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I Determine P(t)=(Px(t),Py(t)) given (x0,y0) and v(x,y,t).

  1. Aug 1, 2016 #1
    Dear all,

    I'm a bit confused about a 2D problem which I thought had a simple solution...
    I don't know if I'm correct, so:

    I have a 2D space, and time. I know the velocity field v(x,y,t). I would like to know where the particle which start at the position x0,y0 at an istant t0 is at a certain istant t>t0.


    My idea is:
    p(t)=<px(t),py(t)>

    dp/dt=v(p,t)=<vx(px,py,t),vy(px,py,t)>
    %%%
    dpx/dt=vx(px,py,t)
    dpy/dt=vy(px,py,t)
    %%%

    Then I would solve it by part...even if the solution can be not easy due to dependence on time,
    Ex:
    vx=px*t, vy=px*py
    %%% % %%
    dpx/px=t dt %px=c1*exp(0.5*t^2)
    dpy/py=px dt %dpy/py=c1*exp(0.5*t^2)dt etc...
    %%% %%%

    Is this answer formally correct?
     
  2. jcsd
  3. Aug 1, 2016 #2

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    These are the path lines of particles. They are given via the velocity field by solving the 1st-order set of ODE's
    $$\frac{\mathrm{d} \vec{x}(t)}{\mathrm{d} t}=\vec{v}(t,\vec{x}(t))$$
    with the initial condition
    $$\vec{x}(0)=\vec{x}_0.$$
    The solution of the ODE's describes the motion of a fluid element that was located at ##\vec{x}_0## at time ##t=0##. In other words it gives you the transformation from Euler to Lagrange coordinates.
     
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