Going from Eulerian Velocity to Lagrangian motion

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Homework Help Overview

The discussion revolves around transitioning from Eulerian velocity to Lagrangian motion, specifically examining the relationship between the two forms of motion as described by the equations V1 = k*z1 and z1 = x1*e^(k(t-t0)). Participants are trying to understand how to manipulate these equations and the relevant differential equations involved.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Some participants attempt to solve the differential equation (dz1/dt) + z1^2 = 0, while others suggest that the problem may simply be an integration issue. There are questions about the relevance of the equations provided and whether they are contradictory. Participants also express confusion over the notation and the initial conditions given.

Discussion Status

The discussion is ongoing, with various interpretations being explored. Some participants have offered guidance on solving the differential equations, while others are questioning the assumptions made about the equations and their applicability. There is a recognition of the need for clarity in the problem statement and the equations involved.

Contextual Notes

Participants note that the original poster's teacher is the author of the referenced textbook, which may contribute to the confusion due to language barriers and unclear explanations. There is also mention of the initial conditions and how they relate to the equations being discussed.

jon8105
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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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Anyone? I think it is just an integration problem (I hope), but I just can't get it to come out right!
 
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:
\frac{dz_{1}}{dt}=v_{1}=kz_{1}
which indeed has the solution you're after.
 
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:
\frac{dz_{1}}{dt}=v_{1}=kz_{1}
which indeed has the solution you're after.
Okay, here are the formulas used:
\frac{dz_{1}}{dt}+z_{1}^2 = 0 (1)

V_{1}=kz_{1} (2)

z_{1}=x_{1} @ 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:

z_{1}=x_{1}e^k^(^t^-^t^_^{0}^)

This is from my continuum mechanics book. Thanks for your help! I think I am just missing something small, but it is bugging me!
 
Let me know if you need more information to help!
 
Equation (1) and eq. (2) are contradictory equations.
Why do you think eq (1) is relevant here?
 
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.
 
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?
 
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 c_{p}(t)

Let the Eulerian velocity field be given by v(x,t)[/tex]<br /> <br /> Now, the particle velocity is simply \frac{dc_{p}}{dt}<br /> <br /> 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.<br /> <br /> That is, we must have the differential equation:<br /> \frac{dc_{p}}{dt}=v(c_{p}(t),t) (1)<br /> <br /> If you differentiate this, you get, for the particle acceleration:<br /> 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)<br /> <br /> From (1), we can therefore rewrite (2) as:<br /> a_{p}=\frac{\partial{v}}{\partial{t}}+\frac{\partial{v}}{\partial{x}}v (3)<br /> <br /> 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 x^-^2dx, 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 z_{1} from the differential equation (which is equation 2).

Then it says with V_{1}=dz{_1}/dt, subject to the initial condition z_{1}=x_{1} @ 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!
 

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