1. May 12, 2005

### tony_engin

hi all!
I know how to solve the part (i) but not part (ii).
Could anyone help?

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2. May 12, 2005

### arildno

Note that in vector notation, we have, from (i):
$$\nabla{u}=\frac{\partial{u}}{\partial{x}}\vec{i}+\frac{\partial{u}}{\partial{y}}\vec{j}=\frac{\partial{u}}{\partial{r}}\vec{i}_{r}+\frac{\partial{u}}{r\partial\theta}\vec{i}_{\theta}$$
when the polar unit vectors are:
$$\vec{i}_{r}=\cos\theta\vec{i}+\sin\theta\vec{j},\vec{i}_{\theta}=\frac{\partial}{\partial\theta}\vec{i}_{r}=-\sin\theta\vec{i}+\cos\theta\vec{j}$$
Thus, the gradient operator has the form, in polar coordinates:
$$\nabla=\vec{i}_{r}\frac{\partial}{\partial{r}}+\vec{i}_{\theta}\frac{\partial}{r\partial\theta}$$
In agreement thus far?

3. May 12, 2005

### dextercioby

Leave vectors alone and simply differentiate

$$\frac{\partial u}{\partial x}=\cos\theta\frac{\partial u}{\partial r}-\frac{\sin\theta}{r}\frac{\partial u}{\partial \theta}$$

$$\frac{\partial^{2} u}{\partial x^{2}}=\frac{\partial}{\partial x}\left(\frac{\partial u}{\partial x}\right)$$

$$=\cos\theta\frac{\partial}{\partial r}\left(\cos\theta\frac{\partial u}{\partial r}-\frac{\sin\theta}{r}\frac{\partial u}{\partial\theta}\right)-\frac{\sin\theta}{r}\frac{\partial}{\partial\theta}\left(\cos\theta\frac{\partial u}{\partial r}-\frac{\sin\theta}{r}\frac{\partial u}{\partial\theta}\right)$$

and the same way for "y"...

Daniel.

Last edited: May 12, 2005
4. May 12, 2005

### tony_engin

Oh! Thank you very much for your reply!
But I just manage to get the correct answer!
Thank you anayway! ^^

5. May 12, 2005

### dextercioby

So that's the laplacian in polar plane coordinates.BTW,it's customarily to denote these coordinates by $\left(\rho,\theta\right)$ and physicists prefer $\left(\rho,\phi\right)$.

Daniel.