In this section we will state the semi-discrete Crank-Nicolson scheme, recall its well-posedness and available stability bounds, and then use these results to prove optimal \(L^{\infty }(H^1)\)-error estimates in the Hilbert space setting. For that, let T denote the final time of computation, N the number of time-steps, and \(\tau = T/N\) the time step size. By \(t_n\) we shall mean \(t_n=n\tau \). The exact solution at time \(t_n\) shall be denoted by \(u^n:=u(t_n,\cdot )\). We also introduce a short hand notation for discrete time derivatives which is \(D_\tau u^n := (u^{n+1}-u^n)/\tau \) and analogously \(D_\tau u_{\tau _{}}^{n} := ( u_{\tau _{}}^{n+1} - u_{\tau _{}}^{n})/\tau \).
Method formulation and main result
With the notation above, the semi-discrete Crank-Nicolson approximation \(u_{\tau _{}}^{n+1} \in H^1_0(\mathscr {D})\) to \(u^{n+1}\) is given recursively as the solution (in the sense of distributions) to the equation
$$\begin{aligned} \mathrm {i}D_\tau u_{\tau _{}}^{n} = - \varDelta u_{\tau _{}}^{n+\frac{1}{2}} + V u_{\tau _{}}^{n+\frac{1}{2}} + \frac{\varGamma (|u_{\tau _{}}^{n+1}|^2)-\varGamma (|u_{\tau _{}}^{n}|^2)}{|u_{\tau _{}}^{n+1}|^2-|u_{\tau _{}}^{n}|^2} \, u_{\tau _{}}^{n+\frac{1}{2}}, \end{aligned}$$
(3.1)
where \(u_{\tau _{}}^{n+\frac{1}{2}}:=(u_{\tau _{}}^{n}+u_{\tau _{}}^{n+1})/2\). The initial value is selected as \(u_{\tau _{}}^{0}=u^0\). It is easily seen that the discretization conserves both mass and energy, i.e.
$$\begin{aligned} \int _{\mathscr {D}} |u_{\tau _{}}^{n}|^2 \, dx = \int _{\mathscr {D}} |u^0|^2 \, dx \quad \hbox {and} \quad E[u_{\tau _{}}^{n}] = E[u^0] \quad \hbox {for all } n\ge 0. \end{aligned}$$
The scheme (3.1) is well-posed and admits a set of a priori error estimates. The properties are summarized in the following theorem that is proved in [35, Theorem 4.1].
Theorem 3.1
Under the general assumptions of this paper, there exists a constant \(C(u)>0\) and a solution \(u_{\tau _{}}^{n}\in H^1_0(\mathscr {D})\) to the semi-discrete Crank-Nicolson scheme (3.1) that is uniquely characterized by the property that
$$\begin{aligned} \sup _{0\le n \le N} \left( \Vert u_{\tau _{}}^{n} \Vert _{L^{\infty }(\mathscr {D})} + \Vert u_{\tau _{}}^{n} \Vert _{H^{2}(\mathscr {D})}\right) \le C(u), \end{aligned}$$
(3.2)
and the a priori estimate for the \(L^2\)-error
$$\begin{aligned} \sup _{0\le n \le N} \Vert u_{\tau _{}}^{n} - u^n \Vert \lesssim \tau _{}^2, \end{aligned}$$
where u is the (unique) exact solution with the regularity property (2.2).
Our main result on optimal error estimates in the \(L^{\infty }(H^1)\) reads as follows.
Theorem 3.2
(Optimal \(H^1\)-error estimates for the semi-discrete method) Consider the setting of Theorem 3.1, then the \(L^{\infty }(H^1)\)-error converges with optimal order in \(\tau \), i.e.
$$\begin{aligned} \sup _{0\le n \le N} \Vert u_{\tau _{}}^{n} - u^n \Vert _{H^{1}(\mathscr {D})} \lesssim \tau _{}^2. \end{aligned}$$
The theorem is proved in Sect. 3.2 below.
Proof of Theorem 3.2
In this section we will prove Theorem 3.2. Let us introduce some notation that is used throughout the proofs. We recall \(D_\tau e^n = ( e^{n+1} - e^{n})/\tau \). Furthermore, we let \(e^{n+1/2}:=(e^{n+1} + e^n)/2\) and \(u^{n+1/2}:=(u^{n+1} + u^n)/2\). For time derivatives at fixed time \(t^n\), we also write \(\partial _t u^n:= \partial _t u(t^n,\cdot )\).
We begin by establishing a differential equation for the time discrete error \(e^n = u^n-u^n_\tau \). This is stated in the following lemma.
Lemma 3.1
(Consistency error) The error \(e^n = u^n-u_\tau ^n\) fulfills the identity
$$\begin{aligned} \mathrm {i}D_\tau e^n+\varDelta e^{n+1/2}-Ve^{n+1/2}-e^n_\gamma = T^n, \end{aligned}$$
(3.3)
where the the consistency error \(T^n\) is given by
$$\begin{aligned} T^n:= & {} \mathrm {i}\,( D_\tau u^n - \partial _t u(t_{n+1/2}))+\varDelta ( u^{n+1/2}-u(t_{n+1/2})) -V(u^{n+1/2}-u(t_{n+1/2})) \nonumber \\&-\,\bigg (\gamma (\xi ^n) -\gamma (|u(t_{n+1/2})|^2) \bigg ). \end{aligned}$$
(3.4)
Here, \(e^n_\gamma := \gamma (\xi ^n)u^{n+1/2}-\gamma (\xi ^n_\tau )u^{n+1/2}_\tau \) for some bounded functions \(\xi ^n,\xi ^n_\tau \in L^{\infty }(\mathscr {D})\) with the properties that
$$\begin{aligned}&\xi ^n(x) \in [\min (|u^n|^2,|u^{n+1}|^2),\max (|u^n|^2,|u^{n+1}|^2)] \quad \hbox {and} \\&\quad \xi ^n_\tau (x) \in [\min (|u^n_\tau |^2,|u^{n+1}_\tau |^2),\max (|u^n_\tau |^2,|u^{n+1}_\tau |^2)] \end{aligned}$$
for almost all \(x\in \mathscr {D}\).
Proof
It is easily verified that exact solution fulfills
$$\begin{aligned} \mathrm {i}D_\tau u^n + \varDelta u^{n+1/2}- Vu^{n+1/2} -\frac{\varGamma (|u^{n+1}|^2)-\varGamma (|u^n|^2)}{|u^{n+1}|^2-|u^n|^2} = T^n. \end{aligned}$$
(3.5)
By the regularity assumptions we can apply Taylor expansion arguments to \(T^n\) to see:
$$\begin{aligned} \sum _{k=0}^{N}\Vert T^k\Vert ^2 \le C\tau ^3 \end{aligned}$$
(3.6)
The argument that proves (3.6) is elaborated in “Appendix A”, where it becomes also visible how the regularity assumptions enter explicitly in the estimate. Next, subtracting (3.5) from (3.1) we find that \(e^n = u^n-u_\tau ^n\) satisfies:
$$\begin{aligned}\mathrm {i}D_\tau e^n+\varDelta e^{n+1/2}-Ve^{n+1/2}-e^n_\gamma = T^n \end{aligned}$$
where \(e^n_\gamma \) denotes the error coming from the nonlinear term, defined by
$$\begin{aligned} e_\gamma ^n= \frac{\varGamma (|u^{n+1}|^2)-\varGamma (|u^n|^2)}{|u^{n+1}|^2-|u^n|^2}u^{n+1/2}-\frac{\varGamma (|u_\tau ^{n+1}|^2)-\varGamma (|u^n_\tau |^2)}{|u^{n+1}_\tau |^2-|u^n_\tau |^2} u_\tau ^{n+1/2}. \end{aligned}$$
Recalling the definition of \(\varGamma \) we have:
$$\begin{aligned} \frac{\varGamma (|u^{n+1}|^2)-\varGamma (|u^n|^2)}{|u^{n+1}|^2-|u^n|^2} = \frac{1}{|u^{n+1}|^2-|u^n|^2}\int _{|u^n|^2}^{|u^{n+1}|^2} \gamma (r)\,dr =:\gamma (\xi ^n) , \end{aligned}$$
likewise
$$\begin{aligned} \frac{\varGamma (|u^{n+1}_\tau |^2)-\varGamma (|u^n_\tau |^2)}{|u^{n+1}_\tau |^2-|u^n_\tau |^2} = \frac{1}{|u^{n+1}_\tau |^2-|u^n_\tau |^2}\int _{|u^n_\tau |^2}^{|u^{n+1}_\tau |^2} \gamma (r)\,dr =: \gamma (\xi ^n_\tau ). \end{aligned}$$
The expression for \(e^n_\gamma \) is thus simplified to
$$\begin{aligned} e^n_\gamma := \gamma (\xi ^n)u^{n+1/2}-\gamma (\xi ^n_\tau )u^{n+1/2}_\tau , \end{aligned}$$
where \(\xi ^n\) is a function taking values between \(|u^n|^2\) and \(|u^{n+1}|^2\) and \(\xi ^n_\tau \) takes values between \(|u^n_\tau |^2\) and \(|u_{\tau _{}}^{n+1}|^2\). \(\square \)
The differential equation in Lemma 3.1 is now used to derive a recurrence formula for the \(H^1\)-norm of the error. Multiplying (3.3) by \(D_\tau e^n\), integrating and taking the real part yields:
$$\begin{aligned} \frac{\Vert \nabla e^{n+1}\Vert ^2-\Vert \nabla e^n\Vert ^2}{2\tau } = \underbrace{ \text {Re}( \langle e_\gamma ^n, D_\tau e^n\rangle )}_{\mathrm{I}} \ \underbrace{-\,\text {Re}(\langle T^n,D_\tau e^n\rangle ).}_{\mathrm{II}} \end{aligned}$$
(3.7)
The idea is to bound the terms I and II in such a way that Grönwall’s inequality can be used. We proceed to bound term I. Multiplying the error PDE (3.3) by \(e^n_\gamma \) results in:
$$\begin{aligned} \mathrm {i}\langle D_\tau e^n, e_\gamma ^n\rangle = \langle \nabla e^{n+1/2},\nabla e_\gamma ^n\rangle + \langle Ve^{n+1/2},e^n_\gamma \rangle + \Vert e_\gamma ^n\Vert ^2 + \langle T^n,e_\gamma ^n\rangle \end{aligned}$$
and consequently
$$\begin{aligned} |\mathrm{I}|= & {} |\text {Re}( \langle D_\tau e^n, e_\gamma ^n\rangle )| \nonumber \\\le & {} |\text {Im}(\langle \nabla e^{n+1/2}, \nabla e_\gamma ^n)\rangle |+|\text {Im}\langle Ve^{n+1/2},e^n_\gamma \rangle |+ |\text {Im}(\langle T^n, e_\gamma ^n\rangle )| \nonumber \\\lesssim & {} \Vert \nabla e^{n+1/2} \Vert ^2+ \Vert \nabla e^n_\gamma \Vert ^2+ \Vert V\Vert _\infty (\Vert e^{n+1/2}\Vert ^2+ \Vert e^n_\gamma \Vert ^2)+\tau ^4+\Vert e^n_\gamma \Vert ^2+\Vert T^n\Vert ^2 \nonumber \\\lesssim & {} \Vert \nabla e^{n+1}\Vert ^2+\Vert \nabla e^{n}\Vert ^2+\Vert \nabla e^n_\gamma \Vert ^2+\Vert e^n_\gamma \Vert ^2+\Vert T^n\Vert ^2+\tau ^4. \end{aligned}$$
(3.8)
In order to use Grönwall’s inequality we need to bound \(\Vert e^n_\gamma \Vert \) and \(\Vert \nabla e^n_\gamma \Vert \) in terms of \(\Vert e^n\Vert \),\(\Vert \nabla e^n\Vert \) and terms of \(\mathscr {O}(\tau ^2)\). These bounds are formulated in the two following lemmas.
Lemma 3.2
Given the optimal \(L^2\)-convergence of Theorem 3.1 and the uniform bounds (3.2), the error coming from the nonlinear term behaves as \(\tau ^2\), i.e. \(\Vert e^n_\gamma \Vert \lesssim \tau ^2 \).
Proof
We introduce the function f by
$$\begin{aligned} f(a,b) = \frac{1}{b-a}\int _{a}^{b} \gamma (r)\,dr. \end{aligned}$$
(3.9)
The partial derivatives of f with respect to the i’th variable are denoted by \(\partial _{i}f\). A standard application of the mean value theorem yields that
$$\begin{aligned} e^n_\gamma= & {} (\gamma (\xi ^n)-\gamma (\xi ^n_\tau ))u^{n+1/2} + \gamma (\xi ^n_\tau )(u^{n+1/2}-u^{n+1/2}_\tau ) \nonumber \\= & {} (f(|u^n|^2,|u^{n+1}|^2)-f(|u^n_\tau |^2,|u^{n+1}_\tau |^2) ) u^{n+1/2}+\gamma (\xi ^n_\tau )e^{n+1/2} \nonumber \\= & {} \big (\partial _1 f(\eta ^n,\eta ^{n+1})(|u^n|^2-|u^n_\tau |^2)+\partial _2 f(\eta ^n,\eta ^{n+1})(|u^{n+1}| ^2-|u^{n+1}_\tau |^2) \big )u^{n+1/2}\nonumber \\&\qquad +\gamma (\xi ^n_\tau )e^{n+1/2}\nonumber \end{aligned}$$
for some functions \(\eta ^{n}\) and \(\eta ^{n=1}\) with
$$\begin{aligned} \min \{ |u^n(x)|^2 , |u^n_{\tau }(x)|^2 \} \le \eta ^n(x) \le \max \{|u^{n}(x)|^2,|u^n_{\tau }(x)|^2 \}. \end{aligned}$$
(3.10)
A quick sanity check shows that the partial derivatives of f are bounded by the derivative of \(\gamma \):
$$\begin{aligned} |\partial _1 f(a,b)|= & {} | \frac{1}{b-a}(\gamma (c)-\gamma (a))| = |\gamma '(\theta ^-) \frac{c-a}{b-a}| \nonumber \\ |\partial _2 f(a,b)|= & {} |\frac{1}{b-a}(\gamma (b)-\gamma (c)) | =| \gamma '(\theta ^+) \frac{b-c}{b-a}| , \end{aligned}$$
(3.11)
where c and \(\theta \) lie somehwere between a and b. Hence
$$\begin{aligned} \Vert e^n_\gamma \Vert\le & {} {} \Vert \partial _1 f(\eta ^n,\eta ^{n+1})\Vert _{L^\infty } \Vert u^{n+1/2}\Vert _{L^\infty }\Vert |u^n|^2-|u^n_\tau |^2\Vert \\&+\, \Vert \partial _2 f(\eta ^n,\eta ^{n+1})\Vert _{L^\infty } \Vert u^{n+1/2}\Vert _{L^\infty }\Vert |u^{n+1}|^2-|u^{n+1}_\tau |^2\Vert \\&+\, \Vert \gamma (\xi ^n_\tau )\Vert _{L^\infty } \Vert e^{n+1/2}\Vert . \end{aligned}$$
Since \(\Vert |u^n|^2-|u^n_\tau |^2 \Vert \le \Vert | u^{n}|+|u^{n+1}\Vert | _{L^{\infty }}\Vert | e^{n}\Vert |\) it now follows with the max-norm bounds for the exact solution and the semi-discrete solution that \( \Vert e^n_\gamma \Vert \lesssim \Vert e^{n+1}\Vert +\Vert e^{n} \Vert \lesssim \tau ^2\).\(\square \)
Lemma 3.3
Given Theorem 3.1, the gradient of the error coming from the nonlinear term is bounded as \(\Vert \nabla e^n_\gamma \Vert \lesssim \Vert \nabla e^{n+1} \Vert +\Vert \nabla e^n\Vert +\tau ^2.\)
Proof
The steps are much the same as in the previous lemma, with the exception that we need to use max-norm estimates for the gradient of the exact solution, which are however not available for the numerical approximation. We begin with a suitable error splitting of the form
$$\begin{aligned} \nabla e^n_\gamma= & {} \nabla [\gamma (\xi ^n)u^{n+1/2}-\gamma (\xi ^n_\tau ) u^{n+1/2}_\tau ] \\= & {} \nabla \gamma (\xi ^n)u^{n+1/2}+\gamma (\xi ^n)\nabla u^{n+1/2} -\nabla \gamma (\xi ^n_\tau )u^{n+1/2}_\tau -\gamma (\xi ^n_\tau )\nabla u^{n+1/2}_\tau \\= & {} u^{n+1/2}_\tau \nabla (\gamma (\xi ^n)-\gamma (\xi ^n_\tau ))+\nabla \gamma (\xi ^n)(u^{n+1/2}-u^{n+1/2}_\tau )\\&\quad + (\gamma (\xi ^n)-\gamma (\xi ^n_\tau ))\nabla u^{n+1/2} + \gamma (\xi ^n_\tau )\nabla (u^{n+1/2}-u^{n+1/2}_\tau ). \end{aligned}$$
Using the previous lemma and the max-norm bounds for the exact solution we may conclude,
$$\begin{aligned}&\Vert \nabla e^n_\gamma \Vert \le \Vert u^{n+1/2}_\tau \Vert _{L^\infty } \Vert \nabla (\gamma (\xi ^n)-\gamma (\xi ^n))\Vert + \Vert \nabla \gamma (\xi ^n)\Vert _{L^\infty }\Vert e^{n+1/2}\Vert \nonumber \\&\quad +\Vert \nabla u^{n+1/2}\Vert _{L^\infty }\tau ^2 +\Vert \gamma (\xi ^n_\tau )\Vert _{L^\infty }\Vert \nabla e^{n+1/2}\Vert . \end{aligned}$$
(3.12)
What is left to bound is the term \(\Vert \nabla (\gamma (\xi ^n)-\gamma (\xi ^n))\Vert \). We consider its dependence on \(|u^n|^2,|u^{n+1}|^2,|u^n_\tau |^2\) and \(|u^{n+1}_\tau |^2\) and find:
$$\begin{aligned} \nabla (\gamma (\xi ^n)-\gamma (\xi ^n_\tau ))= & {} \nabla (f(|u^n|^2,|u^{n+1}|^2)-f(|u^n_\tau |^2,|u^{n+1}_\tau |^2)) \nonumber \\= & {} \partial _1 f^n \nabla |u^n|^2+\partial _2 f^n \nabla |u^{n+1}|^2 - \partial _1 f_{\tau }^n \nabla |u^n_\tau |^2- \partial _2 f_{\tau }^n \nabla |u^{n+1}_\tau |^2 \nonumber \\= & {} (\partial _1 f^n -\partial _1 f^n_{\tau }) \nabla |u^n|^2 + \partial _1 f_{\tau }^n \nabla (|u^n|^2-|u^n_\tau |^2) \nonumber \\&\quad + (\partial _2 f^n -\partial _2 f^n_{\tau })\nabla |u^{n+1}|^2+\partial _2 f_{\tau }^n \nabla (|u^{n+1}|^2-|u^{n+1}_\tau |^2), \end{aligned}$$
(3.13)
where \(\partial _i f^n := \partial _i f(|u^n|^2,|u^{n+1}|^2) \text{ and } \partial _i f^n_{\tau }:= f_i(|u^n_\tau |^2,|u^{n+1}_\tau |^2)\). Another application of the mean value theorem yields:
$$\begin{aligned} \partial _1 (f^n - f_{\tau }^n)= & {} \partial _{1,1}f(\theta ^n,\theta ^{n+1}) (|u^n|^2-|u^n_\tau |^2) + \partial _{1,2}f(\theta ^n,\theta ^{n+1}) (|u^{n+1}|^2-|u^{n+1}_\tau |^2), \nonumber \\ \partial _2 (f^n - f_{\tau }^n)= & {} \partial _{2,1}f(\vartheta ^n,\vartheta ^{n+1}) (|u^n|^2-|u^n_\tau |^2) + \partial _{2,2}f(\vartheta ^n,\vartheta ^{n+1})(|u^{n+1}|^2-|u^{n+1}_\tau |^2), \nonumber \\ \end{aligned}$$
(3.14)
for suitable mean value functions that can be bounded pointwise by the maximum of the exact solution and the semi-discrete CN-approximation. The following quick calculations show that the second partial derivatives of f are bounded by the second derivative of \(\gamma \):
$$\begin{aligned} \partial _{1,1}f(a,b)= & {} \frac{1}{(b-a)^2}2(\gamma (c)-\gamma (a))-\frac{\gamma '(a)}{b-a}= \frac{\gamma '(\theta ^-)-\gamma '(a)}{b-a}+C_{\gamma ''}, \nonumber \\ \partial _{1,2}f (a,b)= & {} \frac{1}{(b-a)^2}(\gamma (b)-2\gamma (c)+\gamma (a)) = \gamma ''(c)+C_{\gamma ''},\nonumber \\ \partial _{2,2}f(a,b)= & {} \frac{\gamma '(b)}{b-a}-\frac{1}{(b-a)^2}2(\gamma (b)-\gamma (c))= \frac{\gamma '(b)-\gamma '(\theta ^+)}{b-a}+C_{\gamma ''}, \nonumber \\ \end{aligned}$$
(3.15)
where it was used that \(c = (b+a)/2+ C_{\gamma ''}(a-b)^2\). It thus becomes clear that \(\Vert \partial _{1,1} f+ \partial _{1,2}f+\partial _{2,2}f \Vert _{L^\infty } \le C_{\gamma ''}\). This gives us the following \(L^2\)-bound on (3.13)
$$\begin{aligned} \Vert \nabla (\gamma (\xi ^n)-\gamma (\xi ^n_\tau ))\Vert \le C_{\gamma ''}(\Vert e^n\Vert +\Vert e^{n+1}\Vert )+C_{\gamma '}\Vert \nabla (|u^{n}|^2-|u^{n}_\tau |^2+|u^{n+1}|^2-|u^{n+1}_\tau |^2) \Vert . \end{aligned}$$
We conclude that
$$\begin{aligned} \Vert \nabla e^n_\gamma \Vert \lesssim \Vert \nabla (|u^n|^2-|u^n_\tau |^2)\Vert +\Vert \nabla (|u^{n+1}|^2-|u^{n+1}_\tau |^2)\Vert +\tau ^2. \end{aligned}$$
Here it is noted that \(\nabla (|u^n|^2-|u^n_\tau |^2)\) may be written as
$$\begin{aligned} \nabla (|u^n|^2-|u^n_\tau |^2) = 2\text {Re}( (u^n-u^n_\tau )\nabla \overline{u}^n+u^n_\tau \nabla (\overline{u}^n-\overline{u}^n_\tau )). \end{aligned}$$
Hence, using the max-norm bounds for the gradient of the exact solution we have
$$\begin{aligned} ||\nabla e^n_\gamma || \lesssim ||\nabla e^{n+1}||+||\nabla e^n||+\tau ^2. \end{aligned}$$
\(\square \)
With Lemmas 3.2 and 3.3 we now have the following bound on term I.
Lemma 3.4
For term \(\mathrm{I}\) which is given by (3.8), we have the estimate
$$\begin{aligned} |\mathrm{I}|\, \lesssim \, \Vert \nabla e^{n+1} \Vert ^2+\Vert \nabla e^n \Vert ^2 + \Vert T^n\Vert ^2+\tau ^4. \end{aligned}$$
(3.16)
We can now proceed to bound term II. Here we explicate the Taylor term using (3.4) to see
$$\begin{aligned} \mathrm{II}= & {} - \text {Re}(\langle T^n , D_\tau e^n \rangle ) \\\le & {} \underbrace{ | \langle ( D_\tau u^n - \partial _t u(t_{n+1/2})),D_\tau e^n\rangle |}_{\mathrm{IIa}}+\underbrace{\text {Re}\langle -\varDelta ( u^{n+1/2}-u(t_{n+1/2})),D_\tau e^n\rangle }_{\mathrm{IIb}} \\&+\,\underbrace{|\langle V(u^{n+1/2}-u(t_{n+1/2})),D_\tau e^n\rangle |}_{\mathrm{IIc}} \\&+\,\underbrace{|\langle \bigg ( \frac{\varGamma (|u^{n+1}|^2)-\varGamma (|u^n|^2)}{|u^{n+1}|^2-|u^n|^2}-\gamma (|u(t_{n+1/2})|^2),D_\tau e^n\rangle \bigg )|}_{\mathrm{IId}}. \end{aligned}$$
We start with estimating \(\mathrm{IIa},\mathrm{IIc}\) and \(\mathrm{IId}\), which can be bounded in a similar way.
Step 1, bounding \(\mathrm{IIa}\):
By replacing \(D_\tau e^n\) using (3.3) (i.e. time derivative is replaced by regularity in space) we have
$$\begin{aligned}&{|\langle D_\tau u^n-\partial _t u(t_{n+1/2}) , D_\tau e^n \ \rangle |}\\&\quad \le |\langle D_\tau u^n-\partial _t u(t_{n+1/2}) ,\varDelta e^{n+1/2}\rangle |+|\langle D_\tau u^n-\partial _t u(t_{n+1/2}) ,V e^{n+1/2}\rangle | \\&\qquad +\, | \langle D_\tau u^n-\partial _t u(t_{n+1/2}) , e_\gamma ^n\rangle | + | \langle D_\tau u^n-\partial _t u(t_{n+1/2}) ,T^n \rangle | \\&\quad \overset{(3.6)}{\lesssim } \Vert \nabla e^{n+1}\Vert ^2 + \Vert \nabla e^n\Vert ^2 + \Vert D_\tau u^n-\partial _t u(t_{n+1/2})\Vert _{H^1}^2 + \Vert T^n\Vert ^2 + \tau ^4 . \end{aligned}$$
Here we see how the assumption that \(\partial _{ttt}u\in L^2(0,T;H^1(\mathscr {D}))\) will enter, as it allows us to conclude that \(\sum \Vert D_\tau u^n-\partial _t u(t_{n+1/2})\Vert _{H^1}^2\lesssim \tau ^3\).
Step 2, bounding \(\mathrm{IIc}\):
We use the same idea as for \(\mathrm{IIa}\) to get
$$\begin{aligned}&{|\langle V(u^{n+1/2}-u(t_{n+1/2})),D_\tau e^n\rangle | \le \Vert V\Vert _{L^\infty }|\langle u^{n+1/2}-u(t_{n+1/2}),D_\tau e^n\rangle |} \\&\quad \le \Vert V\Vert _{L^\infty }\big ( |\langle u^{n+1/2}-u(t_{n+1/2}) ,\varDelta e^{n+1/2}\rangle |+\\&\qquad +\,|\langle u^{n+1/2}-u(t_{n+1/2}) ,V e^{n+1/2}\rangle | +| \langle u^{n+1/2}-u(t_{n+1/2}) , e_\gamma ^n\rangle | \\&\qquad + \, | \langle u^{n+1/2}-u(t_{n+1/2}) ,T^n \rangle | \big ) \\&\quad \lesssim \Vert \nabla e^{n+1}\Vert ^2 + \Vert \nabla e^n\Vert ^2 + \Vert u^{n+1/2}-u(t_{n+1/2})\Vert _{H^1}^2+ \Vert T^n\Vert ^2+ \tau ^4. \end{aligned}$$
Step 3, bounding IId:
We start from
$$\begin{aligned}&\big |\big \langle \frac{\varGamma (|u^{n+1}|^2)-\varGamma (|u^n|^2)}{|u^{n+1}|^2-|u^n|^2}-\gamma (|u(t_{n+1/2})|^2),D_\tau e^n\big \rangle \big | \\&\quad = |\langle \gamma (\xi ^n)-\gamma (|u(t_{n+1/2})|^2), D_\tau e^n\rangle | \end{aligned}$$
replace \(D_\tau e^n\) again using (3.3). Furthermore, in virtue of the assumptions it holds that \(\Vert \nabla (\gamma (\xi ^n)-\gamma (|u(t_{n+1/2})|^2))\Vert \lesssim \Vert u^{n+1/2}-u (t_{n+1/2})\Vert _{H^{1}}\). This is made explicit in the appendix. We thus obtain
$$\begin{aligned} |\langle \gamma (\xi ^n)-\gamma (|u(t_{n+1/2})|^2), D_\tau e^n\rangle |= & {} |\langle \gamma (\xi ^n)-\gamma (|u(t_{n+1/2})|^2), -\varDelta e^{n+1/2} + Ve^{n+1/2}+e^n_\gamma +T^n \rangle | \\\le & {} |\langle \nabla (\gamma (\xi ^n)-\gamma (|u(t_{n+1/2})|^2)),\nabla e^{n+1/2}\rangle | +\Vert T^n\Vert ^2 + \mathscr {O}(\tau ^4) \\\lesssim & {} \Vert \nabla e^{n+1}\Vert ^2+\Vert \nabla e^n\Vert ^2+ \Vert T^n\Vert ^2+\Vert u^{n+1/2}-u(t_{n+1/2})\Vert _{H^1}^2+ \tau ^4. \end{aligned}$$
Step 4, bounding \(\mathrm{IIb}\):
The previous technique does not work on this term since replacing the discrete time derivative with regularity in space would give rise to the term \(\nabla \varDelta (u^{n+1/2}-u(t_{n+1/2}))\), which we can not afford. Instead we use summation by parts in time to get the factor \(D_\tau \varDelta (u^{n+1/2}-u(t_{n+1/2}))\), which when integrated against \(e^{n+1/2}\) can be handled. First we recall:
$$\begin{aligned}D_\tau [a^k b^k] = a^kD_\tau b^k + b^{k+1}D_\tau a^k \quad \Longleftrightarrow \quad -a^kD_\tau b^k = D_\tau [a^k] b^{k+1} - D_\tau [a^kb^k]. \end{aligned}$$
Using this on term \(\mathrm{IIb}\) yields:
$$\begin{aligned}&{\langle -\varDelta (u^{n+1/2}-u), D_\tau e^n\rangle {= \langle \nabla (u^{n+1/2}-u),D_\tau \nabla e^n \rangle }} \\&\quad = D_\tau [\langle \nabla (u^{n+1/2}-u(t_{n+1/2})), \nabla e^{n}\rangle ] - \langle D_\tau \nabla (u^{n+1/2}-u(t_{n+1/2})),\nabla e^{n+1}\rangle \\&\quad \le D_\tau [\langle \nabla (u^{n+1/2}-u(t_{n+1/2})), \nabla e^{n}\rangle ] + |\langle D_\tau \nabla (u^{n+1/2}-u(t_{n+1/2})),\nabla e^{n+1}\rangle | \\&\quad \le D_\tau [\langle \nabla (u^{n+1/2}-u(t_{n+1/2})), \nabla e^{n}\rangle ] + \Vert D_\tau (u^{n+1/2}-u(t_{n+1/2}))\Vert _{H^1}^2 + \Vert \nabla e ^{n+1}\Vert ^2. \end{aligned}$$
Collecting the estimates we have the following estimate for term \(\mathrm{II}\).
Lemma 3.5
For term \(\mathrm{II} = -\text {Re}(\langle T^n , D_\tau e^n \rangle )\) it holds the estimate
$$\begin{aligned} \mathrm{II}\le & {} D_\tau [\langle \nabla (u^{n+1/2}-u(t_{n+1/2})), \nabla e^{n}\rangle ] + C \big ( \Vert \nabla e ^{n+1}\Vert ^2 +\Vert \nabla e^n\Vert ^2 + \tau ^4 +\Vert T^n\Vert ^2 \nonumber \\&+\, \Vert D_\tau u^n-\partial _t u(t_{n+1/2})\Vert _{H^1}^2 +\Vert u^{n+1/2}-u(t_{n+1/2})\Vert _{H^1}^2\big ). \nonumber \\ \end{aligned}$$
(3.17)
Here we note the importance of not estimating the absolute value of the first term since it is necessary to use the fact that n of these terms cancel when summed up, i.e. \(\sum _k D_\tau a^k = \frac{1}{\tau }(a^{n+1}-a^0)\). We are now ready to finish the proof of the first main result.
Proof of Theorem 3.2
We pick off where we left (3.7) and find by using Lemmas 3.4 and 3.5:
$$\begin{aligned}&\frac{\Vert \nabla e^{n+1}\Vert ^2 -\Vert \nabla e^n\Vert ^2}{2\tau }\\&\quad \le D_\tau [\langle \nabla (u^{n+1/2}-u(t_{n+1/2})), \nabla e^{n}\rangle ] +C\big (\Vert \nabla e^{n+1}\Vert ^2 +\Vert \nabla e^n\Vert ^2 +\tau ^4 \\&\qquad +\, \Vert D_\tau u^n -\partial _t u(t_{n+1/2})\Vert ^2_{H^1}+ \Vert T^n\Vert ^2 + \Vert u^{n+1/2}-u(t_{n+1/2})\Vert ^2_{H^1} \big ). \end{aligned}$$
Summing this up and using \(e^0=0\) gives
$$\begin{aligned} \frac{\Vert \nabla e^{n+1}\Vert ^2}{2\tau }&\le C\left( \sum _{k=0}^n \Vert \nabla e^{k}\Vert ^2 \right) + \frac{1}{\tau }\langle \nabla (u^{n+3/2}-u(t_{n+3/2})),\nabla e^{n+1} \rangle \\&\quad +\, C\tau ^3 + \sum _{k=0}^{n} \Vert T^k\Vert ^2+\Vert u^{k+1/2}-u(t_{k+1/2}) \Vert ^2_{H^1}+ \Vert D_\tau u^k\\&\quad -\,\partial _t u(t_{k+1/2})\Vert ^2_{H^1} \end{aligned}$$
and therefore, recalling (3.6),
$$\begin{aligned} \Vert \nabla e^{n+1}\Vert ^2&\le C\left( \sum _{k=0}^n \tau \Vert \nabla e^{k}\Vert ^2 \right) + C \tau ^4 +|\langle \nabla (u^{n+3/2}-u(t_{n+3/2})),\nabla e^{n+1} \rangle |. \end{aligned}$$
Young’s inequality with \(\epsilon >0\) is used on the last term:
$$\begin{aligned} |\langle \nabla (u^{n+3/2}-u(t_{n+3/2})),\nabla e^{n+1} \rangle | \le C\left( \frac{\tau ^4}{\epsilon } + \epsilon \Vert \nabla e^{n+1}\Vert ^2\right) . \end{aligned}$$
(3.18)
Which holds since,
$$\begin{aligned} \Vert \nabla (u^{n+3/2}-u(t_{n+3/2}))\Vert \lesssim \tau ^2 \Vert \partial _{tt}u\Vert _{L^\infty (H^1)}, \end{aligned}$$
where we have \( \Vert \partial _{tt}u\Vert _{L^\infty (H^1)} \lesssim \Vert \partial _{tt}u\Vert _{L^2(H^1)} + \Vert \partial _{ttt}u\Vert _{L^2(H^1)}\) by Sobolev embeddings. Finally we arrive at
$$\begin{aligned} \Vert \nabla e^{n+1}\Vert ^2 \le C\left( \sum _{k=0}^n \tau \Vert \nabla e^k\Vert ^2 \right) + C \tau ^4 + \frac{\tau ^4}{\epsilon } + \epsilon \Vert \nabla e^{n+1}\Vert ^2 \end{aligned}$$
and for e.g. \(\epsilon =1/2\) we can absorb \( \epsilon \Vert \nabla e^{n+1}\Vert ^2\) in the left hand side and conclude
$$\begin{aligned} \Vert \nabla e^{n+1}\Vert ^2 \le C \left( \tau ^4 + \sum _{k=0}^n \tau \Vert \nabla e^k\Vert ^2 \right) . \end{aligned}$$
Grönwall’s inequality now yields:
$$\begin{aligned} \Vert \nabla e^{n+1}\Vert \lesssim \tau ^2. \end{aligned}$$
(3.19)
\(\square \)