The acceleration in terms of the eulerian velocity

Question

I'm having trouble in deriving the the acceleration in terms of the eulerian velocity. How do I apply the chain rule for partial derivatives to achieve the result highlighted in green?

Answer 1

For clarity, let $k(\vec \xi, t) = (\vec x^L(\vec \xi, t), t)$. Then $\vec u^L(\vec \xi,t) = (\vec u^E \circ k)(\vec \xi, t)$. Now that both sides have the same arguments, the meaning of a partial derivative becomes unambiguous, and we can use the chain rule:

$$\partial_t \vec u^L = (\partial_t k \cdot \nabla_{\vec x^L, t}) \vec u^E$$

I'm being a little compact here with notation, so let me explain: $\nabla_{\vec x^L, t}$ means a vector $(\partial_{x^L}, \partial_{y^L}, \partial_{z^L}, \partial_t)$. This is a four-component vector of partial derivative operators.

Let's start with $\partial_t k$:

$$\partial_t k = (\partial_t x^L, \partial_t y^L, \partial_t z^L, \partial_t t)$$

The last component, $\partial_t t$, is just $1$.

Now, just dot it with the partial derivative vector $\nabla_{\vec x^L, t}$:

$$\partial_t k \cdot \nabla_{\vec x^L, t} = \frac{\partial x^L}{\partial t} \frac{\partial}{\partial x^L} + \frac{\partial y^L}{\partial t} \frac{\partial}{\partial y^L} + \frac{\partial z^L}{\partial t} \frac{\partial}{\partial z^L} + (1) \frac{\partial}{\partial t}$$

That the time derivative is implicitly with $\xi$ held constant follows from how I originally wrote the arguments: as functions of $\xi, t$.

It's common to write this in terms of the 3d vector derivative $\nabla = \nabla_{\vec x^L}$ instead, which is where you get the more compact notation $\partial_t \vec u^L = (\vec u^L \cdot \nabla )\vec u^E + \partial_t u^E |_\xi$.

If the Eulerian and Lagrangian velocities are the same, then you can just choose a new symbol $\vec u = \vec u^L = \vec u^E$ and the resulting simplification of the notation follows pretty quickly. It's just a substitution for the sake of clarity.

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Edit: a general statement of the chain rule can be helpful here. Let $Y = g(X)$, where $X, Y$ are position vectors in potentially-different vector spaces. Now consider a function $f(Y) = (f \circ g)(X)$. The chain rule can be stated as follows, for a given vector $a$ that lies in the same vector space as $X$:

$$(a \cdot \nabla_X) (f \circ g)(X) = \{[(a \cdot \nabla_X) g(X) ] \cdot \nabla_Y \}f(Y)$$

For this problem, the vector space $X$ belongs to is the vector space of initial positions and current times, the vector space of $(\xi, t)$. $a$ corresponds to some vector in the direction of time, such that $(a \cdot \nabla_X)$ reduces to $\partial_t$.