Problem with 1D integral while using movemesh for Poisson Problem with Neumann Boundary Condition

Hello,
I have tried the Poisson problem with Neumann boundary. I am facing difficulty in the line 1D boundary integral under transformation. If du/dn=q given then qv ds is transformed to QV dS. If du/dn is not given then what about discretized du/dn. Please look at this-
exp1.edp (939 Bytes)

In any case you have to solve a well-posed problem. On the boundary, possibilities are Dirichlet conditions, Neumann conditions, or a mix of both.
You can look at Finite element
section “Weak Form and Boundary Condition”

Your code corrected:
exp1.edp (1.5 KB)

Thank you for your effort Sir, I have a doubt on how did you get the extra term sqrt(a^2y^2+b^2x^2) in
G=-18.sqrt(x^2+y^24./9.)sqrt(a^2y^2+b^2*x^2);

For the formulation on the unit disc with variables p,q you have the equation
-\partial_p(\frac{b}{a}\partial_p U)-\partial_q(\frac{a}{b}\partial_q U)-fab=0.
When you integrate the two first terms by parts (after multiplication by V) you get a boundary term
-\int_\Gamma V[\frac{b}{a}\partial_p U,\frac{a}{b}\partial_q U]\cdot N, thus
G=[\frac{b}{a}\partial_p U_{ex},\frac{a}{b}\partial_q U_{ex}]\cdot N
with U_{ex}=-18(p^2+q^2) you get the G.

Another way is to change variables on the boundary


In your example you have P=(p,q) with p^2+q^2=1, \vec\tau'=(-q,p), d\varphi^{-1}\vec\tau'=(-aq,bp),
thus w=\sqrt{a^2q^2+b^2p^2}.

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Thank you sir for the explanation.