Converting the Laplacian into polar coordinates

[Xyius]

I need to convert the Laplacian in two dimensions to polar coordinates.

\nabla^2 u=\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}

I am having problems with computing the second derivatives using the chain rule. For example, the first derivative with respect to x would be..
\frac{\partial u}{\partial r}\frac{\partial r}{\partial x}+\frac{\partial u}{\partial \phi}\frac{\partial \phi}{\partial x}

I don't know where to go from here to compute the second derivative. Would I need to use the product rule on each term when differentiating again? By the way, the answer is..

\nabla^2 u=\frac{\partial^2 u}{\partial r^2}+\frac{1}{r}\frac{\partial u}{\partial r}+\frac{1}{r^2}\frac{\partial^2 u}{\partial\phi^2}

 

[Xyius]

Can anyone help me with this??

 

[Mark44]

I need to convert the Laplacian in two dimensions to polar coordinates.

\nabla^2 u=\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}

I am having problems with computing the second derivatives using the chain rule. For example, the first derivative with respect to x would be..
\frac{\partial u}{\partial r}\frac{\partial r}{\partial x}+\frac{\partial u}{\partial \phi}\frac{\partial \phi}{\partial x}

That should be
\frac{\partial u}{\partial x}=\frac{\partial u}{\partial r}\frac{\partial r}{\partial x}+\frac{\partial u}{\partial \theta}\frac{\partial \theta}{\partial x}~~(1)

Polar coordinates are given in r and θ (theta) rather than r and φ (phi).

And the other partial would be
\frac{\partial u}{\partial y}=\frac{\partial u}{\partial r}\frac{\partial r}{\partial y}+\frac{\partial u}{\partial \theta}\frac{\partial \theta}{\partial y}~~(2)

To get \frac{\partial^2 u}{\partial x^2}
and
\frac{\partial^2 u}{\partial y^2}

take the partial with respect to x of (1), above, and take the partial with respect to y of (2), above. To do this you'll need to use the product rule and the chain rule.

It might be helpful to make a diagram showing the relationship between u, r, θ, and x and y.

...u
../.\ 
r...θ
|...|
x...y

What I'm attempting to show here is that the relationship between u and x involves paths through both r and θ; namely, u to r, then r to x, and also u to θ, and then θ to x.

\frac{\partial u}{\partial x}=\frac{\partial u}{\partial r}\frac{\partial r}{\partial x}+\frac{\partial u}{\partial \theta}\frac{\partial \theta}{\partial x}

It's similar for the partial of u with respect to y.

I don't know where to go from here to compute the second derivative. Would I need to use the product rule on each term when differentiating again? By the way, the answer is..

\nabla^2 u=\frac{\partial^2 u}{\partial r^2}+\frac{1}{r}\frac{\partial u}{\partial r}+\frac{1}{r^2}\frac{\partial^2 u}{\partial\phi^2}

 

[Xyius]

Theta and Phi are just dummy variables, my chain rule isn't less correct because of the fact that I used a different variable. I just happen to like Phi better than theta.

Anyway, I haven't gotten a chance to try this again, as soon as I try it ill report back.