Commit f5a65b82 by Tianbai Xiao

### Finish theory part


Former-commit-id: 2fbcd7a2
parent 4133e938
 ... ... @@ -134,7 +134,7 @@ over all angles to give .. math:: \frac{1}{v} \partial_{t} u+\nabla_{x} \cdot\langle\Omega m \psi\rangle=\langle m \mathcal{C}(\psi)\rangle \frac{1}{v} \partial_{t} u+\nabla_{x} \cdot\langle\Omega m \psi\rangle=\langle m \mathcal{C}(\psi)\rangle. The system above is not closed; a recipe, or closure, must be prescribed to express unknown quantities in terms of the given moments. Often this is done via an ... ... @@ -142,7 +142,7 @@ approximation for :math:\psi that depends on :math:u, .. math:: \psi(x, \Omega, t) \simeq \mathcal{E}(u(x, t))(\Omega) \psi(x, \Omega, t) \simeq \mathcal{E}(u(x, t))(\Omega). A general strategy for prescribing a closure is to use the solution of a constrained optimization problem ... ... @@ -151,8 +151,7 @@ use the solution of a constrained optimization problem :label: closure \min_{g \in \operatorname{Dom}(\mathcal{H})} & \mathcal{H}(g) \\ \quad \text { s.t. } & \langle\mathbf{m} g\rangle=\langle\mathbf{m} \psi\rangle=u \end{array} \quad \text { s.t. } & \langle\mathbf{m} g\rangle=\langle\mathbf{m} \psi\rangle=u, where :math:\mathcal H(g)=\langle \eta(g) \rangle and $\eta: \mathbb R \rightarrow \mathbb R$ is a convex function that is related to ... ... @@ -161,21 +160,27 @@ Bose-Einstein statistics .. math:: \eta(g)=\frac{2 k \nu^{2}}{c^{3}}\left[n_{g} \log \left(n_{g}\right)-\left(n_{g}+1\right) \log \left(n_{g}+1\right)\right] \eta(g)=\frac{2 k \nu^{2}}{v^{3}}\left[n_{g} \log \left(n_{g}\right)-\left(n_{g}+1\right) \log \left(n_{g}+1\right)\right], where :math:n_g is the occupation number associated with g, .. math:: n_{g}:=\frac{v^{2}}{2 h \nu^{3}} g. The solution of :eq:closure is expressed in terms of the Legendre dual .. math:: \eta_{*}(f)=-\frac{2 k \nu^{2}}{c^{3}} \log \left(1-\exp \left(-\frac{h \nu c}{k} f\right)\right) \eta_{*}(f)=-\frac{2 k \nu^{2}}{v^{3}} \log \left(1-\exp \left(-\frac{h \nu c}{k} f\right)\right). Let .. math:: \mathcal{B}(\boldsymbol{\alpha}):=\eta_{*}^{\prime}\left(\boldsymbol{\alpha}^{T} \mathbf{m}\right)=\frac{2 h \nu^{3}}{c^{2}} \frac{1}{\exp \left(-\frac{h \nu c}{k} \boldsymbol{\alpha}^{T} \mathbf{m}\right)-1} \mathcal{B}(\boldsymbol{\alpha}):=\eta_{*}^{\prime}\left(\boldsymbol{\alpha}^{T} \mathbf{m}\right)=\frac{2 h \nu^{3}}{v^{2}} \frac{1}{\exp \left(-\frac{h \nu c}{k} \boldsymbol{\alpha}^{T} \mathbf{m}\right)-1}, The solution of :eq:closure is given by :math:\mathcal B(\hat \alpha), where :math:\hat \alpha= \hat \alpha(u) solves the then the solution of :eq:closure is given by :math:\mathcal B(\hat \alpha), where :math:\hat \alpha= \hat \alpha(u) solves the dual problem .. math:: ... ...
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