Appendix 1:Variational Operators
In the preceding sections, we very often considered the same problem on different spaces, e.g., we switched from an operator equation defined on \(V\) to the same equation defined on \(U\). In this section, we want to clarify in more detail why this is justified.
Let \((V,\langle \cdot ,\cdot \rangle _{V})\) be a separable real Hilbert space. Furthermore, let
$$\begin{aligned} a(\cdot ,\cdot ): V\times V\rightarrow \mathbb {R}\end{aligned}$$
be a continuous, symmetric, and elliptic bilinear form. This means that there exist two constants \(c_{\text {ell}}, C_{\text {ell}}>0\) such that for arbitrary \(u,v\in V\) the bilinear form satisfies the following conditions:
$$\begin{aligned} c_{\text {ell}}\, \Vert u\Vert _{V}^2\le a(u,u),\quad a(u,v)=a(v,u),\quad |a(u,v)|\le C_{\text {ell}}\, \Vert u\Vert _{V} \Vert v\Vert _{V}. \end{aligned}$$
(70)
Then, by the Lax–Milgram theorem, the operator
$$\begin{aligned} A:V&\rightarrow V^*\nonumber \\ v&\mapsto Av:= -a(v,\cdot ) \end{aligned}$$
(71)
is boundedly invertible. Let us now assume that \(V\) is densely embedded into a real Hilbert space \((U,\langle \cdot ,\cdot \rangle _{U})\) via a linear embedding \(j\). We write
$$\begin{aligned} V\overset{j}{\hookrightarrow } U. \end{aligned}$$
Furthermore, we identify the Hilbert space \(U\) with its topological dual space \(U^*\) via the Riesz isomorphism \( U\ni u\mapsto \varPhi u:=\langle u,\cdot \rangle _U\in U^*\). The adjoint map \(j^*:U^*\rightarrow V^*\) of \(j\) embeds \(U^*\) densely into the topological dual \(V^*\) of \(V\). All in all we have a so-called Gelfand triple \((V,U,V^*)\):
$$\begin{aligned} V\overset{j}{\hookrightarrow } U \overset{\varPhi }{\equiv } U^*\overset{\,\,\,j^*}{\hookrightarrow }V^*. \end{aligned}$$
Using \(\langle \cdot ,\cdot \rangle _{V^*\times V}\) to denote the dual pairs of \(V\) and \(V^*\), we have
$$\begin{aligned} \langle j(v_1),j(v_2)\rangle _U = \langle j^* \, \varPhi \, j(v_1),v_2\rangle _{V^*\times V} \qquad \text {for all} \quad v_1,v_2\in V. \end{aligned}$$
(72)
In this setting, we can consider the operator \(A:V\rightarrow V^*\) as an unbounded operator on the intermediate space \(U\). More precisely, set
$$\begin{aligned} D(A):=D(A;U):=\{u\in V\,:\, Au \in j^*\varPhi (U)\}, \end{aligned}$$
and define the operator
$$\begin{aligned} \tilde{A}:D(\tilde{A}):=j(D(A;U))\subseteq U&\rightarrow U\\ u&\mapsto \tilde{A}u:= \varPhi ^{-1} {j^*}^{-1} A \, j^{-1} u. \end{aligned}$$
Such an (unbounded) linear operator is sometimes called variational. It is densely defined since \(U^*\) is densely embedded in \(V^*\). Furthermore, the symmetry of the bilinear form \(a(\cdot ,\cdot )\) implies that \(\tilde{A}\) is self-adjoint. At the same time, it is strictly negative definite because of the ellipticity of \(a\). Moreover, since \(A:V\rightarrow V^*\) is boundedly invertible, the operator \(\tilde{A}^{-1}:U\rightarrow U\), defined by \(\tilde{A}^{-1}:= j A^{-1} j^*\varPhi \), is the bounded inverse of \(\tilde{A}\). It is compact if the embedding \(j\) of \(V\) in \(U\) is compact.
Let us now fix \(\tau >0\) and consider the bilinear form
$$\begin{aligned} a_\tau : V\times V&\rightarrow \mathbb {R}\\* (u,v)&\mapsto a_\tau (u,v):=\tau \langle j(u),j(v)\rangle _U + a(u,v), \end{aligned}$$
which is also continuous, symmetric, and elliptic in the sense of (72). Obviously, for \(u,v\in V\) we have the identity
$$\begin{aligned} a_\tau (u,v)&= \tau \langle j^* \varPhi j (u),v\rangle _{V^*\times V}- \langle A u,v\rangle _{V^*\times V}\\&=\langle (\tau j^*\varPhi j - A)u,v\rangle _{V^*\times V}, \end{aligned}$$
so that applying again the Lax–Milgram theorem, we can conclude that \((\tau j^* \varPhi j - A):V\rightarrow V^*\) is boundedly invertible. Therefore, the operator
$$\begin{aligned} (\tau I- \tilde{A}):D(\tilde{A})\subseteq U&\rightarrow U\\ u&\mapsto (\tau I-\tilde{A})u:=\tau u- \tilde{A}u, \end{aligned}$$
which coincides with \(\varPhi ^{-1}{j^*}^{-1} (\tau j^*\varPhi j- A)j^{-1}\) on \(D(\tilde{A})\), possesses a bounded inverse \((\tau I- \tilde{A})^{-1}=j (\tau j^*\varPhi j - A)^{-1} j^*\varPhi : U \rightarrow U\). Thus, the resolvent set \(\varrho (\tilde{A})\) of \(\tilde{A}\) contains all \(\tau \ge 0\). In particular, for any \(\tau >0\) the range of the operator \((\tau I-\tilde{A})\) is all of \(U\). Since, furthermore, \(\tilde{A}\) is dissipative, the Lumer–Phillips theorem implies that \(\tilde{A}\) generates a strongly continuous semigroup \(\{e^{t\tilde{A}}\}_{t\ge 0}\) of contractions on \(U\) (e.g., [35, Theorem 1.4.3]). Thus, an application of the Hille–Yosida theorem (e.g., [35, Theorem 1.3.1]) shows that the operator \(L_{\tau }^{-1}:= (I-\tau \tilde{A})^{-1} = \tau (\tau I-\tilde{A})^{-1}:U\rightarrow U\) is a contraction for each \(\tau >0\).
By an abuse of notation, we sometimes write \(A\) instead of \(\tilde{A}\).
Appendix 2: Proofs of Lemmas 3.9 and 4.6
Proof of Lemma 3.9
By (38) and (39) the stage equations (30) read as
$$\begin{aligned} (I-\tau \gamma _{1,1} A) w_{k,1} =&\ A u_k+f(t_k), \\ (I-\tau \gamma _{2,2} A) w_{k,2} =&\ A (u_k+\tau a_{2,1}w_{k,1})+f(t_k+a_2\tau )+c_{2,1}w_{k,1}. \end{aligned}$$
We begin with an application of the basic observation that
$$\begin{aligned} I=(I-CA)^{-1}(I-CA) \end{aligned}$$
implies
$$\begin{aligned} (I-C A)^{-1} A = -\frac{1}{C} I + \frac{1}{C} (I-C A)^{-1}. \end{aligned}$$
It follows that
$$\begin{aligned} w_{k,1}&=\big ((-{\textstyle \frac{1}{\tau \gamma _{1,1}}} I + {\textstyle \frac{1}{\tau \gamma _{1,1}}} (I-\tau \gamma _{1,1}A)^{-1}\big ) u_k +(I-\tau \gamma _{1,1}A)^{-1}f(t_k)\\&= -{\textstyle \frac{1}{\tau \gamma _{1,1}}} u_k +L_{\tau ,1}^{-1} \big ( {\textstyle \frac{1}{\tau \gamma _{1,1}}} u_k + f(t_k)\big ). \end{aligned}$$
We denote
$$\begin{aligned} v_{k,1} = L_{\tau ,1}^{-1} \big ({\textstyle \frac{1}{\tau \gamma _{1,1}}} u_k + f(t_k)\big ). \end{aligned}$$
A similar computation for the second-stage equation yields
$$\begin{aligned} w_{k,2}&= \big (-{\textstyle \frac{1}{\tau \gamma _{2,2}}} I + {\textstyle \frac{1}{\tau \gamma _{2,2}}} (I-\tau \gamma _{2,2}A)^{-1}\big ) (u_k+\tau a_{2,1}w_{k,1})\\&\quad + (I-\tau \gamma _{2,2}A)^{-1} \big (f(t_k+a_2\tau ) +c_{2,1}w_{k,1} \big )\\&= -{\textstyle \frac{1}{\tau \gamma _{2,2}}} \big ( (1-{\textstyle \frac{a_{2,1}}{\gamma _{1,1}}}) u_k +\tau a_{2,1} v_{k,1} \big )\\&\quad + L_{\tau ,2}^{-1} \Big ( {\textstyle \frac{1}{\tau \gamma _{2,2}}} \big ( (1-{\textstyle \frac{a_{2,1}}{\gamma _{1,1}}}) u_k +\tau a_{2,1} v_{k,1} \big ) \\&\quad + f(t_k+a_2\tau ) +c_{2,1}(-{\textstyle \frac{1}{\tau \gamma _{1,1}}}u_k+v_{k,1}) \Big ) \\&= -{\textstyle \frac{1}{\tau \gamma _{2,2}}} (1-{\textstyle \frac{a_{2,1}}{\gamma _{1,1}}}) u_k - {\textstyle \frac{a_{2,1}}{\gamma _{2,2}} }v_{k,1}\\&\qquad + L_{\tau ,2}^{-1} \Big ( \big ( {\textstyle \frac{1}{\tau \gamma _{2,2}}} (1-{\textstyle \frac{a_{2,1}}{\gamma _{1,1}}}) - {\textstyle \frac{c_{2,1}}{\tau \gamma _{1,1}}} \big ) u_k \\&\qquad + ( {\textstyle \frac{a_{2,1}}{\gamma _{2,2}}} +c_{2,1}) v_{k,1} +f(t_k+a_2\tau ) \Big ). \end{aligned}$$
We denote
$$\begin{aligned} v_{k,2} = L_{\tau ,2}^{-1} \Big ( \big ( {\textstyle \frac{1}{\tau \gamma _{2,2}}} (1-{\textstyle \frac{a_{2,1}}{\gamma _{1,1}}}) - {\textstyle \frac{c_{2,1}}{\tau \gamma _{1,1}}} \big ) u_k + ({\textstyle \frac{a_{2,1}}{\gamma _{2,2}}} +c_{2,1}) v_{k,1} +f(t_k+a_2\tau ) \Big ) \end{aligned}$$
and arrive at
$$\begin{aligned} u_{k+1}&= u_k + \tau m_1(-{\textstyle \frac{1}{\tau \gamma _{1,1}}}u_k+v_{k,1}) \\&\quad + \tau m_2\big ( -{\textstyle \frac{1}{\tau \gamma _{2,2}}} (1-{\textstyle \frac{a_{2,1}}{\gamma _{1,1}}}) u_k - {\textstyle \frac{a_{2,1}}{\gamma _{2,2}}} v_{k,1} +v_{k,2} \big )\\&= \big (1-{\textstyle \frac{m_1}{\gamma _{1,1}}}-{\textstyle \frac{m_2}{\gamma _{2,2}}}(1-{\textstyle \frac{a_{2,1}}{\gamma _{1,1}}})\big )u_k +(\tau m_1-\tau m_2{\textstyle \frac{a_{2,1}}{\gamma _{2,2}}})v_{k,1} + \tau m_2 v_{k,2}. \end{aligned}$$
\(\square \)
Proof of Lemma 4.6
We start with the estimate
$$\begin{aligned} \Vert \hat{w}_{k,i}\Vert _{B^s_{q}(L_{q}(\mathcal O))}&= \big \Vert L_{\tau ,i}^{-1}R_{\tau ,k,i}(\tilde{u}_k,\tilde{w}_{k,1},...,\tilde{w}_{k,i-1})\big \Vert _{B^s_{q}(L_{q}(\mathcal O))} \\&\le \Vert L_{\tau ,i}^{-1}\Vert _{\mathcal {L}(L_2(\mathcal O),B^s_{q}(L_{q}(\mathcal O)))}\, \big \Vert R_{\tau ,k,i}(\tilde{u}_k,\tilde{w}_{k,1},...,\tilde{w}_{k,i-1})\big \Vert _{L_2(\mathcal O)}. \end{aligned}$$
The Lipschitz continuity of \(R_{\tau ,k,i}\) implies the linear growth property
$$\begin{aligned}&\big \Vert R_{\tau ,k,i}(\tilde{u}_k,\tilde{w}_{k,1},...,\tilde{w}_{k,i-1})\big \Vert _{L_2(\mathcal O)}\\&\quad \le C^{\text {Lip,R}}_{\tau ,k,(i)} \Big ( \Vert \tilde{u}_k\Vert _{L_2(\mathcal O)} + \sum _{j=1}^{i-1}\Vert \tilde{w}_{k,j}\Vert _{L_2(\mathcal O)} \Big ) + \big \Vert R_{\tau ,k,i}(0,...,0)\big \Vert _{L_2(\mathcal O)}\\&\quad \le \max \! \Big \{ C^{\text {Lip,R}}_{\tau ,k,(i)}, \big \Vert R_{\tau ,k,i}(0,...,0)\big \Vert _{L_2(\mathcal O)} \Big \} \times \Big (1\! +\! \Vert \tilde{u}_k\Vert _{L_2(\mathcal O)}\! + \!\sum _{j=1}^{i-1}\!\Vert \tilde{w}_{k,j}\Vert _{L_2(\mathcal O)} \Big )\\&\quad \le \max \! \Big \{ C^{\text {Lip,R}}_{\tau ,k,(i)}, \big \Vert R_{\tau ,k,i}(0,...,0)\big \Vert _{L_2(\mathcal O)} \Big \} \times \Big ( 1\! +\! \Vert u_k\Vert _{L_2(\mathcal O)}\! + \!\sum _{j=1}^{i-1}\! \Vert w_{k,j}\Vert _{L_2(\mathcal O)}\\&\qquad + \Vert u_k-\tilde{u}_k\Vert _{L_2(\mathcal O)} + \sum _{j=1}^{i-1}\Vert w_{k,j}-\tilde{w}_{k,j}\Vert _{L_2(\mathcal O)} \Big ). \end{aligned}$$
As before, the Lipschitz continuity of \(L_{\tau ,i}^{-1}R_{\tau ,k,i}\) implies
$$\begin{aligned}&\Vert w_{k,i}\Vert _{L_2(\mathcal O)}= \big \Vert L_{\tau ,i}^{-1}R_{\tau ,k,i}(u_k,w_{k,1},\ldots ,w_{k,i-1})\big \Vert _{L_2(\mathcal O)}\\&\quad \le \max \Big \{ C^{\text {Lip}}_{\tau , {k},({i})}, \big \Vert L_{\tau ,i}^{-1}R_{\tau ,k,i}(0,...,0)\big \Vert _{L_2(\mathcal O)} \Big \} \Big (1\! + \!\Vert u_k\Vert _{L_2(\mathcal O)}\! + \! \sum _{j=1}^{i-1}\! \Vert w_{k,j}\Vert _{L_2(\mathcal O)}\Big ). \end{aligned}$$
By induction, we estimate
$$\begin{aligned}&1 + \Vert u_k\Vert _{L_2(\mathcal O)} +\sum _{j=1}^{i-1} \Vert w_{k,j}\Vert _{L_2(\mathcal O)}\\&\quad \le \prod _{l=1}^{i-1} \Big (1+\max \Big \{ C^{\text {Lip}}_{\tau , {k},({l})}, \big \Vert L_{\tau ,l}^{-1}R_{\tau ,k,l}(0,\ldots ,0)\big \Vert _{L_2(\mathcal O)} \Big \} \Big ) \big (1+\Vert u_k\Vert _{L_2(\mathcal O)}\big ). \end{aligned}$$
Note that
$$\begin{aligned} \Vert \tilde{w}_{k,i}-\hat{w}_{k,i}\Vert _{L_2(\mathcal O)} \le \Vert \tilde{w}_{k,i}-\hat{w}_{k,i}\Vert _{H^{\nu }(\mathcal O)} \le \varepsilon _{k,i}. \end{aligned}$$
This enables us to follow similar lines as in the proof of Theorem 2.21. We estimate
$$\begin{aligned}&\Vert u_k-\tilde{u}_k\Vert _{L_2(\mathcal O)} + \sum _{j=1}^{i-1}\Vert w_{k,j}-\tilde{w}_{k,j}\Vert _{L_2(\mathcal O)} \\&\quad \le (1+C^{\text {Lip}}_{\tau , {k},({i-1})}) \Big (\Vert u_k-\tilde{u}_k\Vert _{L_2(\mathcal O)}+\sum _{j=1}^{i-2}\Vert w_{k,j}-\tilde{w}_{k,j}\Vert _{L_2(\mathcal O)} \Big ) \\&\qquad + \Big \Vert L_{\tau ,{i-1}}^{-1}R_{\tau ,k,{i-1}}(\tilde{u}_k, \tilde{w}_{k,1}, \ldots , \tilde{w}_{k,i-2}) \\&\qquad - \big [L_{\tau ,{i-1}}^{-1}R_{\tau ,k,{i-1}}(\tilde{u}_k, \tilde{w}_{k,1}, \ldots , \tilde{w}_{k,i-2})\big ]_{\varepsilon _{k,{i-1}}} \Big \Vert _{L_2(\mathcal O)}\\&\quad \le (1+C^{\text {Lip}}_{\tau , {k},({i-1})}) \Big ( \Vert u_k-\tilde{u}_k\Vert _{L_2(\mathcal O)}+\sum _{j=1}^{i-2}\Vert w_{k,j}-\tilde{w}_{k,j}\Vert _{L_2(\mathcal O)} \Big ) + \varepsilon _{k,{i-1}} \end{aligned}$$
and conclude, by induction,
$$\begin{aligned}&\Vert u_k-\tilde{u}_k\Vert _{L_2(\mathcal O)} + \sum _{j=1}^{i-1}\Vert w_{k,j}-\tilde{w}_{k,j}\Vert _{L_2(\mathcal O)} \\&\quad \le \left( \,\prod _{l=1}^{i-1}(1+C^{\text {Lip}}_{\tau , {k},({l})})\right) \Vert u_k-\tilde{u}_k\Vert _{L_2(\mathcal O)} + \sum _{j=1}^{i-1}\varepsilon _{k,j}\prod _{l=j+1}^{i-1}(1+C^{\text {Lip}}_{\tau , {k},({l})}). \end{aligned}$$
The proof is completed by
$$\begin{aligned} \Vert u_k-\tilde{u}_k\Vert _{L_2(\mathcal O)} \le \sum _{j=0}^{k-1} \Big (\prod _{l=j+1}^{k-1} (C^\prime _{\tau ,l,(0)}-1)\Big ) \sum _{i=1}^S C^\prime _{\tau ,j,(i)} \varepsilon _{j,i}, \end{aligned}$$
which is shown as in Theorem 2.21. \(\square \)