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I need to do an inverse Laplace transform of a function with essential singularities for a specific problem. I find it is very similar to an equation J. Noolandi worked out in one of his papers in 1977. The paper is at this page, and I have taken a snapshot of the Appendix page where the math problem and the solution were given (view it here). But I don't quite get some part of the solving process which I need help on deciphering.

If I can understand what he was doing, my specific problem can be solved. So now let's take the problem in the Appendix of Noolandi's paper for example. The function is

$\displaystyle \tilde I(s) = \frac{1-\exp(-a(s)t_0)}{a(s)}$

where $\displaystyle a(s)t_0 = \Sigma_{i=1}^n \frac{s\omega_i}{s+r_i} t_0 = \Sigma_{i=1}^n \frac{sM_i}{s+r_i}$. "$s$" is the variable. $\omega_i$, $r_i$, $t_0$, and $M_i$ are all real numbers greater than zero.

The question is how to get inverse Laplace transform $L^{-1}[\tilde I(s)]$. I think the author used Residue Theorem to evaluate the contour integral $\int_C \tilde I(s) \exp(st) ds$, but the hard part is the singularities are essential singularities, thus the residues can not be got conventionally.

However, it seems that the author constructed a contour and solved it anyway. I don't get Equation A5-A8 in the Appendix, and don't understand how to determine the parameter $R$ and $R_1$ he introduced in.

Could anyone please help point out what he was doing? Thanks very much!

By the way, the author also said the inverse Laplace transform could also be evaluated using Laplace transform tables. And the result involved a convolution of the modified Bessel function of the first order. I still haven't worked it out either...

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Hi, this question will probably be closed as not being "research level". But I would say: (1) what is the exact function you want to invert? You haven't said in your question; (2) Inverse Laplace transforms don't always exist, you need conditions on the analytic function; (3) The Residue Theorem works for essential singularities also, only the coefficient of 1/z is important. Whether the negative powers of z are finite or infinite in number is irrelevant, since the Laurent series always converges nicely enough to interchange contour integration with the infinite sum. –  Zen Harper May 31 '12 at 11:20
    
Thanks for your reply. (1) Sorry about the ambiguity. What I was trying to say was I needed to understand how to do the inverse Laplace transform of function like $\tilde I(s)$ in that paper. Once I understood that, my real question would be solved consequently. (2) The answer doesn't have to be analytic. I think Noolandi gave the guideline for doing the transform numerically. But the problem is I don't understand his approach on the paper. (3) Yes. But expanding the function in Laurent series and filtering out the z^{-1} term is not easy. I am hoping for a more clear. –  user24116 May 31 '12 at 15:36
    
I've added the Fourier Analysis tag, since I think it is appropriate [although that usually goes without saying, when Laplace transforms are used!] –  Zen Harper Jun 1 '12 at 10:41
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2 Answers

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After taking a quick look at the paper, I agree with Robert Israel.

Almost no detailed justification is given (perhaps not surprising for a Physics journal...), and it seems to be not so easy to justify.

Take the special case $N=1$ for simplicity; so up to constants, we consider (note the use of (A3) in the paper, this constant is subtracted from the original $\tilde{I}$)

$$\tilde{I}(s) = \frac{(1-\exp(-Ks/(s+r))}{s/(s+r)} - (1-\exp(-K)) $$

for $K$, $r$ constants. Now consider what happens as $|s| \to \infty$, then we get [assuming, dangerously, that my calculations are error free]

$$\tilde{I}(s) \sim -Kr e^{-K}/(s+r) \sim A/s$$

as $|s| \to \infty$, for some constant $A$. The contour is originally a vertical line, but $1/s$ is not in $L^1$ so the integral is not a standard Lebesgue integral (i.e., is not absolutely convergent). However, it is is $L^2$, i.e. is square integrable, so you can use the $L^2$ theory of the Fourier transform to give a meaning.

Now, if you deform the original vertical line contour over { Re(s) = c } by Cauchy's Theorem (push part of it to the left), you get the integral of $\tilde{I}(s) e^{st}$ over the circle $C$ plus an error term, being the sum of integrals over the contours

C1 = { $L + iy : |y| < R $ },

C2 = { $x \pm i R : L < x < c $ },

C3 = { $c + iy : |y|>R$ }

where $c$ is fixed; now let $L \to -\infty$ and $R \to +\infty$ appropriately to get the error term going to zero.

[More detail: as $R \to \infty$ the integral over C3 goes to 0 uniformly in $t$, by the $L^2$ properties of the Fourier transform; also the integral over C2 goes to zero because $|1/s| < 1/R$ on C2.

The integral over C1 goes to zero as $L \to -\infty$ because $| e^{st} / s | < e^{Lt} $ on C1, which clearly $\to 0$ rapidly, at least for $t>0$. ]

There are still 2 things to be justified in this special case: (i) what about $t<0$, and (ii) how do we know the analytic function $\tilde{I}(s)$ really can be represented as the Laplace transform of something?

I think (ii) follows from a general result in my last paper (about Laplace transform representation theorems, which is in Documenta Mathematica 2010, or on my website); but (i) I am not sure about.

[Strangely enough, the contours C1, C2, C3 are exactly the ones I used in my paper, although I got them from an earlier paper by C.Batty and M.D.Blake; and I suspect the original use of these contours dates back at least 50 years, since similar stuff has been used in Tauberian theory and analytic number theory for a long time].

So, even in this special case with just one singularity, there are a lot of extra details to fill in if you want to make it properly rigorous. Since the original paper is from a Physics journal, I would guess it's not the best place to look for rigour(!) Approach with caution...

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The paper is not using residue theory, he is evaluating the contour integral numerically. I don't see a justification given for deforming the path of integration from a vertical line to a circle, but assuming sufficiently nice properties at infinity that should be OK.

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