In this section we deal with the existence of solution to the problem (1.1). It is important to note that the results of this section can easily be adapted to more general semi-linear equations.

Firstly, we should note that *Au* = (*au*_{x})_{x}, in a suitable domain, generates a analytic semigroup. For this purpose, we introduce the following weighted spaces (sometimes, we use ′ to denote the derivative with respect to *x*):

$$\begin{array}{}{\displaystyle {H}_{a}^{1}(0,1):=\{u\in {L}^{2};\phantom{\rule{thinmathspace}{0ex}}u\text{\hspace{0.17em}absolutely continuous in\hspace{0.17em}}[0,1]\text{\hspace{0.17em}and\hspace{0.17em}}\phantom{\rule{thinmathspace}{0ex}}\sqrt{a}{u}^{\prime}\in {L}^{2}(0,1)\}}\end{array}$$(2.1)

with the norm

$$\begin{array}{}{\displaystyle |u|{|}_{{H}_{a}^{1}(0,1)}^{2}:=||u|{|}_{{L}^{2}(0,1)}^{2}+||\sqrt{a}{u}^{\prime}|{|}_{{L}^{2}(0,1)}^{2}}\end{array}$$(2.2)

and

$$\begin{array}{}{\displaystyle {H}_{a}^{2}(0,1):=\{u\in {H}_{a}^{1}(0,1);a{u}^{\prime}\in {H}^{1}(0,1)\}}\end{array}$$(2.3)

with

$$\begin{array}{}{\displaystyle ||u|{|}_{{H}_{a}^{2}(0,1)}^{2}:=||u|{|}_{{H}_{a}^{1}(0,1)}^{2}+||(a{u}^{\prime}{)}^{\prime}|{|}_{{L}^{2}(0,1)}^{2}.}\end{array}$$

#### Definition 2.1

If *u*_{0} ∈ *L*^{2}(0, 1), a function *u* is said to be a weak solution of (1.1) if

$$\begin{array}{}{\displaystyle u\in C([0,T];{L}^{2}(0,1))\cap {L}^{2}(0,T;{H}_{a}^{1}(0,1))}\end{array}$$

and

$$\begin{array}{}{\displaystyle \underset{0}{\overset{1}{\int}}u(T,x)\varphi (T,x)dx-\underset{0}{\overset{1}{\int}}{u}_{0}(x)\varphi (0,x)dx-\underset{(0,T)\times (0,1)}{\int}u{\varphi}_{t}dxdt=-\underset{(0,T)\times (0,1)}{\int}a{u}_{x}{\varphi}_{x}dxdt+\underset{(0,T)\times (0,1)}{\int}f(u)\varphi dxdt}\end{array}$$

for all *ϕ* ∈ *H*^{1}(0, *T*; *L*^{2}(0, 1)) ∩ *L*^{2}(0, *T*; $\begin{array}{}{H}_{a}^{1}\end{array}$(0, 1)).

Now we define the operator *A* by *D*(*A*) := {*u* ∈ $\begin{array}{}{H}_{a}^{2}\end{array}$(0, 1); *u*′(0) = *u*′(1) = 0} and for any *u* ∈ *D*(*A*), *Au* = (*au*′)′.

The proof of the next two results can be found in [1, Lemma 2.1] (see also [7, Lemma 2.1]) and [7, Theorem 2.1], respectively.

#### Lemma 2.2

*For all* (*u*, *v*) ∈ *D*(*A*) × $\begin{array}{}{H}_{a}^{1}\end{array}$(0, 1) *one has*

$$\begin{array}{}{\displaystyle \underset{0}{\overset{1}{\int}}(a{u}^{\prime}{)}^{\prime}vdx=-\underset{0}{\overset{1}{\int}}a{u}^{\prime}{v}^{\prime}dx.}\end{array}$$

#### Theorem 2.3

*The operator* *A* : *D*(*A*) → *L*^{2}(0, 1) *is self*-*adjoint*, *nonpositive on* *L*^{2}(0, 1) *and it generates an analytic contraction semigroup*.

Now we proceed as in [1]. Since *A* is a generator, and setting *B*(*t*)*u* := *u*, working in the spaces considered above, we can prove that the problem below (with *c* ∈ *L*^{∞}(ℝ_{+} × (0, 1))) is well-posed in the sense of semigroup theory using some well-known perturbation technique (see [16], for instance).

$$\begin{array}{}{\displaystyle \left\{\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\begin{array}{l}{u}_{t}(t,x)=(a(x){u}_{x}(t,x){)}_{x}+c(t,x)u(t,x),\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}(t,x)\in {\mathbb{R}}^{+}\times (0,1)\\ {u}_{x}(t,0)={u}_{x}(t,1)=0,\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}t\in {\mathbb{R}}^{+}\\ u(0,x)={u}_{0}(x),\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}x\in (0,1)\end{array}\right.}\end{array}$$(2.4)

Hence, for a fixed *T* > 0 we get the following result.

#### Theorem 2.4

*If* *c*(⋅, *x*) ∈ *C*^{1}(ℝ_{+}) *for all* *x* ∈ [0, 1] *and* *u*_{0} ∈ *D*(*A*) *then there is a unique weak solution*

$$\begin{array}{}{\displaystyle u\in {H}^{1}(0,T;{L}^{2}(0,1))\cap {L}^{2}(0,T;{H}_{a}^{2}(0,1))\cap C([0,T];{H}_{a}^{1}(0,1))}\end{array}$$(2.5)

*of* (2.4) *and*

$$\begin{array}{}{\displaystyle \underset{t\in [0,T]}{sup}||u(t)|{|}_{{H}_{a}^{1}(0,1)}^{2}+\underset{0}{\overset{T}{\int}}\left({\u2225\frac{\mathrm{\partial}u}{\mathrm{\partial}t}\u2225}_{{L}^{2}(0,1)}^{2}+{\u2225\frac{\mathrm{\partial}}{\mathrm{\partial}x}\left(a\frac{\mathrm{\partial}u}{\mathrm{\partial}x}\right)\u2225}_{{L}^{2}(0,1)}^{2}\right)dt\le C||{u}_{0}|{|}_{{H}_{a}^{1}(0,1)}^{2},}\end{array}$$(2.6)

*for a positive constant* *C*.

Next result can be found in [1, Theorem 5.4]

#### Lemma 2.5

*The set* *H*^{1}(0, *T*; *L*^{2}(0, 1)) ∩ *L*^{2}(0, *T*; $\begin{array}{}{H}_{a}^{2}\end{array}$(0, 1)) *is compactly imbedded in* *C*([0, *T*]; *L*^{2}(0, 1)) ∩ *L*^{2}(0, *T*; $\begin{array}{}{H}_{a}^{1}\end{array}$(0, 1)).

We are now in position to state the main result of this section. The proof follows the steps of [1, Theorem 4.12], however, some modifications are necessary because we consider Neumann boundary conditions and the specific nonlinear term *f*(*u*) = *u* − *u*^{3}.

#### Theorem 2.6

*If* *u*_{0}(*x*) ∈ $\begin{array}{}{H}_{a}^{1}\end{array}$(0, 1) *then* (1.1) *has a solution*

$$\begin{array}{}{\displaystyle u\in {H}^{1}(0,T;{L}^{2}(0,1))\cap {L}^{2}(0,T;{H}_{a}^{2}(0,1)).}\end{array}$$

#### Proof

We set *X* := *C*([0, *T*]; *L*^{2}(0, 1)) ∩ *L*^{2}(0, *T*; $\begin{array}{}{H}_{a}^{1}\end{array}$(0, 1)) and for any (*x*, *v*) ∈ (0, 1) × *X*, *c*^{v}(*t*, *x*) := *d*(*t*, *x*, *v*(*t*, *x*)) where *d*(*t*, *x*, *u*) = 1-*u*^{2}. Now, we consider the function

$$\begin{array}{}{\displaystyle \mathcal{T}:v\in X\to {u}^{v}\in X,}\end{array}$$

where *u*^{v} is the unique solution of

$$\begin{array}{}{\displaystyle \left\{\phantom{\rule{thinmathspace}{0ex}}\begin{array}{l}{u}_{t}(t,x)=(a(x){u}_{x}(t,x){)}_{x}+{c}^{v}(t,x)u(t,x),\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}(t,x)\in {\mathbb{R}}^{+}\times (0,1)\\ {u}_{x}(t,0)={u}_{x}(t,1)=0,\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}t\in {\mathbb{R}}^{+}\\ u(0,x)={u}_{0}(x),\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}x\in (0,1).\end{array}\right.}\end{array}$$(2.7)

We use Theorem 2.4 to ensure that (2.7) has a unique weak solution *u* ∈ *X*. Now, we will prove that 𝓣 has a fixed point *u*^{v} (that is, 𝓣(*u*^{v}) = *u*^{v}) to conclude that *u*^{v} is a solution of (1.1).

By Schauder’s Theorem, it is sufficient to prove that

𝓣:*B*_{X} → *B*_{X},

𝓣 is a compact function and

𝓣 is a continuous function,

where

$$\begin{array}{}{\displaystyle {B}_{X}:=\{v\in X;||v|{|}_{X}\le {C}_{T}||{u}_{0}|{|}_{{L}^{2}(0,1)}^{2}\},}\end{array}$$

*C*_{T} it is the same constant of Theorem 2.4 and

$$\begin{array}{}{\displaystyle ||v|{|}_{X}:=\underset{t\in [0,T]}{sup}\left(||u(t)|{|}_{{L}^{2}(0,1)}^{2}\right)+\underset{0}{\overset{T}{\int}}||\sqrt{a}{u}_{x}|{|}_{{L}^{2}(0,1)}^{2}dt.}\end{array}$$

The items (i) and (ii) are consequence of Theorem 2.4 and Lemma 2.5, respectively.

To prove (iii) we take *v*_{k} ∈ *X* such that *v*_{k} → *v* in *X*, as *k* → ∞. We will prove that 𝓣(*v*_{k}) = *u*^{vk} := *u*^{k} → 𝓣(*v*) = *u*^{v} in *X*, as *k* → ∞. Recall that *u*^{k} and *u*^{v} are the solutions of (2.7) associated to *v*_{k} and *v*, respectively. As *D*(*A*) is dense in $\begin{array}{}{H}_{a}^{1}\end{array}$(0, 1) ([7]), (2.5) and (2.6) occurs for *u*_{0} ∈ $\begin{array}{}{H}_{a}^{1}\end{array}$(0, 1). Hence, *u*^{k} ∈ *B*_{Y} where *Y* = *H*^{1}(0, *T*; *L*^{2}(0, 1)) ∩ *L*^{2}(0, *T*; $\begin{array}{}{H}_{a}^{2}\end{array}$(0, 1)) and, up to a sub-sequence, *u*^{k} converges weakly to some *ū* in *Y*. By Lemma 2.5, *u*^{k} converges strongly to *ū* in *X*.

Multiplying the equation

$$\begin{array}{}{\displaystyle {u}_{t}^{k}(t,x)=(a(x){u}_{x}^{k}(t,x){)}_{x}+{c}^{{v}^{k}}(t,x){u}^{k}(t,x)}\end{array}$$

by a test function *ϕ* ∈ *H*^{1}(0, *T*; *L*^{2}(0, 1)) ∩ *L*^{2}(0, *T*; $\begin{array}{}{H}_{a}^{1}\end{array}$(0, 1)) and integrating over (0, *T*) × (0, 1) (recall the Lemma 2.2) we get

$$\begin{array}{c}{\displaystyle \underset{0}{\overset{1}{\int}}{u}^{k}(T,x)\varphi (T,x)dx-\underset{0}{\overset{1}{\int}}{u}_{0}(x)\varphi (0,x)dx-\underset{0}{\overset{T}{\int}}\underset{0}{\overset{1}{\int}}{\varphi}_{t}(t,x){u}^{k}(t,x)dxdt}\\ \\ {\displaystyle =\underset{0}{\overset{T}{\int}}a(x){u}_{x}^{k}(t,1)\varphi (t,1)dt-\underset{0}{\overset{T}{\int}}a(x){u}_{x}^{k}(t,0)\varphi (t,0)dt}\\ \\ {\displaystyle -\underset{0}{\overset{T}{\int}}\underset{0}{\overset{1}{\int}}a(x){u}_{x}^{k}(t,x){\varphi}_{x}(t,x)dxdt+\underset{0}{\overset{T}{\int}}\underset{0}{\overset{1}{\int}}{c}^{{v}^{k}}(t,x){u}^{k}(t,x)\varphi (t,x)dxdt.}\end{array}$$

We recall that $\begin{array}{}{u}_{x}^{k}(t,1)={u}_{x}^{k}(t,0)=0\end{array}$ and our next step is to prove that

$\begin{array}{}{\displaystyle \underset{k\to \mathrm{\infty}}{lim}\underset{0}{\overset{1}{\int}}{u}^{k}(T,x)\varphi (T,x)dx=\underset{0}{\overset{1}{\int}}\overline{u}(T,x)\varphi (T,x)dx;}\end{array}$

$\begin{array}{}{\displaystyle \underset{k\to \mathrm{\infty}}{lim}\underset{0}{\overset{T}{\int}}\underset{0}{\overset{1}{\int}}{\varphi}_{t}(t,x){u}^{k}(t,x)dxdt=\underset{0}{\overset{T}{\int}}\underset{0}{\overset{1}{\int}}{\varphi}_{t}(t,x)\overline{u}(t,x)dxdt;}\end{array}$

$\begin{array}{}{\displaystyle \underset{k\to \mathrm{\infty}}{lim}\underset{0}{\overset{T}{\int}}\underset{0}{\overset{1}{\int}}a(x){u}_{x}^{k}(t,x){\varphi}_{x}(t,x)dxdt=\underset{0}{\overset{T}{\int}}\underset{0}{\overset{1}{\int}}a(x)\overline{{u}_{x}}(t,x){\varphi}_{x}(t,x)dxdt;}\end{array}$

$\begin{array}{}{\displaystyle \underset{k\to \mathrm{\infty}}{lim}\underset{0}{\overset{T}{\int}}\underset{0}{\overset{1}{\int}}{c}^{{v}^{k}}(t,x){u}^{k}(t,x)\varphi (t,x)dxdt=\underset{0}{\overset{T}{\int}}\underset{0}{\overset{1}{\int}}{c}^{v}(t,x)\overline{u}(t,x)\varphi (t,x)dxdt.}\end{array}$

Since *u*^{k} converges strongly to *ū* in *X*, it is immediate to prove (a)-(c). In order to prove (d) we recall that *c*^{vk}(*t*, *x*) = 1 − (*v*^{k})^{2}(*t*, *x*) and *v*^{k} converges strongly to *v* in *X*. Therefore, we can conclude that *v*^{k} converges to *v* a.e. as well as *u*^{k} converges to *ū* a.e.. Thus,

$$\begin{array}{c}{\displaystyle \left|\underset{0}{\overset{T}{\int}}\underset{0}{\overset{1}{\int}}{c}^{{v}^{k}}(t,x){u}^{k}(t,x)\varphi (t,x)-{c}^{v}(t,x)\overline{u}(t,x)\varphi (t,x)dxdt\right|}\\ \\ {\displaystyle \le \underset{0}{\overset{T}{\int}}\underset{0}{\overset{1}{\int}}\left|({u}^{k}(t,x)-\overline{u}(t,x)){c}^{{v}^{k}}(t,x)\varphi (t,x)\right|dxdt}\\ \\ {\displaystyle +\underset{0}{\overset{T}{\int}}\underset{0}{\overset{1}{\int}}\left|({c}^{{v}^{k}}(t,x)-{c}^{v}(t,x))\overline{u}(t,x)\varphi (t,x)\right|dxdt}\\ \\ {\displaystyle =\underset{0}{\overset{T}{\int}}\underset{0}{\overset{1}{\int}}\left|{u}^{k}(t,x)-\overline{u}(t,x)\right|\left|(1-({v}^{k}{)}^{2}(t,x))\varphi (t,x)\right|dxdt}\\ \\ {\displaystyle +\underset{0}{\overset{T}{\int}}\underset{0}{\overset{1}{\int}}\left|({v}^{k}{)}^{2}(t,x)-{v}^{2}(t,x)\right|\left|\overline{u}(t,x)\varphi (t,x)\right|dxdt,}\end{array}$$

and (d) holds by an application of Lebesgue Theorem.

We proved that *ū* is the unique weak solution of (2.4) in *Y* ⊂ *X* associated to *v*; that is *ū* = *u*^{v} and (iii) is proved. It follows that 𝓣 has a fixed point *u*^{v} ∈ *Y* which is a solution of (1.1). The theorem is proved.□

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