- #1
dakold
- 15
- 0
As show below the Fourier Transform of Naiver-Stokes equation. I wonder if the pressure should be in the Fourier transform? In the below transformation there is no pressure.
N-S
[tex]\frac{\partial\vec{u}}{\partial t}\+(\vec{u}\bullet\nabla)\vec{u}=-\frac{\nabla P}{\rho}+\nu\nabla^{2}[/tex]
N-S in Fourier space [tex]\frac{\partial u_{\alpha}}{\partial t}=-i\int(k_{\beta}-p_{\beta}u_{\beta}(\vec{p})u_{\alpha}(\vec{k}-\vec{p})d^{3}p+i\frac{k_{\alpha}}{k^{2}}\int(p_{\gamma}(k_{\beta}-p_{\beta})u_{\beta}(\vec{p})u_{\gamma}(\vec{k}-\vec{p})
d^{3}p-\nu k^{2} u_{\alpha}(\vec{k})[/tex]
I got another question also, how can one interpretate the second term on the left hand side?
[tex](\vec{u}\bullet\nabla)\vec{u}[/tex]
Is this acceleration?
N-S
[tex]\frac{\partial\vec{u}}{\partial t}\+(\vec{u}\bullet\nabla)\vec{u}=-\frac{\nabla P}{\rho}+\nu\nabla^{2}[/tex]
N-S in Fourier space [tex]\frac{\partial u_{\alpha}}{\partial t}=-i\int(k_{\beta}-p_{\beta}u_{\beta}(\vec{p})u_{\alpha}(\vec{k}-\vec{p})d^{3}p+i\frac{k_{\alpha}}{k^{2}}\int(p_{\gamma}(k_{\beta}-p_{\beta})u_{\beta}(\vec{p})u_{\gamma}(\vec{k}-\vec{p})
d^{3}p-\nu k^{2} u_{\alpha}(\vec{k})[/tex]
I got another question also, how can one interpretate the second term on the left hand side?
[tex](\vec{u}\bullet\nabla)\vec{u}[/tex]
Is this acceleration?