A Uzawa-Type Iterative Algorithm for the Stationary Natural Convection Model

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. 2022 Apr 13;24(4):543. doi: 10.3390/e24040543

Algorithm 2: Uzawa-type iterative algorithm.

Step 1. Obtain the initial guess $(u_{h}^{0}, p_{h}^{0}, T_{h}^{0}) \in M_{h} \times W_{h} \times Z_{h}$ from step 1 of Algorithm 1.
Step 2. Find $({\hat{u}}_{h}^{n + 1}, T_{h}^{n + 1}) \in M_{h} \times Z_{h}$ as the solution of
$\begin{matrix} κ (\nabla T_{h}^{n + 1}, \nabla s) + b_{2} ({\hat{u}}_{h}^{n}; T_{h}^{n + 1}, s) = (γ, s), \forall s \in Z_{h}, \end{matrix}$ (19)

$\begin{matrix} P r (\nabla {\hat{u}}_{h}^{n + 1}, \nabla v) + b_{1} ({\hat{u}}_{h}^{n}; {\hat{u}}_{h}^{n + 1}, v) - (p_{h}^{n}, \nabla \cdot v) = P r R a (j T_{h}^{n + 1}, v), \forall v \in M_{h} . \end{matrix}$ (20)
Step 3. Find $ℏ_{h}^{n + 1} \in W_{h}$ as the solution of
$\begin{matrix} \begin{matrix} (\nabla ℏ_{h}^{n + 1}, \nabla q) = (\nabla \cdot {\hat{u}}_{h}^{n + 1}, q), q \in W_{h} . \end{matrix} \end{matrix}$ (21)
Step 4. Compute $u_{h}^{n + 1}$ with $u_{h}^{n + 1} = {\hat{u}}_{h}^{n + 1} + \nabla ℏ_{h}^{n + 1}$ .
Step 5. Given a relaxation parameter $ρ > 0$ , find $p_{h}^{n + 1} \in W_{h}$ from the Richardson update
$\begin{matrix} \begin{matrix} (p_{h}^{n + 1}, q) = (p_{h}^{n}, q) - P r ρ (\nabla ℏ_{h}^{n + 1}, \nabla q), \forall q \in W_{h} . \end{matrix} \end{matrix}$ (22)
From (21) and Step 4 of Algorithm 2, we obtain $(\nabla \cdot u_{h}^{n + 1}, q) = (\nabla \cdot {\hat{u}}_{h}^{n + 1}, q) - (\nabla ℏ_{h}^{n + 1}, \nabla q) = 0$ . So the velocity obtained by Algorithm 2 satisfies the weakly divergence-free condition. Moreover, we expect to show the iterative errors between the finite element solutions to (10)–(12) and the Uzawa-type iterative solutions to Algorithm 2. For convenience, assume that $E_{h}^{n} = u_{h} - u_{h}^{n}$ , ${\hat{E}}_{h}^{n} = u_{h} - {\hat{u}}_{h}^{n}$ , $η_{h}^{n} = p_{h} - p_{h}^{n}$ and $θ_{h}^{n} = T_{h} - T_{h}^{n}$ . Then, we have ${\hat{E}}_{h}^{n} = E_{h}^{n} + \nabla ℏ_{h}^{n}$ .