Axisymmetry- 3D Rod with circular section :

Let us now deal with a cylindrical rod instead of a flat plate. For simplicity we take \(\kappa=1\). In cylindrical coordinates, the Laplace operator becomes (\(r\) is the distance to the axis, \(z\) is the distance along the axis, \(\theta\) polar angle in a fixed plane perpendicular to the axis):

$$ \Delta u = {1\over r}\partial _r(r\partial _r u) + {1\over r^2}\partial ^2_{\theta\theta} u + \partial ^2_{z z}. $$

Symmetry implies that we loose the dependence with respect to \(\theta\); so the domain \(\Omega\) is again a rectangle \(]0,R[\times]0,|[\) . We take the convention of numbering of the edges as in \tt square() (1 for the bottom horizontal ...); the problem is now:

\( r\partial_t u-\partial _r(r\partial _r u) - \partial _z(r\partial _z u) = 0 \hbox{ in } \Omega, \)
\( u(t=0) = u_0 + \frac z{L_z} (u_1-u) \)
\( u|_{\Gamma_4} = u_0,\quad u|_{\Gamma_2} = u_1, \quad \alpha(u-u_e) + {\partial u\over \partial n} |_{\Gamma_1\cup\Gamma_3} = 0. \)

Note that the PDE has been multiplied by \(r\).

After discretization in time with an implicit scheme, with time steps dt, in the FreeFem++ syntax \(r\) becomes \(x\) and \(z\) becomes \(y\) and the problem is:

 problem thermaxi(u,v)=int2d(Th)((u*v/dt + dx(u)*dx(v) + dy(u)*dy(v))*x)
                 + int1d(Th,3)(alpha*x*u*v) - int1d(Th,3)(alpha*x*ue*v)
                 - int2d(Th)(uold*v*x/dt) + on(2,4,u=u0);

Notice that the bilinear form degenerates at \(x=0\). Still one can prove existence and uniqueness for \(u\) and because of this degeneracy no boundary conditions need to be imposed on \(\Gamma_1\).