Proof of identities of divergence of vector fields
You could just expand both sides and see whether they are equal. For example (i),
$$\nabla \cdot (\phi F)=\nabla\cdot (\phi f_1, \phi f_2, \phi f_3)=\frac{\partial (\phi f_1)}{\partial x}+\frac{\partial (\phi f_2)}{\partial y}+\frac{\partial (\phi f_3)}{\partial z}$$
On the other hand,
$$\nabla\phi \cdot F+\phi(\nabla \cdot F)=\left(\frac{\partial \phi}{\partial x}, \frac{\partial \phi}{\partial y}, \frac{\partial \phi}{\partial z}\right)\cdot (f_1, f_2, f_3)+\phi \left(\frac{\partial f_1}{\partial x}+\frac{\partial f_2}{\partial y}+\frac{\partial f_3}{\partial z}\right)$$
You can use product rule to see they are equal. (ii) can be done similarly.
The additional ones are also similar, for example (ii),
$$\nabla \times (\phi F)=\left(\frac{\partial (\phi f_3)}{\partial y}-\frac{\partial (\phi f_2)}{\partial z}, \frac{\partial (\phi f_1)}{\partial z}-\frac{\partial (\phi f_3)}{\partial x}, \frac{\partial (\phi f_2)}{\partial x}-\frac{\partial (\phi f_1)}{\partial y}\right)$$,
and the right hand side,
$$\nabla \phi \times F +\phi (\nabla \times F)=\left(\frac{\partial \phi}{\partial x}, \frac{\partial \phi}{\partial y}, \frac{\partial \phi}{\partial z}\right) \times F + \left(\phi \left(\frac{\partial f_3}{\partial y}-\frac{\partial f_2}{\partial z}\right),\, \phi \left(\frac{\partial f_3}{\partial x}-\frac{\partial f_1}{\partial z}\right),\, \phi \left(\frac{\partial f_2}{\partial x}-\frac{\partial f_1}{\partial y}\right)\right)$$
Again, using product rule and definition of curl, you can see they are the same.
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user189013
Updated on August 01, 2022Comments
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user189013 over 1 year
I want to prove some identities but I don't know how to do this.
First of all,
$φ : R^3 → R$ and vector fields $F = (f_1, f_2, f_3), G = (g_1, g_2, g_3) : R^3 → R^3$
the two identities are:
(i)$ ∇ · (φF) = ∇φ · F + φ(∇ · F) $ (ii) $∇ · (F × G) = G · (∇ × F) − F · (∇ × G)$
Additional identities to prove:
continuously differentiable scalar fields $φ, ψ : R^3 → R$ and vector field $F : R^3 → R^3$:
(i) $∇(φ ψ) = φ∇ψ + ψ∇φ$ (ii) $∇ × (φF) = ∇φ × F + φ(∇ × F)$
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user189013 almost 9 yearsI have added some additional identities I want to prove if you can give me any hint.