Fully discrete convergence analysis of non-linear hyperbolic equations based on finite element analysis

The following mixed problems are considered:

hx,uutt−<!-- - -->∑<!-- ? -->i,j=1d∂<!-- ? -->∂<!-- ? -->xiaijx,u∂<!-- ? -->u∂<!-- ? -->xj−<!-- - -->∑<!-- ? -->i=1dbix.uuxi=fx,ux,t∈<!-- ? -->K×<!-- ? -->0,Tux,0=0,utx,0=0ux,t=0x,t∈<!-- ? -->∂<!-- ? -->K×<!-- ? -->0,T $$\begin{array}{} \displaystyle \left\{ {\begin{array}{*{20}{c}} {h\left( {x,u} \right){u_{tt}} - \sum\limits_{i,j = 1}^d {\frac{\partial }{{\partial {x_i}}}\left( {{a_{ij}}\left( {x,u} \right)\frac{{\partial u}}{{\partial {x_j}}}} \right) - \sum\limits_{i = 1}^d {{b_i}\left( {x.u} \right)u{x_i} = } f\left( {x,u} \right)\left( {x,t} \right) \in K \times \left[ {0,T} \right]} }\\ {u\left( {x,0} \right) = 0,{u_t}\left( {x,0} \right) = 0}\\ {u\left( {x,t} \right) = 0_{}^{}\left( {x,t} \right) \in \partial K \times \left[ {0,T} \right]} \end{array}} \right. \end{array}$$(1)

where utt=∂<!-- ? -->2u∂<!-- ? -->t2,uxi=∂<!-- ? -->u∂<!-- ? -->xi; $\begin{array}{} {u_{tt}} = \frac{{{\partial ^2}u}}{{\partial {t^2}}}, u{x_i} = \frac{{\partial u}}{{\partial {x_i}}}; \end{array}$ K is a fully smooth bounded open domain in R^d, and the boundary ∂ K is smooth [11].

For the semi-discrete or fully discrete finite element method of the non-linear hyperbolic equation with only x or h(x, u) ≡ 1 in h(x, u), there are some research results [12, 13]. If u is included in h(x, u), the error estimation will suffer or fail to reach the convergence order [14] when defining the non-linear or predictor–corrector scheme, and the error equation cannot be obtained by direct weighting method. In this paper, the finite element scheme of second-order nonsexual hyperbolic equation [15] is defined when h contains u. Question (1) is assumed as the following: for (x, p) ∈ K × R,

aij(⋅, ⋅) ∈ C2(K × R); |aij(x, p)| ≤ C1, [aij(x, p)] ′p, [aij(x, p)]p2 ″, it is bounded to P. aij (x, p) = aji(x, p), ∑<!-- ? -->i,h=1daijx,prirj≥<!-- = -->C0∑<!-- ? -->i=1dri2, $\begin{array}{} \displaystyle \sum\limits_{i,h = 1}^d {{a_{ij}}\left( {x,p} \right){r_i}{r_j} \ge {C_0}} \sum\limits_{i = 1}^d {{{\left| {{r_i}} \right|}^2}}, \end{array}$ among them, ∀ r = (r1, r2, …, rd) ∈ Rd.

Let w = u_t, then the original question (1) becomes:

hx,uwt−<!-- - -->∑<!-- ? -->i,j=1d∂<!-- ? -->∂<!-- ? -->xiaijx,u∂<!-- ? -->u∂<!-- ? -->xj−<!-- - -->∑<!-- ? -->i=1dbix.uuxi=fx,ux,t∈<!-- ? -->K×<!-- ? -->0,Twx,0=0,x∈<!-- ? -->Kwx,t=0x,t∈<!-- ? -->∂<!-- ? -->K×<!-- ? -->0,T $$\begin{array}{} \displaystyle \left\{ {\begin{array}{*{20}{}} {h\left( {x,u} \right){w_t} - \sum\limits_{i,j = 1}^d {\frac{\partial }{{\partial {x_i}}}\left( {{a_{ij}}\left( {x,u} \right)\frac{{\partial u}}{{\partial {x_j}}}} \right) - \sum\limits_{i = 1}^d {{b_i}\left( {x.u} \right)u{x_i} = } f\left( {x,u} \right)\left( {x,t} \right) \in K \times \left[ {0,T} \right]} }\\ \qquad\qquad\qquad\qquad\qquad\qquad\qquad~~{w\left( {x,0} \right) = 0,x \in K}\\ \qquad\qquad\qquad\qquad\qquad\qquad{w\left( {x,t} \right) = 0_{}^{}\left( {x,t} \right) \in \partial K \times \left[ {0,T} \right]} \end{array}} \right. \end{array}$$(2)

The variational equations corresponding to question equation (1) and question equation (2) are:

huutt,V+au5u,5V=bu5u,5V+fu,V∀<!-- ? -->V∈<!-- ? -->H01K,t∈<!-- ? -->0,Tu0,V=ut0,V=0⁡<!-- ? -->x∈<!-- ? -->Ku⋅<!-- ? -->,t∈<!-- ? -->H01K0,T $$\begin{array}{} \displaystyle \left\{ {\begin{array}{*{20}{c}} {\left( {h\left( u \right){u_{tt}},V} \right) + \left( {a\left( u \right)5u,5V} \right) = \left( {b\left( u \right)5u,5V} \right) + \left( {f\left( u \right),V} \right)\forall V \in H_0^1\left( K \right),t \in \left[ {0,T} \right]}\\ {\left( {u\left( 0 \right),V} \right) = \left( {{u_t}\left( 0 \right),V} \right) = 0\mathop {}\nolimits_{}^{} x \in K}\\ {u\left( { \cdot ,t} \right) \in H_0^1\left( K \right)\mathop {}\nolimits_{}^{} \left[ {0,T} \right]} \end{array}} \right. \end{array}$$(3)

huwtt,V+au5u,5V=bu5u,5V+fu,V∀<!-- ? -->V∈<!-- ? -->H01K,t∈<!-- ? -->0,Tw0,V=0⁡<!-- ? -->x∈<!-- ? -->Kw⋅<!-- ? -->,t∈<!-- ? -->H01K⁡<!-- ? -->t∈<!-- ? -->0,T $$\begin{array}{} \displaystyle \left\{ {\begin{array}{*{20}{c}} {\left( {h\left( u \right){w_{tt}},V} \right) + \left( {a\left( u \right)5u,5V} \right) = \left( {b\left( u \right)5u,5V} \right) + \left( {f\left( u \right),V} \right)\forall V \in H_0^1\left( K \right),t \in \left[ {0,T} \right]}\\ {\left( {w\left( 0 \right),V} \right) = 0\mathop {}\nolimits_{}^{} x \in K}\\ {w\left( { \cdot ,t} \right) \in H_0^1\left( K \right)\mathop {}\nolimits_{}^{} t \in \left[ {0,T} \right]} \end{array}} \right. \end{array}$$(4)

where f,g=∫<!-- ? -->Kfxgxdx;ap5f,5g=∑<!-- ? -->i,h=1d∫<!-- ? -->Kaijx,p∂<!-- ? -->f∂<!-- ? -->xi∂<!-- ? -->g∂<!-- ? -->xjdx,bp5f,g=∑<!-- ? -->i=1dbix,p∂<!-- ? -->f∂<!-- ? -->xi,g. $\begin{array}{} \left( {f,g} \right) = \int_K {f\left( x \right)g\left( x \right)dx} ; \left( {a\left( p \right)5f,5g} \right) = \sum\limits_{i,h = 1}^d {\int_K {{a_{ij}}\left( {x,p} \right)\frac{{\partial f}}{{\partial {x_i}}}} } \frac{{\partial g}}{{\partial {x_j}}}dx, \left( {b\left( p \right)5f,g} \right) = \sum\limits_{i = 1}^d {\left( {{b_i}\left( {x,p} \right)\frac{{\partial f}}{{\partial {x_i}}},g} \right)} . \end{array}$ For the convenience of calculation, x appearing in the function is omitted, and the intervals [0, T] and region K appearing in the space are also omitted. Also, f2=f,f,f12=∑<!-- ? -->i=1d∂<!-- ? -->f∂<!-- ? -->xi2,f12=f2+f12⋅<!-- ? -->1 $\begin{array}{} {\left\| f \right\|^2} = \left( {f,f} \right), \left| f \right|_1^2 = {\sum\limits_{i = 1}^d {\left\| {\frac{{\partial f}}{{\partial {x_i}}}} \right\|} ^2}, \left\| f \right\|_1^2 = {\left\| f \right\|^2} + \left| f \right|_1^2 {\left\| \cdot \right\|_1} \end{array}$ and |⋅|₁ on H01 $\begin{array}{} H_0^1 \end{array}$(K) are norms of the same order.

Let S_h ⊂ H01 $\begin{array}{} H_0^1 \end{array}$ be a finite dimensional subspace with an approximation order of m + 1. For ∀ V ∪ S_h, it satisfies the usual approximation properties and inverse estimates of VL∞<!-- 8 -->≤<!-- = -->C4h−<!-- - -->d2V $\begin{array}{} \left\| V \right\|{L^\infty } \le {C_4}{h^{ - \frac{d}{2}}}\left\| V \right\| \end{array}$ and ∥V∥₁ ≤ Ch⁻¹ ∥V∥. Elliptic projection is considered: ũ(x, t) ∈ S_h and t ∈ [0, T] are solved to satisfy:

au5u,5V=au5u~<!-- ~ -->,5V⁡<!-- ? -->∀<!-- ? -->V∈<!-- ? -->Sh $$\begin{array}{} \displaystyle \left( {a\left( u \right)5u,5V} \right) = \left( {a\left( u \right)5\tilde u,5V} \right)\mathop {}\nolimits_{}^{} \forall V \in {S_h} \end{array}$$(5)

For the projection function ũ(x, t), we assume that [3] ∥ũ∥L^∞, ∥5ũ∥L^∞ and 5∂<!-- ? -->u~<!-- ~ -->∂<!-- ? -->tL∞<!-- 8 --> $\begin{array}{} \left\| {5\frac{{\partial \tilde u}}{{\partial t}}} \right\|{L^\infty } \end{array}$ are uniformly bounded. At the same time, the regularity results of elliptic equation and the properties of S_h can be obtained [4, 6, 7].

Lemma: if the above assumptions are satisfied, then for p = 2, ∞, s = 0, 1 there are:

u−<!-- - -->u~<!-- ~ -->LpHS+u−<!-- - -->u~<!-- ~ -->tLpHS+u−<!-- - -->u~<!-- ~ -->ttLpHS≤<!-- = -->Chm+1−<!-- - -->s $$\begin{array}{} \displaystyle {\left\| {\left( {u - \tilde u} \right)} \right\|_{{L^p}\left( {{H^S}} \right)}} + {\left\| {{{\left( {u - \tilde u} \right)}_t}} \right\|_{{L^p}\left( {{H^S}} \right)}} + {\left\| {{{\left( {u - \tilde u} \right)}_{tt}}} \right\|_{{L^p}\left( {{H^S}} \right)}} \le C{h^{m + 1 - s}} \end{array}$$(6)

The interval [0, T] is divided into N equal subintervals: 0 = t₀ < t₁ < … < t_N−1 < t_N = T ⋅ t_n+1 − t_n = Δ t, Uⁿ = U(t_n), for the sake of simplicity of writing, the following marks are introduced:

Un+12=12Un+1+Un,5fn,5V=∑<!-- ? -->i=1d∂<!-- ? -->fn∂<!-- ? -->xi,∂<!-- ? -->V∂<!-- ? -->xi $$\begin{array}{} \displaystyle {U^{n + \frac{1}{2}}} = \frac{1}{2}\left( {{U^{n + 1}} + {U^n}} \right),\left( {5{f^n},5V} \right) = \sum\limits_{i = 1}^d {\left( {\frac{{\partial {f^n}}}{{\partial {x_i}}},\frac{{\partial V}}{{\partial {x_i}}}} \right)} \end{array}$$(7)

dtUn=1Δ<!-- ? -->tUn+1−<!-- - -->Un,∂<!-- ? -->Un=Un+1−<!-- - -->Un=Δ<!-- ? -->tdtUn $$\begin{array}{} \displaystyle {d_t}{U^n} = \frac{1}{{\Delta t}}\left( {{U^{n + 1}} - {U^n}} \right),\partial {U^n} = {U^{n + 1}} - {U^n} = \Delta t{d_t}{U^n} \end{array}$$(8)

∂<!-- ? -->t2Un=1Δ<!-- ? -->t2Un+1−<!-- - -->2Un+Un+1,hnU=hUn $$\begin{array}{} \displaystyle \partial _t^2{U^n} = \frac{1}{{{{\left( {\Delta t} \right)}^2}}}\left( {{U^{n + 1}} - 2U{}^n + {U^{n + 1}}} \right),{h^n}\left( U \right) = h\left( {U{}^n} \right) \end{array}$$(9)

hn+12U=12hn+1U+hnU,EUn+1=2Un−<!-- - -->Un−<!-- - -->1 $$\begin{array}{} \displaystyle {h^{n + \frac{1}{2}}}\left( U \right) = \frac{1}{2}\left( {{h^{n + 1}}\left( U \right) + {h^n}\left( U \right)} \right),E{U^{n + 1}} = 2{U^n} - {U^{n - 1}} \end{array}$$(10)

h⌢<!-- ? -->n+1U=hEUn+1,h⌢<!-- ? -->n+12U=12h⌢<!-- ? -->n+1U+hnu $$\begin{array}{} \displaystyle {{{{h\!\!\!^{^{\frown}}}} } ^{n + 1}}\left( U \right) = h\left( {E{U^{n + 1}}} \right),{{{{h\!\!\!^{^{\frown}}}} } ^{n + \frac{1}{2}}}\left( U \right) = \frac{1}{2}\left( {{{{{{h\!\!\!^{^{\frown}}}} } }^{n + 1}}\left( U \right) + {h^n}\left( u \right)} \right) \end{array}$$(11)

Also, let U − u = U − ũ + ũ − u = a + Z, a = U − ũ, Z = ũ – u, w = u_t, w̃ = ũ_t, W − w = W – w̃ + w̃ − w = θ + d, θ = W – w̃, d = w̃ – w.

Question (1) is defined as:

hUn∂<!-- ? -->t2+aUn5Un,5V+λ<!-- ? -->5Un+1−<!-- - -->2Un+Un−<!-- - -->1,5V=bUn5Un,V+fUn,V∀<!-- ? -->V∈<!-- ? -->Sh $$\begin{array}{} \displaystyle \left( {h{{\left( U \right)}^n}} \right)\partial _t^2 + \left( {a\left( {{U^n}} \right)5{U^n},5V} \right) + \lambda \left( {5\left( {{U^{n + 1}} - 2{U^n} + {U^{n - 1}}} \right),5V} \right) = \left( {b\left( {{U^n}} \right)5{U^n},V} \right) + \left( {f\left( {{U^n}} \right),V} \right)\forall V \in {S_h} \end{array}$$(12)

It is known that the solution of the equation (12) is unique. The error equation is obtained from equations (12), (10), (8) and (6):

hUn∂<!-- ? -->t2an,V+aUn5an,5V+λ<!-- ? -->5an+1−<!-- - -->2an+an−<!-- - -->1,5V=An,V−<!-- - -->Bn,5V−<!-- - -->Cn,5V+Dn,V $$\begin{array}{} \displaystyle \left( {h\left( {{U^n}} \right)\partial _t^2{a^n},V} \right) + \left( {a\left( {{U^n}} \right)5{a^n},5V} \right) + \lambda \left( {5\left( {{a^{n + 1}} - 2{a^n} + {a^{n - 1}}} \right),5V} \right) = \left( {{A^n},V} \right) - \left( {{B^n},5V} \right) - \left( {{C^n},5V} \right) + \left( {{D^n},V} \right) \end{array}$$(13)

where

An=hUnuttn0−<!-- - -->∂<!-- ? -->t2un+uttnhun−<!-- - -->hUn∂<!-- ? -->t2Zn+fUn−<!-- - -->fun $$\begin{array}{} \displaystyle {A^n} = h\left( {{U^n}} \right)\left( {u_{tt}^n0 - \partial _t^2{u^n}} \right) + u_{tt}^n\left[ {h\left( {{u^n}} \right) - h\left( {{U^n}} \right)\partial _t^2{Z_n} + f\left( {{U^n}} \right) - f\left( {{u^n}} \right)} \right] \end{array}$$(14)

Bn=aUn−<!-- - -->aun5u~<!-- ~ -->n $$\begin{array}{} \displaystyle {B^n} = \left[ {a\left( {{U^n}} \right) - a\left( {{u^n}} \right)} \right]5{\tilde u^n} \end{array}$$(15)

Cn=λ<!-- ? -->5Zn+1−<!-- - -->2Zn+Zn−<!-- - -->1+λ<!-- ? -->5un+1−<!-- - -->2un+un−<!-- - -->1 $$\begin{array}{} \displaystyle {C^n} = \lambda 5\left( {{Z^{n + 1}} - 2{Z^n} + {Z^{n - 1}}} \right) + \lambda 5\left( {{u^{n + 1}} - 2{u^n} + {u^{n - 1}}} \right) \end{array}$$(16)

Dn=bUn5an+bUn−<!-- - -->bun5u~<!-- ~ -->n+bun5Zn=D1n+D2n $$\begin{array}{} \displaystyle {D^n} = \left[ {b\left( {{U^n}} \right)5{a^n} + \left( {b\left( {{U^n}} \right) - b\left( {{u^n}} \right)} \right)5{{\tilde u}^n}} \right] + b\left( {{u^n}} \right)5{Z^n} = D_1^n + D_2^n \end{array}$$(17)

Let Q=∂<!-- ? -->u~<!-- ~ -->∂<!-- ? -->tL∞<!-- 8 -->L∞<!-- 8 -->+1,Δ<!-- ? -->t,h $\begin{array}{} Q = \left\| {\frac{{\partial \tilde u}}{{\partial t}}} \right\|{L^\infty }\left( {{L^\infty }} \right) + 1, \Delta t,h \end{array}$ are taken to satisfy C4C5Δ<!-- ? -->t2h−<!-- - -->d2≤<!-- = -->Q. $\begin{array}{} {C_4}{C_5}{\left( {\Delta t} \right)^2}{h^{ - \frac{d}{2}}} \le Q. \end{array}$ According to (1.10) and inverse estimates, it can be seen that ∥d_ta⁰∥_L^∞ ≤ Q, ∥d_tU⁰∥_L^∞ ≤ 2Q.

If inductive assumption max0≤<!-- = -->n≤<!-- = -->M−<!-- - -->2dtanL∞<!-- 8 -->≤<!-- = -->Q, $\begin{array}{} \displaystyle \mathop {\max }\limits_{0 \le n \le M - 2} {\left\| {dt{a^n}} \right\|_{{L^\infty }}} \le Q, \end{array}$ then max0≤<!-- = -->n≤<!-- = -->M−<!-- - -->2dtUnL∞<!-- 8 -->≤<!-- = -->2Q. $\begin{array}{} \displaystyle \mathop {\max }\limits_{0 \le n \le M - 2} {\left\| {dt{U^n}} \right\|_{{L^\infty }}} \le 2Q. \end{array}$

Taking the test function V=an+1−<!-- - -->an−<!-- - -->1=∂<!-- ? -->t2an+∂<!-- ? -->t2an−<!-- - -->1=Δ<!-- ? -->tdtan+dtan−<!-- - -->1, $\begin{array}{} V = {a^{n + 1}} - {a^{n - 1}} = \partial _t^2{a^n} + \partial _t^2{a^{n - 1}} = \Delta t\left( {dt{a^n} + dt{a^{n - 1}}} \right), \end{array}$ it can rewrite or estimate the two ends of the error equation.

hUn∂<!-- ? -->t2an,V=hUndtan,dtan−<!-- - -->hUndtan−<!-- - -->1,dtan−<!-- - -->1=hUndtan−<!-- - -->1,dtan−<!-- - -->1−<!-- - -->hUn−<!-- - -->1dtan−<!-- - -->1,dtan−<!-- - -->1−<!-- - -->hUn−<!-- - -->hUn−<!-- - -->1dtan−<!-- - -->1,dtan−<!-- - -->1 $$\begin{array}{} \displaystyle \left( {h\left( {{U^n}} \right)\partial _t^2{a^n},V} \right) = \left( {h\left( {{U^n}} \right)dt{a^n},dt{a^n}} \right) - \left( {h\left( {{U^n}} \right)dt{a^{n - 1}},dt{a^{n - 1}}} \right) = \left[ {\left( {h\left( {{U^n}} \right)dt{a^{n - 1}},dt{a^{n - 1}}} \right)} \right.\\ \left. { - \left( {h\left( {{U^{n - 1}}} \right)dt{a^{n - 1}},dt{a^{n - 1}}} \right)} \right] - \left( {h\left( {{U^n}} \right) - h\left( {{U^{n - 1}}} \right)dt{a^{n - 1}},dt{a^{n - 1}}} \right) \end{array}$$(18)

It is known

|An+D1n|≤<!-- = -->CΔ<!-- ? -->t(||an||12+||Zn||2)+|||dtan||2+||dtan−<!-- - -->1||2+||∂<!-- ? -->t2Zn||2 $$\begin{array}{} \displaystyle | A^n + D_1^n | \leq C \Delta t ( || a^n ||_1^2 + || Z^n ||^2 ) + |||dta^n ||^2 + ||dta^{n-1}||^2 + || \partial_t^2 Z^n ||^2 \end{array}$$(19)

The following estimates are highlighted:

B=aUn−<!-- - -->aun5u~<!-- ~ -->n.5an+1−<!-- - -->an−<!-- - -->1 $$\begin{array}{} \displaystyle B = \left( {\left[ {a\left( {{U^n}} \right) - a\left( {{u^n}} \right)} \right]5{{\tilde u}^n}.5\left( {{a^{n + 1}} - {a^{n - 1}}} \right)} \right) \end{array}$$(20)

Be aware:

∑<!-- ? -->n=1M−<!-- - -->1B11=aUM−<!-- - -->auM5u~<!-- ~ -->M,5aM−<!-- - -->aU1−<!-- - -->au15u~<!-- ~ -->1,5a1 $$\begin{array}{} \displaystyle \sum\limits_{n = 1}^{M - 1} {B_1^{\left( 1 \right)}} = \left( {\left[ {a\left( {{U^M}} \right) - a\left( {{u^M}} \right)} \right]5{{\tilde u}^M},5{a^M}} \right) - \left( {\left[ {a\left( {{U^1}} \right) - a\left( {{u^1}} \right)} \right]5{{\tilde u}^1},5{a^1}} \right) \end{array}$$(21)

And UM−<!-- - -->uM=a1+Z1+Δ<!-- ? -->t∑<!-- ? -->n=1M−<!-- - -->1dtan+dtZn. $\begin{array}{} \displaystyle {U^M} - {u^M} = {a^1} + {Z^1} + \Delta t\sum\limits_{n = 1}^{M - 1} {\left( {dt{a^n} + dt{Z^n}} \right)} . \end{array}$ We can get:

∑<!-- ? -->n=1M−<!-- - -->1B11≤<!-- = -->CΔ<!-- ? -->t∑<!-- ? -->n=1M−<!-- - -->1dtan2+dtZn2+XaM12+CZ12+a112 $$\begin{array}{} \displaystyle \left| {\sum\limits_{n = 1}^{M - 1} {B_1^{\left( 1 \right)}} } \right| \le C\Delta t\sum\limits_{n = 1}^{M - 1} {\left( {{{\left\| {dt{a^n}} \right\|}^2} + {{\left\| {dt{Z^n}} \right\|}^2}} \right)} + X\left\| {{a^M}} \right\|_1^2 + C\left( {\left\| {{Z^1}} \right\|_{}^2 + \left\| {{a^1}} \right\|_1^2} \right) \end{array}$$(22)

Similar estimates can be obtained as follows:

∑<!-- ? -->n=1M−<!-- - -->1B11≤<!-- = -->CΔ<!-- ? -->t∑<!-- ? -->n=1M−<!-- - -->1dtan2+dtZn2+an12+an+112+Zn2+Zn+12+XaM12+Ca12+Z12 $$\begin{array}{} \displaystyle \left| {\sum\limits_{n = 1}^{M - 1} {B_1^{\left( 1 \right)}} } \right| \le C\Delta t\sum\limits_{n = 1}^{M - 1} {\left( {{{\left\| {dt{a^n}} \right\|}^2} + {{\left\| {dt{Z^n}} \right\|}^2} + \left\| {{a^n}} \right\|_1^2 + \left\| {{a^{n + 1}}} \right\|_1^2 + \left\| {{Z^n}} \right\|_{}^2 + \left\| {{Z^{n + 1}}} \right\|_{}^2} \right)} + X\left\| {{a^M}} \right\|_1^2 + C\left( {\left\| {{a^1}} \right\|_{}^2 + \left\| {{Z^1}} \right\|_{}^2} \right) \end{array}$$(23)

Using inverse estimation and distribution integral, it can get:

Cn,5an+1−<!-- - -->an+1≤<!-- = -->CΔ<!-- ? -->tΔ<!-- ? -->t4h−<!-- - -->2∂<!-- ? -->t2Zn12+Δ<!-- ? -->t4+dtan2+dtan−<!-- - -->1∑<!-- ? -->n=1MbiunZxn,an+1−<!-- - -->an−<!-- - -->1≤<!-- = -->∑<!-- ? -->n=1MbiunZxn,an+1−<!-- - -->an−<!-- - -->1+2biuM−<!-- - -->1ZM−<!-- - -->1,axM−<!-- - -->12−<!-- - -->biu1Z1,ax12−<!-- - -->Δ<!-- ? -->t∑<!-- ? -->n=2M−<!-- - -->1biunZn−<!-- - -->Zn−<!-- - -->1Δ<!-- ? -->t+Zn−<!-- - -->1btun−<!-- - -->btun−<!-- - -->1Δ<!-- ? -->t,axn−<!-- - -->12≤<!-- = -->CΔ<!-- ? -->t∑<!-- ? -->n=1MZn2+dtan2+dtan−<!-- - -->12+XaM12+aM−<!-- - -->112+CZM−<!-- - -->12+Z12+a1212+CΔ<!-- ? -->t∑<!-- ? -->n=2M−<!-- - -->1dtZn−<!-- - -->12+Zn−<!-- - -->12+a1212 $$\begin{array}{} \displaystyle \left| {\left( {{C^n},5\left( {{a^{n + 1}} - {a^{n + 1}}} \right)} \right)} \right| \le C\Delta t\left[ {{{\left( {\Delta t} \right)}^4}{h^{ - 2}}} \right]\left\| {\partial _t^2{Z^n}} \right\|_1^2 + {\left( {\Delta t} \right)^4} + {\left\| {dt{a^n}} \right\|^2} + \left\| {dt{a^{n - 1}}} \right\|\\ \left| {\sum\limits_{n = 1}^M {\left( {{b_i}\left( {{u^n}} \right)Z_x^n,{a^{n + 1}} - {a^{n - 1}}} \right)} } \right| \le \left| {\sum\limits_{n = 1}^M {\left( {{b_i}\left( {{u^n}} \right)Z_x^n,{a^{n + 1}} - {a^{n - 1}}} \right)} } \right| + 2\left| {\left( {{b_i}\left( {{u^{M - 1}}} \right){Z^{M - 1}},a_x^{M - \frac{1}{2}}} \right)} \right.\\ \left. { - \left( {{b_i}\left( {{u^1}} \right){Z^1},a_x^{\frac{1}{2}}} \right) - \Delta t\sum\limits_{n = 2}^{M - 1} {\left( {{b_i}\left( {{u^n}} \right)\frac{{{Z^n} - {Z^{n - 1}}}}{{\Delta t}}} \right)} } \right| + {Z^{n - 1}}\frac{{bt\left( {{u^n}} \right) - bt\left( {{u^{n - 1}}} \right)}}{{\Delta t}},\left. {a_x^{n - \frac{1}{2}}} \right|\\ \le C\Delta t\sum\limits_{n = 1}^M {\left( {{{\left\| {{Z^n}} \right\|}^2} + {{\left\| {dt{a^n}} \right\|}^2} + {{\left\| {dt{a^{n - 1}}} \right\|}^2} + X\left( {\left\| {{a^M}} \right\|_1^2 + \left\| {{a^{M - 1}}} \right\|_1^2} \right)} \right)} \\ + C\left( {{{\left\| {{Z^{M - 1}}} \right\|}^2} + {{\left\| {{Z^1}} \right\|}^2} + \left\| {{a^{\frac{1}{2}}}} \right\|_1^2 + C\Delta t\sum\limits_{n = 2}^{M - 1} {\left( {{{\left\| {dt{Z^{n - 1}}} \right\|}^2}} \right) + {{\left\| {{Z^{n - 1}}} \right\|}^2} + \left\| {{a^{\frac{1}{2}}}} \right\|_1^2} } \right) \end{array}$$(24)

It is noticed that m ≥ 1, (Δ t)² = O(h^m+1), the lemma shows that:

ZM−<!-- - -->12+ZM2+CΔ<!-- ? -->t∑<!-- ? -->n=1M−<!-- - -->1Zn2+Zn+12+dtZn2+Zn−<!-- - -->12+Δ<!-- ? -->t4h−<!-- - -->2∂<!-- ? -->t2Zn12≤<!-- = -->Ch2m+2 $$\begin{array}{} \displaystyle {\left\| {{Z^{M - 1}}} \right\|^2} + {\left\| {{Z^M}} \right\|^2} + C\Delta t\sum\limits_{n = 1}^{M - 1} {\left( {{{\left\| {{Z^n}} \right\|}^2} + {{\left\| {{Z^{n + 1}}} \right\|}^2} + {{\left\| {dt{Z^n}} \right\|}^2} + {{\left\| {{Z^{n - 1}}} \right\|}^2} + {{\left( {\Delta t} \right)}^4}{h^{ - 2}}\left\| {\partial _t^2{Z^n}} \right\|_1^2} \right) \le C{h^{2m + 2}}} \end{array}$$(25)

For error equation, the sum can be solved from n = 1, 2, …, M − 1, and it is noted that ∥a⁰∥₁ = 0, ∥a¹∥₁ + ∥dta⁰∥ ≤ C(Δ t)². Using inductive hypothesis and the above estimates, we can get:

C2dtaM−<!-- - -->12+aUM−<!-- - -->15aM−<!-- - -->1,5aM+λ<!-- ? -->aM−<!-- - -->aM−<!-- - -->112≤<!-- = -->Ch2m+2+Δ<!-- ? -->t4+2XaM2+aM−<!-- - -->12+CΔ<!-- ? -->t∑<!-- ? -->n=1M−<!-- - -->1dtan2+an12+an+112 $$\begin{array}{} \displaystyle {C_2}{\left\| {dt{a^{M - 1}}} \right\|^2} + \left( {a\left( {{U^{M - 1}}} \right)5{a^{M - 1}},5{a^M}} \right) + \lambda \left| {{a^M} - {a^{M - 1}}} \right|_1^2 \le C\left( {{h^{2m + 2}} + {{\left( {\Delta t} \right)}^4}} \right)\\ + 2X\left( {{{\left\| {{a^M}} \right\|}^2} + {{\left\| {{a^{M - 1}}} \right\|}^2}} \right) + C\Delta t\sum\limits_{n = 1}^{M - 1} {\left( {{{\left\| {dt{a^n}} \right\|}^2} + \left\| {{a^n}} \right\|_1^2 + \left\| {{a^{n + 1}}} \right\|_1^2} \right)} \end{array}$$(26)

If λ<!-- ? -->><!-- > -->dC14,V=minλ<!-- ? -->−<!-- - -->dC14,C04><!-- > -->0, $\begin{array}{} \lambda \gt \frac{{d{C_1}}}{4}, V = \min \left\{ {\lambda - \frac{{d{C_1}}}{4},\frac{{{C_0}}}{4}} \right\} \gt 0, \end{array}$ so:

λ<!-- ? -->aM−<!-- - -->aM−<!-- - -->112+aUM−<!-- - -->15aM≥<!-- = -->VaM−<!-- - -->aM−<!-- - -->112+aM+aM−<!-- - -->112≥<!-- = -->CaM12+aM−<!-- - -->112 $$\begin{array}{} \displaystyle \lambda \left| {{a^M} - {a^{M - 1}}} \right|_1^2 + \left( {a\left( {{U^{M - 1}}} \right)5{a^M}} \right) \ge V\left( {\left| {{a^M} - {a^{M - 1}}} \right|_1^2 + \left| {{a^M} + {a^{M - 1}}} \right|_1^2} \right) \ge C\left( {\left\| {{a^M}} \right\|_1^2 + \left\| {{a^{M - 1}}} \right\|_1^2} \right) \end{array}$$(27)

If Δ t, X is appropriately small, the Gronwall inequality can be applied to equation (27):

dtaM−<!-- - -->12+aM12+aM−<!-- - -->112≤<!-- = -->CΔ<!-- ? -->t4+h2m+2 $$\begin{array}{} \displaystyle {\left\| {dt{a^{M - 1}}} \right\|^2} + \left\| {{a^M}} \right\|_1^2 + \left\| {{a^{M - 1}}} \right\|_1^2 \le C\left( {{{\left( {\Delta t} \right)}^4} + {h^{2m + 2}}} \right) \end{array}$$(28)

It can be seen immediately that h, Δ t is sufficiently small and max0≤<!-- = -->n≤<!-- = -->M−<!-- - -->1dtanL∞<!-- 8 -->≤<!-- = -->Q, $\begin{array}{} \displaystyle \mathop {\max }\limits_{0 \le n \le M - 1} {\left\| {dt{a^n}} \right\|_{{L^\infty }}} \le Q, \end{array}$ so the inductive hypothesis holds for m = N − 1 [18].

Theorem: If a_ij, b_i, f, h and u satisfy the above conditions, m+1><!-- > -->d2,m≥<!-- = -->1,λ<!-- ? -->≥<!-- = -->dC14, $\begin{array}{} m + 1 \gt \frac{d}{2}, m \ge 1, \lambda \ge \frac{{d{C_1}}}{4}, \end{array}$ then when h, Δ t are sufficiently small:

max0≤<!-- = -->n≤<!-- = -->M−<!-- - -->1dtU−<!-- - -->un−<!-- - -->1+U−<!-- - -->un−<!-- - -->12+hU−<!-- - -->un−<!-- - -->121≤<!-- = -->Chm+1+Δ<!-- ? -->t2 $$\begin{array}{} \displaystyle \mathop {\max }\limits_{0 \le n \le M - 1} \left\{ {\left\| {dt{{\left( {U - u} \right)}^{n - 1}}} \right\| + \left\| {{{\left( {U - u} \right)}^{n - \frac{1}{2}}}} \right\| + h{{\left\| {{{\left( {U - u} \right)}^{n - \frac{1}{2}}}} \right\|}_1}} \right\} \le C\left( {{h^{m + 1}} + {{\left( {\Delta t} \right)}^2}} \right) \end{array}$$(29)

If the super-convergence analysis is completed by V = Uⁿ⁺¹ − Uⁿ⁻¹ in the question (1), then the question (1) is stable.

The following fourth-order non-linear hyperbolic equations are considered:

utt+γ<!-- ? -->Δ<!-- ? -->2u−<!-- - -->Δ<!-- ? -->ut+fu=0,X,t∈<!-- ? -->Ω<!-- O -->×<!-- ? -->0,TuX,t=Δ<!-- ? -->uX,t=0,X,t∈<!-- ? -->∂<!-- ? -->Ω<!-- O -->×<!-- ? -->0,TuX,0=u0X,utX,0=u1X,X∈<!-- ? -->Ω<!-- O --> $$\begin{array}{} \displaystyle \left\{ {\begin{array}{*{20}{c}} {{u_{tt}} + \gamma {\Delta ^2}u - \Delta {u_t} + f\left( u \right) = 0,\left( {X,t} \right) \in \Omega \times \left( {0,T} \right]}\\ {u\left( {X,t} \right) = \Delta u\left( {X,t} \right) = 0,\left( {X,t} \right) \in \partial \Omega \times \left( {0,T} \right]}\\ {u\left( {X,0} \right) = {u_0}\left( X \right),{u_t}\left( {X,0} \right) = {u_1}\left( X \right),X \in \Omega } \end{array}} \right. \end{array}$$(30)

where Ω ∈ R² is a bounded convex polygon region with Lipschitz continuous boundary, ∂ Ω is the boundary of Ω, T ∈ (0, + ∞), γ is a positive fixed value, X = (x, y), f(u) is a global Lipschitz continuous about u, that is, there exists a constant C greater than 0. Let

fu1−<!-- - -->fu2≤<!-- = -->Cu1−<!-- - -->u2 $$\begin{array}{} \displaystyle \left| {f\left( {{u_1}} \right) - f\left( {{u_2}} \right)} \right| \le C\left| {{u_1} - {u_2}} \right| \end{array}$$(31)

In this paper, W^m,p is used to denote the usual Sobolev spaces, whose norms and seminorms are denoted as ∥⋅∥_m,p and |⋅|_m,p, respectively. Especially when p = 2, W^m,p is denoted as H^m(Ω), and the corresponding norms and seminorms are denoted as ∥⋅∥_m and |⋅|_m.

φ<!-- f -->L∞<!-- 8 -->0,Y;HKΩ<!-- O -->◃<!-- ? -->φ<!-- f -->HKΩ<!-- O -->,φ<!-- f -->L20,Y;HKΩ<!-- O -->◃<!-- ? -->∫<!-- ? -->0tφ<!-- f -->2HKΩ<!-- O -->ds12 $$\begin{array}{} \displaystyle \left\| \varphi \right\|{}_{{L^\infty }\left( {0,Y;{H^K}\left( \Omega \right)} \right)} \triangleleft {\left\| \varphi \right\|_{{H^K}\left( \Omega \right)}},{\left\| \varphi \right\|_{{L^2}\left( {0,Y;{H^K}\left( \Omega \right)} \right)}} \triangleleft {\left( {\int_0^t {{{\left\| \varphi \right\|}^2}_{{H^K}\left( \Omega \right)}ds} } \right)^{\frac{1}{2}}} \end{array}$$(32)

Let p⃗ = − ∇ u, v = ∇ ⋅ p⃗, then the problem equation (30) is equivalent to the following problem:

utt+γ<!-- ? -->Δ<!-- ? -->v+v+vt+fu=0,X,t∈<!-- ? -->Ω<!-- O -->×<!-- ? -->0,Tv−<!-- - -->∇<!-- ? -->⋅<!-- ? -->p→<!-- ? -->=0,X,t∈<!-- ? -->Ω<!-- O -->×<!-- ? -->0,Tp→<!-- ? -->+∇<!-- ? -->u=0,X,t∈<!-- ? -->Ω<!-- O -->×<!-- ? -->0,TuX,t=vX,t=0,X,t∈<!-- ? -->∂<!-- ? -->Ω<!-- O -->×<!-- ? -->0,TuX,0=u0,utX,0=u1,X∈<!-- ? -->Ω<!-- O --> $$\begin{array}{} \displaystyle \left\{ {\begin{array}{*{20}{c}} {{u_{tt}} + \gamma \Delta v + v + {v_t} + f\left( u \right) = 0,\left( {X,t} \right) \in \Omega \times \left( {0,T} \right]}\\ {v - \nabla \cdot \vec p = 0,\left( {X,t} \right) \in \Omega \times \left( {0,T} \right]}\\ {\vec p + \nabla u = 0,\left( {X,t} \right) \in \Omega \times \left( {0,T} \right]}\\ {u\left( {X,t} \right) = v\left( {X,t} \right) = 0,\left( {X,t} \right) \in \partial \Omega \times \left( {0,T} \right]}\\ {u\left( {X,0} \right) = {u_0},{u_t}\left( {X,0} \right) = {u_1},X \in \Omega } \end{array}} \right. \end{array}$$(33)

The variational question of equation (30) is to find {u, v, p⃗}[0, T] → H01 $\begin{array}{} H_0^1 \end{array}$(Ω) × H01 $\begin{array}{} H_0^1 \end{array}$(Ω) × (L²(Ω))² so that:

utt,ϕ<!-- ? -->=γ<!-- ? -->∇<!-- ? -->v,∇<!-- ? -->ϕ<!-- ? -->+v,ϕ<!-- ? -->+fu,ϕ<!-- ? -->=0,∀<!-- ? -->ϕ<!-- ? -->∈<!-- ? -->H01Ω<!-- O -->v,χ<!-- ? -->+p→<!-- ? -->,∇<!-- ? -->χ<!-- ? -->=0,∀<!-- ? -->ϕ<!-- ? -->∈<!-- ? -->H01Ω<!-- O -->p→<!-- ? -->,w→<!-- ? -->+∇<!-- ? -->u,w→<!-- ? -->=0,∀<!-- ? -->w→<!-- ? -->∈<!-- ? -->L2Ω<!-- O -->2uX,0=u0,utX,0=u1,X∈<!-- ? -->Ω<!-- O --> $$\begin{array}{} \displaystyle \left\{ {\begin{array}{*{20}{c}} {\left( {{u_{tt}},\phi } \right) = \gamma \left( {\nabla v,\nabla \phi } \right) + \left( {v,\phi } \right) + \left( {f\left( u \right),\phi } \right) = 0,\forall \phi \in H_0^1\left( \Omega \right)}\\ {\left( {v,\chi } \right) + \left( {\vec p,\nabla \chi } \right) = 0,\forall \phi \in H_0^1\left( \Omega \right)}\\ {\left( {\vec p,\vec w} \right) + \left( {\nabla u,\vec w} \right) = 0,\forall \vec w \in {{\left( {{L^2}\left( \Omega \right)} \right)}^2}}\\ {u\left( {X,0} \right) = {u_0},{u_t}\left( {X,0} \right) = {u_1},X \in \Omega } \end{array}} \right. \end{array}$$(34)

Considering the semi-discrete scheme of equation (34), {u_h, v_h, p⃗_h} : [0, T] → M_h × M_h × W⃗_h is obtained, so that:

uhtt,ϕ<!-- ? -->h+γ<!-- ? -->∇<!-- ? -->vh,∇<!-- ? -->ϕ<!-- ? -->h+vh,ϕ<!-- ? -->h+vht,ϕ<!-- ? -->h+fuh,ϕ<!-- ? -->h=0,∀<!-- ? -->ϕ<!-- ? -->h∈<!-- ? -->Mhvh,χ<!-- ? -->h+p→<!-- ? -->h,∇<!-- ? -->χ<!-- ? -->h=0,∀<!-- ? -->χ<!-- ? -->h∈<!-- ? -->Mhp→<!-- ? -->h,w→<!-- ? -->h+∇<!-- ? -->uh,w→<!-- ? -->h=0,∀<!-- ? -->w→<!-- ? -->h∈<!-- ? -->W→<!-- ? -->huh0=Rhu0,uht0=Rhu1,X∈<!-- ? -->Ω<!-- O -->uh0=R⌢<!-- ? -->h−<!-- - -->∇<!-- ? -->u0,X∈<!-- ? -->Ω<!-- O --> $$\begin{array}{} \displaystyle \left\{ {\begin{array}{*{20}{c}} {\left( {{u_{htt}},{\phi _h}} \right) + \gamma \left( {\nabla {v_h},\nabla {\phi _h}} \right) + \left( {{v_h},{\phi _h}} \right) + \left( {{v_{ht}},{\phi _h}} \right) + \left( {f\left( {{u_h}} \right),{\phi _h}} \right) = 0,\forall {\phi _h} \in {M_h}}\\ {\left( {{v_h},{\chi _h}} \right) + \left( {{{\vec p}_h},\nabla {\chi _h}} \right) = 0,\forall {\chi _h} \in {M_h}}\\ {\left( {{{\vec p}_h},{{\vec w}_h}} \right) + \left( {\nabla {u_h},{{\vec w}_h}} \right) = 0,\forall {{\vec w}_h} \in {{\vec W}_h}}\\ {{u_h}\left( 0 \right) = {R_h}{u_0},{u_{ht}}\left( 0 \right) = {R_h}{u_1},X \in \Omega }\\ {{u_h}\left( 0 \right) = {{R\!\!\!^{^{\frown}} }_h}\left( { - \nabla {u_0}} \right),X \in \Omega } \end{array}} \right. \end{array}$$(35)

It is easy to verify that equation (35) has unique solutions. The super-approximation and super-convergence results of mixed element solutions are given in the semi-discrete scheme [19].

Let:

u−<!-- - -->uh=u−<!-- - -->Rhu+Rhu−<!-- - -->uh=η<!-- ? -->+ξ<!-- ? --> $$\begin{array}{} \displaystyle u - {u_h} = \left( {u - {R_h}u} \right) + \left( {{R_h}u - {u_h}} \right) = \eta + \xi \end{array}$$(39)

v−<!-- - -->vh=v−<!-- - -->R⌢<!-- ? -->hv+R⌢<!-- ? -->hv−<!-- - -->vh=τ<!-- t -->+τ<!-- t --> $$\begin{array}{} \displaystyle v - {v_h} = \left( {v - {R\!\!\!^{^{\frown}}}_h}v \right) + \left( {{{ R\!\!\!^{^{\frown}}}_h}v - {v_h}} \right) = \tau + \tau \end{array}$$(40)

p→<!-- ? -->−<!-- - -->p→<!-- ? -->h=p→<!-- ? -->−<!-- - -->R⌢<!-- ? -->hp→<!-- ? -->+R⌢<!-- ? -->hp→<!-- ? -->−<!-- - -->p→<!-- ? -->h=p→<!-- ? -->+θ<!-- ? -->→<!-- ? --> $$\begin{array}{} \displaystyle \vec p - {\vec p_h} = \left( {\vec p - {{R\!\!\!^{^{\frown}} }_h}\vec p} \right) + \left( {{{R\!\!\!^{^{\frown}} }_h}\vec p - {{\vec p}_h}} \right){\rm{ = }}\vec p + \vec \theta \end{array}$$(41)

From equations (30) and (35), the following error equation can be obtained:

ξ<!-- ? -->tt,ϕ<!-- ? -->h+γ<!-- ? -->∇<!-- ? -->τ<!-- t -->,∇<!-- ? -->ϕ<!-- ? -->h+τ<!-- t -->,ϕ<!-- ? -->h+τ<!-- t -->t,ϕ<!-- ? -->h=η<!-- ? -->tt,ϕ<!-- ? -->h−<!-- - -->τ<!-- t -->,ϕ<!-- ? -->h−<!-- - -->τ<!-- t -->t,ϕ<!-- ? -->h−<!-- - -->fu−<!-- - -->fuh,ϕ<!-- ? -->hτ<!-- t -->,χ<!-- ? -->h+θ<!-- ? -->→<!-- ? -->,∇<!-- ? -->χ<!-- ? -->h=−<!-- - -->τ<!-- t -->,χ<!-- ? -->hθ<!-- ? -->→<!-- ? -->,w→<!-- ? -->h+∇<!-- ? -->ξ<!-- ? -->,w→<!-- ? -->h=0 $$\begin{array}{} \displaystyle \left\{ {\begin{array}{*{20}{c}} {\left( {{\xi _{tt}},{\phi _h}} \right) + \gamma \left( {\nabla \tau ,\nabla {\phi _h}} \right) + \left( {\tau ,{\phi _h}} \right) + \left( {{\tau _t},{\phi _h}} \right)}\\ { = \left( {{\eta _{tt}},{\phi _h}} \right) - \left( {\tau ,{\phi _h}} \right) - \left( {{\tau _t},{\phi _h}} \right) - \left( {f\left( u \right) - f\left( {{u_h}} \right),{\phi _h}} \right)}\\ {\left( {\tau ,{\chi _h}} \right) + \left( {\vec \theta ,\nabla {\chi _h}} \right) = - \left( {\tau ,{\chi _h}} \right)}\\ {\left( {\vec \theta ,{{\vec w}_h}} \right) + \left( {\nabla \xi ,{{\vec w}_h}} \right) = 0} \end{array}} \right. \end{array}$$(42)

In equation (42), ϕ_h = τ_t in formula 1, and for t in the second and third formulas, derivatives are obtained. And then χ _h = ξ_tt and w⃗_h = ∇ ξ_tt, respectively, there are:

ξ<!-- ? -->tt,τ<!-- t -->t+γ<!-- ? -->∇<!-- ? -->τ<!-- t -->,∇<!-- ? -->τ<!-- t -->t+τ<!-- t -->,τ<!-- t -->t+τ<!-- t -->t,τ<!-- t -->t=−<!-- - -->η<!-- ? -->tt,τ<!-- t -->t−<!-- - -->τ<!-- t -->,τ<!-- t -->t−<!-- - -->τ<!-- t -->t,τ<!-- t -->t−<!-- - -->fu−<!-- - -->fuh,τ<!-- t -->tτ<!-- t -->t,ξ<!-- ? -->tt+θ<!-- ? -->→<!-- ? -->t,∇<!-- ? -->ξ<!-- ? -->tt=−<!-- - -->τ<!-- t -->,ξ<!-- ? -->ttθ<!-- ? -->→<!-- ? -->t,∇<!-- ? -->ξ<!-- ? -->tt+∇<!-- ? -->ξ<!-- ? -->t,∇<!-- ? -->ξ<!-- ? -->tt=0 $$\begin{array}{} \displaystyle \left\{ {\begin{array}{*{20}{c}} {\left( {{\xi _{tt}},{\tau _t}} \right) + \gamma \left( {\nabla \tau ,\nabla {\tau _t}} \right) + \left( {\tau ,{\tau _t}} \right) + \left( {{\tau _t},{\tau _t}} \right)}\\ { = - \left( {{\eta _{tt}},{\tau _t}} \right) - \left( {\tau ,{\tau _t}} \right) - \left( {{\tau _t},{\tau _t}} \right) - \left( {f\left( u \right) - f\left( {{u_h}} \right),{\tau _t}} \right)}\\ {\left( {{\tau _t},{\xi _{tt}}} \right) + \left( {{{\vec \theta }_t},\nabla {\xi _{tt}}} \right) = - \left( {\tau ,{\xi _{tt}}} \right)}\\ {\left( {{{\vec \theta }_t},\nabla {\xi _{tt}}} \right) + \left( {\nabla {\xi _t},\nabla {\xi _{tt}}} \right) = 0} \end{array}} \right. \end{array}$$(43)

According to equation (43):

12ddt∇<!-- ? -->ξ<!-- ? -->t02+γ<!-- ? -->2ddt∇<!-- ? -->τ<!-- t -->02+12ddtτ<!-- t -->02+τ<!-- t -->02=−<!-- - -->η<!-- ? -->tt,τ<!-- t -->t−<!-- - -->r,τ<!-- t -->t−<!-- - -->rt,τ<!-- t -->t−<!-- - -->fu−<!-- - -->fuh,τ<!-- t -->t+τ<!-- t -->t,ξ<!-- ? -->tt=∑<!-- ? -->i=15Ai $$\begin{array}{} \displaystyle \frac{1}{2}\frac{d}{{dt}}\left\| {\nabla {\xi _t}} \right\|_0^2 + \frac{\gamma }{2}\frac{d}{{dt}}\left\| {\nabla \tau } \right\|_0^2 + \frac{1}{2}\frac{d}{{dt}}\left\| \tau \right\|_0^2 + \left\| \tau \right\|_0^2\\ = - \left( {{\eta _{tt}},{\tau _t}} \right) - \left( {r,{\tau _t}} \right) - \left( {{r_t},{\tau _t}} \right) - \left( {f\left( u \right) - f\left( {{u_h}} \right),{\tau _t}} \right) + \left( {{\tau _t},{\xi _{tt}}} \right) = \sum\limits_{i = 1}^5 {{A_i}} \end{array}$$(44)

Let estimate A_i(i = 1, 2, …, 5) in turn.

Using Schwarz inequality, Young inequality and interpolation theory, the following conclusions are obtained:

∑<!-- ? -->i=12Ai≤<!-- = -->Cη<!-- ? -->tt0+r0τ<!-- t -->t0≤<!-- = -->Ch4utt22+vtt22+12τ<!-- t -->t22 $$\begin{array}{} \displaystyle \sum\limits_{i = 1}^2 {{A_i}} \le C\left( {{{\left\| {{\eta _{tt}}} \right\|}_0} + {{\left\| r \right\|}_0}} \right){\left\| {{\tau _t}} \right\|_0} \le C{h^4}\left( {\left\| {{u_{tt}}} \right\|_2^2 + \left\| {{v_{tt}}} \right\|_2^2} \right) + \frac{1}{2}\left\| {{\tau _t}} \right\|_2^2 \end{array}$$(45)

Since f satisfies the Lipschitz condition, there are:

A4≤<!-- = -->Cu−<!-- - -->uh0τ<!-- t -->t0≤<!-- = -->Cξ<!-- ? -->02+h4u22+12τ<!-- t -->t02 $$\begin{array}{} \displaystyle {A_4} \le C{\left\| {u - {u_h}} \right\|_0}{\left\| {{\tau _t}} \right\|_0} \le C\left( {\left\| \xi \right\|_0^2 + {h^4}\left\| u \right\|_2^2 + \frac{1}{2}\left\| {{\tau _t}} \right\|_0^2} \right) \end{array}$$(46)

Using derivative transfer techniques, there are:

A3=rtt,τ<!-- t -->−<!-- - -->ddtrt,τ<!-- t -->≤<!-- = -->Ch4vtt22+τ<!-- t -->02−<!-- - -->ddtrt,τ<!-- t --> $$\begin{array}{} \displaystyle {A_3} = \left( {{r_{tt}},\tau } \right) - \frac{d}{{dt}}\left( {{r_t},\tau } \right) \le C\left( {{h^4}\left\| {{v_{tt}}} \right\|_2^2 + \left\| \tau \right\|_0^2} \right) - \frac{d}{{dt}}\left( {{r_t},\tau } \right) \end{array}$$(47)

A5=rtt,ξ<!-- ? -->t+ddtrt,ξ<!-- ? -->t≤<!-- = -->Ch4vtt22+ξ<!-- ? -->t02−<!-- - -->ddtrt,ξ<!-- ? -->t $$\begin{array}{} \displaystyle {A_5} = \left( {{r_{tt}},{\xi _t}} \right) + \frac{d}{{dt}}\left( {{r_t},{\xi _t}} \right) \le C\left( {{h^4}\left\| {{v_{tt}}} \right\|_2^2 + \left\| {{\xi _t}} \right\|_0^2} \right) - \frac{d}{{dt}}\left( {{r_t},{\xi _t}} \right) \end{array}$$(48)

By introducing the above estimates of A_i into equation (44), we can obtain that:

min1,γ<!-- ? -->2ddtξ<!-- ? -->t12+τ<!-- t -->12≤<!-- = -->Ch4u22+utt22+v22+utt22+Cξ<!-- ? -->02+ξ<!-- ? -->t02+τ<!-- t -->02−<!-- - -->ddtrt,τ<!-- t -->−<!-- - -->rt,ξ<!-- ? -->t $$\begin{array}{} \displaystyle \frac{{\min \left\{ {1,\gamma } \right\}}}{2}\frac{d}{{dt}}\left( {\left\| {{\xi _t}} \right\|_1^2 + \left\| \tau \right\|_1^2} \right) \le C{h^4}\left( {\left\| u \right\|_2^2 + \left\| {{u_{tt}}} \right\|_2^2 + \left\| v \right\|_2^2 + \left\| {{u_{tt}}} \right\|_2^2} \right)\\ + C\left( {\left\| \xi \right\|_0^2 + \left\| {{\xi _t}} \right\|_0^2 + \left\| \tau \right\|_0^2} \right) - \frac{d}{{dt}}\left[ {\left( {{r_t},\tau } \right) - \left( {{r_t},{\xi _t}} \right)} \right] \end{array}$$(49)

The two ends of the equation are multiplied by 2min1,γ<!-- ? -->, $\begin{array}{} \frac{2}{{\min \left\{ {1,\gamma } \right\}}}, \end{array}$ and then integrated from 0 to t. It is noted that τ (0) = ξ_t(0) = ξ (0) = 0, ξ<!-- ? -->02≤<!-- = -->C∫<!-- ? -->0tξ<!-- ? -->t02ds, $\begin{array}{} \left\| \xi \right\|_0^2 \le C\int_0^t {\left\| {{\xi _t}} \right\|} _0^2ds, \end{array}$ and then according to inequality ∫<!-- ? -->0t∫<!-- ? -->0sφ<!-- f -->2dsdτ<!-- t -->≤<!-- = -->C∫<!-- ? -->0tφ<!-- f -->2ds, $\begin{array}{} \int_0^t {\int_0^s {{\varphi ^2}dsd\tau } } \le C\int_0^t {{\varphi ^2}} ds, \end{array}$ there are:

ξ<!-- ? -->t12+τ<!-- t -->12≤<!-- = -->Ch4∫<!-- ? -->0tu22+utt22+v22+utt22ds+C∫<!-- ? -->0tτ<!-- t -->12+ξ<!-- ? -->t12+ξ<!-- ? -->02ds−<!-- - -->1min1,γ<!-- ? -->rt,τ<!-- t -->−<!-- - -->ξ<!-- ? -->t≤<!-- = -->Ch4vt22+∫<!-- ? -->0tu22+utt22+v22+utt22ds+C∫<!-- ? -->0tτ<!-- t -->12+ξ<!-- ? -->t12ds+12τ<!-- t -->12+ξ<!-- ? -->t12 $$\begin{array}{} \left\| {{\xi _t}} \right\|_1^2 + \left\| \tau \right\|_1^2 \le C{h^4}\int_0^t {\left( {\left\| u \right\|_2^2 + \left\| {{u_{tt}}} \right\|_2^2 + \left\| v \right\|_2^2 + \left\| {{u_{tt}}} \right\|_2^2} \right)ds} \\ + C\int_0^t {\left( {\left\| \tau \right\|_1^2 + \left\| {{\xi _t}} \right\|_1^2 + \left\| \xi \right\|_0^2} \right)ds - \frac{1}{{\min \left\{ {1,\gamma } \right\}}}\left[ {\left( {{r_t},\tau } \right) - \left( {{\xi _t}} \right)} \right]} \\ \le C{h^4}\left[ {\left\| {{v_t}} \right\|_2^2 + \int_0^t {\left( {\left\| u \right\|_2^2 + \left\| {{u_{tt}}} \right\|_2^2 + \left\| v \right\|_2^2 + \left\| {{u_{tt}}} \right\|_2^2} \right)ds} } \right]\\ + C\int_0^t {\left( {\left\| \tau \right\|_1^2 + \left\| {{\xi _t}} \right\|_1^2} \right)ds + \frac{1}{2}\left( {\left\| \tau \right\|_1^2 + \left\| {{\xi _t}} \right\|_1^2} \right)} \end{array}$$(50)

The Gronwall inequality is used to obtain:

ξ<!-- ? -->t1+τ<!-- t -->1≤<!-- = -->Ch2vt22+∫<!-- ? -->0tR1ds12 $$\begin{array}{} \displaystyle {\left\| {{\xi _t}} \right\|_1} + \left\| \tau \right\|{}_1 \le C{h^2}{\left( {\left\| {{v_t}} \right\|_2^2 + \int_0^t {{R_1}ds} } \right)^{\frac{1}{2}}} \end{array}$$(51)

On the other hand, if it is noticed that ξ (0) = ∇ ξ (0) = 0, it can get from ξ<!-- ? -->12≤<!-- = -->C∫<!-- ? -->0tξ<!-- ? -->t12ds: $\begin{array}{} \left\| \xi \right\|_1^2 \le C\int_0^t {\left\| {{\xi _t}} \right\|} _1^2ds: \end{array}$

ξ<!-- ? -->12≤<!-- = -->C∫<!-- ? -->0tξ<!-- ? -->t12ds≤<!-- = -->Ch4∫<!-- ? -->0tvt22+∫<!-- ? -->0sR1dτ<!-- t -->≤<!-- = -->Ch4∫<!-- ? -->0tR2ds $$\begin{array}{} \displaystyle \left\| \xi \right\|_1^2 \le C\int_0^t {\left\| {{\xi _t}} \right\|} _1^2ds \le C{h^4}\left[ {\int_0^t {\left( {\left\| {{v_t}} \right\|_2^2 + \int_0^s {{R_1}d\tau } } \right) \le C{h^4}\int_0^t {{R_2}ds} } } \right] \end{array}$$(52)

Namely:

ξ<!-- ? -->1≤<!-- = -->Ch2∫<!-- ? -->0tR2ds12 $$\begin{array}{} \displaystyle \left\| \xi \right\|_1^{} \le C{h^2}{\left( {\int_0^t {{R_2}ds} } \right)^{\frac{1}{2}}} \end{array}$$(53)

In the third form of equation (42), w⃗_h = θ⃗ is obtained from Schwarz inequality.

θ<!-- ? -->→<!-- ? -->02=θ<!-- ? -->→<!-- ? -->,θ<!-- ? -->→<!-- ? -->=−<!-- - -->∇<!-- ? -->ξ<!-- ? -->,θ<!-- ? -->→<!-- ? -->≤<!-- = -->C∇<!-- ? -->ξ<!-- ? -->0θ<!-- ? -->→<!-- ? -->0 $$\begin{array}{} \displaystyle \left\| {\vec \theta } \right\|_0^2 = \left( {\vec \theta ,\vec \theta } \right) = - \left( {\nabla \xi ,\vec \theta } \right) \le C\left\| {\nabla \xi \left\| 0 \right\|\left. {\vec \theta } \right\|} \right\|{}_0 \end{array}$$(54)

According to equation (54), there are:

θ<!-- ? -->→<!-- ? -->0<<!-- ? -->Cξ<!-- ? -->1≤<!-- = -->Ch2∫<!-- ? -->0tR2ds12 $$\begin{array}{} \displaystyle \left\| {\vec \theta } \right\|_0^{} \lt C{\left\| \xi \right\|_1} \le C{h^2}{\left( {\int_0^t {{R_2}ds} } \right)^{\frac{1}{2}}} \end{array}$$(55)

By using the trigonometric inequality, it can obtain:

Ihu−<!-- - -->uh1≤<!-- = -->CIhu−<!-- - -->Rhu1+Rhu−<!-- - -->uh1≤<!-- = -->Ch2u3+∫<!-- ? -->0tR2ds12 $$\begin{array}{} \displaystyle {\left\| {{I_h}u - {u_h}} \right\|_1} \le C\left( {{{\left\| {{I_h}u - {R_h}u} \right\|}_1} + {{\left\| {{R_h}u - {u_h}} \right\|}_1}} \right) \le C{h^2}\left[ {{{\left\| u \right\|}_3} + {{\left( {\int_0^t {{R_2}ds} } \right)}^{\frac{1}{2}}}} \right] \end{array}$$(56)

The theorem is proved.

In order to obtain global super-convergence, the adjacent four elements are merged into one large element processing operators I2h2andΠ<!-- ? -->2h2: $\begin{array}{} \displaystyle I_{2h}^2~~ \text{and}~~ \Pi _{2h}^2: \end{array}$

I2h2Ihu=I2h2u,∀<!-- ? -->u∈<!-- ? -->H2Ω<!-- O -->I2h2u−<!-- - -->u1≤<!-- = -->Ch2u3,∀<!-- ? -->u∈<!-- ? -->H3Ω<!-- O -->I2h2uh1≤<!-- = -->Cuh1,∀<!-- ? -->uh∈<!-- ? -->Mh $$\begin{array}{} \displaystyle \left\{ {\begin{array}{*{20}{c}} {I_{2h}^2{I_h}u = I_{2h}^2u,\forall u \in {H^2}\left( \Omega \right)}\\ {{{\left\| {I_{2h}^2u - u} \right\|}_1} \le C{h^2}{{\left\| u \right\|}_3},\forall u \in {H^3}\left( \Omega \right)}\\ {{{\left\| {I_{2h}^2{u_h}} \right\|}_1} \le C{{\left\| {{u_h}} \right\|}_1},\forall {u_h} \in {M_h}} \end{array}} \right. \end{array}$$(57)

I2h2Ihu=I2h2u,∀<!-- ? -->u∈<!-- ? -->H2Ω<!-- O -->I2h2u−<!-- - -->u1≤<!-- = -->Ch2u3,∀<!-- ? -->u∈<!-- ? -->H3Ω<!-- O -->I2h2uh1≤<!-- = -->Cuh1,∀<!-- ? -->uh∈<!-- ? -->Mh $$\begin{array}{} \begin{array}{} \displaystyle \left\{ {\begin{array}{*{20}{c}} {I_{2h}^2{I_h}u = I_{2h}^2u,\forall u \in {H^2}\left( \Omega \right)}\\ {{{\left\| {I_{2h}^2u - u} \right\|}_1} \le C{h^2}{{\left\| u \right\|}_3},\forall u \in {H^3}\left( \Omega \right)}\\ {{{\left\| {I_{2h}^2{u_h}} \right\|}_1} \le C{{\left\| {{u_h}} \right\|}_1},\forall {u_h} \in {M_h}} \end{array}} \right. \end{array} \end{array}$$(58)

Combining theorem 1 and the properties of operators above, the global super-convergence results can be obtained as follows.