# Velocity Field In Navier-Stokes Equation

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: $$\frac{\partial \textbf{u}}{\partial x} \dot{x} + \frac{\partial \textbf{u}}{\partial y} \dot{y} + \frac{\partial \textbf{u}}{\partial z} \dot{z} + \frac{\partial \textbf{u}}{\partial t} = \textbf{u} \cdot \nabla \textbf{u} + \partial_t \textbf{u}$$

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 $$\dfrac{d\vec{u}}{dt} + \left(\vec{u}\cdot\vec{\nabla}\right)\vec{u}$$.

$$\textbf{u}$$ 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

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

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.