Vector calculus identities navigation

[Dafe]

Homework Statement

I'm reading in a fluid dynamics book and in it the author shortens an equation using identities my rusty vector calculus brain cannot reproduce.

Homework Equations

$$\vec{e} \cdot \frac{\partial}{\partial t}(\rho \vec{u}) = -\nabla\cdot (\rho\vec{u})\cdot\vec{e} - \rho(\vec{u}\cdot\nabla)\vec{u}\cdot\vec{e} - (\nabla p)\cdot\vec{e} + \rho\vec{b}\cdot\vec{e}$$

The author turns the left side of the equation into:

$$-\nabla\cdot(p\vec{e} + \rho\vec{u}(\vec{u}\cdot\vec{e})) + \rho\vec{b}\cdot\vec{e}$$

Just to be clear;
$$\vec{u}$$ is a vector valued function,
$$\vec{e}$$ is a fixed vector,
$$\rho, p$$ are scalar valued functions.

The Attempt at a Solution

The first part is fine:
$$\nabla\cdot (p\vec{e}) = \nabla(p)\cdot\vec{e} + p(\nabla\cdot\vec{e})$$
The divergence of a fixed vector is zero and so,
$$\nabla\cdot (p\vec{e}) = \nabla(p)\cdot\vec{e}$$

Next I need to find
$$\nabla\cdot (\rho\vec{u}(\vec{u}\cdot\vec{e}))$$

I am not sure what to do with is. $$\rho$$ is a scalar valued function, but so is I think $$\vec{u}\cdot\vec{e}$$. I know of the product rule between a scalar and a vector valued function, but what happens when there are two scalar valued functions?

Any suggestions are welcome, thanks.

 

[mighty2000]

Hi there,

Thank you for sharing your question with us. It seems like the author has used some vector calculus identities to simplify the given equation. Let me try to explain how they might have arrived at the shortened version.

Firstly, let's look at the left side of the original equation:

\vec{e} \cdot \frac{\partial}{\partial t}(\rho \vec{u})

We can use the chain rule to expand this as follows:

\frac{\partial}{\partial t}(\rho \vec{u}) = \frac{\partial \rho}{\partial t} \vec{u} + \rho \frac{\partial \vec{u}}{\partial t}

Substituting this back into the original equation, we get:

\vec{e} \cdot \frac{\partial \rho}{\partial t} \vec{u} + \vec{e} \cdot \rho \frac{\partial \vec{u}}{\partial t}

Now, using the identity \nabla \cdot (\rho \vec{u}) = \vec{u} \cdot \nabla \rho + \rho \nabla \cdot \vec{u}, we can rewrite the first term as:

\vec{e} \cdot \frac{\partial \rho}{\partial t} \vec{u} = \nabla \cdot (\rho \vec{u}) \cdot \vec{e} - \rho \nabla \cdot \vec{u} \cdot \vec{e}

And the second term as:

\vec{e} \cdot \rho \frac{\partial \vec{u}}{\partial t} = \rho \vec{e} \cdot \frac{\partial \vec{u}}{\partial t} = \rho \vec{u} \cdot \nabla \vec{e} + \rho \vec{u} \cdot (\nabla \cdot \vec{e})

Now, using the identity \vec{u} \cdot (\nabla \cdot \vec{e}) = \nabla \cdot (\vec{u} \cdot \vec{e}) - (\nabla \times \vec{u}) \times \vec{e}, we can rewrite the second term as:

\rho \vec{u} \cdot (\nabla \cdot \vec{