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 $$\Gamma_1$$.