Please help A question about partial differentiation

[tony_engin]

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

 

[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?

 

[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.

 

[tony_engin]

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

 

[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.