On 4/23/24 07:59, Andreas Müsing wrote:
a) Is the weak form above correct for the cylinder symmetric case?
Yes, this looks correct. It also correctly leads to a symmetric bilinear form,
which is reassuring for the laplace equation.
b) How do I implement this in deal.ii?
for (cell=...)
for (q=...)
for (i=...)
for (j=...)
local_matrix(i,j) += fe_values.shape_grad(i,q) *
fe_values.shape_grad(j,q) *
fe_values.quadrature_point(q)[0] * // r
fe_values.JxW(q);
c) When I look at the first term in the PDE's strong form, I can carry out the
first term r derivative by applying the derivative product rule, which results
in terms having a first and a second derivative in r, respectively. Therefore,
the solution should be in any case a linear combination of first derivative
and second derivative values.
Can you elaborate? It's true that if you multiply everything out, the equation
is of the form
d/dr ... + d^2/dr^2 ... = ...
but I don't understand how you infer something about the solution then?
I used tutorial step-63 with success as a starting point, where the
advection-diffusion equation has both a second derivative (diffusion) term,
and a first derivative (=advection) term. Further, I used an analytic known
solution of my problem to compare the results, and I get reasonable solutions
for e = 1 and b = 2. But why beta = 2 and not beta = 1?
What is the form of the advection equation you are trying to solve, in
cylindrical coordinates?
Best
W.
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Wolfgang Bangerth email: bange...@colostate.edu
www: http://www.math.colostate.edu/~bangerth/
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