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Acceleration of fluid particle

  1. Feb 16, 2016 #1
    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. jcsd
  3. Feb 16, 2016 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    No, what your lecturer did is correct. Perhaps what you are missing is that [itex]\vec{u}\cdot\nabla[/itex], with [itex]\vec{u}= u\vec{i}+ v\vec{j}[/itex] is equal to [itex]u\frac{\partial }{\partial x}+ v\frac{\partial}{\partial y}[/itex] and that, being a scalar operator "multiplied" by [itex]\vec{u}= u\vec{I}+ v\vec{j}[/itex] is "multiplied" by both u and v: [tex](\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)[/tex]
    [tex]= \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)[/tex].
     
  4. Feb 16, 2016 #3
    Thanks for the reply.

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

    [tex](\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)[/tex]


    And, regarding the the final result:

    [tex]= \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)[/tex]

    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:

    538885b73da17a0e7211a97108354461.png

    Thanks.
     
  5. Feb 16, 2016 #4

    pasmith

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    Homework Helper

    Something's gone wrong here. If [itex]\mathbf{u} = (u,v)[/itex] with respect to cartesian coordinates [itex](x,y)[/itex] then indeed [tex]\mathbf{u} \cdot \nabla = u \frac{\partial}{\partial x} + v \frac{\partial}{\partial y}[/tex] but [tex]
    (\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)[/tex] and hence [tex]
    \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).[/tex]
     
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