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I have to prove that,

$$\nabla \cdot (A_1\mathbf e_1) = \frac{1}{h_1h_2h_3}\frac{\partial (A_1h_2h_3)}{\partial u_1} $$

My approach:

$$\begin{align} \nabla \cdot (A_1\mathbf e_1) &= \nabla \cdot (A_1 h_1 \nabla u_1)\\ &= \underbrace{\nabla (A_1h_1) \cdot \nabla u_1} + \underbrace{A_1h_1 \nabla \cdot \nabla u_1}\\ &= \underbrace {\frac{1}{h_1^{2}} \frac{\partial(A_1h_1)}{\partial u_1}} + \underbrace{\nabla^{2}u_1}\\ &= \underbrace {\frac{1}{h_1^{2}} \frac{\partial(A_1h_1)}{\partial u_1}} + \underbrace{\frac{1}{h_1h_2h_3}\frac{\partial(\frac{h_2h_3}{h_1})}{\partial u_1}}\\ \end{align} $$

Notes:

To expand the first step I used $\nabla \cdot (\phi \mathbf A) = \nabla \phi \cdot \mathbf A + \phi \nabla \cdot \mathbf A$

To expand the third step I used the expansion formula for Laplace operator i.e.,

\begin{eqnarray} \nabla ^2 \phi = \frac{1}{h_1 h_2 h_3}\left(\frac{\partial }{\partial u_1}\left[\frac{h_2h_3}{h_1}\frac{\partial \phi}{\partial u_1}\right] + \ldots \right) \end{eqnarray}

Is there any mistake in any of the above steps? Could it be simplified further from here?

Please note that I do have the right solution for which in the first step I'll have to substitute $\mathbf e_1=h_2h_3\nabla u_2 \times \nabla u_3$ and proceed.

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  • 1
    $\begingroup$ The coefficient $A_1 h_1$ in front of $\nabla^2 u_1$ got lost, everything else is correct. But you're computing a first derivative by first computing a second derivative. If you know $\nabla A_1$ and $\nabla \cdot \mathbf e_1$, you can use the formula for $\nabla \cdot (\phi \mathbf A)$ to obtain $$\nabla \cdot (A_1 \mathbf e_1) = \frac 1 {h_1} \frac {\partial A_1} {\partial u_1} + \frac {A_1} {h_1 h_2 h_3} \frac {\partial(h_2 h_3)} {\partial u_1}.$$ It's the same as expanding the derivative of $h_2 h_3/h_1$ in the last brace or the derivative of $A_1 h_2 h_3$ in the first formula. $\endgroup$ – Maxim Nov 19 '18 at 17:54
  • $\begingroup$ @Maxim Gosh! I get it now. Thanks a lot! I clearly need a lot more practice on this topic now. $\endgroup$ – yathish Nov 19 '18 at 18:08

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