# Acceleration of fluid particle

1. Feb 16, 2016

### sa1988

1. The problem statement, all variables and given/known data

Find acceleration for a general point (x, y) at time t, in the velocity field u = (y, t-x)

2. Relevant equations

Du/Dt = ∂u/∂t + (u⋅∇)u

3. The attempt at a solution

u/∂t = (0, 1)

(u⋅∇)u = [(y, t-x)⋅(∂/∂x, ∂/∂y)](y, t-x)

= (0 + 0)(y, t-x) = (0, 0)

Thus Du/Dt = ∂u/∂t + (u⋅∇)u = (0, 1) + (0, 0)

= (0, 1)

It seems correct but my lecturer has given a different example, for a different problem, in which he draws out the entire process (presumably for the sake of letting students follow every tiny step) but I think there are typos in his work and it's really confusing things for me. So I've stuck with what I believe to be correct, but would like to run it by you guys to see if things are all ok.

Thanks!

For reference, my lecturer's example is here - http://www1.maths.leeds.ac.uk/~kersale/2620/Examples/sol1.pdf - The part on the final page.

It looks like he's written ∂/∂x in far too many places...?

2. Feb 16, 2016

### HallsofIvy

No, what your lecturer did is correct. Perhaps what you are missing is that $\vec{u}\cdot\nabla$, with $\vec{u}= u\vec{i}+ v\vec{j}$ is equal to $u\frac{\partial }{\partial x}+ v\frac{\partial}{\partial y}$ and that, being a scalar operator "multiplied" by $\vec{u}= u\vec{I}+ v\vec{j}$ is "multiplied" by both u and v: $$(\vec{u}\cdot\nabla)\vec{u}= \left(\vec{u}\cdot\nabla(u\vec{i}+ v\vec{j})\right)+ \left(\vec{v}\cdot\nabla(u\vec{i}+ v\vec{j})\right)$$
$$= \left(u\frac{\partial u}{\partial x}+ u\frac{\partial u}{\partial y}\right)+ \left(v\frac{\partial u}{\partial x}+ v\frac{\partial u}{\partial y}\right)+ \left(u\frac{\partial v}{\partial x}+ u\frac{\partial v}{\partial y}\right)+ \left(v\frac{\partial v}{\partial x}+ v\frac{\partial v}{\partial y}\right)$$.

3. Feb 16, 2016

### sa1988

I'm a little confused though. I can't see how you get this bit:

$$(\vec{u}\cdot\nabla)\vec{u}= \left(\vec{u}\cdot\nabla(u\vec{i}+ v\vec{j})\right)+ \left(\vec{v}\cdot\nabla(u\vec{i}+ v\vec{j})\right)$$

And, regarding the the final result:

$$= \left(u\frac{\partial u}{\partial x}+ u\frac{\partial u}{\partial y}\right)+ \left(v\frac{\partial u}{\partial x}+ v\frac{\partial u}{\partial y}\right)+ \left(u\frac{\partial v}{\partial x}+ u\frac{\partial v}{\partial y}\right)+ \left(v\frac{\partial v}{\partial x}+ v\frac{\partial v}{\partial y}\right)$$

Is there a way of writing that in vector form?

One thing I did notice is that I messed up by forgetting that in (u⋅∇) , ∂/∂x is not commutative, so I was doing (∇⋅u)u, which is different.

So now I'm getting

(u⋅∇)u = (ux, uu)⋅(∂/∂x, ∂/∂y) (ux + uy)
= (ux∂/∂x + uy∂/∂y) (ux + uy)
= (ux∂(ux)/∂x + ux∂(uy)/∂x + uy∂(ux)/∂y + uy∂(uy)/∂y)

Which I then apply to the given velocity field u = (y, t-x), to get:

Du/Dt = ∂u/∂t + (u⋅∇)u

= (0, 1) + (y ∂/∂x + (t-x) ∂/∂y) * (y, t-x)
= (0, 1) + (y ∂(y)/∂x + (t-x)∂(y)/∂y, y ∂(t-x)/∂x + (t-x)∂(t-x)/∂y)
= (0, 1) + ( 0 + t-x , -y + 0)

= (t-x, 1-y)

Have I gone wrong here?

I've found this on wikipedia, which seems to reflect the same process I've just used:

Thanks.

4. Feb 16, 2016

### pasmith

Something's gone wrong here. If $\mathbf{u} = (u,v)$ with respect to cartesian coordinates $(x,y)$ then indeed $$\mathbf{u} \cdot \nabla = u \frac{\partial}{\partial x} + v \frac{\partial}{\partial y}$$ but $$(\mathbf{u} \cdot \nabla)\mathbf{u} = \left( u \frac{\partial u}{\partial x} + v \frac{\partial u}{\partial y}, u \frac{\partial v}{\partial x} + v \frac{\partial v}{\partial y} \right)$$ and hence $$\frac{D\mathbf{u}}{Dt} = \left(\frac{\partial u}{\partial t} + u \frac{\partial u}{\partial x} + v \frac{\partial u}{\partial y}, \frac{\partial v}{\partial t} + u \frac{\partial v}{\partial x} + v \frac{\partial v}{\partial y} \right).$$