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A new approach to convergence analysis of linearized finite element method for nonlinear hyperbolic equation

Boundary Value Problems20192019:48

https://doi.org/10.1186/s13661-019-1161-9

  • Received: 29 October 2018
  • Accepted: 26 February 2019
  • Published:

Abstract

We study a new way to convergence results for a nonlinear hyperbolic equation with bilinear element. Such equation is transformed into a parabolic system by setting the original solution u as \(u_{t}=q\). A linearized backward Euler finite element method (FEM) is introduced, and the splitting skill is exploited to get rid of the restriction on the ratio between h and τ. The boundedness of the solutions about the time-discrete system in \(H^{2}\)-norm is proved skillfully through temporal error. The spatial error is derived without the mesh-ratio, where some new techniques are utilized to deal with the problems caused by the new parabolic system. The final unconditional optimal error results of u and q are deduced at the same time. Finally, a numerical example is provided to support the theoretical analysis. Here h is the subdivision parameter, and τ is the time step.

Keywords

  • Nonlinear hyperbolic equation
  • Parabolic system
  • Bilinear element
  • Linearized FEM
  • Optimal error results

1 Introduction

Consider the following nonlinear hyperbolic equation:
$$ \textstyle\begin{cases} u_{tt}-\nabla\cdot(a(u)\nabla u)=f(u), & (X, t)\in\varOmega\times(0, T], \\ u=0,& (X, t)\in\partial\varOmega\times(0, T], \\ u(X, 0)=u_{0}(X), \qquad u_{t}(X, 0)=u_{1}(X),& X\in\varOmega, \end{cases} $$
(1)
where \(\varOmega\subset\mathbb{R}^{2}\) is a rectangle with boundary ∂Ω parallel to the coordinate axes, \(0< T<\infty\), \(X=(x, y)\), and \(a(u)\) and \(f(u)\) are known smooth functions on \(\mathbb{R}\), for which we assume that \(0< a_{0} \leq a(u)\leq a_{1}\).

A nonlinear hyperbolic equation is a kind of important problems on nonlinear vibration, the permeation fluid mechanics, and so on. Indeed, such partial differential equations (PDEs) have attracted lots of attention to various methods, especially numerical methods. For example, the two-grid method was studied for solving a type of nonlinear hyperbolic equations, and the error estimate in \(H^{1}\)-norm was deduced in [1]. Newton’s modified method was utilized to a nonlinear wave equation depending on different norms of the initial conditions in [2], and optimal error results were given in the \(L^{2}\)- and \(H^{1}\)-norms. The interpolation theory and integral identity skill were used to obtain a superclose result for the nonlinear hyperbolic equations with nonlinear boundary condition in [3]. Moreover, the global superconvergence was also obtained through the interpolated postprocessing technique. The Galerkin alternating-direction method was applied to a three-dimensional nonlinear hyperbolic equation in [4], and the error estimates in the \(H^{1}\)- and \(L^{2}\)-norms were deduced. A mixed FEM was discussed in [5] and [6], and optimal error estimates were derived.

The inverse inequality is usually employed to discuss the boundedness of numerical solution \(U_{h}^{n}\) in a nonlinear evolution equation, and such an issue usually results in some time-step restrictions, such as \(\tau=O(h)\), \(h^{r}=O(\tau)\) (\(1\leq r\leq k+1\), \(k\geq0\)), and \(\tau=O(h^{2})\) in [4] and [6], respectively. To get rid of such a restriction, [7, 8] took advantage of a special inequality for getting unconditional superclose results for nonlinear Sobolev equations. In [9] a corresponding time-discrete system to split the error into two parts, the temporal error and the spatial error, is introduced. Then the spatial error leads to the unconditional boundedness of a numerical solution in the \(L^{\infty}\)-norm. Subsequently, this so-called splitting technique was also applied to the other nonlinear parabolic type equations in [1018]. Later, in [19] and [20] a second-order scheme for the nonlinear hyperbolic equation and the unconditional superconvergence analysis by using the splitting skill were given. It can be seen that constructing a linearized form for a nonlinear hyperbolic equation is not an easy task in comparison with nonlinear parabolic equations. In fact, there are lots of literature referring to parabolic equations [2124]. In [24] a special technique to change sine-Gordon equation into a parabolic system through \(u_{t}=q\) was used, and optimal order error estimates of the Crank–Nicolson fully discrete scheme were obtained.

Inspired by [24], in this paper, we consider the unconditional convergent estimates for (1), which is a much more general nonlinear model than that in [24], with a bilinear element. First of all, we change a nonlinear hyperbolic equation into a nonlinear parabolic system. Such a practice can be used to avoid the difficulty in constructing a linearized scheme for a nonlinear hyperbolic equation and also give the error analysis for u and \(q=u_{t}\) at the same time. Then we develop a linearized backward Euler FE scheme for the nonlinear parabolic system and apply the idea of splitting technique in [1020] to split the error into the temporal and spatial errors. We obtain a temporal error, which implies the regularities of the solutions about the time-discrete equations. The spatial error result is exploited to get rid of the restriction on the ratio between h and τ. The unconditional optimal error results of u and q are simultaneously deduced. Note that, differently from [17, 18], we utilize some new tricks such as rewriting some error terms, the new mean-value technique, and some other skills to handle new difficulties brought by the special nonlinear parabolic system during the process. Further, the results in this paper also hold for linear conforming triangular elements but do not hold for some other particular elements; for example, the biquadratic finite element for \(\Delta v_{h}|_{k}=0\) cannot be true, where \(v_{h}\) belongs to the FE space. Some numerical results in the last section also show the validity of the theoretical analysis.

Throughout this paper, we denote the natural inner production in \(L^{2}(\varOmega)\) by \((\cdot,\cdot)\) and the norm by \(\|\cdot\|_{0}\), and let \(H_{0}^{1}(\varOmega)=\{u\in H^{1}(\varOmega):u|_{\partial\varOmega}=0\}\). Further, we use the classical Sobolev spaces \(W^{m,p}(\varOmega)\), \(1\leq p\leq\infty\), denoted by \(W^{m,p}\), with norm \(\|\cdot\|_{m,p}\). When \(p=2\), we simply write \(\|\cdot\|_{m,p}\) as \(\|\cdot\|_{m}\). Besides, we define the space \(L^{p}(a,b;Y)\) with norm \(\|f\|_{L^{p}(a,b;Y)}=(\int_{a}^{b}\|f(\cdot,t)\| _{Y}^{p}\,dt)^{\frac{1}{p}}\), and if \(p=\infty\), the integral is replaced by the essential supremum.

2 Conforming FE approximation scheme

Let Ω be a rectangle in the \((x,y)\) plane with edges parallel to the coordinate axes, and let \(\varGamma_{h}\) be a regular rectangular subdivision. Given \(K\in\varGamma_{h}\), let the four vertices and edges be \(a_{i}\), \(i=1\sim4\), and \(l_{i}=\overline{a_{i}a_{i+1}}\), \(i=1\sim4\pmod{4}\), respectively. Let \(V_{h}\) be the usual bilinear FE space, and let \(V_{h0}=\{v_{h}\in V_{h}, v_{h}|_{\partial\varOmega}=0\}\). Also, it can be found in [25] that if \(u\in H^{2}(\varOmega)\), then
$$ \bigl(\nabla(u-I_{h}u), \nabla v_{h}\bigr)=0, \quad v_{h}\in V_{h0}, $$
(2)
where \(I_{h}\) be the so-called Ritz projection operator on \(V_{h0}\).
Set \(\{t_{n}:t_{n}=n\tau;0\leq n\leq N\}\) be a uniform partition of \([0,T]\) with time step \(\tau=T/N\). We denote \(\sigma^{n}=\sigma(X,t_{n})\). For a sequence of functions \(\{\sigma^{n}\}_{n=0}^{N}\), we denote \(\bar{\partial }_{t}\sigma^{n}=\frac{\sigma^{n}-\sigma^{n-1}}{\tau}\), \(n=1,2,\ldots,N\). With these notations, setting \(u_{t}=q\), the weak form of (1) is seeking \(u,q \in H_{0}^{1}(\varOmega)\) such that, for all \(v\in H_{0}^{1}(\varOmega)\),
$$ \textstyle\begin{cases} (\bar{\partial}_{t}u^{n},v)=(q^{n}, v)+(R_{1}^{n},v), & v\in H_{0}^{1}(\varOmega), \\ (\bar{\partial}_{t}q^{n}, v)+(a(u^{n-1})\tau\sum_{i=1}^{n}\nabla q^{i}, \nabla v)+(a(u^{n-1})\nabla u^{0}, \nabla v) \\ \quad =(f(u^{n-1}), v) +(R_{2}^{n}+ R_{3}^{n}+R_{4}^{n}, v), & v\in H_{0}^{1}(\varOmega), \end{cases} $$
(3)
where
$$\begin{aligned}& R_{1}^{n}=\bar{\partial}_{t}u^{n}-u_{t}^{n},\qquad R_{2}^{n}= \bar{\partial }_{t}q^{n}-q_{t}^{n},\qquad R_{4}^{n}=- \bigl(f\bigl(u^{n-1}\bigr)-f\bigl(u^{n}\bigr)\bigr), \\& R_{3}^{n}=-\nabla\cdot\Biggl(a\bigl(u^{n-1}\bigr) \tau\sum_{i=1}^{n}\nabla q^{i}-a \bigl(u^{n}\bigr) \int _{0}^{t_{n}}\nabla q \,ds\Biggr)-\nabla\cdot\bigl( \nabla u^{0}\bigl(a\bigl(u^{n-1}\bigr)-a\bigl(u^{n} \bigr)\bigr)\bigr). \end{aligned}$$
We develop the linearized Galerkin FEM to problem (3): seek \(U_{h}^{n},Q_{h}^{n}\in V_{h0} \) such that
$$ \textstyle\begin{cases} (\bar{\partial}_{t}U_{h}^{n},v_{h})=(Q^{n}_{h}, v_{h}), & v_{h}\in V_{h0}, \\ (\bar{\partial}_{t}Q_{h}^{n}, v_{h})+(a(U_{h}^{n-1})\tau\sum_{i=1}^{n}\nabla Q_{h}^{i}, \nabla v_{h}) \\ \quad {}+(a(U_{h}^{n-1})\nabla U_{h}^{0}, \nabla v_{h})=(f(U^{n-1}_{h}), v_{h}), & v_{h}\in V_{h0}, \end{cases} $$
(4)
where \(U_{h}^{0}=I_{h}u_{0}\) and \(Q_{h}^{0}=I_{h}u_{1}\). A well-known consequence is that the linear system (4) may always be solved for \(U_{h}^{n}\) and \(Q_{h}^{n}\); see [26].

3 Error estimates for the time-discrete system

To get rid of the ratio restriction between h and τ, we introduce a time-discrete system as follows:
$$ \textstyle\begin{cases} \bar{\partial}_{t}U^{n}=Q^{n}, & (X, t)\in\varOmega, \\ \bar{\partial}_{t}Q^{n}-\nabla\cdot(a(U^{n-1})\tau\sum_{i=1}^{n}\nabla Q^{i}) -\nabla\cdot(a(U^{n-1})\nabla U^{0}) \\ \quad =f(U^{n-1}), & (X, t)\in\varOmega, \\ U^{n}=0,\qquad Q^{n}=0,& (X, t)\in\partial\varOmega, \\ U^{0}=u_{0}(X),\qquad Q^{0}=u_{1}(X), & X\in\varOmega. \end{cases} $$
(5)
The existence and uniqueness of solutions for this linear elliptic system (5) are obvious. To show the unconditional results, the regularities of \(U^{n}\) and \(Q^{n}\) are inevitable, and we therefore need some estimates for \(u^{n}-U^{n}\) and \(q^{n}-Q^{n}\). In what follows, we set \(e^{n}\triangleq u^{n}-U^{n}\), \(\delta^{n}\triangleq q^{n}-Q^{n} \) (\(n=1,2,\ldots, N\)), analyze the temporal errors and give the regularity results for \(U^{n}\) and \(Q^{n}\). It is easy to see that \(e^{0}=\delta^{0}=0\).

Theorem 1

Let \(u^{m}\) and \(U^{m}\) (\(m=0,1,2,\ldots, N\)) be solutions of (1) and (5), respectively, \(u,q\in L^{2}(0,T;H^{3}(\varOmega))\), \(u_{t},q_{t}\in L^{\infty }(0,T;H^{2}(\varOmega))\), and \(u_{tt}\in L^{2}(0,T;L^{2}(\varOmega))\). Then for \(m=1,\ldots, N\), there exists \(\tau_{0}\) such that when \(\tau\leq\tau _{0}\), we have
$$\begin{aligned}& \bigl\Vert e^{m} \bigr\Vert _{2}+\tau\Biggl(\sum _{i=2}^{m} \bigl\Vert \bar{ \partial}_{t}e^{i} \bigr\Vert _{2}^{2} \Biggr)^{\frac{1}{2}}+ \bigl\Vert \delta^{m} \bigr\Vert _{1}+\tau \Biggl\Vert \sum_{i=1}^{m} \delta^{i} \Biggr\Vert _{2}+\tau\Biggl(\sum _{i=2}^{m} \bigl\Vert \delta^{i} \bigr\Vert _{2}^{2}\Biggr)^{\frac{1}{2}} \leq C_{0} \tau, \end{aligned}$$
(6)
$$\begin{aligned}& \bigl\Vert \bar{\partial}_{t}U^{m} \bigr\Vert _{2}+ \bigl\Vert Q^{m} \bigr\Vert _{2}\leq C_{0}, \end{aligned}$$
(7)
where \(C_{0}\) is a positive constant independent of m, h, and τ.

Proof

Setting \(K_{0}\triangleq1+\max_{1\leq m\leq N}(\|u^{m}\|_{0,\infty}+\sqrt {\tau}(\sum_{i=1}^{m}\|\bar{\partial}_{t}u^{i}\|_{0,\infty}^{2})^{\frac{1}{2}})\), we begin to prove (6)–(7) by mathematical induction. When \(m=1\), by (1) and (5) we have the error equation
$$ \textstyle\begin{cases} \bar{\partial}_{t}e^{1}=\delta^{1}+R_{1}^{1}, \\ \bar{\partial}_{t}\delta^{1}-\nabla\cdot(a(u^{0})\tau\nabla\delta^{1})=R_{2}^{1}+R_{3}^{1}. \end{cases} $$
(8)
With \(\delta^{0}=0\), multiplying the second equation of (8) by \(\Delta \delta^{1}\) and integrating it over Ω, we get
$$\begin{aligned} \frac{1}{\tau} \bigl\Vert \nabla\delta^{1} \bigr\Vert _{0}^{2}+\tau \bigl\Vert a^{\frac{1}{2}} \bigl(u^{0}\bigr)\Delta \delta^{1} \bigr\Vert _{0}^{2} &=-\bigl(a_{u}\bigl(u^{0} \bigr)\nabla u^{0}\tau\nabla\delta^{1},\Delta\delta ^{1}\bigr)-\bigl(R_{2}^{1}+R_{3}^{1}, \Delta\delta^{1}\bigr) \\ &\leq C\tau \bigl\Vert \nabla\delta^{1} \bigr\Vert _{0} \bigl\Vert \Delta\delta^{1} \bigr\Vert _{0}+C\tau \bigl\Vert \Delta \delta^{1} \bigr\Vert _{0}. \end{aligned}$$
(9)
Further, since \(e^{1}\in H^{2}(\varOmega)\cap H_{0}^{1}(\varOmega)\), using the first equation of (8), we get
$$ \bigl\Vert e^{1} \bigr\Vert _{2}=\tau \bigl\Vert \bar{\partial}_{t}e^{1} \bigr\Vert _{2}\leq C\tau \bigl\Vert \Delta\delta^{1} \bigr\Vert _{0}+C\tau \bigl\Vert R_{1}^{1} \bigr\Vert _{2}. $$
(10)
Thus there exist positive constants \(\tau_{1}\), \(\tau_{2}\), \(C_{1}\), \(C_{2}\) such that when \(\tau\leq\tau_{1}\), we have
$$ \bigl\Vert e^{1} \bigr\Vert _{2}+\tau \bigl\Vert \bar{\partial}_{t}e^{1} \bigr\Vert _{2}+ \bigl\Vert \delta^{1} \bigr\Vert _{1}+ \tau \bigl\Vert \delta ^{1} \bigr\Vert _{2}\leq C_{1}\tau, $$
(11)
which implies
$$\begin{aligned}& \biggl\Vert \frac{U^{1}-U^{0}}{\tau} \biggr\Vert _{2}+ \bigl\Vert Q^{1} \bigr\Vert _{2}\leq C_{2}, \end{aligned}$$
(12)
$$\begin{aligned}& \bigl\Vert U^{1} \bigr\Vert _{0,\infty}\leq \bigl\Vert e^{1} \bigr\Vert _{0,\infty}+ \bigl\Vert u^{1} \bigr\Vert _{0,\infty }\leq CC_{1}\tau+ \bigl\Vert u^{1} \bigr\Vert _{0,\infty} \leq K_{0}, \end{aligned}$$
(13)
where \(\tau\leq\tau_{2}\leq1/CC_{1}\).
By mathematical induction we assume that (6) and (7) hold for \(m\leq n-1\). Then there exists \(\tau_{3}\) such that
$$\begin{aligned} & \bigl\Vert U^{m} \bigr\Vert _{0,\infty}+\sqrt{\tau}\Biggl( \sum_{i=1}^{m} \bigl\Vert \bar{\partial }_{t}U^{i} \bigr\Vert _{0,\infty}^{2} \Biggr)^{\frac{1}{2}} \\ &\quad \leq C \bigl\Vert e^{m} \bigr\Vert _{2}+C\sqrt{ \tau}\Biggl(\sum_{i=1}^{m} \bigl\Vert \bar{\partial}_{t}e^{i} \bigr\Vert _{2}^{2} \Biggr)^{\frac{1}{2}}+ \bigl\Vert u^{m} \bigr\Vert _{0,\infty}+\sqrt{\tau}\Biggl(\sum_{i=1}^{m} \bigl\Vert \bar{\partial}_{t}u^{i} \bigr\Vert _{0,\infty}^{2}\Biggr)^{\frac{1}{2}} \\ &\quad \leq CC_{0}\tau+CC_{0}\sqrt{\tau}+ \bigl\Vert u^{m} \bigr\Vert _{0,\infty}+\sqrt{\tau}\Biggl(\sum _{i=1}^{m} \bigl\Vert \bar{\partial}_{t}u^{i} \bigr\Vert _{0,\infty}^{2}\Biggr)^{\frac{1}{2}} \leq K_{0}, \end{aligned}$$
(14)
where \(\tau\leq\tau_{3}=\min\{1/2CC_{0},1/4C^{2}C_{0}^{2}\}\).
Then we begin to prove (6) and (7) for \(m=n\). Subtracting (5) from (1), we obtain
$$ \textstyle\begin{cases} \bar{\partial}_{t}e^{n}=\delta^{n}+R_{1}^{n}, \\ \bar{\partial}_{t}\delta^{n}-\nabla\cdot(a(U^{n-1})\tau\sum_{i=1}^{n}\nabla\delta^{i}) -\nabla\cdot(\tau\sum_{i=1}^{n}\nabla q^{i} (a(u^{n-1})-a(U^{n-1}))) \\ \qquad {}-\nabla\cdot(\nabla u^{0} (a(u^{n-1})-a(U^{n-1}))) \\ \quad =f(u^{n-1})-f(U^{n-1})+R_{2}^{n}+R_{3}^{n}+R_{4}^{n}. \end{cases} $$
(15)
Multiplying the second equation of (15) by \(\Delta\delta^{n}\) and integrating, we get
$$\begin{aligned} &\frac{1}{2\tau}\bigl( \bigl\Vert \nabla\delta^{n} \bigr\Vert _{0}^{2}- \bigl\Vert \nabla\delta^{n-1} \bigr\Vert _{0}^{2}\bigr) +\Biggl(a\bigl(U_{h}^{n-1} \bigr)\tau\sum_{i=1}^{n}\Delta \delta^{i},\Delta\delta ^{n}\Biggr) \\ & \quad =-\Biggl(a_{u}\bigl(U^{n-1}\bigr)\nabla U^{n-1}\Biggl(\tau\sum_{i=1}^{n} \nabla\delta^{i}\Biggr),\Delta \delta^{n}\Biggr) \\ & \qquad {}-\Biggl(\nabla\cdot\Biggl(\tau\sum_{i=1}^{n} \nabla q^{i} \bigl(a\bigl(u^{n-1}\bigr)-a\bigl(U^{n-1} \bigr)\bigr)\Biggr),\Delta\delta^{n}\Biggr) \\ & \qquad {}-\bigl(\nabla\cdot\bigl(\nabla u^{0} \bigl(a \bigl(u^{n-1}\bigr)-a\bigl(U^{n-1}\bigr)\bigr)\bigr),\Delta \delta ^{n}\bigr) \\ & \qquad {}-\bigl(f\bigl(u^{n-1}\bigr)-f\bigl(U^{n-1}\bigr), \Delta\delta^{n}\bigr) -\bigl(R_{2}^{n}+R_{3}^{n}+R_{4}^{n}, \Delta\delta^{n}\bigr). \end{aligned}$$
(16)
Observe that \((a(U^{n-1})\tau\sum_{i=1}^{n}\Delta\delta^{i},\Delta\delta ^{n})\) cannot be bounded directly; we rewrite it as
$$\begin{aligned} &\Biggl(a\bigl(U^{n-1}\bigr)\tau\sum_{i=1}^{n} \Delta\delta^{i},\Delta\delta^{n}\Biggr) \\ &\quad =\tau \int_{\varOmega}a\bigl(U^{n-1}\bigr)\sum _{i=1}^{n-1}\Delta\delta^{i}\cdot\Delta \delta^{n}+\tau \bigl\Vert a^{\frac{1}{2}}\bigl(U^{n-1} \bigr)\Delta\delta^{n} \bigr\Vert _{0}^{2} \\ &\quad =\frac{1}{2}\tau \int_{\varOmega}a\bigl(U^{n-1}\bigr) \Biggl(\sum _{i=1}^{n}\Delta\delta ^{i} \Biggr)^{2}-\frac{1}{2}\tau \int_{\varOmega}a\bigl(U^{n-1}\bigr) \Biggl(\sum _{i=1}^{n-1}\Delta\delta^{i} \Biggr)^{2} \\ &\qquad {}+\frac{1}{2}\tau \bigl\Vert a^{\frac{1}{2}} \bigl(U^{n-1}\bigr)\Delta\delta^{n} \bigr\Vert _{0}^{2}. \end{aligned}$$
Then we have
$$\begin{aligned} &\frac{1}{2\tau}\bigl( \bigl\Vert \nabla\delta^{n} \bigr\Vert _{0}^{2}- \bigl\Vert \nabla\delta^{n-1} \bigr\Vert _{0}^{2}\bigr) +\frac{1}{2}\tau \Biggl\Vert a^{\frac{1}{2}}\bigl(U^{n-1}\bigr)\sum_{i=1}^{n} \Delta\delta^{i} \Biggr\Vert _{0}^{2}- \frac{1}{2}\tau \Biggl\Vert a^{\frac{1}{2}}\bigl(U^{n-2}\bigr) \sum_{i=1}^{n-1}\Delta\delta^{i} \Biggr\Vert _{0}^{2} \\ &\qquad {} +\frac{1}{2}\tau \bigl\Vert a^{\frac{1}{2}} \bigl(U^{n-1}\bigr)\Delta\delta^{n} \bigr\Vert _{0}^{2} \\ &\quad \leq C\tau^{2} \bigl\Vert \bar{\partial}_{t}U^{n-1} \bigr\Vert _{2} \Biggl\Vert \sum_{i=1}^{n-1} \Delta\delta^{i} \Biggr\Vert _{0}^{2}- \Biggl(a_{u}\bigl(U^{n-1}\bigr)\nabla U^{n-1}\Biggl( \tau\sum_{i=1}^{n}\nabla\delta^{i} \Biggr),\Delta \delta^{n}\Biggr) \\ &\qquad {} -\Biggl(\tau\sum_{i=1}^{n} \Delta q^{i} \bigl(a\bigl(u^{n-1}\bigr)-a\bigl(U^{n-1} \bigr)\bigr),\Delta \delta^{n}\Biggr) -\Biggl(\tau\sum _{i=1}^{n}\nabla q^{i} a_{u} \bigl(U^{n-1}\bigr)\nabla e^{n-1},\Delta\delta ^{n} \Biggr) \\ &\qquad {} -\Biggl(\tau\sum_{i=1}^{n} \nabla q^{i} \nabla u^{n-1}\bigl(a_{u} \bigl(u^{n-1}\bigr)-a_{u}\bigl(U^{n-1}\bigr)\bigr), \Delta\delta^{n}\Biggr) \\ &\qquad {} -\bigl(\Delta u^{0} \bigl(a\bigl(u^{n-1}\bigr)-a \bigl(U^{n-1}\bigr)\bigr),\Delta\delta^{n}\bigr) -\bigl(\nabla u^{0} \bigl(a_{u}\bigl(U^{n-1}\bigr)\nabla e^{n-1}\bigr),\Delta\delta^{n}\bigr) \\ &\qquad {}-\bigl(\nabla u^{0} \nabla u^{n-1} \bigl(a_{u}\bigl(u^{n-1}\bigr)-a_{u} \bigl(U^{n-1}\bigr)\bigr),\Delta\delta^{n}\bigr) -\bigl(f\bigl(u^{n-1}\bigr)-f\bigl(U^{n-1}\bigr),\Delta \delta^{n}\bigr) \\ &\qquad {} -\bigl(R_{2}^{n}+R_{3}^{n}+R_{4}^{n}, \Delta\delta^{n}\bigr)\triangleq\sum_{i=1}^{10}A_{i}. \end{aligned}$$
In what follows, we will bound \(A_{i}\), \(i=2\sim10\), one by one. Note the particularity of \(\Delta\delta^{n}\) on the left-hand side, so we have to use new ways to handle \(\Delta\delta^{n}\) on the right-hand side instead of applying the Young inequality directly. In view of Green’s formula, it follows that
$$\begin{aligned} A_{9} &=-\bigl(f_{u}\bigl(\mu_{1}^{n-1} \bigr)e^{n-1},\Delta\delta^{n}\bigr) \\ &=\bigl(f_{uu}\bigl(\mu_{1}^{n-1}\bigr)\nabla \mu_{1}^{n-1}e^{n-1},\nabla\delta^{n}\bigr) +\bigl(f_{u}\bigl(\mu_{1}^{n-1}\bigr)\nabla e^{n-1},\nabla\delta^{n}\bigr) \\ &\leq C \bigl\Vert \nabla e^{n-1} \bigr\Vert _{0}^{2}+C \bigl\Vert \nabla\delta^{n} \bigr\Vert _{0}^{2}, \end{aligned}$$
where \(\mu_{1}^{n-1}=U^{n-1}+\lambda_{1}^{n-1} e^{n-1}\) and \(0<\lambda^{n-1}_{1}<1\).
For \(A_{2} \sim A_{8}\), \(A_{10}\), it is not so obvious to be dealt with. We choose to rewrite \(\Delta\delta^{n}\) by \(\tau\sum_{i=1}^{n}\Delta\bar{\partial}_{t}\delta ^{i}\) and then try to transfer τ from one side in the inner product to the other; more precisely,
$$\begin{aligned} A_{4} &=-\Biggl(\tau\sum_{i=1}^{n} \nabla q^{i} a_{u}\bigl(U^{n-1}\bigr)\nabla e^{n-1},\tau\sum_{i=1}^{n}\bar{ \partial}_{t}\Delta\delta^{i}\Biggr) \\ & =\Biggl(a_{u}\bigl(U^{n-1}\bigr)\nabla e^{n-1} \nabla q^{n},\tau\sum_{i=1}^{n-1} \Delta\delta ^{i}\Biggr)+\Biggl(\tau\sum_{i=1}^{n-1} \nabla q^{i} a_{u}\bigl(U^{n-2}\bigr)\bar{\partial }_{t}\nabla e^{n-1},\tau\sum_{i=1}^{n-1} \Delta\delta^{i}\Biggr) \\ &\quad {} +\Biggl(\tau\sum_{i=1}^{n-1}\nabla q^{i}\nabla e^{n-1}\frac{a_{u}(U^{n-1})- a_{u}(U^{n-2})}{\tau},\tau\sum _{i=1}^{n-1}\Delta\delta^{i}\Biggr) \\ &\quad {} -\bar{\partial}_{t}\Biggl(\tau\sum _{i=1}^{n}\nabla q^{i} a_{u} \bigl(U^{n-1}\bigr)\nabla e^{n-1},\tau\sum _{i=1}^{n}\Delta\delta^{i}\Biggr) \\ & \leq C \bigl\Vert \bar{\partial}_{t}\nabla e^{n-1} \bigr\Vert _{0} \Biggl\Vert \tau\sum_{i=1}^{n-1} \Delta \delta^{i} \Biggr\Vert _{0} +C \bigl\Vert \nabla e^{n-1} \bigr\Vert _{0} \bigl\Vert \bar{ \partial}_{t}U^{n-1} \bigr\Vert _{0,\infty} \Biggl\Vert \tau \sum_{i=1}^{n-1}\Delta \delta^{i} \Biggr\Vert _{0} \\ &\quad {}+C \bigl\Vert \nabla e^{n-1} \bigr\Vert _{0} \Biggl\Vert \tau\sum_{i=1}^{n-1}\Delta \delta^{i} \Biggr\Vert _{0}-\bar {\partial}_{t} \Biggl(\tau\sum_{i=1}^{n}\nabla q^{i} a_{u}\bigl(U^{n-1}\bigr)\nabla e^{n-1},\tau\sum_{i=1}^{n-1}\Delta \delta^{i}\Biggr) \\ & \leq C \bigl\Vert \bar{\partial}_{t}\nabla e^{n-1} \bigr\Vert _{0}^{2}+C \bigl\Vert \nabla e^{n-1} \bigr\Vert _{0}^{2}+C \bigl\Vert \bar{\partial}_{t}U^{n-1} \bigr\Vert _{0,\infty}^{2} \Biggl\Vert \tau\sum _{i=1}^{n-1}\Delta \delta^{i} \Biggr\Vert _{0}^{2} \\ &\quad {}+C \Biggl\Vert \tau\sum_{i=1}^{n-1} \Delta\delta^{i} \Biggr\Vert _{0}^{2} -\bar{ \partial}_{t}\Biggl(\tau\sum_{i=1}^{n} \nabla q^{i} a_{u}\bigl(U^{n-1}\bigr)\nabla e^{n-1},\tau\sum_{i=1}^{n}\Delta \delta^{i}\Biggr). \end{aligned}$$
Similarly,
$$\begin{aligned} A_{10} &=\Biggl(\bar{\partial}_{t}R_{2}^{n}+ \bar{\partial}_{t}R_{3}^{n}+\bar{\partial }_{t}R_{4}^{n},\tau\sum _{i=1}^{n-1}\Delta\delta^{i}\Biggr) -\bar{ \partial}_{t}\Biggl(R_{2}^{n}+R_{3}^{n}+R_{4}^{n}, \tau\sum_{i=1}^{n}\Delta\delta ^{i}\Biggr) \\ & \leq C\tau^{2}+C \Biggl\Vert \tau\sum _{i=1}^{n-1}\Delta\delta^{i} \Biggr\Vert _{0}^{2}-\bar {\partial}_{t}\Biggl(R_{2}^{n}+R_{3}^{n}+R_{4}^{n}, \tau\sum_{i=1}^{n}\Delta\delta^{i} \Biggr). \end{aligned}$$
For \(A_{2}\), we rewrite it as follows:
$$\begin{aligned} A_{2} &=-\Biggl(a_{u}\bigl(U^{n-1}\bigr)\nabla U^{n-1}\Biggl(\tau\sum_{i=1}^{n} \nabla\delta^{i}\Biggr),\tau \sum_{i=1}^{n} \Delta\bar{\partial}_{t}\delta^{i}\Biggr) \\ &=\Biggl(a_{u}\bigl(U^{n-2}\bigr)\nabla U^{n-2} \nabla\delta^{n},\tau\sum_{i=1}^{n-1} \Delta\delta^{i}\Biggr) \\ &\quad {} +\Biggl(\Biggl(\tau\sum_{i=1}^{n} \nabla\delta^{i}\Biggr)a_{u}\bigl(U^{n-2}\bigr)\bar{ \partial}_{t}\nabla U^{n-1},\tau\sum _{i=1}^{n-1}\Delta\delta^{i}\Biggr) \\ &\quad {} +\Biggl(\Biggl(\tau\sum_{i=1}^{n} \nabla\delta^{i}\Biggr)\nabla U^{n-1}\frac {a_{u}(U^{n-1})-a_{u}(U^{n-2})}{\tau},\tau \sum_{i=1}^{n-1}\Delta\delta ^{i} \Biggr) \\ &\quad {} -\bar{\partial}_{t}\Biggl(a_{u} \bigl(U^{n-1}\bigr)\nabla U^{n-1}\Biggl(\tau\sum _{i=1}^{n}\nabla \delta^{i}\Biggr),\tau\sum _{i=1}^{n}\Delta\delta^{i}\Biggr) \triangleq A_{2i}. \end{aligned}$$
In view of the embedding theorem, this yields
$$\begin{aligned} A_{22} &\leq C \Biggl\Vert \tau\sum_{i=1}^{n} \Delta\delta^{i} \Biggr\Vert _{0} \bigl\Vert \bar{ \partial}_{t}\Delta U^{n-1} \bigr\Vert _{0} \Biggl\Vert \tau\sum_{i=1}^{n-1}\Delta \delta^{i} \Biggr\Vert _{0} \\ & \leq C \Biggl\Vert \tau\sum_{i=1}^{n} \Delta\delta^{i} \Biggr\Vert _{0}^{2}+C \bigl\Vert \bar{\partial }_{t}\Delta U^{n-1} \bigr\Vert _{0}^{2} \Biggl\Vert \tau\sum_{i=1}^{n-1} \Delta\delta^{i} \Biggr\Vert _{0}^{2}. \end{aligned}$$
To get round the need of \(U^{i}\in H^{3}(\varOmega)\), \(i=1,2,\ldots, n-1\), we split \(U^{i}\), \(i=1,2,\ldots, n-1\), into two parts; with inductive assumption (14), it reduces to
$$\begin{aligned} A_{21} &=-\Biggl(a_{u}\bigl(U^{n-2}\bigr)\nabla e^{n-2}\nabla\delta^{n},\tau\sum_{i=1}^{n-1} \Delta \delta^{i}\Biggr)+\Biggl(a_{u}\bigl(U^{n-2} \bigr)\nabla u^{n-2}\nabla\delta^{n},\tau\sum _{i=1}^{n-1}\Delta\delta^{i}\Biggr) \\ & \leq C \bigl\Vert \Delta e^{n-2} \bigr\Vert _{0} \bigl\Vert \Delta\delta^{n} \bigr\Vert _{0} \Biggl\Vert \tau \sum_{i=1}^{n-1}\Delta\delta^{i} \Biggr\Vert _{0} +C \bigl\Vert \nabla\delta^{n} \bigr\Vert _{0} \Biggl\Vert \tau\sum_{i=1}^{n-1} \Delta\delta^{i} \Biggr\Vert _{0} \\ & \leq C \Biggl\Vert \tau\sum_{i=1}^{n-1} \Delta\delta^{i} \Biggr\Vert _{0}^{2}+ \frac{a_{0}}{4} \bigl\Vert \Delta e^{n-2} \bigr\Vert _{0}^{2} \bigl\Vert \Delta\delta^{n} \bigr\Vert _{0}^{2} +C \bigl\Vert \nabla\delta^{n} \bigr\Vert _{0}^{2} \\ &\leq C \Biggl\Vert \tau\sum_{i=1}^{n-1} \Delta\delta^{i} \Biggr\Vert _{0}^{2}+ \frac{a_{0}}{4}\tau \bigl\Vert \Delta\delta^{n} \bigr\Vert _{0}^{2} +C \bigl\Vert \nabla\delta^{n} \bigr\Vert _{0}^{2}. \end{aligned}$$
Similarly, we have
$$\begin{aligned} A_{23} &\leq C \bigl\Vert \Delta U^{n-1} \bigr\Vert _{0} \bigl\Vert \bar{\partial}_{t}\Delta U^{n-1} \bigr\Vert _{0} \Biggl\Vert \tau \sum_{i=1}^{n} \Delta\delta^{i} \Biggr\Vert _{0} \Biggl\Vert \tau\sum _{i=1}^{n-1}\Delta\delta^{i} \Biggr\Vert _{0} \\ &\leq C \Biggl\Vert \tau\sum_{i=1}^{n} \Delta\delta^{i} \Biggr\Vert _{0}^{2}+C \bigl\Vert \bar{\partial }_{t}\Delta U^{n-1} \bigr\Vert _{0}^{2} \Biggl\Vert \tau\sum_{i=1}^{n-1} \Delta\delta^{i} \Biggr\Vert _{0}^{2}. \end{aligned}$$
We split \(A_{3}\) as
$$\begin{aligned} A_{3} &=-\Biggl(\tau\sum_{i=1}^{n} \Delta q^{i} \bigl(a\bigl(u^{n-1}\bigr)-a\bigl(U^{n-1} \bigr)\bigr),\tau\sum_{i=1}^{n}\bar{ \partial}_{t}\Delta\delta^{i}\Biggr) \\ &=\Biggl(\tau\sum_{i=1}^{n-1}\Delta q^{i}\frac {(a(u^{n-1})-a(U^{n-1}))-(a(u^{n-2})-a(U^{n-2}))}{\tau},\tau\sum_{i=1}^{n-1} \Delta\delta^{i}\Biggr) \\ &\quad {} +\Biggl(\bigl(a\bigl(u^{n-1}\bigr)-a\bigl(U^{n-1} \bigr)\bigr)\Delta q^{n},\tau\sum_{i=1}^{n-1} \Delta\delta ^{i}\Biggr) \\ &\quad {}-\bar{\partial}_{t}\Biggl(\tau\sum _{i=1}^{n}\Delta q^{i} \bigl(a \bigl(u^{n-1}\bigr)-a\bigl(U^{n-1}\bigr)\bigr),\tau\sum _{i=1}^{n}\Delta\delta^{i}\Biggr)\triangleq \sum_{i=1}^{3}A_{3i}. \end{aligned}$$
We can see that
$$ A_{32}=\Biggl(\bigl(a\bigl(u^{n-1}\bigr)-a \bigl(U^{n-1}\bigr)\bigr)\Delta q^{n},\tau\sum _{i=1}^{n-1}\Delta \delta^{i}\Biggr)\leq C \bigl\Vert \nabla e^{n-1} \bigr\Vert _{0}^{2} +C \Biggl\Vert \tau\sum_{i=1}^{n-1}\Delta \delta^{i} \Biggr\Vert _{0}^{2}. $$
Since
$$\begin{aligned} &\frac{(a(u^{n-1})-a(U^{n-1}))-(a(u^{n-2})-a(U^{n-2}))}{\tau} \\ &\quad =a'\bigl(\mu_{2}^{n-1}\bigr)\bar{ \partial}_{t}e^{n-1} +\bar{\partial}_{t}u^{n-1} \bigl(a'\bigl(\mu_{3}^{n-1}\bigr)-a' \bigl(\mu_{2}^{n-1}\bigr)\bigr), \end{aligned}$$
(17)
where
$$\begin{aligned}& \mu_{3}^{n-1}=u^{n-2}+\tau\lambda_{3}^{n-1} \bar{\partial}_{t}u^{n-1},\qquad \mu _{2}^{n-1} =U^{n-2}+\tau\lambda_{2}^{n-1}\bar{ \partial}_{t}U^{n-1}, \\& \quad 0< \lambda_{2}^{n-1}< 1,0< \lambda_{3}^{n-1}< 1, \end{aligned}$$
and
$$ \mu_{3}^{n-1}-\mu_{2}^{n-1}=e^{n-2} +\tau\lambda_{2}^{n-1}\bar{\partial}_{t}e^{n-1} +\bar{\partial}_{t}u^{n-1}\tau\bigl(\lambda_{3}^{n-1}- \lambda_{2}^{n-1}\bigr), $$
we see that
$$\begin{aligned} A_{31}&\leq \Biggl\Vert \tau\sum_{i=1}^{n-1} \Delta q^{i} \Biggr\Vert _{0,4} \biggl\Vert \frac {(a(u^{n-1})-a(U^{n-1}))-(a(u^{n-2})-a(U^{n-2}))}{\tau} \biggr\Vert _{0,4} \Biggl\Vert \tau \sum _{i=1}^{n-1}\Delta\delta^{i} \Biggr\Vert _{0} \\ & \leq C \bigl\Vert \bar{\partial}_{t}\nabla e^{n-1} \bigr\Vert _{0} \Biggl\Vert \tau\sum_{i=1}^{n-1} \Delta \delta^{i} \Biggr\Vert _{0} +C \bigl\Vert \nabla e^{n-2} \bigr\Vert _{0} \Biggl\Vert \tau\sum _{i=1}^{n-1}\Delta\delta^{i} \Biggr\Vert _{0} +C\tau \Biggl\Vert \tau\sum_{i=1}^{n-1} \Delta\delta^{i} \Biggr\Vert _{0} \\ & \leq C\tau^{2}+C \bigl\Vert \bar{\partial}_{t}\nabla e^{n-1} \bigr\Vert _{0}^{2}+C \bigl\Vert \nabla e^{n-2} \bigr\Vert _{0}^{2} +C \Biggl\Vert \tau \sum_{i=1}^{n-1}\Delta\delta^{i} \Biggr\Vert _{0}^{2}, \end{aligned}$$
whence
$$\begin{aligned} A_{3} & \leq C\tau^{2}+C \bigl\Vert \bar{ \partial}_{t}\nabla e^{n-1} \bigr\Vert _{0}^{2}+C \bigl\Vert \nabla e^{n-2} \bigr\Vert _{0}^{2}+C \bigl\Vert \nabla e^{n-1} \bigr\Vert _{0}^{2} +C \Biggl\Vert \tau\sum_{i=1}^{n-1}\Delta \delta^{i} \Biggr\Vert _{0}^{2} \\ &\quad {} -\bar{\partial}_{t}\Biggl(\tau\sum _{i=1}^{n}\Delta q^{i} \bigl(a \bigl(u^{n-1}\bigr)-a\bigl(U^{n-1}\bigr)\bigr),\tau\sum _{i=1}^{n}\Delta\delta^{i}\Biggr). \end{aligned}$$
Rewriting \(A_{5}\), \(A_{6}\), \(A_{8}\), with (17), we obtain
$$\begin{aligned}& A_{6}=\Biggl(\Delta u^{0}\frac{ (a(u^{n-1})-a(U^{n-1}))-(a(u^{n-2})-a(U^{n-2}))}{\tau },\tau\sum _{i=1}^{n-1}\Delta\delta^{i}\Biggr) \\& \hphantom{A_{6}={}}{}-\bar{\partial}_{t}\Biggl(\Delta u^{0} \bigl(a\bigl(u^{n-1}\bigr)-a\bigl(U^{n-1}\bigr)\bigr),\tau\sum _{i=1}^{n}\Delta\delta^{i}\Biggr) \\& \hphantom{A_{6}}\leq \bigl\Vert \Delta u^{0} \bigr\Vert _{0,4} \biggl\Vert \frac {(a(u^{n-1})-a(U^{n-1}))-(a(u^{n-2})-a(U^{n-2}))}{\tau} \biggr\Vert _{0,4} \Biggl\Vert \tau \sum_{i=1}^{n-1}\Delta \delta^{i} \Biggr\Vert _{0} \\& \hphantom{A_{6}={}}{}-\bar{\partial}_{t}\Biggl(\Delta u^{0} \bigl(a\bigl(u^{n-1}\bigr)-a\bigl(U^{n-1}\bigr)\bigr),\tau\sum _{i=1}^{n}\Delta\delta^{i}\Biggr) \\& \hphantom{A_{6}}\leq C \bigl\Vert \bar{\partial}_{t}\nabla e^{n-1} \bigr\Vert _{0} \Biggl\Vert \tau\sum _{i=1}^{n-1}\Delta \delta^{i} \Biggr\Vert _{0} +C \bigl\Vert \nabla e^{n-2} \bigr\Vert _{0} \Biggl\Vert \tau\sum_{i=1}^{n-1} \Delta\delta^{i} \Biggr\Vert _{0} \\& \hphantom{A_{6}={}}{}+C\tau \Biggl\Vert \tau\sum_{i=1}^{n-1} \Delta\delta^{i} \Biggr\Vert _{0}-\bar{\partial }_{t}\Biggl(\Delta u^{0} \bigl(a\bigl(u^{n-1} \bigr)-a\bigl(U^{n-1}\bigr)\bigr),\tau\sum_{i=1}^{n} \Delta\delta^{i}\Biggr) \\& \hphantom{A_{6}}\leq C\tau^{2}+C \bigl\Vert \bar{ \partial}_{t}\nabla e^{n-1} \bigr\Vert _{0}^{2}+C \bigl\Vert \nabla e^{n-2} \bigr\Vert _{0}^{2}+C \Biggl\Vert \tau\sum_{i=1}^{n-1}\Delta \delta^{i} \Biggr\Vert _{0}^{2} \\& \hphantom{A_{6}={}}{} -\bar{\partial}_{t}\Biggl(\Delta u^{0} \bigl(a\bigl(u^{n-1}\bigr)-a\bigl(U^{n-1}\bigr)\bigr),\tau\sum _{i=1}^{n}\Delta\delta^{i}\Biggr), \\& A_{5} =-\Biggl(\tau\sum_{i=1}^{n} \nabla q^{i} \nabla u^{n-1}\bigl(a_{u} \bigl(u^{n-1}\bigr)-a_{u}\bigl(U^{n-1}\bigr)\bigr), \tau\sum_{i=1}^{n}\bar{\partial }_{t}\Delta\delta^{i}\Biggr) \\& \hphantom{A_{5}}=\Biggl(\tau\sum_{i=1}^{n-1} \nabla q^{i} \nabla u^{n-2}\frac {(a_{u}(u^{n-1})-a_{u}(U^{n-1}))-(a_{u}(u^{n-2})-a_{u}(U^{n-2}))}{\tau},\tau \sum _{i=1}^{n-1}\Delta\delta^{i}\Biggr) \\& \hphantom{A_{5} ={}}{} +\Biggl(\bigl(a_{u}\bigl(u^{n-1} \bigr)-a_{u}\bigl(U^{n-1}\bigr)\bigr)\tau\sum _{i=1}^{n-1}\nabla q^{i}\bar { \partial}_{t}\nabla u^{n-1},\tau\sum _{i=1}^{n-1}\Delta\delta ^{i}\Biggr) \\& \hphantom{A_{5} ={}}{} +\Biggl(\bigl(a_{u}\bigl(u^{n-1} \bigr)-a_{u}\bigl(U^{n-1}\bigr)\bigr)\nabla u^{n-1} \nabla q^{n},\tau\sum_{i=1}^{n-1} \Delta\delta^{i}\Biggr) \\& \hphantom{A_{5} ={}}{} -\bar{\partial}_{t}\Biggl(\tau\sum _{i=1}^{n}\nabla q^{i} \nabla u^{n-1}\bigl(a_{u}\bigl(u^{n-1}\bigr)-a_{u} \bigl(U^{n-1}\bigr)\bigr),\tau\sum_{i=1}^{n} \Delta\delta ^{i}\Biggr) \\& \hphantom{A_{5} } \leq C \biggl\Vert \frac {(a_{u}(u^{n-1})-a_{u}(U^{n-1}))-(a_{u}(u^{n-2})-a_{u}(U^{n-2}))}{\tau} \biggr\Vert _{0} \Biggl\Vert \tau\sum_{i=1}^{n-1} \Delta\delta^{i} \Biggr\Vert _{0} \\& \hphantom{A_{5} ={}}{} +C \bigl\Vert e^{n-1} \bigr\Vert _{0} \Biggl\Vert \tau\sum_{i=1}^{n-1}\Delta \delta^{i} \Biggr\Vert _{0}-\bar{\partial }_{t} \Biggl(\tau\sum_{i=1}^{n}\nabla q^{i} \nabla u^{n-1}\bigl(a_{u} \bigl(u^{n-1}\bigr)-a_{u}\bigl(U^{n-1}\bigr)\bigr), \tau\sum_{i=1}^{n}\Delta\delta ^{i}\Biggr) \\& \hphantom{A_{5}} \leq C\tau^{2}+C \bigl\Vert \bar{ \partial}_{t}\nabla e^{n-1} \bigr\Vert _{0}^{2}+C \bigl\Vert \nabla e^{n-1} \bigr\Vert _{0}^{2}+C \bigl\Vert \nabla e^{n-2} \bigr\Vert _{0}^{2}+C \Biggl\Vert \tau\sum_{i=1}^{n-1}\Delta \delta^{i} \Biggr\Vert _{0}^{2} \\& \hphantom{A_{5} ={}}{} -\bar{\partial}_{t}\Biggl(\tau\sum _{i=1}^{n}\nabla q^{i} \nabla u^{n-1}\bigl(a_{u}\bigl(u^{n-1}\bigr)-a_{u} \bigl(U^{n-1}\bigr)\bigr),\tau\sum_{i=1}^{n} \Delta\delta^{i}\Biggr), \end{aligned}$$
and
$$\begin{aligned} A_{8} &=\Biggl(\nabla u^{0} \nabla u^{n-2} \frac {(a_{u}(u^{n-1})-a_{u}(U^{n-1}))-(a_{u}(u^{n-2})-a_{u}(U^{n-2}))}{\tau},\tau \sum_{i=1}^{n-1}\Delta \delta^{i}\Biggr) \\ &\quad {} +\Biggl(\nabla u^{0} \bigl(a_{u} \bigl(u^{n-1}\bigr)-a_{u}\bigl(U^{n-1}\bigr)\bigr) \bar{\partial}_{t}\nabla u^{n-1},\tau\sum _{i=1}^{n-1}\Delta\delta^{i}\Biggr) \\ &\quad {} -\bar{\partial}_{t}\Biggl(\nabla u^{0} \nabla u^{n-1}\bigl(a_{u}\bigl(u^{n-1}\bigr)-a_{u} \bigl(U^{n-1}\bigr)\bigr),\tau\sum_{i=1}^{n} \Delta\delta ^{i}\Biggr) \\ &\leq C\tau^{2}+C \bigl\Vert \bar{\partial}_{t}\nabla e^{n-1} \bigr\Vert _{0}^{2}+C \bigl\Vert \nabla e^{n-1} \bigr\Vert _{0}^{2}+C \bigl\Vert \nabla e^{n-2} \bigr\Vert _{0}^{2} \\ &\quad {}+C \Biggl\Vert \tau\sum_{i=1}^{n-1} \Delta\delta^{i} \Biggr\Vert _{0}^{2}-\bar{ \partial }_{t}\Biggl(\nabla u^{0} \nabla u^{n-1} \bigl(a_{u}\bigl(u^{n-1}\bigr)-a_{u} \bigl(U^{n-1}\bigr)\bigr),\tau\sum_{i=1}^{n} \Delta\delta^{i}\Biggr). \end{aligned}$$
Finally, \(A_{7}\) can be bounded as
$$\begin{aligned} A_{7} &=\Biggl(\nabla u^{0} a_{u} \bigl(U^{n-2}\bigr)\bar{\partial}_{t}\nabla e^{n-1}, \tau\sum_{i=1}^{n-1}\Delta \delta^{i}\Biggr) \\ &\quad {}+\Biggl(\nabla u^{0} \nabla e^{n-1} \frac{a_{u}(U^{n-1})-a_{u}(U^{n-2})}{\tau},\tau \sum_{i=1}^{n-1}\Delta \delta^{i}\Biggr) \\ &\quad {}-\bar{\partial}_{t}\Biggl(\nabla u^{0} a_{u}\bigl(U^{n-1}\bigr)\nabla e^{n-1},\tau\sum_{i=1}^{n}\Delta \delta^{i}\Biggr) \\ & \leq C \bigl\Vert \bar{\partial}_{t}\nabla e^{n-1} \bigr\Vert _{0} \Biggl\Vert \tau\sum_{i=1}^{n-1} \Delta \delta^{i} \Biggr\Vert _{0} +C \bigl\Vert \nabla e^{n-1} \bigr\Vert _{0} \bigl\Vert \bar{ \partial}_{t}U^{n-1} \bigr\Vert _{0,\infty} \Biggl\Vert \tau \sum_{i=1}^{n-1}\Delta \delta^{i} \Biggr\Vert _{0} \\ &\quad {} -\bar{\partial}_{t}\Biggl(\nabla u^{0} a_{u}\bigl(U^{n-1}\bigr)\nabla e^{n-1},\tau\sum _{i=1}^{n}\Delta\delta^{i}\Biggr) \\ & \leq C \bigl\Vert \bar{\partial}_{t}\nabla e^{n-1} \bigr\Vert _{0}^{2} +C \bigl\Vert \nabla e^{n-1} \bigr\Vert _{0}^{2}+C \Biggl\Vert \tau\sum _{i=1}^{n-1}\Delta\delta^{i} \Biggr\Vert _{0}^{2} \\ &\quad {} +C \bigl\Vert \bar{\partial}_{t}U^{n-1} \bigr\Vert _{0,\infty}^{2} \Biggl\Vert \tau\sum _{i=1}^{n-1}\Delta \delta^{i} \Biggr\Vert _{0}^{2} -\bar{\partial}_{t}\Biggl(\nabla u^{0} a_{u}\bigl(U^{n-1}\bigr)\nabla e^{n-1},\tau\sum_{i=1}^{n}\Delta \delta^{i}\Biggr). \end{aligned}$$
Moreover, because
$$\begin{aligned} & \bigl\Vert \nabla\bar{\partial}_{t}e^{n} \bigr\Vert _{0}= \bigl\Vert \nabla\bigl(\delta^{n}+R_{1}^{n} \bigr) \bigr\Vert _{0}\leq C \bigl\Vert \nabla\delta^{n} \bigr\Vert _{0}+C \bigl\Vert \nabla R_{1}^{n} \bigr\Vert _{0} \leq C \bigl\Vert \nabla\delta^{n} \bigr\Vert _{0}+C\tau, \\ & \bigl\Vert \Delta\bar{\partial}_{t}e^{n} \bigr\Vert _{0}= \bigl\Vert \Delta\bigl(\delta^{n}+R_{1}^{n} \bigr) \bigr\Vert _{0}\leq C \bigl\Vert \Delta\delta^{n} \bigr\Vert _{0}+C \bigl\Vert \Delta R_{1}^{n} \bigr\Vert _{0} \leq C \bigl\Vert \Delta\delta^{n} \bigr\Vert _{0}+C\tau, \\ & \bigl\Vert \nabla e^{n} \bigr\Vert _{0}\leq C\sqrt{ \tau}\Biggl(\sum_{i=1}^{n} \bigl\Vert \nabla\bar{\partial }_{t}e^{i} \bigr\Vert _{0}^{2}\Biggr)^{\frac{1}{2}}\leq C\sqrt{\tau}\Biggl(\sum _{i=1}^{n} \bigl\Vert \nabla \delta^{i} \bigr\Vert _{0}^{2} \Biggr)^{\frac{1}{2}}+C\tau, \end{aligned}$$
(18)
we have
$$\begin{aligned} &\frac{1}{\tau}\bigl( \bigl\Vert \nabla\delta^{n} \bigr\Vert _{0}^{2}- \bigl\Vert \nabla\delta^{n-1} \bigr\Vert _{0}^{2}\bigr) +\tau \Biggl\Vert a^{\frac{1}{2}} \bigl(U^{n-1}\bigr)\sum_{i=1}^{n} \Delta\delta^{i} \Biggr\Vert _{0}^{2}-\tau \Biggl\Vert a^{\frac{1}{2}}\bigl(U^{n-2}\bigr)\sum _{i=1}^{n-1}\Delta\delta^{i} \Biggr\Vert _{0}^{2}+\tau \bigl\Vert \Delta\delta^{n} \bigr\Vert _{0}^{2} \\ &\quad \leq C\tau^{2}+C \bigl\Vert \nabla\delta^{n} \bigr\Vert _{0}^{2}+C \bigl\Vert \nabla\delta^{n-1} \bigr\Vert _{0}^{2}+C\tau\sum_{i=1}^{n} \bigl\Vert \nabla\delta^{i} \bigr\Vert _{0}^{2} \\ &\qquad {} +C \Biggl\Vert \tau\sum_{i=1}^{n} \Delta\delta^{i} \Biggr\Vert _{0}^{2}+C \bigl\Vert \bar{\partial}_{t}U^{n-1} \bigr\Vert _{2}^{2} \Biggl\Vert \tau\sum_{i=1}^{n-1} \Delta\delta^{i} \Biggr\Vert _{0}^{2} \\ &\qquad {} +C \Biggl\Vert \tau\sum_{i=1}^{n-1} \Delta\delta^{i} \Biggr\Vert _{0}^{2} -\bar{ \partial}_{t}\Biggl(a_{u}\bigl(U^{n-1}\bigr)\nabla U^{n-1}\Biggl(\tau\sum_{i=1}^{n} \nabla \delta^{i}\Biggr),\tau\sum_{i=1}^{n} \Delta\delta^{i}\Biggr) \\ &\qquad {} -\bar{\partial}_{t}\Biggl(\tau\sum _{i=1}^{n}\nabla q^{i} a_{u} \bigl(U^{n-1}\bigr)\nabla e^{n-1},\tau\sum _{i=1}^{n}\Delta\delta^{i}\Biggr) \\ &\qquad {} -\bar{\partial}_{t}\Biggl(\tau\sum _{i=1}^{n}\Delta q^{i} \bigl(a \bigl(u^{n-1}\bigr)-a\bigl(U^{n-1}\bigr)\bigr),\tau\sum _{i=1}^{n}\Delta\delta^{i}\Biggr) \\ &\qquad {} -\bar{\partial}_{t}\Biggl(\Delta u^{0} \bigl(a \bigl(u^{n-1}\bigr)-a\bigl(U^{n-1}\bigr)\bigr),\tau\sum _{i=1}^{n}\Delta\delta^{i}\Biggr) \\ &\qquad {} -\bar{\partial}_{t}\Biggl(\nabla u^{0} a_{u}\bigl(U^{n-1}\bigr)\nabla e^{n-1},\tau\sum _{i=1}^{n}\Delta\delta^{i}\Biggr)- \bar{\partial}_{t}\Biggl(R_{2}^{n}+R_{3}^{n}+R_{4}^{n}, \tau\sum_{i=1}^{n}\Delta\delta^{i} \Biggr) \\ &\qquad {} -\bar{\partial}_{t}\Biggl(\tau\sum _{i=1}^{n}\nabla q^{i} \nabla u^{n-1}\bigl(a_{u}\bigl(u^{n-1}\bigr)-a_{u} \bigl(U^{n-1}\bigr)\bigr),\tau\sum_{i=1}^{n} \Delta\delta ^{i}\Biggr) \\ &\qquad {} -\bar{\partial}_{t}\Biggl(\nabla u^{0} \nabla u^{n-1}\bigl(a_{u}\bigl(u^{n-1}\bigr)-a_{u} \bigl(U^{n-1}\bigr)\bigr),\tau\sum_{i=1}^{n} \Delta\delta^{i}\Biggr). \end{aligned}$$
Summing this inequality from 2 to n, we get
$$\begin{aligned} & \bigl\Vert \nabla\delta^{n} \bigr\Vert _{0}^{2} + \Biggl\Vert \tau\sum_{i=1}^{n}\Delta \delta^{i} \Biggr\Vert _{0}^{2} +\sum _{i=2}^{n} \bigl\Vert \tau\Delta \delta^{i} \bigr\Vert _{0}^{2} \\ &\quad \leq \bigl\Vert \nabla\delta^{1} \bigr\Vert _{0}^{2}+\tau^{2} \bigl\Vert \Delta \delta^{1} \bigr\Vert _{0}^{2} +C \tau^{2}+C\tau\sum_{i=1}^{n} \bigl\Vert \nabla\delta^{i} \bigr\Vert _{0}^{2}+C\tau \sum_{i=1}^{n} \Biggl\Vert \tau\sum _{j=1}^{i-1}\Delta\delta^{j} \Biggr\Vert _{0}^{2} \\ &\qquad {} +C\tau^{2}\sum _{i=2}^{n}\sum_{j=1}^{i-1} \bigl\Vert \nabla\delta^{j} \bigr\Vert _{0}^{2}+C\tau\sum_{i=1}^{n} \bigl\Vert \bar{\partial}_{t}\Delta U^{i-1} \bigr\Vert _{0}^{2} \Biggl\Vert \tau\sum_{j=1}^{i-1} \Delta\delta^{j} \Biggr\Vert _{0}^{2} \\ &\qquad {}- \Biggl(a_{u}\bigl(U^{n-1}\bigr)\nabla U^{n-1}\Biggl( \tau\sum_{i=1}^{n}\nabla\delta^{i} \Biggr),\tau\sum_{i=1}^{n}\Delta \delta^{i}\Biggr) \\ &\qquad {} +\bigl(a_{u}\bigl(U^{0}\bigr)\nabla U^{0}\bigl(\tau\nabla\delta^{1}\bigr),\tau\Delta \delta^{1}\bigr) -\Biggl(\tau\sum_{i=1}^{n} \Delta q^{i} \bigl(a\bigl(u^{n-1}\bigr)-a\bigl(U^{n-1} \bigr)\bigr),\tau\sum_{i=1}^{n}\Delta \delta^{i}\Biggr) \\ &\qquad {} -\Biggl(\tau\sum_{i=1}^{n} \nabla q^{i} a_{u}\bigl(U^{n-1}\bigr)\nabla e^{n-1},\tau\sum_{i=1}^{n}\Delta \delta^{i}\Biggr) \\ &\qquad {}-\Biggl(\Delta u^{0} \bigl(a \bigl(u^{n-1}\bigr)-a\bigl(U^{n-1}\bigr)\bigr),\tau\sum _{i=1}^{n}\Delta\delta^{i}\Biggr) \\ &\qquad {} -\Biggl(\nabla u^{0} a_{u}\bigl(U^{n-1} \bigr)\nabla e^{n-1},\tau\sum_{i=1}^{n} \Delta\delta^{i}\Biggr) -\Biggl(R_{2}^{n}+R_{3}^{n}+R_{4}^{n}, \tau\sum_{i=1}^{n}\Delta\delta^{i} \Biggr) \\ &\qquad {} +\bigl(R_{2}^{1}+R_{3}^{1}+R_{4}^{1}, \tau\Delta\delta^{1}\bigr)-\Biggl(\tau\sum_{i=1}^{n} \nabla q^{i} \nabla u^{n-1}\bigl(a_{u} \bigl(u^{n-1}\bigr)-a_{u}\bigl(U^{n-1}\bigr)\bigr), \tau\sum_{i=1}^{n}\Delta \delta^{i}\Biggr) \\ &\qquad {}-\Biggl(\nabla u^{0} \nabla u^{n-1} \bigl(a_{u}\bigl(u^{n-1}\bigr)-a_{u} \bigl(U^{n-1}\bigr)\bigr),\tau\sum_{i=1}^{n} \Delta\delta^{i}\Biggr). \end{aligned}$$
(19)
Due to
$$\begin{aligned} &\Biggl(a_{u}\bigl(U^{n-1}\bigr)\nabla U^{n-1} \Biggl(\tau\sum_{i=1}^{n}\nabla \delta^{i}\Biggr),\tau \sum_{i=1}^{n} \Delta\delta^{i}\Biggr) \\ &\quad = \Biggl(a_{u}\bigl(U^{n-1}\bigr)\nabla u^{n-1}\Biggl(\tau\sum_{i=1}^{n} \nabla\delta^{i}\Biggr),\tau\sum_{i=1}^{n} \Delta\delta^{i}\Biggr) \\ &\qquad {} -\Biggl(a_{u}\bigl(U^{n-1}\bigr)\nabla e^{n-1}\Biggl(\tau\sum_{i=1}^{n} \nabla\delta^{i}\Biggr),\tau \sum_{i=1}^{n} \Delta\delta^{i}\Biggr) \\ &\quad \leq C\tau^{\frac{1}{4}} \Biggl\Vert \tau\sum _{i=1}^{n}\Delta\delta^{i} \Biggr\Vert _{0}^{2}+C \Biggl\Vert \tau \sum _{i=1}^{n}\nabla\delta^{i} \Biggr\Vert _{0}^{2} +\frac{1}{4} \Biggl\Vert \tau\sum _{i=1}^{n}\Delta\delta^{i} \Biggr\Vert _{0}^{2} \\ &\quad \leq C\tau^{\frac{1}{2}} \Biggl\Vert \tau\sum _{i=1}^{n}\Delta\delta^{i} \Biggr\Vert _{0}^{2} +\frac{1}{2} \Biggl\Vert \tau\sum _{i=1}^{n}\Delta\delta^{i} \Biggr\Vert _{0}^{2}+C\tau\sum_{i=1}^{n} \bigl\Vert \nabla\delta^{i} \bigr\Vert _{0}^{2}, \end{aligned}$$
(20)
after obvious estimates and a kickback of \(\tau\sum_{i=1}^{n}\|\nabla \delta^{i}\|_{0}^{2}\), together with our earlier estimate for \(n=1\), we obtain
$$ \bigl\Vert \nabla\delta^{n} \bigr\Vert _{0}^{2}+\tau^{2} \Biggl\Vert \sum _{i=1}^{n}\Delta\delta^{i} \Biggr\Vert _{0}^{2}+\tau^{2}\sum_{i=2}^{n} \bigl\Vert \Delta\delta^{i} \bigr\Vert _{0}^{2} \leq C\tau^{2}. $$
(21)
Here by (18) we have
$$ \tau \Biggl\Vert \sum_{i=1}^{n} \bar{\partial}_{t}e^{i} \Biggr\Vert _{2} +\tau \Biggl(\sum_{i=2}^{n} \bigl\Vert \bar{ \partial}_{t}e^{i} \bigr\Vert _{2}^{2} \Biggr)^{\frac{1}{2}} \leq C\tau $$
(22)
and, further,
$$ \bigl\Vert e^{n} \bigr\Vert _{2}=\tau \Biggl\Vert \sum_{i=1}^{n}\bar{ \partial}_{t}e^{i} \Biggr\Vert _{2}\leq C\tau. $$
(23)
Then we conclude that there exist \(\tau_{4}\), \(\tau_{5}\), \(C_{3}\), \(C_{4}\) such that when \(\tau\leq\tau_{4}\), we have
$$ \bigl\Vert e^{n} \bigr\Vert _{2}+\tau \Biggl(\sum_{i=2}^{n} \bigl\Vert \bar{ \partial}_{t}e^{i} \bigr\Vert _{2}^{2} \Biggr)^{\frac{1}{2}}+ \bigl\Vert \delta^{n} \bigr\Vert _{1}+\tau \Biggl\Vert \sum_{i=1}^{n} \delta^{i} \Biggr\Vert _{2}+\tau\Biggl(\sum _{i=2}^{n} \bigl\Vert \delta^{i} \bigr\Vert _{2}^{2}\Biggr)^{\frac{1}{2}}\leq C_{3} \tau, $$
(24)
which leads to
$$\begin{aligned} & \bigl\Vert e^{n} \bigr\Vert _{2}\leq \tau^{\frac{1}{4}},\qquad \bigl\Vert \bar{\partial}_{t}U^{n} \bigr\Vert _{2}\leq C_{4}, \end{aligned}$$
(25)
$$\begin{aligned} & \bigl\Vert U^{n} \bigr\Vert _{0,\infty}+\sqrt{\tau}\Biggl( \sum_{i=1}^{n} \bigl\Vert \bar{\partial }_{t}U^{i} \bigr\Vert _{0,\infty}^{2} \Biggr)^{\frac{1}{2}} \\ &\quad \leq C \bigl\Vert e^{n} \bigr\Vert _{2}+C\sqrt{\tau} \Biggl(\sum_{i=1}^{n} \bigl\Vert \bar{ \partial}_{t}e^{i} \bigr\Vert _{2}^{2} \Biggr)^{\frac{1}{2}}+ \bigl\Vert u^{n} \bigr\Vert _{0,\infty}+\sqrt{\tau}\Biggl(\sum_{i=1}^{n} \bigl\Vert \bar{\partial}_{t}u^{i} \bigr\Vert _{0,\infty}^{2}\Biggr)^{\frac{1}{2}} \\ &\quad \leq CC_{3}\tau+CC_{3}\sqrt{\tau}+ \bigl\Vert u^{n} \bigr\Vert _{0,\infty}+\sqrt{\tau}\Biggl(\sum _{i=1}^{n} \bigl\Vert \bar{\partial}_{t}u^{i} \bigr\Vert _{0,\infty}^{2}\Biggr)^{\frac{1}{2}} \leq K_{0}, \end{aligned}$$
(26)
where \(\tau\leq\tau_{5}=\min\{1/2CC_{3},1/4C^{2}C_{3}^{2}\}\). Clearly, \(C_{3}, C_{4}\) have nothing to do with \(C_{0}\), and thus (6) and (7) hold for \(m=n\) if we take \(C_{0}\geq\sum_{i=1}^{4}C_{i}\) and \(\tau_{0}\leq \min_{1\leq\tau\leq5}\tau_{i}\). Then the induction is closed. The proof is completed. □

Remark 1

The special method used to tackle the left-hand side of (16) is important to deduce the regularities of \(U^{n}\) and \(Q^{n}\) in the \(H^{2}\)-norm. Further, the terms including \(\Delta\delta^{n}\) on the right-hand side needs innovative technologies to treat.

4 Error estimates for spatial-discrete system and optimal error results

In this section, we will establish τ-independent optimal error results for \(u^{n}\) and \(q^{n}\) through the spatial results. We decompose the errors as follows:
$$\begin{aligned} &U^{i}-U^{i}_{h}=U^{i}-I_{h}U^{i}+I_{h}U^{i}-U^{i}_{h} \triangleq\eta^{i}+\xi^{i}, \\ &Q^{i}-Q^{i}_{h}=Q^{i}-I_{h}Q^{i}+I_{h}Q^{i}-Q^{i}_{h} \triangleq r^{i}+\theta^{i},\quad i=1,2,\ldots, n, \end{aligned}$$
and we are now ready for the unconditional spatial results.

Theorem 2

Let \(u^{m}\) and \(U_{h}^{m}\) be solutions of (3) and (4), respectively, for \(m=1,2,\ldots, N\). Under the conditions of Theorem 1, there exist \(\tau _{0}',h_{0}'\) such that, for \(\tau\leq\tau_{0}'\) and \(h\leq h_{0}'\), we have
$$ \bigl\Vert u^{m}-U_{h}^{m} \bigr\Vert _{0}+ \bigl\Vert q^{m}-Q_{h}^{m} \bigr\Vert _{0}=O\bigl(h^{2}+\tau\bigr) $$
(27)
and
$$ \bigl\Vert \nabla\bigl(u^{m}-U_{h}^{m} \bigr) \bigr\Vert _{0}+ \bigl\Vert \nabla\bigl( q^{m}-Q_{h}^{m} \bigr) \bigr\Vert _{0}=O(h+\tau). $$
(28)

Proof

Before discussing (27) and (28), we shall pause to give the results
$$ \bigl\Vert \xi^{m} \bigr\Vert _{0}+ \bigl\Vert \theta^{m} \bigr\Vert _{0}+\tau\Biggl(\sum _{i=1}^{m} \bigl\Vert \nabla\theta^{i} \bigr\Vert _{0}^{2}\Biggr)^{\frac{1}{2}}\leq C_{0}'h\bigl(h+\tau^{\frac{1}{2}}\bigr) $$
(29)
by mathematical induction, where \(C_{0}'\) is a positive constant independent of m, τ, and h. Since \(\|I_{h}U^{m}\|_{0,\infty}+\sqrt{\tau}(\sum_{i=2}^{m}\|\bar{\partial }_{t}I_{h}U^{i}\|_{0,\infty}^{2})^{\frac{1}{2}}\leq C\), let \(K_{0}'\triangleq 1+\|I_{h}U^{m}\|_{0,\infty}+\sqrt{\tau}\times (\sum_{i=2}^{m}\|\bar{\partial }_{t}I_{h}U^{i}\|_{0,\infty}^{2})^{\frac{1}{2}}\). We begin with \(m=1\):
$$\begin{aligned} &\bigl(\bar{\partial}_{t}\theta^{1},v_{h}\bigr)+ \bigl(a\bigl(U^{0}\bigr)\tau\nabla\theta^{1},\nabla v_{h}\bigr) \\ &\quad = -\bigl(\bar{\partial}_{t}r^{1},v_{h} \bigr)-\bigl(a\bigl(U^{0}\bigr)\tau\nabla r^{1},\nabla v_{h}\bigr) \\ &\qquad {}-\bigl(\bigl(a\bigl(U^{0}\bigr)-a\bigl(U_{h}^{0} \bigr)\bigr)\tau\nabla Q_{h}^{1},\nabla v_{h} \bigr)-\bigl(a\bigl(U_{h}^{0}\bigr)\nabla\eta^{0}, \nabla v\bigr) \\ &\qquad {} -\bigl(\nabla U^{0} \bigl(a\bigl(U^{0}\bigr)-a \bigl(U_{h}^{0}\bigr)\bigr), \nabla v\bigr)+\bigl(f \bigl(U^{0}\bigr)-f\bigl(U_{h}^{0} \bigr),v_{h}\bigr). \end{aligned}$$
(30)
Taking \(v_{h}=\theta^{1}\) in (30), we get
$$\begin{aligned} &\frac{1}{\tau} \bigl\Vert \theta^{1} \bigr\Vert _{0}^{2}+\tau \bigl\Vert a^{\frac{1}{2}} \bigl(U^{0}\bigr)\nabla\theta^{1} \bigr\Vert _{0}^{2} \\ &\quad = -\bigl(\bar{\partial}_{t}r^{1}, \theta^{1}\bigr) -\bigl(a\bigl(U^{0}\bigr)\tau\nabla r^{1},\nabla\theta^{1}\bigr) \\ &\qquad {}-\bigl(\bigl(a \bigl(U^{0}\bigr)-a\bigl(U_{h}^{0}\bigr)\bigr) \tau \nabla Q_{h}^{1},\nabla\theta^{1}\bigr)- \bigl(a\bigl(U_{h}^{0}\bigr)\nabla\eta^{0}, \nabla \theta ^{1}\bigr) \\ &\qquad {} -\bigl(\nabla U^{0} \bigl(a\bigl(U^{0}\bigr)-a \bigl(U_{h}^{0}\bigr)\bigr), \nabla\theta ^{1} \bigr)+\bigl(f\bigl(U^{0}\bigr)-f\bigl(U_{h}^{0} \bigr),\theta^{1}\bigr). \end{aligned}$$
(31)
It is easy to see that
$$\begin{aligned} &\bigl(\bar{\partial}_{t}r^{1},\theta^{1}\bigr) \leq Ch^{2} \bigl\Vert \bar{\partial}_{t}U^{1} \bigr\Vert _{2} \bigl\Vert \theta^{1} \bigr\Vert _{0}\leq Ch^{4}+C \bigl\Vert \theta^{1} \bigr\Vert _{0}^{2}, \\ &\bigl(\nabla U^{0} \bigl(a\bigl(U^{0}\bigr)-a \bigl(U_{h}^{0}\bigr)\bigr), \nabla\theta^{1} \bigr)\leq C \bigl\Vert r^{0} \bigr\Vert _{0} \bigl\Vert \nabla\theta^{1} \bigr\Vert _{0} \leq Ch^{2}\tau+\frac{1}{8\tau} \bigl\Vert \theta^{1} \bigr\Vert _{0}^{2}, \\ &\bigl(f\bigl(U^{0}\bigr)-f\bigl(U_{h}^{0} \bigr),\theta^{1}\bigr)\leq C \bigl\Vert r^{0} \bigr\Vert _{0} \bigl\Vert \theta^{1} \bigr\Vert _{0} \leq Ch^{4}+C \bigl\Vert \theta^{1} \bigr\Vert _{0}^{2}. \end{aligned}$$
Denoting \(\overline{\gamma(X)}|_{K}=\frac{1}{|K|}\int_{K}\gamma(X)\, dX\) and then using the mean-value technique, we obtain
$$\begin{aligned}& \bigl(a\bigl(U^{0}\bigr)\tau\nabla r^{1},\nabla \theta^{1}\bigr) \\& \quad =\sum_{K}\bigl(\bigl(a\bigl(U^{0} \bigr)-\overline{a\bigl(U^{0}\bigr)}\bigr)\tau\nabla r^{1}, \nabla\theta ^{1}\bigr)_{K} \\& \qquad {}-\sum_{K}\overline{a\bigl(U^{0} \bigr)}|_{K}\bigl(\tau\bigl(\nabla e^{1}-\nabla I_{h}e^{1}\bigr),\nabla \theta^{1} \bigr)_{K}+\sum_{K}\overline{a \bigl(U^{0}\bigr)}|_{K}\bigl(\tau\bigl(\nabla u^{1}-\nabla I_{h}u^{1}\bigr),\nabla \theta^{1}\bigr)_{K} \\& \quad \leq Ch^{2}\tau \bigl\Vert U^{1} \bigr\Vert _{2} \bigl\Vert \nabla\theta^{1} \bigr\Vert _{0}+Ch\tau \bigl\Vert e^{1} \bigr\Vert _{2} \bigl\Vert \nabla \theta^{1} \bigr\Vert _{0}\leq Ch^{4}+Ch^{2}\tau^{2}+C\tau^{2} \bigl\Vert \nabla\theta^{1} \bigr\Vert _{0}^{2}, \\& \bigl(a\bigl(U_{h}^{0}\bigr)\nabla\eta^{0}, \nabla\theta^{1}\bigr) \\& \quad =\sum_{K}\bigl(\bigl(a\bigl(U_{h}^{0} \bigr)-\overline{a\bigl(U_{h}^{0}\bigr)}\bigr)\nabla \eta^{0}, \nabla\theta^{1}\bigr)_{K} +\sum _{K}\overline{a\bigl(U_{h}^{0} \bigr)}|_{K}\bigl(\nabla\eta^{0}, \nabla\theta ^{1} \bigr)_{K} \\& \quad \leq Ch^{2} \bigl\Vert u^{0} \bigr\Vert _{2} \bigl\Vert \nabla\theta^{1} \bigr\Vert _{0} \leq Ch\sqrt{\tau}\frac{1}{\sqrt{\tau}} \bigl\Vert \theta^{1} \bigr\Vert _{0}\leq Ch^{2}\tau+ \frac {1}{8\tau} \bigl\Vert \theta^{1} \bigr\Vert _{0}^{2}. \end{aligned}$$
By Theorem 1 we have
$$\begin{aligned}& \bigl(\bigl(a\bigl(U^{0}\bigr)-a\bigl(U_{h}^{0} \bigr)\bigr)\tau\nabla Q_{h}^{1},\nabla\theta^{1} \bigr) \\& \quad =-\bigl(\bigl(a\bigl(U^{0}\bigr)-a\bigl(U_{h}^{0} \bigr)\bigr)\tau\nabla \theta^{1},\nabla\theta^{1}\bigr) \\& \qquad {}-\bigl(\bigl(a\bigl(U^{0}\bigr)-a\bigl(U_{h}^{0} \bigr)\bigr)\tau\nabla r^{1},\nabla\theta^{1}\bigr) -\bigl( \bigl(a\bigl(U^{0}\bigr)-a\bigl(U_{h}^{0}\bigr) \bigr)\tau\nabla \delta^{1},\nabla\theta^{1}\bigr) \\& \qquad {}+\bigl(\bigl(a\bigl(U^{0}\bigr)-a\bigl(U_{h}^{0} \bigr)\bigr)\tau\nabla q^{1},\nabla\theta^{1}\bigr) \\& \quad \leq Ch^{2}\tau \bigl\Vert U^{0} \bigr\Vert _{2} \bigl\Vert \nabla \theta^{1} \bigr\Vert _{0,\infty} \bigl\Vert \nabla\theta^{1} \bigr\Vert _{0} \\& \qquad {}+Ch^{3}\tau \bigl\Vert U^{0} \bigr\Vert _{2} \bigl\Vert U^{1} \bigr\Vert _{2} \bigl\Vert \nabla \theta^{1} \bigr\Vert _{0,\infty}+Ch^{2} \tau \bigl\Vert \nabla \delta^{1} \bigr\Vert _{0} \bigl\Vert \nabla \theta^{1} \bigr\Vert _{0,\infty} \\& \qquad {}+Ch^{2} \tau \bigl\Vert \nabla q^{1} \bigr\Vert _{0,\infty} \bigl\Vert \nabla \theta^{1} \bigr\Vert _{0} \\& \quad \leq Ch^{4}+Ch^{2}\tau^{2}+Ch\tau \bigl\Vert \nabla\theta^{1} \bigr\Vert _{0}^{2} +C \tau^{2} \bigl\Vert \nabla \theta^{1} \bigr\Vert _{0}^{2}. \end{aligned}$$
Allocating all the estimates obtained, we have
$$\begin{aligned} &\frac{1}{\tau} \bigl\Vert \theta^{1} \bigr\Vert _{0}^{2}+\tau \bigl\Vert \nabla\theta^{1} \bigr\Vert _{0}^{2} \\ &\quad \leq Ch^{4}+Ch^{2}\tau+Ch\tau \bigl\Vert \nabla \theta^{1} \bigr\Vert _{0}^{2}+C \bigl\Vert \theta^{1} \bigr\Vert _{0}^{2}+C\tau ^{2} \bigl\Vert \nabla\theta^{1} \bigr\Vert _{0}^{2}. \end{aligned}$$
(32)
Thus there exist \(\tau_{1}', \tau_{2}', h_{1}', h_{2}', C_{1}'\) such that, for \(\tau\leq\tau_{1}'\) and \(h\leq h_{1}'\), we have
$$ \bigl\Vert \theta^{1} \bigr\Vert _{0}+\tau \bigl\Vert \nabla\theta^{1} \bigr\Vert _{0}\leq C_{1}'h(h+\sqrt{\tau}), $$
(33)
which implies
$$\begin{aligned} \bigl\Vert U_{h}^{1} \bigr\Vert _{0,\infty} &\leq Ch^{-1} \bigl\Vert \xi^{1} \bigr\Vert _{0}+ \bigl\Vert I_{h}U^{1} \bigr\Vert _{0,\infty} \\ &\leq CC_{1}'h+CC_{1}' \sqrt{\tau}+ \bigl\Vert I_{h}U^{1} \bigr\Vert _{0,\infty}\leq K_{0}', \end{aligned}$$
(34)
where \(h\leq h_{2}'\leq1/2CC_{1}'\) and \(\tau\leq\tau_{2}'\leq1/2CC_{1}'\).
By mathematical induction we assume that (29) holds for \(m\leq n-1\). Then there exist \(\tau_{3}'\) and \(h_{3}'\) such that
$$\begin{aligned}& \bigl\Vert U_{h}^{m} \bigr\Vert _{0,\infty}+\sqrt{ \tau}\Biggl(\sum_{i=2}^{m} \bigl\Vert \bar{ \partial }_{t}U_{h}^{i} \bigr\Vert _{0,\infty}^{2}\Biggr)^{\frac{1}{2}} \\& \quad \leq Ch^{-1}\Biggl( \bigl\Vert \xi^{m} \bigr\Vert _{0}+\sqrt{\tau}\Biggl(\sum_{i=2}^{m} \bigl\Vert \bar{\partial }_{t}\xi^{i} \bigr\Vert _{0}^{2}\Biggr)^{\frac{1}{2}}\Biggr) \\& \qquad {}+ \Biggl( \bigl\Vert I_{h}U^{m} \bigr\Vert _{0,\infty}+\sqrt{\tau}\Biggl( \sum_{i=2}^{m} \bigl\Vert \bar{\partial }_{t}I_{h}U^{i} \bigr\Vert _{0,\infty}^{2}\Biggr)^{\frac{1}{2}}\Biggr) \\& \quad \leq2CC_{0}'h+2CC_{0}' \sqrt{\tau}+ \Biggl( \bigl\Vert I_{h}U^{m} \bigr\Vert _{0,\infty}+\sqrt{\tau}\Biggl(\sum_{i=2}^{m} \bigl\Vert \bar{\partial }_{t}I_{h}U^{i} \bigr\Vert _{0,\infty}^{2}\Biggr)^{\frac{1}{2}}\Biggr)\leq K_{0}', \end{aligned}$$
(35)
where \(h\leq h_{3}'\leq1/4CC_{0}'\) and \(\tau\leq\tau_{3}'\leq 1/6(CC_{0}')^{2}\).
Then we prove that (29) also holds for \(m=n\). By (4) and (5) we derive the error equations
$$ \textstyle\begin{cases} (\bar{\partial}_{t}\xi^{n},v_{h})=-(\bar{\partial}_{t}\eta^{n},v_{h})+(\theta ^{n},v_{h})+(r^{n},v_{h}), \\ (\bar{\partial}_{t}\theta^{n},v_{h})+(a(U_{h}^{n-1})\tau\sum_{i=1}^{n}\nabla\theta^{i},\nabla v_{h}) \\ \quad = -(\bar{\partial}_{t}r^{n},v_{h})-(a(U^{n-1})\tau\sum_{i=1}^{n}\nabla r^{i},\nabla v_{h}) \\ \qquad {}-((a(U^{n-1})-a(U_{h}^{n-1}))\tau\sum_{i=1}^{n}\nabla I_{h}Q^{i},\nabla v_{h})-(a(U_{h}^{n-1})\nabla\eta^{0}, \nabla v) \\ \qquad {}-(\nabla U^{0} (a(U^{n-1})-a(U_{h}^{n-1})), \nabla v)+(f(U^{n-1})-f(U_{h}^{n-1}),v_{h}). \end{cases} $$
(36)
For \(v_{h}=\theta^{n}\) in the second equation of (36), we have
$$\begin{aligned}& \tau\Biggl(a\bigl(U_{h}^{n-1}\bigr)\sum _{i=1}^{n}\nabla\theta^{i},\nabla \theta^{n}\Biggr) \\& \quad =\tau \int_{\varOmega}a\bigl(U_{h}^{n-1}\bigr)\sum _{i=1}^{n-1}\nabla\theta^{i}\cdot \nabla\theta^{n}+\tau \bigl\Vert a^{\frac{1}{2}}\bigl(U_{h}^{n-1} \bigr)\nabla\theta^{n} \bigr\Vert _{0}^{2} \\& \quad =\frac{1}{2}\tau \int_{\varOmega}a\bigl(U_{h}^{n-1}\bigr) \Biggl( \sum_{i=1}^{n}\nabla\theta ^{i} \Biggr)^{2}-\frac{1}{2}\tau \int_{\varOmega}a\bigl(U_{h}^{n-1}\bigr) \Biggl( \sum_{i=1}^{n-1}\nabla \theta^{i} \Biggr)^{2} \\& \qquad {}+\frac{1}{2}\tau \bigl\Vert a^{\frac{1}{2}} \bigl(U_{h}^{n-1}\bigr)\nabla\theta^{n} \bigr\Vert _{0}^{2}, \end{aligned}$$
and hence we find
$$\begin{aligned}& \frac{1}{2\tau}\bigl( \bigl\Vert {\theta}^{n} \bigr\Vert _{0}^{2}- \bigl\Vert {\theta}^{n-1} \bigr\Vert _{0}^{2}\bigr) +\frac{1}{2}\tau \bigl\Vert a^{\frac{1}{2}}\bigl(U_{h}^{n-1}\bigr)\nabla \theta^{n} \bigr\Vert _{0}^{2} \\& \qquad {} +\frac{1}{2}\tau \Biggl\Vert a^{\frac{1}{2}} \bigl(U_{h}^{n-1}\bigr)\sum_{i=1}^{n} \nabla\theta^{i} \Biggr\Vert _{0}^{2} - \frac{1}{2}\tau \Biggl\Vert a^{\frac{1}{2}}\bigl(U_{h}^{n-2} \bigr)\sum_{i=1}^{n-1}\nabla \theta^{i} \Biggr\Vert _{0}^{2} \\& \quad \leq C\tau^{2} \bigl\Vert \bar{\partial}_{t}U_{h}^{n-1} \bigr\Vert _{0,\infty} \Biggl\Vert \sum_{i=1}^{n-1} \nabla\theta^{i} \Biggr\Vert _{0}^{2}-\bigl(\bar{ \partial}_{t}r^{n},\theta ^{n}\bigr)-\Biggl(a \bigl(U^{n-1}\bigr)\tau\sum_{i=1}^{n} \nabla r^{i},\nabla\theta^{n}\Biggr) \\& \qquad {}-\Biggl(\bigl(a\bigl(U^{n-1}\bigr)-a\bigl(U_{h}^{n-1} \bigr)\bigr)\tau\sum_{i=1}^{n}\nabla I_{h}Q^{i},\nabla \theta^{n}\Biggr)-\bigl(a \bigl(U_{h}^{n-1}\bigr)\nabla\eta^{0}, \nabla \theta^{n}\bigr) \\& \qquad {}-\bigl(\nabla U^{0} \bigl(a\bigl(U^{n-1}\bigr)-a \bigl(U_{h}^{n-1}\bigr)\bigr), \nabla\theta ^{n} \bigr)-\bigl(f\bigl(U^{n-1}\bigr)-f\bigl(U_{h}^{n-1} \bigr),\theta^{n}\bigr)\triangleq\sum_{i=1}^{7}B_{i}. \end{aligned}$$
(37)
Obviously,
$$\begin{aligned}& B_{1}\leq C\tau^{2} \Biggl\Vert \sum _{i=1}^{n-1}\nabla\theta^{i} \Biggr\Vert _{0}^{2}, \\& B_{2}\leq Ch^{2} \bigl\Vert \bar{\partial}_{t}U_{h}^{n} \bigr\Vert _{2} \bigl\Vert \theta^{n} \bigr\Vert _{0}\leq Ch^{4}+C \bigl\Vert \theta^{n} \bigr\Vert _{0}^{2}, \\& B_{7}\leq Ch^{4}+C \bigl\Vert \xi^{n-1} \bigr\Vert _{0}^{2}+C \bigl\Vert \theta^{n} \bigr\Vert _{0}^{2}. \end{aligned}$$
(38)
Similarly to the proof of \(A_{2}\sim A_{8}\) and \(A_{10}\), we rewrite \(\theta^{n}\) by \(\tau\sum_{i=1}^{n}\bar{\partial}_{t} \theta^{i}\) and then try to transfer τ from one side to the other in the inner product. For simplicity and concreteness, with the help of (2), we show that
$$\begin{aligned} B_{5}&=\Biggl(a_{u}\bigl(\mu_{4}^{n-1} \bigr)\bar{\partial}_{t}U_{h}^{n-1}\nabla \eta^{0}, \tau\sum_{i=1}^{n-1}\nabla \theta^{i}\Biggr) \\ &\quad {} -\frac{1}{\tau} \int_{\varOmega}\nabla\eta^{0} \Biggl(a\bigl(U_{h}^{n-1} \bigr)\tau\sum_{i=1}^{n}\nabla \theta^{i}-a\bigl(U_{h}^{n-2}\bigr) \tau\sum _{i=1}^{n-1}\nabla\theta ^{i}\Biggr) \\ &=\sum_{K}\Biggl(\bigl(a_{u}\bigl( \mu_{4}^{n-1}\bigr)\bar{\partial}_{t} \xi^{n-1}-\overline{a_{u}\bigl(\mu _{4}^{n-1} \bigr)\bar{\partial}_{t}\xi^{n-1}}\bigr)\nabla \eta^{0}, \tau\sum_{i=1}^{n-1}\nabla \theta^{i}\Biggr)_{K} \\ &\quad {} +\sum_{K}\overline{a_{u} \bigl(\mu_{4}^{n-1}\bigr)\bar{\partial}_{t}\xi ^{n-1}}|_{K}\Biggl(\nabla\eta^{0}, \tau\sum _{i=1}^{n-1}\nabla\theta^{i}\Biggr) -\Biggl(a_{u}\bigl(\mu_{4}^{n-1}\bigr)\bar{ \partial}_{t}\eta^{n-1}\nabla\eta^{0}, \tau\sum _{i=1}^{n-1}\nabla\theta^{i}\Biggr) \\ &\quad {} +\sum_{K}\Biggl(\bigl(a_{u} \bigl(\mu_{4}^{n-1}\bigr)\bar{\partial}_{t}U^{n-1}- \overline{a_{u}\bigl(\mu _{4}^{n-1}\bigr)\bar{ \partial}_{t}U^{n-1}}\bigr)\nabla\eta^{0}, \tau\sum _{i=1}^{n-1}\nabla\theta^{i} \Biggr)_{K} \\ &\quad {} +\sum_{K}\overline{a_{u} \bigl(\mu_{4}^{n-1}\bigr)\bar{\partial}_{t}U^{n-1}}|_{K} \Biggl(\nabla \eta^{0}, \tau\sum_{i=1}^{n-1} \nabla\theta^{i}\Biggr) \\ &\quad {} -\frac{1}{\tau} \int_{\varOmega}\nabla\eta^{0} \Biggl(a\bigl(U_{h}^{n-1} \bigr)\tau\sum_{i=1}^{n}\nabla \theta^{i}-a\bigl(U_{h}^{n-2}\bigr) \tau\sum _{i=1}^{n-1}\nabla\theta ^{i}\Biggr) \\ &\leq Ch^{2} \bigl\Vert \bar{\partial}_{t} \xi^{n-1} \bigr\Vert _{1} \bigl\Vert u^{0} \bigr\Vert _{2,4} \Biggl\Vert \tau\sum_{i=1}^{n-1} \nabla\theta^{i} \Biggr\Vert _{0,4} +Ch^{2} \Biggl\Vert \tau\sum_{i=1}^{n-1}\nabla \theta^{i} \Biggr\Vert _{0} \\ &\quad {}-\frac{1}{\tau} \int_{\varOmega}\nabla\eta^{0} \Biggl(a\bigl(U_{h}^{n-1} \bigr)\tau\sum_{i=1}^{n}\nabla \theta^{i}-a\bigl(U_{h}^{n-2}\bigr) \tau\sum _{i=1}^{n-1}\nabla\theta ^{i}\Biggr) \\ &\leq Ch^{4}+C \bigl\Vert \bar{\partial}_{t} \xi^{n-1} \bigr\Vert _{0}^{2}+C \Biggl\Vert \tau \sum_{i=1}^{n-1}\nabla\theta^{i} \Biggr\Vert _{0}^{2} \\ &\quad {} -\frac{1}{\tau} \int_{\varOmega}\nabla\eta^{0} \Biggl(a\bigl(U_{h}^{n-1} \bigr)\tau\sum_{i=1}^{n}\nabla \theta^{i}-a\bigl(U_{h}^{n-2}\bigr) \tau\sum _{i=1}^{n-1}\nabla\theta^{i}\Biggr), \end{aligned}$$
where \(\mu_{4}^{n-1}=U_{h}^{n-2}+\tau\lambda_{4}^{n-1}\bar{\partial }_{t}U_{h}^{n-1}\) and \(0< \lambda_{4}^{n-1}< 1\). Again transferring τ from one part of the inner product to the other, we have
$$\begin{aligned} B_{3}&=-\Biggl(a\bigl(U^{n-1}\bigr)\tau\sum _{i=1}^{n}\nabla r^{i},\tau\sum _{i=1}^{n}\nabla \bar{\partial}_{t} \theta^{i}\Biggr) \\ &=\Biggl(a\bigl(U^{n-2}\bigr)\nabla r^{n},\tau\sum_{i=1}^{n-1}\nabla \theta^{i}\Biggr)+\Biggl(\frac{a(U^{n-1})-a(U^{n-2})}{\tau}\tau\sum_{i=1}^{n} \nabla r^{i},\tau \sum_{i=1}^{n-1} \nabla\theta^{i}\Biggr) \\ &\quad {}-\bar{\partial}_{t}\Biggl(a \bigl(U^{n-1}\bigr)\tau\sum_{i=1}^{n} \nabla r^{i},\tau\sum_{i=1}^{n} \nabla\theta^{i}\Biggr) \\ &\triangleq\sum_{i=1}^{3}B_{3i}. \end{aligned}$$
We split \(B_{31}\) and \(B_{32}\) and estimate them as follows:
$$\begin{aligned}& B_{31}= \sum_{K}\Biggl(\bigl(a \bigl(U^{n-2}\bigr)-\overline{a\bigl(U^{n-2}\bigr)}\bigr)\nabla r^{n},\tau\sum_{i=1}^{n-1}\nabla \theta^{i}\Biggr)_{K} \\& \hphantom{B_{31}={}}{} -\sum_{K}\overline{a \bigl(U^{n-2}\bigr)}|_{K}\Biggl(\nabla e^{i}-\nabla I_{h}e^{i},\tau\sum_{i=1}^{n-1} \nabla\theta^{i}\Biggr)_{K} \\& \hphantom{B_{31}={}}{} +\sum_{K}\overline{a \bigl(U^{n-2}\bigr)}|_{K}\Biggl(\nabla u^{i}-\nabla I_{h}u^{i},\tau\sum_{i=1}^{n-1} \nabla\theta^{i}\Biggr)_{K} \\& \hphantom{B_{31}} \leq Ch^{2} \Biggl\Vert \tau\sum _{i=1}^{n-1}\nabla\theta^{i} \Biggr\Vert _{0}+Ch\sqrt{\tau} \Biggl\Vert \tau \sum _{i=1}^{n-1}\nabla\theta^{i} \Biggr\Vert _{0} \\& \hphantom{B_{31}}\leq Ch^{4}+Ch^{2}\tau+C \Biggl\Vert \tau \sum_{i=1}^{n-1}\nabla\theta^{i} \Biggr\Vert _{0}^{2}, \\& B_{32}=\Biggl(\frac{a(U^{n-1})-a(U^{n-2})}{\tau}\tau\sum_{i=1}^{n} \bigl(\nabla e^{i}-\nabla I_{h}e^{i}\bigr),\tau\sum _{i=1}^{n-1}\nabla\theta^{i}\Biggr) \\& \hphantom{B_{32}={}}{} +\sum_{K}\Biggl(\biggl( \frac{a(U^{n-1})-a(U^{n-2})}{\tau}-\overline{\frac {a(U^{n-1})-a(U^{n-2})}{\tau}}\biggr) \\& \hphantom{B_{32}={}}{}\times\tau\sum _{i=1}^{n}\bigl(\nabla u^{i}-\nabla I_{h}u^{i}\bigr),\tau\sum_{i=1}^{n-1} \nabla\theta^{i}\Biggr)_{K} \\& \hphantom{B_{32}={}}{} +\sum_{K}\overline{ \frac{a(U^{n-1})-a(U^{n-2})}{\tau}}\bigg|_{K}\Biggl(\tau\sum_{i=1}^{n} \bigl(\nabla u^{i}-\nabla I_{h}u^{i}\bigr),\tau\sum _{i=1}^{n-1}\nabla\theta ^{i} \Biggr)_{K} \\& \hphantom{B_{32}} \leq Ch\sqrt{\tau} \Biggl\Vert \tau\sum _{i=1}^{n-1}\nabla\theta^{i} \Biggr\Vert _{0}+ Ch^{2} \Biggl\Vert \tau\sum _{i=1}^{n-1}\nabla\theta^{i} \Biggr\Vert _{0} \\& \hphantom{B_{32}}\leq Ch^{4}+Ch^{2}\tau+C \Biggl\Vert \tau \sum_{i=1}^{n-1}\nabla\theta^{i} \Biggr\Vert _{0}^{2}. \end{aligned}$$
Then we have
$$ B_{3}\leq Ch^{4}+Ch^{2}\tau+C \Biggl\Vert \tau \sum_{i=1}^{n-1}\nabla\theta^{i} \Biggr\Vert _{0}^{2} -\bar{\partial}_{t}\Biggl(a \bigl(U^{n-1}\bigr)\tau\sum_{i=1}^{n} \nabla r^{i},\tau\sum_{i=1}^{n} \nabla\theta^{i}\Biggr). $$
Note that
$$\begin{aligned} B_{6}&=\Biggl(\nabla U^{0} a_{u}\bigl( \mu_{5}^{n-2}\bigr) \bigl(\bar{\partial}_{t} \xi^{n-1}+\bar{\partial }_{t}\eta^{n-1}\bigr), \tau \sum_{i=1}^{n-1}\nabla\theta^{i} \Biggr) \\ &\quad {} +\Biggl(\nabla U^{0}\bigl(\xi^{n-1}+ \eta^{n-1}\bigr)\frac{ a_{u}(\mu_{5}^{n-1})- a_{u}(\mu _{5}^{n-2})}{\tau}, \tau\sum _{i=1}^{n-1}\nabla\theta^{i}\Biggr) \\ &\quad {} -\bar{\partial}_{t}\Biggl(\nabla U^{0} a_{u}\bigl(\mu_{5}^{n-1}\bigr) \bigl( \xi^{n-1}+\eta^{n-1}\bigr), \tau\sum _{i=1}^{n}\nabla\theta^{i}\Biggr) \\ & \leq Ch^{4}+C \bigl\Vert \xi^{n-1} \bigr\Vert _{0}^{2}+C \bigl\Vert \bar{\partial}_{t} \xi^{n-1} \bigr\Vert _{0}^{2}+C \Biggl\Vert \tau \sum_{i=1}^{n-1}\nabla\theta^{i} \Biggr\Vert _{0}^{2} \\ &\quad {} -\bar{\partial}_{t}\Biggl(\nabla U^{0} a_{u}\bigl(\mu_{5}^{n-1}\bigr) \bigl( \xi^{n-1}+\eta^{n-1}\bigr), \tau\sum _{i=1}^{n}\nabla\theta^{i}\Biggr), \end{aligned}$$
where
$$ \mu_{5}^{n-1}=U^{n-1} +\lambda_{5}^{n-1} \bigl(\xi^{n-1}+\eta^{n-1}\bigr),\quad 0< \lambda_{5}^{n-1}< 1, $$
and
$$ \biggl\vert \frac{a_{u}(\mu_{5}^{n-1})-a_{u}(\mu_{5}^{n-2})}{\tau} \biggr\vert \leq \bigl\vert \bar{ \partial}_{t}U^{n-1} \bigr\vert +\lambda_{5}^{n-1} \bigl( \bigl\vert \bar{\partial}_{t}\xi^{n-1} \bigr\vert + \bigl\vert \bar{\partial}_{t}\eta ^{n-1} \bigr\vert \bigr). $$
Rewriting \(B_{4}\) and splitting it into several parts, we obtain
$$\begin{aligned} B_{4} &=\Biggl(\bigl(a\bigl(U^{n-1}\bigr)-a \bigl(U_{h}^{n-1}\bigr)\bigr)\tau\sum _{i=1}^{n}\nabla r^{i},\tau\sum _{i=1}^{n}\nabla\bar{\partial}_{t} \theta^{i}\Biggr) \\ &\quad {} -\Biggl(\bigl(a\bigl(U^{n-1}\bigr)-a\bigl(U_{h}^{n-1} \bigr)\bigr)\tau\sum_{i=1}^{n}\nabla Q^{i},\nabla\theta ^{n}\Biggr) \\ &= -\Biggl(\bigl(a\bigl(U^{n-2}\bigr)-a\bigl(U_{h}^{n-2} \bigr)\bigr)\nabla r^{n},\tau\sum_{i=1}^{n-1} \nabla\theta ^{i}\Biggr) \\ &\quad {}-\Biggl(\tau\sum_{i=1}^{n}\nabla r^{i}\frac {(a(U^{n-1})-a(U_{h}^{n-1}))-(a(U^{n-2})-a(U_{h}^{n-2}))}{\tau},\tau\sum_{i=1}^{n-1} \nabla\theta^{i}\Biggr) \\ &\quad {}+\bar{\partial}_{t}\Biggl(\bigl(a\bigl(U^{n-1} \bigr)-a\bigl(U_{h}^{n-1}\bigr)\bigr)\tau\sum _{i=1}^{n}\nabla r^{i},\tau\sum _{i=1}^{n}\nabla\theta^{i}\Biggr) \\ &\quad {} -\Biggl(\bigl(a\bigl(U^{n-1}\bigr)-a\bigl(U_{h}^{n-1} \bigr)\bigr)\tau\sum_{i=1}^{n}\nabla Q^{i},\nabla\theta ^{n}\Biggr). \end{aligned}$$
It is obvious that
$$\begin{aligned} &-\Biggl(\bigl(a\bigl(U^{n-2}\bigr)-a\bigl(U_{h}^{n-2} \bigr)\bigr)\nabla r^{n},\tau\sum_{i=1}^{n-1} \nabla \theta^{i}\Biggr) \\ &\quad \leq Ch \bigl\Vert U^{n} \bigr\Vert _{2} \bigl(Ch^{2} \bigl\Vert U^{n-2} \bigr\Vert _{2}+ \bigl\Vert \xi^{n-2} \bigr\Vert _{0}\bigr) \Biggl\Vert \tau\sum_{i=1}^{n-1}\nabla\theta^{i} \Biggr\Vert _{0,\infty} \\ &\quad \leq Ch^{4}+C \bigl\Vert \xi^{n-2} \bigr\Vert _{0}^{2}+C \Biggl\Vert \tau\sum _{i=1}^{n-1}\nabla\theta^{i} \Biggr\Vert _{0}^{2}. \end{aligned}$$
Because
$$\begin{aligned} &\frac{(a(U^{n-1})-a(U_{h}^{n-1}))-(a(U^{n-2})-a(U_{h}^{n-2}))}{\tau} \\ &\quad =a_{u}\bigl(\mu_{6}^{n-1}\bigr)\bar{ \partial}_{t}U^{n-1}-a_{u}\bigl( \mu_{7}^{n-1}\bigr)\bar{\partial }_{t}U_{h}^{n-1} \\ &\quad =a_{u}\bigl(\mu_{7}^{n-1}\bigr) \bigl( \bar{\partial}_{t}\xi^{n-1}+\bar{\partial}_{t} \eta^{n-1}\bigr) \\ &\qquad {}+\bar{\partial}_{t}U^{n-1} \bigl(a_{u}\bigl(\mu_{6}^{n-1}\bigr)-a_{u} \bigl(\mu_{7}^{n-1}\bigr)\bigr), \end{aligned}$$
(39)
where
$$\begin{aligned}& \mu_{6}^{n-1}=U^{n-2} +\tau\lambda_{6}^{n-1} \bar{\partial}_{t}U^{n-1},\qquad \mu_{7}^{n-1}= U_{h}^{n-2}+\tau \lambda_{7}^{n-1}\bar{ \partial}_{t}U_{h}^{n-1}, \\& \quad 0< \mu_{6}^{n-1},\mu_{7}^{n-1}< 1, \end{aligned}$$
and
$$\begin{aligned} \mu_{6}^{n-1}-\mu_{7}^{n-1}&= \xi^{n-2}+\eta^{n-2} +\tau\lambda_{7}^{n-1} \bigl(\bar{\partial}_{t}\xi^{n-1}+\bar{\partial}_{t} \eta ^{n-1}\bigr) \\ &\quad {} +\tau\bar{\partial}_{t}U^{n-1}\bigl( \lambda_{6}^{n-1}-\lambda_{7}^{n-1}\bigr), \end{aligned}$$
it follows that
$$\begin{aligned} &\Biggl(\tau\sum_{i=1}^{n}\nabla r^{i}\frac {(a(U^{n-1})-a(U_{h}^{n-1}))-(a(U^{n-2})-a(U_{h}^{n-2}))}{\tau},\tau\sum_{i=1}^{n-1} \nabla\theta^{i}\Biggr) \\ &\quad \leq Ch^{2} \Biggl\Vert \tau\sum _{i=1}^{n}Q^{i} \Biggr\Vert _{2} \biggl\Vert \frac {(a(U^{n-1})-a(U_{h}^{n-1}))-(a(U^{n-2})-a(U_{h}^{n-2}))}{\tau} \biggr\Vert _{0} \Biggl\Vert \tau\sum_{i=1}^{n-1}\nabla \theta^{i} \Biggr\Vert _{0,\infty} \\ &\quad \leq \bigl( \bigl\Vert \bar{\partial}_{t}\xi^{n-1} \bigr\Vert _{0}+ \bigl\Vert \bar{\partial}_{t} \eta^{n-1} \bigr\Vert _{0} + \bigl\Vert \xi^{n-2} \bigr\Vert _{0}+ \bigl\Vert \eta^{n-2} \bigr\Vert _{0} +\tau\bigr)Ch \Biggl\Vert \tau\sum_{i=1}^{n-1} \nabla\theta^{i} \Biggr\Vert _{0} \\ &\quad \leq Ch^{4}+Ch^{2}\tau+ \bigl\Vert \bar{ \partial}_{t}\xi^{n-1} \bigr\Vert _{0}^{2} + \bigl\Vert \xi^{n-2} \bigr\Vert _{0}^{2}+C \Biggl\Vert \tau\sum_{i=1}^{n-1}\nabla \theta^{i} \Biggr\Vert _{0}^{2}. \end{aligned}$$
The fourth part of \(B_{4}\) can be found:
$$\begin{aligned} &\Biggl(\bigl(a\bigl(U^{n-1}\bigr)-a\bigl(U_{h}^{n-1} \bigr)\bigr)\tau\sum_{i=1}^{n}\nabla Q^{i},\nabla\theta^{n}\Biggr) \\ &\quad =\Biggl(a_{u}\bigl(\mu_{8}^{n-1}\bigr) \eta^{n-1}\tau\sum_{i=1}^{n}\nabla Q^{i},\nabla\theta ^{n}\Biggr)+\Biggl(a_{u}\bigl( \mu_{8}^{n-1}\bigr)\xi^{n-1}\tau\sum _{i=1}^{n}\nabla Q^{i},\nabla \theta^{n}\Biggr), \end{aligned}$$
(40)
where \(\mu_{8}^{n-1}=U^{n-1}+\lambda_{8}^{n-1}(U_{h}^{n-1}-U^{n-1})\) and \(0<\lambda_{8}^{n-1}<1\).
On one hand,
$$\begin{aligned}& \Biggl(a_{u}\bigl(\mu_{8}^{n-1}\bigr) \eta^{n-1}\tau\sum_{i=1}^{n}\nabla Q^{i},\nabla\theta^{n}\Biggr) \\& \quad =\Biggl(a_{u}\bigl(\mu_{8}^{n-1}\bigr) \eta^{n-1}\tau\sum_{i=1}^{n}\nabla Q^{i},\tau\sum_{i=1}^{n}\bar{ \partial}_{t}\nabla\theta^{i}\Biggr) \\& \quad = \Biggl(a_{u}\bigl(\mu_{8}^{n-2}\bigr) \eta^{n-2}\nabla\delta^{n},\tau\sum _{i=1}^{n-1}\nabla\theta^{i}\Biggr) + \Biggl(a_{u}\bigl(\mu_{8}^{n-2}\bigr)\tau\sum _{i=1}^{n}\nabla\delta^{i}\bar{ \partial}_{t}\eta ^{n-1},\tau\sum _{i=1}^{n-1}\nabla\theta^{i}\Biggr) \\& \qquad {}+\Biggl(\eta^{n-1}\tau\sum_{i=1}^{n} \nabla\delta^{i}\frac{a_{u}(\mu _{8}^{n-1})-a_{u}(\mu_{6}^{n-2})}{\tau},\tau\sum _{i=1}^{n-1}\nabla\theta^{i}\Biggr) \\& \qquad {}-\Biggl(a_{u}\bigl(\mu_{8}^{n-2}\bigr) \eta^{n-2}\nabla q^{n},\tau\sum_{i=1}^{n-1} \nabla\theta ^{i}\Biggr)-\Biggl(a_{u}\bigl( \mu_{8}^{n-2}\bigr)\tau\sum_{i=1}^{n} \nabla q^{i}\bar{\partial}_{t}\eta ^{n-1},\tau\sum _{i=1}^{n-1}\nabla\theta^{i}\Biggr) \\& \qquad {}-\Biggl(\eta^{n-1}\tau\sum_{i=1}^{n} \nabla q^{i}\frac{a_{u}(\mu_{8}^{n-1})-a_{u}(\mu _{6}^{n-2})}{\tau},\tau\sum_{i=1}^{n-1} \nabla\theta^{i}\Biggr) \\& \qquad {}+\bar{\partial}_{t}\Biggl(a_{u}\bigl( \mu_{8}^{n}\bigr)\eta^{n}\tau\sum _{i=1}^{n}\nabla Q^{i},\tau\sum _{i=1}^{n}\nabla\theta^{i}\Biggr) \\& \quad \leq Ch^{4}+Ch^{2}\tau+C \Biggl\Vert \tau\sum _{i=1}^{n-1}\nabla\theta^{i} \Biggr\Vert _{0}^{2}+\bar {\partial}_{t} \Biggl(a_{u}\bigl(\mu_{8}^{n}\bigr) \eta^{n}\tau\sum_{i=1}^{n}\nabla Q^{i},\tau \sum_{i=1}^{n}\nabla \theta^{i}\Biggr). \end{aligned}$$
On the other hand,
$$\begin{aligned}& -\Biggl(a_{u}\bigl(\mu_{8}^{n-1}\bigr) \xi^{n-1}\tau\sum_{i=1}^{n}\nabla Q^{i},\tau\sum_{i=1}^{n-1}\bar{ \partial}_{t}\nabla\theta^{i}\Biggr) \\& \quad =-\Biggl(a_{u}\bigl(\mu_{8}^{n-2}\bigr) \xi^{n-2}\nabla Q^{n},\tau\sum_{i=1}^{n-1} \nabla\theta^{i}\Biggr) -\Biggl(a_{u}\bigl( \mu_{8}^{n-2}\bigr)\tau\sum_{i=1}^{n} \nabla Q^{i}\bar{\partial}_{t}\xi ^{n-1},\tau\sum _{i=1}^{n-1}\nabla\theta^{i}\Biggr) \\& \qquad {} -\Biggl(\tau\sum_{i=1}^{n}\nabla Q^{i}\xi^{n-1}\frac{a_{u}(\mu_{8}^{n-1})-a_{u}(\mu _{8}^{n-2})}{\tau},\tau\sum _{i=1}^{n-1}\nabla\theta^{i}\Biggr) \\& \qquad {} +\bar{\partial}_{t}\Biggl(a_{u}\bigl( \mu_{8}^{n-1}\bigr)\xi^{n-1}\tau\sum _{i=1}^{n}\nabla Q^{i},\tau\sum _{i=1}^{n}\nabla\theta^{i}\Biggr) \triangleq\sum_{i=1}^{4}D_{i}. \end{aligned}$$
(41)
Now we make use of the mean-value technique:
$$\begin{aligned} D_{1} &=\sum_{K}\Biggl( \bigl(a_{u}\bigl(\mu_{8}^{n-2}\bigr)- \overline{a_{u}\bigl(\mu_{8}^{n-2}\bigr)}\bigr)\xi ^{n-2}\nabla Q^{n}, \tau\sum_{i=1}^{n-1} \nabla\theta^{i}\Biggr)_{K} \\ &\quad {} +\sum_{K}\overline{a_{u} \bigl(\mu_{8}^{n-2}\bigr)}|_{K}\Biggl( \xi^{n-2}\nabla Q^{n}, \tau\sum_{i=1}^{n-1} \nabla\theta^{i}\Biggr)_{K}. \end{aligned}$$
Since \(Q^{n}\in H^{2}\), we cannot use the mean-value of \(\nabla Q^{n}\) directly as before, thus with the first equation of (36), we try to tackle it as follows:
$$\begin{aligned}& \sum_{K}\overline{a_{u}\bigl( \mu_{8}^{n-2}\bigr)}|_{K}\Biggl(\xi^{n-2} \nabla Q^{n}, \tau\sum_{i=1}^{n-1} \nabla\theta^{i}\Biggr)_{K} \\& \quad =\sum_{K}\overline{a_{u}\bigl( \mu_{8}^{n-2}\bigr)}|_{K}\Biggl(\nabla Q^{n}\tau\sum_{i=1}^{n-1}\nabla \theta^{i}-\overline{\nabla Q^{n}\tau\sum _{i=1}^{n-1}\nabla\theta^{i}}, \xi^{n-2}\Biggr)_{K} \\& \qquad {}-\sum_{K}\Biggl(\overline{a_{u} \bigl(\mu_{8}^{n-2}\bigr)}\overline{\nabla Q^{n}\tau \sum_{i=1}^{n-1}\nabla\theta^{i}} \Biggr)\Bigg|_{K}\tau\sum_{i=1}^{n-2} \bigl(1, \bar{\partial}_{t}\eta^{i}\bigr)_{K} \\& \qquad {}-\sum_{K}\Biggl(\overline{a_{u} \bigl(\mu_{8}^{n-2}\bigr)}\overline{\nabla Q^{n}\tau \sum_{i=1}^{n-1}\nabla\theta^{i}} \Biggr)\Bigg|_{K}\tau\sum_{i=1}^{n-2} \bigl(1, \theta^{i}\bigr)_{K} \\& \qquad {}-\sum_{K}\Biggl(\overline{a_{u} \bigl(\mu_{8}^{n-2}\bigr)}\overline{\nabla Q^{n}\tau \sum_{i=1}^{n-1}\nabla\theta^{i}} \Biggr)\Bigg|_{K}\tau\sum_{i=1}^{n-2} \bigl(1,r^{i}\bigr)_{K}. \end{aligned}$$
Because \(\Delta\theta^{i}|_{K}=0\), with the help of Theorem 1, we have
$$\begin{aligned} &\sum_{K}\overline{a_{u}\bigl( \mu_{8}^{n-2}\bigr)}|_{K}\Biggl(\nabla Q^{n}\tau\sum_{i=1}^{n-1}\nabla \theta^{i}-\overline{\nabla Q^{n}\tau\sum _{i=1}^{n-1}\nabla\theta^{i}}, \xi^{n-2}\Biggr)_{K} \\ &\quad \leq Ch \Biggl\Vert \nabla Q^{n}\tau\sum _{i=1}^{n-1}\nabla\theta^{i} \Biggr\Vert _{1} \bigl\Vert \xi^{n-2} \bigr\Vert _{0} \leq Ch \bigl\Vert Q^{n} \bigr\Vert _{2} \Biggl\Vert \tau \sum_{i=1}^{n-1}\nabla\theta^{i} \Biggr\Vert _{0,\infty} \bigl\Vert \xi ^{n-2} \bigr\Vert _{0} \\ &\quad \leq C \Biggl\Vert \tau\sum_{i=1}^{n-1} \nabla\theta^{i} \Biggr\Vert _{0}^{2}+C \bigl\Vert \xi^{n-2} \bigr\Vert _{0}^{2}. \end{aligned}$$
(42)
Further,
$$\begin{aligned} &\sum_{K}\Biggl(\overline{a_{u}\bigl( \mu_{8}^{n-2}\bigr)}\overline{\nabla Q^{n}\tau\sum _{i=1}^{n-1}\nabla\theta^{i}} \Biggr)\Bigg|_{K}\tau\sum_{i=1}^{n-2} \bigl(1, \bar{\partial}_{t}\eta^{i}\bigr)_{K} \\ &\quad =\sum_{K}\overline{a_{u}\bigl( \mu_{8}^{n-2}\bigr)}\frac{1}{|K|} \int_{K}\nabla\delta ^{n}\tau\sum _{i=1}^{n-1}\nabla\theta^{i} \,dx\,dy \tau \sum_{i=1}^{n-2} \int_{K} \bar{\partial}_{t}\eta^{i}\,dx \,dy \\ &\qquad {} +\sum_{K}\overline{a_{u} \bigl(\mu_{8}^{n-2}\bigr)}\frac{1}{|K|} \int_{K}\nabla q^{n}\tau \sum _{i=1}^{n-1}\nabla\theta^{i} \,dx\,dy \tau \sum_{i=1}^{n-2} \int_{K} \bar{\partial}_{t}\eta^{i}\,dx \,dy \\ &\quad \leq C\sum_{K}\frac{1}{|K|} \bigl\Vert \nabla\delta^{n} \bigr\Vert _{0,4} \Biggl\Vert \tau \sum_{i=1}^{n-1}\nabla\theta^{i} \Biggr\Vert _{0,4}|K|^{\frac{1}{2}} \tau\sum _{i=1}^{n-2}\biggl( \int_{K} \bigl\vert \bar{\partial}_{t} \eta^{i} \bigr\vert ^{2}\,dx\,dy\biggr)^{\frac{1}{2}}|K|^{\frac{1}{2}} \\ &\qquad {} + C\sum_{K}\frac{1}{|K|} \Biggl\Vert \tau\sum_{i=1}^{n-1}\nabla \theta^{i} \Biggr\Vert _{0}|K|^{\frac{1}{2}} \tau\sum _{i=1}^{n-2}\biggl( \int_{K} \bigl\vert \bar{\partial}_{t} \eta^{i} \bigr\vert ^{2}\,dx\,dy\biggr)^{\frac{1}{2}}|K|^{\frac{1}{2}} \\ &\quad \leq C\sum_{K} \bigl\Vert \delta^{n} \bigr\Vert _{2} \Biggl\Vert \tau\sum _{i=1}^{n-1}\nabla\theta^{i} \Biggr\Vert _{0,4} \tau\sum_{i=1}^{n-2} \bigl\Vert \bar{\partial}_{t}\eta^{i} \bigr\Vert _{0} + C\sum_{K} \Biggl\Vert \tau\sum _{i=1}^{n-1}\nabla\theta^{i} \Biggr\Vert _{0} \tau\sum_{i=1}^{n-2} \bigl\Vert \bar{\partial}_{t}\eta^{i} \bigr\Vert _{0} \\ &\quad \leq Ch^{\frac{3}{2}} \bigl\Vert \delta^{n} \bigr\Vert _{2} \Biggl\Vert \tau\sum_{i=1}^{n-1} \nabla\theta ^{i} \Biggr\Vert _{0} + Ch^{2} \Biggl\Vert \tau\sum_{i=1}^{n-1}\nabla \theta^{i} \Biggr\Vert _{0} \\ &\quad \leq Ch^{4}+Ch^{3} \bigl\Vert \delta^{n} \bigr\Vert _{2}^{2}+C \Biggl\Vert \tau\sum_{i=1}^{n-1}\nabla \theta^{i} \Biggr\Vert _{0}^{2}. \end{aligned}$$
Similarly, we have
$$\begin{aligned} &\sum_{K}\Biggl(\overline{a_{u}\bigl( \mu_{8}^{n-2}\bigr)}\overline{\nabla Q^{n}\tau\sum _{i=1}^{n-1}\nabla\theta^{i}} \Biggr)\Bigg|_{K}\tau\sum_{i=1}^{n-2} \bigl(1,r^{i}\bigr)_{K} \leq Ch^{4}+Ch^{3} \bigl\Vert \delta^{n} \bigr\Vert _{2}^{2}+C \Biggl\Vert \tau\sum_{i=1}^{n-1}\nabla \theta^{i} \Biggr\Vert _{0}^{2}, \\ & \sum_{K}\Biggl(\overline{a_{u}\bigl( \mu_{8}^{n-2}\bigr)} \overline{\nabla Q^{n}\tau\sum _{i=1}^{n-1}\nabla\theta^{i}} \Biggr)\Bigg|_{K}\tau\sum_{i=1}^{n-2} \bigl(1, \theta^{i}\bigr)_{K} \\ &\quad \leq C\sum_{K}\frac{1}{|K|} \bigl\Vert \nabla Q^{n} \bigr\Vert _{0,4} \Biggl\Vert \tau\sum _{i=1}^{n-1}\nabla\theta^{i} \Biggr\Vert _{0}|K|^{\frac{1}{4}} \tau\sum_{i=1}^{n-2} \biggl( \int_{K}\bigl|\theta^{i}\bigr|^{4}\,dx\,dy \biggr)^{\frac{1}{4}}|K|^{\frac{3}{4}} \\ &\quad \leq C \Biggl\Vert \tau\sum_{i=1}^{n-1} \nabla\theta^{i} \Biggr\Vert _{0} \tau\sum _{i=1}^{n-2} \bigl\Vert \theta^{i} \bigr\Vert _{0,4} \leq C \Biggl\Vert \tau\sum _{i=1}^{n-1}\nabla\theta^{i} \Biggr\Vert _{0}^{2}+ C\tau\sum_{i=1}^{n-2} \bigl\Vert \nabla\theta^{i} \bigr\Vert _{0}^{2}. \end{aligned}$$
Thus we have
$$ D_{1} \leq C \Biggl\Vert \tau\sum_{i=1}^{n-1} \nabla\theta^{i} \Biggr\Vert _{0}^{2}+C \bigl\Vert \xi^{n-2} \bigr\Vert _{0}^{2}+Ch^{4}+Ch^{3} \bigl\Vert \delta^{n} \bigr\Vert _{2}^{2}+C \tau\sum_{i=1}^{n-2} \bigl\Vert \nabla\theta ^{i} \bigr\Vert _{0}^{2}. $$
By a similar method we have
$$\begin{aligned} D_{2}+D_{3}&\leq C \Biggl\Vert \tau\sum _{i=1}^{n-1}\nabla\theta^{i} \Biggr\Vert _{0}^{2}+C \bigl\Vert \bar {\partial}_{t} \xi^{n-1} \bigr\Vert _{0}^{2}+C \bigl\Vert \xi^{n-1} \bigr\Vert _{0}^{2} \\ &\quad {} +Ch^{4}+Ch^{3} \bigl\Vert \delta^{n} \bigr\Vert _{2}^{2}+C\tau\sum _{i=1}^{n-2} \bigl\Vert \nabla\theta^{i} \bigr\Vert _{0}^{2}. \end{aligned}$$
Altogether,
$$\begin{aligned} &\frac{1}{2\tau}\bigl( \bigl\Vert {\theta}^{n} \bigr\Vert _{0}^{2}- \bigl\Vert {\theta}^{n-1} \bigr\Vert _{0}^{2}\bigr) +\frac{1}{2}\tau \bigl\Vert a^{\frac{1}{2}}\bigl(U^{n-1}\bigr)\nabla\theta^{n} \bigr\Vert _{0}^{2} \\ &\qquad {}+\frac{1}{2}\tau \Biggl\Vert a^{\frac{1}{2}} \bigl(U^{n-1}\bigr)\sum_{i=1}^{n} \nabla\theta^{i} \Biggr\Vert _{0}^{2} - \frac{1}{2}\tau \Biggl\Vert a^{\frac{1}{2}}\bigl(U^{n-2} \bigr)\sum_{i=1}^{n-1}\nabla \theta^{i} \Biggr\Vert _{0}^{2} \\ &\quad \leq Ch^{4}+Ch^{2}\tau+C \bigl\Vert \xi^{n-1} \bigr\Vert _{0}^{2}+C \bigl\Vert \xi^{n-2} \bigr\Vert _{0}^{2}+C \bigl\Vert \theta^{n} \bigr\Vert _{0}^{2}+C \bigl\Vert \bar{\partial}_{t}\xi^{n-1} \bigr\Vert _{0}^{2} \\ &\qquad {} +C\tau^{2} \Biggl\Vert \sum_{i=1}^{n-1} \nabla\theta^{i} \Biggr\Vert _{0}^{2}+Ch\tau \bigl\Vert \nabla\theta ^{n} \bigr\Vert _{0}^{2}+C \tau\bigl(h^{\frac{1}{4}}+\tau^{\frac{1}{4}}\bigr) \bigl\Vert \nabla \theta^{n} \bigr\Vert _{0}^{2} \\ &\qquad {}-\frac{1}{\tau} \int_{\varOmega}\nabla\eta^{0} \Biggl(a\bigl(U_{h}^{n-1} \bigr)\tau\sum_{i=1}^{n}\nabla \theta^{i}-a\bigl(U_{h}^{n-2}\bigr) \tau\sum _{i=1}^{n-1}\nabla\theta^{i}\Biggr) \\ &\qquad {} -\bar{\partial}_{t}\Biggl(a\bigl(U^{n-1}\bigr) \tau\sum_{i=1}^{n}\nabla r^{i}, \tau\sum_{i=1}^{n}\nabla\theta^{i} \Biggr) \\ &\qquad {}+\bar{\partial }_{t}\Biggl(\bigl(a\bigl(U^{n-1}\bigr)-a \bigl(U_{h}^{n-1}\bigr)\bigr)\tau\sum _{i=1}^{n}\nabla r^{i},\tau\sum _{i=1}^{n}\nabla\theta^{i}\Biggr) \\ &\qquad {} +\frac{1}{2} \int_{\varOmega}\bigl(a\bigl(U^{n-1}\bigr)-a \bigl(U_{h}^{n-1}\bigr)\bigr)\tau\Biggl(\sum _{i=1}^{n}\nabla\theta^{i} \Biggr)^{2} \\ &\qquad {}-\frac{1}{2} \int_{\varOmega }\bigl(a\bigl(U^{n-2}\bigr)-a \bigl(U_{h}^{n-2}\bigr)\bigr)\tau\Biggl(\sum _{i=1}^{n-1}\nabla\theta ^{i} \Biggr)^{2} \\ &\qquad {}-\bar{\partial}_{t}\Biggl(a_{u}\bigl( \mu_{8}^{n-1}\bigr) \bigl(\xi^{n-1}+ \eta^{n-1}\bigr)\tau\sum_{i=1}^{n} \nabla Q^{i},\tau\sum_{i=1}^{n-1} \nabla\theta^{i}\Biggr) \\ &\qquad {} -\bar{\partial}_{t}\Biggl(\nabla U^{0} a_{u}\bigl(\mu_{8}^{n-1}\bigr) \bigl( \xi^{n-1}+\eta^{n-1}\bigr), \tau\sum _{i=1}^{n}\nabla\theta^{i}\Biggr) \end{aligned}$$
with \(\theta^{1}\) estimated earlier. Using the Gronwall lemma, we have
$$ \bigl\Vert {\theta}^{n} \bigr\Vert _{0}^{2}+\tau^{2} \Biggl\Vert \sum _{i=1}^{n}\nabla\theta^{i} \Biggr\Vert _{0}^{2}+\sum_{i=2}^{n} \tau^{2} \bigl\Vert \nabla\theta^{i} \bigr\Vert _{0}^{2}\leq Ch^{4}+Ch^{2}\tau. $$
(43)
Again using the first equation of (36) with \(v_{h}=\bar{\partial}_{t}\xi ^{n}\), for \(\xi^{n}\), we obtain
$$\begin{aligned} \bigl\Vert \bar{\partial}_{t}\xi^{n} \bigr\Vert _{0}^{2} &=-\bigl(\bar{\partial}_{t} \eta^{n},\bar {\partial}_{t}\xi^{n}\bigr) +\bigl( \theta^{n},\bar{\partial}_{t}\xi^{n}\bigr)+ \bigl(r^{n},\bar{\partial}_{t}\xi ^{n}\bigr) \\ &\leq Ch^{4} \bigl\Vert U^{n} \bigr\Vert _{2}^{2}+Ch^{4} \bigl\Vert \bar{ \partial}_{t}U^{n} \bigr\Vert _{2}^{2}+C \bigl\Vert \theta^{n} \bigr\Vert _{0}^{2}+ \frac{1}{2} \bigl\Vert \bar{\partial}_{t}\xi^{n} \bigr\Vert _{0}^{2}, \end{aligned}$$
(44)
which implies
$$ \tau\sum_{i=2}^{n} \bigl\Vert \bar{\partial}_{t}\xi^{i} \bigr\Vert _{0}^{2}\leq Ch^{4}+Ch^{4}\tau \sum _{i=2}^{n} \bigl\Vert \bar{ \partial}_{t}U^{i} \bigr\Vert _{2}^{2}+ \tau\sum_{i=2}^{n} \bigl\Vert \theta^{i} \bigr\Vert _{0}^{2} \leq Ch^{4}+Ch^{2}\tau, $$
(45)
or
$$ \bigl\Vert \xi^{n} \bigr\Vert _{0}^{2}= \tau^{2} \Biggl\Vert \sum_{i=2}^{n} \bar{\partial}_{t}\xi^{i} \Biggr\Vert _{0}^{2} \leq C\tau^{2}\Biggl(\sum_{i=2}^{n} \bigl\Vert \bar{\partial}_{t}\xi^{i} \bigr\Vert _{0}\Biggr)^{2} \leq Ch^{4}+Ch^{2}\tau. $$
(46)
Then there exist \(\tau_{4}'\), \(\tau_{5}'\), \(h_{4}'\), \(h_{5}'\), \(C_{2}'\) such that, for \(\tau\leq\tau_{4}'\) and \(h\leq h_{4}'\), we have
$$ \sqrt{\tau}\Biggl(\sum_{i=2}^{n} \bigl\Vert \bar{\partial}_{t}\xi^{i} \bigr\Vert _{0}^{2}\Biggr)^{\frac{1}{2}}+ \bigl\Vert \xi^{n} \bigr\Vert _{0}+ \bigl\Vert \theta^{n} \bigr\Vert _{0}+\tau\Biggl(\sum_{i=1}^{n} \bigl\Vert \nabla\theta^{i} \bigr\Vert _{0}^{2} \Biggr)^{\frac{1}{2}}\leq C_{2}'h\bigl(h+ \tau^{\frac{1}{2}}\bigr), $$
(47)
from which we deduce
$$\begin{aligned} & \bigl\Vert U_{h}^{n} \bigr\Vert _{0,\infty}+ \sqrt{\tau}\Biggl(\sum_{i=2}^{n} \bigl\Vert \bar{\partial }_{t}U_{h}^{n} \bigr\Vert _{0,\infty}^{2}\Biggr)^{\frac{1}{2}} \\ &\quad \leq Ch^{-1}\Biggl( \bigl\Vert \xi^{n} \bigr\Vert _{0}+\sqrt{\tau}\Biggl(\sum_{i=2}^{n} \bigl\Vert \bar{\partial }_{t}\xi^{i} \bigr\Vert _{0,\infty}^{2}\Biggr)^{\frac{1}{2}}\Biggr) \\ &\qquad {}+ \Biggl( \bigl\Vert I_{h}U^{n} \bigr\Vert _{0,\infty}+\sqrt{\tau}\Biggl( \sum_{i=2}^{n} \bigl\Vert \bar{\partial }_{t}I_{h}U^{i} \bigr\Vert _{0,\infty}^{2}\Biggr)^{\frac{1}{2}}\Biggr) \\ &\quad \leq2CC_{2}'h+2CC_{2}' \sqrt{\tau}+ \Biggl( \bigl\Vert I_{h}U^{n} \bigr\Vert _{0,\infty}+\sqrt{\tau}\Biggl(\sum_{i=2}^{n} \bigl\Vert \bar{\partial }_{t}I_{h}U^{i} \bigr\Vert _{0,\infty}^{2}\Biggr)^{\frac{1}{2}}\Biggr)\leq K_{0}', \end{aligned}$$
(48)
where \(h\leq h_{5}'\leq1/2CC_{2}'\) and \(\tau\leq\tau_{5}'\leq 1/4(CC_{2}')^{2}\). Clearly, \(C_{2}'\) has nothing to do with \(C_{0}'\), and thus (29) holds for \(m=n\) if we take \(C_{0}'\geq\sum_{i=1}^{2}C_{i}'\), \(\tau _{0}'\leq\min_{1\leq\tau\leq5}\tau_{i}'\), and \(h_{0}'\leq\min_{1\leq\tau\leq5}h_{i}'\). Then the induction is closed.

The desired estimate for \(u^{n}\) and \(q^{n}\) in (27) and (28) are thus consequences of (29) combined with the triangle inequality. The proof is completed. □

Remark 2

It is precious to point out that to avoid the restriction involved by the regularities of \(Q^{n}\), we try to use the new mean-value technique in the proof of \(D_{1}\sim D_{3}\).

Remark 3

It can be seen that (27) and (28) do not hold for the elements dissatisfying (42), such as the biquadratic finite element.

5 Numerical results

In this section, we consider the hyperbolic equation
$$ \textstyle\begin{cases} u_{tt}-\nabla\cdot(a(u)\nabla u)-f(u)=g(X,t), & (X, t)\in\varOmega\times (0, T], \\ u=0,& (X, t)\in\partial\varOmega\times(0, T], \\ u(X, 0)=u_{0}(X),\qquad u_{t}(X, 0)=u_{1}(X), & X\in\varOmega, \end{cases} $$
(49)
with \(\varOmega=[0, 1]\times[0, 1]\), \(a(u)=\sin u+0.1\), \(f(u)=u^{2}\), and \(g(X,t)\) chosen corresponding to the exact solution \(u=e^{t}xy(1-x)(1-y)\). Setting \(q=u_{t}\), (49) is changed into a parabolic system. A uniform rectangular partition with \(m+1\) nodes in each direction is used in our computation. We solve the system by the linearized Galerkin method with bilinear element.
To confirm our error analysis for (27) and (28), we choose \(\tau=5h^{2}\) for the backward Euler FEM with bilinear FE. Therefore, from our theoretical analysis, the \(L^{2}\)-norm errors for u and q are \(O(h^{2}+\tau)\sim O(h^{2})\), and the \(H^{1}\)-norm errors for u and q are \(O(h+\tau)\sim O(h)\). We present the numerical results with respect to time \(t=0.25, 0.5, 0.75, 1.0\) in Tables 14, respectively. It can be seen that \(\|u^{n}-U_{h}^{n}\|_{0}\) and \(\|q^{n}-Q_{h}^{n}\|_{0}\) are convergent at rate \(O(h^{2})\) and \(\| u^{n}-U_{h}^{n}\|_{1}\) and \(\|q^{n}-Q_{h}^{n}\|_{1}\) are convergent at rate \(O(h)\), which indicate the optimal convergence rates of the methods. Further, to show the unconditional convergence results, we test the FEM with \(h=1/160\) and the large time steps \(\tau=h, 5h, 10h, 20h, 40h\), respectively. We present the numerical results in Tables 56, which suggest that the scheme is stable for large time steps. All these results are in good agreement with our theoretical analysis.
Table 1

Results for \(U^{n}_{h}\) and \(Q^{n}_{h}\) when \(t=0.25\) (\(\tau=5h^{2}\))

m × m

\(\|u^{n}-U_{h}^{n}\|_{0}\)

Order

\(\|u^{n}-U_{h}^{n}\|_{1}\)

Order

\(\|q^{n}-Q_{h}^{n}\|_{0}\)

Order

\(\|q^{n}-Q_{h}^{n}\|_{1}\)

Order

5 × 5

2.9760 × 10−3

3.9028 × 10−2

2.9757 × 10−3

3.9028 × 10−2

10 × 10

7.3473 × 10−4

2.0181

1.9230 × 10−2

1.0211

7.3473 × 10−4

2.0179

1.9230 × 10−2

1.0211

20 × 20

1.8429 × 10−4

1.9952

9.5818 × 10−3

1.0050

1.8429 × 10−4

1.9953

9.5818 × 10−3

1.0050

40 × 40

4.6110 × 10−5

1.9988

4.7867 × 10−3

1.0013

4.6110 × 10−5

1.9988

4.7867 × 10−3

1.0013

Table 2

Results for \(U^{n}_{h}\) and \(Q^{n}_{h}\) when \(t=0.5\) (\(\tau=5h^{2}\))

m × m

\(\|u^{n}-U_{h}^{n}\|_{0}\)

Order

\(\|u^{n}-U_{h}^{n}\|_{1}\)

Order

\(\|q^{n}-Q_{h}^{n}\|_{0}\)

Order

\(\|q^{n}-Q_{h}^{n}\|_{1}\)

Order

5 × 5

3.8700 × 10−3

5.0155 × 10−2

3.8696 × 10−3

5.0155 × 10−2

10 × 10

9.3055 × 10−4

2.0562

2.4686 × 10−2

1.0227

9.3055 × 10−4

2.0560

2.4686 × 10−2

1.0227

20 × 20

2.3338 × 10−4

1.9954

1.2303 × 10−2

1.0048

2.3338 × 10−4

1.9954

1.2303 × 10−2

1.0048

40 × 40

5.8391 × 10−5

1.9989

6.1461 × 10−3

1.0012

5.8391 × 10−5

1.9989

6.1461 × 10−3

1.0012

Table 3

Results for \(U^{n}_{h}\) and \(Q^{n}_{h}\) when \(t=0.75\) (\(\tau=5h^{2}\))

m × m

\(\|u^{n}-U_{h}^{n}\|_{0}\)

Order

\(\|u^{n}-U_{h}^{n}\|_{1}\)

Order

\(\|q^{n}-Q_{h}^{n}\|_{0}\)

Order

\(\|q^{n}-Q_{h}^{n}\|_{1}\)

Order

5 × 5

5.0318 × 10−3

6.4456 × 10−2

5.0313 × 10−3

6.4456 × 10−2

10 × 10

1.1784 × 10−3

2.0943

3.1692 × 10−2

1.0242

1.1784 × 10−3

2.0941

3.1692 × 10−2

1.0242

20 × 20

2.9550 × 10−4

1.9956

1.5796 × 10−2

1.0045

2.9549 × 10−4

1.9956

1.5796 × 10−2

1.0045

40 × 40

7.3929 × 10−5

1.9989

7.8917 × 10−3

1.0012

7.3929 × 10−5

1.9989

7.8917 × 10−3

1.0011

Table 4

Results for \(U^{n}_{h}\) and \(Q^{n}_{h}\) when \(t=1.0\) (\(\tau=5h^{2}\))

m × m

\(\|u^{n}-U_{h}^{n}\|_{0}\)

Order

\(\|u^{n}-U_{h}^{n}\|_{1}\)

Order

\(\|q^{n}-Q_{h}^{n}\|_{0}\)

Order

\(\|q^{n}-Q_{h}^{n}\|_{1}\)

Order

5 × 5

5.8966 × 10−3

8.2298 × 10−2

5.8959 × 10−3

8.2297 × 10−2

10 × 10

1.4919 × 10−3

1.9827

4.0685 × 10−2

1.0164

1.4919 × 10−3

1.9826

4.0685 × 10−2

1.0164

20 × 20

3.7407 × 10−4

1.9958

2.0281 × 10−2

1.0043

3.7407 × 10−4

1.9958

2.0281 × 10−2

1.0043

40 × 40

9.3585 × 10−5

1.9989

1.0133 × 10−2

1.0011

9.3585 × 10−5

1.9989

1.0133 × 10−2

1.0011

Table 5

Results for \(\|u^{n}-U^{n}_{h}\|_{1}\) (\(h=\frac{1}{160}\), \(\tau=kh\))

t

k = 1

k = 5

k = 10

k = 20

k = 40

0.25

2.396051 × 10−3

2.509776 × 10−3

2.868899 × 10−3

4.142662 × 10−3

7.852834 × 10−3

0.50

3.093389 × 10−3

3.652433 × 10−3

5.078443 × 10−3

8.767703 × 10−3

1.632563 × 10−2

0.75

4.009582 × 10−3

5.496080 × 10−3

8.725847 × 10−3

1.598647 × 10−2

2.828099 × 10−2

1.00

5.216721 × 10−3

8.300234 × 10−3

1.426017 × 10−2

2.670072 × 10−2

4.519938 × 10−2

Table 6

Results for \(\|q^{n}-Q^{n}_{h}\|_{1}\) (\(h=\frac{1}{160}\), \(\tau=kh\))

t

k = 1

k = 5

k = 10

k = 20

k = 40

0.25

2.396041 × 10−3

2.502978 × 10−3

2.774917 × 10−3

3.325243 × 10−3

7.807021 × 10−3

0.50

3.093358 × 10−3

3.634416 × 10−3

4.873726 × 10−3

7.011551 × 10−3

6.395837 × 10−3

0.75

4.009512 × 10−3

5.462226 × 10−3

8.391812 × 10−3

1.334560 × 10−2

1.239224 × 10−2

1.00

5.216582 × 10−3

8.244615 × 10−3

1.376092 × 10−2

2.292862 × 10−2

2.288225 × 10−2

6 Conclusion

In this paper, we have established unconditional error estimates for a nonlinear hyperbolic equation. A striking feature of our analysis is that we transform the nonlinear hyperbolic equation into a parabolic system. Then a linearized backward Euler FEM is constructed for the nonlinear parabolic equation. It is shown in this paper that such an idea avoids the difficulty in constructing a linearized first-order scheme for a nonlinear hyperbolic equation, and we can also give the error analysis for u and \(q = u_{t}\) at the same time. Splitting skill is exploited to derive the final unconditional convergent results. Some special methods are utilized to derive the boundedness of the solutions about the time-discrete system in \(H^{2}\)-norm, which may play a crucial role for getting rid of the restriction on the ratio between h and τ. Since the new parabolic system caused lots of problems for our the spatial errors analysis, several new techniques, such as rewriting the error equations, are introduced. It should be noted that the results in this paper also hold for linear conforming triangular elements but not hold for some other particular elements, such as the biquadratic finite element.

Declarations

Acknowledgements

The authors would like to thank the referees for their valuable suggestions, which helped to improve this work.

Availability of data and materials

Not applicable.

Funding

This work was supported by the National Natural Science Foundation of China (No. 11671369), the Doctoral Starting Foundation of Pingdingshan University (No. PXY-BSQD-2019001), and the University Cultivation Foundation of Pingdingshan (No. PXY-PYJJ-2019006).

Authors’ contributions

JW carried out theoretical calculation, participated in the design of the study, and drafted the manuscript. LG participated in its design and helped to draft the manuscript. Both authors read and approved the final manuscript.

Competing interests

The authors declare that they have no competing interests.

Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

Authors’ Affiliations

(1)
School of Mathematics and Statistics, Pingdingshan University, Pingdingshan, P.R. China

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