Let us fix a 𝔗 > 0. For *τ* > 0 let *N* be such as *Nτ* = 𝔗 and let us consider the function

$$\begin{array}{}{\displaystyle {V}^{\tau}(t,S)=\sum _{n=0}^{+\mathrm{\infty}}\sigma (\frac{t}{\tau}-n){V}^{n}(S),}\end{array}$$(25)

where *σ* is defined by *σ*(*t*) = max(0, 1 – |*t*|). We plan to let *N* → +∞ (and so *τ* → 0). Note that in the time interval *I*_{n} = ]*nτ*, (*n* + 1)*τ*[ we have

$$\begin{array}{}{\displaystyle {V}^{\tau}(t,S)=(n+1-\frac{t}{\tau}){V}^{n}(S)+(\frac{t}{\tau}-n){V}^{n+1}(S),}\end{array}$$(26)

$$\begin{array}{}{\displaystyle \frac{d{V}^{\tau}}{dt}(t,S)=\frac{{V}^{n+1}(S)-{V}^{n}(S)}{\tau}.}\end{array}$$(27)

#### Lemma 3.1

*The following assertions hold true*

*The sequence* $\begin{array}{}{\displaystyle \frac{d{V}^{\tau}}{dt}}\end{array}$ *is bounded in* $\begin{array}{}{\displaystyle {L}_{S}^{2}}\end{array}$( *Ω* × [0, 𝔗]) + $\begin{array}{}{\displaystyle {L}_{S}^{\frac{3}{2}}}\end{array}$(*Ω* × [0, 𝔗]).

*The sequence* $\begin{array}{}{\displaystyle {V}_{S}^{\tau}}\end{array}$ *is bounded in* *L*^{∞}([0, 𝔗]; *L*^{2}(*Ω*)).

*The sequence* *β*($\begin{array}{}{\displaystyle \frac{d{V}^{\tau}}{dt}}\end{array}$) *is bounded in* $\begin{array}{}{\displaystyle {L}_{S}^{2}}\end{array}$(*Ω* × [0, 𝔗]) ∩ $\begin{array}{}{\displaystyle {L}_{S}^{3}}\end{array}$(*Ω* × [0, 𝔗]).

*The sequence* *V*^{τ} *is bounded in* *L*^{∞}(*Ω* × [0, 𝔗]).

#### Proof

1. Summing (24) for *V͂*^{n+1} = *V*^{n} with respect to *n* yields

$$\begin{array}{}{\displaystyle \sum _{n=0}^{+\mathrm{\infty}}\tau \underset{\mathit{\Omega}}{\int}\beta \left(\frac{{V}^{n+1}-{V}^{n}}{\tau}\right)\left(\frac{{V}^{n+1}-{V}^{n}}{\tau}\right)\frac{dS}{{S}^{2}}+\frac{1}{2}\underset{\mathit{\Omega}}{\int}|{V}_{S}^{n+1}{|}^{2}dS\le \frac{1}{2}\underset{\mathit{\Omega}}{\int}|{V}_{S}^{0}{|}^{2}dS.}\end{array}$$

Hence 1. and 2. of the Lemma are proved. The point 3. is straightforward due to the properties of *β* that maps *L*^{2} + $\begin{array}{}{\displaystyle {L}^{\frac{3}{2}}}\end{array}$ into *L*^{2} ∩ *L*^{3}. Besides, since *V*^{τ} is a convex combination of *V*^{n} and *V*^{n+1} in *I*_{n} then *V*^{0} ≤ *V*^{τ} ≤ 0 everywhere. This completes the proof of the Lemma.□

We know pass to the limit when *τ* → 0 using a monotonicity argument [19].

For (*t*, *S*) in *I*_{n} × *Ω*, we have

$$\begin{array}{}{\displaystyle \beta \left(\frac{d{V}^{\tau}}{dt}\right)-{S}^{2}{V}_{SS}^{\tau}={S}^{2}(n+1-\frac{t}{\tau})({V}_{SS}^{n+1}-{V}_{SS}^{n}).}\end{array}$$(28)

We first consider the scalar product in $\begin{array}{}{\displaystyle {L}_{S}^{2}}\end{array}$(*Ω*) by a test function *ϕ* ∈ *D*(*Ω*); then we have

$$\begin{array}{}{\displaystyle \underset{\mathit{\Omega}}{\int}\left(\beta \left(\frac{d{V}^{\tau}}{dt}\right)-{S}^{2}{V}_{SS}^{\tau}\right)\frac{\varphi (S)}{{S}^{2}}dS={\u03f5}_{n}.}\end{array}$$(29)

where

$$\begin{array}{}{\displaystyle {\u03f5}_{n}=(n+1-\frac{t}{\tau})\underset{\mathit{\Omega}}{\int}\left(\frac{{V}^{n+1}-{V}^{n}}{\tau}\right)\tau {\varphi}_{SS}.}\end{array}$$

Let us observe that

$$\begin{array}{}{\displaystyle {\u03f5}_{n}\le C\tau ||\frac{d{V}^{\tau}}{dt}|{|}_{{L}_{S}^{2}(\mathit{\Omega})+{L}_{S}^{\frac{3}{2}}(\mathit{\Omega})}||{\varphi}_{SS}|{|}_{{L}_{S}^{2}(\mathit{\Omega})\cap {L}_{S}^{3}(\mathit{\Omega})},}\end{array}$$

and then this term converges towards 0 when *τ* → 0. Considering now a subsequence *V*^{τ} that converges towards *V* in *L*^{∞}(0, 𝔗; 𝕍) and such that $\begin{array}{}{\displaystyle (\frac{d{V}^{\tau}}{dt},\beta (\frac{d{V}^{\tau}}{dt}))}\end{array}$ converges weakly towards $\begin{array}{}{\displaystyle (\frac{dV}{dt},\chi )\text{\hspace{0.17em}}\text{in}\text{\hspace{0.17em}}{L}_{J}\times {L}_{J}^{\prime},}\end{array}$ we have that *χ* = *S*^{2}*V*_{SS}.

We now prove that $\begin{array}{}{\displaystyle \chi =\beta \left(\frac{dV}{dt}\right).}\end{array}$ We then compute

$$\begin{array}{}{\displaystyle 0\le {Q}_{\tau}=\underset{0}{\overset{\mathfrak{T}}{\int}}<\beta \left(\frac{d{V}^{\tau}}{dt}\right)-\beta (v),\frac{d{V}^{\tau}}{dt}-v{>}_{{L}_{S}^{2}},}\end{array}$$(30)

where *v* is a test function. By weak convergence we have

$$\begin{array}{}{\displaystyle \underset{0}{\overset{\mathfrak{T}}{\int}}(<\beta \left(\frac{d{V}^{\tau}}{dt}\right),v{>}_{{L}_{S}^{2}}+<\beta (v),\frac{d{V}^{\tau}}{dt}{>}_{{L}_{S}^{2}})\to \underset{0}{\overset{\mathfrak{T}}{\int}}(<\chi ,v{>}_{{L}_{S}^{2}}+<\beta (v),\frac{dV}{dt}{>}_{{L}_{S}^{2}}).}\end{array}$$(31)

Moreover, observing that

$$\begin{array}{}\begin{array}{rl}\underset{0}{\overset{\mathfrak{T}}{\int}}<\beta \left(\frac{d{V}^{\tau}}{dt}\right),\frac{d{V}^{\tau}}{dt}{>}_{{L}_{S}^{2}}& =\sum _{n}\underset{\mathit{\Omega}}{\int}{V}_{SS}^{n+1}({V}^{n+1}-{V}^{n})dS\\ & \le -\frac{1}{2}\underset{\mathit{\Omega}}{\int}|{V}_{S}^{\tau}(\mathfrak{T}){|}^{2}dS+\frac{1}{2}\underset{\mathit{\Omega}}{\int}|{V}_{S}^{0}(0){|}^{2}dS,\end{array}\end{array}$$

we have, by weak convergence

$$\begin{array}{}\begin{array}{rl}\overline{lim}\underset{0}{\overset{\mathfrak{T}}{\int}}<\beta \left(\frac{d{V}^{\tau}}{dt}\right),\frac{d{V}^{\tau}}{dt}{>}_{{L}_{S}^{2}}& \le \underset{=-{\displaystyle \frac{1}{2}}{\displaystyle \underset{0}{\overset{T}{\int}}\frac{d}{dt}||V||{|}_{\mathbb{V}}^{2}dt}}{\underset{\u23df}{-\frac{1}{2}\underset{\mathit{\Omega}}{\int}|{V}_{S}(T){|}^{2}dS+\frac{1}{2}\underset{\mathit{\Omega}}{\int}|{V}_{S}(0){|}^{2}dS}}.\end{array}\end{array}$$

Therefore

$$\begin{array}{}{\displaystyle \overline{lim}\underset{\mathit{\Omega}}{\int}\underset{0}{\overset{\mathfrak{T}}{\int}}\beta \left(\frac{d{V}^{\tau}}{dt}\right)\frac{d{V}^{\tau}}{dt}\le -\frac{1}{2}\underset{0}{\overset{\mathfrak{T}}{\int}}\frac{d}{dt}||V||{|}_{\mathbb{V}}^{2}=\underset{0}{\overset{\mathfrak{T}}{\int}}<\chi ,\frac{dV}{dt}{>}_{{L}_{S}^{2}}.}\end{array}$$(32)

Thus we have

$$\begin{array}{}{\displaystyle 0\le \overline{lim}{Q}_{\tau}\le \underset{0}{\overset{\mathfrak{T}}{\int}}<\beta (v)-\chi ,v-\frac{dV}{dt}{>}_{{L}_{S}^{2}}.}\end{array}$$(33)

We choose $\begin{array}{}{\displaystyle v=\frac{dV}{dt}+\u03f5\varphi}\end{array}$ where *ϕ* is a given test function; then we have

$$\begin{array}{}{\displaystyle 0\le \u03f5\underset{0}{\overset{T}{\int}}<\beta \left(\frac{dV}{dt}\right)-\chi ,\varphi {>}_{{L}_{S}^{2}}+o(\u03f5).}\end{array}$$(34)

By changing *ϕ* to –*ϕ* and tending *ϵ* to 0^{+} we obtain

$$\begin{array}{}{\displaystyle \beta \left(\frac{dV}{dt}\right)=\chi \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{a.e}\text{\hspace{0.17em}}.}\end{array}$$(35)

This completes the proof of the existence of a solution *V* that belongs to *L*^{∞}(0, 𝔗, 𝕍) such that $\begin{array}{}{\displaystyle \frac{dV}{dt}}\end{array}$ belongs to *L*^{2}(*Ω* × (0, 𝔗)) + $\begin{array}{}{\displaystyle {L}^{\frac{3}{2}}}\end{array}$(*Ω* × (0, 𝔗)). To prove that *V* is continuous in *t* with values in 𝕍 is a consequence of

$$\begin{array}{}{\displaystyle ||V(t)-V(s)|{|}_{\mathbb{V}}\le ||\frac{dV}{dt}|{|}_{{L}^{2}(\mathit{\Omega}\times (0,\mathfrak{T}))+{L}^{\frac{3}{2}}(\mathit{\Omega}\times (0,\mathfrak{T}))}max(|t-s{|}^{\frac{1}{2}},|t-s{|}^{\frac{1}{3}}).}\end{array}$$

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