Velocity Field In Navier-Stokes Equation

Question

In the Book by Chorin and Marsden, the velocity field is written: $$\textbf{u}(x(t),y(t),z(t),t)$$

This does not make sense to me though -- if I think about a velocity field in three-space, it's just a bunch of vectors pointing in different directions at different times i.e. : $$\textbf{u}(x(t),y(t),z(t))$$

Where does this extra time dependence in the vector field come from?

This subtle distinction has major implications: the LHS of Navier-Stokes equation would be completely different in the second case: $$\frac{\partial \textbf{u}}{\partial x} \dot{x} + \frac{\partial \textbf{u}}{\partial y} \dot{y} + \frac{\partial \textbf{u}}{\partial z} \dot{z} = \textbf{u} \cdot \nabla \textbf{u} $$ as opposed to the usual: $$\frac{\partial \textbf{u}}{\partial x} \dot{x} + \frac{\partial \textbf{u}}{\partial y} \dot{y} + \frac{\partial \textbf{u}}{\partial z} \dot{z} + \frac{\partial \textbf{u}}{\partial t} = \textbf{u} \cdot \nabla \textbf{u} + \partial_t \textbf{u}$$

Answer 1

You're right, there are two points of view from which to analyze fluid flow. The first way, called "Eulerian," treats the velocity as a field in space that's changing in time, so that $(x,y,z)$ are fixed coordinates and $\vec{u}(x,y,z,t)$ gives the fluid velocity at each location at time $t$. That's the framework we usually use for the N.S. equations. The other way, called "Lagrangian," treats the fluid as a collection of infinitesimal fluid parcels. Then $(x,y,z)=(x(t),y(t),z(t))$ are the coordinates of a particular fluid parcel at time $t$, and $\vec{u} = \left(\dfrac{dx(t)}{dt},\dfrac{dy(t)}{dt},\dfrac{dz(t)}{dt}\right)$ is the velocity of that parcel.

Physically, the acceleration on the LHS of the NS equation is the acceleration of a fluid parcel, and so in the Lagrangian picture would be just $\dfrac{d\vec{u}}{dt}$, but translating that into Eulerian terms gives the convective derivative $\dfrac{d\vec{u}}{dt} + \left(\vec{u}\cdot\vec{\nabla}\right)\vec{u}$.

Answer 2

$\textbf{u}$ is a function of $x,y,z,t$, but $x,y,z$ are in turn functions of $t$. So $\textbf{u}$ is the velocity experienced by a particle following the path $\textbf{r}(t) = \left(x(t), y(t), z(t) \right)$. So we have

$\displaystyle \frac {d \textbf{u}}{dt} = \left( \frac {d \textbf{r}}{dt}. \nabla \right) \textbf{u} + \frac {\partial \textbf{u}}{ \partial t}$

If we assume that the particle is moving with the fluid (or, more simply, the particle is a small element of the fluid) then we have $\displaystyle \frac {d \textbf{r}}{dt} = \textbf{u}$ and so

$\displaystyle \frac {d \textbf{u}}{dt} = \left( \textbf{u}.\nabla \right) \textbf{u} + \frac {\partial \textbf{u}}{ \partial t}$

Answer 3

Lets say I have two packets of fluid which follow trajectories $\mathbf{r}_1(t)$ and $\mathbf{r}_2(t)$ and for some $t_1 \ne t_2 $ we have $\mathbf{r}_1(t_1) = \mathbf{r}_2(t_2)$, that is that one fluid packet passes through the spot the other occupied some time earlier. Is it necessarily true that $\mathbf{u}(\mathbf{r}_1(t_1)) = \mathbf{u}(\mathbf{r}_2(t_2))$? In other words if I pick a fixed point in space will I always observe the same velocity for the fluid at that point?

Now if you have ever looked at the water in a sink or stirred a cup of coffee you know that the answer to this question is no. Therefore the field $\mathbf{u}$ must have its own time dependence, independent of the time dependence of the point that we are looking at.