We first demonstrate the following unconditional superclose result.
Theorem 3.1
Let \(u^{n}\) and \(U_{h}^{n}\) be solutions of (2) and (3), respectively. Assume that \(u\in L^{\infty}(0,T;H^{3}(\Omega ))\), \(u_{t}\in L^{2}(0,T;H^{3}(\Omega ))\), \(u_{tt},u_{ttt}\in L^{2}(0,T;H^{1}(\Omega ))\), there holds
$$\begin{aligned} \bigl\Vert I_{h}u^{n}-U_{h}^{n} \bigr\Vert _{1}\leq C\bigl(h^{2}+(\Delta t)^{2} \bigr), \end{aligned}$$
(16)
where \(\Delta t>0\) is small enough so that \(1-C\Delta t>0\).
Proof
Let \(u^{n}-U^{n}_{h}=(u^{n}-I_{h}u^{n})+(I_{h}u^{n}-U_{h}^{n}):= \xi ^{n}+ \eta ^{n}\), then the error equation can be derived by (2) and (3):
$$\begin{aligned} &\bigl(\bar{\partial}_{t}\eta ^{n},v_{h}\bigr)+ \alpha \bigl(\nabla \bar{\partial}_{t} \eta ^{n},\nabla v_{h}\bigr)+\beta \bigl(\nabla \eta ^{n-\frac{1}{2}},\nabla v_{h}\bigr) \\ &\quad=-\bigl(\bar{\partial}_{t}\xi ^{n},v_{h}\bigr)- \alpha \bigl(\nabla \bar{\partial}_{t} \xi ^{n},\nabla v_{h}\bigr)-\beta \bigl(\nabla \xi ^{n-\frac{1}{2}},\nabla v_{h}\bigr) \\ &\quad\quad{}+\bigl(\vec{f}\bigl(U^{n-\frac{1}{2}}_{h}\bigr)-\vec{f} \bigl(u(t_{n-\frac{1}{2}})\bigr), \nabla v_{h}\bigr)- \bigl(R_{1}^{n},v_{h}\bigr) \\ &\quad\quad{}-\alpha \bigl(\nabla R_{1}^{n},\nabla v_{h} \bigr)-\beta \bigl(\nabla R_{2}^{n}, \nabla v_{h} \bigr), \end{aligned}$$
(17)
where \(R^{n}_{1}=u_{t}(t_{n-\frac{1}{2}})-\bar{\partial}_{t}u^{n}\), \(R_{2}^{n}=u(t_{n- \frac{1}{2}})-u^{n-\frac{1}{2}}\).
Taking \(v_{h}=\eta ^{n-\frac{1}{2}}\) in (17), there holds
$$\begin{aligned} &\bigl(\bar{\partial}_{t}\eta ^{n},\eta ^{n-\frac{1}{2}} \bigr)+\alpha \bigl(\nabla \bar{\partial}_{t}\eta ^{n},\nabla \eta ^{n-\frac{1}{2}}\bigr)+\beta \bigl( \nabla \eta ^{n-\frac{1}{2}},\nabla \eta ^{n-\frac{1}{2}}\bigr) \\ &\quad=-\bigl(\bar{\partial}_{t}\xi ^{n},\eta ^{n-\frac{1}{2}} \bigr)-\alpha \bigl(\nabla \bar{\partial}_{t}\xi ^{n},\nabla \eta ^{n-\frac{1}{2}}\bigr) \\ &\quad\quad{}-\beta \bigl(\nabla \xi ^{n-\frac{1}{2}},\nabla \eta ^{n-\frac{1}{2}}\bigr) + \bigl( \vec{f}\bigl(U^{n-\frac{1}{2}}_{h}\bigr)-\vec{f} \bigl(u(t_{n-\frac{1}{2}})\bigr),\nabla \eta ^{n-\frac{1}{2}}\bigr) \\ &\quad\quad{}-\bigl(R_{1}^{n},\eta ^{n-\frac{1}{2}}\bigr)-\alpha \bigl( \nabla R_{1}^{n},\nabla \eta ^{n-\frac{1}{2}}\bigr) -\beta \bigl(\nabla R_{2}^{n},\nabla \eta ^{n- \frac{1}{2}}\bigr) . \end{aligned}$$
(18)
The left-hand side of (18) can be rewritten as
$$\begin{aligned} \frac{1}{2\Delta t}\bigl[\bigl( \bigl\Vert \eta ^{n} \bigr\Vert ^{2}_{0}- \bigl\Vert \eta ^{n-1} \bigr\Vert ^{2}_{0}\bigr)+ \alpha \bigl( \bigl\Vert \nabla \eta ^{n} \bigr\Vert ^{2}_{0}- \bigl\Vert \nabla \eta ^{n-1} \bigr\Vert ^{2}_{0}\bigr)\bigr] + \beta \bigl\Vert \nabla \eta ^{n-\frac{1}{2}} \bigr\Vert ^{2}_{0}. \end{aligned}$$
(19)
Now we estimate the right-hand side: By virtue of Lemma 2.1, we arrive at
$$\begin{aligned} &\bigl(\bar{\partial}_{t}\xi ^{n},\eta ^{n-\frac{1}{2}} \bigr)+\alpha \bigl(\nabla \bar{\partial}_{t}\xi ^{n},\nabla \eta ^{n-\frac{1}{2}}\bigr)+\beta \bigl( \nabla \xi ^{n-\frac{1}{2}},\nabla \eta ^{n-\frac{1}{2}}\bigr) \\ &\quad \leq Ch^{4}\biggl(\frac{1}{\Delta t} \int ^{t_{n}}_{t_{n-1}} \Vert u_{t} \Vert ^{2}_{3}\,d \tau + \bigl\Vert u^{n-\frac{1}{2}} \bigr\Vert ^{2}_{3}\biggr)+\frac{1}{2} \bigl\Vert \nabla \eta ^{n- \frac{1}{2}} \bigr\Vert ^{2}_{0}. \end{aligned}$$
(20)
Using the Taylor expansion, the truncation error can be estimated as
$$\begin{aligned} &\bigl( R_{1}^{n},\eta ^{n-\frac{1}{2}}\bigr)+\alpha \bigl( \nabla R_{1}^{n},\nabla \eta ^{n-\frac{1}{2}}\bigr)+\beta \bigl(\nabla R_{2}^{n},\nabla \eta ^{n- \frac{1}{2}}\bigr) \\ &\quad\leq C\bigl( \bigl\Vert R_{1}^{n} \bigr\Vert ^{2}_{1}+ \bigl\Vert R_{2}^{n} \bigr\Vert ^{2}_{1}\bigr)+\frac{1}{2} \bigl\Vert \nabla \eta ^{n-\frac{1}{2}} \bigr\Vert ^{2}_{0} \\ &\quad\leq C(\Delta t)^{3} \int ^{t_{n}}_{t_{n-1}}\bigl( \Vert u_{ttt} \Vert ^{2}_{1}+ \Vert u_{tt} \Vert ^{2}_{1}\bigr)\,d\tau +\frac{1}{2} \bigl\Vert \nabla \eta ^{n-\frac{1}{2}} \bigr\Vert ^{2}_{0}, \end{aligned}$$
(21)
the nonlinear term can be written as
$$\begin{aligned} & \bigl\vert \bigl(\vec{f}\bigl(U^{n-\frac{1}{2}}_{h}\bigr)-\vec{f} \bigl(u(t_{n-\frac{1}{2}})\bigr), \nabla \eta ^{n-\frac{1}{2}}\bigr) \bigr\vert \\ &\quad = \biggl\vert \biggl(\frac{1}{2}\bigl(\bigl(u(t_{n-\frac{1}{2}}) \bigr)^{2}-\bigl(U^{n-\frac{1}{2}}_{h}\bigr)^{2} \bigr), \nabla \eta ^{n-\frac{1}{2}}\biggr)+\bigl(u(t_{n-\frac{1}{2}})-U^{n-\frac{1}{2}}_{h}, \nabla \eta ^{n-\frac{1}{2}}\bigr) \biggr\vert \\ &\quad = \biggl\vert \biggl(\frac{1}{2}\bigl(\bigl(u(t_{n-\frac{1}{2}}) \bigr)^{2}-\bigl(u^{n-\frac{1}{2}}\bigr)^{2}+ \bigl(u^{n- \frac{1}{2}}\bigr)^{2}-\bigl(U^{n-\frac{1}{2}}_{h} \bigr)^{2}\bigr), \nabla \eta ^{n- \frac{1}{2}}\biggr) \\ &\quad\quad{}+\bigl(u(t_{n-\frac{1}{2}})-u^{n-\frac{1}{2}}+u^{n-\frac{1}{2}}-U^{n- \frac{1}{2}}_{h}, \nabla \eta ^{n-\frac{1}{2}}\bigr) \biggr\vert \\ &\quad = \biggl\vert \frac{1}{2}\bigl(u(t_{n-\frac{1}{2}})^{2}- \bigl(u^{n-\frac{1}{2}}\bigr)^{2}, \nabla \eta ^{n-\frac{1}{2}}\bigr)+ \frac{1}{2}\bigl(\bigl(u^{n-\frac{1}{2}}\bigr)^{2}- \bigl(U^{n- \frac{1}{2}}_{h}\bigr)^{2}, \nabla \eta ^{n-\frac{1}{2}}\bigr) \\ &\quad\quad{}+\bigl(u(t_{n-\frac{1}{2}})-u^{n-\frac{1}{2}}+u^{n-\frac{1}{2}}-U^{n- \frac{1}{2}}_{h}, \nabla \eta ^{n-\frac{1}{2}}\bigr) \biggr\vert \\ &\quad = \biggl\vert \frac{1}{2}\bigl(R^{n}_{2} \bigl(u(t_{n-\frac{1}{2}})+u^{n-\frac{1}{2}}\bigr), \nabla \eta ^{n-\frac{1}{2}} \bigr)+\frac{1}{2}\bigl(\bigl(\xi ^{n-\frac{1}{2}}+\eta ^{n- \frac{1}{2}}\bigr) \bigl(u^{n-\frac{1}{2}}+U^{n-\frac{1}{2}}_{h}\bigr), \nabla \eta ^{n- \frac{1}{2}}\bigr) \\ &\quad\quad{}+\bigl(R^{n}_{2}, \nabla \eta ^{n-\frac{1}{2}}\bigr)+\bigl( \xi ^{n-\frac{1}{2}}+ \eta ^{n-\frac{1}{2}},\nabla \eta ^{n-\frac{1}{2}}\bigr) \biggr\vert \\ &\quad = \Biggl\vert \sum^{4}_{i=1}A_{i} \Biggr\vert . \end{aligned}$$
(22)
By the estimation of truncation error, there holds
$$\begin{aligned} A_{1}+A_{3}&\leq C \bigl\Vert R^{n}_{2} \bigr\Vert _{0} \bigl\Vert \eta ^{n-\frac{1}{2}} \bigr\Vert _{1}\bigl( \bigl\Vert u(t_{n- \frac{1}{2}})+u^{n-\frac{1}{2}} \bigr\Vert _{0,\infty}+1\bigr) \\ &\leq C \bigl\Vert R^{n}_{2} \bigr\Vert _{0} \bigl\Vert \eta ^{n-\frac{1}{2}} \bigr\Vert _{1} \\ &\leq C(\Delta t)^{3} \int ^{t_{n}}_{t_{n-1}} \Vert u_{tt} \Vert ^{2}_{0}\,d\tau +C \bigl\Vert \eta ^{n-\frac{1}{2}} \bigr\Vert ^{2}_{1}. \end{aligned}$$
(23)
From the Sobolev imbedding theorem, (4) and (7), we have
$$\begin{aligned} A_{2}& \leq C \bigl\Vert \xi ^{n-\frac{1}{2}}+\eta ^{n-\frac{1}{2}} \bigr\Vert _{0,4} \bigl\Vert u^{n- \frac{1}{2}}+U^{n-\frac{1}{2}}_{h} \bigr\Vert _{0,4} \bigl\Vert \eta ^{n-\frac{1}{2}} \bigr\Vert _{1} \\ & \leq C \bigl\Vert \xi ^{n-\frac{1}{2}}+\eta ^{n-\frac{1}{2}} \bigr\Vert _{0,4}\bigl( \bigl\Vert u^{n- \frac{1}{2}} \bigr\Vert _{0,4}+ \bigl\Vert U^{n-\frac{1}{2}}_{h} \bigr\Vert _{0,4}\bigr) \bigl\Vert \eta ^{n- \frac{1}{2}} \bigr\Vert _{1} \\ & \leq C \bigl\Vert \xi ^{n-\frac{1}{2}}+\eta ^{n-\frac{1}{2}} \bigr\Vert _{0,4}\bigl( \bigl\Vert u^{n- \frac{1}{2}} \bigr\Vert _{0,\infty}+ \bigl\Vert U^{n-\frac{1}{2}}_{h} \bigr\Vert _{1}\bigr) \bigl\Vert \eta ^{n- \frac{1}{2}} \bigr\Vert _{1} \\ & \leq C \bigl\Vert \xi ^{n-\frac{1}{2}}+\eta ^{n-\frac{1}{2}} \bigr\Vert _{0,4} \bigl\Vert \eta ^{n- \frac{1}{2}} \bigr\Vert _{1} \\ & \leq Ch^{4} \bigl\Vert u^{n-\frac{1}{2}} \bigr\Vert ^{2}_{2,4}+C \bigl\Vert \eta ^{n-\frac{1}{2}} \bigr\Vert ^{2}_{1} \\ & \leq Ch^{4} \bigl\Vert u^{n-\frac{1}{2}} \bigr\Vert ^{2}_{3}+C \bigl\Vert \eta ^{n-\frac{1}{2}} \bigr\Vert ^{2}_{1} \end{aligned}$$
(24)
and
$$\begin{aligned} A_{4}\leq C\bigl( \bigl\Vert \xi ^{n-\frac{1}{2}} \bigr\Vert _{0}+ \bigl\Vert \eta ^{n-\frac{1}{2}} \bigr\Vert _{0} \bigr) \bigl\Vert \eta ^{n-\frac{1}{2}} \bigr\Vert _{1} \leq Ch^{4} \bigl\Vert u^{n-\frac{1}{2}} \bigr\Vert ^{2}_{2}+C \bigl\Vert \eta ^{n-\frac{1}{2}} \bigr\Vert ^{2}_{1}. \end{aligned}$$
(25)
Substituting (23)–(25) into (22), we get
$$\begin{aligned} &\bigl\vert \bigl(\vec{f}\bigl(U^{n-\frac{1}{2}}_{h}\bigr)-\vec{f} \bigl(u(t_{n-\frac{1}{2}})\bigr), \nabla \eta ^{n-\frac{1}{2}}\bigr) \bigr\vert \\ &\quad \leq C(\Delta t)^{3} \int ^{t_{n}}_{t_{n-1}} \Vert u_{tt} \Vert ^{2}_{0}\,d\tau +Ch^{4} \bigl\Vert u^{n-\frac{1}{2}} \bigr\Vert ^{2}_{3}+C \bigl\Vert \eta ^{n- \frac{1}{2}} \bigr\Vert ^{2}_{1}. \end{aligned}$$
(26)
Hence, from (18)–(21) and (26), we have
$$\begin{aligned} &\frac{1}{2\Delta t}\bigl[\bigl( \bigl\Vert \eta ^{n} \bigr\Vert ^{2}_{0}- \bigl\Vert \eta ^{n-1} \bigr\Vert ^{2}_{0}\bigr)+ \alpha \bigl( \bigl\vert \eta ^{n} \bigr\vert ^{2}_{1}- \bigl\vert \eta ^{n-1} \bigr\vert ^{2}_{1}\bigr)\bigr] \\ &\quad \leq Ch^{4}+C(\Delta t)^{3} \int ^{t_{n}}_{t_{n-1}}\bigl( \Vert u_{tt} \Vert ^{2}_{1}+ \Vert u_{ttt} \Vert ^{2}_{1}\bigr)\,d\tau +C \bigl\Vert \eta ^{n-\frac{1}{2}} \bigr\Vert _{1}^{2}, \end{aligned}$$
multiplying by \(2\Delta t\), then summing up the above inequality and noting that \(\eta ^{0}=0\), we can obtain
$$\begin{aligned} \bigl\Vert \eta ^{n} \bigr\Vert ^{2}_{1} \leq Ch^{4}+C(\Delta t)^{4}+C\Delta t\sum ^{n}_{i=1} \bigl\Vert \eta ^{i} \bigr\Vert ^{2}_{1}. \end{aligned}$$
(27)
Choosing Δt small enough so that \(1-C\Delta t>0\) and applying discrete Gronwall’s lemma, there holds
$$\begin{aligned} \bigl\Vert \eta ^{n} \bigr\Vert ^{2}_{1}\leq C \bigl(h^{4}+(\Delta t)^{4}\bigr), \end{aligned}$$
the proof is completed. □
To obtain the global superconvergence estimate, we combine the adjacent four small elements \(K_{1}\), \(K_{2}\), \(K_{3}\), \(K_{4}\) into a big element K̃, i.e., \(\tilde{K}=\bigcup_{i=1}^{4}K_{i}\) (see Fig. 1), the corresponding subdivision is defined by \(T_{2h}\).
As in [19], we define the following interpolation postprocessing operator \(I_{2h}\):
$$\begin{aligned} I_{2h}u| _{\tilde{K}}\in Q_{2}(\tilde{K}), \quad \forall u\in C( \tilde{K}), \qquad I_{2h}u(Z_{i})=u(Z_{i}), \quad i=1,2,\ldots ,9, \end{aligned}$$
(28)
where \(Q_{2}(\tilde{K})\) and \(C(\tilde{K})\) denote the spaces of biquadratic piecewise polynomial and continuous function on K̃, respectively, \(Z_{i}\) are all vertices of four small elements (see Fig. 1).
Meanwhile, \(I_{2h}\) has the following properties (see [19]):
$$\begin{aligned} I_{2h}I_{h}u=I_{2h}u, \qquad \Vert u-I_{2h}u \Vert _{1}\leq Ch^{2} \Vert u \Vert _{3}, \qquad \Vert I_{2h}v_{h} \Vert _{1}\leq C \Vert v_{h} \Vert _{1}, \quad \forall v_{h}\in V_{h}. \end{aligned}$$
(29)
Theorem 3.2
Under the assumption of Theorem 3.1, there holds the global superconvergence result as follows:
$$\begin{aligned} \bigl\Vert u^{n}-I_{2h}U^{n}_{h} \bigr\Vert _{1}\leq C\bigl(h^{2}+(\Delta t)^{2} \bigr). \end{aligned}$$
Proof
Let \(u^{n}-I_{2h}U^{n}_{h}=u^{n}-I_{2h}I_{h}u^{n}+I_{2h}I_{h}u^{n}-I_{2h}U^{n}_{h}\). By (16), (29), and the triangular inequality, we can get
$$\begin{aligned} \bigl\Vert u^{n}-I_{2h}U^{n}_{h} \bigr\Vert _{1}&\leq \bigl\Vert u^{n}-I_{2h}I_{h}u^{n} \bigr\Vert _{1}+ \bigl\Vert I_{2h}I_{h}u^{n}-I_{2h}U^{n}_{h} \bigr\Vert _{1} \\ &\leq \bigl\Vert u^{n}-I_{2h}u^{n} \bigr\Vert _{1}+ \bigl\Vert I_{2h}\bigl(I_{h}u^{n}-U^{n}_{h} \bigr) \bigr\Vert _{1} \\ &\leq Ch^{2} \bigl\Vert u^{n} \bigr\Vert _{3}+C \bigl\Vert I_{h}u^{n}-U^{n}_{h} \bigr\Vert _{1} \\ &\leq C\bigl(h^{2}+(\Delta t)^{2}\bigr), \end{aligned}$$
(30)
the proof is completed. □
Remark 3.1
In this paper, the boundedness of \(\|U^{n}_{h}\|_{1}\) is crucial to the unconditional superclose and superconvergence results, the technique we used is more simple and direct than those in [22] and [23] ([22] applied an error splitting technique and [23] employed a complicated mathematical inductive hypothesis method to derive the boundedness of numerical solution).