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After the null-result in my previous question I would like someone to demonstrate this function is it's own Fourier transform:

If we set $f(x,y) = (x^4 - 6x^2 y^2 + y^4) e^{-\pi (x^2 + y^2) t} $ when we'd like to show that $\mathcal{F}(f) $ is $f$ itself. The answer in this previous question states (but does not show) that this is always true for harmonic function. ("Harmonic" here means something like $\nabla^2 u = 0$ since $f = u(x)\, e^{-x^2 }$ basically.)

$$ \widehat{f}(m,n) = \int_0^\infty dx\int_0^\infty dy \; \left[ e^{2\pi i \,(\,mx + ny\,)} \big(x^4 - 6 x^2 y^2 + y^4\big)\,e^{-\pi\,(x^2 + y^2)\,t} \right] $$

You don't have to do the double-Fourier integral, but can very quickly come up with the answer in some other way. These rules seem the way to go:

  • $f(x+h) \longrightarrow \hat{f}(\xi) e^{2\pi i h \xi}$
  • $f(x) e^{-2\pi i xh} \longrightarrow \hat{f}(\xi+h)$
  • $f(\delta x) \longrightarrow \delta^{-1}\hat{f}(\delta^{-1}\xi)$
  • $f'(x) \longrightarrow 2\pi i \xi \hat{f}(\xi)$
  • $-2\pi i x f(x) \longrightarrow \frac{d}{d\xi}\hat{f}(\xi)$

Hopefully now I have a nonzero result . If you wish show that:

$$ \big[ \nabla^2 u(x,y) = 0 \big] \longrightarrow \Bigg[ \mathcal{F}: u(x,y) \, e^{-\pi(x^2 + y^2)t} \to \frac{1}{\sqrt{t}}\,u(a,b) \, e^{-\pi(a^2 + b^2)/t} \Bigg] $$ The original question:

Poisson summation example $ f\big(x,y \big) = \big(x^2 - y^2\big)\,e^{-\pi\,(x^2 + y^2)\,t} $

cactus314
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    The polynomial $\phi(x,y)=x^4-6x^2y^2+y^4$ is harmonic of degree $4$. Then $\overline f=(\pm i)^4f$ where $f(x,y)=\phi(x,y) \exp(-\pi(x^2+y^2))$. – Angina Seng Oct 19 '17 at 14:31
  • @LordSharktheUnknown that's not obvious to me, could you sketch a proof that result? I'd put a check mark to that... – cactus314 Oct 19 '17 at 14:32
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    Show that $(\partial_x+i\partial_y)^k e^{-\pi (x^2+y^2)} = (-2\pi)^k (x+iy)^k e^{-\pi (x^2+y^2)}$ – reuns Oct 19 '17 at 15:28
  • @reuns is that a pseudodifferential operator ? I guess if $k \in \mathbb{N}$ it's OK. – cactus314 Oct 19 '17 at 15:30
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    Of course $k \in \mathbb{N}$ since we are treating polynomials. Since $\partial_x\partial_y = \partial_y\partial_x$, $(\partial_x+i\partial_y)^k$ is well-defined. – reuns Oct 19 '17 at 15:32
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    If $f(x,y) = \sum_{k=0}^d c_k (x+iy)^k$ then $(\partial_x+i\partial_y) f(x,y) = 0$. Thus $(\partial_x+i\partial_y) f(x,y) e^{-\pi (x^2+y^2)} = ?$ Finally write your harmonic polynomial as $ \sum_{k=0}^d c_k (x+iy)^k+ \sum_{k=0}^d b_k (x-iy)^k$ use what I said and conclude with the FT of the Gaussian and the derivative property of the FT. – reuns Oct 19 '17 at 15:38
  • Can you put the general harmonic polynomial case in an answer now ? – reuns Oct 19 '17 at 19:23
  • @reuns I'll answer my own questions. – cactus314 Oct 19 '17 at 21:18

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