Superconvergent Postprocessing of the Continuous Galerkin Time Stepping Method for Nonlinear Initial Value Problems with Application to Parabolic Problems

Abstract

We introduce and analyze a very simple but efficient postprocessing technique for improving the accuracy of the continuous Galerkin (CG) time stepping method for nonlinear first-order initial value problems with smooth and singular solutions. The key idea of the postprocessing technique is to add a higher order Lobatto polynomial of degree $k+1$ to the CG solution of degree k. We first establish global superconvergent error bounds for the postprocessed CG approximations over arbitrary time partitions, and as a by-product, it is shown that the convergence rates of the $L^2$-, $H^1$- and $L^\infty$-estimates for the CG method with quasi-uniform meshes for smooth solutions are improved by one order. Moreover, for solutions with initial singularities, we prove that the optimal global error estimates and nodal superconvergent estimate can be obtained for the CG method with graded meshes, and after postprocessing, the convergence rates of the $L^2$-, $H^1$- and $L^\infty$-estimates are also improved by one order. As an application, we apply the superconvergent postprocessing technique to the CG time discretization of nonlinear parabolic equations. Numerical examples are presented to verify the theoretical results.

Appendix A. Some Proofs

1.1 A.1. Proof of Lemma 2.5

Proof

We generalize the idea in [5] (where nodal superconvergence of the DG method has been studied) to the CG method. For convenience, we set $e:=u-U$. By (1.1) and (2.2), we have

$$\begin{aligned} \int _{I_n} e'\varphi dt=\int _{I_n}(f(t,u(t))-f(t,U(t))) \varphi dt,\quad \forall \varphi \in P_{k-1}(I_n). \end{aligned}$$

(A.1)

Assume that there is a positive constant M such that $|f_{uu}(t, u)|\le M$ on ${\bar{I}}\times {\mathbb {R}}$. Using the classical Taylor’s theorem with Lagrange remainder in the variable u, there exists a function $\chi$, whose value $\chi (t)$ at t is between u(t) and U(t), such that

$$\begin{aligned} f(t,u)-f(t,U)=\theta e +Re^2, \quad t\in I, \end{aligned}$$

(A.2)

where

$$\begin{aligned} \theta (t):=f_u(t,u(t))\quad \text{ and } \quad R(t):=-\frac{f_{uu}(t,\chi (t))}{2}. \end{aligned}$$

(A.3)

Clearly, there holds

$$\begin{aligned} |R|\le \frac{M}{2}, \quad t\in I. \end{aligned}$$

(A.4)

Inserting (A.2) into (A.1), yields

$$\begin{aligned} \int _{I_n} (e'-\theta e)\varphi dt={\int _{I_n}Re^2\varphi dt} ,\quad \forall \varphi \in P_{k-1}(I_n). \end{aligned}$$

(A.5)

We next construct an auxiliary problem: find v such that

$$\begin{aligned} \left\{ \begin{aligned}&v'+\theta v=0,\quad t\in [0,t_n),\\&v(t_n)=e(t_n). \end{aligned} \right. \end{aligned}$$

(A.6)

Obviously, the exact solution of (A.6) can be expressed as

$$\begin{aligned} v (t)=e(t_n) \exp \left( \int _{t}^{t_n}\theta (y)dy\right) {:=e(t_n)w(t),} \end{aligned}$$

where the function $w(t)=\exp \left( \int _{t}^{t_n}\theta (y)dy\right)$ with $\theta (y)= f_u(y,u(y))$. Assume that $f_u(t,u(t))\in C^{k-1}({\bar{I}})$. Then we have $w \in C^{k}([0,t_n))$, which implies that

$$\begin{aligned} \Vert v\Vert _{H^{k}(0,t_n)}\le C|e(t_n)|. \end{aligned}$$

(A.7)

Using integration by parts, the fact $e(0)=0$ and (A.6), gives

$$\begin{aligned} \int _{0}^{t_n}(e'-\theta e)v dt=e(t_n)v(t_n)-e(0)v(0)-\int _{0}^{t_n}(v '+\theta v ) e dt=e^2(t_n). \end{aligned}$$

(A.8)

For any function $w\in L^2(I_n)$, we denote by $\pi _{I_n}^{k-1,*}w\in P_{k-1}(I_n)$ the usual $L^2$-projection of w onto $P_{k-1}(I_n)$, namely,

$$\begin{aligned} \displaystyle \int _{I_n} (w- \pi _{I_n}^{k-1,*}w)\varphi dt =0, \quad \forall \varphi \in P_{k-1}(I_n) \end{aligned}$$

(A.9)

for $k \ge 1$. Then, for $w\in H^k(I_n)$ there holds (cf. [24])

$$\begin{aligned} \Vert w-\pi ^{k-1,*}_{I_n} w\Vert _{L^2(I_n)}\le C h_n^k\Vert w\Vert _{H^k(I_n)}. \end{aligned}$$

(A.10)

Using (A.5), (A.4), the Cauchy–Schwarz inequality and $L^2$-stability of the $L^2$-projection operator $\pi _{I_n}^{k-1,*}$, we get

$$\begin{aligned} \begin{aligned} \int _{0}^{t_n}(e'-\theta e)v dt&=\displaystyle \sum _{i=1}^n \int _{I_i}(e'-\theta e)(v-\pi ^{k-1,*}_{I_i} v) dt{+\sum _{i=1}^n \int _{I_i}Re^2 \pi ^{k-1,*}_{I_i} v dt}\\&\le \displaystyle \sum _{i=1}^n \Vert e'-\theta e\Vert _{L^2(I_i)}\Vert v-\pi ^{k-1,*}_{I_i} v\Vert _{L^2(I_i)}{+C\sum _{i=1}^n \Vert e^2\Vert _{L^2(I_i)} \Vert \pi ^{k-1,*}_{I_i} v\Vert _{L^2(I_i)} }\\&{ \le C\Vert e\Vert _{H^1(0,t_n)}\left( \displaystyle \sum _{i=1}^n \Vert v-\pi ^{k-1,*}_{I_i} v\Vert ^2_{L^2(I_i)}\right) ^{\frac{1}{2}}+C \Vert e^2\Vert _{L^2(0,t_n)} \Vert v\Vert _{L^2(0,t_n)} }. \end{aligned} \end{aligned}$$

(A.11)

Noticing the well-known embedding inequality, namely, $\Vert e\Vert _{L^\infty (0,t_n)}\le C \Vert e\Vert _{H^1(0,t_n)}$, we obtain

$$\begin{aligned} \Vert e^2\Vert _{L^2(0,t_n)}=\left( \int _{0}^{t_n}e^4dt\right) ^{\frac{1}{2}}\le \Vert e\Vert _{L^\infty (0,t_n)}\left( \int _{0}^{t_n}e^2dt\right) ^{\frac{1}{2}}\le C\Vert e\Vert _{H^1(0,t_n)}\Vert e\Vert _{L^2(0,t_n)}. \end{aligned}$$

(A.12)

Combining (A.8), (A.11), (A.10), (A.12), (2.23) and (A.7), yields

$$\begin{aligned} \begin{aligned} |e(t_n)|^4&\le C\Vert e\Vert _{H^1(0,t_n)}^2\sum _{i=1}^{n}h_i^{2k}\Vert v\Vert ^2_{H^{k}(I_i)}{+C \Vert e \Vert ^2_{H^1(0,t_n)}\Vert e\Vert ^2_{L^2(0,t_n)} \Vert v\Vert ^2_{L^2(0,t_n)} }\\&\le C h^{2k}\Vert e\Vert _{H^1(I)}^2 \Vert v\Vert ^2_{H^{k}(0,t_n)}{+C h^{2k+2} \Vert e \Vert ^2_{H^1(I)} \Vert v\Vert ^2_{L^2(0,t_n)} }\\&\le C h^{2k} \Vert e\Vert _{H^1(I)}^2 |e(t_n)|^2, \end{aligned} \end{aligned}$$

which implies that

$$\begin{aligned} |e(t_n)|^2\le Ch^{2k} \Vert e\Vert ^2_{H^1(I)}. \end{aligned}$$

(A.13)

Using (A.13) and the $H^1$-error estimate (2.24) in Lemma 2.4, gives (2.29). As a direct consequence of (2.29), we get (2.30) easily. This ends the proof. $\square$

1.2 A.2. Proof of Lemma 3.6

Proof

We start by dividing the error $e:=u-U$ into two parts that can be analyzed separately:

$$\begin{aligned} e=\eta +\xi :=(u-\Pi ^k u)+(\Pi ^k u-U), \end{aligned}$$

(A.14)

where U is the CG solution given by (2.1) and $\Pi ^ku$ is the global projection defined by (2.16). Since Lemma 2.2 can be used to bound $\eta$, the main task reduces to estimate the term $\xi$.

We first prove (3.4). Since $e=\eta +\xi$, using (A.1) we get

$$\begin{aligned} \int _{I_n}\eta '\varphi dt+\int _{I_n}\xi '\varphi dt={\int _{I_n}(f(t,u(t)-f(t,U(t))\varphi dt},\quad \forall \varphi \in P_{k-1}(I_n). \end{aligned}$$

In view of (2.15), we have

$$\begin{aligned} \int _{I_n}\xi '\varphi dt ={ \int _{I_n}(f(t,u(t)-f(t,U(t))\varphi dt}, \quad \forall \varphi \in P_{k-1}(I_n). \end{aligned}$$

(A.15)

Selecting $\varphi =\xi '$ in (A.15), then using (1.2) and the Cauchy–Schwarz inequality, we obtain

$$\begin{aligned} \begin{aligned} \int _{I_n}|\xi '|^2 dt \le { L\int _{I_n}|e\xi '| dt } \le { L}\Vert e\Vert _{L^2(I_n)}\Vert \xi '\Vert _{L^2(I_n)}, \end{aligned} \end{aligned}$$

which implies that

$$\begin{aligned} \Vert \xi ' \Vert _{L^2(I_n)} \le { L}\Vert e\Vert _{L^2(I_n)}. \end{aligned}$$

(A.16)

Summing up (A.16) over all element $I_n,\ 1\le n\le N$, yields

$$\begin{aligned} \Vert \xi '\Vert ^2_{L^2(I)}=\sum _{n=1}^{N}\Vert \xi ' \Vert ^2_{L^2(I_n)}\le { L^2}\sum _{n=1}^{N}\Vert e\Vert ^2_{L^2(I_n)}={ L^2}\Vert e\Vert ^2_{L^2(I)}, \end{aligned}$$

(A.17)

which together with (2.23) leads to (3.4).

We next prove (3.5). Since

$$\begin{aligned} \xi (t)=\xi (t_n)-\int _{t}^{t_ n}\xi '(s) ds, \quad t\in [t_{n-1},t_n], \end{aligned}$$

then using the Cauchy–Schwarz inequality, gives

$$\begin{aligned} \begin{aligned} |\xi (t)|^2=\left( \xi (t_n)-\int _{t}^{t_n}\xi '(s) ds\right) ^2 \le 2\xi ^2(t_n)+2h_n\Vert \xi '\Vert _{L^2(I_n)}^2. \end{aligned} \end{aligned}$$

(A.18)

Integrating (A.18) with respect to t on $I_n$, yields

$$\begin{aligned} \Vert \xi \Vert _{L^2(I_n)}^2\le \int _{I_n}2|\xi (t_n)|^2dt+\int _{I_n}2h_n\Vert \xi '\Vert _{L^2(I_n)}^2 dt \le 2h_n|\xi (t_n)|^2+2h_n^2\Vert \xi '\Vert _{L^2(I_n)}^2. \end{aligned}$$

(A.19)

From (2.15), (2.16) and (A.13), we find that

$$\begin{aligned} |\xi (t_n)|^2=|\Pi ^k u(t_n)-U(t_n)|^2=|u(t_n)-U(t_n)|^2=|e(t_n)|^2 \le Ch^{2k} \Vert e\Vert ^2_{H^1(I)}, \end{aligned}$$

(A.20)

Then, summing up (A.19) over all elements $I_n,\ 1\le n\le N$, using (A.20) and (A.16), we obtain

$$\begin{aligned} \begin{aligned} \Vert \xi \Vert ^2_{L^2(I)}&\le 2\sum _{n=1}^{N}h_n|\xi (t_n)|^2+ 2\sum _{n=1}^{N} h_n^2\Vert \xi '\Vert _{L^2(I_n)}^2\\&\le C\sum _{n=1}^{N}\left( h_n h^{2k} \Vert e\Vert ^2_{H^1(I)}\right) +C\sum _{n=1}^{N}\left( h_n^2\Vert e\Vert ^2_{L^2(I_n)}\right) \\&\le C h^{2k}\Vert e\Vert ^2_{H^1(I)} \sum _{n=1}^{N}h_n +Ch^2 \sum _{n=1}^{N}\Vert e\Vert ^2_{L^2(I_n)}\\&\le Ch^{2k}\Vert e\Vert ^2_{H^1(I)}+Ch^2\Vert e\Vert ^2_{L^2(I)}. \end{aligned} \end{aligned}$$

(A.21)

Inserting (2.23) and (2.24) into (A.21), gives (3.5).

It remains to prove (3.6). According to the Sobolev inequality (see (3.9) of [13])

$$\begin{aligned} \Vert v\Vert ^2_{L^{\infty }(a,b)}\le \frac{2}{b-a}\Vert v\Vert ^2_{L^2(a,b)}+2(b-a)\Vert v'\Vert ^2_{L^2(a,b)},\quad \forall v\in H^1(a,b), \end{aligned}$$

(A.22)

we have

$$\begin{aligned} \Vert \xi \Vert ^2_{L^{\infty }(I_n)}\le \frac{2}{h_n}\Vert \xi \Vert ^2_{L^2(I_n)}+2h_n\Vert \xi '\Vert ^2_{L^2(I_n)}. \end{aligned}$$

(A.23)

Inserting (A.19) into (A.23), then using (A.20) and (A.16) we get

$$\begin{aligned} \Vert \xi \Vert ^2_{L^{\infty }(I_n)} \le C |\xi (t_n)|^2+C h_n\Vert \xi '\Vert _{L^2(I_n)}^2 \le Ch^{2k} \Vert e\Vert ^2_{H^1(I)}+Ch_n\Vert e\Vert _{L^2(I_n)}^2. \end{aligned}$$

(A.24)

Noting the fact that (see pages 532–533 of [26])

$$\begin{aligned} \Vert e\Vert ^2_{L^2(I_n)}\le Ch_n\Vert u-\Pi ^k u\Vert ^2_{L^2(0,t_n)}+C\Vert u-\Pi ^k u\Vert ^2_{L^2(I_n)}, \end{aligned}$$

(A.25)

and using (A.24), we get

$$\begin{aligned} \Vert \xi \Vert ^2_{L^{\infty }(I_n)}\le Ch^{2k} \Vert e\Vert ^2_{H^1(I)}+Ch_n^2\Vert u-\Pi ^k u\Vert ^2_{L^2(I)}+Ch_n\Vert u-\Pi ^k u\Vert ^2_{L^2(I_n)}. \end{aligned}$$

(A.26)

Consequently, using (2.17), (2.24) and the fact $\Vert u\Vert ^2_{H^{k+1}(I_n)} \le C h_n \Vert u\Vert ^2_{W^{k+1,\infty }(I_n)}$, yields

$$\begin{aligned} \begin{aligned} \Vert \xi \Vert ^2_{L^{\infty }(I)}&\le Ch^{2k} \Vert e\Vert ^2_{H^1(I)}+Ch^2\Vert u-\Pi ^k u\Vert ^2_{L^2(I)}+C\max _{1\le n\le N} \left\{ h_n\Vert u-\Pi ^k u\Vert ^2_{L^2(I_n)}\right\} \\&\le Ch^{2k} \sum _{n=1}^{N}h_n^{2k}\Vert u\Vert ^2_{H^{k+1}(I_n)}+ Ch^2 \sum _{n=1}^{N}h_n^{2k+2}\Vert u\Vert ^2_{H^{k+1}(I_n)}\\&\quad + C\max \limits _{1 \le n\le N}\left\{ h_n^{2k+3}\Vert u\Vert ^2_{H^{k+1}(I_n)}\right\} \\&\le Ch^{2k} \sum _{n=1}^{N}h_n^{2k+1}\Vert u\Vert ^2_{W^{k+1,\infty }(I_n)}+ Ch^2 \sum _{n=1}^{N}h_n^{2k+3}\Vert u\Vert ^2_{W^{k+1,\infty }(I_n)}\\&\quad + C\max \limits _{1 \le n\le N}\left\{ h_n^{2k+4}\Vert u\Vert ^2_{W^{k+1,\infty }(I_n)}\right\} . \end{aligned} \end{aligned}$$

This completes the proof of (3.6).

Finally, using (3.4)–(3.6) and the quasi-uniformity of the time partition, we obtain (3.7)–(3.9). $\square$

1.3 A.3. Proof of Theorem 3.2

Proof

We proceed in several steps.

Step 1 We first prove (3.30). Due to (2.20) and (2.16), there holds

$$\begin{aligned} \Vert u-U\Vert ^2_{L^2(I)} \le C\Vert u-\pi ^k_{I_1}u\Vert ^2_{L^2(I_1)}+C\sum _{n=2}^{N}\Vert u-\pi ^k_{I_n}u\Vert ^2_{L^2(I_n)}:=A_1+A_2. \end{aligned}$$

(A.27)

By (3.27), we have

$$\begin{aligned} \Vert u\Vert ^2_{H^1(I_1)}\le C\int _{0}^{t_1}\left( t^{2\sigma }+t^{2(\sigma -1)}\right) dt \le C t_1^{2\sigma -1} \end{aligned}$$

(A.28)

for $\sigma >\frac{1}{2}$. Then, using (2.17) with $s_1=0$, (A.28) and (3.28), we get

$$\begin{aligned} A_1=C\Vert u-\pi ^k_{I_1}u\Vert ^2_{L^2(I_1)}\le Ch_1^2\Vert u\Vert ^2_{H^1(I_1)} \le Ch_1^2t_1^{2\sigma -1} \le C\tau ^{(2\sigma +1)\beta }. \end{aligned}$$

(A.29)

Moreover, using (2.17) with $s_n=k$, (3.27) and (3.28), yields

$$\begin{aligned} A_2&=C\sum _{n=2}^{N}\Vert u-\pi ^k_{I_n}u\Vert ^2_{L^2(I_n)} \le C\sum _{n=2}^{N}h_n^{2k+2}\Vert u\Vert ^2_{H^{k+1}(I_n)}\nonumber \\&\le C\sum _{n=2}^{N}\tau ^{2k+2}t_n^{(2k+2)(1-\frac{1}{\beta })} \int _{t_{n-1}}^{t_n}\left( t^{2\sigma }+t^{2(\sigma -1)}+\cdots +t^{2(\sigma -k-1)}\right) dt \nonumber \\&\le C\tau ^{2k+2}\sum _{n=2}^{N}t_n^{(2k+2)(1-\frac{1}{\beta })}t_{n-1}^{-(2k+2)(1-\frac{1}{\beta })} \int _{t_{n-1}}^{t_n} t^{(2k+2)(1-\frac{1}{\beta })}\left( t^{2\sigma }+t^{2(\sigma -1)}+\cdots +t^{2(\sigma -k-1)}\right) dt\nonumber \\&\le C\tau ^{2k+2}\int _{t_1}^{T}\left( t^{2\sigma -\frac{2k+2}{\beta }}t^{2k+2}+t^{2\sigma -\frac{2k+2}{\beta }}t^{2k}+\cdots +t^{2\sigma -\frac{2k+2}{\beta }}\right) dt \nonumber \\&\le C\tau ^{2k+2}\int _{t_1}^{T}\left( t^{2\sigma -\frac{2k+2}{\beta }}T^{2k+2}+t^{2\sigma -\frac{2k+2}{\beta }}T^{2k}+\cdots +t^{2\sigma -\frac{2k+2}{\beta }}\right) dt \nonumber \\&\le C\tau ^{2k+2}\int _{t_1}^{T}t^{2\sigma -\frac{2k+2}{\beta }} dt. \end{aligned}$$

(A.30)

Suppose that $2\sigma -\frac{2k+2}{\beta }>-1$, i.e., $\beta >\frac{k+1}{\sigma +1/2}$. From (A.29) and (A.30), we obtain

$$\begin{aligned}{} & {} A_1 \le C\tau ^{(2\sigma +1)\beta } \le C\tau ^{2k+2}, \end{aligned}$$

(A.31)

$$\begin{aligned}{} & {} A_2 \le C\tau ^{2k+2}\int _{t_1}^{T}t^{2\sigma -\frac{2k+2}{\beta }} dt \le C\tau ^{2k+2}T^{2\sigma +1-\frac{2k+2}{\beta }}\le C\tau ^{2k+2}. \end{aligned}$$

(A.32)

Combining (A.27), (A.31) and (A.32) gives (3.30).

Step 2 We next prove (3.31). Due to (2.21) and (2.16), there holds

$$\begin{aligned} \Vert u-U\Vert ^2_{H^1(I)} \le C\Vert u-\pi ^k_{I_1}u\Vert ^2_{H^1(I_1)}+C\sum _{n=2}^{N}\Vert u-\pi ^k_{I_n}u\Vert ^2_{H^1(I_n)}:=B_1+B_2. \end{aligned}$$

(A.33)

Using (2.18) with $s_1=0$, (A.28) and (3.28), we get

$$\begin{aligned} B_1=C\Vert u-\pi ^k_{I_1}u\Vert ^2_{H^1(I_1)} \le C\Vert u\Vert ^2_{H^1(I_1)}\le Ct_1^{2\sigma -1} \le C\tau ^{(2\sigma -1)\beta }. \end{aligned}$$

(A.34)

Moreover, using (2.18) with $s_n=k$, (3.27) and (3.28), gives

$$\begin{aligned} B_2&=\sum _{n=2}^{N}\Vert u-\pi ^k_{I_n}u\Vert ^2_{H^1(I_n)} \le C\sum _{n=2}^{N}h_n^{2k}\Vert u\Vert ^2_{H^{k+1}(I_n)}\nonumber \\&\le C\sum _{n=2}^{N}\tau ^{2k}t_n^{2k(1-\frac{1}{\beta })} \int _{t_{n-1}}^{t_n}\left( t^{2\sigma }+t^{2(\sigma -1)}+\cdots +t^{2(\sigma -k-1)}\right) dt\nonumber \\&\le C\sum _{n=2}^{N}\tau ^{2k}t_n^{2k(1-\frac{1}{\beta })}t_{n-1}^{-2k(1-\frac{1}{\beta })} \int _{t_{n-1}}^{t_n} t^{2k(1-\frac{1}{\beta })} \left( t^{2\sigma }+t^{2(\sigma -1)}+\cdots +t^{2(\sigma -k-1)}\right) dt\nonumber \\&\le C\tau ^{2k}\int _{t_1}^{T}\left( t^{2\sigma -2-\frac{2k}{\beta }}t^{2k+2}\right. \left. +t^{2\sigma -2-\frac{2k}{\beta }}t^{2k}+\cdots +t^{2\sigma -2-\frac{2k}{\beta }}\right) dt\nonumber \\&\le C\tau ^{2k}\int _{t_1}^{T}t^{2\sigma -2-\frac{2k}{\beta }} dt. \end{aligned}$$

(A.35)

Suppose that $2\sigma -2-\frac{2k}{\beta }>-1$, i.e., $\beta >\frac{k}{\sigma -1/2}$. From (A.34) and (A.35), we have

$$\begin{aligned}{} & {} B_1 \le C\tau ^{(2\sigma -1)\beta } \le C\tau ^{2k}, \end{aligned}$$

(A.36)

$$\begin{aligned}{} & {} B_2 \le C\tau ^{2k}\int _{t_1}^{T}t^{2\sigma -2-\frac{2k}{\beta }} dt \le C\tau ^{2k}T^{2\sigma -1-\frac{2k}{\beta }}\le C\tau ^{2k}. \end{aligned}$$

(A.37)

Combining (A.33), (A.36) and (A.37) gives (3.31).

Step 3 We further show (3.32). In view of Lemma 2.3, we have

$$\begin{aligned} \Vert u-U\Vert ^2_{L^\infty (I)}\le C\Vert u-\Pi ^k u\Vert ^2_{L^\infty (I)}\le C \max _{1\le n\le N} \left\{ \Vert u-\pi ^k_{I_n}u\Vert ^2_{L^\infty (I_n)}\right\} . \end{aligned}$$

(A.38)

On the one hand, using the Sobolev inequality (A.22), (3.28), (A.29) and (A.34), we get

$$\begin{aligned} \Vert u-\pi ^k_{I_1}u\Vert ^2_{L^\infty (I_1)} \le \frac{2}{h_1}\Vert u-\pi ^k_{I_1}u\Vert ^2_{L^2(I_1)}+2h_1\Vert u-\pi ^k_{I_1}u\Vert ^2_{H^1(I_1)} \le C\tau ^{2\sigma \beta }. \end{aligned}$$

(A.39)

On the other hand, using (2.19) with $s_n=k$, (3.27) and (3.28), we obtain

$$\begin{aligned} \begin{aligned} \Vert u-\pi ^k_{I_n}u\Vert ^2_{L^\infty (I_n)}&\le Ch_n^{2k+2}\Vert u\Vert ^2_{W^{k+1,\infty }(I_n)} \\&\le C\tau ^{2k+2}t_n^{(2k+2)(1-\frac{1}{\beta })} \max _{t\in {\bar{I}}_n}\left\{ t^{2\sigma }, t^{2(\sigma -1)},\cdots , t^{2(\sigma -k-1)}\right\} \\&\le C\tau ^{2k+2}t_n^{(2k+2)(1-\frac{1}{\beta })}t_{n-1}^{-(2k+2)(1-\frac{1}{\beta })} \max _{t\in {\bar{I}}_n} \left\{ t^{2\sigma -\frac{2k+2}{\beta }+2k+2}, \cdots , t^{2\sigma -\frac{2k+2}{\beta }}\right\} \\&\le C\tau ^{2k+2} \max _{t_1\le t\le T}\left\{ t^{2\sigma -\frac{2k+2}{\beta }}t^{2k+2},t^{2\sigma -\frac{2k+2}{\beta }}t^{2k},\cdots , t^{2\sigma -\frac{2k+2}{\beta }}\right\} \\&\le C\tau ^{2k+2}\max _{t_1\le t\le T}\left\{ t^{2\sigma -\frac{2k+2}{\beta }}\right\} \end{aligned} \end{aligned}$$

(A.40)

for $2\le n\le N$. Suppose that $2\sigma -\frac{2k+2}{\beta }\ge 0$, i.e., $\beta \ge \frac{k+1}{\sigma }$. From (A.39) and (A.40), we have

$$\begin{aligned}{} & {} \Vert u-\pi ^k_{I_1}u\Vert ^2_{L^\infty (I_1)} \le C\tau ^{2\sigma \beta }\le C\tau ^{2k+2}, \end{aligned}$$

(A.41)

$$\begin{aligned}{} & {} \Vert u-\pi ^k_{I_n}u\Vert ^2_{L^\infty (I_n)} \le C\tau ^{2k+2}T^{2\sigma -\frac{2k+2}{\beta }}\le C\tau ^{2k+2}, \quad 2\le n\le N. \end{aligned}$$

(A.42)

Combining (A.38), (A.41) and (A.42) gives (3.32).

Step 4 It remains to show (3.33). Similar to the proof of (2.30) for regular solutions, we consider the same auxiliary problem (A.6) with exact solution

$$\begin{aligned} v (t)= e(t_n) \exp \left( \int _{t}^{t_n}\theta (y)dy\right) :=e(t_n)w(t). \end{aligned}$$

(A.43)

Here, the function $w(t)=\exp \left( \int _{t}^{t_n}\theta (y)dy\right)$ with $\theta (y)= {f_u(y,u(y))}$. Assume that $f_u(t,u(t))\in C({\bar{I}})$. In view of the definition of the function $\theta$, we may assume that the solution v of the auxiliary problem (A.6) exhibits the similar regularity as the solution of the original problem (1.1), namely,

$$\begin{aligned} |v^{(s)}(t)|= |e(t_n)w^{(s)}(t)|\le C|e(t_n)|t^{\sigma -s},\quad t\in (0,T],\quad s\in {\mathbb {N}}_0,\quad {\sigma \ge 1} \end{aligned}$$

(A.44)

for some positive constants C.

Combining (A.8), (A.43) and (A.11), gives

$$\begin{aligned} \begin{aligned} e^2(t_n)&=\int _{0}^{t_n}(e'-\theta e)v dt = e(t_n)\displaystyle \sum _{i=1}^n \int _{I_i}(e'-\theta e)(w-\pi ^{k-1,*}_{I_i} w) dt\\&\quad {+e(t_n)\displaystyle \sum _{i=1}^n \int _{I_i}Re^2 \pi ^{k-1,*}_{I_i} w dt}\\&\le C|e(t_n)| \cdot \Vert e\Vert _{H^1(0,t_n)}\left( \displaystyle \sum _{i=1}^n \Vert w-\pi ^{k-1,*}_{I_i} w\Vert ^2_{L^2(I_i)}\right) ^{\frac{1}{2}}\\&\quad {+ C|e(t_n)|\Vert e^2\Vert _{L^2(0,t_n)}\Vert w\Vert _{L^2(0,t_n)}}, \end{aligned} \end{aligned}$$

which together with (A.12) leads to

$$\begin{aligned} |e(t_n)| \le C \Vert e\Vert _{H^1(0,t_n)}\left( \displaystyle \sum _{i=1}^n \Vert w-\pi ^{k-1,*}_{I_i} w\Vert ^2_{L^2(I_i)}\right) ^{\frac{1}{2}}{+ C \Vert e \Vert _{L^2(0,t_n)}\Vert e \Vert _{H^1(0,t_n)}\Vert w\Vert _{L^2(0,t_n)}}. \end{aligned}$$

(A.45)

Here, $\pi _{I_n}^{k-1,*}w$ is the $L^2$-projection of w onto $P_{k-1}(I_n)$ as defined in (A.9).

Using the $L^2$-stability of the projection operator $\pi _{I_n}^{k-1,*}$, (A.44) and (3.28), we have

$$\begin{aligned} \Vert w-\pi ^{k-1,*}_{I_1} w\Vert ^2_{L^2(I_1)}\le 4\Vert w\Vert ^2_{L^2(I_1)} \le C\int _{0}^{t_1} t^{2\sigma } dt \le C h_1^{2\sigma +1} \le C \tau ^{\beta (2\sigma +1)} {\le C\tau ^{2k}} \end{aligned}$$

(A.46)

for $\beta >\frac{k}{\sigma +1/2}$. Similarly to (A.30), using (A.10), (A.44) and (3.28), we have

$$\begin{aligned} \sum _{i=2}^{n}\Vert w-\pi ^{k-1,*}_{I_1} w\Vert ^2_{L^2(I_i)}\le & {} C\sum _{i=2}^{n}h_i^{2k}\Vert w\Vert ^2_{H^{k}(I_i)} \le C\tau ^{2k}\int _{t_1}^{t_n}t^{2\sigma -\frac{2k}{\beta }} dt\nonumber \\\le & {} C \tau ^{2k}\int _{t_1}^{T}t^{2\sigma -\frac{2k}{\beta }} dt {\le C\tau ^{2k}} \end{aligned}$$

(A.47)

for $\beta >\frac{k}{\sigma +1/2}$. In view of (A.46) and (A.44), we get

$$\begin{aligned} \Vert w\Vert ^2_{L^2(0,t_n)}= \Vert w\Vert ^2_{L^2(I_1)}+\sum _{i=2}^{n}\Vert w\Vert ^2_{L^2(I_i)}\le C\tau ^{\beta (2\sigma +1)} +\int _{t_1}^{t_n}t^{2\sigma }dt\le CT^{2\sigma +1}\le C. \end{aligned}$$

(A.48)

Inserting (A.46)–(A.48) into (A.45), gives

$$\begin{aligned} |e(t_n)| \le C \tau ^{k} \Vert e\Vert _{H^1(I)} + { C \Vert e \Vert _{L^2(I)}\Vert e \Vert _{H^1(I)}} \end{aligned}$$

for $\beta >\frac{k}{\sigma +1/2}$, which together with (3.31) and (3.30) leads to

$$\begin{aligned} |e(t_n)| \le { C \tau ^{2k}+C\tau ^{2k+1}} \le C \tau ^{2k} \end{aligned}$$

for $\beta > \max \{\frac{k+1}{\sigma +1/2}, \frac{k}{\sigma -1/2}\}$. This proves (3.33). $\square$

1.4 A.4. Proof of Lemma 3.7

Proof

We first recall the error splitting (A.14) that $e=\xi +\eta$, where $\eta =u-\Pi ^k u$ and $\xi =\Pi ^k u-U$. In view of (A.17) and (3.30), there holds

$$\begin{aligned} \Vert \xi '\Vert _{L^2(I)} \le C\Vert e\Vert _{L^2(I)}\le C\tau ^{k+1}, \quad \text{ if }~~ \beta >\frac{k+1}{\sigma +1/2}. \end{aligned}$$

This proves (3.34).

We next show (3.35). Noting the facts that $\xi (t_n)=e(t_n)$ and $h_n\le C \tau$, then using (A.21), (3.33) and (3.34), we have

$$\begin{aligned} \begin{aligned} \Vert \xi \Vert ^2_{L^2(I)}&\le 2\sum _{n=1}^{N}h_n|\xi (t_n)|^2+ 2\sum _{n=1}^{N} h_n^2\Vert \xi '\Vert _{L^2(I_n)}^2\\&\le 2 \max _{1\le n\le N}|e(t_n)|^2 \sum _{n=1}^{N}h_n +C \tau ^2 \Vert \xi '\Vert ^2_{L^2(I)}\\&\le C\tau ^{4k} +C\tau ^{2k+4}\\&\le C\tau ^{\min \{4k, 2k+4\}} \end{aligned} \end{aligned}$$

for $\beta > \max \{\frac{k+1}{\sigma +1/2}, \frac{k}{\sigma -1/2}\}$, which implies (3.35).

It remains to prove (3.36). Using again the facts that $\xi (t_n)=e(t_n)$ and $h_n\le C \tau$, combining (A.24), (3.33), (A.16), (A.25), (A.27), (A.31) and (A.32), results in

$$\begin{aligned} \begin{aligned} \Vert \xi \Vert ^2_{L^{\infty }(I_n)}&\le C |\xi (t_n)|^2+C h_n\Vert \xi '\Vert _{L^2(I_n)}^2 \le C\tau ^{4k} +Ch_n\Vert e\Vert _{L^2(I_n)}^2\\&\le C \tau ^{4k}+Ch_n^2\Vert u-\Pi ^k u\Vert ^2_{L^2(I)}+Ch_n\Vert u-\pi ^k_{I_n} u\Vert ^2_{L^2(I_n)}\\&\le C \tau ^{4k}+C\tau ^{2k+4}+Ch_n\Vert u-\pi ^k_{I_n} u\Vert ^2_{L^2(I_n)} \end{aligned} \end{aligned}$$

for $\beta > \max \{\frac{k+1}{\sigma +1/2}, \frac{k}{\sigma -1/2}\}$. Therefore, we have

$$\begin{aligned} \Vert \xi \Vert ^2_{L^{\infty }(I)}\le \max _{1\le n\le N}\left\{ \Vert \xi \Vert ^2_{L^{\infty }(I_n)} \right\} \le C \tau ^{4k}+C\tau ^{2k+4}+ C\max _{1\le n\le N}\left\{ h_n\Vert u-\pi ^k_{I_n} u\Vert ^2_{L^2(I_n)}\right\} \end{aligned}$$

(A.49)

for $\beta > \max \{\frac{k+1}{\sigma +1/2}, \frac{k}{\sigma -1/2}\}$.

For $n=1$, using (2.17) with $s_1=0$, (A.28) and (3.28), gives

$$\begin{aligned} h_1\Vert u-\pi ^{k}_{I_1} u\Vert ^2_{L^2(I_1)}\le Ch_1^3\Vert u\Vert ^2_{H^1(I_1)}\le Ch_1^3t_1^{2\sigma -1}\le C\tau ^{(2\sigma +2)\beta } \le C\tau ^{2k+4} \end{aligned}$$

(A.50)

for $\beta \ge \frac{k+2}{\sigma +1}$. For $2\le n\le N$, using (2.17) with $s_n=k$, the fact $\Vert u\Vert ^2_{H^{k+1}(I_n)} \le C h_n \Vert u\Vert ^2_{W^{k+1,\infty }(I_n)}$, (3.27) and (3.28), results in

$$\begin{aligned} \begin{aligned}&h_n\Vert u-\pi ^{k}_{I_n} u\Vert ^2_{L^2(I_n)}\\&\quad \le C h_n^{2k+3}\Vert u\Vert ^2_{H^{k+1}(I_n)} \le C h_n^{2k+4}\Vert u\Vert ^2_{W^{k+1,\infty }(I_n)} \\&\quad \le C\tau ^{2k+4} t_n^{(2k+4)(1-\frac{1}{\beta })} \max _{t\in {\bar{I}}_n}\left\{ t^{2\sigma }, t^{2(\sigma -1)},\cdots , t^{2(\sigma -k-1)}\right\} \\&\quad \le C\tau ^{2k+4} t_n^{(2k+4)(1-\frac{1}{\beta })} t_{n-1}^{-(2k+4)(1-\frac{1}{\beta })} \max _{t\in {\bar{I}}_n}\left\{ t^{2\sigma -\frac{2k+4}{\beta }+2k+4}, \cdots , t^{2\sigma -\frac{2k+4}{\beta }+2}\right\} \\&\quad \le C\tau ^{2k+4} \max _{t_1 \le t \le T}\left\{ t^{2\sigma -\frac{2k+4}{\beta }+2k+4}, t^{2\sigma -\frac{2k+4}{\beta }+2k+2}, \cdots , t^{2\sigma -\frac{2k+4}{\beta }+2}\right\} \\&\quad \le C\tau ^{2k+4} \max _{t_1\le t\le T}\left\{ t^{2\sigma -\frac{2k+4}{\beta }+2}\right\} \\&\quad \le C\tau ^{2k+4} T^{2\sigma -\frac{2k+4}{\beta }+2}\\&\quad \le C\tau ^{2k+4}, \end{aligned} \end{aligned}$$

(A.51)

provided that $2\sigma -\frac{2k+4}{\beta }+2\ge 0$, i.e., $\beta \ge \frac{k+2}{\sigma +1}$.

Combining (A.49), (A.50) and (A.51), yields

$$\begin{aligned} \Vert \xi \Vert ^2_{L^{\infty }(I)} \le C \tau ^{4k}+C\tau ^{2k+4}\le C\tau ^{ \min \{4k, 2k+4\}}, \end{aligned}$$

provided that $\beta > \max \{\frac{k+2}{\sigma +1}, \frac{k+1}{\sigma +1/2}, \frac{k}{\sigma -1/2}\}$. This ends the proof of (3.36). $\square$