# Proving the 3D Diffusion Equation from the 3D Fourier Transform

I am trying to prove the 3D Diffusion Equation $$\begin{cases}u_t(\vec x,t) &= c\nabla u\\ u(\vec x,0)&=g(\vec x)\end{cases}$$ From the 3D Fourier Transform, where $\vec k,\vec x\in\mathbb R^3$.

I took the Fourier Transform of both sides of the 3D Diffusion to get:

$$\widehat u_t(k,t)+c|\vec k|^2\widehat u(k,t) = 0$$

Now, this gives that the solution to the ODE is: $$\widehat u(k,t) = \widehat g(k) e^{-ct|\vec k|^2} = \widehat g(\vec k)\cdot\widehat f(\vec k)$$

Now, to find the solution, I want to find the Inverse FT of $f$, whcih leads to: $$\frac{1}{(2\pi)^3}\iiint_{\mathbb R^3}e^{-ct|\vec k|^2+ikx} d\vec k$$ So isn't $|\vec k|^2$ a scalar? If not, how would I take the integral of the integrand?

$|k|^2$ is a scalar, but $\vec k$ is a vector. Note that also, in your last equation, $x$ should be $\vec x$, so the last expression is $$\frac{1}{2\pi}\iiint_{\mathbb R^3}e^{-ct|\vec k|^2+i\vec k\vec x} d\vec k$$ You can either solve the integral in Cartesian coordinates, or change it to spherical coordinates. $|\vec k|^2$ will transform into $k^2$, $\vec k \vec x$ is $kx\cos\theta$, and $d\vec k$ is $k^2\sin\theta dk d\theta d\phi$. The integration limits for $\phi$ is from $0$ to $2\pi$, $theta$ goes from $0$ to $\pi$, and $k$ is from $0$ to $\infty$
• Wouldn't it be a lot harder in spherical...? And so wouldn't I just have to integrate with respect to the vector $k$? Or do I have to consider the scalar $|\vec k|$ there as well? Nov 6 '17 at 2:37
• The scalar $k$ is the "radius" in spherical coordinates. In Cartesian coordinates $|\vec k|^2=k_x^2+k_y^2+k_z^2$, and $\vec k\vec x=k_x x$ if you choose the appropriate coordinate system. And $d\vec k=dk_x dk_y dk_z$ Nov 6 '17 at 2:41
• Which one is the easier way to integrate this? Both of them seem like a lot of work... The Cartesian you would have you integrate over $k_1, k_2,k_3$ given $\vec k=<k_1,k_2,k_3>$. But Spherical you would have to do crazy integration by parts to solve this, both of them seem very ridiculous. Is there a trick I am missing here? Nov 6 '17 at 2:51