t f(t)) = - \frac{d}{ds}F(s).$ Suppose $\frac{d}{ds} F(s) = G(s).$ Then $g(t) = {\mathcal L}^{-1} G(s) = {\mathcal L}^{-1} \frac{d}{ds} F(s) = - t f(t).$ Therefore,

   $${\mathcal L}^{-1} \left(\frac{d^2}{ds^2} F(s) \right) = {\mathcal L}^{-1} \left( \frac{d}{ds} G(s) \right) = - t g(t) = t^2 f(t).$$
   Thus we get $f(t) = 2 t^2 e^t.$

LEMMA 10.3.12 (Laplace Transform of an Integral)   If $F(s) = {\mathcal L}(f(t))$ then

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

Equivalently, ${\mathcal L}^{-1} \left( \displaystyle\frac{F(s)}{s} \right) = \int_0^t f(\tau) d \tau.$

Proof. By definition,

$${\mathcal L} \bigl( \int_0^t f(\tau) \; d \tau \bigr) = \int_0^\... ... \; d \tau \right) d t = \int_0^\infty \int_0^t e^{-st} f(\tau) \; d \tau d t.$$

We don't go into the details of the proof of the change in the order of integration. We assume that the order of the integrations can be changed and therefore

$$\int_0^\infty \int_0^t e^{-st} f(\tau) \; d \tau d t = \int_0^\infty \int_{\tau}^\infty e^{-st} f(\tau) \; d t \; d\tau.$$

Thus,

$${\mathcal L} \bigl( \int_0^t f(\tau) \; d \tau \bigr) = \int_0^\infty \int_0^t e^{-st} f(\tau) \; d \tau d t$$

$$= \int_0^\infty \int_{\tau}^\infty e^{-st} f(\tau) \; d t \; d\tau ... ...nt_0^\infty \int_{\tau}^\infty e^{-s(t-\tau) - s \tau} f(\tau) \; d t \; d \tau$$

$$= \int_0^\infty e^{-s \tau} f (\tau)d \tau \left(\int_\tau^\infty e^{-s(t - \tau)} d t \right)\;$$

$$= \int_0^\infty e^{-s \tau} f (\tau)d \tau \left(\int_0^\infty e^{-sz} d z \right)\; = F(s) \frac{1}{s}.$$

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EXAMPLE 10.3.13  

1. Find ${\mathcal L} (\int_0^t \sin (az) dz).$
   Solution: We know ${\mathcal L} (\sin (at)) = \displaystyle\frac{a}{s^2 + a^2}.$ Hence
   $${\mathcal L} (\int_0^t \sin (az) dz ) = \displaystyle\frac{1}{s} \cdot \frac{a }{(s^2 + a^2)} = \frac{a}{s(s^2+a^2)}.$$

2. Find ${\mathcal L}\left(\displaystyle\int_0^t \tau^2 d\tau \right).$
   Solution: By Lemma 10.3.12
   $${\mathcal L}\left(\int_0^t \tau^2 d\tau \right) = \frac{ {\mathcal L}\left(t^2 \right)}{s} = \frac{1}{s} \cdot \frac{2!}{s^3} = \frac{2}{s^4}.$$

3. Find the function $f(t)$ such that $F(s) = \displaystyle\frac{4}{s(s-1)}.$
   Solution: We know ${\mathcal L} (e^t) = \displaystyle\frac{1}{s-1}.$ So,
   $${\mathcal L}^{-1}\left(\frac{4}{s(s-1)} \right)= 4 {\mathcal L}^... ...t(\frac{1}{s}\frac{1}{s-1} \right)= 4 \int_0^t e^{\tau} d \tau = 4 ( e^t - 1).$$

LEMMA 10.3.14 ($s$ -Shifting)   Let ${\mathcal L}(f(t)) = F(s).$ Then ${\mathcal L}(e^{at} f(t)) = F(s-a)$ for $s > a.$

Proof.

$${\mathcal L}(e^{at} f(t)) = \int_0^\infty e^{at} f(t) e^{-s t} dt = \int_0^\infty f(t) e^{-(s-a) t} dt$$

$$= F(s-a) \hspace{.5in} s > a.$$

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EXAMPLE 10.3.15  

1. Find ${\mathcal L} (e^{at} \sin (bt)).$
   Solution: We know ${\mathcal L} (\sin (bt)) = \displaystyle\frac{b}{s^2 + b^2}.$ Hence ${\mathcal L} (e^{at} \sin (bt)) = \displaystyle\frac{b}{(s-a)^2 + b^2}.$
2. Find ${\mathcal L}^{-1}\left( \frac{s-5}{(s-5)^2 + 36}\right).$
   Solution: By $s$ -Shifting, if ${\mathcal L}(f(t)) = F(s)$ then ${\mathcal L}(e^{at} f(t)) = F(s-a)$ . Here, $a = 5$ and
   $${\mathcal L}^{-1} \left(\frac{s}{s^2+ 36} \right) = {\mathcal L}^{-1} \left(\frac{s}{s^2+ 6^2}\right) = \cos (6t).$$
   Hence, $f(t) = e^{5t} \cos (6t).$