Need Help Proving Vector Calculus Formula? Look No Further!

[vas85]

Vector Calculus Proof Help Please :)

Heya Ppl i have a problem i am trying to solve.

Prove that


(Delta) . ( (fi)F) = (fi)(Delta) . F + F . (Delta)(fi)

were these contain GRAD DIV in my opinion but i seem to not be able to get the answer.

F = Vector F where F = F1i + F2j + F3k is a vector field in R3 and (fi) the Greek symbol that looks similar to Theta is a fuction of x, y and z.

 

[vas85]

oh and also guys the . is like the DOT PRODUCT dot not a MULTIPLICATION

 

[TALewis]

Do you mean:

$$\nabla \cdot (\phi \mathbf{F}) = (\phi \nabla) \cdot \mathbf{F} + \mathbf{F} \cdot (\nabla \phi)$$

If you're interested, you can click on the equation image to see what code was used to make it.

Also, the upside-down triangle representing the del operator is called "nabla," and the greek letter that looks like theta is called "phi."

 

[vas85]

Yeh TALewis that is how it looked like except that the BRACKETS wernt around the NABLA and PHI which are in the Right Hand Side, but i think that is a okay way to group it. Now if any1 has ideas on how to solve it i would much appreciate it Thanks

 

[vas85]

And um the question now after editing your Code looked like this

$$\nabla \cdot (\phi \mathbf{F}) = \phi \nabla \cdot \mathbf{F} + \mathbf{F} \cdot \nabla \phi$$

But i think the way grouped in your rewrite should be the same thing, now if sum1 knows how to solve that proof :)

 

[TALewis]

I will try to prove it considering the x-direction only. The full result should follow easily in the other two dimensions.

First, the left hand side:

$$\begin{align*}<br /> \nabla \cdot (\phi \mathbf{F}) &= <br /> \frac{\partial}{\partial x}\mathbf{i} \cdot \phi F_1 \mathbf{i}\\<br /> &= \frac{\partial}{\partial x}(\phi F_1)\\<br /> &= \phi\frac{\partial F_1}{\partial x} + F_1\frac{\partial \phi}{\partial x} \quad \mbox{(product rule)}<br /> \end{align}$$

Now, the first term of the right hand side:

$$\begin{align*}<br /> (\phi\nabla)\cdot\mathbf{F} &=<br /> \phi\frac{\partial}{\partial x}\mathbf{i} \cdot F_1\mathbf{i}\\<br /> &= \phi\frac{\partial F_1}{\partial x}<br /> \end{align*}$$

The second term of the right hand side:

$$\begin{align*}<br /> \mathbf{F}\cdot(\nabla\phi) &=<br /> F_1\mathbf{i}\cdot\frac{\partial\phi}{\partial x}\mathbf{i}\\<br /> &= F_1\frac{\partial\phi}{\partial x}<br /> \end{align*}$$

I think you should be able to see now how it all comes together.

 

[vas85]

Thanks Heaps TALewis, i get the picture now! :):)