L^2(0,T;V).\tag{2.19}\end{align*}$$2a ESTIMATIVAComo o operador $\tilde{A}:X\to H$ tem inversa simetrica (ver Secao 2.3), segue do item (a) do Lema 2.7 que$$(A^{-1}h,g)_V=\lambda_j(\tilde{A}^{-1}h,g)=\lambda_j(h,\tilde{A}^{-1}g)=(h,A^{-1}g)_V,\quad \forall\ h,g\in H.$$Logo, por (2.8),\begin{align*}\|u'_m(t)\|_{V'}^2=\|AA^{-1}u'_m(t)\|_{V'}^2&\leq c\|A^{-1}u'_m(t)\|_V^2=c\sum_{i=1}^\infty\frac{1}{\lambda_j}\left|\left(A^{-1}u'_m(t),w_j\right)_V\right|^2\\&=c\sum_{i=1}^\infty\frac{1}{\lambda_j}\left|\left(u'_m(t),A^{-1}w_j\right)_V\right|^2=c\sum_{i=1}^\infty\frac{1}{\lambda_j^2}|(u'_m(t),w_j)_V|^2\\&=c\sum_{i=1}^m\frac{1}{\lambda_j}|(u'_m(t),w_j)|^2=c\sum_{i=1}^m\frac{1}{\lambda_j}|\langle u'_m(t),w_j\rangle_{V'\times V}|^2.\end{align*}Escrevendo $h_m(t)=f(t)-\nu Au_m(t)-B(u_m(t),u_m(t))$, segue de (2.14)$_1$, (2.9) e (2.10) que\begin{align*}\|u'_m(t)\|_{V'}^2&\leq c\sum_{i=1}^m\frac{1}{\lambda_j}|\langle h_m(t),w_j\rangle _{V'\times V}|^2=c\|{A^{-1}}h_m(t)\|_{V}^2\leq c_1\|h_m(t)\|_{V'}^2\\&\leq c_1\big(\|f(t)\|_{V'}+\nu \|Au_m(t)\|_{V'}+\|B(u_m(t),u_m(t))\|_{V'}\big)^2\\&\leq c_2\big(\|f(t)\|_{V'}^2+\|u_m(t)\|^2_{V}+\|u_m(t)\|^2_{V}\big).\end{align*}Como $f\in L^2(0,T;V')$, segue de (2.19) que$$(u'_m)\quad\text{e limitada em}\quad L^2(0,T;V').\tag{2.20}$$PASSAGEM AO LIMITEResulta de (2.18), (2.19) e (2.20) que existe $u\in L^\infty(0,T;H)\cap L^2(0,T;V)$ com $u'\in L^2(0,T;V')$ tal que$$\begin{align*}u_m&\overset{*}{\rightharpoonup} u\quad\text{ em}\quad L^\infty(0,T;H),\tag{2.21}\\u_m&\rightharpoonup u\quad\text{ em}\quad L^2(0,T;V),\tag{2.22}\\u'_m&\rightharpoonup u'\quad\text{em}\quad L^2(0,T;V').\tag{2.23}\end{align*}$$De (2.14)$_1$ resulta que, para $1\leq j\leq m$,$$\label{}(u'_m(\cdot),w_j)+\nu a(u_m(\cdot),w_j)+b(u_m(\cdot), u_m(\cdot),w_j)=\langle f(\cdot),w_j\rangle_{V'\times V}\text{ em }\mathcal{D}'(0,T)\tag{2.24}.$$Vejamos que$$\left.\begin{aligned}(u'_m(\cdot),w_j)&\to \frac{d}{dt}(u(\cdot),w_j)\\a(u_m(\cdot),w_j)&\to a(u(\cdot),w_j)\\b(u_m(\cdot), u_m(\cdot),w_j)&\to b(u(\cdot), u(\cdot),w_j)\end{aligned}\right\}\quad \text{em}\quad \mathcal{D}'(0,T)^{\text{(viii)}}.\tag{2.25}$$ A seguir provaremos a terceira convergencia, que envolve o termo nao linear da equacao (gracas a linearidade, as duas primeiras convergencia sao obtidas de maneira mais simples e, por isso, omitiremos os detalhes). Das limitacoes (2.19) e (2.20), concluimos que$$(u_m)\quad\text{e limitada em}\quad W:= \{g\in L^2(0,T;V)\mid g'\in L^2(0,T;V')\}.\tag{2.26}$$Pelo Teorema de Aubin-Lions, $W\overset{\mathrm{c}}{\hookrightarrow} L^2(0,T; H)$ (aqui estamos usando os itens (b), (c) e (d) do Lema 2.6). Logo, passando a uma subsequencia,$$u_m\to u\quad\text{em}\quad L^2(0,T;H).$$Consequentemente,$$u_m^i\to u^i\quad\text{em}\quad L^2(0,T;L^2(\Omega))\cong L^2(Q),\qquad i=1,2$$onde $Q=\Omega\times(0,T)$ (para detalhes sobre o isomorfismo acima, ver p. 24 em [13]). Resulta disto que, passando a uma subsequencia,$$u_m^iu_m^k\to u^iu^k\quad\text{q.s. em}\quad Q,\qquad i,k=1,2.\tag{2.27}$$Como $n=2$, temos $H_0^1(\Omega)\hookrightarrow L^4(\Omega)$ (ver Teorema 4.12, p. 85, em [1]). Consequentemente, $L^2(0,T;V)\hookrightarrow L^2(0,T;[L^4(\Omega)]^2)$ e\begin{align*}\|u_m^iu_m^k\|_{L^2}^2&=\int_\Omega|u_m^i|^2|u_m^k|^2\;dx\leq\left(\int_\Omega|u_m^i|^4\;dx\right)^{1/2}\left(\int_\Omega|u_m^k|^4\;dx\right)^{1/2}=\|u^i_m\|_{L^4}^2\|u^k_m\|_{L^4}^2\\&\leq \frac{1}{2}\|u^i_m\|_{L^4}^4+\frac{1}{2}\|u^k_m\|_{L^4}^4.\end{align*}Logo, de (2.19),$$(u_m^iu_m^k)\quad\text{e limitada em}\quad L^2(0,T;L^2(\Omega))\cong L^2(Q),\qquad i,k=1,2.\tag{2.28}$$De (2.27), de (2.28) e do Lema de Lions, segue que$$u_m^iu_m^k\rightharpoonup u^iu^k\quad\text{em}\quad L^2(Q)\cong L^2(0,T;L^2(\Omega)),\qquad i,k=1,2.$$Isto implica que$$\int_0^T(u^k_m(t)v(\cdot,t),u_m^i(t))_{L^2}\;dt=\int_Q u^i_mu^k_mv\;dz\to\int_Q u^iu^kv\;dz=\int_0^T(u^k(t)v(\cdot,t),u^i(t))_{L^2}\;dt$$para toda $ v\in L^2(Q)$. Tomando $v(x,t)=\theta(t)\frac{d}{dx_k}w_j^i(x)$ com $\theta\in L^2(0,T)$ e somando com respeito aos indices $i,k$, segue que$$\int_0^Tb(u_m(t),\theta(t)w_j,u_m(t))\;dx\to\int_0^Tb(u(t),\theta(t)w_j,u(t))\;dx.$$Do Lema 2.8, obtemos$$\int_0^Tb(u_m(t),u_m(t),w_j)\theta(t)\;dt\to\int_0^Tb(u(t),u(t),w_j)\theta(t)\;dt,\quad\forall\ \theta\in L^2(0,T).$$A ultima convergencia vale, em particular, para todo $\theta\in\mathcal{D}(0,T)$ e, portanto,$$b(u_m(\cdot),u_m(\cdot),w_j)\to b(u(\cdot),u(\cdot),w_j)\quad\text{em}\quad \mathcal{D}'(0,T).$$Isto conclui a prova de (2.25). Assim, fixando $j\in\mathbb{N}$ e tomando o limite em (2.24) com $m\to\infty$, concluimos que$$\frac{d}{dt}(u(\cdot),w_j)+\nu a(u(\cdot),w_j)+b(u(\cdot), u(\cdot),w_j)=\langle f(\cdot),w_j\rangle_{V'\times V}\quad\text{em}\quad\mathcal{D}'(0,T),\quad\forall\ j\in\mathbb{N}.$$Multiplicando por $\frac{1}{\sqrt{\lambda_j}}$, segue do item (b) do Lema 2.7 (e do Teorema 2.3) que vale (2.12)$_1$.CONDICAO INICIALJa vimos que $u$ pertence ao espaco $W$ definido em (2.26). Logo, o Teorema 2.5 implica que $u\in C([0,T];H)$ e, portanto, $u(0)$ tem sentido. De (2.21) e (2.23),\begin{align*}\int_0^T(u_m(t),v(t))\;dt&\to \int_0^T(u(t),v(t))\;dt,\quad\forall\ v\in L^1(0,T;H),\\\int_0^T(u_m'(t),z(t))\;dt=\int_0^T\langle u_m'(t),z(t)\rangle_{V'\times V}\;dt&\to \int_0^T\langle u'(t),z(t)\rangle_{V'\times V}\;dt,\quad\forall\ z\in L^2(0,T;V).\end{align*}Tome $\theta\in C^1[0,T]$ tal que $\theta(0)=1$ e $\theta(T)=0$. Das ultimas duas convergencias com $v(t)=\theta'(t)w_j$ e $z(t)=\theta(t)w_j$ segue que\begin{align*}\int_0^T(u_m(t),\theta'(t)w_j)\;dt&\to \int_0^T(u(t),\theta'(t)w_j)\;dt=\int_0^T\langle u(t),\theta'(t)w_j\rangle_{V'\times V}\;dt,\quad\forall\ j\in\mathbb{N},\\\int_0^T(u'_m(t),\theta(t)w_j)\;dt&\to \int_0^T\langle u'(t),\theta(t)w_j\rangle_{V'\times V}\;dt,\quad\forall\ j\in\mathbb{N}.\end{align*}Somando as duas ultimas convergencias, segue da formula de integracao por partes (2.6) que$$(u_m(0),w_j)\to (u(0),w_j),\quad\forall\ j\in\mathbb{N}.$$Como $\operatorname{span}\{w_j\mid j\in\mathbb{N}\}$ e denso em $H$ (veja Teorema 2.3), segue de (2.14)$_2$ e da ultima convergencia que$${(u_0}_m,v)=(u_m(0),v)\to (u(0),v),\quad\forall\ v\in H.$$Mas, de (2.14)$_3$,$$({u_0}_m,v)\to (u_0,v),\quad\forall\ v\in H.$$Logo, vale (2.12)$_2$ porque, pela unicidade do limite, as duas ultimas convergencias implicam que$$(u(0),v)=(u_0,v),\quad\forall\ v\in H.$$REGULARIDADE EXTRAA igualdade ja provada (2.12)$_1$ significa que, para quaisquer $\theta\in\mathcal{D}(0,T)$ e $v\in V$,$$\left\langle \frac{d}{dt}(u(\cdot),v),\theta\right\rangle+ \big\langle \nu a(u(\cdot),v),\theta\big\rangle+\big\langle b(u(\cdot),u(\cdot),v),\theta\big\rangle=\big\langle \langle f(\cdot),v\rangle_{V'\times V},\theta\big\rangle.%,\quad\forall\ \theta\in\mathcal{D}(0,T),\;v\in V.\tag{2.29}$$Mas,\begin{align*}\big\langle b(u(\cdot),u(\cdot),v),\theta\big\rangle&=\int_0^T b(u(t),u(t),v)\theta(t)\;dt=\int_0^T\langle B(u(t),u(t)),v\rangle_{V'\times V}\theta(t)\;dt\\&=\int_0^T\langle B(u(t),u(t))\theta(t),v\rangle_{V'\times V}\;dt=\left\langle \int_0^TB(u(t),u(t))\theta(t)\;dt,v\right\rangle_{V'\times V}\\&=\Big\langle \langle B(u(\cdot),u(\cdot)),\theta\rangle,v\Big\rangle_{V'\times V}.\end{align*}Observe que na penultima igualdade estamos usando o Lema 2.10, de acordo com o qual a integral que aparece dentro da dualidade esta bem definida. Analogamente,\begin{gather*}\left\langle \frac{d}{dt}(u(\cdot),v),\theta\right\rangle=\Big\langle \langle u'(\cdot),\theta\rangle,v\Big\rangle_{V'\times V}\qquad\qquad\big\langle \nu a(u(\cdot),v),\theta\big\rangle=\Big\langle \langle\nu Au(\cdot),\theta\rangle,v\Big\rangle_{V'\times V}\\\big\langle \langle f(\cdot),v\rangle_{V'\times V},\theta\big\rangle=\Big\langle \langle f(\cdot),\theta\rangle,v\Big\rangle_{V'\times V}\end{gather*}Logo, por (2.29),$$\Big\langle \langle u'(\cdot),\theta\rangle+\langle\nu Au(\cdot),\theta\rangle+\langle B(u(\cdot),u(\cdot)),\theta\rangle-\langle