Line 6: Line 6:
 
<math>
 
<math>
 
\Delta f=\frac{\partial^{2} f}{\partial x^{2}}+\frac{\partial^{2} f}{\partial y^{2}}+\frac{\partial^{2} f}{\partial z^{2}}\
 
\Delta f=\frac{\partial^{2} f}{\partial x^{2}}+\frac{\partial^{2} f}{\partial y^{2}}+\frac{\partial^{2} f}{\partial z^{2}}\
<math>
+
</math>
 +
 
 +
The basis for this conversion is, in the opinion of the authors, quite clear. The dot product of the vectors: <math> \nabla f = \Bigg[\frac{\partial f}{\partial x_{1}},\cdots,\frac{\partial f}{\partial x_{n}}\Bigg] </math>

Revision as of 14:10, 6 December 2020

Coordinate Conversions for the Laplace Operator

It is most common to use the Laplace Operator $ \Delta $ in three-dimensions, as that is the dimensionality of our physical universe. Thus, the Laplace Operator is often used in 3-D Cartesian coordinates, cylindrical coordinates, and spherical coordinates.

$ \Delta f=\frac{\partial^{2} f}{\partial x^{2}}+\frac{\partial^{2} f}{\partial y^{2}}+\frac{\partial^{2} f}{\partial z^{2}}\ $

The basis for this conversion is, in the opinion of the authors, quite clear. The dot product of the vectors: $ \nabla f = \Bigg[\frac{\partial f}{\partial x_{1}},\cdots,\frac{\partial f}{\partial x_{n}}\Bigg] $

Alumni Liaison

Basic linear algebra uncovers and clarifies very important geometry and algebra.

Dr. Paul Garrett