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Fluid dynamics, solving ODE to find particle path

  1. Feb 9, 2016 #1
    1. The problem statement, all variables and given/known data

    A time-dependent two-dimensional fluid flow is given, in a Cartesian coordinates system (x, y), by the velocity field:

    u = (y, t-x)

    Show that, at time t = 0, the streamlines of this flow are circles centred on the origin.
    Find equation of the streamline that passes through the point (0, 1)
    Find the equation of the path of a particle released at the point (0,1) at t=0.

    2. Relevant equations

    For streamlines in u = (u, v):
    dx(t)/u = dy(t)/w

    Particle path:
    dx/dt = u
    dy/dt = v

    3. The attempt at a solution

    I think I'm okay with the streamline part but here it is anyway:

    Streamline:
    Using the given equation for finding streamlines, plug in the values and rearrange to get:

    (t-x) dx = y dy

    y2/2 = tx - x2/2 + C

    Rearrange to show x2+y2 = 2tx + C

    For t = 0 , it reduces to x2+y2 = + C,
    which is a circle.

    Streamline passing though (0, 1) is given in:
    x2+y2 = C = 02+12 + C
    C = 1

    Streamline is x2+y2 = 1
    _______________________________________________________________

    Now, onto the particle path. This is the bit I'm stumped on.

    I've plotted the stream flow in mathematica and intuitively it seems the path should be:
    (x, y) = (sin[ t ] + t, cos[ t ])
    But my answer is coming out slightly different.
    Here goes...

    Velocity u = (y, t-x)
    Particle path:
    dx/dt = u
    dy/dt = v

    So
    dx/dt = y
    dy/dt = t-x

    Taking the 2nd derivative of the first part, we get
    d2x/dt2 = dy/dt = t - x
    This gives a differential equation for x(t):

    d2x/dt2 + x = t

    Solving the left hand side:
    d2x/dt2 + x = 0

    Using characteristic method to find particular solution, setting xc(t) = Aeλt
    Thus:
    λ = ±i
    Giving:
    xc(t) = Aeit + Be-it

    Then to solve for the full differential equation, d2x/dt2 + x = t
    Try particular solution xp(t) = t
    This appears to work, giving finally:
    x(t) = Aeit + Be-it + t

    Then, from the information above, we have:
    dx/dt = y
    giving
    y(t) = Aieit - Bie-it + 1

    So, final solution for positions x(t) and y(t) are:

    x(t) = Aeit + Be-it + t
    y(t) = Aieit - Bie-it + 1


    The information states that the particle is released at (x, y) = (0, 1) at time t=0
    So this can be used to find constants A and B:

    x(0) = Ae0 + Be0 + 0 = 0
    A + B = 0
    A = -B

    y(0) = Aie0 - Bie0 + 1 = 1
    Substitute A = -B
    -Bi - Bi + 1 = 1
    -2Bi = 0
    B = 0

    Therefore:
    A = -B = 0

    So both functions x(t) and y(t) are reduced to zero for the given boundary conditions.

    And this is where I am lost.

    :oldconfused::oldconfused::oldconfused:

    Thanks for any help anyone can give, and thanks for taking the time to look at this because I know it's a large post!
     
  2. jcsd
  3. Feb 9, 2016 #2

    BvU

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    I read (x,y) = (t,1) , not (0,0) :smile: ?

    And you can check that it satisfies the equation ! And the initial conditions ! Well done :rolleyes: !
     
    Last edited: Feb 9, 2016
  4. Feb 9, 2016 #3
    I'm sorry but I don't understand what you mean?

    Where have you got (x,y) = (t,1) from? Where did you read that? And where did you see any mention of (0,0)?

    Thanks
     
  5. Feb 9, 2016 #4

    BvU

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    So if A and B are 0 then x = t and y = 1, in other words (x,y) = (t, 1)
    Which I interpret as x = 0 and y = 0, in other words (x,y) = (0, 0)​
     
  6. Feb 9, 2016 #5
    Ahh I see what you mean now.

    So it does satisfy the equation for the given boundary conditions (woohoo!)

    BUT, the problem is that I'm certain it's wrong.

    The original velocity vector, u = (y, t-x), describes a circular vector field for any fixed time t. Like this:
    http://student.ulb.ac.be/~lclaesse/phystricks-documentation/_build/html/_images/Picture_FIGLabelFigChampVecteursPICTChampVecteurs-for_eps.png [Broken]

    As time increases, the field drifts to the right.

    This would surely mean that the particle path describes something a little bit 'loopy'.

    As I said, intuitively I'd have though the particle path is (x, y) = (sin[ t ] + t, cos[ t ]) . This would satisfy the start point (0,1) at t=0, and would describe a path that nicely follows the rightward-moving circular vector field that the particle exists in.

    However the answer I've got is (t, 1), which is a basically a straight line at y=1 ! :oldfrown:

    Good to know I got the ODE right but it seems I'm still wrong somewhere. What are your thoughts?

    Thanks.

    EDIT: Hang on a minute... Maybe I'm not wrong??? If the vector field moves to the right at the same rate of its circular rotation, maybe the particle will sit at that y=1 spot continually...?
     
    Last edited by a moderator: May 7, 2017
  7. Feb 9, 2016 #6
    Actually... it makes more sense for it to stay on that straight line y=1...

    For every moment of time that passes, the vector field moves a step to the right, meaning that there is never an 'opportunity' for the y-component of the velocity to change its value away from 0, because the component uy = t - x , with the initial condition x = 0, t = 0, so of course t - x = 0 for the whole time.

    Fantastic. I think I've done it.

    Happy days.

    Thanks for the help!
     
  8. Feb 9, 2016 #7

    BvU

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    You have indeed! Well done (again) :rolleyes: !

    And you'll note that this surprising trajectory comes out for this particular starting point only.
     
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