1. May 23, 2004

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.

2. May 23, 2004

vas85

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

3. May 23, 2004

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."

Last edited: May 23, 2004
4. May 23, 2004

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

5. May 23, 2004

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 :)

6. May 24, 2004

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

Now, the first term of the right hand side:

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

The second term of the right hand side:

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

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

7. May 24, 2004

vas85

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