[Math] Fourier transform of heat equation

Question

I need to solve following partial differential equation with Fourier transform numerically.

$
\frac{\partial T}{\partial t} = \nabla(c\nabla T)
$

where T is temperature, c heat conductivity and t is time.

Now the problem is c itself has space dependence. Had it not been after Fourier transform equation would look like
$
\frac{\partial \tilde T}{\partial t} = -k^2c\tilde T
$

How should Fourier transform of first equation look like?

What I am doing is as follows:

1. Take Fourier transform of T. Multiply corresponding values of c(in real space) and T (in Fourier space). i.e. evaluate $g = k\cdot i \cdot c \cdot\tilde T$

2. Take $g$ back to real space. Now $g = c\nabla T$

3. Take $g$ back to Fourier space . Evaulate $f = k \cdot i \cdot \tilde g$

4. Take $f$ to real space. Now $f$ should be $\nabla c \nabla T$

But results of the above procedure are not matching with Finite Difference approach. What am I missing here? Using convolution theorem seems difficult. Is using convolution theorem the only option?

Thanks for any help in advance

Answer 1

To evaluate $\nabla(c\nabla T)$ in Fourier space, you need to do the following. Suppose that you are given $\hat T$, which is the Fourier image of $T$.

1. Compute $\hat g(k)=ik\hat T(k)$. This corresponds to real space gradient.
2. $g= \mathrm{IFT}\,\hat g$, the inverse Fourier transform.
3. $f=cg$.
4. $\hat f = \mathrm{FT}\,f$, the Fourier transform.
5. Compute $\hat r(k)=ik\cdot\hat f(k)$. Note the scalar product. This corresponds to real space divergence.
6. $r= \mathrm{IFT}\,\hat r$, the inverse Fourier transform.

Now you have $r = \nabla\cdot(c\nabla T)$. I think in practice, you don't need step 6, because the left hand side $\partial T/\partial t$ can be computed in Fourier space directly from $\hat T$. You can also write all the steps in one formula $$ \frac{\partial\hat T}{\partial t} = ik\cdot\mathrm{FT} (c\cdot\mathrm{IFT}(ik\hat{T})). $$