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Going from Eulerian Velocity to Lagrangian motion

  1. Sep 14, 2007 #1
    1. The problem statement, all variables and given/known data
    Let the motion in Eulerian equal = V1=k*z1
    Show that the motion in Lagrangian equals = z1=x1*e^k(t-t0)


    2. Relevant equations
    Supposedly the solution to the equation (dz1/dt) + z1^2 = 0, with V1 = dz/dt at initial condition of z1=x1 @ t=0, will give me the answer, but I am not getting it.


    3. The attempt at a solution
    (dz1/dt) + z1^2 = 0, so

    dz1/dt = -z1^2, which makes (-1/z1^2)dz1 = dt

    Am I even on the right track!! Thanks for any help
     
  2. jcsd
  3. Sep 14, 2007 #2
    Anyone? I think it is just an integration problem (I hope), but I just can't get it to come out right!
     
  4. Sep 14, 2007 #3

    arildno

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    Your notation is all crazy and incomprehensible.
    Please post the question EXACTLY as it was written.

    Most likely, you are to solve the diff.eq:
    [tex]\frac{dz_{1}}{dt}=v_{1}=kz_{1}[/tex]
    which indeed has the solution you're after.
     
  5. Sep 14, 2007 #4
    Okay, here are the formulas used:
    [tex]\frac{dz_{1}}{dt}+z_{1}^2 = 0[/tex] (1)

    [tex]V_{1}=kz_{1}[/tex] (2)

    [tex]z_{1}=x_{1}[/tex] @ t=0 Initial conditions ( I am assuming this is to solve for C in the integral)

    Now I am supposed to solve formula 1 using info from formula 2, which must result in:

    [tex]z_{1}=x_{1}e^k^(^t^-^t^_^{0}^)[/tex]

    This is from my continuum mechanics book. Thanks for your help!! I think I am just missing something small, but it is bugging me!
     
  6. Sep 14, 2007 #5
    Let me know if you need more information to help!
     
  7. Sep 15, 2007 #6

    arildno

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    Equation (1) and eq. (2) are contradictory equations.
    Why do you think eq (1) is relevant here?
     
  8. Sep 15, 2007 #7
    Equation 1 is what the teacher (and several books) say I have to use to switch from Eulerian Velocity to Lagrangian motion. What is the solution to equation 1? I tried it, but I don't know if it is right.
     
  9. Sep 15, 2007 #8

    HallsofIvy

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    Are you really serious? I think every one here is trying figure out what your REAL question is because it just doesn't make sense that you would be doing problems involving "Lagrangians" and differential equations if you don't know how to integrate x-2dx!
    Do you know the derivative rule for xn? If so what is the derivative of x-1?
     
  10. Sep 15, 2007 #9

    arildno

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    No, it is not.

    You are most likely mixing together the expression for the acceleration as given by the Eulerian velocity field.

    Given a fluid field, in which a particular particle has the trajectory [tex]c_{p}(t)[/tex]

    Let the Eulerian velocity field be given by [itex]v(x,t)[/tex]

    Now, the particle velocity is simply [tex]\frac{dc_{p}}{dt}[/tex]

    The Eulerian velocity field must now at any given point coincide with the velocity to the particle that happens to be at that particular point at time t.

    That is, we must have the differential equation:
    [tex]\frac{dc_{p}}{dt}=v(c_{p}(t),t) (1)[/tex]

    If you differentiate this, you get, for the particle acceleration:
    [tex]a_{p}=\frac{d^{2}c_{p}}{dt^{2}}=\frac{\partial{v}}{\partial{t}}+\frac{\partial{v}}{\partial{x}}\mid_{x=c_{p}(t)}\frac{dc_{p}}{dt} (2)[/tex]

    From (1), we can therefore rewrite (2) as:
    [tex]a_{p}=\frac{\partial{v}}{\partial{t}}+\frac{\partial{v}}{\partial{x}}v (3)[/tex]

    The right hand side of (3) is the general expression of particle acceleration as expressed in terms of the Eulerian velocity field and its partial derivatives.
     
    Last edited: Sep 15, 2007
  11. Sep 15, 2007 #10

    HallsofIvy

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    And after all that you are asking how to integrate x-2dx?
     
  12. Sep 15, 2007 #11
    Yes, I KNOW how to integrate [tex]x^-^2dx[/tex], but what I am getting just isn't matching up, once I try to implement the second equation.

    Arildno, YES it is. Here is the wording directly from my Continuum Mechanics book:

    The Eulerian velocities (which is equation 1) may be transformed back into the Lagrangian velocities if we solve for [tex]z_{1}[/tex] from the differential equation (which is equation 2).

    Then it says with [tex]V_{1}=dz{_1}/dt[/tex], subject to the initial condition [tex]z_{1}=x_{1}[/tex] @ t=0

    HallsofIvy, I am just trying to ask a question!!! There is no reason to be rude!!
     
  13. Sep 15, 2007 #12

    arildno

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    You have gravely misunderstood your book.

    The differential equation defining the Eulerian velocity field is given by my 1, which equals in your particular case (2).

    You have given two distinct diff.eqs for the SAME trajectory, which is not correct.

    You are mixing together two distinct problems or something else.
     
  14. Sep 15, 2007 #13

    Gokul43201

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    What text is this (title, author, edition, page number)?

    What are "equation 1" and "equation 2"? Clearly, they are not the same (1) and (2) as post #4.

    Or else, forget all that and start here:

    That's a straightforward problem. Just plug in v1 = dz1/dt and solve the separable equation.
     
  15. Sep 15, 2007 #14
    That is IT!! Thanks!

    You guys have to understand that my teacher is the author of the book (Continuum Mechanics, auther TJ CHung) and he does not speak English very well. Also, he doesn't explain things well and for some reason told me to take equation 2 (entire equation) and solve for z, then implement with V. However, like arildno said they contradict each other and that is where I was getting confused.

    Anyways, thanks for all of your help, and maybe you want think I am as stupid as I made myself seem!!
     
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