Carleman estimates and controllability results for fully discrete approximations of 1D parabolic equations

Abstract

In this paper, we prove a Carleman estimate for fully discrete approximations of one-dimensional parabolic operators in which the discrete parameters h and △t are connected to the large Carleman parameter. We use this estimate to obtain relaxed observability inequalities which yield, by duality, controllability results for fully discrete linear and semilinear parabolic equations.

Appendices

Appendix A: Discrete calculus results

1.1 A.1 Results for space-discrete variables

The goal of this first section is to provide a self-contained summary of calculus rules for space-discrete operators like ∂_h, $\overline {\partial _h}$, and to provide estimates for the successive applications of such operators on the weight functions. We present the results without a proof and refer the reader to [11] (see also [13]) for a complete discussion.

To avoid introducing cumbersome notation, we introduce the following continuous difference and averaging operators

$$ \begin{array}{@{}rcl@{}} &&\textsf{s}_{h}^{+} f(x):=f(x+\tfrac{h}{2}), \quad \textsf{s}_{h}^{-}f(x)=f(x-\tfrac{h}{2}), \\ &&\textsf{d}_{h} f:=\frac{1}{h}\left( \textsf{s}_{h}^{+}-\textsf{s}_{h}^{-}\right)f, \quad \textsf{m}_{h}{f}=\widehat{f}:=\frac{1}{2}\left( \textsf{s}_{h}^{+}+\textsf{s}_{ h}^{-}\right)f. \end{array} $$

With this notation, the results given below will be naturally translated to discrete versions. More precisely, for a function f continuously defined on ${\mathbb {R}}$, the discrete function ∂_hf is in fact d_hf sampled on the dual mesh $\overline {\mathfrak {M}}$, and $\overline {\partial _h} f$ is d_hf sampled on the primal mesh $\mathfrak {M}$. Similar meanings will be used for the averaging symbols $\tilde {f}$ and $\overline {f}$ (see Eqs. ?? and ??, respectively) and for more intricate combinations: for instance, $\widetilde {{\Delta }_h f}=\widetilde {\overline {\partial _h}\partial _h f}$ is the function $\widehat {\textsf {d}_{h}\textsf {d}_{h} f}$ sampled on $\overline {\mathfrak {M}}$.

1.1.1 A.1.1 Discrete calculus formulas

Lemma A.1

Let the functions f₁,f₂ be continuously defined over ${\mathbb {R}}$. We have

$$ \textsf{d}_{h}(f_{1}f_{2})=\textsf{d}_{h}(f_{1})\widehat{f_{2}}+\widehat{f_{1}}\textsf{d}_{h}(f_{2}). $$

The translation of the result to discrete functions $f_{1},f_{2}\in {\mathbb {R}}^{\mathfrak {M}}$ (resp. $g_{1},g_{2}\in {\mathbb {R}}^{\overline {\mathfrak {M}}}$) is

$$ \partial_h(f_{1}f_{2})=\partial_h(f_{1}) \widetilde{f_{2}} + \widetilde{f_{1}}\partial_h(f_{2}), \quad \Big(\text{resp. } \overline{\partial_h}(g_{1}g_{2})=\overline{\partial_h}(g_{1}) \overline{g_{2}}+\overline{g_{1}} \overline{\partial_h}(g_{2})\Big). $$

(1)

Lemma A.2

Let the functions f₁,f₂ be continuously defined over ${\mathbb {R}}$. We have

$$ \widehat{f_{1}f_{2}}=\widehat{f_{1}}\widehat{f_{2}}+\frac{h^{2}}{4}\textsf{d}_{h}(f_{1})\textsf{d}_{h}(f_{2}). $$

The translation of the result to discrete functions $f_{1},f_{2}\in {\mathbb {R}}^{\mathfrak {M}}$ (resp. $g_{1},g_{2}\in {\mathbb {R}}^{\overline {\mathfrak {M}}}$) is

$$ \widetilde{f_{1}f_{2}}=\widetilde{f_{1}}\widetilde{f_{2}} + \frac{h^{2}}{4}\partial_h(f_{1})\partial_h(f_{2}), \quad \Big(\text{resp. } \overline{f_{1}f_{2}}=\overline{f_{1}} \overline{f_{2}} + \frac{h^{2}}{4}\overline{\partial_h}(f_{1})\overline{\partial_h}(f_{2}))\Big). $$

Lemma A.3

Let the function f be continuously defined over ${\mathbb {R}}$. We have

$$ \textsf{m}_{h}^{2} f:= \hat{\hat{f}}=f+\frac{h^{2}}{4}\textsf{d}_{h}\textsf{d}_{h} f. $$

The following result provides a discrete integration-by-parts formula and a related identity for averaged functions.

Proposition A.4

Let $f\in {\mathbb {R}}^{\mathfrak {M}\cup \partial \mathfrak {M}}$ and $g\in {\mathbb {R}}^{\overline {\mathfrak {M}}}$. Then

$$ \begin{array}{@{}rcl@{}} &\displaystyle {\int}_{\Omega} f (\overline{\partial_h} g)=-{\int}_{\Omega}(\partial_h f)g+f_{N+1}g_{N+\frac{1}{2}}-f_{0} g_{\frac{1}{2}}, \end{array} $$

(2)

$$ \begin{array}{@{}rcl@{}} &\displaystyle {\int}_{\Omega} f\overline{g}={\int}_{\Omega}\tilde{f} g-\frac{h}{2}f_{N+1}g_{N+\frac{1}{2}}-\frac{h}{2}f_{0}g_{\frac{1}{2}}. \end{array} $$

(3)

Lemma A.5

Let f be a sufficiently smooth function defined in a neighborhood of $\overline {\Omega }$. We have

• (i) $\displaystyle \textsf {s}_{h}^{\pm } f = f \pm \frac {h}{2}{{\int \limits }_{0}^{1}} \partial _{x} f(\cdot \pm \sigma h/2)\text {d}{\sigma }$,

• (ii) $\displaystyle \textsf {m}_{h}^{j} f = f + \frac {h^{2}}{4^{2-j}} {\int \limits }_{-1}^{1} (1-|\sigma |) {\partial _{x}^{2}} f(\cdot + l_{j}\sigma h)\text {d}{\sigma }$,

• (iii) $\displaystyle \textsf {d}_{h}^{j}f ={\partial _{x}^{j}} f + \frac {h^{2}}{8^{2-j}(j+1)!}{\int \limits }_{-1}^{1}(1-|\sigma |)^{j+1}\partial _{x}^{j+2}f(\cdot +l_{j} \sigma h)\text {d}{\sigma }$, j = 1, 2, $l_{1}=\frac {1}{2}$, l₂ = 1.

Proof

The results follow from Taylor formula

$$ f(x+y)={\sum}_{j=0}^{n-1}\frac{y^{j}}{j!}f^{(j)}(x)+y^{n}{{\int}_{0}^{1}}\frac{(1-\sigma)^{n-1}}{(n-1)!}f^{(n)}(x+\sigma y)\text{d}\sigma $$

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at order n = 1 for item (i), n = 2 for item (ii) with j = 1, 2, n = 3 for item (iii) with j = 1 and n = 4 for item (iii) with j = 2. □

1.1.2 A.1.2 Space-discrete computations related to Carleman weights

We present here a summary of results related to space-discrete operations performed on the Carleman weigh functions. The estimates presented below are at the heart of our fully discrete Carleman estimate. The proof of such results can be found on [11].

We set r = e^sφ and ρ = r^− 1. The positive parameters s and h will be large and small, respectively. We highlight the dependence on s,h and λ in the following results. We always assume that s ≥ 1 and λ ≥ 1.

Lemma A.6

Let $\alpha ,\beta \in \mathbb N$. We have

$$ \begin{array}{@{}rcl@{}} \partial_{x}^{\beta}(r\partial_{x}^{\alpha} \rho )&=&\alpha^{\beta}(-s\phi)^{\alpha}\lambda^{\alpha+\beta}(\partial_{x} \psi)^{\alpha+\beta} \\ &\quad +& \alpha\beta (s\phi)^{\alpha} \lambda^{\alpha+\beta-1}\mathcal O(1)+\alpha(\alpha-1)s^{\alpha-1}\mathcal O_{\lambda}(1)= s^{\alpha}\mathcal O_{\lambda}(1). \end{array} $$

Let σ ∈ [− 1, 1]. We have

$$ \partial_{x}^{\beta}\left( r(t,\cdot)(\partial_{x}^{\alpha} \rho)(t,\cdot+\sigma h)\right)=\mathcal O_{\lambda}(s^{\alpha}(1+(sh)^{\beta}))e^{\mathcal O_{\lambda}(sh)}. $$

Provided $\frac {\tau h}{\delta T^{2}}\leq 1$, we have $\partial _{x}^{\beta }\left (r(t,\cdot )(\partial _{x}^{\alpha } \rho )(t,\cdot +\sigma h)\right )=s^{\alpha } \mathcal O_{\lambda }(1)$. The same expressions hold with r and ρ interchanged if we replace s by − s everywhere.

Proposition A.7

Let $\alpha \in \mathbb N$. Provided $\frac {\tau h}{\delta T^{2}}\leq 1$, we have

• (i) $r\textsf {m}_{h}^{j} \partial _{x}^{\alpha } \rho =r\partial _{x}^{\alpha } \rho +s^{\alpha }(sh)^{2}\mathcal O_{\lambda }(1)=s^{\alpha }\mathcal O_{\lambda }(1)$, j = 1, 2,

• (ii) $r \textsf {m}_{h}^{j} \textsf {d}_{h} \rho =r\partial _{x} \rho +s(sh)^{2}\mathcal O_{\lambda }(1)$, j = 0, 1,

• (iii) $r\textsf {d}_{h}^{2}\rho =r{\partial _{x}^{2}}\rho +s^{2}(sh)^{2}\mathcal O_{\lambda }(1)=s^{2}\mathcal O_{\lambda }(1)$.

The same estimates hold with ρ and r interchanged.

Proposition A.8

Let $\alpha \in \mathbb N$. For k = 0, 1, 2, and provided $\frac {\tau h}{\delta T^{2}}\leq 1$ we have

• (i) $\textsf {d~}_{h}^{k}(r\textsf {d}_{h}^{2} \rho )={\partial _{x}^{k}}(r{\partial _{x}^{2}}\rho )+s^{2}(sh)^{2}\mathcal O_{\lambda }(1)=s^{2}\mathcal O_{\lambda }(1)$,

• (ii) $\textsf {d~}_{h}^{k}(r\textsf {m}_{h}^{2} \rho )=(sh)^{2}\mathcal O_{\lambda }(1)$.

The same estimates hold with ρ and r interchanged.

Proposition A.9

Let $\alpha ,\beta \in \mathbb N$ and k,j = 0, 1, 2. Provided $\frac {\tau h}{\delta T^{2}}\leq 1$, we have

• (i) $\textsf {m}_{h}^{j}\textsf {d~}_{h}^{k} \partial _{x}^{\alpha }\left (r^{2}\widehat {\textsf {d}_{h}\rho } \textsf {d}_{ h}^{2}\rho \right )={\partial _{x}^{k}}\partial _{x}^{\alpha } (r^{2}(\partial _{x}\rho ){\partial _{x}^{2}}\rho )+s^{3}(sh)^{2}\mathcal O_{\lambda }(1)=s^{3}\mathcal O_{\lambda }(1)$,

• (ii) $\textsf {m}_{h}^{j} \textsf {d~}_{h}^{k} \partial _{x}^{\alpha }\left (r^{2}\widehat {\textsf {d}_{h} \rho } \textsf {m}_{ h}^{2}\rho \right )={\partial _{x}^{k}}\partial _{x}^{\alpha }(r\partial _{x} \rho )+s(sh)^{2}\mathcal O_{\lambda }(1)=s\mathcal O_{\lambda }(1)$.

The same estimates hold with ρ and r interchanged.

Remark A.10

We set $\textsf {d}_{h2}:=((\textsf {s}_{h}^{+})^{2}-(\textsf {s}_{h}^{-})^{2})/2h=\textsf {m}_{h}\textsf {d}_{h}$ and $\textsf {m}_{h2}:=((\textsf {s}_{h}^{+})^{2}+(\textsf {s}_{ h}^{-})^{2})/2$. The estimates presented in the previous results are then preserved when we replace some of the d_h by d_h2 and some of the m_h by m_h2.

1.2 A.2 Results for time-discrete variables

Following the spirit of the previous section, here, we devote to provide a summary of calculus rules manipulating the time-discrete operators D_t and $\overline D_t$, and also to provide estimates for the application of such operators on the weight functions.

As we did before, to avoid introducing additional notation, we present the following continuous difference operator. For a function f defined on ${\mathbb {R}}$, we set

$$ \begin{array}{@{}rcl@{}} &&\textsf{t}^{+} f(t):=f(t+\tfrac{\triangle t}{2}), \quad \textsf{t}^{-} f(t):=f(t-\tfrac{\triangle t}{2}), \\ &&\textsf{D}_t f:=\frac{1}{\triangle t}(\textsf{t}^{+}-\textsf{t}^{-})f. \end{array} $$

As for the the space variable, we can obtain discrete versions of the results presented below quite naturally. Indeed, for a function f continuously defined on ${\mathbb {R}}$, the discrete function D_tf amounts to evaluate D_tf at the mesh points $\mathcal D$ and $\overline D_t f$ is D_tf sampled at the mesh points $\mathcal P$.

1.2.1 A.2.1 Time-discrete calculus formulas

Lemma A.11

Let the functions f₁ and f₂ be continuously defined over $\mathbb R$. We have

$$ \textsf{D}_t(f_{1}f_{2})={\textsf t}^+f_{1} \textsf{D}_t f_{2}+\textsf{D}_t f_{1} {\textsf t}^-f_{2}. $$

The same holds for

$$ \textsf{D}_t(f_{1}f_{2})={\textsf t}^-f_{1} \textsf{D}_t f_{2}+\textsf{D}_t f_{1} {\textsf t}^+f_{2}. $$

From the above formulas, if f₁ = f₂ = f, we have the useful identities

$$ \begin{array}{@{}rcl@{}} &{\textsf t}^+f \textsf{D}_t f=\frac{1}{2}\textsf{D}_t \left( f^{2}\right)+\frac{1}{2}\triangle t(\textsf{D}_t f)^{2},\qquad {\textsf t}^-f \textsf{D}_t f=\frac{1}{2}\textsf{D}_t\left( f^{2}\right)-\frac{1}{2}\triangle t(\textsf{D}_t f)^{2}. \end{array} $$

The translation of the result to discrete functions $f,g_{1},g_{2}\in H^{\overline {\mathcal D}}$ is

$$ \begin{array}{ll} &\overline D_t(g_{1}g_{2})=(\bar{\mathtt t}^+g_{1})\overline D_t g_{2}+\overline D_t g_{1}(\bar{\mathtt t}^-g_{2}), \\ &\overline D_t(g_{1}g_{2})=(\bar{\mathtt t}^-g_{1})\overline D_t g_{2}+\overline D_t g_{1}(\bar{\mathtt t}^+g_{2}), \end{array} $$

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and

$$ \begin{array}{@{}rcl@{}} &(\bar{\mathtt t}^+f)\overline D_t f=\frac{1}{2}\overline D_t \left( f^{2}\right)+\frac{1}{2}\triangle t(\overline D_t f)^{2}, \end{array} $$

(6)

$$ \begin{array}{@{}rcl@{}} &(\bar{\mathtt t}^-f)\overline D_t f=\frac{1}{2}\overline D_t\left( f^{2}\right)-\frac{1}{2}\triangle t(\overline D_t f)^{2}. \end{array} $$

(7)

Of course, the above identities also hold for functions $f,g_{1},g_{2}\in H^{\overline {\mathcal P}}$ and their respective translation operators and difference operator t^± and D_t.

The following result covers discrete integration by parts and some useful related formulas.

Proposition A.12

Let {H, (⋅,⋅)_H} be a real Hilbert space and consider $u\in H^{\overline {\mathcal P}}$ and $v\in H^{\overline {\mathcal D}}$. We have the following:

(8)

(9)

Moreover, combining the above identities, we have the following discrete integration by parts formula

(10)

Remark A.13

If we consider two functions $f,g\in H^{\overline {\mathcal D}}$, we can combine Eqs. 8 and 10 to obtain the formula

$$ {{\int}_{0}^{T}}\left( \overline D_t f,\bar{\mathtt{t}}^{-}g\right)_{H}=-(f^{\frac12},g^{\frac12})_{H}+(f^{M+\frac12},g^{M+\frac12})_{H}-{{\int}_{0}^{T}}\left( \bar{\mathtt{t}}^{+} f,\overline D_t g\right)_{H}. $$

(11)

Analogously, for $f,g\in H^{\overline {\mathcal P}}$, the following holds

(12)

Observe that in these formulas, the integrals are taken over the same discrete points. These will be particularly useful during the derivation of the Carleman estimates Eqs. ?? and ??.

1.2.2 A.2.2 Time-discrete computations related to Carleman weights

We present some lemmas related to time-discrete operations applied to the Carleman weights. The proof of these results can be found in [22, Appendix B]. We recall that r = e^sφ and ρ = r^− 1. We highlight the dependence on τ, δ, △t and λ in the following estimates.

Lemma A.14 (Time-discrete derivative of the Carleman weight)

Provided △tτ(T³δ²)^− 1 ≤ 1, we have

$$ {\textsf t}^-(r) \textsf{D}_t \rho=-\tau {\textsf t}^-(\theta^{\prime})\varphi +\triangle t\left( \frac{\tau}{\delta^{3}T^{4}}+\frac{\tau^{2}}{\delta^{4}T^{6}}\right)\mathcal O_{\lambda}(1). $$

Lemma A.15 (Discrete operations on the weight 𝜃)

There exists a universal constant C > 0 uniform with respect to △t, δ and T such that

• (i) $|\textsf {D}_t (\theta ^{\ell })| \leq \ell T{\textsf t}^-(\theta ^{\ell +1}) + C \frac {\triangle t}{\delta ^{\ell +2}T^{2\ell +2}} , \quad \ell =1,2,\ldots$

• (ii) $\textsf {D}_t(\theta ^{\prime }) \leq C T^{2}{\textsf t}^-(\theta ^{3})+C\frac {\triangle t}{\delta ^{4}T^{5}}$.

In addition to the results presented above, since we are dealing with a fully discrete case, we need to give an additional lemma concerning the effect of the time-discrete operator D_t over some discrete operations in the space variable. This is a new result as compared to [13] and [22] but follows the arguments there. The result reads as follows.

Lemma A.16 (Mixed derivatives)

Provided τh(δT²) ≤ 1 and σ is bounded, we have

• (i) $\partial _{t}^{\beta }\left (r(t,x)\partial _{x}^{\alpha }\rho (t,x+\sigma h)\right )=T^{\beta } s^{\alpha }(t) \theta ^{\beta }(t)\mathcal O_{\lambda }(1)$, β = 1, 2 and $\alpha \in \mathbb N$.

If in addition $\frac {\triangle t \tau }{T^{3}\delta ^{2}}\leq \frac {1}{2}$, the following estimates hold

• (ii) $\textsf {D}_t(r \textsf {m}_{h}^{2} \rho )= T \textsf {t}^{-}(\theta [sh]^{2})\mathcal O_{\lambda }(1)+\left (\frac {\tau \triangle t}{\delta ^{3} T^{4}}\right )\left (\frac {\tau h}{\delta ^{} T^{2}}\right )\mathcal O_{\lambda }(1)$.

• (iii) $\textsf {D}_t(r \textsf {d}_{h}^{2} \rho )= T \textsf {t}^{-}(s^{2}\theta )\mathcal O_{\lambda }(1)+ \left (\frac {\tau ^{2} \triangle t}{\delta ^{4} T^{6}}\right )\mathcal O_{\lambda }(1)+\left (\frac {\tau \triangle t}{\delta ^{3} T^{4}}\right )\left (\frac {\tau h}{\delta ^{} T^{2}}\right )^{3}\mathcal O_{\lambda }(1)$.

Proof

For β = 1, the proof of item (i) can be found on [13, Proof of Proposition 2.14]. For β = 2, the proof follows exactly as in that work just by noting that ${\partial _{t}^{2}} (r\partial _{x}^{\alpha } \rho )= T^{2} s^{\alpha } \theta ^{2} \mathcal O_{\lambda }(1)$.

To prove item (ii), we shall exploit that the weights ρ and r can be written in separated variables. Indeed, by Lemma A.5(ii), we have

$$ r(t,x)\textsf{m}_{h}^{2}\rho(t,x)=1+C h^{2} {\int}_{-1}^{1}(1-|\sigma|)r(t,x){\partial_{x}^{2}} \rho(t,x+\sigma h)d\sigma. $$

(13)

hence

$$ \begin{array}{@{}rcl@{}} &&\frac{r(t+\triangle t,x)\textsf{m}_{h}^{2}\rho(t+\triangle t,x)-r(t,x)\textsf{m}_{h}^{2}\rho(t,x)}{\triangle t}\\ && =\partial_{t}(r(t,x)\textsf{m}_{h}^{2} \rho(t,x))+{\triangle t{\int}_{0}^{1}}(1-\gamma){\partial_{t}^{2}}\left( r(t+\gamma\triangle t,x)\textsf{m}_{h}^{2} \rho(t+\gamma \triangle t,x)\right)d \gamma \\ & & = A_{1}+A_{2}. \end{array} $$

(14)

by a first order Taylor formula in the time variable (see formula Eq. 4). Differentiating with respect to t in Eq. 13 and item (i) yield

$$ A_{1}=T[s(t)h]^{2}\theta(t) \mathcal O_{\lambda}(1). $$

(15)

For estimating A₂, we use the change of variable t↦t + γ△t for γ ∈ [0, 1] in item (i) and observe that provided $\frac {\triangle t \tau }{T^{3}\delta ^{2}} \leq \frac {1}{2}$, we have $\max \limits _{t\in [0,T+\triangle t]} \theta (t)\leq \frac {2}{\delta T^{2}}$. Therefore,

$$ A_{2}=\triangle t T^{2} s^{2}(t+ \triangle t)\theta^{2}(t+ \triangle t) h^{2} \mathcal O_{\lambda}(1) = \left( \frac{\tau \triangle t}{\delta^{3} T^{4}}\right)\left( \frac{\tau h}{\delta^{} T^{2}}\right)\mathcal O_{\lambda}(1), $$

(16)

where we have used that h ≪ 1 to remove one power of h. Putting together Eqs. 14–16 and performing the change of variable $t\mapsto t-\frac {\triangle t}{2}$ yield the desired result.

Finally, to prove item (iii), we have from Lemma A.5(iii) that

$$ r(t,x)\textsf{d}_{h}^{2}\rho(t,x)=r(t,x){\partial_{x}^{2}} \rho(t,x)+C h^{4} {\int}_{-1}^{1}(1-|\sigma|)^{3} r(t,x){\partial_{x}^{4}} \rho(t,x+\sigma h)d\sigma. $$

(17)

Therefore, arguing as above

$$ \begin{array}{@{}rcl@{}} &&\frac{r(t+\triangle t,x)\textsf{d}_{h}^{2}\rho(t+\triangle t,x)-r(t,x)\textsf{d}_{h}^{2}\rho(t,x)}{\triangle t}\\ && =\partial_{t}(r(t,x)\textsf{d}_{h}^{2} \rho(t,x))+{\triangle t{\int}_{0}^{1}}(1-\gamma){\partial_{t}^{2}}\left( r(t+\gamma\triangle t,x)\textsf{d}_{h}^{2} \rho(t+\gamma \triangle t,x)\right)d \gamma \\ && = B_{1}+B_{2}. \end{array} $$

(18)

Differentiating with respect to t in Eq. 17 and using item (i) yield

$$ B_{1}=Ts(t)^{2}\theta(t) \mathcal O_{\lambda}(1)+ h^{4}\left[Ts(t)^{4}\theta(t)\mathcal O_{\lambda}(1)\right]. $$

Using that $\frac {\tau h}{\delta T^{2}}\leq 1$ and h ≪ 1, we can adjust some powers in the above expression and obtain

$$ B_{1}= Ts(t)^{2}\theta(t)\mathcal O_{\lambda}(1). $$

(19)

For the term B₂, we argue as for A₂. We use the change of variable t↦t + γ△t, γ ∈ [0, 1], in item (i) and observe that provided $\frac {\triangle t \tau }{T^{3}\delta ^{2}}\leq \frac {1}{2}$, we have $\max \limits _{t\in [0,t+\triangle t]}\theta (t)\leq \frac {2}{\delta T^{2}}$. In this way, we get

$$ \begin{array}{@{}rcl@{}} B_{2}&=&\triangle t T^{2} s^{2}(t+\triangle t)\mathcal\theta^{2}(t+\triangle t) \mathcal O_{\lambda}(1)+\triangle t T^{2}h^{4} s^{4}(t+\triangle t)\theta^{2}(t+\triangle t)\mathcal O_{\lambda}(1)\\ &=& \frac{\tau^{2}\triangle t}{\delta^{4} T^{6}}\mathcal O_{\lambda}(1)+\left( \frac{\tau\triangle t}{\delta^{3} T^{4}}\right)\left( \frac{\tau h}{\delta T^{2}}\right)^{3}\mathcal O_{\lambda}(1). \end{array} $$

(20)

Collecting the estimates Eq. 19–Eq. 20 in Eq. 18 and setting the change of variable $t\mapsto t-\frac {\triangle t}{2}$ gives the desired result. □

Appendix B: Estimates of the cross product in the Carleman estimate

1.1 B.1 Estimates that only require space discrete computations

1.1.1 B.1.1 Proof of Lemma 2.1

The proof is straightforward. First, using Eq. 8, we can relax a little bit the notation by hiding the operator $\bar {\mathtt {t}}^{-}$. More precisely we have

Now, we can focus on the space variable. Noting that ${\Delta }_h=\overline {\partial _h}\partial _h$ and $\overline {\partial _h}([\partial _h z]^{2})=2\overline {\partial _h}(\partial _h z)\overline {\partial _h z}$ thanks to Lemma A.1, we can integrate by parts (see formula Eq. 2) and obtain

Using that $\partial _h(r^{2} \overline {\tilde {\rho }} \overline {\partial _h\rho })=-s\lambda ^{2}(\partial _{x}\psi )^{2}\phi +s\lambda \phi \mathcal {O}(1)+s(sh)^{2}\mathcal O_{\lambda }(1)$ (obtained from Lemma A.9(ii) and A.6) and $r\overline {\tilde {\rho }}=1+(sh)^{2}\mathcal O_{\lambda }(1)$ (see Lemma A.7(i)), we get

The result follows by shifting back the time integral with formula Eq. 8.

1.1.2 B.1.2 Proof of Lemma 2.2

We proceed as in the previous proof. First, we shift the time integral and then using integration by parts in space, we obtain

(21)

Here, we have used that $(z_{|\partial {\Omega }})^{n-\frac {1}{2}}=0$ for n ∈⟦1,M⟧ so no boundary conditions appear.

Using Lemma A.5(iii), we see that $\widetilde {\partial _{xx}\phi }=\partial _{xx}\phi +h\mathcal O_{\lambda }(1)$; thus, from Proposition A.7(i), we get

$$ \widetilde{(r\overline{\tilde{\rho}}\partial_{xx}\phi)}=\partial_{xx}\phi+\left[h+(sh)^{2}\right]\mathcal O_{\lambda}(1). $$

(22)

On the other hand, by Propositions A.7(i) and A.8, we get

$$ \partial_h(r\overline{\tilde{\rho}}\partial_{xx}\phi)= \left[1+(sh)^{2}\right]\mathcal O_{\lambda}(1). $$

(23)

Using estimates Eq. 22–Eq. 23 in Eq. 21, we see that I₁₂ can be written as

(24)

where

Using that

$$ \partial_{xx}\phi=\lambda^{2}|\partial_{x}\psi|^{2}\phi+\lambda\phi \mathcal O(1) $$

(25)

in expression Eq. 24, the result follows by Cauchy-Schwarz and Young inequalities and shifting the time integral. Notice that we have adjusted some powers of the product (sh) by using that s ≥ 1 and (sh) ≤ τh(δT²)^− 1 ≤ 1.

1.1.3 B.1.3 Proof of Lemma 2.3

Define $p:=r^{2}{\Delta }_h \rho \overline {\partial _h \rho }$. From Eq. 8 and noting that $\overline {\partial _h z}=\overline {\partial _h}(\tilde {z})$, we see that

where we have used Lemma A.1. Integrating by parts in space, we get

We observe that $\tilde {z}_{\frac {1}{2}}=\frac {h}{2}(\partial _h z)_{\frac {1}{2}}$ and $\tilde {z}_{N+\frac {1}{2}}=-\frac {h}{2}(\partial _h z)_{N+\frac {1}{2}}$. Thanks to A.6 and Proposition A.9(i) notice that $p=\left [s^{2}\mathcal O_{\lambda }(1)+s^{2}(sh)^{2}\mathcal O_{\lambda }(1)\right ]r\overline {\partial _h\rho }$. Thus,

(26)

by using that sh ≤ τh(δT²)^− 1 ≤ 1.

From Lemma A.2, formula Eq. 3 and recalling that $(z_{|\partial {\Omega }})^{n-\frac {1}{2}}$ for n ∈⟦1,M⟧, we have that

(27)

We claim the following.

Claim B.1

Provided τh(δT²)^− 1 ≤ 1, we have

$$ \begin{array}{@{}rcl@{}} &&\partial_h p= s^{3}\mathcal O_{\lambda}(1)+s^{3}(sh)^{2}\mathcal O_{\lambda}(1), \\ &&\overline{\partial_h p}=-s^{3}\lambda^{4}\phi^{3}(\partial_{x}\psi)^{4}+(s\lambda\phi)^{3}\mathcal O_{\lambda}(1)+s^{2}\mathcal O_{\lambda}(1)+s^{3}(sh)^{2}\mathcal O_{\lambda}(1). \end{array} $$

Proof

From Proposition A.9(i), we have $\partial _h p=\partial _{x}(r^{2}(\partial _{x} \rho )\partial _{xx}\rho )+s^{3}(sh)^{2}\mathcal O_{\lambda }(1)$. On the other hand, a straightforward computation gives

$$ \partial_{x}(r^{2}(\partial_{x}\rho)\partial_{xx}\rho)=-3s^{3}\lambda^{4}(\partial_{x} \psi)^{4}\phi^{3}+(s\lambda\phi)^{3}\mathcal O(1)+s^{2}\mathcal O_{\lambda}(1). $$

(28)

Thus, the first result follows immediately.

For the second one, we note that $\overline {\partial _h p}={\partial _h}_{\scriptscriptstyle 2} p$ (see Remark A.10), whence $\overline {\partial _h p}= {\partial _h}_{\scriptscriptstyle 2}(r^{2}{\Delta }_h \rho \overline {\partial _h \rho })$ and using Proposition A.9(i) the results follows from Eq. 28. □

Using Claim B.1 to estimate in Eq. 27 and recalling Eq. 26, we readily get

where

$$ \mu_{21}=(s\lambda\phi)^{3}\mathcal O(1)+s^{2}\mathcal O_{\lambda}(1)+s^{3}(sh)^{2}\mathcal O_{\lambda}(1), \quad \nu_{21}=s(sh)^{2}\mathcal O_{\lambda}(1). $$

As before, we have adjusted some powers of (sh) in ν₂₁ by recalling that (sh) ≤ τh(δT²)^− 1 ≤ 1 and h ≪ 1. The desired result then follows by shifting the integral in time.

1.1.4 B.1.3 Proof of Lemma 2.4

By Lemma A.3, we have that $\overline {\tilde {z}}=z+\frac {h^{2}}{4}{\Delta }_h z$. Thus

(29)

From Proposition A.7(iii) and Lemma A.6, we readily have

$$ r{\Delta}_h \rho= r\partial_{xx}\rho+s^{2}(sh)^{2}\mathcal O_{\lambda}(1)=s^{2}\lambda^{2}(\partial_{x}\psi)^{2}\phi^{2}+s\mathcal O_{\lambda}(1)+s^{2}(sh)^{2}\mathcal O_{\lambda}(1), $$

(30)

whence, combining with Eq. 25, we obtain that

(31)

where $\mu =s^{3}\lambda ^{3}\phi ^{3}\mathcal O(1)+s^{2}\mathcal O_{\lambda }(1)+s^{3}(sh)^{2}\mathcal O_{\lambda }(1)$.

For the term $I_{22}^{(2)}$, we proceed as follows. We set p_ϕ := rΔ_hρ∂_xxϕ and by using that $(z_{|\partial {\Omega }})^{n-\frac {1}{2}}=0$ for n ∈⟦1,M⟧, we get after integration by parts in space

Noting that $\partial _h(z^{2})=2\tilde {z}\partial _h z$, we can integrate once again in the last term of the above to obtain

(32)

Let us estimate J₁ and J₂. For the first term, we have from Eq. 25 that $\partial _{xx}\phi =\mathcal O_{\lambda }(1)$ and from estimate Eq. 30, we have $p_{\phi }=s^{2}\mathcal O_{\lambda }(1)+s^{2}(sh)^{2}\O _{\lambda }(1)$. The same estimate holds for $\widetilde {p_{\phi }}$. From this fact and adjusting some powers of (sh) yields

(33)

For estimating J₂, we claim the following

Claim B.2

Provided τh(δT²)^− 1 ≤ 1 we have $h^{2}{\Delta }_h p_{\phi }=s^{2}(h+h^{2})\mathcal O_{\lambda }(1)+(sh)^{4}\mathcal O_{\lambda }(1)$.

Proof

From Eq. 25, notice that

$$ \left\lVert\partial_{xx}\phi\right\rVert_{\infty}=\mathcal O_{\lambda}(1), \quad \|\partial_h(\partial_{xx}\phi)\|_{\infty}=\mathcal O_{\lambda}(1), \quad \left\lVert h{\Delta}_h(\partial_{xx}\phi)\right\rVert_{\infty}=\mathcal O_{\lambda}(1). $$

(34)

On the other hand, a direct computation gives

$$ h^{2}{\Delta}_h p_{\phi}= h^{2} {\Delta}_h(\partial_{xx}\phi) \overline{\widetilde{r{\Delta}_h \rho}}+2 h^{2} \overline{\partial_h{\partial_{xx}\phi}} \overline{\partial_h(r{\Delta}_h \rho)}+h \overline{\widetilde{\partial_{xx}\phi}} {\Delta}_h(r{\Delta}_h \rho). $$

From Propositions A.7(i) and A.8, estimates Eq. 34, and the fact that $(r\partial _{xx}\phi )=\partial _{x}(r\partial _{xx}\phi )=\partial _{xx}(r\partial _{xx}\phi )=s^{2} \mathcal O_{\lambda }(1)$, we obtain the desired result. □

Using Claim B.2, we readily see that

(35)

Putting together Eqs. 29, 31–32, Eqs. 33 and 35, the desired result follows by adjusting some powers of (sh).

1.2 B.2 Estimates involving time-discrete computations

1.2.1 B.2.1 Proof of Lemma 2.5

We begin by shifting the integral in time; hence,

(36)

where we have used formula Eq. 3 in the second equality. Observe that no boundary conditions appear since $z_{|\partial {\Omega }}^{n-\frac {1}{2}}=0$ for n ∈⟦1,M⟧. Noting that

$$ \widetilde{(\varphi r \overline{\partial_h \rho}z)}= \widetilde{(\varphi r\overline{\partial_h \rho})} \widetilde{z} +\frac{h^{2}}{4} \partial_h(\varphi r\overline{\partial_h \rho})\partial_h z $$

thanks to Lemma A.2, we rewrite Eq. 36 as

Integrating by parts in space the first term in the above expression and using that $\overline {\partial _h} \tilde {p}=\overline {\partial _h p}$ for $p\in {\mathbb {R}}^{\mathfrak {M}}$, we have

Using that $\|\partial _h \varphi \|_{\infty }=\mathcal O_{\lambda }(1)$ and $\partial _{x}(r\partial _{x} \rho )=s\mathcal O_{\lambda }(1)$, we can prove as in Claim B.1 that

$$ \partial_h(\varphi r\overline{\partial_h \rho})= \overline{\partial_h(\varphi r\overline{\partial_h \rho})} =s\mathcal O_{\lambda}(1)+s(sh)^{2}\mathcal O_{\lambda}(1). $$

Notice that since (sh) ≤ τh(δT²)^− 1 ≤ 1, we can further simplify the above estimate and obtain that both derivatives are of order $s\mathcal O_{\lambda }(1)$. Thus, from this remark, we have

$$ I_{31}=\int\!\!\!{\int}_{Q} \tau(\bar{\mathtt t}^-s \theta^{\prime}) \mathcal O_{\lambda}(1) (\bar{\mathtt t}^-z)^{2}+h^{2}\int\!\!\!{\int}_{Q} \tau(\bar{\mathtt t}^-s\theta^{\prime}) \mathcal{O}_{\lambda}(1) |\partial_h (\bar{\mathtt t}^-z)|^{2}. $$

Using that $\theta ^{\prime }=(2t-T)\theta ^{2}$ for t ∈ [0,T], we obtain

$$ \begin{array}{@{}rcl@{}} I_{31}&=\int\!\!\!{\int}_{Q} T\bar{\mathtt{t}}^{-}(s^{2}\theta)\mathcal O_{\lambda}(1)(\bar{\mathtt t}^-z)^{2} + \int\!\!\!{\int}_{Q} T\bar{\mathtt{t}}^{-}(\theta[sh]^{2})\mathcal O_{\lambda}(1)|\partial_h (\bar{\mathtt t}^-z)|^{2}. \end{array} $$

This ends the proof.

1.2.2 B.2.2 Proof of Lemma 2.7

The proof of this term can be carried out exactly as in [22], since the space variable does not play any major role. For completeness, we sketch it briefly.

Using formula Eq. 7, we write

$$ \begin{array}{@{}rcl@{}} I_{33}&=&-\int\!\!\!{\int}_{Q}\tau\varphi (\bar{\mathtt t}^-\theta^{\prime})(\bar{\mathtt t}^-z)\overline D_t z =-\frac{1}{2}\int\!\!\!{\int}_{Q}\tau \varphi (\bar{\mathtt t}^-\theta^{\prime})\overline D_t(z^{2})\\&&+\frac{\triangle t}{2} \int\!\!\!{\int}_{Q}\tau\varphi(\bar{\mathtt t}^-\theta^{\prime})(\overline D_t z)^{2}, \end{array} $$

and integrating by parts in time using Eq. 11, we get

$$ \begin{array}{@{}rcl@{}} I_{33}&=&\frac{1}{2}{\int}_{\Omega} \tau \varphi (\theta^{\prime})^{\frac{1}{2}}(z^{\frac{1}{2}})^{2}-\frac{1}{2}{\int}_{\Omega}\tau\varphi (\theta^{\prime})^{M+\frac{1}{2}}(z^{M+\frac{1}{2}})^{2}\\ &\quad &+\frac{1}{2}\int\!\!\!{\int}_{Q}\tau\varphi \overline D_t(\theta^{\prime})(\bar{\mathtt t}^-z)^{2}+\frac{\triangle t}{2} \int\!\!\!{\int}_{Q}\tau\varphi(\bar{\mathtt t}^-\theta^{\prime})(\overline D_t z)^{2}. \end{array} $$

(37)

By definition, we have that

$$ \theta^{\prime}=(2t-T)\theta^{2}, $$

(38)

thus $(\theta ^{\prime })^{\frac {1}{2}}<0$ and $(\theta ^{\prime })^{M+\frac {1}{2}}>0$. Therefore, recalling that φ < 0 for all x ∈Ω, we see that the first two terms in Eq. 37 are positive and therefore can be dropped. A further computation using Eq. 38 and Lemma A.15(ii) and yields

$$ I_{33}\geq -\int\!\!\!{\int}_{Q} (\bar{\mathtt t}^-\mu_{33}) (\bar{\mathtt t}^-z)^{2}- \int\!\!\!{\int}_{Q} (\bar{\mathtt t}^-\gamma_{33})(\overline D_t z)^{2} $$

with $\mu _{33}=(\tau T^{2}\theta ^{3}+\frac {\tau \triangle t}{\delta ^{4}T^{5}})\mathcal O_{\lambda }(1)$ and $\gamma _{33}=\triangle t \tau T \theta ^{2}\mathcal O_{\lambda }(1)$, where we have used that $\varphi =\mathcal O_{\lambda }(1)$.

1.3 B.3 A new estimate

1.3.1 B.3.1 Proof of Lemma 2.8

This is the most delicate and cumbersome estimate since it combines the action of both space and time discrete results. For clarity, we have divided the proof in three steps.

Step 1. An estimate for I₁₃. Using integration by parts in space, we get

$$ \begin{array}{@{}rcl@{}} I_{13}&=&\int\!\!\!{\int}_{Q}\bar{\mathtt{t}}^{-}\left( r\overline{\tilde{\rho}}{\Delta}_h z\right)\overline D_t{z}\\ &=& -\int\!\!\!{\int}_{Q} \overline D_t(\partial_h z) \partial_h(\bar{\mathtt t}^-z)\bar{\mathtt t}^-\widetilde{(r\overline{\tilde{\rho}})}-\int\!\!\!{\int}_{Q} \bar{\mathtt{t}}^{-}\left( \partial_h\left[r\overline{\tilde{\rho}} \right]\right)\partial_h(\bar{\mathtt t}^- z) \overline D_t\left( \widetilde{z}\right)\\ &=:& I_{13}^{(1)}+I_{13}^{(2)}, \end{array} $$

(39)

where we have used that $\overline D_t$ commutes with ∂_h and m_h in the first and second terms, respectively. Using formula Eq. 7, we have that the $I_{13}^{(1)}$ can be written as

$$ I_{13}^{(1)}=-\frac{1}{2}\int\!\!\!{\int}_{Q}\bar{\mathtt{t}}^{-}\widetilde{(r\overline{\tilde{\rho}})}\overline D_t\left( |\partial_h z|^{2}\right)+\frac{\triangle t}{2}\int\!\!\!{\int}_{Q} \bar{\mathtt{t}}^{-}\widetilde{(r\overline{\tilde{\rho}})}\left( \overline D_t(\partial_h z)\right)^{2} $$

and integrating by parts in time in the first integral, we get

$$ \begin{array}{@{}rcl@{}} I_{13}^{(1)}&=&\int\!\!\!{\int}_{Q}|\partial_h(\bar{\mathtt t}^+z)|^{2} \overline D_t \widetilde{(r\overline{\tilde{\rho}})}-{\int}_{\Omega} \widetilde{(r\overline{\tilde{\rho}})}^{M+\frac{1}{2}} \left|(\partial_h z)^{M+\frac{1}{2}}\right|^{2} \\ &\quad +& {\int}_{\Omega} \widetilde{(r\overline{\tilde{\rho}})}^{\frac{1}{2}} \left|(\partial_h z)^{\frac{1}{2}}\right|^{2}+\frac{\triangle t}{2}\int\!\!\!{\int}_{Q} \bar{\mathtt{t}}^{-}\widetilde{(r\overline{\tilde{\rho}})}\left( \overline D_t(\partial_h z)\right)^{2}. \end{array} $$

From Proposition A.7(i), we observe that for $0<(sh)\leq \frac {\tau h}{\delta T^{2}}<\epsilon _{1}(\lambda )$ with 𝜖₁(λ) small enough, we have that $\left (r\overline {\tilde {\rho }}\right )\geq c_{\lambda }>0$; thus, the last three terms of the above equation have prescribed signs (we remark that the extra average does not affect the form of that estimate). That is,

$$ \begin{array}{@{}rcl@{}} I_{13}^{(1)}&\geq& K - c_{\lambda} {\int}_{\Omega} \left|(\partial_h z)^{M+\frac{1}{2}}\right|^{2} \end{array} $$

$$ \begin{array}{@{}rcl@{}} &\quad +&c_{\lambda} {\int}_{\Omega} \left|(\partial_h z)^{\frac{1}{2}}\right|^{2}+c_{\lambda}\triangle t \int\!\!\!{\int}_{Q} \left( \overline D_t(\partial_h z)\right)^{2}, \end{array} $$

(40)

where $K:=\int \limits \!\!\!{\int \limits }_{Q}|\partial _h(\bar {\mathtt t}^+z)|^{2} \overline D_t\widetilde {(r\overline {\tilde {\rho }})}$. We claim the following.

Claim B.3

Provided $\frac {\triangle t \tau }{\delta ^{2} T^{3}}\leq \frac {1}{2}$, we have

$$ \overline D_t\widetilde{(r\overline{\tilde{\rho}})}=T \bar{\mathtt{t}}^{-}(\theta[sh]^{2})\mathcal O_{\lambda}(1)+\left( \frac{\tau \triangle t}{\delta^{3} T^{4}}\right)\left( \frac{\tau h}{\delta^{} T^{2}}\right)\mathcal O_{\lambda}(1). $$

(41)

Proof

Notice that we can write $2 \overline D_t\widetilde {(r\overline {\tilde {\rho }})}=(\mathtt s^{+}+\mathtt s^{-}) \overline D_t (r\overline {\tilde {\rho }})$. The result follows by applying Lemma A.16(ii). □

Observe that the condition on △t of Claim B.3 is in agreement with our initial hypothesis. Actually, the assumption of Lemma 2.8 is stronger than this condition. Using that $|\partial _h(\bar {\mathtt t}^+z)|^{2}\leq C |\partial _h(\bar {\mathtt t}^-z)|^{2}+C(\triangle t)^{2}(\overline D_t(\partial _h z))^{2}$ and Eq. 41 we get

$$ \begin{array}{@{}rcl@{}} K&\geq - \int\!\!\!{\int}_{Q}|\partial_h (\bar{\mathtt t}^-z)|^{2} (\bar{\mathtt t}^-\nu_{K}) - C_{\lambda} \triangle t \int\!\!\!{\int}_{Q} \left( \frac{\tau \triangle t}{\delta^{3} T^{4}}\right)\left( \frac{\tau h}{\delta^{} T^{2}} \right) \left( \overline D_t(\partial_h z)\right)^{2}, \end{array} $$

(42)

for some C_λ > 0 uniform with respect to △t and where $\nu _{K}:=T \theta (sh)^{2} \mathcal O_{\lambda }(1)+\left (\frac {\tau \triangle t}{\delta ^{3} T^{4}}\right )\left (\frac {\tau h}{\delta ^{} T^{2}}\right )\mathcal O_{\lambda }(1)$. Notice that taking 𝜖₁(λ) small enough in our initial hypothesis the last term of Eq. 40 controls the last term of Eq. 42. So, overall, $I_{13}^{(1)}$ can be bounded as

$$ I_{13}^{(1)}\geq - c_{\lambda} {\int}_{\Omega} \left|(\partial_h z)^{M+\frac{1}{2}}\right|^{2} - \int\!\!\!{\int}_{Q}|\partial_h (\bar{\mathtt t}^-z)|^{2} (\bar{\mathtt t}^-\nu_{K})+ \tilde{c}_{\lambda}\triangle t \int\!\!\!{\int}_{Q} \left( \overline D_t(\partial_h z)\right)^{2}, $$

(43)

for a constant $\tilde c_{\lambda }>0$ uniform with respect to △t.

Let us comeback to the term $I_{13}^{(2)}$. From Proposition A.8(ii), we have that $\partial _h[r\overline {\tilde {\rho }}]=(sh)^{2}\mathcal O_{\lambda }(1)$ and using Cauchy-Schwarz and Young inequalities, we get

$$ |I_{13}^{(2)}| \leq C \left( \int\!\!\!{\int}_{Q} \bar{\mathtt{t}}^{-}\left( s^{-1}[sh]^{2}\right) |\overline D_t(\tilde{z})|^{2}+ \int\!\!\!{\int}_{Q} \bar{\mathtt{t}}^{-}(s[sh]^{2})|\partial_h(\bar{\mathtt t}^-z)|^{2}\right), $$

for some C > 0 only depending on λ. A direct computation shows that $|\overline D_t(\widetilde {z})|^{2}=|\widetilde {(\overline D_t z)}|^{2}\leq \widetilde {(\overline D_t z)^{2}}$ by convexity. This, together the fact that $z_{|\partial {\Omega }}^{n-\frac {1}{2}}=0$ for n ∈⟦1,M⟧, we can use Eq. 3 to deduce

$$ |I_{13}^{(2)}| \leq C \left( \int\!\!\!{\int}_{Q} \bar{\mathtt{t}}^{-}\left( s^{-1}[sh]^{2}\right) |\overline D_t(z)|^{2}+ \int\!\!\!{\int}_{Q} \bar{\mathtt{t}}^{-}(s[sh]^{2})|\partial_h(\bar{\mathtt t}^-z)|^{2}\right). $$

(44)

Combining Eqs. 43 and 44 will provide a lower bound for I₁₃. This concludes Step 1.

Step 2. An estimate for I₂₃. Using that $\overline {\tilde {z}}=z+\frac {h^{2}}{4}\overline {\partial _h}\partial _h z$, we can rewrite I₂₃ as

$$ \begin{array}{@{}rcl@{}} I_{23}&=& \int\!\!\!{\int}_{Q}\bar{\mathtt{t}}^{-}\left( r{\Delta}_h \rho \overline{\tilde{z}}\right) \overline D_t{z} \\ &=&\int\!\!\!{\int}_{Q}\bar{\mathtt{t}}^{-}\left( r{\Delta}_h \rho z\right)\overline D_t z+\frac{h^{2}}{4}\int\!\!\!{\int}_{Q}\bar{\mathtt{t}}^{-}\left( r{\Delta}_h \rho \right){\Delta}_h(\bar{\mathtt t}^-z)\overline D_t z =: I_{32}^{(1)}+I_{32}^{(2)}. \end{array} $$

Let us estimate $I_{23}^{(1)}$. Using identity Eq. 7 and the integration-by parts formula in Eq. 11, we see that

$$ \begin{array}{@{}rcl@{}} I_{23}^{(1)}&=&\frac{1}{2}\int\!\!\!{\int}_{Q} \bar{\mathtt{t}}^{-}\left( r{\Delta}_h \rho\right)\overline D_t(|z|^{2})-\frac{\triangle t}{2}\int\!\!\!{\int}_{Q}\bar{\mathtt{t}}^{-}\left( r{\Delta}_h \rho\right) (\overline D_t z)^{2} \\ &=& -\frac{1}{2}\int\!\!\!{\int}_{Q}\overline D_t(r{\Delta}_h \rho)(\bar{\mathtt t}^-z)^{2}+\frac{1}{2}{\int}_{\Omega} (r{\Delta}_h \rho)^{M+\frac{1}{2}} |z^{M+\frac{1}{2}}|^{2}\\&&-\frac{1}{2}{\int}_{\Omega} (r{\Delta}_h \rho)^{\frac{1}{2}} |z^{\frac{1}{2}}|^{2} \\ &\quad-&\frac{\triangle t}{2}\int\!\!\!{\int}_{Q}\bar{\mathtt{t}}^{-}\left( r{\Delta}_h \rho\right) (\overline D_t z)^{2}. \end{array} $$

To estimate the last three terms of the above expression, we can use that

$$ r{\Delta}_h \rho=s^{2}\mathcal O_{\lambda}(1)+s\lambda\phi \mathcal O(1)+s^{2}(sh)^{2}\mathcal O_{\lambda}(1)=s^{2}\mathcal O_{\lambda}(1). $$

(45)

For the first one, we can use directly Lemma A.16(iii) sampled on the primal mesh $\mathcal P$. Thus,

$$ \begin{array}{@{}rcl@{}} I_{23}^{(1)}&=& \int\!\!\!{\int}_{Q}(\bar{\mathtt t}^-\mu_{P})(\bar{\mathtt t}^-z)^{2}+{\int}_{\Omega} (s^{M+\frac{1}{2}})^{2}\mathcal O_{\lambda}(1)|z^{M+\frac{1}{2}}|^{2}+{\int}_{\Omega} (s^{\frac{1}{2}})^{2}\mathcal O_{\lambda}(1)|z^{\frac{1}{2}}|^{2}\\ && + \triangle t \int\!\!\!{\int}_{Q} (\bar{\mathtt t}^-s^{2})\mathcal O_{\lambda}(1)(\overline D_t z)^{2}, \end{array} $$

(46)

where $\mu _{P}=T s^{2}\theta \mathcal O_{\lambda }(1)+ \left (\frac {\tau ^{2} \triangle t}{\delta ^{4} T^{6}}\right )\mathcal O_{\lambda }(1)+\left (\frac {\tau \triangle t}{\delta ^{3} T^{4}}\right )\left (\frac {\tau h}{\delta ^{} T^{2}}\right )^{3}\mathcal O_{\lambda }(1)$.

We focus now on the term $I_{23}^{(2)}$. Using that ${\Delta }_h z=\overline {\partial _h}(\partial _h z)$, we can integrate by parts in space and get

$$ \begin{array}{@{}rcl@{}} I_{23}^{(2)}&=&\frac{h^{2}}{4}\int\!\!\!{\int}_{Q}\bar{\mathtt{t}}^{-}\left( r {\Delta}_h \rho \right) {\Delta}_h(\bar{\mathtt t}^-z)\overline D_t z\\ &=&-\frac{h^{2}}{4}\int\!\!\!{\int}_{Q} \bar{\mathtt{t}}^{-}\widetilde{\left( r {\Delta}_h \rho\right)}\partial_h(\bar{\mathtt t}^-z)\overline D_t(\partial_h z)\\ &&-\frac{h^{2}}{4}\int\!\!\!{\int}_{Q} \bar{\mathtt{t}}^{-}\left( \partial_h(r{\Delta}_h \rho)\right)\partial_h(\bar{\mathtt t}^-z) \overline D_t \left( \widetilde{z}\right)\\ &=:& \mathcal J_{1}+\mathcal J_{2}, \end{array} $$

(47)

where we have used once again that $\overline D_t$ commutes with the space-discrete operations ∂_h and m_h.

Arguing as we did for $I_{12}^{(1)}$, we see that

$$ \begin{array}{@{}rcl@{}} \mathcal J_{1} &=& -\frac{h^{2}}{8}\int\!\!\!{\int}_{Q}\bar{\mathtt{t}}^{-}\widetilde{\left( r{\Delta}_h \rho\right)}\overline D_t(|\partial_h z|^{2})+\triangle t\frac{h^{2}}{8}\int\!\!\!{\int}_{Q}\bar{\mathtt{t}}^{-}\widetilde{\left( r{\Delta}_h \rho \right)}|\overline D_t(\partial_h z)|^{2} \\ &=&-\frac{h^{2}}{8}{\int}_{\Omega}\widetilde{(r{\Delta}_h \rho)}^{M+\frac{1}{2}} |(\partial_h z)^{M+\frac{1}{2}}|^{2}+\frac{h^{2}}{8}{\int}_{\Omega}\widetilde{(r{\Delta}_h \rho)}^{\frac{1}{2}} |(\partial_h z)^{\frac{1}{2}}|^{2} \\ &&+ \frac{h^{2}}{8}\int\!\!\!{\int}_{Q}\overline D_t\widetilde{(r{\Delta}_h \rho)}|\partial_h(\bar{\mathtt t}^+z)|^{2}+ \triangle t \frac{h^{2}}{8}\int\!\!\!{\int}_{Q}\bar{\mathtt{t}}^{-}\widetilde{\left( r{\Delta}_h \rho \right)}|\overline D_t(\partial_h z)|^{2}. \end{array} $$

We have the following.

Claim B.4

Provided $\frac {\triangle t \tau }{\delta ^{2} T^{3}}\leq \frac {1}{2}$, we have

$$ \overline D_t\widetilde{(r{\Delta}_h \rho)}=T \textsf{t}^{-}(s^{2}\theta )\mathcal O_{\lambda}(1)+ \left( \frac{\tau^{2} \triangle t}{\delta^{4} T^{6}}\right)\mathcal O_{\lambda}(1)+\left( \frac{\tau \triangle t}{\delta^{3} T^{4}}\right)\left( \frac{\tau h}{\delta^{} T^{2}}\right)^{3}\mathcal O_{\lambda}(1). $$

(48)

The proof is exactly as in Claim B.3 but taking into account the estimate in Lemma A.16(iii).

Using Eq. 45 (and noting that the estimate does not change with the extra average in space), Claim B.4 and the fact that $|\partial _h(\bar {\mathtt t}^+z)|^{2}\leq C |\partial _h(\bar {\mathtt t}^-z)|^{2}+C(\triangle t)^{2}(\overline D_t(\partial _h z))^{2}$, we compute after a long but straightforward computation that

$$ \begin{array}{@{}rcl@{}} |\mathcal J_{1}| & \leq& C{\int}_{\Omega} (s^{M+\frac{1}{2}}h)^{2}|(\partial_h z)^{M+\frac{1}{2}}|^{2}+C{\int}_{\Omega} (s^{\frac{1}{2}}h)^{2}|(\partial_h z)^{\frac{1}{2}}|^{2}\\ &&+ C\int\!\!\!{\int}_{Q} T \bar{\mathtt t}^-\theta[sh]^{2}|\partial_h (\bar{\mathtt t}^-z)|^{2}\\ &&+ C\int\!\!\!{\int}_{Q}\left[ \left( \frac{\tau^{2} \triangle t}{\delta^{4} T^{6}}\right)+\left( \frac{\tau \triangle t}{\delta^{3} T^{4}}\right)\left( \frac{\tau h}{\delta^{} T^{2}}\right)^{3} \right] |\partial_h (\bar{\mathtt t}^-z)|^{2} \end{array} $$

$$ \begin{array}{@{}rcl@{}} && +C\triangle t \int\!\!\!{\int}_{Q} \left( \frac{\tau\triangle t}{\delta^{2}T^{3}}\right)\left( \frac{\tau h}{\delta T^{2}}\right)|\overline D_t(\partial_h z)|^{2}+C\triangle t \int\!\!\!{\int}_{Q}\left( \frac{\tau h}{\delta T^{2}}\right)^{2}|\overline D_t(\partial_h z)|^{2} \\ && + C \triangle t \int\!\!\!{\int}_{Q}\left[ \left( \frac{\tau^{2} \triangle t}{\delta^{4} T^{6}}\right)+\left( \frac{\tau \triangle t}{\delta^{3} T^{4}}\right)\left( \frac{\tau h}{\delta^{} T^{2}}\right)^{3} \right] |\overline D_t(\partial_h{z})|^{2}, \end{array} $$

(49)

for some positive C only depending on λ. Above we used that h ≪ 1 to adjust some powers of h. We also notice that in the above equation, all of the terms containing $|\overline D_t(\partial _h z)|^{2}$ can be made small enough by the initial hypothesis of the lemma. This will be important in the next step since those terms will be absorbed by the corresponding one in Eq. 43.

We turn our attention to the term $\mathcal J_{2}$. Estimate Eq. 45 and a quick computation yield

$$ \begin{array}{@{}rcl@{}} |\mathcal J_{2}| &\leq C\left( \int\!\!\!{\int}_{Q}\bar{\mathtt{t}}^{-}\left( s[sh]^{2}\right)|\partial_h(\bar{\mathtt t}^-z)|^{2}+\int\!\!\!{\int}_{Q}\bar{\mathtt{t}}^{-}\left( s^{-1}[sh]^{2}\right)|\overline D_t(\tilde{z})|^{2}\right)\\ &\leq C\left( \int\!\!\!{\int}_{Q}\bar{\mathtt{t}}^{-}\left( s[sh]^{2}\right)|\partial_h(\bar{\mathtt t}^-z)|^{2}+\int\!\!\!{\int}_{Q}\bar{\mathtt{t}}^{-}\left( s^{-1}[sh]^{2}\right)|\overline D_t z|^{2}\right), \end{array} $$

(50)

where the second inequality is obtained by arguing exactly as for the term $I_{13}^{(2)}$.

Step 3. Conclusion. Now, we are in position to find a lower bound for I₁₃ + I₂₃. First, we shall notice that the last three terms of Eq. 49 can be controlled by the last term of Eq. 43 by decreasing (if necessary) the parameter 𝜖₁(λ) in our initial hypothesis. Then, we just need to combine Eq. 43, Eq. 44, Eq. 46, Eq. 47, Eq. 49, and Eq. 50 to obtain

$$ \begin{array}{@{}rcl@{}} I_{13}+I_{23}&\geq& - c_{\lambda} {\int}_{\Omega} \left|(\partial_h z)^{M+\frac{1}{2}}\right|^{2} - \int\!\!\!{\int}_{Q} (\bar{\mathtt t}^-\mu_{+})(\bar{\mathtt t}^-z)^{2} - \int\!\!\!{\int}_{Q}(\bar{\mathtt t}^-\nu_{+}) |\partial_h (\bar{\mathtt t}^-z)|^{2} \\ &&+ c^{\prime}_{\lambda}\triangle t \int\!\!\!{\int}_{Q} \left( \overline D_t(\partial_h z)\right)^{2} - \int\!\!\!{\int}_{Q} (\bar{\mathtt t}^- \gamma_{+})|\overline D_t z|^{2} \\ && - {\int}_{\Omega} (s^{M+\frac{1}{2}})^{2}\mathcal O_{\lambda}(1)|z^{M+\frac{1}{2}}|^{2}-{\int}_{\Omega} (s^{\frac{1}{2}})^{2}\mathcal O_{\lambda}(1)|z^{\frac{1}{2}}|^{2} \\ && - {\int}_{\Omega} (s^{M+\frac{1}{2}}h)^{2}\mathcal O_{\lambda}(1)|(\partial_h z)^{M+\frac{1}{2}}|^{2}-{\int}_{\Omega} (s^{\frac{1}{2}}h)^{2}\mathcal O_{\lambda}(1)|(\partial_h z)^{\frac{1}{2}}|^{2}, \end{array} $$

for some $c_{\lambda },c^{\prime }_{\lambda }>0$ uniform with respect to h and △t and where

$$ \begin{array}{@{}rcl@{}} \nu_{+}&:=& \left\{T \theta (sh)^{2} +s(sh)^{2}+\left( \frac{\tau \triangle t}{\delta^{3} T^{4}}\right)\left( \frac{\tau h}{\delta^{} T^{2}}\right) + \left( \frac{\tau^{2} \triangle t}{\delta^{4} T^{6}}\right)+\left( \frac{\tau \triangle t}{\delta^{3} T^{4}}\right)\left( \frac{\tau h}{\delta^{} T^{2}}\right)^{3}\right\}\\&&\mathcal O_{\lambda}(1), \\ \mu_{+}&:=&\left\{T s^{2}\theta + \left( \frac{\tau^{2} \triangle t}{\delta^{4} T^{6}}\right)+\left( \frac{\tau \triangle t}{\delta^{3} T^{4}}\right)\left( \frac{\tau h}{\delta^{} T^{2}}\right)^{3}\right\}\mathcal O_{\lambda}(1), \\ \gamma_{+}&:= &\left\{s^{-1}(sh)^{2}+\triangle t s^{2} \right\}\mathcal O_{\lambda}(1). \end{array} $$

This ends the proof.

Appendix C: Some intermediate lemmas

1.1 C.1 Proof of Lemma 2.12

By shifting the integral in time (see Eq. 8) and then using Eq. 3, we have

(51)

where we recall that s = τ𝜃. Since ϕ is a positive function, notice that the last two terms of the above expression are positive.

Let us focus on the term $\mathcal D$. From Lemma A.2, we get

For the term $\mathcal D_{1}$, using once again Lemma A.2 gives

where we have used that $\overline {\partial _h}\partial _h = {\Delta }_h$. On the other hand, integrating by parts in the space variable, from $\mathcal D_{2}$, we get

(52)

Notice that $\overline {\partial _h} (\phi )=\partial _{x}\phi +h^{2}\mathcal O_{\lambda }(1)=\mathcal O_{\lambda }(1)$ thanks to Lemma A.5(iii). Thus, once the parameter λ is fixed, we can choose h sufficiently small such that the last two terms of Eq. 51 control the last two terms of Eq. 52. Moreover, using items (ii) and (iii) of Lemma A.5, we see that $\partial _h(\overline {\partial _h} \phi )={\partial _{x}^{2}}\phi +h^{2}\mathcal O_{\lambda }(1)=\O _{\lambda }(1)$ and $\overline {\phi }=\phi +h\mathcal O_{\lambda }(1)$. From these and putting together Eq. 51–Eq. 52, we obtain

Shifting the integrals in time in the right-hand side of the above expression yields the desired result.

1.2 C.2 Proof of Lemma 2.14

We begin by increasing, if necessary, the parameter τ₁ such that τ₁ ≥ 1 and τ ≥ 1. Notice that

$$ 1\leq \tau_{1}\left( \tfrac{1}{T}+1\right)\leq \tau\theta(t)=s(t), \quad t\in[0,T]. $$

(53)

We repeat the definition of $\underline {X_{2}}$ for convenience. We have

$$ \begin{array}{@{}rcl@{}} \!\!\!\!\!\!\!\!\!\!\underline{X_{2}}\!&=&\! \left\{\left( \frac{\tau \triangle t}{\delta^{4}T^{5}}\right) + \left( \frac{\tau^{2} \triangle t}{\delta^{4} T^{6}}\right) + \left( \frac{\triangle t \tau}{\delta^{3}T^{4}}\right)^{2}+\left( \frac{\triangle t \tau^{2}}{\delta^{4} T^{6}}\right)^{2} + \left( \frac{\tau \triangle t}{\delta^{3} T^{4}}\right)\left( \frac{\tau h}{\delta^{} T^{2}}\right)^{3}\right\}\\ && \int\!\!\!{\int}_{Q}(\bar{\mathtt t}^-z^{2})\\ && \!+ \left\{\!\left( \!\frac{\tau^{2} \triangle t}{\delta^{4} T^{6}}\!\right) + \left( \frac{\tau \triangle t}{\delta^{3} T^{4}}\right)\!\left( \frac{\tau h}{\delta^{} T^{2}}\right) + \left( \frac{\tau \triangle t}{\delta^{3} T^{4}}\right)\!\left( \frac{\tau h}{\delta^{} T^{2}}\right)^{3}\! \right\}\int\!\!\!{\int}_{Q}|\partial_h(\bar{\mathtt t}^-z)|^{2}. \end{array} $$

(54)

We remark that at this point, we have the condition $\frac {\tau h}{\delta T^{2}}\leq \epsilon _{3}$ for some 𝜖₃ = 𝜖₃(λ) small enough (see Eq. ??). Recalling that δ ≤ 1/2, 0 < T < 1 and Eq. 53 allows us to see that provided

$$ \frac{\tau^{2}\triangle t}{\delta^{4}T^{6}}\leq \kappa_{1} $$

(55)

for some κ₁ > 0 small enough, the term $\underline {X_{2}}$ can be bounded as

$$ \underline{X_{2}}\leq \left( 2\kappa_{1}+2{\kappa_{1}^{2}}+\kappa_{1}{\epsilon_{3}^{2}}\right)\int\!\!\!{\int}_{Q}(\bar{\mathtt t}^-s)^{3}(\bar{\mathtt t}^-z)^{2} \leq 5 \kappa_{1}\int\!\!\!{\int}_{Q}(\bar{\mathtt t}^-s)^{3}(\bar{\mathtt t}^-z)^{2}. $$

(56)

Notice that in the first inequality we have the product of the small parameters κ₁ and 𝜖₃. Since 𝜖₃ has been already chosen small we can bound it uniformly by 1.

On the other hand, from definition Eq. ?? and rewriting it as $\underline {W}={\sum }_{i=1}^{3}W^{(i)}$, we proceed to bound each of the terms. For the first one, using that $\max \limits _{t\in [0,T]}\theta \leq (\delta T^{2})^{-1}$ we get

$$ \begin{array}{@{}rcl@{}} W^{(1)}&=&\int\!\!\!{\int}_{Q} \triangle t\left( \tau T (\bar{\mathtt t}^-\theta)^{2} + \frac{\tau \triangle t}{\delta^{3} T^{4}}\right)(\overline D_t z)^{2}\\ &\leq& \left\{\frac{\triangle t \tau^{2}}{\delta^{3} T^{5}}+ \left( \frac{\tau \triangle t }{\delta^{2}T^{3}}\right)^{2}\right\}\int\!\!\!{\int}_{Q} (\bar{\mathtt t}^-s)^{-1}(\overline D_t z)^{2}\\ & \leq& (\kappa_{1}+{\kappa_{1}^{2}})\int\!\!\!{\int}_{Q}(\bar{\mathtt t}^-s)^{-1}(\overline D_t z)^{2}. \end{array} $$

(57)

For W⁽²⁾, we have

$$ \begin{array}{@{}rcl@{}} W^{(2)}&=&\int\!\!\!{\int}_{Q} \triangle t (\bar{\mathtt t}^-s)^{2} (\overline D_t z)^{2}=\int\!\!\!{\int}_{Q}\triangle t (\bar{\mathtt t}^-s)^{3} (\bar{\mathtt t}^-s)^{-1} (\overline D_t z)^{2} \\ & \leq& \kappa_{2} \int\!\!\!{\int}_{Q} (\bar{\mathtt t}^-s)^{-1}(\overline D_t z)^{2}, \end{array} $$

(58)

where the condition

$$ \frac{\tau^{3}\triangle t}{\delta^{3} T^{6}}\leq \kappa_{2} $$

(59)

holds for some κ₂ > 0 small enough. Finally, using Eq. 53, we see that

$$ \begin{array}{@{}rcl@{}} W^{(3)}&=&\int\!\!\!{\int}_{Q}\left( T^{2}(\tau\triangle t)^{2}\theta^{4}+\frac{\tau^{2}(\triangle t)^{4}}{\delta^{6} T^{8}}\right) (\overline D_t z)^{2}\\ &\leq& \tau^{4}(\triangle t^{2})T^{2}\int\!\!\!{\int}_{Q}(\bar{\mathtt t}^-\theta)^{6}(\bar{\mathtt t}^-s)^{-1}(\overline D_t z)^{2} + \left( \frac{\tau \triangle t}{\delta^{3} T^{4}}\right)^{4} \int\!\!\!{\int}_{Q}(\bar{\mathtt t}^-s)^{-1}(\overline D_t z)^{2}\\ & \leq& \left\{\left( \frac{\tau^{2} \triangle t}{\delta^{3} T^{5}}\right)^{2}+\left( \frac{\tau \triangle t}{\delta^{3} T^{4}}\right)^{4}\right\} \int\!\!\!{\int}_{Q}(\bar{\mathtt t}^-s)^{-1}(\overline D_t z)^{2}\\ & \leq& \left( {\kappa_{1}^{2}}+{\kappa_{1}^{4}}\right)\int\!\!\!{\int}_{Q} (\bar{\mathtt t}^-s)^{-1}(\overline D_t z)^{2}. \end{array} $$

(60)

Since δ ≤ 1/2, τ ≥ 1 and 0 < T < 1, we can combine conditions Eqs. 55 and 59 into a single one verifying

$$ \frac{\tau^{4}\triangle t }{\delta^{4} T^{6}}\leq \epsilon_{5} $$

for some 𝜖₅ = 𝜖₅(λ) small enough. Collecting estimates Eq. 57, Eqs. 58, and 60 gives the desired result.

1.3 C.3 Proof of Lemma 2.16

We adapt the procedure used in the continuous setting. In particular, we follow [27, pp. 1409]. Let us consider $\eta \in C_{c}^{\infty }({\Omega })$ such that

$$ 0\leq \eta\leq 1\text{ in } {\Omega}, \quad \eta=1 \text{ in a neighborhood of } \mathcal B_{0}, \quad \text{supp} \eta \subset\subset \mathcal B. $$

(61)

By the properties of discretization, we can ensure the additional property

$$ \frac{\overline{\partial_h}(\eta)}{\eta^{1/2}}\in L^{\infty}({\Omega}) $$

(62)

uniformly with respect to h. Obviously, the above function is supported within $\mathcal B$.

By shifting the time integral and the definition of the function η, we have

(63)

For the first term in the above expression, we have by using Eq. 62 and Cauchy-Schwarz and Young inequalities

(64)

where we have used that s(t) ≥ 1 for all t to adjust the power of s in the last term of the above equation.

For the term $\mathcal L_{2}$, using that $\overline {\eta }=\eta +h\mathcal O(1)$ together with Cauchy-Schwarz and Young inequalities

(65)

for any γ > 0.

Using estimates Eq. 64–Eq. 65 in Eq. 63 and recalling the properties of η in Eq. 61 gives the desired result after shifting the integral in time.