# Fluid dynamics, solving ODE to find particle path

1. Feb 9, 2016

### sa1988

1. The problem statement, all variables and given/known data

A time-dependent two-dimensional fluid flow is given, in a Cartesian coordinates system (x, y), by the velocity field:

u = (y, t-x)

Show that, at time t = 0, the streamlines of this flow are circles centred on the origin.
Find equation of the streamline that passes through the point (0, 1)
Find the equation of the path of a particle released at the point (0,1) at t=0.

2. Relevant equations

For streamlines in u = (u, v):
dx(t)/u = dy(t)/w

Particle path:
dx/dt = u
dy/dt = v

3. The attempt at a solution

I think I'm okay with the streamline part but here it is anyway:

Streamline:
Using the given equation for finding streamlines, plug in the values and rearrange to get:

(t-x) dx = y dy

y2/2 = tx - x2/2 + C

Rearrange to show x2+y2 = 2tx + C

For t = 0 , it reduces to x2+y2 = + C,
which is a circle.

Streamline passing though (0, 1) is given in:
x2+y2 = C = 02+12 + C
C = 1

Streamline is x2+y2 = 1
_______________________________________________________________

Now, onto the particle path. This is the bit I'm stumped on.

I've plotted the stream flow in mathematica and intuitively it seems the path should be:
(x, y) = (sin[ t ] + t, cos[ t ])
But my answer is coming out slightly different.
Here goes...

Velocity u = (y, t-x)
Particle path:
dx/dt = u
dy/dt = v

So
dx/dt = y
dy/dt = t-x

Taking the 2nd derivative of the first part, we get
d2x/dt2 = dy/dt = t - x
This gives a differential equation for x(t):

d2x/dt2 + x = t

Solving the left hand side:
d2x/dt2 + x = 0

Using characteristic method to find particular solution, setting xc(t) = Aeλt
Thus:
λ = ±i
Giving:
xc(t) = Aeit + Be-it

Then to solve for the full differential equation, d2x/dt2 + x = t
Try particular solution xp(t) = t
This appears to work, giving finally:
x(t) = Aeit + Be-it + t

Then, from the information above, we have:
dx/dt = y
giving
y(t) = Aieit - Bie-it + 1

So, final solution for positions x(t) and y(t) are:

x(t) = Aeit + Be-it + t
y(t) = Aieit - Bie-it + 1

The information states that the particle is released at (x, y) = (0, 1) at time t=0
So this can be used to find constants A and B:

x(0) = Ae0 + Be0 + 0 = 0
A + B = 0
A = -B

y(0) = Aie0 - Bie0 + 1 = 1
Substitute A = -B
-Bi - Bi + 1 = 1
-2Bi = 0
B = 0

Therefore:
A = -B = 0

So both functions x(t) and y(t) are reduced to zero for the given boundary conditions.

And this is where I am lost.

Thanks for any help anyone can give, and thanks for taking the time to look at this because I know it's a large post!

2. Feb 9, 2016

### BvU

I read (x,y) = (t,1) , not (0,0) ?

And you can check that it satisfies the equation ! And the initial conditions ! Well done !

Last edited: Feb 9, 2016
3. Feb 9, 2016

### sa1988

I'm sorry but I don't understand what you mean?

Where have you got (x,y) = (t,1) from? Where did you read that? And where did you see any mention of (0,0)?

Thanks

4. Feb 9, 2016

### BvU

So if A and B are 0 then x = t and y = 1, in other words (x,y) = (t, 1)
Which I interpret as x = 0 and y = 0, in other words (x,y) = (0, 0)​

5. Feb 9, 2016

### sa1988

Ahh I see what you mean now.

So it does satisfy the equation for the given boundary conditions (woohoo!)

BUT, the problem is that I'm certain it's wrong.

The original velocity vector, u = (y, t-x), describes a circular vector field for any fixed time t. Like this:
http://student.ulb.ac.be/~lclaesse/phystricks-documentation/_build/html/_images/Picture_FIGLabelFigChampVecteursPICTChampVecteurs-for_eps.png [Broken]

As time increases, the field drifts to the right.

This would surely mean that the particle path describes something a little bit 'loopy'.

As I said, intuitively I'd have though the particle path is (x, y) = (sin[ t ] + t, cos[ t ]) . This would satisfy the start point (0,1) at t=0, and would describe a path that nicely follows the rightward-moving circular vector field that the particle exists in.

However the answer I've got is (t, 1), which is a basically a straight line at y=1 !

Good to know I got the ODE right but it seems I'm still wrong somewhere. What are your thoughts?

Thanks.

EDIT: Hang on a minute... Maybe I'm not wrong??? If the vector field moves to the right at the same rate of its circular rotation, maybe the particle will sit at that y=1 spot continually...?

Last edited by a moderator: May 7, 2017
6. Feb 9, 2016

### sa1988

Actually... it makes more sense for it to stay on that straight line y=1...

For every moment of time that passes, the vector field moves a step to the right, meaning that there is never an 'opportunity' for the y-component of the velocity to change its value away from 0, because the component uy = t - x , with the initial condition x = 0, t = 0, so of course t - x = 0 for the whole time.

Fantastic. I think I've done it.

Happy days.

Thanks for the help!

7. Feb 9, 2016

### BvU

You have indeed! Well done (again) !

And you'll note that this surprising trajectory comes out for this particular starting point only.