# Eulerian vs. Lagrangian description

1. May 30, 2010

### coverband

1. The problem statement, all variables and given/known data
A particle moves so that $$\vec{x} \equiv [x_0 exp(2t^2), y_0 exp(-t^2), z_0 exp(-t^2)]$$. Find the velocity of the particle in terms of $$x_0$$ and t (the LAgrangian description) and show it can be written as $$\vec{u} \equiv (4xt, -2yt, -2zt),$$ the Eulerian description

2. Relevant equations

3. The attempt at a solution
$$\vec {u} = [4tx_0 exp(2t^2), -2ty_0 exp(-t^2), -2tz_0 exp(-t^2)]$$

The exp's go to 1 !!?

Thanks

2. May 30, 2010

### irycio

No, i suppose it's just that in your velocity vecotr you can substitute the term $$x_0exp(2t^2)$$ with x - "first part" from your $$\vec{x}$$ vector and so on

3. May 30, 2010

### coverband

Yeah but why/how can you do that

4. May 30, 2010

### irycio

Well, i guess because they're equal. I see no physical meaning of such operation, it's just the mathematical identity that allows you to subsitute it.
Let's consider a single function $$f(x)=exp(2x)$$. Obviously, $$\frac{df}{dx}=2exp(2x)$$. And you can write $$\frac{df}{dx}=2f(x)$$, which is true for all x, but doesn't really mean anything.

E:and so, in your example, you just get $$x_x=x_0exp(2t^2)$$, $$u_x=4tx_0exp(2t^2)=4tx$$

5. May 30, 2010

Anyone else?