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How I could evaluate Gabor's Uncertainty relation for a system with finite duration?

Intro____________

In this another question I think I found an ansatz for solving the PDE $$ \Delta |u|^{\frac12} = 0$$ through a particular singular solution given by $$u(x,t) = \frac{\text{sgn}(u(0,0))}{4}\left(2\sqrt{|u(0,0)|}-(x+t)\right)^2\cdot\theta\!\left(2\sqrt{|u(0,0)|}-(x+t)\right)$$ with $\theta(t)$ the Heaviside step function (I am still not sure if it is completely right).

animation of the solution as time pases

Question______

How I could evaluate Gabor's Uncertainty relation for the system of $u(x,t)$?

My initial plan was to find its Fourier Transform, then take the second moment in time and in the frequencies, and evaluate Gabor's relation, but I tried to take the bidimensional Fourier Transform but got lost on the variable integration limits. Also, I am conceptually confused about if I have to take the Bidimensional Fourier Transform, or instead only a Fourier Transform in the time variable... I would like to emulate what is done in quantum muchanics but for this classical system (if it makes sense).

Gabor's paper section I would like to reproduce

Gabor's paper

Gabor's paper could be found for free in Google as PDF file: "Theory of communication", D. Gabor (1947)

Joako
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    You have a function of two variables, while Gabor is dealing with uncertainty principles for functions of one variable. It stands to reason that you are confused. Just as there are generalisations of Fourier theory to functions of many variables, there are also such generalisations of uncertainty principles, and you should probably study one of those for you situation? If you want to evaluate exactly the relation Gabor mentions, then you're restricted to dealing with functions of one variable, so you could study the behaviour of $f_t(x) := u(x,t)$ for a fixed $t$, or similarly for a fixed $x$. – stochasticboy321 Nov 01 '24 at 00:19
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    An example of a multivariate uncertainty principle is the following: for a probability density $\pi$, let $V(\pi)$ be the covariance matrix induced by $\pi$. Take some $f \in L_2(\mathbb{R}^n)$ such that $\int |f|^2 = 1$ (so $|f|^2$ is density on $\mathbb{R}^n$). If $\hat{f}$ denotes the Fourier transform of $f$, then $\det V(|f|^2) \cdot \det V(|\hat{f}|^2) \ge (16\pi^2)^{-n}$ (see, e.g., Corollary 5.7 here for this, and the rest of the paper for more). – stochasticboy321 Nov 01 '24 at 00:19
  • @stochasticboy321 about your first comment, you are right, I am confused... understanding your point about the paper, I think it is still valid for more variables, since it is used in quantum mechanics to relate position (something related to space) with momentum (something related to time), and I was trying to do a similar evaluation for the function $u(x,t)$, but I don't know how, neither if it is possible or not and why – Joako Nov 01 '24 at 01:39
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    The linked question is for $\Delta$ but here you write $\nabla$. $\nabla$ is gradient. $\Delta$ is the Laplacian. Note that at every point where $u\neq0$, $\nabla |u|^{1/2} = \frac12|u|^{-3/2} u \nabla u$ which is the zero vector iff $u\partial_xu = u\partial_yu=0$. Since $u\neq 0$ one requires $\partial_x u=0=\partial_y u = 0$. So $u$ is piecewise constant on the components of ${(x,y): u \neq 0}$. So if $u=0$ somewhere and is continuous, it is 0 everywhere – Calvin Khor Nov 02 '24 at 19:45
  • @CalvinKhor Thanks for noticing it, it was a typo. – Joako Nov 02 '24 at 19:55
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    OK. Well IDK about Gabor but your function does not satisfy $\Delta |u|^{1/2}=0$. If it did, then of course $v := |u|^{1/2}$ solves $\Delta v = 0$, i.e. $v$ would be harmonic. But (non-zero) harmonic functions cannot be compactly supported or have discontinuities. – Calvin Khor Nov 02 '24 at 19:59
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    Indeed, following @CalvinKhor comments, even for tempered distributions $v$, applying Fourier transform, $r^2\cdot \widehat{v}=0$, so $\widehat{v}$ is a distribution supported on ${0}$, so $v$ must have been a (harmonic) polynomial. :) – paul garrett Nov 02 '24 at 20:10
  • @CalvinKhor In the other question using Wolfram-Alpha I think I show that the piecewise function I found it is indeed a solution... please take a look to see where I messed it up and give an answer accordingly. – Joako Nov 02 '24 at 20:15
  • @paulgarrett In the other question using Wolfram-Alpha I think I show that the piecewise function I found it is indeed a solution... please take a look to see where I messed it up and give an answer accordingly. – Joako Nov 02 '24 at 20:16
  • Please do not rely on pictures of text. – Shaun Nov 02 '24 at 21:35
  • @Shaun I have added the DOI direction of the paper, and also it's name since you could find plenty of copies in pdf file if you Google it (is a very famous paper on basics of telecommunication) – Joako Nov 03 '24 at 00:58
  • @CalvinKhor I have been thinking in your comment: I know that non-piecewise power series cannot become a constant value, that is why the function I made it is defined piecewise, but I don't know if the same applies or not for harmonic functions (at least is not clear for me from what I see in Wikipedia: since the trivial zero function is harmonic, (...) – Joako Nov 03 '24 at 01:09
  • @CalvinKhor (...) and a power series of the form $f(\eta)=\sum_{k=0} a_k (T-\eta)^k\theta(T-\eta)$ with $a_0=a_1=a_2=0$ will becoming zero at $\eta=T$ with continuous $f,\ f',\ f''$ I don't understand why $u(x,t)=f(x+t)$ would not be harmonic: when its zero it is harmonic, and where it is not zero it should be also harmonic (since it is an entire function), and in the boundaries near $x+t=T$ I believe it is also harmonic since the first derivatives are also continuous. – Joako Nov 03 '24 at 01:12

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