To begin with, we define a weak solution of (1.10).

#### Definition 2.1.

By a weak solution of problem (1.10) we mean a pair $(u,\mathrm{\Phi})$ such that

$u\in {L}^{\mathrm{\infty}}({\mathbb{R}}_{+},{L}_{x}^{2}(\mathbb{R}))\cap {C}_{w}({\mathbb{R}}_{+},{L}_{x}^{2}(\mathbb{R}))\cap {L}^{2}((0,T),{H}_{x}^{1}(\mathbb{R})),$$\mathrm{\Phi}\in {L}^{\mathrm{\infty}}({\mathbb{R}}_{+},{\mathbb{H}}_{x,y}^{1}({\mathbb{R}}^{2}))\cap C({\mathbb{R}}_{+},{\mathbb{L}}_{x,y}^{2}({\mathbb{R}}^{2})),$

and which verifies

$\{\begin{array}{cc}& \frac{d}{dt}\left[{(u(t),\stackrel{~}{u})}_{{L}^{2}(\mathbb{R})}+2{(\mathrm{\Gamma}\mathrm{\Phi}(t),\stackrel{~}{u})}_{{L}^{2}(\mathbb{R})}\right]+{({u}_{x}(t),\stackrel{~}{u})}_{{L}^{2}(\mathbb{R})}+{({u}_{x}(t),{\stackrel{~}{u}}_{x})}_{{L}^{2}(\mathbb{R})}\hfill \\ & =-{(u(t){u}_{x}(t),\stackrel{~}{u})}_{{L}^{2}(\mathbb{R})}\hspace{1em}\text{for all}\stackrel{~}{u}\in {H}^{1}(\mathbb{R}),\hfill \\ & \frac{d}{dt}{[\mathrm{\Phi}(t),\stackrel{~}{\mathrm{\Phi}}]}_{{\mathbb{L}}^{2}({\mathbb{R}}^{2})}+{[{\mathrm{\Phi}}_{y}(t),{\stackrel{~}{\mathrm{\Phi}}}_{y}]}_{{\mathbb{L}}^{2}({\mathbb{R}}^{2})}+\epsilon {[{\mathrm{\Phi}}_{x}(t),{\stackrel{~}{\mathrm{\Phi}}}_{x}]}_{{\mathbb{L}}^{2}({\mathbb{R}}^{2})}\hfill \\ & ={(u(t),\stackrel{~}{\mathrm{\Phi}}(x,0))}_{{L}^{2}(\mathbb{R})}\hspace{1em}\text{for all}\stackrel{~}{\mathrm{\Phi}}\in {\mathbb{H}}^{1}({\mathbb{R}}^{2}),\hfill \\ & u(0)={u}_{0}\in {L}^{2}(\mathbb{R}),\mathrm{\Phi}(0)=0.\hfill \end{array}$

Before going to the proof of Theorem 1.1, we state some preliminary inequalities.

#### Lemma 2.2.

*For all $\mathrm{\Phi}\mathrm{\in}X$, we have*

${|\mathrm{\Gamma}(\mathrm{\Phi})|}_{{L}_{x}^{2}(\mathbb{R})}\le {\parallel \mathrm{\Phi}\parallel}_{{\mathbb{L}}^{2}({\mathbb{R}}^{2})}^{1/2}{\parallel {\mathrm{\Phi}}_{y}\parallel}_{{\mathbb{L}}^{2}({\mathbb{R}}^{2})}^{1/2}.$(2.1)

*Moreover, for all $\mathrm{\Phi}\mathrm{\in}{\mathrm{H}}^{\mathrm{1}}\mathit{}\mathrm{(}{\mathrm{R}}^{\mathrm{2}}\mathrm{)}$, we have the straightforward consequence*

${|\mathrm{\Gamma}(\mathrm{\Phi})|}_{{L}_{x}^{2}(\mathbb{R})}\le {\parallel \mathrm{\Phi}\parallel}_{{\mathbb{L}}^{2}({\mathbb{R}}^{2})}^{1/2}{\parallel \mathrm{\Phi}\parallel}_{{\mathbb{H}}^{1}({\mathbb{R}}^{2})}^{1/2}.$

#### Proof.

We prove (2.1) for $\mathrm{\Phi}\in \mathcal{\mathcal{D}}({\mathbb{R}}^{2})$ smooth, and then proceed to a limit argument.
For such a Φ, we have

${\int}_{0}^{+\mathrm{\infty}}}\mathrm{\Phi}(x,y){\partial}_{y}\mathrm{\Phi}(x,y)\mathit{d}y={\displaystyle \frac{1}{2}}{\displaystyle {\int}_{0}^{+\mathrm{\infty}}}{\partial}_{y}{|\mathrm{\Phi}(x,y)|}^{2}dy=-{\displaystyle \frac{1}{2}}{|\mathrm{\Phi}(x,0)|}^{2},$${\int}_{-\mathrm{\infty}}^{0}}\mathrm{\Phi}(x,y){\partial}_{y}\mathrm{\Phi}(x,y)\mathit{d}y={\displaystyle \frac{1}{2}}{\displaystyle {\int}_{-\mathrm{\infty}}^{0}}{\partial}_{y}{|\mathrm{\Phi}(x,y)|}^{2}dy={\displaystyle \frac{1}{2}}{|\mathrm{\Phi}(x,0)|}^{2}.$

It follows that

$|\mathrm{\Phi}(x,0){|}^{2}=-{\int}_{0}^{+\mathrm{\infty}}\mathrm{\Phi}(x,y){\partial}_{y}\mathrm{\Phi}(x,y)dy+{\int}_{-\mathrm{\infty}}^{0}\mathrm{\Phi}(x,y){\partial}_{y}\mathrm{\Phi}(x,y)dy\le {\int}_{\mathbb{R}}|\mathrm{\Phi}(x,y)||{\partial}_{y}\mathrm{\Phi}(x,y)|dy.$

Integrating this equation with respect to *x* and using the Cauchy–Schwarz inequality, we get

$|\mathrm{\Gamma}(\mathrm{\Phi})(x){|}_{{L}_{x}^{2}}={\int}_{\mathbb{R}}|\mathrm{\Phi}(x,0){|}^{2}dx\le {\int}_{{\mathbb{R}}^{2}}|\mathrm{\Phi}(x,y)||{\partial}_{y}\mathrm{\Phi}(x,y)|dxdy\le \parallel \mathrm{\Phi}{\parallel}_{{\mathbb{L}}_{x,y}^{2}}\parallel {\partial}_{y}\mathrm{\Phi}{\parallel}_{{\mathbb{L}}_{x,y}^{2}}.$

Using the density of $\mathcal{\mathcal{D}}({\mathbb{R}}^{2})$ into *X* completes the proof of the lemma.
∎

We now proceed to the core of the proof of Theorem 1.1.
For convenience, we omit the subscript ε throughout this proof.
The main idea here is to construct an approximate solution of the perturbed system (1.10) by a suitable Galerkin approximation, and then pass to the limit.
Hence, let $({({e}_{i})}_{i\ge 1})$ be an orthogonal basis of ${L}_{x}^{2}(\mathbb{R})$.
We suppose that every element ${e}_{i}$ is compactly supported.
We may take, for example, a fractal wavelet basis of Daubechies (see [17]).
Let ${V}_{n}=\mathrm{Span}({e}_{1},\mathrm{\dots},{e}_{n})$ be the subspace of ${L}_{x}^{2}(\mathbb{R})$ spanned by ${e}_{1},\mathrm{\dots},{e}_{n}$ and let ${W}_{n}={V}_{n}\otimes {V}_{n}$ be the corresponding subspace of ${\mathbb{L}}^{2}({\mathbb{R}}^{2})$.
We then seek an approximate solution ${({u}_{n},{\mathrm{\Phi}}_{n})}_{n}\in {V}_{n}\times {W}_{n}$ that verifies the following approximate problem:

$\{\begin{array}{cc}& \frac{d}{dt}\left[{({u}_{n}(t),\stackrel{~}{u})}_{{L}^{2}(\mathbb{R})}+2{(\mathrm{\Gamma}{\mathrm{\Phi}}_{n}(t),\stackrel{~}{u})}_{{L}^{2}(\mathbb{R})}\right]+{({u}_{nx}(t),\stackrel{~}{u})}_{{L}^{2}(\mathbb{R})}+{({u}_{nx}(t),{\stackrel{~}{u}}_{x})}_{{L}^{2}(\mathbb{R})}\hfill \\ & =-{({u}_{n}(t){u}_{nx}(t),\stackrel{~}{u})}_{{L}^{2}(\mathbb{R})}\hspace{1em}\text{for all}\stackrel{~}{u}\in {V}_{n},\hfill \\ & \frac{d}{dt}{[{\mathrm{\Phi}}_{n}(t),\stackrel{~}{\mathrm{\Phi}}]}_{{\mathbb{L}}^{2}({\mathbb{R}}^{2})}+{[{\mathrm{\Phi}}_{ny}(t),{\stackrel{~}{\mathrm{\Phi}}}_{y}]}_{{\mathbb{L}}^{2}({\mathbb{R}}^{2})}+\epsilon {[{\mathrm{\Phi}}_{nx}(t),{\stackrel{~}{\mathrm{\Phi}}}_{x}]}_{{\mathbb{L}}^{2}({\mathbb{R}}^{2})}\hfill \\ & ={({u}_{n}(t),\stackrel{~}{\mathrm{\Phi}}(\cdot ,0))}_{{L}^{2}(\mathbb{R})}\hspace{1em}\text{for all}\stackrel{~}{\mathrm{\Phi}}\in {W}_{n},\hfill \\ & {u}_{n}(0)={P}_{n}({u}_{0})\in {L}^{2}(\mathbb{R}),{\mathrm{\Phi}}_{n}(0)=0,\hfill \end{array}$(2.2)

where ${P}_{n}$ is the orthogonal projector onto ${V}_{n}$.

#### Proposition 2.3.

*The approximate problem (2.2) has a unique maximal solution defined on the interval $\mathrm{[}\mathrm{0}\mathrm{,}{T}_{\mathrm{max}}\mathrm{[}$, where $\mathrm{0}\mathrm{<}{T}_{\mathrm{max}}\mathrm{\le}\mathrm{+}\mathrm{\infty}$.*

#### Proof.

First, we denote by ${p}_{n}(t)={({p}_{k}^{n}(t))}_{1\le k\le n}$ and ${q}_{n}(t)={({q}_{ij}^{n}(t))}_{1\le i,j\le n}$,
defined, respectively, by ${u}_{n}=\sum {p}_{k}^{n}{e}_{k}$ and ${\mathrm{\Phi}}_{n}=\sum {q}_{ij}^{n}{e}_{i}\otimes {e}_{j}$.
Then the problem (2.2) can be written in the matrix form

$\mathcal{\mathcal{M}}\left(\begin{array}{c}\hfill {\dot{p}}_{n}\hfill \\ \hfill {\dot{q}}_{n}\hfill \end{array}\right)+\mathcal{\mathcal{N}}\left(\begin{array}{c}\hfill {p}_{n}\hfill \\ \hfill {q}_{n}\hfill \end{array}\right)=\left(\begin{array}{c}\hfill {F}_{1}({p}_{n}(t))\hfill \\ \hfill {0}_{{\mathbb{R}}^{{n}^{2}}}\hfill \end{array}\right).$(2.3)

where

$\mathcal{\mathcal{M}}=\left(\begin{array}{cc}\hfill {\mathrm{I}}_{n}\hfill & \hfill {\mathcal{\mathcal{M}}}_{1}\hfill \\ \hfill 0\hfill & \hfill {\mathrm{I}}_{{n}^{2}}\hfill \end{array}\right),$

is a square matrix of order $(n+{n}^{2})$, ${\mathrm{I}}_{n}$ is the $n\times n$ identity matrix and ${\mathcal{\mathcal{M}}}_{1}$ is a square matrix of order ${n}^{2}$. Moreover, $\mathcal{\mathcal{N}}$ is a square matrix of order $(n+{n}^{2})$.
Finally, ${F}_{1}$ is a polynomial vectorial function with respect to ${p}_{n}(t)$.
Since $\mathcal{\mathcal{M}}$ is an invertible matrix, problem (2.3) can be written as

$\frac{d}{dt}\left(\begin{array}{c}\hfill {p}_{n}\hfill \\ \hfill {q}_{n}\hfill \end{array}\right)=F\left(\begin{array}{c}\hfill {p}_{n}\hfill \\ \hfill {q}_{n}\hfill \end{array}\right),$

where $F:{\mathbb{R}}^{n}\times {\mathbb{R}}^{{n}^{2}}\to {\mathbb{R}}^{n+{n}^{2}}$ is a locally lipschitzian map.
The Cauchy–Lipschitz theorem applies and we deduce that there exists a unique maximal solution $({p}_{n},{q}_{n})$ of class ${C}^{1}$ from $[0,{T}_{\mathrm{max}}[$ to ${\mathbb{R}}^{n+{n}^{2}}$, where $0<{T}_{\mathrm{max}}\le +\mathrm{\infty}$.
∎

We now proceed to the limit as *n* tends to infinity. For this purpose, we need following a priori estimates.

#### Lemma 2.4.

*Let $T\mathrm{>}\mathrm{0}$. Then we have the following a priori estimates, uniformly with respect to **n*:

${({u}_{n})}_{n}\mathit{\text{is bounded in}}{L}^{\mathrm{\infty}}(0,T,{L}_{x}^{2}(\mathbb{R}))\cap {L}^{2}(0,T,{H}_{x}^{1}(\mathbb{R})),$(2.4)${({\mathrm{\Phi}}_{n})}_{n}\mathit{\text{is bounded in}}{L}^{\mathrm{\infty}}(0,T,{\mathbb{H}}_{x,y}^{1}({\mathbb{R}}^{2})),$(2.5)${({\partial}_{t}{\mathrm{\Phi}}_{n})}_{n}\mathit{\text{is bounded in}}{L}^{2}(0,T,{\mathbb{L}}_{x,y}^{2}({\mathbb{R}}^{2})),$(2.6)${(\mathrm{\Gamma}({\mathrm{\Phi}}_{n}))}_{n}\mathit{\text{is bounded in}}{L}^{\mathrm{\infty}}(0,T,{L}_{x}^{2}(\mathbb{R})).$(2.7)

#### Proof.

Set $(\stackrel{~}{u},\stackrel{~}{\mathrm{\Phi}})=({u}_{n},{\mathrm{\Phi}}_{nt})$ in (2.2).
We then have

$\frac{d}{dt}\left(\frac{1}{2}{|{u}_{n}(t)|}_{{L}_{x}^{2}}^{2}+{\parallel {\mathrm{\Phi}}_{ny}(t)\parallel}_{{L}_{x,y}^{2}}^{2}+\epsilon {\parallel {\mathrm{\Phi}}_{nx}(t)\parallel}_{{L}_{x,y}^{2}}^{2}\right)+\left({|{u}_{nx}(t)|}_{{L}_{x}^{2}}^{2}+2{\parallel {\mathrm{\Phi}}_{nt}(t)\parallel}_{{L}_{x,y}^{2}}^{2}\right)=0.$

Integrating in time, this leads to (2.4), (2.6) and to an upper bound
for the gradient of ${\mathrm{\Phi}}_{n}$.
To complete the proof of (2.5), a bound on the ${L}^{2}$ norm of ${\mathrm{\Phi}}_{n}$ is required.
We set $\stackrel{~}{\mathrm{\Phi}}={\mathrm{\Phi}}_{n}$ in the second equation in (2.2) and, thanks to the previous estimates and Lemma 2.2, we obtain

$\frac{d}{dt}{\parallel {\mathrm{\Phi}}_{n}(t)\parallel}_{{L}_{x,y}^{2}}^{2}+{\parallel {\mathrm{\Phi}}_{ny}(t)\parallel}_{{L}_{x,y}^{2}}^{2}\le {({u}_{n}(t),\stackrel{~}{\mathrm{\Phi}}(\cdot ,0))}_{{L}^{2}(\mathbb{R})}\le C{\parallel {\mathrm{\Phi}}_{n}(t)\parallel}_{{L}_{x,y}^{2}}.$

Then (2.5) follows promptly.
Eventually, (2.7) is a consequence of the previous estimates and Lemma 2.2.
∎

Thanks to Lemma 2.4, we deduce that ${T}_{\mathrm{max}}=+\mathrm{\infty}$.

The limit process requires some compactness argument.
For this purpose, we state the following lemma.

#### Lemma 2.5.

*Let $T\mathrm{>}\mathrm{0}$ and $\delta \mathrm{\in}\mathrm{(}\mathrm{0}\mathrm{,}T\mathrm{)}$.
There exists a constant $C\mathrm{>}\mathrm{0}$ that may depends on **T* but that is independent of δ and *n* such that

${\int}_{\delta}^{T-\delta}}\parallel {\mathrm{\Phi}}_{n}(t+\delta )-{\mathrm{\Phi}}_{n}(t){\parallel}_{{\mathbb{L}}^{2}({\mathbb{R}}^{2})}^{2}\le C{\delta}^{2$$\mathit{\text{for all}}n\ge 0,$(2.8)${\int}_{\delta}^{T-\delta}}{|{u}_{n}(t+\delta )-{u}_{n}(t)|}_{{L}^{2}(\mathbb{R})}^{2}\le C\sqrt{\delta$$\mathit{\text{for all}}n\ge 0.$(2.9)

#### Proof of Lemma 2.5.

Let $T>0$ and $\delta \in (0,T)$.
Using the mean value theorem, for all $n\in \mathbb{N}$, we get

${\mathrm{\Phi}}_{n}(t+\delta )-{\mathrm{\Phi}}_{n}(t)={\int}_{t}^{t+\delta}{\partial}_{t}{\mathrm{\Phi}}_{n}(\tau )\mathit{d}\tau .$

Using the Cauchy–Schwarz inequality yields

$\parallel {\mathrm{\Phi}}_{n}(t+\delta )-{\mathrm{\Phi}}_{n}(t){\parallel}_{{\mathbb{L}}^{2}({\mathbb{R}}^{2})}^{2}\le \delta {\int}_{t}^{t+\delta}\parallel {\partial}_{t}{\mathrm{\Phi}}_{n}(\tau ){\parallel}_{{\mathbb{L}}^{2}({\mathbb{R}}^{2})}^{2}d\tau .$(2.10)

Now, we integrate (2.10) between 0 and $T-\delta $.
Then, using the Fubini–Tonelli theorem, we get

${\int}_{\delta}^{T-\delta}\parallel {\mathrm{\Phi}}_{n}(t+\delta )-{\mathrm{\Phi}}_{n}(t){\parallel}_{{\mathbb{L}}^{2}({\mathbb{R}}^{2})}^{2}dt\le \delta {\int}_{\delta}^{T-\delta}{\int}_{t}^{t+\delta}\parallel {\partial}_{t}{\mathrm{\Phi}}_{n}(\tau ){\parallel}_{{\mathbb{L}}^{2}({\mathbb{R}}^{2})}^{2}d\tau dt.$

Using the change of variables $\tau =t+\delta $, we deduce

${\int}_{\delta}^{T-\delta}\parallel {\mathrm{\Phi}}_{n}(t+\delta )-{\mathrm{\Phi}}_{n}(t){\parallel}_{{\mathbb{L}}^{2}({\mathbb{R}}^{2})}^{2}dt\le {\delta}^{2}{\int}_{\delta}^{T}\parallel {\partial}_{t}{\mathrm{\Phi}}_{n}(\tau ){\parallel}_{{\mathbb{L}}^{2}({\mathbb{R}}^{2})}^{2}d\tau .$

Hence, using (2.6), we deduce (2.8).

In order to establish (2.9), we recall that ${u}_{n}$ satisfies, for any $v\in {V}_{n}$, the following equation:

$\frac{d}{d\tau}\left({\int}_{\mathbb{R}}{u}_{n}(\tau )v\mathit{d}x+2{\int}_{\mathbb{R}}\mathrm{\Gamma}({\mathrm{\Phi}}_{n})(\tau )v\mathit{d}x\right)+{\int}_{\mathbb{R}}{u}_{nx}(\tau )v\mathit{d}x+{\int}_{\mathbb{R}}{u}_{nx}(\tau ){v}_{x}\mathit{d}x=-{\int}_{\mathbb{R}}{u}_{n}(\tau ){u}_{nx}(\tau )v\mathit{d}x.$(2.11)

We now integrate (2.11) with respect to τ from *t* to $t+\delta $.
Then, setting $v={u}_{n}(t+\delta )-{u}_{n}(t)$, we get

${|{u}_{n}(t+\delta )-{u}_{n}(t)|}_{{L}^{2}(\mathbb{R})}^{2}+2{(\mathrm{\Gamma}({\mathrm{\Phi}}_{n})(t+\delta )-\mathrm{\Gamma}({\mathrm{\Phi}}_{n})(t),{u}_{n}(t+\delta )-{u}_{n}(t))}_{{L}^{2}(\mathbb{R})}$$+{\displaystyle {\int}_{t}^{t+\delta}}{({u}_{nx}(\tau ),{u}_{n}(t+\delta )-{u}_{n}(t))}_{{L}^{2}(\mathbb{R})}\mathit{d}\tau +{\displaystyle {\int}_{t}^{t+\delta}}{({u}_{nx}(\tau ),{u}_{nx}(t+\delta )-{u}_{nx}(t))}_{{L}^{2}(\mathbb{R})}\mathit{d}\tau $$=-{\displaystyle {\int}_{t}^{t+\delta}}{({u}_{n}(\tau ){u}_{nx}(\tau ),{u}_{n}(t+\delta )-{u}_{n}(t))}_{{L}^{2}(\mathbb{R})}\mathit{d}\tau .$(2.12)

Using first the Cauchy–Schwarz inequality and then the Young inequality, we get

$2{\displaystyle {\int}_{\delta}^{T-\delta}}{(\mathrm{\Gamma}({\mathrm{\Phi}}_{n})(t+\delta )-\mathrm{\Gamma}({\mathrm{\Phi}}_{n})(t),{u}_{n}(t+\delta )-{u}_{n}(t))}_{{L}^{2}(\mathbb{R})}\mathit{d}t$$\le C{\displaystyle {\int}_{\delta}^{T-\delta}}{\left|\mathrm{\Gamma}({\mathrm{\Phi}}_{n}(t+\delta )-{\mathrm{\Phi}}_{n}(t+\delta ))\right|}_{{L}^{2}(\mathbb{R})}^{2}dt+{\displaystyle {\int}_{\delta}^{T-\delta}}|{u}_{n}(t+\delta )-{u}_{n}(t){|}_{{L}^{2}(\mathbb{R})}^{2}dt.$(2.13)

From Lemma 2.2, we conclude that

${\int}_{\delta}^{T-\delta}|\mathrm{\Gamma}({\mathrm{\Phi}}_{n}(t+\delta )-{\mathrm{\Phi}}_{n}(t)){|}_{{L}^{2}(\mathbb{R})}^{2}dt\le C{\int}_{\delta}^{T-\delta}\parallel {\mathrm{\Phi}}_{n}(t+\delta )-{\mathrm{\Phi}}_{n}(t){\parallel}_{{\mathbb{L}}^{2}({\mathbb{R}}^{2})}\parallel {\mathrm{\Phi}}_{n}(t+\delta )-{\mathrm{\Phi}}_{n}(t){\parallel}_{{\mathbb{H}}^{1}({\mathbb{R}}^{2})}dt.$

Moreover, using the Cauchy–Schwarz inequality, and estimations (2.5) and (2.8), we get

${\int}_{\delta}^{T-\delta}|\mathrm{\Gamma}({\mathrm{\Phi}}_{n}(t+\delta )-{\mathrm{\Phi}}_{n}(t)){|}_{{L}^{2}(\mathbb{R})}^{2}dt\le C\delta .$(2.14)

Then, gathering (2.14) and (2.13), we obtain

$2{\int}_{\delta}^{T-\delta}{(\mathrm{\Gamma}({\mathrm{\Phi}}_{n})(t+\delta )-\mathrm{\Gamma}({\mathrm{\Phi}}_{n})(t),{u}_{n}(t+\delta )-{u}_{n}(t))}_{{L}^{2}(\mathbb{R})}dt\le C\delta +{\int}_{\delta}^{T-\delta}|{u}_{n}(t+\delta )-{u}_{n}(t){|}_{{L}^{2}(\mathbb{R})}^{2}dt.$(2.15)

In addition, using the Cauchy–Schwarz inequality and estimation (2.4), we get

${\int}_{t}^{t+\delta}{({u}_{nx}(\tau ),{u}_{n}(t+\delta )-{u}_{n}(t))}_{{L}^{2}(\mathbb{R})}\mathit{d}\tau \le C\sqrt{\delta}.$(2.16)

Finally, we integrate (2.16) with respect to *t* from δ to $T-\delta $ and get

${\int}_{\delta}^{T-\delta}{\int}_{t}^{t+\delta}{({u}_{nx}(\tau ),{u}_{n}(t+\delta )-{u}_{n}(t))}_{{L}^{2}(\mathbb{R})}\mathit{d}\tau \mathit{d}t\le C\sqrt{\delta}.$(2.17)

In the same way, we establish that

${\int}_{t}^{t+\delta}{({u}_{nx}(\tau ),{u}_{nx}(t+\delta )-{u}_{nx}(t))}_{{L}^{2}(\mathbb{R})}\mathit{d}\tau \le C\sqrt{\delta}{|{u}_{nx}(t+\delta )-{u}_{nx}(t)|}_{{L}^{2}(\mathbb{R})}.$

From (2.5), we deduce

${\int}_{\delta}^{T-\delta}{\int}_{t}^{t+\delta}{({u}_{nx}(\tau ),{u}_{nx}(t+\delta )-{u}_{nx}(t))}_{{L}^{2}(\mathbb{R})}\mathit{d}\tau \mathit{d}t\le C\sqrt{\delta}.$(2.18)

Then, using the Hölder and Agmon inequalities, we have

${\int}_{t}^{t+\delta}}{({u}_{n}(\tau ){u}_{nx}(\tau ),{u}_{n}(t+\delta )-{u}_{n}(t))}_{{L}^{2}(\mathbb{R})}d\tau \le ({\displaystyle {\int}_{t}^{t+\delta}}|{u}_{n}(\tau ){|}_{{L}^{\mathrm{\infty}}(\mathbb{R})}|{u}_{nx}(\tau ){|}_{{L}^{2}(\mathbb{R})}dt)|{u}_{n}(t+\delta )-{u}_{n}(t){|}_{{L}^{2}(\mathbb{R})$$\le ({\displaystyle {\int}_{t}^{t+\delta}}|{u}_{n}(\tau ){|}_{{L}^{2}(\mathbb{R})}^{1/2}|{u}_{nx}(\tau ){|}_{{L}^{2}(\mathbb{R})}^{3/2}dt)|{u}_{n}(t+\delta )-{u}_{n}(t){|}_{{L}^{2}(\mathbb{R})}.$

Moreover, using Lemma 2.4, we deduce that there exists a constant $C>0$ such that

${\int}_{t}^{t+\delta}{({u}_{n}(\tau ){u}_{nx}(\tau ),{u}_{n}(t+\delta )-{u}_{n}(t))}_{{L}^{2}(\mathbb{R})}\mathit{d}\tau \le C{\delta}^{1/4}{|{u}_{n}(t+\delta )-{u}_{n}(t)|}_{{L}^{2}(\mathbb{R})}.$

Finally, using the Young inequality, we get

${\int}_{\delta}^{T-\delta}{\int}_{t}^{t+\delta}{({u}_{n}(\tau ){u}_{nx}(\tau ),{u}_{n}(t+\delta )-{u}_{n}(t))}_{{L}^{2}(\mathbb{R})}d\tau dt\le C\sqrt{\delta}+\frac{1}{4}{\int}_{\delta}^{T-\delta}|{u}_{n}(t+\delta )-{u}_{n}(t){|}_{{L}^{2}(\mathbb{R})}^{2}dt.$(2.19)

In conclusion, gathering (2.15), (2.17), (2.18), (2.19) and (2.12) yields (2.9).
∎

We rephrase Lemma 2.5 as follows: the sequences ${u}_{n}$ and ${\mathrm{\Phi}}_{n}$ are, respectively, bounded in ${H}^{1/4}(0,T;{L}^{2}(\mathbb{R}))$ and ${H}^{1}(0,T;{L}^{2}({\mathbb{R}}^{2}))$.
Combining this with the a priori bounds in Lemma 2.4 allow us to deduce that ${u}_{n}$ and ${\mathrm{\Phi}}_{n}$ converge strongly in ${L}_{\mathrm{loc}}^{2}(\mathbb{R})$ and ${L}_{\mathrm{loc}}^{2}({\mathbb{R}}^{2})$, respectively.
Gathering these and the weak convergence that comes from the a priori estimates, it is standard to pass to the limit when *n* tends towards $+\mathrm{\infty}$.
We omit the details.

To complete the proof of the theorem, it remains to establish the continuity of the solution $(u,\mathrm{\Phi})$ of (1.10).
First, we know that
$\mathrm{\Phi}\in {L}^{\mathrm{\infty}}(0,T,{\mathbb{H}}^{1}({\mathbb{R}}^{2}))$ and ${\mathrm{\Phi}}_{t}\in {L}^{2}(0,T,{\mathbb{L}}^{2}({\mathbb{R}}^{2})).$
Then, due to [24, Lemma 1.1, p. 250], we deduce that

$\mathrm{\Phi}\in C([0,T],{\mathbb{L}}_{x,y}^{2}({\mathbb{R}}^{2})).$

In the following, we establish new estimations on the approximate solution ${u}_{n}$ in order to establish the weak continuity of the solution over $[0,T]$ with values in ${L}_{x}^{2}(\mathbb{R})$.

#### Lemma 2.6.

*Let $\delta \mathrm{\in}\mathrm{(}\mathrm{0}\mathrm{,}\mathrm{1}\mathrm{)}$.
There exists a constant $C\mathrm{>}\mathrm{0}$ that may depend on **T* but that is independent of δ and *n* such that

${|{u}_{n}(t+\delta )-{u}_{n}(t)|}_{{H}_{x}^{-1}}\le C{\delta}^{1/4}.$

Due to Lemma 2.6, we have that

$|u(t+\delta )-u(t){|}_{{H}_{x}^{-1}}\le \underset{n\to +\mathrm{\infty}}{lim\; inf}|{u}_{n}(t+\delta )-{u}_{n}(t){|}_{{H}_{x}^{-1}}\le C{\delta}^{1/4},$

It follows that

$\underset{\delta \to 0}{lim}|u(t+\delta )-u(t){|}_{{H}_{x}^{-1}}=0.$

This implies that *u* is continuous $[0,T]$ with values in ${H}_{x}^{-1}(\mathbb{R})$.
Also, we know that $u\in {L}^{\mathrm{\infty}}(0,T,{L}_{x}^{2}(\mathbb{R}))$.
Hence, thanks to [24, Lemma 1.4, p. 263] and using the continuous injection ${L}^{2}(\mathbb{R})\hookrightarrow {H}^{-1}(\mathbb{R})$, we deduce that *u* is weak continuous with values in ${L}_{x}^{2}(\mathbb{R})$.

The proof of the uniqueness of a solution is left to the reader.

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