While streamlines, streaklines, and pathlines are primarily used
as *visualizers*, clearly one needs to be able to calculate
the actual lines in order to plot them.

Taking the following mathematical definitions, we can see how one could calculate these lines from knowing the underlying Eulerian description of the flow:

A **pathline** is the easiest, being given as:

$\displaystyle{\frac{d\vec{x}}{dt} = \vec{v}(\vec{x},t)}$

This means that getting the x trajectory of the pathline is
simply obtained by integrating the v_{x} expression. A
**line** is generated from this data by eliminating the t
variable from the x, y, (and z) trajectory expressions.

A **streamline** is almost as easy, but we must recall that
here we want the **instantaneous** flow field. Clearly, we
cannot integrate with respect to t then (because we would then get
trajectories moving into the future), instead we have:

$\displaystyle{\frac{d\vec{x}}{ds} = \vec{v}(\vec{x},t)}$

This means that integrating the v_{x} expression, but
based on s! In other words, we treat t like a constant. This way we
get x trajectories for a particular time as a function of s.
Eliminating the s from our x, y, (and z) expressions again yields a
**line**.

**Streaklines** are the hardest. recall here that we need to
"connect the dots" for all points (x, y, z) that correspond to
fluid elements that passed through a specific point (xo, yo, zo)
for times before now. So, we need to use the pathlines....

- Rearrange the pathline trajectory expressions to solve for the "initial" positions (xo, yo, zo) so that we have an expression for all of the positions x' of that particle at future times, t'.
- Plug this new expression for initial conditions in to replace the generic (xo, yo, zo) that we had in the original pathline expressions
- Eliminate the t' in order to order to obtain the
**lines**corresponding to streaklines

Mathematically derive streaklines, streamlines, and pathlines from an Eulerian velocity field.

Calculate the pathlines, streamlines, and streaklines for the applet example used earlier. That is, for a velocity field of $v_x = sin(t)$ and $v_y = 1$.

Let's start with the pathlines ...

**Pathlines** are calculated using:

$\displaystyle{\frac{d\vec{x}}{dt} = \vec{v}(\vec{x},t)}$

In our case this means that we need to integrate the $v_x$ and $v_y$ equations:

$\displaystyle{\frac{dx}{dt} = sin(t)}$ and $\displaystyle{\frac{dy}{dt} = 1}$ give

$\displaystyle{\int dx = \int sin(t) dt}$ so that $\displaystyle{x = x_o + 1 - cos(t)}$ and

$\displaystyle{\int dy = \int dt}$ so that $\displaystyle{y = y_o + t}$

we then eliminate $t$ from the equations to give the curve:

$\displaystyle{x = x_o + 1 - cos(y-y_o)}$

**Streamlines** are calculated in much the same way except
that we **fix** the time at some value, $t$, and then integrate
against a "dummy" time, $s$:

$\displaystyle{\frac{d\vec{x}}{ds} = \vec{v}(\vec{x},t)}$

In our case this means that we need to integrate the $v_x$ and $v_y$ equations:

$\displaystyle{\frac{dx}{ds} = sin(t)}$ and $\displaystyle{\frac{dy}{ds} = 1}$ give

$\displaystyle{\int dx = \int sin(t) ds}$ so that $\displaystyle{x = x_o + sin(t)s}$ and

$\displaystyle{\int dy = \int ds}$ so that $\displaystyle{y = y_o + s}$

we then eliminate $s$ from the equations to give the curve:

$\displaystyle{x = x_o +(y-y_o)sin(t)}$

For **streaklines** we atart with our solution to the
**pathline trajectories** (before we eliminated $t$):

$\displaystyle{x = x_o + 1 - cos(t)}$ and $\displaystyle{y = y_o + t}$

We need to rearrange them so that they tell us what initial position ($x_o,y_o$) was occupied by some particle that is currently at position (x', y') at time t':

$\displaystyle{x_o = x' - 1 + cos(t')}$ and $\displaystyle{y_o = y' - t'}$

If we then use these "values" as our initial positions in the pathline trajectories, we can see where (else) that particle went (for all the other values of time $t$):

$\displaystyle{x = [x' - 1 + cos(t')] + 1 - cos(t)}$ and $\displaystyle{y = [y' - t'] + t}$

Last, we eliminate the t' so that we have streaklines for particles that have visited point (x', y'):

$\displaystyle{x = x' + cos(y'-y+t) - cos(t)}$

In order to prove to ourselves that these three lines are the same for steady flows we need to note that (x', y') could easily be chosen to be ($x_o, y_o$), since clearly if we had chosen t' = 0 that would fit our definition of (x', y').