Question

• A. $\nabla \cdot (\phi \mathbf{u}) = \phi \, \nabla \cdot \mathbf{u} + \mathbf{u} \cdot \nabla \phi$
• B. $\nabla \cdot (\phi \mathbf{u}) = \phi \, \nabla \cdot \mathbf{u} + \mathbf{u} \times \nabla \phi$
• C. $\nabla \cdot (\phi \mathbf{u}) = \phi \, \nabla \phi + \mathbf{u} \cdot \nabla \phi$
• D. $\nabla \cdot (\phi \mathbf{u}) = \phi \, \nabla (\mathbf{u}) + \mathbf{u} \times \nabla \phi$

Hint:

Remember: For vector calculus, $\nabla \cdot (\phi \mathbf{A}) = \phi (\nabla \cdot \mathbf{A}) + \mathbf{A} \cdot \nabla \phi$, analogous to the product rule in differentiation.

Answer 1

The Correct Option is A

Step 1: Recall the vector calculus product rule.
For a scalar field $\phi$ and a vector field $\mathbf{u}$, the product rule for divergence is: \[ \nabla \cdot (\phi \mathbf{u}) = \phi \, (\nabla \cdot \mathbf{u}) + (\nabla \phi) \cdot \mathbf{u}. \]

Step 2: Interpret terms.
- The first term $\phi \, (\nabla \cdot \mathbf{u})$ scales the divergence of $\mathbf{u}$ by $\phi$.
- The second term $(\nabla \phi) \cdot \mathbf{u}$ is the dot product of the gradient of $\phi$ with $\mathbf{u}$.

Step 3: Compare with options.
This matches exactly with Option (A): \[ \nabla \cdot (\phi \mathbf{u}) = \phi \, \nabla \cdot \mathbf{u} + \mathbf{u} \cdot \nabla \phi. \]
\[ \boxed{\nabla \cdot (\phi \mathbf{u}) = \phi \, (\nabla \cdot \mathbf{u}) + \mathbf{u} \cdot \nabla \phi} \]