Dear Dr. Yves, Yes, you are right. I make the mistake. Thank you so much for time and your appreciated help.
2016-03-22 12:03 GMT+00:00 Yves Renard <[email protected]>: > > > 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 > > --------- > > -- *ZAIM Yassine * *PhD Student in Applied Mathematics*
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