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

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1. Aug 1, 2016

### Wiccan

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...
%%% %%%

2. Aug 1, 2016

### vanhees71

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.