On 04/22/2018 01:20 AM, Pawan Kumar wrote:
So, by multiplying with an additional weight factor it simply becomes a 1D
coupled system with homogeneous Neumann.
It *looks* like a Neumann boundary condition, but isn't technically speaking.
If you take the Laplace term,
v_{rr} + \frac{1}{r} v_r
and multiply it by 2*pi*r and the test function w, then you get
int_0^R w (v_{rr} + \frac{1}{r} v_r) 2 pi r dr
=
2*pi int_0^R r w v_{rr} + w v_r dr
The first term you can integrate by parts:
2*pi int_0^R -(r w)_r v_r + w v_r dr
=
2*pi int_0^R - w v_r - r w_r v_r + w v_r dr
=
2*pi int_0^R - r w_r v_r dr
which nicely shows the same structure you get when starting with
w (Delta v)
and integrating by parts.
Best
W.
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Wolfgang Bangerth email: [email protected]
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