4.1 The existence and uniqueness and the stability as well as the convergence of the SMFEROE solutions
The existence and uniqueness and the stability as well as the convergence of the solutions for the SMFEROE formulation of the 2D unsteady conduction-convection problem have the following main conclusions.
Theorem 3
Under the conditions of Theorem
2, Problem
IV
has only a set of solutions
\((\boldsymbol {u}_{d}^{n}, p_{d}^{n},Q_{d}^{n})\in X^{d}\times M^{d}\times W^{d}\)
such that
$$ \bigl\Vert \boldsymbol {u}_{d}^{n} \bigr\Vert _{0}+ \bigl\Vert Q_{d}^{n} \bigr\Vert _{0}+ k\sum_{i=L+1}^{n}\bigl( \bigl\Vert \nabla\boldsymbol {u}_{d}^{i} \bigr\Vert _{0}+ \bigl\Vert p_{d}^{i} \bigr\Vert _{0}+ \bigl\Vert \nabla Q_{d}^{i} \bigr\Vert _{0}\bigr) \le C \Vert \omega \Vert _{0},\quad 1\le n\le N, $$
(41)
which implies that the SMFEROE solutions
\((\boldsymbol {u}_{d}^{n}, p_{d}^{n},Q_{d}^{n})\) (\(1\le n\le N\)) of Problem
IV
are stable. When
\(k=O(h)\)
and
\({N}_{0}\mu^{-1} \Vert \nabla{\boldsymbol {u}}_{d}^{n} \Vert _{0}\le1/4\) (\(L+1\le n\le N\)), we have the error estimations
$$\begin{aligned}& \bigl\Vert \boldsymbol {u}_{h}^{n}- \boldsymbol {u}_{d}^{n} \bigr\Vert _{0}+ \bigl\Vert Q_{h}^{n}-Q_{d}^{n} \bigr\Vert _{0}+k \bigl\Vert \nabla \bigl(\boldsymbol {u}_{h}^{n}- \boldsymbol {u}_{d}^{n}\bigr) \bigr\Vert _{0}+k \bigl\Vert \nabla \bigl(Q_{h}^{n}-Q_{d}^{n} \bigr) \bigr\Vert _{0} \\& \quad {}+\sqrt{k} \bigl\Vert p_{h}^{n}-p_{d}^{n} \bigr\Vert _{0}\le CLk \Biggl(\sum_{j=d+1}^{l} \lambda_{j} \Biggr)^{1/2}, \quad 1\le n\le L; \end{aligned}$$
(42)
$$\begin{aligned}& \bigl\Vert \boldsymbol {u}_{h}^{n}- \boldsymbol {u}_{d}^{n} \bigr\Vert _{0}+ \bigl\Vert Q_{h}^{n}-Q_{d}^{n} \bigr\Vert _{0}+k \bigl\Vert \nabla \bigl(\boldsymbol {u}_{h}^{n}- \boldsymbol {u}_{d}^{n}\bigr) \bigr\Vert _{0}+k \bigl\Vert \nabla \bigl(Q_{h}^{n}-Q_{d}^{n} \bigr) \bigr\Vert _{0} \\& \quad {}+\sqrt{k} \bigl\Vert p_{h}^{n}-p_{d}^{n} \bigr\Vert _{0}\leq C\bigl(k+h^{2}\bigr)+ CLk \Biggl(\sum _{j=d+1}^{l}\lambda_{j} \Biggr)^{1/2},\quad L+1\le n\le N. \end{aligned}$$
(43)
Proof
When \(1\le n\le L\), from (37), we immediately gain unique \((\boldsymbol {u}_{d}^{n},p_{d}^{n}, Q_{d}^{n})\in X^{d}\times M^{d}\times W^{d}\) (\(1\le n\le L\)). When \(L+1\le n\le N\), by using the same approaches as proving Theorem 2 in [7], from (38)-(40) we can gain unique \((\boldsymbol {u}_{d}^{n},p_{d}^{n}, Q_{d}^{n})\in X^{d}\times M^{d}\times W^{d}\) (\(L+1\le n\le N\)). Thus, Problem IV has only a set of solutions \((\boldsymbol {u}_{d}^{n},p_{d}^{n}, Q_{d}^{n})\in X^{d}\times M^{d}\times W^{d}\) (\(1\le n\le N\)).
Next, we devote ourselves to proving that (41) holds.
When \(1\le n\le L\), by (24)–(26) and Theorem 2, there holds (41).
When \(L+1\le n\le N\), by choosing \(\boldsymbol{\psi}_{d}=\boldsymbol {u}_{d}^{n}\) in (38) and \(q_{d}=p_{d}^{n}\) in (39), noting that there hold \((\rho_{h}p_{d},p_{d})= \Vert \rho_{h}p_{d} \Vert _{0}^{2}\) and \((p_{d}-\rho_{h}p_{d},p_{d}-\rho_{h}p_{d})= \Vert p_{d} \Vert _{0}^{2}- \Vert \rho_{h}p_{d} \Vert _{0}^{2}\ge0\) from (11), and using (3) and Hölder’s and Cauchy’s inequalities, we obtain
$$\begin{aligned}& \bigl\Vert \boldsymbol {u}_{d}^{n} \bigr\Vert _{0}^{2}+ k\mu \bigl\Vert \nabla{\boldsymbol {u}}_{d}^{n} \bigr\Vert _{0}^{2}+k \varepsilon\bigl( \bigl\Vert p_{d}^{n} \bigr\Vert _{0}^{2}- \bigl\Vert \varrho_{h}p_{h}^{d} \bigr\Vert _{0}^{2}\bigr) \\& \quad =\bigl(\boldsymbol {u}_{d}^{n-1},\boldsymbol {u}_{d}^{n}\bigr)+ \bigl( Q_{d}^{n}\boldsymbol {j},\boldsymbol {u}_{d}^{n}\bigr) \\& \quad \leq \frac{1}{2}\bigl( \bigl\Vert \boldsymbol {u}_{d}^{n} \bigr\Vert _{0}^{2}+ \bigl\Vert \boldsymbol {u}_{d}^{n-1} \bigr\Vert _{0}^{2}\bigr)+ C{k} \bigl\Vert Q_{d}^{n} \bigr\Vert _{-1}^{2}+ \frac{k\mu}{2} \bigl\Vert \nabla {\boldsymbol {u}}_{d}^{n} \bigr\Vert _{0}^{2}. \end{aligned}$$
(44)
It follows from (44) that
$$ \bigl\Vert \boldsymbol {u}_{d}^{n} \bigr\Vert _{0}^{2}- \bigl\Vert \boldsymbol {u}_{d}^{n-1} \bigr\Vert _{0}^{2}+ 2k\mu \bigl\Vert \nabla {\boldsymbol {u}}_{d}^{n} \bigr\Vert _{0}^{2}+2k\varepsilon\bigl( \bigl\Vert p_{d}^{n} \bigr\Vert _{0}^{2}- \bigl\Vert \varrho_{h}p_{h}^{d} \bigr\Vert _{0}^{2}\bigr)\leq Ck \bigl\Vert Q_{d}^{n} \bigr\Vert _{-1}^{2}. $$
(45)
If \(p_{d}^{n}\neq0\), then it is easily known from (11) that \(\Vert p_{d}^{n} \Vert _{0}^{2}> \Vert \varrho_{h}p_{d}^{n} \Vert _{0}^{2}\). Thus, there is a positive real number \(\delta\in(0,1)\) that satisfies \(\delta \Vert p_{d}^{n} \Vert _{0}^{2}\ge \Vert \varrho_{h}p_{d}^{n} \Vert _{0}^{2}\). By summing (45) from \(L+1\) to n simplified, we have
$$ \bigl\Vert \boldsymbol {u}_{d}^{n} \bigr\Vert _{0}^{2}+ k\sum_{i=L+1}^{n} \bigl( \bigl\Vert \nabla\boldsymbol {u}_{d}^{i} \bigr\Vert _{0}^{2}+ \bigl\Vert p_{d}^{i} \bigr\Vert _{0}^{2}\bigr) \le C \bigl\Vert \boldsymbol {u}_{d}^{L} \bigr\Vert _{0}^{2}+ Ck\sum _{i=L+1}^{n} \bigl\Vert Q_{d}^{i} \bigr\Vert _{-1}^{2}. $$
(46)
Taking a square root for (46) and utilizing \((\sum_{i=1}^{n}a_{i}^{2} )^{1/2}\ge\sum_{i=1}^{n} \vert a_{i} \vert /\sqrt{n}\) yield
$$ \bigl\Vert \boldsymbol {u}_{d}^{n} \bigr\Vert _{0}+ k\sum_{i=L+1}^{n} \bigl( \bigl\Vert \nabla\boldsymbol {u}_{d}^{i} \bigr\Vert _{0}+ \bigl\Vert p_{d}^{i} \bigr\Vert _{0}\bigr) \le C \Biggl( \bigl\Vert \boldsymbol {u}_{d}^{L} \bigr\Vert _{0}^{2}+ k\sum_{i=L+1}^{n} \bigl\Vert Q_{d}^{i} \bigr\Vert _{-1}^{2} \Biggr)^{1/2}. $$
(47)
By choosing \(\varphi_{d}=Q_{d}^{n}\) in (40) and by making use of (4) and Hölder ’s and Cauchy’s inequalities, we obtain
$$ \bigl\Vert Q_{d}^{n} \bigr\Vert _{0}^{2}+ \frac{2k}{\gamma_{0}} \bigl\Vert \nabla Q_{d}^{n} \bigr\Vert _{0}^{2}\le \bigl\Vert Q_{d}^{n-1} \bigr\Vert _{0}^{2}. $$
(48)
Summing (48) from L+1 to n yields
$$ \bigl\Vert Q_{d}^{n} \bigr\Vert _{0}^{2}+ \frac{2k}{\gamma_{0}}\sum _{i=L+1}^{n} \bigl\Vert \nabla Q_{d}^{i} \bigr\Vert _{0}^{2}\le \bigl\Vert Q^{L} \bigr\Vert _{0}^{2}. $$
(49)
By extracting a square root for (49), making use of \((\sum_{i=1}^{n}a_{i}^{2} )^{1/2}\ge\sum_{i=1}^{n} \vert a_{i} \vert /\sqrt{n}\) and (41) when \(n=L\), and then simplifying, we obtain
$$ \bigl\Vert Q_{d}^{n} \bigr\Vert _{0}+ k\sum_{i=L+1}^{n} \bigl\Vert \nabla Q_{d}^{i} \bigr\Vert _{0} \le C \Vert \omega \Vert _{0}. $$
(50)
By noting that \(\Vert \cdot \Vert _{-1}\le C \Vert \cdot \Vert _{0}\) and by using (41) when \(n=L\), from (47) and (49), we obtain
$$ \bigl\Vert \boldsymbol {u}_{d}^{n} \bigr\Vert _{0}+ k\sum_{i=L+1}^{n} \bigl( \bigl\Vert \nabla\boldsymbol {u}_{d}^{i} \bigr\Vert _{0}+ \bigl\Vert p_{d}^{i} \bigr\Vert _{0}\bigr) \le C \Vert \omega \Vert _{0}. $$
(51)
Combining (50) with (51) yields that (41) holds when \(L+1\le n\le N\). If \(p_{d}^{n}=0\), (41) is distinctly correct.
When \(1\le n\le L\), with Lemma 2 and (37), we immediately obtain (42).
When \(L+1\le n\le N\), by subtracting Problem IV from Problem III choosing \(\boldsymbol{\psi}_{h}=\boldsymbol{\psi}_{d}\), \(q_{h}=q_{d}\), and \(\varphi_{h}=\varphi_{d}\), we acquire
$$\begin{aligned}& \bigl(\boldsymbol {u}_{h}^{n}- \boldsymbol {u}_{d}^{n}, \boldsymbol{\psi}_{d}\bigr)+ kA \bigl({\boldsymbol {u}}_{h}^{n}- {\boldsymbol {u}}_{d}^{n}, \boldsymbol{\psi}_{d} \bigr)+kA_{1}\bigl( {\boldsymbol {u}}_{h}^{n}, { \boldsymbol {u}}_{h}^{n},\boldsymbol {\psi}_{d}\bigr) -kA_{1}\bigl( {\boldsymbol {u}}_{d}^{n}, { \boldsymbol {u}}_{d}^{n}, \boldsymbol{\psi}_{d}\bigr) \\& \quad {}-kB\bigl(p_{h}^{n}-p_{d}^{n},\boldsymbol {\psi}_{d}\bigr)= k\bigl(\bigl( Q_{h}^{n}- Q_{d}^{n}\bigr)\boldsymbol {j}, \boldsymbol{\psi}_{d} \bigr) +\bigl(\boldsymbol {u}_{h}^{n-1}-\boldsymbol {u}_{d}^{n-1}, \boldsymbol{\psi}_{d}\bigr),\quad \forall \boldsymbol{\psi}_{d}\in X^{d}, \end{aligned}$$
(52)
$$\begin{aligned}& b\bigl(q_{d},\boldsymbol {u}_{h}^{n}- \boldsymbol {u}_{d}^{n}\bigr)+\varepsilon \bigl(p_{h}^{n}-p_{d}^{n}- \varrho_{h}\bigl(p_{h}^{n}-p_{d}^{n} \bigr), q_{d}-\varrho_{h}q_{d}\bigr) =0,\quad \forall q_{d}\in M^{d}, \end{aligned}$$
(53)
$$\begin{aligned}& \bigl(Q_{h}^{n}-Q_{d}^{n}, \varphi_{d}\bigr)+kD_{0}\bigl(Q_{h}^{n}-Q_{d}^{n}, \varphi_{d}\bigr)+kA_{2}\bigl({\boldsymbol {u}}_{h}^{n}, Q_{h}^{n}, \varphi_{d}\bigr)-kA_{2}\bigl( {\boldsymbol {u}}_{d}^{n}, Q_{d}^{n}, \varphi_{d}\bigr) \\& \quad =\bigl(Q_{h}^{n-1}-Q_{d}^{n-1}, \varphi_{d}\bigr),\quad\forall\varphi_{d}\in W^{d}, L+1\le n\le N. \end{aligned}$$
(54)
Let \(\boldsymbol {e}^{n}=P^{d}\boldsymbol {u}_{h}^{n}-\boldsymbol {u}_{d}^{n}\), \(\boldsymbol {f}^{n}=\boldsymbol {u}_{h}^{n}-P^{d}\boldsymbol {u}_{h}^{n}\), \(\eta ^{n}=Z^{d}p_{h}^{n}-p_{d}^{n}\), and \(\xi^{n}=p_{h}^{n}-Z^{d}p_{h}^{n}\). First, from (18), (52), and (53), we obtain
$$ \begin{aligned}[b] &\bigl\Vert \boldsymbol {e}^{n} \bigr\Vert _{0}^{2}+k \mu \bigl\Vert \nabla\boldsymbol {e}^{n} \bigr\Vert _{0}^{2}\\ &\quad =\bigl(P^{d} \boldsymbol {u}_{h}^{n}-\boldsymbol {u}_{d}^{n}, \boldsymbol {e}^{n}\bigr)+kA\bigl(P^{d}\boldsymbol {u}_{h}^{n}-\boldsymbol {u}_{d}^{n}, \boldsymbol {e}^{n}\bigr) \\ &\quad =-\bigl(\boldsymbol {f}^{n},\boldsymbol {e}^{n}\bigr)+\bigl( \boldsymbol {u}_{h}^{n}-\boldsymbol {u}_{d}^{n}, \boldsymbol {e}^{n}\bigr)+kA\bigl(\boldsymbol {u}_{h}^{n}- \boldsymbol {u}_{d}^{n},\boldsymbol {e}^{n}\bigr) \\ &\quad =\bigl(\boldsymbol {f}^{n-1}-\boldsymbol {f}^{n},\boldsymbol {e}^{n}\bigr)+kB\bigl(p_{h}^{n}-p_{d}^{n}, \boldsymbol {e}^{n}\bigr)-kA_{1}\bigl( {\boldsymbol {u}}_{h}^{n}, {\boldsymbol {u}}_{h}^{n}, \boldsymbol {e}^{n}\bigr) \\ &\qquad {}+kA_{1}\bigl( {\boldsymbol {u}}_{d}^{n},{ \boldsymbol {u}}_{d}^{n}, \boldsymbol {e}^{n}\bigr)+ \bigl(\boldsymbol {e}^{n-1}, \boldsymbol {e}^{n}\bigr) +k\bigl( \bigl( Q_{h}^{n}- Q_{d}^{n}\bigr) \boldsymbol {j}, \boldsymbol {e}^{n}\bigr) \\ &\quad =\bigl(\boldsymbol {f}^{n-1}-\boldsymbol {f}^{n},\boldsymbol {e}^{n}\bigr)-kA_{1}\bigl({\boldsymbol {u}}_{h}^{n}, {\boldsymbol {u}}_{h}^{n}, \boldsymbol {e}^{n} \bigr)+kA_{1}\bigl({\boldsymbol {u}}_{d}^{n},{ \boldsymbol {u}}_{d}^{n}, \boldsymbol {e}^{n}\bigr) \\ &\qquad {}+k\bigl(\bigl( Q_{h}^{n}- Q_{d}^{n} \bigr)\boldsymbol {j}, \boldsymbol {e}^{n}\bigr) +\bigl(\boldsymbol {e}^{n-1},\boldsymbol {e}^{n}\bigr)+k B\bigl(\xi^{n}, \boldsymbol {e}^{n}\bigr)+kB\bigl(\eta,\boldsymbol {e}^{n}\bigr) \\ &\qquad {}-2k\varepsilon\bigl(p_{h}^{n}-p_{d}^{n}- \varrho _{h}\bigl(p_{h}^{n}-p_{d}^{n} \bigr), \eta^{n}-\varrho_{h}\eta\bigr) \\ &\quad \le C\bigl(k^{-1} \bigl\Vert \boldsymbol {f}^{n-1}- \boldsymbol {f}^{n} \bigr\Vert _{-1}^{2}\bigr)+4k \mu^{-1} \bigl\Vert \eta^{n} \bigr\Vert _{0}^{2}+Ck \bigl\Vert \xi^{n} \bigr\Vert _{0}^{2} \\ &\qquad {}+\frac{k\mu}{8} \bigl\Vert \nabla\boldsymbol {e}^{n} \bigr\Vert _{0}^{2}+\frac{1}{2} \bigl\Vert \boldsymbol {e}^{n-1} \bigr\Vert _{0}^{2}+\frac{1}{2} \bigl\Vert \boldsymbol {e}^{n} \bigr\Vert _{0}^{2}-2k \varepsilon\bigl( \bigl\Vert \eta^{n} \bigr\Vert _{0}^{2}- \Vert \varrho_{h}\eta \Vert _{0}^{2}\bigr) \\ &\qquad {}-kA_{1}\bigl( {\boldsymbol {u}}_{h}^{n}, { \boldsymbol {u}}_{h}^{n}, \boldsymbol {e}^{n}\bigr) +kA_{1}\bigl( {\boldsymbol {u}}_{d}^{n}, { \boldsymbol {u}}_{d}^{n}, \boldsymbol {e}^{n}\bigr) +k \bigl(\bigl( Q_{h}^{n}- Q_{d}^{n}\bigr) \boldsymbol {j}, \boldsymbol {e}^{n}\bigr). \end{aligned} $$
(55)
Next, when \(N_{0}\mu^{-1} \Vert \nabla\boldsymbol {\boldsymbol {u}}_{h}^{n} \Vert _{0}\le1/4\) and \(N_{0}\mu^{-1} \Vert \nabla {\boldsymbol {u}}_{d}^{n} \Vert _{0}\le1/4\) (\(L+1\le n\le N\)), with the properties of \(A_{1}(\cdot,\cdot,\cdot)\), Hölder’s and Cauchy’s inequalities, and Lemma 2, we gain
$$ kA_{1}\bigl( {\boldsymbol {u}}_{d}^{n}, {\boldsymbol {u}}_{d}^{n}, \boldsymbol {e}^{n} \bigr)-kA_{1}\bigl( {\boldsymbol {u}}_{h}^{n}, { \boldsymbol {u}}_{h}^{n}, \boldsymbol {e}^{n}\bigr) \le Ck \bigl\Vert \nabla\boldsymbol {f}^{n} \bigr\Vert _{0}^{2}+\frac{k\mu }{4} \bigl\Vert \nabla\boldsymbol {e}^{n} \bigr\Vert _{0}^{2}. $$
(56)
And then, with Hölder’s and Cauchy’s inequalities, we gain
$$ k\bigl(\bigl( Q_{h}^{n}- Q_{d}^{n}\bigr)\boldsymbol {j}, \boldsymbol {e}^{n} \bigr)\le Ck \bigl\Vert Q_{h}^{n}- Q_{d}^{n} \bigr\Vert _{-1}^{2}+ \frac{k\mu}{8} \bigl\Vert \nabla\boldsymbol {e}^{n} \bigr\Vert _{0}^{2}. $$
(57)
If \(\eta^{n}\neq0\), it is accessible to get \(\Vert \eta^{n} \Vert _{0}^{2}> \Vert \varrho_{h}\eta \Vert _{0}^{2}\) from (11). Thus, there exists a positive real number \(\delta\in(0,1)\) that satisfies \(\delta \Vert \eta^{n} \Vert _{0}^{2}\ge \Vert \varrho_{h}\eta \Vert _{0}^{2}\). When \(k=O(h)\), by choosing \(\varepsilon=5\mu^{-1}(1-\delta)^{-1}\), combining (55) with (56) and (57), using (28), and then simplifying, we acquire
$$ \begin{gathered}[b] \bigl\Vert \boldsymbol {e}^{n} \bigr\Vert _{0}^{2}- \bigl\Vert \boldsymbol {e}^{n-1} \bigr\Vert _{0}^{2}+k \bigl\Vert \nabla \boldsymbol {e}^{n} \bigr\Vert _{0}^{2}+k \bigl\Vert \eta^{n} \bigr\Vert _{0}^{2} \\ \quad \le Ck\bigl( \bigl\Vert \nabla\boldsymbol {f}^{n} \bigr\Vert _{0}^{2}+ \bigl\Vert \boldsymbol {f}^{n-1} \bigr\Vert _{0}^{2}+ \bigl\Vert \xi^{n} \bigr\Vert _{0}^{2}\bigr) + Ck \bigl\Vert Q_{h}^{n}- Q_{d}^{n} \bigr\Vert _{-1}^{2}. \end{gathered} $$
(58)
Summing (58) from \(L+1\) to n yields
$$\begin{aligned}& \bigl\Vert \boldsymbol {e}^{n} \bigr\Vert _{0}^{2}+k\sum_{i=L+1}^{n} \bigl( \bigl\Vert \nabla\boldsymbol {e}^{i} \bigr\Vert _{0}^{2}+ \bigl\Vert \eta^{n} \bigr\Vert _{0}^{2}\bigr) \\& \quad \le C \bigl\Vert \boldsymbol {e}^{L} \bigr\Vert _{0}^{2}+Ck \sum_{i=L}^{n} \bigl( \bigl\Vert \nabla\boldsymbol {f}^{i} \bigr\Vert _{0}^{2}+ \bigl\Vert \xi^{i} \bigr\Vert _{0}^{2}+ \bigl\Vert Q_{h}^{i}- Q_{d}^{i} \bigr\Vert _{-1}^{2}\bigr). \end{aligned}$$
(59)
By extraction of a square root to (59) and making use of \((\sum_{i=1}^{n}a_{i}^{2} )^{1/2}\ge\sum_{n=1}^{n} \vert a_{i} \vert /\sqrt{n}\), we gain
$$\begin{aligned}& \bigl\Vert \boldsymbol {e}^{n} \bigr\Vert _{0} +k \sum_{i=L+1}^{n}\bigl( \bigl\Vert \nabla\boldsymbol {e}^{i} \bigr\Vert _{0}+ \bigl\Vert \eta^{i} \bigr\Vert _{0}\bigr) \\& \quad \le C \Biggl[ \bigl\Vert \boldsymbol {e}^{L} \bigr\Vert _{0}^{2}+ k\sum_{i=L}^{n} \bigl( \bigl\Vert \nabla\boldsymbol {f}^{i} \bigr\Vert _{0}^{2}+ \bigl\Vert \xi^{i} \bigr\Vert _{0}^{2}+ \bigl\Vert Q_{h}^{i}- Q_{d}^{i} \bigr\Vert _{-1}^{2}\bigr) \Biggr]^{1/2}. \end{aligned}$$
(60)
Moreover, from Lemma 2 as well as Theorem 2, we acquire
$$\begin{aligned}& \begin{aligned}[b] k\sum_{i=L}^{n} \bigl\Vert \nabla\boldsymbol {f}^{i} \bigr\Vert _{0} &\le k \sum_{i=L}^{n}\bigl[ \bigl\Vert \nabla \bigl(\boldsymbol {u}_{h}^{i}-\boldsymbol {u}(t_{i}) \bigr) \bigr\Vert _{0}+ \bigl\Vert \nabla \bigl(\boldsymbol {u}(t_{i})-\boldsymbol {u}^{i}\bigr) \bigr\Vert _{0} \\ &\quad {}+ \bigl\Vert \nabla\bigl(\boldsymbol {u}^{i}-P^{h} \boldsymbol {u}^{i}\bigr) \bigr\Vert _{0}+ \bigl\Vert \nabla( P^{h}\bigl(\boldsymbol {u}^{i}-\boldsymbol {u}_{h}^{i}\bigr) \bigr\Vert _{0}\bigr] \\ &\le C\bigl(h^{2}+k\bigr), \end{aligned} \end{aligned}$$
(61)
$$\begin{aligned}& \begin{aligned}[b]k\sum_{i=L}^{n} \bigl\Vert \xi^{i} \bigr\Vert _{0} &\le k\sum _{i=L}^{n} \bigl[ \bigl\Vert p_{h}^{i}-p(t_{i}) \bigr\Vert _{0}+ \bigl\Vert p(t_{i})-p^{i} \bigr\Vert _{0}\\ &\quad {}+ \bigl\Vert p^{i}-Z^{h}p^{i} \bigr\Vert _{0}+ \bigl\Vert Z^{h}\bigl(p^{i}-p_{h}^{i} \bigr) \bigr\Vert _{0}\bigr] \\ &\le C\bigl(h^{2}+k\bigr). \end{aligned} \end{aligned}$$
(62)
Combining (61) and (62) with (60) and using Lemma 2 and (42) when \(n=L\) yield
$$\begin{aligned}& \bigl\Vert \boldsymbol {e}^{n} \bigr\Vert _{0} +k \sum _{i=L+1}^{n}\bigl( \bigl\Vert \nabla \boldsymbol {e}^{i} \bigr\Vert _{0}+ \bigl\Vert \eta^{i} \bigr\Vert _{0}\bigr) \\& \quad \le C\bigl(k+h^{2}\bigr)+CLk \Biggl(\sum_{j=d+1}^{l} \lambda_{j} \Biggr)^{1/2}+C \Biggl[ k\sum _{i=L}^{n} \bigl\Vert Q_{h}^{i}- Q_{d}^{i} \bigr\Vert _{-1}^{2} \Biggr]^{1/2}. \end{aligned}$$
(63)
Let \(F_{n}=Q_{h}^{n}-Z^{d}Q_{h}^{n}\), \(E_{n}=Z^{d}Q_{h}^{n}-Q_{d}^{n}\). First, by making use of (54) and Lemma 2, we acquire
$$ \begin{gathered}[b] \Vert E_{n} \Vert _{0}^{2}+k \gamma_{0}^{-1} \Vert \nabla {E}_{n} \Vert _{0}^{2} \\ \quad =(E_{n}, {E}_{n})+kD_{0}( {E}_{n}, {E}_{n}) \\ \quad=-(F_{n}, {E}_{n})+kD_{0}\bigl(Z^{d} Q_{h}^{n}- Q_{h}^{n}, {E}_{n} \bigr) +\bigl[\bigl(Q_{h}^{n}-Q_{d}^{n}, {E}_{n}\bigr)+kD_{0}\bigl( Q_{h}^{n}- Q_{d}^{n}, {E}_{n}\bigr)\bigr] \\ \quad=-(F_{n}, {E}_{n})+kA_{2}\bigl( {\boldsymbol {u}}_{d}^{n}, Q_{d}^{n}, {E}_{n} \bigr)-kA_{2}\bigl({\boldsymbol {u}}_{h}^{n}, Q_{h}^{n}, {E}_{n}\bigr) + \bigl(Q_{h}^{n-1}-Q_{d}^{n-1}, {E}_{n}\bigr) \\ \quad=(F_{n-1}-F_{n}, {E}_{n})+kA_{2}\bigl( {\boldsymbol {u}}_{d}^{n}, Q_{d}^{n}, {E}_{n}\bigr)-kA_{2}\bigl( {\boldsymbol {u}}_{h}^{n}, Q_{h}^{n}, {E}_{n}\bigr) +(E_{n-1}, {E}_{n}) \\ \quad\le Ck^{-1}\bigl( \Vert F_{n} \Vert _{-1}^{2}+ \Vert F_{n-1} \Vert _{-1}^{2}\bigr)+Ck \Vert F_{n} \Vert _{0}^{2})+\frac{k}{4\gamma_{0}} \Vert \nabla {E}_{n} \Vert _{0}^{2} \\ \qquad{}+kA_{2}\bigl( {\boldsymbol {u}}_{d}^{n}, Q_{d}^{n}, {E}_{n}\bigr)-kA_{2}\bigl( { \boldsymbol {u}}_{h}^{n}, Q_{h}^{n}, {E}_{n}\bigr)+ \frac{1}{2} \Vert E_{n-1} \Vert _{0}^{2}+\frac{1}{2} \Vert E_{n} \Vert _{0}^{2}. \end{gathered} $$
(64)
And then, when \(N_{0}\mu^{-1} \Vert \nabla{\boldsymbol {u}}_{d}^{n} \Vert _{0}\le1/4\) (\(n=1, 2, \ldots, N\)), with Lemma 2, (4), and Hölder’s and Cauchy’s inequalities, we have
$$ kA_{2}\bigl( {\boldsymbol {u}}_{d}^{n}, Q_{d}^{n}, {E}_{n} \bigr)-kA_{2}\bigl( {\boldsymbol {u}}_{h}^{n}, Q_{h}^{n}, {E}_{n}\bigr)\le\frac{k}{4\gamma_{0}} \Vert \nabla {E}_{n} \Vert _{0}^{2}+Ck \Vert \nabla F_{n} \Vert _{0}^{2}. $$
(65)
Combining (64) with (65) and using Lemma 2, Theorems 1 and 2, the same technique as (61) yield that
$$ \Vert E_{n} \Vert _{0}^{2}+k \gamma_{0}^{-1} \Vert \nabla {E}_{n} \Vert _{0}^{2} \le Ck\bigl(h^{4}+k^{2}\bigr)+ \Vert E_{n-1} \Vert _{0}^{2}. $$
(66)
Summing (66) from \(L+1\) to n yields that
$$ \Vert E_{n} \Vert _{0}^{2}+k \gamma_{0}^{-1}\sum_{i=L+1}^{n} \Vert \nabla {E}_{i} \Vert _{0}^{2} \le Cnk \bigl(h^{4}+k^{2}\bigr)+ C \Vert E_{L} \Vert _{0}^{2}. $$
(67)
By extraction of a square root to (67) and making use of \((\sum_{i=1}^{n}a_{i}^{2} )^{1/2}\ge\sum_{i=1}^{n} \vert a_{i} \vert /\sqrt{n}\) and (42), we acquire
$$ \Vert E_{n} \Vert _{0}+k\sum _{i=L+1}^{n} \Vert \nabla {E}_{i} \Vert _{0} \le C\bigl(h^{2}+k\bigr)+CLk \Biggl(\sum _{j=d+1}^{l}\lambda_{j} \Biggr)^{1/2}. $$
(68)
With the triangle inequality of norm, (68), and Lemma 2, we acquire
$$ \bigl\Vert Q_{h}^{n}-Q_{d}^{n} \bigr\Vert _{0}+k\sum_{i=L+1}^{n} \bigl\Vert \nabla\bigl(Q_{h}^{i}-Q_{d}^{i} \bigr) \bigr\Vert _{0}\le C\bigl(h^{2}+k\bigr)+CLk \Biggl( \sum_{j=d+1}^{l}\lambda_{j} \Biggr)^{1/2}. $$
(69)
By combining (63) with (69) and making use of Lemma 2, we acquire
$$ \begin{gathered}[b] \bigl\Vert \boldsymbol {u}_{h}^{n}-\boldsymbol {u}_{d}^{n} \bigr\Vert _{0}+k\sum_{i=L+1}^{n}\bigl( \bigl\Vert \nabla\bigl(\boldsymbol {u}_{h}^{i}-\boldsymbol {u}_{d}^{i}\bigr) \bigr\Vert _{0}+ \bigl\Vert p_{h}^{i}-p_{d}^{i} \bigr\Vert _{0}\bigr) \\ \quad \le C\bigl(k+h^{2}\bigr) +CLk \Biggl(\sum _{j=d+1}^{l}\lambda_{j} \Biggr)^{1/2}. \end{gathered} $$
(70)
Combining (69) with (70) yields (43). When \(\eta^{n}=0\), (43) is distinctly correct. Thus, the argument of Theorem 3 is accomplished. □
By combining Theorem 2 with Theorem 3, we immediately acquire the following conclusion.
Theorem 4
Under the conditions of Theorems
2
and
3, the SMFEROE solutions
\((\boldsymbol {u}_{d}^{n}, p_{d}^{n}, Q_{d}^{n})\)
for Problem
IV
hold the error estimations
$$\begin{gathered} k\sum_{i=1}^{n} \bigl[ \bigl\Vert \nabla\bigl(\boldsymbol {u}(t_{i})-\boldsymbol {u}_{d}^{i}\bigr) \bigr\Vert _{0}+ \bigl\Vert \nabla \bigl(Q(t_{i})-Q_{d}^{i}\bigr) \bigr\Vert _{0}\bigr]+ \bigl\Vert p(t_{i})-p_{d}^{i} \bigr\Vert _{0}] \\ \quad {}+ \bigl\Vert \boldsymbol {u}(t_{n})-\boldsymbol {u}_{d}^{n} \bigr\Vert _{0}+ \bigl\Vert Q(t_{n})-Q_{d}^{n} \bigr\Vert _{0}\le C\bigl(k+h^{2}\bigr)+ CLk \Biggl(\sum _{j=d+1}^{l}\lambda_{j} \Biggr)^{1/2}, \end{gathered} $$
where
\((\boldsymbol {u},p,T)\)
represents the generalized solution for the 2D unsteady conduction-convection problem.
Remark 3
The factor \(Lk (\sum_{j=d+1}^{l}\lambda _{j} )^{1/2}\) in Theorems 3 and 4 is caused by reduced-order for Problem II, it can be used as a suggestion choosing the amount of POD basis, that is, we only need to choose d that satisfies \(k^{2}L^{2}\sum_{j=d+1}^{l}\lambda_{j} = O(k^{2},h^{4})\), we can acquire the SMFROE solutions satisfying the accuracy requirement.
4.2 The algorithm process for the SMFEROE model
The algorithm process for the SMFEROE model can be carried out according to the next seven steps.
- Step 1:
-
Extract the snapshots \(\boldsymbol {U}_{n}(x, y)=(\boldsymbol {u}_{h}^{n},p_{h}^{n},Q_{h}^{i})\) (\(1\le n\le L\) and \(L\ll N\)) from the initial L SMFE solutions.
- Step 2:
-
Compile the snapshot matrix \(\tilde {\boldsymbol {A}}=(\tilde{{A}}_{ij})_{L\times L}\), where \(\tilde{{A}}_{ij}=[(\nabla\boldsymbol {u}_{h}^{i},\nabla \boldsymbol {u}_{h}^{j}) +(p_{h}^{i}, p_{h}^{j}) +(\nabla Q_{h}^{i},\nabla Q_{h}^{j})]/L\).
- Step 3:
-
Find the positive eigenvalues \(\lambda_{1}\ge \lambda_{2}\ge\cdots\ge\lambda_{l}>0\) (\(l=\dim\{\boldsymbol {U}_{1}, \boldsymbol {U}_{2}, \ldots, \boldsymbol {U}_{L}\}\)) of \(\tilde {\boldsymbol {A}}\) and the corresponding eigenvectors \(\boldsymbol {v}^{j}=(a_{1}^{j},a_{2}^{j},\ldots,a_{L}^{j})^{\tau}\) (\(j=1,2,\ldots,l\)).
- Step 4:
-
For h, k, and error ν needed, determine the amount d of POD basis that satisfies \(k^{2}+h^{4}+L^{2}k^{2}\sum_{j=d+1}^{l}\lambda_{j}\le\nu^{2}\).
- Step 5:
-
Constitute the POD basis \(\boldsymbol{\omega}_{j}(x,y)=(\boldsymbol{\omega}_{uj}(x,y),\omega _{pj}(x,y),\omega_{Qj}(x,y)) =\sum_{j=1}^{L}a_{i}^{j} (\boldsymbol {u}_{h}^{i}, p_{h}^{i}, Q_{h}^{i} )/\sqrt{L\lambda_{j}}\) (\(1\le j\le d\)).
- Step 6:
-
Let \(X^{d}= \operatorname{span}\{\boldsymbol{\omega}_{u1}(x, y), \boldsymbol{\omega}_{u2}(x,y), \ldots, \boldsymbol {\omega}_{u d}(x,y)\}\), \(M^{d}=\operatorname{span} \{\omega_{p1}(x, y), \omega _{p2}(x,y), \ldots, \omega_{pd}(x,y)\}\), and \(W^{d}=\operatorname{span}\{\omega_{Q1}(x, y), \omega_{Q2}(x,y), \ldots, \omega_{Qd}(x,y)\}\). Solving Problem IV gives the SMFEROE solutions \((\boldsymbol {u}_{d}^{n},p_{d}^{n}, Q_{d}^{n})\) (\(1\le n\le N\)).
- Step 7:
-
If \(\Vert\boldsymbol {u}_{d}^{n-1}-\boldsymbol {u}_{d}^{n} \Vert_{0}\ge\Vert\boldsymbol {u}_{d}^{n}-\boldsymbol {u}_{d}^{n+1} \Vert_{0}\), \(\Vert p_{d}^{n-1} -p_{d}^{n} \Vert_{0}\ge\Vert p_{d}^{n}-p_{d}^{n+1} \Vert_{0}\), and \(\Vert Q_{d}^{n-1}-Q_{d}^{n} \Vert_{0}\ge\Vert Q_{d}^{n}-Q_{d}^{n+1} \Vert_{0}\) (\(L\le n\le N-1\)), then \((\boldsymbol {u}_{d}^{n},p_{d}^{n}, Q_{d}^{n})\) (\(1\le n\le N\)) are the SMFEROE solutions satisfying the accuracy requirement. Else, namely, if \(\Vert\boldsymbol {u}_{d}^{n-1}-\boldsymbol {u}_{d}^{n} \Vert_{0}<\Vert\boldsymbol {u}_{d}^{n}-\boldsymbol {u}_{d}^{n+1}\Vert_{0}\) or \(\Vert p_{d}^{n-1}-p_{d}^{n} \Vert_{0}< \Vert p_{d}^{n}-p_{d}^{n+1} \Vert_{0}\) or \(\Vert Q_{d}^{n-1}-Q_{d}^{n} \Vert_{0}< \Vert Q_{d}^{n}-Q_{d}^{n+1} \Vert_{0} \) (\(n=L, L+1,\ldots, N-1\)), put \(\boldsymbol {U}_{n+j-L}=(\boldsymbol {u}_{d}^{j},p_{d}^{j}, Q_{d}^{j})\) (\(j=0, 1, \ldots, L-1\)), return to Step 2.