maximal existence time for quasilinear parabolic PDEs This should be a classical result, but I wasn't able to find a reference for it:
If $\Omega\subseteq\mathbb R^n$ is bounded and open and I already know that for all $T>0$, $a,f\in C^1(\Omega\times(0,T)\times\mathbb R)\cap L^\infty(\overline\Omega\times[0,T]\times\mathbb R)$, $b\in C^1(\Omega\times(0,T)\times\mathbb R;\mathbb R^d)\cap L^\infty(\overline\Omega\times[0,T]\times\mathbb R;\mathbb R^d)$ and $u_0\in C(\overline\Omega)$ with $\left.u_0\right|_{\partial\Omega}=0$, there is a solution $u\in C(\Omega\times[0,T])\cap C^{2,\:1}(\Omega\times(0,T))$ of \begin{equation}\left\{\begin{split}u_t&=\nabla\cdot a(x,t,u)\nabla u+b(x,t,u)\cdot\nabla u+f(x,t,u)&\;\;\;\text{in }(0,T)\times\Omega\\u&=0&\;\;\;\text{on }(0,T)\times\partial\Omega\\u(0,\;\cdot\;)&=u_0,\end{split}\right.\tag1\end{equation} how can we conclude that there is a $T_{\text{max}}\in(0,\infty]$ such that for all $T\in(0,T_{\text{max}})$ problem $(1)$ has a solution in $C(\Omega\times[0,T])\cap C^{2,\:1}(\Omega\times(0,T))$ and either $T_{\text{max}}<\infty$ or $\limsup_{t\to T_{\text{max}}-}\left\|u(t)\right\|_{L^\infty(\Omega)}=\infty$?
 A: Here is an idea on how you could proceed. This is adapted from a proposition in chapter 3.1.3 of "Fourier Analysis and Nonlinear PDEs" by H. Bahouri.
I will write $u_x = (\partial_{x_1}u, ..., \partial_{x_n}u) \in \mathbb R^n$ and $u_{xx} = (\partial_{x_1x_1}u, ..., \partial_{x_nx_n}u)  \in \mathbb R^n$ for simplicity.
Assume that $u(x,t) \in C(\overline{\Omega} \times [0,T^*[) \cap C^2(\Omega \times ]0,T^*[)$ is a solution of $(1)$ on $\Omega \times ]0,T^*[$ such that $T^* < + \infty$ and $||(u(\cdot, t), u_x(\cdot, t), u_{xx}(\cdot, t))||_{L^\infty(\Omega)}$ is bounded in a neighborhood of $T^*$. Then we can extend $u(x,t)$ beyond $T^*$.
Let $F \in C^0(\Omega \times [0,T^*[ \times \mathbb R \times \mathbb R^n  \times \mathbb R^n)$ be the function
$$F(x,t,u,u_x,u_{xx}) = \nabla_x a(x,t,u) \cdot \nabla_x u + a(x,t,u) \Delta_x u + b(x,t,u) \cdot \nabla_x u + f(x,t,u)$$
By integrating $(1)$, we have
$$u(x,t) = u(x,0) + \int_0^t F(x,t,u(x,t),u_x(x,t),u_{xx}(x,t)) dt,  \ \forall x \in \Omega \ \forall t \in [0,T^*[$$
Assume that
$$|F(x,t,u,u_{x}, u_{xx})| \leq \beta(t) M(||(u,u_x,u_{xx})||_2), \ \forall t \in [0,T^*[ \ \forall (x,u,u_x,u_{xx}) \in \Omega \times \mathbb R^{1 + 2n}$$
where $\beta \in L^1_{loc}([0,T^*[, \mathbb R_{\geq 0})$ and $M : \mathbb R_{\geq 0} \rightarrow \mathbb R_{\geq 0}$ is locally bounded.
For example, if $f(x,t,u)$, $b(x,t,u)$ and $\nabla_x a(x,t,u)$ are bounded on $\Omega \times [0,T^*[ \times \mathbb R$, then the above inequality is satisfied.
Fix $0 < T_0 < T^*$ such that $||(u(\cdot, t), u_x(\cdot, t), u_{xx}(\cdot, t))||_{L^\infty(\Omega)}$ is bounded on $[T_0, T^*[$, say by a constant $C_0$. Let $C = \max_{0 \leq r \leq C_0}M(r)$.
Let $\varepsilon > 0$ be fixed. Since $\beta \in L^1([T_0, T^*[, \mathbb R_{\geq 0})$, there exists $\delta > 0$ such that for all measurable subsets $J \subset [T_0,T^*[$ :
$$\mu(J) \leq \delta \Longrightarrow \int_J \beta(t) d\mu(t) \leq \dfrac{\varepsilon}{C + 1}$$
Hence, for all $t_1, t_2 \in [T_0,T^*[$, $|t_1-t_2| \leq \delta$, we have
$$|u(x,t_1) - u(x, t_2)| \leq \int_{[t_1,t_2]}|F(x,t,u(x,t),u_x(x,t),u_{xx}(x,t))| dt \leq C\int_{[t_1,t_2]} \beta(t) dt \leq \varepsilon$$
In particular, for all $t_1, t_2 \in [T_0,T^*[$, $|t_1-t_2| \leq \delta$, we have
$$||u(\cdot ,t_1) - u(\cdot , t_2)||_{L^\infty(\Omega)} \leq \varepsilon$$
Therefore, $\lim \limits_{t \to T^*}u(x,t)$ converges uniformly in $x$ to a limit $u^* \in C(\overline{\Omega})$ and we have $u^*(x) = 0$ on $\partial \Omega$, since $u(x,t) = 0$ on $[T_0,T^*[ \times \partial \Omega$.
To extend $u(x,t)$ beyond $T^*$, solve locally the PDE :
\begin{equation}\left\{\begin{split}u_t&=\nabla\cdot a(x,t,u)\nabla u+b(x,t,u)\cdot\nabla u+f(x,t,u)&\;\;\;\text{in }(T^* - \xi,T^* + \xi)\times\Omega\\u&=0&\;\;\;\text{on }(T^* - \xi,T^* + \xi)\times\partial\Omega\\u(\;\cdot\;,T^*)&=u^*,\end{split}\right.\tag2\end{equation}
However, you need uniqueness of the local solution to conclude.
It follows from the above claim that if $u(x,t)$ is a maximal solution of $(1)$ on $\Omega \times ]0,T_{\max}[$, then either $T_{\max} = +\infty$, or one of $u(x,t), u_x(x,t), u_{xx}(x,t)$ blows up when $t \to T_{\max}$.
