Eulerian velocities to Lagrangian velocities

[paccali]

Homework Statement

Eulerian velocity: V_{1}=-z_{1}^{2}

V_{1}=\frac{dz_{1}}{dt}

z_{1}(t=0)=x_{1}

This is supposed to become the Lagrangian velocity of:

z_{1}=\frac{x_{1}}{1+tx_{1}}

I don't understand how to take the Eulerian velocity and transform it to Lagrangian.

Homework Equations

The Attempt at a Solution

\frac{dz_{1}}{dt}+z_{1}^{2}=0

After this, I don't know how to take this and move forward.

I've been working the problem for a day, and I still can't get any closer. I can take Lagrangian and transform it to Eulerian, but I don't know how to do the reverse. TJ Chung's General Continuum Mechanics book is poorly developed for examples and proofs.

 

[fzero]

The Attempt at a Solution

\frac{dz_{1}}{dt}+z_{1}^{2}=0

After this, I don't know how to take this and move forward.

This equation is separable, since it can be rewritten as

-\frac{dz_{1}}{z_{1}^{2}}=dt.

You can integrate both sides and fix the integration constant with

z_{1}(t=0)=x_{1}.

 

[paccali]

Nevermind, I figured out how to solve the problem using my old differential equations textbook. In case someone is curious, here's what I did:

I rearranged the terms so that like terms were on the same side:

\frac{dz_{1}}{dt}=-z_{1}^{2}
\frac{dz_{1}}{-z_{1}^{2}}=dt

I then integrated each side:

\int -\frac{1}{z_{1}^{2}}dz=\int dt
\frac{1}{z_{1}}+C_{1}=t+C_{2}

Since both sides had constants, I dropped one, and I then used the initial condition of z_{1}(t=0)=x_{1} to solve for C

\frac{1}{z_{1}}+C=t
\frac{1}{z_{1}}=t-C
z_{1}=\frac{1}{t-C}
z_{1}(0)=x_{1}=\frac{1}{0-C}
C=-\frac{1}{x_{1}}

I plugged in C above and got this equation for Lagrangian position:

z_{1}=\frac{1}{t+\frac{1}{x_{1}}}=\frac{x_{1}}{1+tx_{1}}

 

[paccali]

I just found what I did wrong, but thanks for the help nonetheless.