College Math Teaching

October 25, 2013

A Laplace Transform of a function of non-exponential order

Many differential equations textbooks (“First course” books) limit themselves to taking Laplace transforms of functions of exponential order. That is a reasonable thing to do. However I’ll present an example of a function NOT of exponential order that has a valid (if not very useful) Laplace transform.

Consider the following function: n \in \{1, 2, 3,...\}

g(t)= \begin{cases}      1,& \text{if } 0 \leq t \leq 1\\      10^n,              & \text{if } n \leq t \leq n+\frac{1}{100^n} \\  0,  & \text{otherwise}  \end{cases}

Now note the following: g is unbounded on [0, \infty) , lim_{t \rightarrow \infty} g(t) does not exist and
\int^{\infty}_0 g(t)dt = 1 + \frac{1}{10} + \frac{1}{100^2} + .... = \frac{1}{1 - \frac{1}{10}} = \frac{10}{9}

One can think of the graph of g as a series of disjoint “rectangles”, each of width \frac{1}{100^n} and height 10^n The rectangles get skinnier and taller as n goes to infinity and there is a LOT of zero height in between the rectangles.

notexponentialorder

Needless to say, the “boxes” would be taller and skinnier.

Note: this is an example can be easily modified to provide an example of a function which is l^2 (square integrable) which is unbounded on [0, \infty) . Hat tip to Ariel who caught the error.

It is easy to compute the Laplace transform of g :

G(s) = \int^{\infty}_0 g(t)e^{-st} dt . The transform exists if, say, s \geq 0 by routine comparison test as |e^{-st}| \leq 1 for that range of s and the calculation is easy:

G(s) = \int^{\infty}_0 g(t)e^{-st} dt = \frac{1}{s} (1-e^{-s}) + \frac{1}{s} \sum^{\infty}_{n=1} (\frac{10}{e^s})^n(1-e^{\frac{-s}{100^n}})

Note: if one wants to, one can see that the given series representation converges for s \geq 0 by using the ratio test and L’Hoptial’s rule.

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