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I "stick" in velocity field

  1. Oct 11, 2016 #1

    joshmccraney

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    Hi PF!

    A professor drew a velocity field on the board and he placed a line segment in the field at some time ##t_1## and then drew the stick at some new time ##t_2##. My question, let's say we're given a velocity field ##\vec{V} = v_x \hat{i} + v_y \hat{j}## where ##v_x## and ##v_y## are the magnitudes of velocity. Given a point in space ##(p,q)##, after a certain amount of time, how would we tell where the point would be?

    Thinking about this, if all we had was ##\vec{V} = f(x,y) \hat{i} + g(x,y) \hat{j}## then to determine where ##p## would go, we would say ##\partial_t p = f(x,y) \implies \int_{p(0)}^{p(t)} \, dp = \int_0^t f(x,y) \, dt## but here I am stuck.
     
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  3. Oct 12, 2016 #2

    andrewkirk

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    The problem you are trying to solve is called the computation of flow curves of a vector field. It involves solving the differential equation:
    $$\frac{d}{dt}\vec r(t)=\vec V(\vec r(t))$$
    where ##\vec r(t)## is the vector position of the particle at time ##t##.

    There is no standard method to solve this differential equation. Given a particular vector field ##\vec V##, an explicit solution may be able to be obtained if ##\vec V## is particularly nice. Otherwise numerical techniques would be needed.

    Take the example ##\vec V(\vec r)=\vec r##. Then the above DE is:
    $$\frac{d}{dt}(x(t)\hat i+y(t)\hat j)=(x(t)\hat i+y(t)\hat j)$$
    which separates out into two equations:
    $$\frac{dx}{dt}(t)=x(t);\ \ \frac{dy}{dt}(t)=y(t)$$
    which has solutions ##x(t)=x_0e^t;\ y(t)=y_0e^t## (where ##x_0,y_0## are the coordinates of the point at ##t=0##), which can be expressed neatly as ##\vec r(t)=\vec r_0e^t##.

    Most vector fields do not yield to analysis as readily as this one though.
     
  4. Oct 12, 2016 #3

    joshmccraney

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    Cool, so it seems the first ODE you wrote is the one to follow for general cases? Good to know in case this ever arises. Thanks!
     
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