Fluid dynamics, solving ODE to find particle path

[sa1988]

Homework Statement

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.

Homework Equations

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

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

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 x_c(t) = Ae^λt
Thus:
λ = ±i
Giving:
x_c(t) = Ae^it + Be^-it

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

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

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

x(t) = Ae^it + Be^-it + t
y(t) = Aie^it - 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!

 

[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 !

 

[sa1988]

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

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

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

 

[BvU]

x(t) = Ae^it + Be^-it + t
y(t) = Aie^it - Bie^-it + 1​

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

Therefore: A = -B = 0
So both functions x(t) and y(t) are reduced to zero for the given boundary conditions

Which I interpret as x = 0 and y = 0, in other words (x,y) = (0, 0)​

 

[sa1988]

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)​

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

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...?

 

[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 u_y = 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!

 

[BvU]

You have indeed! Well done (again) !

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