I want to numerically solve the diffusion equation $\partial_t u = D \partial_x^2 u$ in Fourier space, and can think of multiple ways to do it.
Setup
Option 1
Differentiating $u$ twice in Fourier space brings down two factors of $ik$ from the exponential $e^{i k x}$.
$$\partial_t \tilde{u}_k = D (ik)^2 \tilde{u}_k$$
which can be discretised, for example, like:
$$\frac{\tilde{u}^{n+1} - \tilde{u}^n}{\Delta t} = -Dk^2\left(\frac{\tilde{u}^{n+1} + \tilde{u}^{n}}{2} \right)$$
Option 2
Taking the second order finite difference equation in real space:
$$\partial_x^2 u \approx \frac{u_{i+1} - 2 u_i + u_{i-1}}{\Delta x^2}$$
where $u_{i+1}$ and $u_i$ are $\Delta x $ apart. Fourier transforming this, remembering that phase shift of $\Delta x$ is given by $e^{\Delta x k}$ we get:
$$\partial_x^2 \tilde{u}_k \approx \frac{e^{+\Delta x k} -2 + e^{-\Delta x k} }{\Delta x^2} \tilde{u}_k = 2 \frac{\cos(\Delta x k) - 1}{\Delta x^2} \tilde{u}_k\qquad(1)$$
Applying again to the diffusion equation:
$$\frac{\tilde{u}^{n+1}_k - \tilde{u}^n_k}{\Delta t} = D\frac{2 (\cos(\Delta x k)-1)}{\Delta x^2} \left(\frac{\tilde{u}^{n+1}_k + \tilde{u}^{n}_k}{2} \right)$$
I came across Option 2 in a few places, for example, in this paper.
Question
Which of these (if either) should I be using, and why? I'm guessing the discrete nature of the FFT is important, as the approaches are equivalent as $\Delta x \to 0$, seen by Taylor expanding Eq. (1):
$$2 \frac{\cos(\Delta x k) - 1}{\Delta x^2} = 2 \frac{1 - \frac{(\Delta x k)^2}{2} + \mathcal{O}(\Delta x^4) - 1}{\Delta x^2} = -k^2 + \mathcal{O}(\Delta x^2)$$
Any help is greatly appreciated!
Update
Thinking more about this, the properties of the DFT are actually important. The maximum $k$ is $k_{max} =2\pi (N/2)$ where $\Delta x = L / N$. So as $\Delta x \to 0$, $k_{max} \to \infty$. Therefore the two methods are different for large $k$ at any $\Delta x$:
This looks like the latter option is applying some smoothing at high $k$.