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I have been unable to find the equivalent of the 5-point stencil finite differences for the Laplacian operator.

In 2 dimensions for me it is clear that, using the finite difference method: $$ \nabla_{2D}^2u = \frac{1}{h^2} \left( u_{1,0} + u_{-1,0} + u_{0,1} + u_{0,-1} -4 u_{0,0} \right) $$ (h being the size grid/step)

But I am not sure if it is completely symmetric for the 3-dimensional case. Can I just add the terms referring to the 3rd dimension? $$ \nabla_{3D}^2u = \frac{1}{h^2} \left( u_{1,0,0} + u_{-1,0,0} + u_{0,1,0} + u_{0,-1,0} + u_{0,0,1} + u_{0,0,-1} -6 u_{0,0,0} \right) $$

A source where I could find the different accuracy terms for the 3D Laplacian would be also helpful.

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    $\begingroup$ Just a comment: In Finite Differences, it can be much convenient to write it in terms of Kronecker products: if $A$ is the classical $[1,-2,1]$ matrix, you have that the 3D Laplacian is $$A = I \otimes I \otimes A + I \otimes A \otimes I + A \otimes I \otimes I$$ where $I$ is the identity matrix. $\endgroup$ – VoB Sep 14 at 17:19
  • $\begingroup$ @VoB, I think that you could expand your comment into an answer. $\endgroup$ – nicoguaro Sep 15 at 14:05
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Yes, that finite difference is correct. You can obtain it using a finite difference in each direction for the Laplace operator in each coordinate.

\begin{align} \nabla^2 u =& \frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} + \frac{\partial^2 u}{\partial z^2}\\ \approx& \frac{1}{h^2}[u(x + h, y, z) - 2u(x, y, z) + u(x -h, y, z)]\\ &+ \frac{1}{h^2}[u(x, y + h, z) - 2u(x, y, z) + u(x, y - h, z)]\\ &+ \frac{1}{h^2}[u(x, y, z + h) - 2u(x, y, z) + u(x, y, z - h)]\\ =& \frac{1}{h^2}[u(x + h, y, z) + u(x -h, y, z) + u(x, y + h, z) + u(x, y - h, z)\\ &+ u(x, y, z + h) + u(x, y, z - h) - 6u(x, y, z)]\, . \end{align}

Alternatively, you could interpolate a polynomial that passes through the points in your stencil and then compute its Laplacian and evaluate it at $(x, y, z)$.

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