For context:
I want to calculate the Fundamental solution for $\frac{\partial^2}{\partial t^2} – \Delta^2$.
My normalization convention is ($\mathcal{F_x}(\varphi))(\xi) := {\frac{1}{(2 \pi)^{n/2}}} \int_{x\in\mathcal{R}^n} e^{-i x \xi} \varphi(x) dx $ as Fourier transform. So the problem reduces to finding the inverse Fourier transform of $(\mathcal{F_{(x,t)}}(f))(\xi,s) := \frac{1}{|\xi|^4+s^2}$ . With $\mathcal{P}(\varphi)(x) := \varphi(-x)$ I get that $ \mathcal{F}^{-1}(\varphi) = \mathcal{F}^3(\varphi) = (\mathcal{F} \circ \mathcal{P}) (\varphi) = \mathcal{F} (\varphi)$ as my function is symmetric, so it doesn't matter whether to calculate the Fourier transform or the inverse.
I can calculate the time-inverse fourier transform of $(\mathcal{F_{(x,t)}}(f))$ and get $g(\xi) := (\mathcal{F_{x}}(f)) = \sqrt{\frac{\pi}{2}} \frac{e^{-|\xi|^2 |t|}}{|\xi|^2}$.
So far so good, but now I run into problems:
I can not directly compute $ \mathcal{F_x}^{-1}(g) = \mathcal{F_x}^{-1} (\sqrt{\frac{\pi}{2}} \frac{e^{-|\xi|^2 |t|}}{|\xi|^2})$ as the $1/|\xi|^2$ part makes it very complicated to me.
I also tried it via $$\mathcal{F_x}^{-1}(u \ v) = \frac{1}{(2\pi)^{n/2}} \mathcal{F_x}^{-1}(u) \ * \mathcal{F_x}^{-1}(v) \\ = \frac{1}{(2\pi)^{n/2}} \sqrt{\frac{\pi}{2}} \ \mathcal{F_x}^{-1}(e^{-|\xi|^2 |t|}) \ * \mathcal{F_x}^{-1}(\frac{1}{|\xi|^2}) \\ = \sqrt{\frac{\pi}{2}} \ \frac{1}{(4 \pi |t|)^{n/2}} e^{-\frac{|x|^2}{4|t|}} * E_{\Delta} $$ with $ E_{\Delta} $ being the fundamental solution to the laplace equation, but also couldn't solve it.
Finally I tried to manually find a function $u$ that has $$\Delta u(x) = \sqrt{\frac{\pi}{2}} \ \frac{1}{(4 \pi |t|)^{n/2}} e^{-\frac{|x|^2}{4|t|}}$$ .
By assuming $u(x) = v(|x|^2)$ I got $$ \Delta u(x) = 2 n v'(|x|^2) + 4 |x|^2 v''(|x|^2) $$ with $n$ the Dimension.
According to wolframalpha my calculations this gives:
$$
v(|x|^2) = \frac{2 \ |t| \ c \ e^{\frac{|x|^2}{4 |t|}}}{n – 2} + \frac{c 2^{n – 1} |x|^2 (-\frac{|x|^2}{|t|})^{-n/2} \ \Gamma(\frac{n}{2}, -\frac{|x|^2}{4 |t|})}{n – 2} + \frac{k_1 |x|^{(2 – n)}}{1 – n/2} + k_2
$$
But this does not seem practical. Is there an easier way to calculate the Fourier transform or some way to simplify the solution?
Edit: I also tried to take the inverse fourier transform of $x_1$ before the time-inverse fourier transform using the residue theorem, but got to something like:
$$
\frac{\pi}{|t| \ \sqrt{b^4+|t|^2}} \frac{e^{ix_1 \ \sqrt{-b^2+i|t|}}+e^{-ix_1 \ \sqrt{-b^2-i|t|}}}{2}
$$ with $b^2 = \sum_{j=2}^n \xi_j^2$ , which is not quite a cosine, so I'm totally lost.
Best Answer
Let us look at the case $n=3$. we use a polar coordinate system where $\vec{\eta}\vec{x}=|\eta||x|\cos(\phi)$. After the simple integral over the second angle we obtain (note the singularity cancelations due to the Jacobian):
$$ I=2\pi\int_{0}^{\infty}d|\eta|\int_0^{\pi}d\phi\sin(\phi)e^{-i\cos(\phi)|\eta||x|}e^{-t|\eta|^2} $$
doing the remaining angular integral:
$$ I=4\pi\int_{0}^{\infty}d|\eta|\frac{\sin{|x||\eta|}}{|x||\eta|} e^{-t|\eta|^2} $$
Now you can show (for example using Feynman's trick) that the last integral equals $$ I=4 \pi \text{erf}\left(\frac{|x|}{2 \sqrt{t}}\right) $$
with $\text{erf}$ being the error function. I don't have the time to work out higher dimensions, but i guess this should generalize at least for odd dimensions.In even dimension Bessel integrals occur which might be also doable.