I'm trying to find the Kernel of the 3D Diffusion Equation but my lecture notes are unclear on the procedure.
$$\frac{\partial u}{\partial t} = D \nabla ^2u$$ $$u(\vec{x}, 0) = R(|\vec{x}|)$$
I need to show that the solutions might be written as: $$u(|\vec{x}|, t) = \int ^{+\infty}_0 dr r^2 R(r)K(r, |\vec{x}|; t)$$ where $K$ is the kernel which I also need to find explicitly.
THIS IS WHAT I HAVE TRIED:
I started by transforming the equation: $\frac{\partial \tilde{u}}{\partial t} = -Dk^2\tilde{u} \Leftrightarrow \tilde{u}(k,t) = \tilde{u}(k,0)\exp(-Dk^2t)$
Then using Radial / Spherical Polar Fourier Transform: $$u(x,t) = \sqrt{\frac{2}{\pi}}\int^{+\infty}_0k^2\frac{\sin(kr)}{kr}\tilde{u}(k,t)dk = \sqrt{\frac{2}{\pi}}\int^{+\infty}_0k^2\frac{\sin(kr)}{kr}\tilde{u}(k,0)\exp{(-Dk^2t)}dk$$ $$u(x,t) = \frac{2}{\pi}\int^{+\infty}_0k^2\frac{\sin(kr)}{kr}\exp{(-Dk^2t)}dk\int^{+\infty}_0r^2\frac{\sin(kr)}{kr}R(r)dr$$ $$u(x,t) = \int^{+\infty}_0r^2R(r)K(r, |\vec{x}|;t)dr$$
where here the kernel is: $\frac{2}{\pi}\int^{+\infty}_{0}k^2\frac{\sin(kr)}{k^2r^2}\exp{(-Dk^2t)}dk$
I don't know how to handle the integral and I don't even know if what I am doing is good.
Any ideas?
Many thanks!