I want to solve the Poisson equation for a 2D polar system:

$$\Delta_r f(r) = u(r)$$ with the Laplace operator: $ \Delta_r f(r) = \frac{1}{r}\frac{\partial}{\partial r} \left[r \frac{\partial}{\partial r} f(r) \right] $

I have $u(r)$ given as a vector for a non-uniform (quasi-logarithmic) grid. Any ideas how to solve this?


closed as too broad by Bill Barth, Christian Clason, Geoff Oxberry May 10 '14 at 19:56

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    $\begingroup$ What are the boundary conditions? What numerical methods are you familiar with? Finite differences? Spectral methods? Finite element? As Bill Barth says there are many ways to tackle this problem. Please provide more information so that the community can help you. $\endgroup$ – James May 8 '14 at 13:44

You can use finite differences (this is from GD Smith page 213)

$$\frac{\partial^2 }{\partial r^2}u+\frac{1}{r}\frac{\partial}{\partial r}u=f$$

Define the mesh points in the $r-\theta$ as $r=i \delta r$ so now at point $(i)$ the the equation is approximated as

$$\frac{u_{i+1}-2u_{i}+u_{1-i}}{(r(i+1)-r(i-1))^2}+\frac{1}{r(i)} \frac{u_{i+1}-u_{i-1}}{2(r(i+1)-r(i))} =f(i)$$

At this point you should be able to see how to construct this into a matrix formulation and include your BC's.

  • $\begingroup$ Those difference formulas are not correct for a non-uniform grid since the left and right spacing are generally different. Also worth mentioning that you've given a 2D (r,θ) case, but it looks like OP has a 1D (cylindrically symmetric) problem. $\endgroup$ – Doug Lipinski May 8 '14 at 17:50
  • $\begingroup$ Doug, you are correct. When I get the chance later I can rewrite the FD equation with explicit differencing in the denominators. Also, yes I see now that the OP's post has no $\partial\theta$ term. $\endgroup$ – user7257 May 8 '14 at 18:05

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