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} $