# What restrictions on time boundary conditions does it have to use Fourier transform to solve wave equation?

The wave equation can be solved using Fourier transform, by assuming a solution of the form of $$\mathbf{E}(x,y,z,t)~=~\mathbf{E}(x,y,z)e^{j\omega t}$$ and then reducing the equation to the Helmholtz equation.

• What are the presumed restrictions on the solution, when solving the equation this way? (e.g., on time boundary condition) I mean can this method give the most general solution (given some boundary conditions)? What features does the solution obtained this way have?

• Does it have any difference with solutions obtained using Laplace transform? (The very same above questions, for Laplace transform.)

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You need give a little more detail if you want to know 'presumed restrictions' and 'time boundary' conditions. I believe it will be more helpful for you to post this question on Math StackExchange and give it the partial-differential equations' tag. –  dj_mummy Sep 17 '13 at 11:57
Would math.stackexchange.com be a better home for this question? –  Qmechanic Sep 17 '13 at 15:58

The simple answer to your question is this: given reasonable assumptions, the Fourier transform puts no practical restrictions on the solutions of the wave equation. Or:

Any time variation and its Fourier transform constitute the same information

for all physically reasonable signals. Fourier transforms can be losslessly (in the sense of uniquely i.e. losing no information) inverted. A Fourier transform is like a perfect mechanic who can work with any physical machine. He or she can strip the machine down into the tiniest bits that make it up, examine each of these bits in detail, and put the machine back together again perfectly. Thus the Fourier transform allows us to reversibly split any time variation into its constituent time harmonic variations, study the effect of these variations, then perfectly put all these pieces back together again to build the original solution.

Now for some technicalities.

I think the true limitations on what kind of time variation is currently construed to be representable a Fourier transform is a Math SE or even Math Overflow question: the FT's ken of applicability has been steadily widened by successive new concepts in mathematics over the last hundred years (e.g. through theory of distributions (generalized functions), advanced measure theory and theory of hyperfunctions) so I can't claim to know the current "state of the art".

However, a good, pithy summary is that Fourier transforms work with tempered distributions, which is the dual space of Schwarz space.

What does this mean practically? Probably that any time variation you want to represent, with any initial value, can be represented by a Fourier superposition. I shall explain these tempered distributions more fully below and give you some references.

Laplace transforms are, somewhat bizarrely, actually more restricted in what kinds of functions the can represent (aside from in a contrived sense that they can handle one sided exponential divergence). This is because the kernel $\exp(-s\,t)$ in the integral transform is now unbounded, and one cannot in general define a Laplace transform over the interval $(\infty,\,\infty)$ as one can do with a Fourier transform. For a Fourier transform this kernel does not diverge, so the Fourier transform is a unitary (power preserving) map. Laplace transforms are therefore restricted to causal functions, i.e. those functions $f:\mathbb{C}\to\mathbb{C}$ for which there is some $t_0\in\mathbb{R}$ such that $f(t) =0,\;\forall t < t_0$. Then the integral transform kernel $\exp(-s\,t)$ is bounded over this function's support.

Fourier transforms, of course, automatically take causal functions in their stride: now they are integrals of the form $\int_{t_0}^\infty \,e^{-i\,\omega\,t} f(t)\,\mathrm{d}t$. Whenever a function is causal, and its Laplace transform exists, the Laplace transform is then the Fourier transform (modulo multiplicative constant) with its domain of definition broadened from $\mathbb{R}$ out into the complex plane (aside from where the function has poles) by analytic continuation. So, where they are applicable, Laplace transforms are almost the same as Fourier transforms. Strictly speaking, a Laplace transform can regularize a function that diverges exponentially as $t\to\infty$ (we just make the real part of the transform variable $s$ big and positive enough to cancel this divergence) whereas the Fourier transform cannot handle such divergence. But this is not likely to be a practical problem. Certainly it is almost never a problem in electrodynamics. Time variations are assumed to have finite energy, and, whilst we're at it, Fourier transforms are also very useful spatially to split an electromagnetic field up into constituent plane waves. There is almost never occasion to study fields that diverge exponentially at infinity. Also, the theory of Laplace transforms also comes kitted with specific procedures to handle initial conditions which are not wontedly seen as being part and parcel of the Fourier transform, but the separation is mainly a matter of custom: one can equally well handle such conditions with the Fourier transform.

Causality, or the potential for causality, can be detected through the Paley-Wiener criterion; see my answer to the question More extensions of the wave equation for dispersion for more information on this criterion.

## Fourier Transforms and Tempered Distributions

We define tempered distributions as follows. Firstly, we consider the Schwartz space $\mathscr{S}\left(\mathbb{R}^N, \mathbb{C}\right)$ of complex valued functions defined on $\mathbb{R}^N$ (in this case you have a time variation) which are (i) smooth (i.e. have derivatives in all directions of all orders) and (ii) which themselves as well as all their derivatives dwindle "more swiftly than polynomial speed" to nought at infinity; these two conditions can be summarised by $\left\|\, \left| \mathbf{x} \right|^\alpha D^\beta \psi\left(\mathbf{x}\right) \right\| < \infty,\,\forall \alpha, \beta \in \mathbb{Z}^+$ and $D$ is any first order differential operator - there is only one such operator - to wit - $\mathrm{d}_t$ when we're dealing with time variations (i.e. $N=1$) but the ideas hold equally well in any number of dimensions.. The Fourier transform $\mathfrak{F}\,\psi$ of any $\psi \in \mathscr{S}\left(\mathbb{R}^N, \mathbb{C}\right)$ is then defined and also belongs to the Schwartz space i.e. $\psi \in \mathscr{S}\left(\mathbb{R}^N, \mathbb{C}\right) \Leftrightarrow \mathfrak{F}\, \psi \in \mathscr{S}\left(\mathbb{R}^N, \mathbb{C}\right)$.

Furthermore, the kernel of $\mathfrak{F}:\mathscr{S}\left(\mathbb{R}^N, \mathbb{C}\right)\mapsto \mathscr{S}\left(\mathbb{R}^N, \mathbb{C}\right)$ is trivial, to wit $\mathfrak{F}\psi = 0, \psi \in \mathscr{S}\left(\mathbb{R}^N, \mathbb{C}\right) \Rightarrow \psi = 0$. Lastly, every member of the Schwartz space is the Fourier transform of some other member of the Schwartz space. Thus, in the Scwartz space:

Schwartz functions on $\mathbb{R}^N$ and their Fourier transforms constitute exactly the same information

However, the Schwartz space does not include all fields of interest to us: we may wish to define piecewise continuous boundary conditions, nondiminishing time variations (e.g. pure sinusoid) which, although smooth, do not fulfill the swift decay criterion but only the much weaker $\left|\mathbf{r}\right| \psi\left(\mathbf{r}\right) \rightarrow 0$ as $\left|\mathbf{r}\right| \rightarrow \infty$ and indeed $\left|\mathbf{r}\right|^2 \psi\left(\mathbf{r}\right)$ generally diverges. Therefore, we consider the topological dual space $\mathscr{S}^\prime\left(\mathbb{R}^N, \mathbb{C}\right)$ of all complex-valued linear functionals on the Schwartz space; the topological dual is defined by a stronger topology than simply the $\mathbf{L}^2$ norm of the original Hilbert space $\mathbf{L}^2(\mathbb{R}^N)$. This stronger topology is the one induced by the family of norms:

$$\rho_{\alpha,\,\beta}(f) \stackrel{def}{=} \sup\limits_{\mathbf{u}\in\mathbb{R}^N} \left.||\mathbf{x}|^\alpha\,D^\beta f(\mathbf{x})|\right|_{\mathbf{x}=\mathbf{u}}$$

Thus, for example, the Dirac delta is a continuous linear functional on $\mathscr{S}\left(\mathbb{R}^N, \mathbb{C}\right)$ kitted with this topology, but it is not continuous on the Hilbert space $\mathbf{L}^2(\mathbb{R}^N)$ kitted with the original $\mathbf{L}^2$ norm. So we've brought the stronger topology to bear to ferret out all the linear functionals which are useful to us (Dirac delta, multiplication operator $f(x)\mapsto x\,f(x)$, $f(x)\mapsto e^{i\,k\,x}\,f(x)$ and so forth) which the original topology couldn't "sniff out".

The members of $\mathscr{S}^\prime\left(\mathbb{R}^N, \mathbb{C}\right)$ are known as tempered distributions or sometimes generalised functions. We can think of an ordinary scalar field $\psi\left(\mathbf{r}\right)$ as the linear functional $\Psi : \mathscr{S}^\prime\left(\mathbb{R}^N, \mathbb{C}\right) \rightarrow \mathscr{S}^\prime\left(\mathbb{R}^N, \mathbb{C}\right): \varphi \in \mathscr{S}\left(\mathbb{R}^N, \mathbb{C}\right) \mapsto \int_{\mathbb{R}^N} \psi\left(\mathbf{u}\right) \varphi\left(\mathbf{u}\right) \mathrm{d}^N u$; given a linear functional $\Psi \in \mathscr{S}^\prime\left(\mathbb{R}^N, \mathbb{C}\right)$, we can recover the ordinary function by evaluating, for example, $\Psi\left(\kappa^N \exp\left(-\kappa^2 \left|\mathbf{r} - \mathbf{u}\right|^2\right) / \pi^{3/2}\right)$ and taking the limit as $\kappa\rightarrow\infty$. If the linear functional in question is defined as just described from an ordinary function, the ordinary function's value at $\mathbf{r}\in\mathbb{R}^N$ is recovered by the limit. If not (e.g. if the functional is the delta function), the limit will not exist at all points. The tempered distributions also have the useful properties that:

1. A tempered distribution's Fourier transform is also a tempered distribution;

2. Any tempered distribution is the Fourier transform of a tempered distribution; and

3. The Fourier transform's kernel is trivial. More pithily, $\mathfrak{F}$ is then a unitary bijection (one-to-one, onto map) from $\mathscr{S}^\prime\left(\mathbb{R}^n, \mathbb{C}\right))$ onto itself.

So now we have again:

Tempered distributions on $\mathbb{R}^N$ and their Fourier transforms constitute exactly the same information

and almost anything dreamt up in practical problems can be represented by a tempered distributions.

Some good references for these concepts:

1. E. M. Stein and G. L. Weiss, Introduction to fourier analysis on euclidean spaces, Princeton University Press, 1990, Chapter 1, especially Sections 2 and 3 in that chapter.

2. M. J. Lighthill, "Introduction to fourier analysis and generalised functions", Cambridge University Press, Cambridge, U.K., 1996, Chapter 4, especially section 4.2 and Theorem 17. Lighthill uses the slightly offbeat name "good" function for any Schwartz space member; aside from the unwonted nomenclature, this is an excellent readable reference.

3. J. K. Hunter and B. Nachtergaele, "Applied analysis", World Scientific Publishing Company Incorporated, Singapore, 2005, Chapter 11, section 11.2.

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@user215721 Just added some more "physical" comments –  WetSavannaAnimal aka Rod Vance Sep 22 '13 at 1:47
In scattering situations the behaviour of the fields for all $t\in (-\infty,+\infty)$ is relevant. Consider the case where an electromagnetic wave-packet is scattered from a finite dielectric object such as a sphere. Initially, for large negative times it is moving freely towards the scatterer and after scattering the scattered wave-packet moves freely again, $$\mathbf{E}(\mathbf{x},t)\overset{t\rightarrow\pm\infty}{\sim}\mathbf{E}_{\pm }(\mathbf{x},t),\;\mathbf{B}(\mathbf{x},t)\rightarrow\overset{t\rightarrow \pm\infty}{\sim}\mathbf{B}_{\pm}(\mathbf{x},t),$$
where $\mathbf{E}_{\pm}(\mathbf{x},t)$ and $\mathbf{B}_{\pm}(\mathbf{x},t)$ satisfy Maxwell's equations in vacuum. The standard procedure is then to introduce wave operators, leading to a scattering operator, from which all properties of the scattering process can be obtained. Now there is no given initial condition but rather an expression for the asymptotic motion of the system as $t\rightarrow-\infty$ that fixes the behaviour of the fields. The actual procedure is rather lengthy, see, for instance, A. Tip, Phys. Rev. 56, 4818 (1998), Section VI.
A further remark about Laplace transforms: They are extremely important for the study of a unitary time evolution in Hilbert space (a linear time evolution in the electromagnetic case can be formulated this way). Then, modulo a constant, the Laplace transform of the evolution equals the resolvent $R(z)=[z-H]^{-1}$of its generator $$\int_{0}^{\infty}dt\exp[izt]\exp[-iHt]=-i[z-H]^{-1}$$ (Im$z>0$) and it has turned out that the study of the properties of $H$ are most conveniently done through its resolvent (see Kato's Linear Operators or the volumes by Reed and Simon).
So when we use Fourier transform, we can not set initial conditions in time and it will be set to zero at $-\infty$ automatically. Right? –  user215721 Sep 19 '13 at 15:58