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Recently I have come across this book of Integral Transforms. At this point I realized for myself that there are so many transforms with different kernel out there. Now to provide a little bit of context I would summarize my thoughts concerning Integral Transforms.

First of all there are I would call the classical transforms such as the Fourier Transform or the Laplace Transform, given by

$$\mathcal{F}\{f(t)\}(\omega)~=~\int_{-\infty}^{\infty}f(t)e^{-i\omega\cdot t}~\mathrm{d}t~~~~~~\mathcal{L}\{f(t)\}(s)~=~\int_{0}^{\infty}f(t)e^{-st}~\mathrm{d}t,$$

which are widely used within the theory of differential equations, probability or at the analysis of signal processing. Beside them there are transforms like the Mellin Transform or the Hankel Transform, given by

$$\mathcal{M}\{f(t)\}(s)~=~\int_{0}^{\infty}f(t)t^{s-1}~\mathrm{d}t~~~~~~H_{\nu}\{f(t)\}(u)~=~\int_{0}^{\infty}f(t)J_{\nu}(ut)t~\mathrm{d}t,$$

which are there for some more specific manner since I did not encountered them so often and it is even harder to find transform tables, like this one which contains a lot of transformed functions, than e.g. for the Fourier Transform. To go even further lately I have found out that there are some transform I honestly speaking never heard of before like the Wavelet Transform

$$\mathcal{W}_{\psi}\{x(t)\}(a,b)~=~\frac1{a}\int_{-\infty}^{\infty}\overline{\psi\left(\frac{t-b}{a}\right)}f(t)\mathrm{d}t,$$

where $\psi$ denotes a wavelet as far as I understand, or the two types of the Narain Transform

$$g(z)~=~\int_{0}^{\infty}k(z,y)f(y)~\mathrm{d}y~~~~~~f(z)~=~\int_{0}^{\infty}h(z,y)g(y)~\mathrm{d}y$$

where the kernels $k(z,y)$ and $h(z,y)$ are given by

$$\begin{align} k(z,y)~&=~2\gamma(zy)^{\gamma-1/2}G^{m,p}_{p+q,m+n}\left(\left.\begin{array} \text{\textbf{a}_\textbf{p}},\textbf{b}_\textbf{q}\\\textbf{c}_\textbf{m},\textbf{d}_\textbf{n}\end{array}\right|(zy)^{2\gamma}\right)\\ k(z,y)~&=~2\gamma(zy)^{\gamma-1/2}G^{n,q}_{p+q,m+n}\left(\left.\begin{array} \text{-}\textbf{b}_\textbf{q},-\textbf{a}_\textbf{p}\\-\textbf{d}_\textbf{n},-\textbf{c}_\textbf{m}\end{array}\right|(zy)^{2\gamma}\right) \end{align}$$

where $G^{m,n}_{p,q}$ denotes the Meijer G-function and $\gamma$ the Euler-Mascheroni-constant. Hence the complexity of the kernel is widely scattered I am not sure why there are so many different transforms.

Since some of them can written in terms of other transforms, for example the Mellin Transform and the Fourier Transform

$$\mathcal{M}_f(is)~=~\sqrt{2\pi}\mathcal{F}^{-1}(s)~~~~~~\mathcal{B}_f(i\omega)~=~\mathcal{F}_f(\omega)$$

I cannot justify the variety of transforms for myself. Surely some of them are way more useful hence one can transform nearly every single function because the properties of the kernel allow it. On the other hand some of them seem to be to specific for a small area of application to be useful somewhere else.

I am interested in a more general approach to this whole topic of integral transforms. Therefore I would be satisfied if someone could explain to me why there are so many different integral transforms and not one single generalized which somehow contains all the other ones as special cases. I guess transforms at all are a topic of functional analysis so maybe one could include this to contribute some important facts. Furthermore I would refer to my last, still unanswered questio,n about this topic, Fix points of integral transforms, hence this was the first time I thought about this whole topic in a more general matter.

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I don't know what sort of answer you expect to such a broad question. I'm about to answer a few of the questions you ask. I suspect in each case when you read my answer you'll be inclined to say "Well duh, I knew that...". You might want to refrain from saying that out loud, because if you do my reply will be "Fine. That is the answer to the question - if you already knew the answer then why did you ask?".

Q: Why are there so many integral transforms?

A: Each of the transforms you mention is used for something. Something that the other transforms on the list don't do.

Of course you knew that.

Q: Why aren't they all special cases of some generalized integral transform?

A: They are. By definition if $T$ is an "integral transform" then $T$ is a special case of $$Tf(x)=\int K(x,t) f(t)\,dt.$$

In that much generality there's nothing interesting that can be said about $T$.

Of course you knew that.

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  • $\begingroup$ I do not know exactly what I have done concerning my question that you appear to be kind of pissed of by it. Nevertheless thank you for your response. You are right with your first assumption: I know that each of the transforms is unique in it own way and that they are there to do something special; but could you maybe give an example where it is the case that a transform is incapable to do something another transform could do? $\endgroup$ – mrtaurho Aug 12 '18 at 2:43
  • $\begingroup$ To be honest I was not exact enough concerning the question of generalisation. I would be more interested in the question of a generalized kernel. Yes, I do know that we can write every single transform as the given definition. But this like assume that every single derivation is just a special case of the differential quotient, which is in fact true, but does not yield anywhere. $\endgroup$ – mrtaurho Aug 12 '18 at 2:46
  • $\begingroup$ @mrtaurho Not pissed off at all, just bemused. "but could you maybe give an example where it is the case that a transform is incapable to do something another transform could do?" The question just makes no sense. The Fourier transform gives the Fourier transform of a function - the Hankel transform does not. $\endgroup$ – David C. Ullrich Aug 13 '18 at 14:00
  • $\begingroup$ I am aware of the fact that every single transform is unique in its own way. My question is more about the application of the transforms. Why it is common for example to use the Laplace Transform to solve ODE's and not the Hankel Transform. Ye, of course the Hankel does not gives me the Laplace Transform of a function but why do we need to use the Laplace Transform here? Is this just a historical manner or is there a reason such as the Hankel Transform cannot give you the wanted things? I do not know for example where to use the Hankel Transform so why does it even exist? $\endgroup$ – mrtaurho Aug 18 '18 at 14:18

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