# Why Fourier transform to solve this PDE

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?

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

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

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.