## Monday Math 129

Continuing from last week, we now consider the Fourier transform of the divergence of a vector field in three dimensions. First, vector integration by parts tells us that for vector field v and scalar field φ
$\iiint_{V}\phi\mathbf{\nabla}\cdot\mathbf{v}\,dV=\iint_{S}\phi\mathbf{v}\cdot{d}\mathbf{A}-\iiint_{V}\mathbf{\nabla}\phi\cdot\mathbf{v}\,dV$.
Letting $\phi(\mathbf{x})=e^{-\imath\mathbf{k}\cdot\mathbf{x}}$, this means
$\iiint_{V}e^{-\imath\mathbf{k}\cdot\mathbf{x}}\mathbf{\nabla}\cdot\mathbf{v}\,dV=\iint_{S}e^{-\imath\mathbf{k}\cdot\mathbf{x}}\mathbf{v}\cdot{d}\mathbf{A}-\iiint_{V}\left(\mathbf{\nabla}e^{-\imath\mathbf{k}\cdot\mathbf{x}}\right)\cdot\mathbf{v}\,dV$,
and since $\mathbf{\nabla}e^{-\imath\mathbf{k}\cdot\mathbf{x}}=-\imath\mathbf{k}e^{-\imath\mathbf{k}\cdot\mathbf{x}}$, as before, we then have
$\iiint_{V}e^{-\imath\mathbf{k}\cdot\mathbf{x}}\mathbf{\nabla}\cdot\mathbf{v}\,dV=\iint_{S}e^{-\imath\mathbf{k}\cdot\mathbf{x}}\mathbf{v}\cdot{d}\mathbf{A}-\iiint_{V}e^{-\imath\mathbf{k}\cdot\mathbf{x}}(-\imath\mathbf{k}\cdot\mathbf{v})\,dV$
$\iiint_{V}e^{-\imath\mathbf{k}\cdot\mathbf{x}}\mathbf{\nabla}\cdot\mathbf{v}\,dV=\iint_{S}e^{-\imath\mathbf{k}\cdot\mathbf{x}}\mathbf{v}\cdot{d}\mathbf{A}+\imath\mathbf{k}\cdot\iiint_{V}e^{-\imath\mathbf{k}\cdot\mathbf{x}}\mathbf{v}\,dV$.
Now, if v goes to zero sufficiently fast as r goes to infinity (|v| goes to zero faster than 1/r), then the surface integral will vanish as the volume is expanded to all space, and so
$\iiint{e}^{-\imath\mathbf{k}\cdot\mathbf{x}}\mathbf{\nabla}\cdot\mathbf{v}\,dV=\imath\mathbf{k}\cdot\iiint{e}^{-\imath\mathbf{k}\cdot\mathbf{x}}\mathbf{v}\,dV$,
and so
$\mathcal{F}\left[\mathbf{\nabla}\cdot\mathbf{v}(\mathbf{x})\right](\mathbf{k})=\imath\mathbf{k}\cdot\mathcal{F}\left[\mathbf{v}(\mathbf{x})\right](\mathbf{k})$.

Now, to find the Fourier transform of the curl of a vector field, we use a form of the divergence theorem which states that
$\iiint_{V}\mathbf{\nabla}\times\mathbf{P}\,dV=\iint_{S}\mathbf{\hat{n}}\times\mathbf{P}\,dA$,
where $\mathbf{\hat{n}}$ is the outward normal to the surface S bounding the volume V.
Now, the product rule for the curl of the product of a scalar field and a vector field is
$\mathbf{\nabla}\times(\psi\mathbf{v})=\mathbf{\nabla}\psi\times\mathbf{v}+\psi(\mathbf{\nabla}\times\mathbf{v})$,
so letting P=ψv, we see
$\iiint_{V}\mathbf{\nabla}\psi\times\mathbf{v}\,dV+\iiint_{V}\psi(\mathbf{\nabla}\times\mathbf{v})\,dV=\iint_{S}\mathbf{\hat{n}}\times(\psi\mathbf{v})\,dA$,
so
$\iiint_{V}\psi(\mathbf{\nabla}\times\mathbf{v})\,dV=\iint_{S}(\mathbf{\hat{n}}\times\mathbf{v})\psi\,dA-\iiint_{V}(\mathbf{\nabla}\psi)\times\mathbf{v}\,dV$.

Now, letting $\psi(\mathbf{x})=e^{-\imath\mathbf{k}\cdot\mathbf{x}}$, and using
$\mathbf{\nabla}\psi=\mathbf{\nabla}e^{-\imath\mathbf{k}\cdot\mathbf{x}}=-\imath\mathbf{k}e^{-\imath\mathbf{k}\cdot\mathbf{x}}$,
we get
$\iiint_{V}e^{-\imath\mathbf{k}\cdot\mathbf{x}}(\mathbf{\nabla}\times\mathbf{v})\,dV=\iint_{S}(\mathbf{\hat{n}}\times\mathbf{v})e^{-\imath\mathbf{k}\cdot\mathbf{x}}\,dA-\iiint_{V}(-\imath\mathbf{k}e^{-\imath\mathbf{k}\cdot\mathbf{x}})\times\mathbf{v}\,dV$,
and thus,
$\iiint_{V}e^{-\imath\mathbf{k}\cdot\mathbf{x}}(\mathbf{\nabla}\times\mathbf{v})\,dV=\iint_{S}(\mathbf{\hat{n}}\times\mathbf{v})e^{-\imath\mathbf{k}\cdot\mathbf{x}}\,dA+\imath\mathbf{k}\times\iiint_{V}e^{-\imath\mathbf{k}\cdot\mathbf{x}}\mathbf{v}\,dV$.
Again, we can expand the volume to all space, and if v goes to zero fast enough as r goes to infinity, then the surface integral vanishes, so
$\iiint{e}^{-\imath\mathbf{k}\cdot\mathbf{x}}(\mathbf{\nabla}\times\mathbf{v})\,dV=\imath\mathbf{k}\times\iiint{e}^{-\imath\mathbf{k}\cdot\mathbf{x}}\mathbf{v}\,dV$,
and thus
$\mathcal{F}\left[\mathbf{\nabla}\times\mathbf{v}(\mathbf{x})\right](\mathbf{k})=\imath\mathbf{k}\times\mathcal{F}\left[\mathbf{v}(\mathbf{x})\right](\mathbf{k})$.

### 4 Responses to “Monday Math 129”

1. Monday Math 130 « Twisted One 151's Weblog Says:

[…] of F(x), its divergence, and its curl. With the Fourier transform of F, then our results from last week tell us that . and . Now, the ability to decompose a vector into components parallel and […]

2. m.diehl Says:

I have a question concerning the calculation of divergence in Fourier space.
I have a 3D field of 3×3 tensors in real space that I transform to Fourier space using FFT. Now I calculate the divergence by multiplying the tensor field in Fourier space with the corresponding frequency vector. The result should be the divergence in fourier space that I back transform.
My Problem is now, that I have imaginary parts in this divergence field. If I do a direct divergence calculation, using finite differences, the divergence is real only. That makes much more sense.
Any Ideas?

• twistedone151 Says:

Unfortunately, I’m not familiar with how the Fourier transform extends to (rank 2 or higher) tensors. However, I would note that FFT is an algorithm for computing a discrete Fourier transform, not the continuous Fourier transform, so I’m not sure it works for computing divergence.

• Martin Diehl Says:

Well, I found the solution.
When differentiating (e.g. calculating the divergence) you shouldn’t use the highest frequencies as they are not correctly represented using the DFT/FFT. Just set the to zero and use the formulas derive for the continuous FT