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can this be a proper Laplace transform?

Started by lucy October 3, 2005
>>>>> "dave" == dave <david_lawrence_petry@yahoo.com> writes:
dave> Raymond Toy wrote: >> >>>>> "dave" == dave <david_lawrence_petry@yahoo.com> writes: >> dave> lucy wrote: >> >> sin(s)/(s^2+4), >> >> >> >> can it be a proper Laplace transform? >> dave> I don't think so. The Laplace Inversion formula doesn't dave> work for it. >> >> What fails? >> >> I took a quick stab at it by replacing sin(s) with its power series. >> Then we have terms of the form s^(2*k+1)/(s^2+4), for which we know >> the inverse Laplace transform. dave> Really? What is the inverse Laplace transform of s^5/(s^2+4) ? I was using the relationship f'(t) -> sF(s) - f(+0). But there are probably conditions that need to be satisfied for this to be true. I dont' have any transform theory book handy, other than a short table of transforms. Ray
Thanks a lot Robert!
Will it possibly be some Laplace Transform of an non-exponentially
bounded function?

In article <1128664482.687884.120780@z14g2000cwz.googlegroups.com>,
lucy <losemind@yahoo.com> wrote:
>Thanks a lot Robert! >Will it possibly be some Laplace Transform of an non-exponentially >bounded function?
The Laplace Transform is defined by F(s) = int_0^infty f(t) exp(-st) dt, when that improper integral converges. f doesn't have to be exponentially bounded, but it helps. Now let's say the integral converges absolutely for s = s_0 (real), i.e. B = int_0^infty |f(t)| exp(-s_0 t) dt < infty. Then |F(s)| <= B whenever Re(s) >= s_0. In your case F(s) is unbounded for Re(s) >= s_0, so the integral can't ever converge absolutely. I guess there's a possibility that the integral could converge conditionally for s in some domain (it's easy to arrange for some particular s, but hard to imagine how to do it for s in an open set). You might also try the case where f is not a function, but a distribution. However, even allowing a finite number of derivatives of functions for which the integral converges absolutely at s_0 will make the Laplace transform bounded by a polynomial in |s| for Re(s) > s_0, so you still can't get sin(s)/(s^2+4) that way. Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada
Robert Israel wrote:

> In article <1128664482.687884.120780@z14g2000cwz.googlegroups.com>, > lucy <losemind@yahoo.com> wrote: > >>Thanks a lot Robert! >>Will it possibly be some Laplace Transform of an non-exponentially >>bounded function? > > > The Laplace Transform is defined by F(s) = int_0^infty f(t) exp(-st) dt, > when that improper integral converges. f doesn't have to be > exponentially bounded, but it helps. > > Now let's say the integral converges absolutely for s = s_0 (real), i.e. > B = int_0^infty |f(t)| exp(-s_0 t) dt < infty. Then |F(s)| <= B > whenever Re(s) >= s_0. > > In your case F(s) is unbounded for Re(s) >= s_0, so the integral can't > ever converge absolutely. I guess there's a possibility that the integral > could converge conditionally for s in some domain (it's easy to arrange > for some particular s, but hard to imagine how to do it for s in an open > set). > > You might also try the case where f is not a function, but a > distribution. However, even allowing a finite number of derivatives > of functions for which the integral converges absolutely at s_0 will > make the Laplace transform bounded by a polynomial in |s| for > Re(s) > s_0, so you still can't get sin(s)/(s^2+4) that way. > > Robert Israel israel@math.ubc.ca > Department of Mathematics http://www.math.ubc.ca/~israel > University of British Columbia Vancouver, BC, Canada >
What's the difference between a function and a distribution? Judging from the Dirac delta distribution I gather that a distribution is something that you can sometimes treat as a function if you hold your mouth right, but that's hardly a rigorous definition. -- Tim Wescott Wescott Design Services http://www.wescottdesign.com
Tim Wescott wrote:
> Robert Israel wrote:
> > You might also try the case where f is not a function, but a > > distribution. However, even allowing a finite number of derivatives > > of functions for which the integral converges absolutely at s_0 will > > make the Laplace transform bounded by a polynomial in |s| for > > Re(s) > s_0, so you still can't get sin(s)/(s^2+4) that way.
> What's the difference between a function and a distribution? Judging > from the Dirac delta distribution I gather that a distribution is > something that you can sometimes treat as a function if you hold your > mouth right, but that's hardly a rigorous definition.
A distribution is a continuous linear functional on the space of C^infinity functions with compact support. For the purposes of Fourier and Laplace transforms, one generally restricts attention to tempered distributions. See e.g. <http://en.wikipedia.org/wiki/Tempered_distribution> Robert Israel israel@math.ubc.ca Department of Mathematics http://www.math.ubc.ca/~israel University of British Columbia Vancouver, BC, Canada
On 7 Oct 2005 16:25:18 GMT, israel@math.ubc.ca (Robert Israel) wrote:

>In article <1128664482.687884.120780@z14g2000cwz.googlegroups.com>, >lucy <losemind@yahoo.com> wrote: >>Thanks a lot Robert! >>Will it possibly be some Laplace Transform of an non-exponentially >>bounded function? > >The Laplace Transform is defined by F(s) = int_0^infty f(t) exp(-st) dt, >when that improper integral converges. f doesn't have to be >exponentially bounded, but it helps. > >Now let's say the integral converges absolutely for s = s_0 (real), i.e. >B = int_0^infty |f(t)| exp(-s_0 t) dt < infty. Then |F(s)| <= B >whenever Re(s) >= s_0. > >In your case F(s) is unbounded for Re(s) >= s_0, so the integral can't >ever converge absolutely. I guess there's a possibility that the integral >could converge conditionally for s in some domain (it's easy to arrange >for some particular s, but hard to imagine how to do it for s in an open >set).
Actually it's not so hard. Take s_0 = 0 for convenience. Suppose that f is locally (Lebesgue) integrable on [0, infinity). Let I(A) = int_0^A f(t) dt, and suppose that I(A) -> L as A -> infinity. Then I is bounded, so an integration by parts shows that (as an improper Riemann integral) F(s) converges for Re(s) > 0, in fact F(s) = s int_0^infinity I(t) exp(-st) dt. If f takes is large enough in absolute value (but wildly oscillating so that I(A) converges) then the integral defining F(s) will fail to be absolutely convergent for any s with Re(s) > 0. Of course the same argument shows that |F(s)| is bounded by c(1 + |Im(s)|), so you don't get sin(s)/(s^2+4) this way. (The bound shows that F _is_ holomorphic in the right half-plane...)
>You might also try the case where f is not a function, but a >distribution.
If f is a distribution it's not clear to me what conditional convergence would even mean - we'd have to define it using a limit of smooth cutoff functions, and it seems like the answer might depend on the choice of cutoff function. If f is a tempered distribution (with support in (0, infinity), say) then I _think_ it's clear that F has polynomial growth - don't quote me on that...
>However, even allowing a finite number of derivatives >of functions for which the integral converges absolutely at s_0 will >make the Laplace transform bounded by a polynomial in |s| for >Re(s) > s_0, so you still can't get sin(s)/(s^2+4) that way. > >Robert Israel israel@math.ubc.ca >Department of Mathematics http://www.math.ubc.ca/~israel >University of British Columbia Vancouver, BC, Canada
************************ David C. Ullrich