Is the Laplacian Operator Different in Radial Coordinates?

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SUMMARY

The Laplacian operator in three dimensions is defined as \(\nabla^2 = \frac{1}{r} \frac{d^2}{dr^2}(r) + \frac{1}{r^2}\left(\frac{1}{\sin \phi} \frac{d}{d \phi}(\sin \phi \frac{d}{d \phi}) + \frac{1}{\sin^2 \phi} \frac{d^2}{d \theta^2}\right)\). In a radial system where the azimuthal angle \(\theta\) is set to zero, the operator simplifies to \(\nabla^2 = \frac{1}{r} \frac{d^2}{dr^2}(r) + \frac{1}{r^2}\left(\frac{1}{\sin \phi} \frac{d}{d \phi}(\sin \phi \frac{d}{d \phi})\right)\). This formulation is valid and confirms that the Laplacian operator retains its structure in polar coordinates when \(\theta\) is absent.

PREREQUISITES
  • Understanding of the Laplacian operator in three-dimensional space
  • Familiarity with polar and spherical coordinate systems
  • Knowledge of partial differential equations
  • Basic calculus, particularly differentiation
NEXT STEPS
  • Study the derivation of the Laplacian operator in spherical coordinates
  • Explore applications of the Laplacian operator in physics, particularly in potential theory
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Mathematicians, physicists, and engineers working with differential equations, particularly those focusing on problems in radial coordinates and potential theory.

SeM
Hi, I have that the Laplacian operator for three dimensions of two orders,

\nabla ^2 is:

1/r* d^2/dr^2 (r) + 1/r^2( 1/sin phi d/d phi sin phi d/d phi + 1/sin^2 phi * d^2/d theta^2)

Can this operator be used for a radial system, where r and phi are still valid, but theta absent, by setting theta = 0 ?

so giving:1/r* d^2/dr^2 (r) + 1/r^2( 1/sin phi d/d phi sin phi d/d phi) ?

Does that make sense or is the Laplacian operator of second order in radial (polar) coordinates different?

Thanks
 
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