In solving a linear system, when would I use a Fourier transform versus a Laplace transform? I am not a mathematician, so the little intuition I have tells me that it could be related to the boundary conditions imposed on the solution I am trying to find, but I am unable to state this rigorously or find a reference that discusses this. Any help would be appreciated. Thanks.
[Math] Fourier vs Laplace transforms
ca.classical-analysis-and-odesfourier analysisfourier transformheuristicslaplace transform
Related Solutions
What is sufficient (though not necessary) is that the Fourier transform decays exponentially at $\infty$ (if you want just analyticity on the line) or faster than any exponent (if you want your original function to be entire). In particular, anything with compact support will do. If this is too restrictive for your construction, you'd better just tell what exactly you are trying to construct.
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...
Best Answer
Here is a heuristic point of view from engineering considerations. I must confess I do not fully know the mathematical reasons.
Suppose you want to consider $f(t)$, a function of time, $t$. Imagine that as we look at the direction of positive $t$-axis, the graph of $f(t)$ s like looking behind to the trail $f$ left in time. If you do not care about the future, ie the case $t < 0$, then it makes sense to use Laplace transform, because the transform integral goes from $0$ to $\infty$. On the other hand, if you care about the future also, it makes more sense to consider the Fourier transform. The transformation integral here goes from $-\infty$ to $\infty$.
So if you want to include future in your analysis, then Fourier transform is the way. This makes sense in electrical engineering applications for example, where you consider sinusoidal signals and you have an idea of what is going to come.
However for some physical systems, you only have the data of what happened until then. And you want all your analysis to be based on this, without predicting the future. Then Laplace transforms is the way.
If you do not care about the future, ie if you can declare $f(t) = 0$ for $t < 0$, then the Laplace and Fourier transforms coincide: The Fourier transform is nothing but the Laplace transform evaluated on the imaginary axis. Such systems are called causal systems: the response depends only on what happened so far. This a terminology from control systems or signal processing.
For control systems engineering, stability of electrical networks, etc., Laplace transformation defines a more natural transfer function, and is easier to deal with, and the poles and zeros would immediately tell you about the stability of the network under consideration. Here we use Laplace transforms rather than Fourier, since its integral is simpler.
For instances where you look at the "frequency components", "spectrum", etc., Fourier analysis is always the best. The Fourier transform is simply the frequency spectrum of a signal. If you know that the sin/cos/complex exponentials would behave nicely, you might as well want to express a function in terms of these and observe how it behaves then.
Another example is solving the wave equation. Fourier himself used Fourier series/transforms for heat conduction problems.