# I "stick" in velocity field

1. Oct 11, 2016

### joshmccraney

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.

2. Oct 12, 2016

### andrewkirk

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.

3. Oct 12, 2016

### joshmccraney

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!