Going from Eulerian Velocity to Lagrangian motion

In summary, the motion in Eulerian equals V1=k*z1 and the motion in Lagrangian equals z1=x1*e^k(t-t0). To solve for the motion in Lagrangian, plug in V1=dz1/dt and solve the separable equation. It is not necessary to solve for z in equation 2.
  • #1
jon8105
12
0

Homework Statement


Let the motion in Eulerian equal = V1=k*z1
Show that the motion in Lagrangian equals = z1=x1*e^k(t-t0)


Homework Equations


Supposedly the solution to the equation (dz1/dt) + z1^2 = 0, with V1 = dz/dt at initial condition of z1=x1 @ t=0, will give me the answer, but I am not getting it.


The Attempt at a Solution


(dz1/dt) + z1^2 = 0, so

dz1/dt = -z1^2, which makes (-1/z1^2)dz1 = dt

Am I even on the right track! Thanks for any help
 
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  • #2
Anyone? I think it is just an integration problem (I hope), but I just can't get it to come out right!
 
  • #3
Your notation is all crazy and incomprehensible.
Please post the question EXACTLY as it was written.

Most likely, you are to solve the diff.eq:
[tex]\frac{dz_{1}}{dt}=v_{1}=kz_{1}[/tex]
which indeed has the solution you're after.
 
  • #4
arildno said:
Your notation is all crazy and incomprehensible.
Please post the question EXACTLY as it was written.

Most likely, you are to solve the diff.eq:
[tex]\frac{dz_{1}}{dt}=v_{1}=kz_{1}[/tex]
which indeed has the solution you're after.
Okay, here are the formulas used:
[tex]\frac{dz_{1}}{dt}+z_{1}^2 = 0[/tex] (1)

[tex]V_{1}=kz_{1}[/tex] (2)

[tex]z_{1}=x_{1}[/tex] @ t=0 Initial conditions ( I am assuming this is to solve for C in the integral)

Now I am supposed to solve formula 1 using info from formula 2, which must result in:

[tex]z_{1}=x_{1}e^k^(^t^-^t^_^{0}^)[/tex]

This is from my continuum mechanics book. Thanks for your help! I think I am just missing something small, but it is bugging me!
 
  • #5
Let me know if you need more information to help!
 
  • #6
Equation (1) and eq. (2) are contradictory equations.
Why do you think eq (1) is relevant here?
 
  • #7
Equation 1 is what the teacher (and several books) say I have to use to switch from Eulerian Velocity to Lagrangian motion. What is the solution to equation 1? I tried it, but I don't know if it is right.
 
  • #8
Are you really serious? I think every one here is trying figure out what your REAL question is because it just doesn't make sense that you would be doing problems involving "Lagrangians" and differential equations if you don't know how to integrate x-2dx!
Do you know the derivative rule for xn? If so what is the derivative of x-1?
 
  • #9
jon8105 said:
Equation 1 is what the teacher (and several books) say I have to use to switch from Eulerian Velocity to Lagrangian motion.
No, it is not.

You are most likely mixing together the expression for the acceleration as given by the Eulerian velocity field.

Given a fluid field, in which a particular particle has the trajectory [tex]c_{p}(t)[/tex]

Let the Eulerian velocity field be given by [itex]v(x,t)[/tex]

Now, the particle velocity is simply [tex]\frac{dc_{p}}{dt}[/tex]

The Eulerian velocity field must now at any given point coincide with the velocity to the particle that happens to be at that particular point at time t.

That is, we must have the differential equation:
[tex]\frac{dc_{p}}{dt}=v(c_{p}(t),t) (1)[/tex]

If you differentiate this, you get, for the particle acceleration:
[tex]a_{p}=\frac{d^{2}c_{p}}{dt^{2}}=\frac{\partial{v}}{\partial{t}}+\frac{\partial{v}}{\partial{x}}\mid_{x=c_{p}(t)}\frac{dc_{p}}{dt} (2)[/tex]

From (1), we can therefore rewrite (2) as:
[tex]a_{p}=\frac{\partial{v}}{\partial{t}}+\frac{\partial{v}}{\partial{x}}v (3)[/tex]

The right hand side of (3) is the general expression of particle acceleration as expressed in terms of the Eulerian velocity field and its partial derivatives.
 
Last edited:
  • #10
And after all that you are asking how to integrate x-2dx?
 
  • #11
Yes, I KNOW how to integrate [tex]x^-^2dx[/tex], but what I am getting just isn't matching up, once I try to implement the second equation.

Arildno, YES it is. Here is the wording directly from my Continuum Mechanics book:

The Eulerian velocities (which is equation 1) may be transformed back into the Lagrangian velocities if we solve for [tex]z_{1}[/tex] from the differential equation (which is equation 2).

Then it says with [tex]V_{1}=dz{_1}/dt[/tex], subject to the initial condition [tex]z_{1}=x_{1}[/tex] @ t=0

HallsofIvy, I am just trying to ask a question! There is no reason to be rude!
 
  • #12
You have gravely misunderstood your book.

The differential equation defining the Eulerian velocity field is given by my 1, which equals in your particular case (2).

You have given two distinct diff.eqs for the SAME trajectory, which is not correct.

You are mixing together two distinct problems or something else.
 
  • #13
What text is this (title, author, edition, page number)?

What are "equation 1" and "equation 2"? Clearly, they are not the same (1) and (2) as post #4.

Or else, forget all that and start here:

Let the motion in Eulerian equal = V1=k*z1
Show that the motion in Lagrangian equals = z1=x1*e^k(t-t0)
That's a straightforward problem. Just plug in v1 = dz1/dt and solve the separable equation.
 
  • #14
Gokul43201 said:
What text is this (title, author, edition, page number)?

What are "equation 1" and "equation 2"? Clearly, they are not the same (1) and (2) as post #4.

Or else, forget all that and start here:

That's a straightforward problem. Just plug in v1 = dz1/dt and solve the separable equation.

That is IT! Thanks!

You guys have to understand that my teacher is the author of the book (Continuum Mechanics, auther TJ CHung) and he does not speak English very well. Also, he doesn't explain things well and for some reason told me to take equation 2 (entire equation) and solve for z, then implement with V. However, like arildno said they contradict each other and that is where I was getting confused.

Anyways, thanks for all of your help, and maybe you want think I am as stupid as I made myself seem!
 

1. What is the difference between Eulerian velocity and Lagrangian motion?

Eulerian velocity refers to the velocity of a fluid at a fixed point in space, while Lagrangian motion refers to the motion of individual fluid particles as they move through space.

2. Why is it important to convert from Eulerian velocity to Lagrangian motion?

Converting from Eulerian velocity to Lagrangian motion allows us to track the motion of individual fluid particles, which can provide valuable information for understanding fluid dynamics and predicting changes in flow patterns.

3. How is the conversion from Eulerian velocity to Lagrangian motion done?

The conversion is done by solving the Lagrangian equations of motion, which describe the position, velocity, and acceleration of a fluid particle as it moves through space. This involves taking into account factors such as fluid viscosity, pressure, and external forces.

4. What are some applications of converting from Eulerian velocity to Lagrangian motion?

This conversion is commonly used in fluid mechanics, meteorology, and oceanography to study and predict fluid flow patterns. It is also useful in understanding the motion of particles in a fluid, such as in chemical reactions or sediment transport.

5. Are there any limitations to converting from Eulerian velocity to Lagrangian motion?

Yes, there are limitations, as the conversion assumes that the fluid is continuous and homogeneous. It also does not take into account factors such as turbulence or changes in flow conditions over time. Additionally, the conversion can become complex and computationally intensive for large-scale systems.

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