Laplace Multipliers

Problem 1. Evaluate $\mathcal L\{\cos^2 t\}$ .

(Click for Solution)

Solution. Applying trigonometric identities,

$\displaystyle \cos^2 t = \frac 12 + \frac 12\cos 2t.$

Taking Laplace transforms on both sides,

$\begin{aligned} \mathcal L\{ \cos^2 t \} &= \mathcal L\left\{ \frac 12 + \frac 12\cos 2t \right\} \\ &= \frac 12\cdot \mathcal L\{1 \} + \frac 12\cdot \mathcal L \{\cos 2t \} \\ &= \frac 12 \cdot \frac 1s + \frac 12 \cdot \frac s{s^2+2^2} \\ &= \frac 1{2s} + \frac {s}{2(s^2+4)}. \end{aligned}$

The point is that calculating the Laplace transform of products is not obvious. Suppose $f$ is any function with Laplace transform $F = \mathcal L \{f\}$ .

Problem 2. Prove that $\mathcal L\{t f(t)\} = -F'(s)$ .

(Click for Solution)

Solution. We need to return to the definition of Laplace transforms:

$\displaystyle \begin{aligned} \mathcal L\{t f(t)\} &= \int_0^\infty t f(t) e^{-st}\, \mathrm dt. \end{aligned}$

The answer is not obvious, but if we pay close attention, we see that differentiating $e^{-st}$ with respect to $s$ yields $e^{-st} \cdot (-t) = -te^{-st}$ . In more technical notation,

$\displaystyle \frac{\partial}{\partial s} (e^{-st}) = -te^{-st},$

where the partial derivative $\partial$ just means that we will treat $t$ as a constant. This means that

$\displaystyle \begin{aligned} \int_0^\infty t f(t) e^{-st}\, \mathrm dt &= \int_0^\infty f(t) \cdot \frac{\partial}{\partial s} (-e^{-st})\, \mathrm dt \\ &= -\int_0^\infty \frac{\partial}{\partial s} (f(t) e^{-st})\, \mathrm dt. \end{aligned}$

We shall now perform a trick: bringing the derivative out of the integral. We will omit the proof of this result as it falls in a far richer study in measure theory and Lebesgue integration. In any case,

$\displaystyle \begin{aligned} \int_0^\infty t f(t) e^{-st}\, \mathrm dt &= -\int_0^\infty \frac{\partial}{\partial s} (f(t) e^{-st})\, \mathrm dt \\ &= -\frac{\mathrm d}{\mathrm ds} \int_0^\infty f(t) e^{-st}\, \mathrm dt\\ &= -F'(s). \end{aligned}$

Problem 3. For any positive integer $n$ , prove that $\mathcal L\{t^n f(t)\} = (-1)^n F^{(n)}(s)$ .

(Click for Solution)

Solution. Applying Problem 2 many more (i.e. $n$ ) times yields

$\mathcal L\{t^n f(t)\} = (-1)^n F^{(n)}(s).$

If multiplying by $t$ gives differentiation, then intuitively, dividing by $t$ ought to yield integration.

Problem 4. Prove that

$\displaystyle \mathcal L\left\{ \frac{f(t)}{t} \right\} = \int_s^\infty F(u)\, \mathrm du,$

assuming that the left-hand side exists.

(Click for Solution)

Solution. We shall return to the definition of Laplace transforms:

$\displaystyle \begin{aligned} \mathcal L\left\{ \frac{f(t)}{t} \right\} &= \int_0^\infty \frac{f(t)}{t} e^{-st}\, \mathrm dt. \end{aligned}$

Similar to Problem 2, if we pay close attention, we see that integrating $e^{-ut}$ with respect to $u$ yields

$\displaystyle \int_s^\infty e^{-ut}\, \mathrm du = \left[ \frac{e^{-ut}}{-t} \right]_s^\infty = 0 - \frac{e^{-st}}{-t} = \frac{e^{-st}}{t}.$

Hence,

$\displaystyle \begin{aligned} \mathcal L\left\{ \frac{f(t)}{t} \right\} &= \int_0^\infty f(t)\int_s^\infty e^{-ut}\, \mathrm du\, \mathrm dt \\ &= \int_0^\infty \int_s^\infty f(t) e^{-ut}\, \mathrm du\, \mathrm dt. \end{aligned}$

We need another tool from multivariable calculus known as Fubini’s theorem, which we will not prove this result in this discussion. This theorem allows us to swap the order of the integrals:

$\displaystyle \begin{aligned} \mathcal L\left\{ \frac{f(t)}{t} \right\} &= \int_0^\infty f(t)\int_s^\infty e^{-ut}\, \mathrm du\, \mathrm dt \\ &= \int_s^\infty \int_0^\infty f(t) e^{-ut}\, \mathrm dt\, \mathrm du \\ &= \int_s^\infty \mathcal L\{f\}(u)\, \mathrm du = \int_s^\infty F(u)\, \mathrm du. \end{aligned}$

Similarly, taking the Laplace transform of an integral ought to transform it into a quotient.

Problem 5. Prove that

$\displaystyle \mathcal L\left\{ \int_0^t f(u)\,\mathrm du \right\} = \frac{F(s)}{s}.$

(Click for Solution)

Solution. Employing the definition of the Laplace transform and Fubini’s theorem,

$\displaystyle \begin{aligned} \mathcal L\left\{ \int_0^t f(u)\,\mathrm du \right\} &= \int_0^\infty \left(\int_0^t f(u)\,\mathrm du \right) \cdot e^{-st}\, \mathrm dt \\ &= \int_0^\infty \int_0^t f(u) e^{-st}\, \mathrm du\, \mathrm dt \\ &= \int_0^\infty \int_u^\infty f(u) e^{-st}\, \mathrm dt\, \mathrm du \\ &= \int_0^\infty f(u) \int_u^\infty e^{-st}\, \mathrm dt\, \mathrm du \\ &= \int_0^\infty f(u) \left[ \frac{e^{-st}}{-s} \right]_u^\infty\, \mathrm du \\ &= \int_0^\infty f(u) \left( 0 - \frac{e^{-su}}{-s} \right)\, \mathrm du \\ &= \frac 1s \int_0^\infty f(u) e^{-su}\, \mathrm du = \frac 1s \cdot F(s) = \frac{F(s)}{s}. \end{aligned}$

Finally for general multiplication problems, while we cannot in general evaluate the Laplace transform of a product, we have a tool to evaluate the inverse Laplace transform of a product.

Problem 6. Given a function $g$ with Laplace transform $G = \mathcal L\{g\}$ , prove that

$\displaystyle \mathcal L^{-1}\{F(s)G(s)\} = \int_0^t f(t-u)g(u)\, \mathrm du =: (f * g)(t).$

(Click for Solution)

Solution. Expanding the definitions of $F$ and $G$ ,

$\begin{aligned} F(s)G(s) &= \int_0^\infty f(v)e^{-sv}\, \mathrm dv \cdot \int_0^\infty g(u)e^{-su}\, \mathrm dv \\ &= \int_0^\infty \int_0^\infty \left( f(v)g(u)\cdot e^{-s(v+u)}\right) \mathrm dv\, \mathrm du \end{aligned}$

Making the substitution $v = t-u \Rightarrow \mathrm dv =\mathrm dt$ ,

$\begin{aligned} F(s)G(s) &= \int_0^\infty \int_{u}^\infty \left( f(t-u)g(u)\cdot e^{-st} \right) \mathrm dt\, \mathrm du \\ &= \int_0^\infty \int_0^t \left( f(t-u)g(u)\cdot e^{-st} \right) \mathrm du\, \mathrm dt \\ &= \int_0^\infty \left( \int_0^t f(t-u)g(u)\, \mathrm du \right) e^{-st} \, \mathrm dt \\ &= \mathcal L\left\{ \int_0^t f(t-u)g(u)\, \mathrm du \right\}, \end{aligned}$

where we used Fubini’s theorem to swap the limits. Hence,

$\displaystyle \mathcal L^{-1}(F(s)G(s)) = \int_0^t f(t-u)g(u)\, \mathrm du =: (f * g)(t)$ ,

where the right-hand side is known as the convolution of the functions $f$ and $g$ .

—Joel Kindiak, 14 Feb 25, 0918H

KindiakMath

Laplace Multipliers

Responses

Leave a comment Cancel reply

Laplace Multipliers

Share this:

Responses

Leave a comment Cancel reply