heat equation with a laplacian operator squared and power of 2 I've seen 2 variations of the laplacian operator fora heat equation.
Some of which has a square on the laplacian operator and some of which does not have any power on the operator.
What is the differences? Do they refer to the n-tuple of the spatial dimension?
 A: Cartesian coordinates
Gradient ($\nabla $scalar), divergence ($\nabla \cdot$ vector), and curl ($\nabla \times$ vector) extend notions of derivative in a unique way to 3 dimensions or 3 degrees of freedom. The Laplace operator requires the square symbol and is a specific notion of repeated gradients, divergence, and curl as the use of the operator on various objects only makes sense if interpreted in certain ways ie. as a gradient or divergence or curl. 
In the Laplace operator, it start's on a scalar field so div and curl do not make sense and so it has to be a gradient. The second iteration could be divergence or curl but for typical notation and it's use in the heat equation it implies the divergence and not curl operator.
I call it "del" operator but it implies the symbol $\nabla$ and has a meaningless definition by itself of 
$$ \nabla = \frac{\partial}{\partial x} \hat i + \frac{\partial}{\partial y} \hat j + \frac{\partial}{\partial z} \hat k$$ 
and is treated like a vector though it is not by itslf. Knowing that the Laplace operator is created from successive iterations, namely a gradient, then divergence. We can create it's operator as well thanks to the divergence of del and thanks to associativity of the dot product over scalar multiplication.
$$ \nabla^2 = \nabla \cdot \nabla =  \left[\frac{\partial}{\partial x} \hat i + \frac{\partial}{\partial y} \hat j + \frac{\partial}{\partial z} \hat k \right] \cdot  \left[ \frac{\partial}{\partial x} \hat i + \frac{\partial}{\partial y} \hat j + \frac{\partial}{\partial z} \hat k\right] $$
$$ \nabla^2 = \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2}$$
Notably we can see it operates on a scalar field.
Cuvilinear coordinates
Because the axis are themselves functions of the coordinates, the del and Laplace operator are not the same in curvilinear coordinate systems. It is as if the derivative now is dependent on the basis vectors themselves and so they do not get ignored.
See http://en.wikipedia.org/wiki/Laplace_operator
