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Reduced Basis

Heat Diffusion on a 3D thermal fin \(\Omega\): Heat flux enter from \(\Gamma_{root}\) and outs \(\Gamma_{top}\). (Download all here : )

Thermal fin is represented as a union of blocks \(\Omega =\cup B_b \) for \(b=0,1,2,...,11\). Each block with conductivity \(\kappa_b>0\)

Thermal FinConductivitiesTriangulation of Thermal Fin

Heat flux Input on root is 1 ...... Output to measure is \( s(u)=\int_{\Gamma_{top}} u(x)dx \)

\begin{array}{rcl} -\kappa_b\Delta u & =& 0 \quad\mbox{on each block } B_b,\quad b=1,2,...\\ -\kappa_b\frac{\partial u}{\partial n}&=& Bi \cdot\,u \quad \mbox{on } \Gamma_b=\partial B_b\cap \partial\Omega,\quad b=1,2,... (Bi>0 \mbox{ Biot number})\\ -\kappa_b\frac{\partial u}{\partial n}&=& -q \quad \mbox{on } \Gamma_{root} \end{array}

Variational formulation:

Find \(u\in H^1(\Omega)\) such that for all test \(v\in H^1(\Omega)\) \begin{array}{rcl} \sum_{b=0}^{11}\kappa_b\int_{B_b}\nabla u \cdot \nabla v+Bi\sum_{b=0}^{11}\int_{\Gamma_b} u \,v& =& \int_{\Gamma_{root}} q \,v. \end{array}

Example of solution (using medit)Example of solution (using plot)

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Page last modified on February 07, 2017, at 05:46 PM