Dear Yassine ZAIM,
The test program "Laplacian" solves the problem
- \Delta u = f
+ boundary conditions
where \Delta is the Laplace operator.
The consequence is that the right-hand side has the wrong sign in your
expression.
Yves.
Le 18/03/2016 17:50, Yassine ZAIM a écrit :
> /* exact solution */
> scalar_type sol_u(const base_node &x) { return (pow(x[0],2) -
> pow(x[0],1))*pow(x[1],1); }
>
> /* righ hand side */
> scalar_type sol_f(const base_node &x)
> { return 2*pow(x[1],1); }
>
> /* gradient of the exact solution */
> base_small_vector sol_grad(const base_node &x)
> { base_small_vector res(2);
> res[0] = (2*pow(x[0],1) - 1)*pow(x[1],1);
> res[1] = pow(x[0],2) - pow(x[0],1);
> return res; }
--
Yves Renard ([email protected]) tel : (33) 04.72.43.87.08
Pole de Mathematiques, INSA-Lyon fax : (33) 04.72.43.85.29
20, rue Albert Einstein
69621 Villeurbanne Cedex, FRANCE
http://math.univ-lyon1.fr/~renard
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