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I'm reading a paper and has the linearized NS equation and follows it by getting the solution through a Fourier transform. What is the thought behind this? Meaning, why use a Fourier transform?

$ \rho\frac{\partial \mathbf{v}}{\partial t} = -\nabla \mathbf{p} + \eta \nabla^2 \mathbf{v}$

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  • $\begingroup$ Because it’s a standard technique in PDE theory. $\endgroup$
    – Alex R.
    Feb 17, 2022 at 5:34
  • $\begingroup$ Because it transforms gradient into $x$ $\endgroup$
    – LL 3.14
    Feb 17, 2022 at 7:18
  • $\begingroup$ $\mathbf{p}$ is the pressure and should not be marked as a vector, should it? $\endgroup$
    – md2perpe
    Feb 17, 2022 at 10:48

1 Answer 1

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Applying the Fourier transform on the space coordinates $\mathbf{x} = (x_1,x_2,x_3)$ turns the PDE into an ODE: $$ \rho\frac{d\hat{\mathbf{v}}}{dt} = -i\mathbf{k} \hat{p} - \eta k^2 \hat{\mathbf{v}}, $$ where $\mathbf{k}=(k_1,k_2,k_3)$ is the new variable (often called wave vector), $k=|\mathbf{k}|,$ and $\hat{}$ denotes transformed quantity. I have here assumed that $\rho$ and $\eta$ are constants. The symbol $i$ is the imaginary unit.

ODE's are usually easier to solve than PDE's. For example, the above equation has solutions $$ \hat{\mathbf{v}} = -i \mathbf{k} e^{-\frac{\eta}{\rho}k^2t} \int e^{\frac{\eta}{\rho}k^2t} \hat{p} \, dt + e^{-\frac{\eta}{\rho}k^2t} C(\mathbf{k}), $$ where $\int \cdot \, dt$ denotes anti-derivative (primitive function) and $C$ is some rather arbitrary function.

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