applying the product rule to a vector analysis question

Question

I have been doing doing this problem $∇ × (\varphi∇\varphi)=0$

I am just having trouble applying the product result i get which is below.

$$i(( \frac {d}{dy} )(\varphi \frac {d}{dz} \varphi) - ((\frac {d}{dz})(\varphi \frac {d}{dy} \varphi)) )$$

if i take the first part

$$(\varphi \frac {d}{dz} \varphi)$$

and use the product rule i get the following

$$\frac {d}{dx}(uv)= ((\varphi \frac {d}{dz} \varphi) + ((\frac {d^2}{d^2z} \varphi^2)))$$

this doesnt seem right, can someone help by going through the how to apply the product rule to this. thank you

Answer 1

You don't need to use product rule. Use the identity $$\nabla\times(\psi\vec{A})=\psi\nabla\times\vec{A}+(\nabla\psi)\times \vec{A}$$

Your equation becomes $$\varphi\nabla\times(\nabla\varphi)+\nabla\varphi\times\nabla\varphi=0$$ because each of the two terms is zero.

For future reference, to apply product rule on $\varphi\frac{\partial \phi}{\partial z}$:

$$\frac{\partial}{\partial y}(\varphi\frac{\partial \phi}{\partial z})=\frac{\partial \varphi}{\partial y}\frac{\partial \varphi}{\partial z}+\varphi\frac{\partial^2 \varphi}{\partial y \partial z}$$

This is by the product rule $$\frac{d}{dy}(uv)=\frac{du}{dy}v+u\frac{dv}{dy}$$

Directly applying this rule should give you the above equation. And by the way, this has nothing to do with your first equation in the question. That question does not need product rule.

Answer 2

Let's start from basic principles.

$$\begin{align} \nabla \times (\phi\nabla \phi)&=\sum_{i=1}^3\left(\hat x_i \frac{\partial}{\partial x_i}\right) \times \sum_{j=1}^3\left(\phi\hat x_j \frac{\partial \phi}{\partial x_j}\right)\\\\ &=\sum_{i=1}^3\sum_{j=1}^3(\hat x_i \times \hat x_j)\frac{\partial}{\partial x_i} \left(\phi\frac{\partial \phi}{\partial x_j}\right)\\\\ &=\sum_{i=1}^3\sum_{j=1}^3(\hat x_i \times \hat x_j)\left(\frac{\partial \phi}{\partial x_i}\frac{\partial \phi}{\partial x_j}+\frac{\partial^2 \phi}{\partial x_i \partial x_j}\right) \end{align}$$

Now note that

$$\hat x_i \times \hat x_j=-\hat x_j \times \hat x_i$$

is anti-symmetrical in $i$ and $j$. However, both

$$\frac{\partial \phi}{\partial x_i}\frac{\partial \phi}{\partial x_j}=\frac{\partial \phi}{\partial x_j}\frac{\partial \phi}{\partial x_i}$$

and

$$\frac{\partial^2 \phi}{\partial x_i \partial x_j}=\frac{\partial^2 \phi}{\partial x_j \partial x_i}$$

are symmetrical in $i$ and $j$.

Therefore, upon summing over $i$ and $j$, it is easy to see that the result is zero. And we are done!