Integrating velocity field to get position

Question

I feel silly for simply being brainstuck, but consider the following integral, physically it would be the solution of $\mathbf{p} = \tfrac{d\mathbf{v}}{dt}$ - the position of a given particle in space with respect to the time and a velocity vector field.

$$\mathbf{p}(x,y) = \int_a^b{\mathbf{v}(x,y)}dt$$

However I have no idea how to describe the $x$ and $y$ components of the velocity vector in t. Or how to convert dt to dx & dy? Say for example $\mathbf{v} = \left \langle 3x, xy \right \rangle$ Which would result in:

$$\mathbf{p}(x,y) = \int_a^b{3x}dt\cdot\mathbf{i}+\int_a^b{xy}dt\cdot\mathbf{j}$$

Buth how then to continue? I should be possible to calculate this right? I know the speed vector at each point in space, so over a given time period I should be able to get the new position right?

Answer 1

I think you might be confused because of the notation. If $\mathbb p$ is supposed to be the position, it should depend only on time, not on another position. That means $\mathbb p$ should be a function of one variable.

The differential equation is supposed to be $\frac{d\mathbb p}{dt} = \mathbb v(\mathbb p)$. The integral form is $$ \mathbb p(t) = \mathbb p(0) + \int_{t_0}^{t} \mathbb v(\mathbb p(s)) ds, $$ but this generally is not the method to find the solution.

To find $\mathbb p$ given $\mathbb p(0)$ and $\mathbb v$, it may help to break $\mathbb p(t)$ into $\mathbb p(t) = \langle p_x(t), p_y(t) \rangle$ and $\mathbb v(x, y)$ into $\mathbb v(x, y) = \langle v_x(x, y), v_y(x, y) \rangle$. The differential equation for the vector $\mathbb p$ can be written as a system of scalar differential equations: \begin{align*} \frac{dp_x}{dt} & = v_x(p_x, p_y) \\ \frac{dp_y}{dt} & = v_y(p_x, p_y). \end{align*}

As an example, suppose $\mathbb v(x, y) = \langle 3x, xy \rangle$, i.e., $v_x(x, y) = 3x$ and $v_y(x, y) = xy$. Then you have the system \begin{align*} \frac{dp_x}{dt} & = 3p_x \\ \frac{dp_y}{dt} & = p_xp_y. \end{align*} The first equation can be solved independently for $p_x$, giving $$p_x(t) = c_1e^{3t}$$ where $c_1$ is a constant. Substitute this into $p_x$ in the second equation to get $$ \frac{dp_y}{dt} = c_1e^{3t}p_y. $$ This equation is separable. The solution is $$ p_y(t) = c_2e^{\frac{c_1}3 e^{3t}} $$ where $c_2$ is another constant.

$c_1$ and $c_2$ can be determined once the initial condition is given. If $\mathbb p(0) = \langle x_0, y_0\rangle$ is given, then $$ \mathbb p(0) = \langle x_0, y_0\rangle = \langle c_1, c_2e^{\frac{c_1}3} \rangle. $$ It is easy to verify that $c_1 = x_0$ and $c_2 = y_0e^{-\frac{x_0}3}$. Therefore, $$ \mathbb p(t) = \langle x_0e^{3t}, y_0e^{\frac{x_0}3\left(e^{3t} - 1\right)} \rangle $$

Answer 2

The first relation is wrong, it should be $$\mathbf{v}=\frac{d\mathbf{p}}{dt}.\tag{1}$$

Next, the solution is $$\mathbf{p}(t)=\mathbf{p}_0+\int_0^t\mathbf{v}(t')dt'$$ where you should known $\mathbf{v}$ as a function of $t$, not as a function of $(x,y)$.

If you have $\mathbf{v}(x,y)$, so you have no need to use an integral representation, but you should solve the pair of differential equation $(1)$ where $\mathbf{p}=(x,y)$, and $x(t), y(t)$ are the unknown functions.