Resultados

$~$

Se enuncia el resultado principal de este trabajo:

Teorema 6   Sea el problema % latex2html id marker 5560
$ (\ref{eq:i_2})$:

\begin{displaymath}\begin{cases}
\displaystyle \frac{\partial u}{\partial t} (x,...
...
u(x, t) = 0, &\mbox x \in \partial \Omega, t>0.\\
\end{cases}\end{displaymath}    

donde $f$ es continua en $(x, t)$ y de Lipschitz en $u$ con constante de Lipschitz $L$, satisfaciendo una de las dos condiciones:
i)
$f(x, t, -u) = - f(x, t, u)$
ii)
$f(x, t, 0) = 0$
y $u_0 \in L^{\infty}(\Omega)$, con $u_0(x) \geq 0$, para $x \in \overline {\Omega}$. Sea $k > 0$ se define la función:

$\displaystyle \phi(x, t): \overline{\Omega} \times (0, +\infty) \longrightarrow {\mathbb{R}}$    

como:

$\displaystyle \phi(x, t) = e^{-k(\vert\vert x\vert\vert^2 + t^2)}.$    

Entonces:
$(I)$
Existe una constante $C > 0$ tal que:
$\displaystyle \int_{\Omega} \int_0^T \frac{\partial u}{\partial t}(x, t)\phi(x, t)dtdx$ $\displaystyle \leq$ $\displaystyle 2C\vert\vert u\vert\vert _{L^1([0, T]; H^1_{0}(\overline {\Omega}))}$  

ó:
$\displaystyle \int_{\Omega} \int_0^T \frac{\partial u}{\partial t}(x, t)\phi(x, t)dtdx$ $\displaystyle \leq$ $\displaystyle 2C\sqrt{T} \vert\vert u\vert\vert _{L^2([0, T]; H_0^1(\overline{\Omega}))}$  

$(II)$
Si existe un $0 < T < +\infty$ tal que:

$\displaystyle \displaystyle \int_{\Omega} u(x, t)\phi(x, t) dx = +\infty$    

cuando $t \to T$. Entonces $u(x, t)$ está definida en $[0, T)$, siendo $T$ el tiempo de explosión. Además se define la solución propia minimal del problema como:

$\displaystyle \bar u(x, t) = \begin{cases}
\displaystyle u(x, t), & t \in [0, T) \\
+ \infty, & t \in [T, +\infty). \\
\end{cases}
$

$(III)$
Sea $T < +\infty$ tiempo de explosión. Si se verifica que:
-
$u \in L^{\infty}((0, T); L^{\infty}(\overline{\Omega})) \cap L^2((0, T); L^{\infty}(\overline{\Omega}))$ y:

$\displaystyle \vert\vert u\vert\vert _{L^2((0, T); L^{\infty}(\overline{\Omega}...
...leq \vert\vert u\vert\vert _{L^{\infty}((0, T); L^{\infty}(\overline{\Omega}))}$

-
existe $\delta > 0$ tal que:
$\displaystyle \vert\vert u\vert\vert _{L^{\infty}((0, T); L^{\infty}(\overline{...
...line{\Omega})} \leq \frac{1}{\vert\Omega\vert}\int_{\Omega} u(x,t) \phi(x, t)dx$      

$\forall \; t \in [0, T)$.
entonces para todo $T_0 \in (0, T)$, existe $\delta_1 = \delta_1(T_0)$ tal que:
$\displaystyle \vert\vert u\vert\vert _{L^2((0, T); L^{\infty}(\overline{\Omega}))}$ $\displaystyle \leq$ $\displaystyle \frac{\vert\Omega\vert}{\vert\Omega\vert - D\delta_1\sqrt{T}}.\vert\vert u_0\vert\vert _{L^{\infty}(\Omega)}$  

$\forall t \in [0, T_0]$. Así se obtiene que:

$\displaystyle T < \left( \frac{\vert\Omega\vert}{D\delta_1}\right)^2,$    

donde $D = C + \sqrt{2k}\vert\Omega\vert^{\frac{1}{2}}e^{-\frac{1}{2}}$.

Prueba:$~$
De la ecuación:

$\displaystyle \frac{\partial u}{\partial t} (x, t)= \Delta u(x, t) + f(x, t, u(x, t))$      

multiplicando por $\phi(x, t)$ e integrando sobre $\Omega \times [0, T)$, donde $T$ es el tiempo máximo de existencia:

$\displaystyle \int_{\Omega} \int_0^T \frac{\partial u}{\partial t} (x, t) \phi(...
... u(x, t)\phi(x, t)dtdx + \int_{\Omega} \int_0^T f(x, t, u(x, t))\phi(x, t)dtdx,$    

Notar que $\phi(., t) \in C^{\infty}(\Omega)$, $\forall \; t \in [0, T]$. Aplicando la identidad de Green e intercambiando el orden de integración:
$\displaystyle \int_{\Omega} \int_0^T \frac{\partial u}{\partial t}(x, t)\phi(x, t)dtdx$ $\displaystyle =$ $\displaystyle \int_0^T \int_{\partial \Omega}\frac{\partial u}{\partial \eta} (x, t)\phi(x, t)dSdt - \int_0^T \int_{\Omega} \nabla u(x, t) \nabla \phi(x, t)dxdt$  
  $\displaystyle +$ $\displaystyle \int_0^T \int_{\Omega}f(x, t, u(x, t))\phi(x, t)dxdt$  

tomando el valor absoluto y usando la desigualdad triangular:
$\displaystyle \left\vert \int_{\Omega} \int_0^T \frac{\partial u}{\partial t} (x, t) \phi(x, t)dtdx \right\vert$ $\displaystyle \leq$ $\displaystyle \int_0^T \int_{\partial \Omega}\left \vert\frac{\partial u}{\part...
..._0^T \int_{\Omega} \left \vert\nabla u(x, t) \nabla \phi(x, t) \right\vert dxdt$  
  $\displaystyle +$ $\displaystyle \int_0^T \int_{\Omega}\vert f(x, t, u(x, t))\phi(x, t)\vert dxdt$  

Estimando las integrales por separado:
$1)$

$\displaystyle \int_0^T \int_{\partial \Omega}\left\vert \frac{\partial u}{\partial \eta} (x, t)\phi(x, t)\right\vert dSdt$ $\displaystyle \leq$ $\displaystyle \int_0^T \left[ \int_{\partial \Omega}\vert\nabla u(x, t). \eta(x)\vert dS \right]. \vert\phi(x, t)\vert dt$  

pero:
$\displaystyle \vert\vert\nabla \phi(x)\vert\vert$ $\displaystyle =$ $\displaystyle 2k\vert\vert x\vert\vert e^{-k(\vert\vert x\vert\vert^2+t^2)} \leq \sqrt{2k}e^{-\frac{1}{2}}$  

luego:
$\displaystyle \int_0^T \int_{\partial \Omega}\left\vert \frac{\partial u}{\partial \eta} (x, t)\phi(x, t)\right\vert dSdt$ $\displaystyle \leq$ $\displaystyle \sqrt{2k}e^{-\frac{1}{2}} \int_0^T \left[ \int_{\partial \Omega}\vert\nabla u(x, t)\vert.\vert\eta(x)\vert dS \right]dt$  
  $\displaystyle \leq$ $\displaystyle \sqrt{2k}e^{-\frac{1}{2}} \int_0^T \left[ \int_{\partial \Omega}\vert\nabla u(x, t)\vert\cdot1dS \right]dt$  
  $\displaystyle \leq$ $\displaystyle \sqrt{2k}e^{-\frac{1}{2}} \int_0^T \left[ \left(\int_{\partial \O...
...{\frac{1}{2}}. \left(\int_{\partial \Omega}1^2dS\right)^{\frac{1}{2}} \right]dt$  
  $\displaystyle =$ $\displaystyle \sqrt{2k}e^{-\frac{1}{2}} \int_0^T \vert\vert\nabla u(., t)\vert\vert _{L^2(\partial \Omega)}.\vert\partial \Omega\vert^{\frac{1}{2}}dt$  
  $\displaystyle \leq$ $\displaystyle \sqrt{2k}e^{-\frac{1}{2}} \int_0^T \vert\vert\nabla u(., t)\vert\vert _{L^2(\Omega)}.\vert\Omega\vert^{\frac{1}{2}}dt$  
  $\displaystyle \leq$ $\displaystyle \sqrt{2k}e^{-\frac{1}{2}} \vert\Omega\vert^{\frac{1}{2}} \int_0^T \vert\vert\nabla u(., t)\vert\vert _{L^2(\overline{\Omega})}dt$  

$2) $

$\displaystyle \int_0^T \left[\int_{\Omega} \vert\nabla u(x, t) \nabla \phi(x)\vert dx\right]dt$ $\displaystyle \leq$ $\displaystyle \int_0^T \left[\int_{\Omega} \vert\nabla u(x, t)\vert.\sqrt{2k}e^{-\frac{1}{2}}dx \right]dt$  
  $\displaystyle \leq$ $\displaystyle \sqrt{2k}e^{-\frac{1}{2}} \int_0^T \left[ \left(\int_{\Omega} \ve...
...right)^{\frac{1}{2}}. \left(\int_{\Omega}1^2dx \right)^{\frac{1}{2}} \right] dt$  
  $\displaystyle =$ $\displaystyle \sqrt{2k}e^{-\frac{1}{2}} \vert\Omega\vert^{\frac{1}{2}} \int_0^T \left(\int_{\Omega} \vert\nabla u(x, t)\vert^2dx\right)^{\frac{1}{2}} dt$  
  $\displaystyle =$ $\displaystyle \sqrt{2k}e^{-\frac{1}{2}} \vert\Omega\vert^{\frac{1}{2}} \int_0^T \vert\vert\nabla u(., t)\vert\vert _{L^{2}(\Omega)}dt$  
  $\displaystyle \leq$ $\displaystyle \sqrt{2k}e^{-\frac{1}{2}} \vert\Omega\vert^{\frac{1}{2}} \int_0^T \vert\vert\nabla u(., t)\vert\vert _{L^{2}(\overline {\Omega})}dt$  

$3)$

$\displaystyle \int_0^T \int_{\Omega}\vert f(x, t, u(x, t))\phi(x, t)\vert dxdt$ $\displaystyle \leq$ $\displaystyle \int_0^T \int_{\Omega}\vert f(x, t, u(x, t))\vert.\vert\phi(x, t)\vert dxdt$  
  $\displaystyle \leq$ $\displaystyle \int_0^T \int_{\Omega}\vert f(x, t, u(x, t))\vert dxdt$  

Por otro lado, de la definición de Lipschitz de $f$ con $v = -u$ y usando $i)$:
$\displaystyle \vert f(x, t, u) - f(x, t, -u)\vert$ $\displaystyle \leq$ $\displaystyle L\vert u - (-u)\vert$  
$\displaystyle \vert 2f(x, t, u)\vert$ $\displaystyle \leq$ $\displaystyle L\vert 2u\vert$  
$\displaystyle \vert f(x, t, u)\vert$ $\displaystyle \leq$ $\displaystyle L\vert u\vert,$  

o de la misma definición de Lipschitz de $f$ con $v = 0$ y usando $ii)$:
$\displaystyle \vert f(x, t, u) - f(x, t, 0)\vert$ $\displaystyle \leq$ $\displaystyle L\vert u - 0\vert$  
$\displaystyle \vert f(x, t, u)\vert$ $\displaystyle \leq$ $\displaystyle L\vert u\vert$  

En ambos casos, resulta que:
$\displaystyle \vert f(x, t, u)\vert \leq L\vert u\vert.$      

Por tanto:
$\displaystyle \int_0^T \int_{\Omega}\vert f(x, t, u(x, t))\phi(x)\vert dxdt$ $\displaystyle \leq$ $\displaystyle \int_0^T \left[ \int_{\Omega}\vert f(x, t, u(x, t))\vert dx \right]dt$  
  $\displaystyle \leq$ $\displaystyle \int_0^T \left[ \int_{\Omega}L\vert u(x, t)\vert dx \right]dt$  
  $\displaystyle =$ $\displaystyle L\int_0^T \left[ \int_{\Omega}\vert u(x, t)\vert . 1dx \right]dt$  
  $\displaystyle \leq$ $\displaystyle L\int_0^T \left[ \left(\int_{\Omega}\vert u(x, t)\vert^2dx\right)^{\frac{1}{2}}.\left(\int_{\Omega}1^2dx\right)^{\frac{1}{2}} \right]dt$  
  $\displaystyle =$ $\displaystyle L\vert\Omega\vert^{\frac{1}{2}} \int_0^T \left(\int_{\Omega}\vert u(x, t)\vert^2dx\right)^{\frac{1}{2}}dt$  
  $\displaystyle =$ $\displaystyle L\vert\Omega\vert^{\frac{1}{2}}\int_0^T \vert\vert u(., t)\vert\vert _{L^2(\Omega)} dt$  
  $\displaystyle \leq$ $\displaystyle L\vert\Omega\vert^{\frac{1}{2}}\int_0^T \vert\vert u(., t)\vert\vert _{L^2(\overline {\Omega})}dt$  

Luego, reuniendo las estimaciones se tiene:
$\displaystyle \left\vert \int_{\Omega} \int_0^T \frac{\partial u}{\partial t} (x, t) \phi(x, t)dtdx \right\vert$ $\displaystyle \leq$ $\displaystyle \int_0^T \int_{\partial \Omega}\left \vert\frac{\partial u}{\part...
..._0^T \int_{\Omega} \left \vert\nabla u(x, t) \nabla \phi(x, t) \right\vert dxdt$  
  $\displaystyle +$ $\displaystyle \int_0^T \int_{\Omega}\vert f(x, t, u(x, t))\phi(x, t)\vert dxdt$  
  $\displaystyle \leq$ $\displaystyle \sqrt{2k}e^{-\frac{1}{2}}\vert\Omega\vert^{\frac{1}{2}} \int_0^T ...
...{2}} \int_0^T \vert\vert\nabla u(., t)\vert\vert _{L^{2}(\overline {\Omega})}dt$  
  $\displaystyle +$ $\displaystyle L\vert\Omega\vert^{\frac{1}{2}}\int_0^T \vert\vert u(., t)\vert\vert _{L^2(\overline {\Omega})}dt$  
  $\displaystyle =$ $\displaystyle \left(2\sqrt{2k}e^{-\frac{1}{2}} \vert\Omega\vert^{\frac{1}{2}} \right) \int_0^T \vert\vert\nabla u(., t)\vert\vert _{L^{2}(\overline {\Omega})}$  
  $\displaystyle +$ $\displaystyle L\vert\Omega\vert^{\frac{1}{2}}\int_0^T \vert\vert u(., t)\vert\vert _{L^2(\overline {\Omega})}dt$  

Sea $B = 2\sqrt{2k}e^{-\frac{1}{2}} \vert\Omega\vert^{\frac{1}{2}}$, entonces:
$\displaystyle \left\vert \int_{\Omega} \int_0^T \frac{\partial u}{\partial t}(x, t) \phi(x)dtdx \right\vert$ $\displaystyle \leq$ $\displaystyle B\int_0^T \vert\vert\nabla u(., t)\vert\vert _{L^2(\overline{\Ome...
...\frac{1}{2}} \int_0^T \vert\vert u(., t)\vert\vert _{L^2(\overline {\Omega})}dt$  

Sea ahora $C = \max \{B, L\vert\Omega\vert^{\frac{1}{2}}\}$, entonces:
$\displaystyle \left\vert \int_{\Omega} \int_0^T \frac{\partial u}{\partial t} (x, t) \phi(x, t)dtdx \right\vert$ $\displaystyle \leq$ $\displaystyle C\left( \int_0^T\vert\vert\nabla u(., t)\vert\vert _{L^2(\overlin...
...)}dt + \int_0^T\vert\vert u(., t)\vert\vert _{L^2(\overline {\Omega})}dt\right)$  
  $\displaystyle =$ $\displaystyle C \int_0^T \left( \vert\vert\nabla u(., t)\vert\vert _{L^2(\overline{\Omega})} + \vert\vert u(., t)\vert\vert _{L^2(\overline {\Omega})} \right)dt$  
  $\displaystyle \leq$ $\displaystyle 2C\int_0^T \vert\vert u(., t)\vert\vert _{H^1_{0}(\overline {\Omega})}dt$  
  $\displaystyle =$ $\displaystyle 2C\vert\vert u\vert\vert _{L^1([0, T]; H^1_{0}(\overline {\Omega}))}$  

o también:
$\displaystyle \left\vert \int_{\Omega} \int_0^T \frac{\partial u}{\partial t} (x, t) \phi(x, t)dtdx \right\vert$ $\displaystyle \leq$ $\displaystyle 2C\int_0^T\vert\vert u(., t)\vert\vert _{H^1_{0}(\overline {\Omeg...
...dt = 2C\int_0^T \vert\vert u(., t)\vert\vert _{H^1_{0}(\overline {\Omega})}.1dt$  
  $\displaystyle \leq$ $\displaystyle 2C \left(\int_0^T \vert\vert u(., t)\vert\vert _{H^1_{0}(\overline {\Omega})}^2 dt\right)^{\frac{1}{2}}. \left(\int_0^T1^2 dt \right)^{\frac{1}{2}}$  
  $\displaystyle =$ $\displaystyle 2C\sqrt{T} \vert\vert u\vert\vert _{L^2([0, T]; H_0^1(\overline{\Omega}))}$  

En consecuencia se tiene que:
$\displaystyle \int_{\Omega} \int_0^T \frac{\partial u}{\partial t}(x, t)\phi(x, t)dtdx$ $\displaystyle \leq$ $\displaystyle 2C\vert\vert u\vert\vert _{L^1([0, T]; H^1_{0}(\overline {\Omega}))}$  
$\displaystyle \int_{\Omega} \int_0^T \frac{\partial u}{\partial t}(x, t)\phi(x, t)dtdx$ $\displaystyle \leq$ $\displaystyle 2C\sqrt{T} \vert\vert u\vert\vert _{L^2([0, T]; H_0^1(\overline{\Omega}))}$  

Con esto queda demostrado (I).
Por otro lado:
$\displaystyle \left\vert \int_{\Omega} \int_0^T \frac{\partial u}{\partial t} (x, t) \phi(x, t)dtdx \right\vert$ $\displaystyle =$ $\displaystyle \left\vert \int_{\Omega} \left[ u(x, t).\phi(x, t)\vert _{0}^{T} - \int_0^T u(x, t).\frac{\partial \phi}{\partial t}(x, t)dt \right]dx \right\vert$  
  $\displaystyle =$ $\displaystyle \left\vert \int_{\Omega} u(x, t).\phi(x, t)\vert _{0}^{T}dx - \int_{\Omega}\int_0^T u(x, t).\frac{\partial \phi}{\partial t}(x, t)dtdx \right\vert$  
  $\displaystyle \geq$ $\displaystyle \left\vert \int_{\Omega} u(x, t).\phi(x, t)\vert _{0}^{T}dx \righ...
...{\Omega}\int_0^T u(x, t).\frac{\partial \phi}{\partial t}(x, t)dtdx \right\vert$  

Entonces:
$\displaystyle \left\vert \int_{\Omega} u(x, t).\phi(x, t)\vert _{0}^{T}dx \righ...
...{\Omega}\int_0^T u(x, t).\frac{\partial \phi}{\partial t}(x, t)dtdx \right\vert$ $\displaystyle \leq$ $\displaystyle \left\vert \int_{\Omega} \int_0^T \frac{\partial u}{\partial t} (x, t) \phi(x, t)dtdx \right\vert$  
  $\displaystyle \leq$ $\displaystyle 2C\int_0^T \vert\vert u(., t)\vert\vert _{H^1_{0}(\overline {\Omega})}dt$  

extrayendo los extremos:
$\displaystyle \left\vert \int_{\Omega} u(x, t).\phi(x, t)\vert _{0}^{T}dx \righ...
...{\Omega}\int_0^T u(x, t).\frac{\partial \phi}{\partial t}(x, t)dtdx \right\vert$ $\displaystyle \leq$ $\displaystyle 2C\int_0^T \vert\vert u(., t)\vert\vert _{H^1_{0}(\overline {\Omega})}dt$  


$\displaystyle \left\vert \int_{\Omega} u(x, t).\phi(x, t)\vert _{0}^{T}dx \right\vert$ $\displaystyle \leq$ $\displaystyle \left\vert \int_{\Omega}\int_0^T u(x, t).\frac{\partial \phi}{\pa...
...\vert +2 C\int_0^T\vert\vert u(., t)\vert\vert _{H^1_{0}(\overline {\Omega})}dt$  

Ahora:
$\displaystyle \left\vert \int_{\Omega} u(x, t).\phi(x, t)\vert _{0}^{T}dx \right\vert$ $\displaystyle =$ $\displaystyle \left\vert \int_{\Omega} (u(x, T).\phi(x, T) - u(x, 0)\phi(x, 0)) dx \right\vert$  
  $\displaystyle \geq$ $\displaystyle \left\vert \int_{\Omega} (u(x, T).\phi(x, T) dx\right\vert - \left\vert \int_{\Omega} u(x, 0)\phi(x, 0)) dx \right\vert$  

y de esto:
$\displaystyle \left\vert \int_{\Omega} (u(x, T).\phi(x, T) dx\right\vert - \left\vert \int_{\Omega} u(x, 0)\phi(x, 0)) dx \right\vert$ $\displaystyle \leq$ $\displaystyle \left\vert \int_{\Omega} u(x, t).\phi(x, t)\vert _{0}^{T}dx \right\vert$  

y además:

$\displaystyle \left\vert \int_{\Omega}\int_0^T u(x, t).\frac{\partial \phi}{\partial t}(x, t)dtdx \right\vert$ $\displaystyle \leq$ $\displaystyle \int_{\Omega}\int_0^T \vert u(x, t)\vert.\vert\frac{\partial \phi}{\partial t}(x, t)\vert dtdx$  
  $\displaystyle \leq$ $\displaystyle \int_{\Omega} \int_0^T \vert u(x, t)\vert.\sqrt{2k}e^{-\frac{1}{2}}dtdx$  
  $\displaystyle =$ $\displaystyle \sqrt{2k}e^{-\frac{1}{2}} \int_0^T \int_{\Omega}\vert u(x, t)\vert.1dxdt$  
  $\displaystyle \leq$ $\displaystyle \sqrt{2k}e^{-\frac{1}{2}} \int_0^T \left[ \left( \int_{\Omega} \v...
... \right)^\frac{1}{2}. \left( \int_{\Omega} 1^2 dx \right)^\frac{1}{2} \right]dt$  
  $\displaystyle =$ $\displaystyle \sqrt{2k}\vert\Omega\vert^{\frac{1}{2}}e^{-\frac{1}{2}} \int_0^T \vert\vert u(., t)\vert\vert _{L^2(\Omega)}dt$  
  $\displaystyle \leq$ $\displaystyle \sqrt{2k}\vert\Omega\vert^{\frac{1}{2}}e^{-\frac{1}{2}} \int_0^T \vert\vert u(., t)\vert\vert _{L^2(\overline{\Omega})}dt$  

Reemplazando en:
$\displaystyle \left\vert \int_{\Omega} \int_0^T \frac{\partial u}{\partial t} (x, t) \phi(x, t)dtdx \right\vert$ $\displaystyle \geq$ $\displaystyle \left\vert \int_{\Omega} u(x, t).\phi(x, t)\vert _{0}^{T}dx \righ...
...{\Omega}\int_0^T u(x, t).\frac{\partial \phi}{\partial t}(x, t)dtdx \right\vert$  

o equivalentemente se tiene:
$\displaystyle \left\vert \int_{\Omega} u(x, t).\phi(x, t)\vert _{0}^{T}dx \right\vert$ $\displaystyle \leq$ $\displaystyle \left\vert \int_{\Omega} \int_0^T \frac{\partial u}{\partial t}(x...
...{\Omega}\int_0^T u(x, t).\frac{\partial \phi}{\partial t}(x, t)dtdx \right\vert$  

pero se sabe que:
$\displaystyle \left\vert \int_{\Omega} u(x, T).\phi(x, T) dx\right\vert - \left\vert \int_{\Omega} u(x, 0)\phi(x, 0)) dx \right\vert$ $\displaystyle \leq$ $\displaystyle \left\vert \int_{\Omega} u(x, t).\phi(x, t)\vert _{0}^{T}dx \right\vert$  
$\displaystyle \left\vert \int_{\Omega}\int_0^T u(x, t).\frac{\partial \phi}{\partial t}(x, t)dtdx \right\vert$ $\displaystyle \leq$ $\displaystyle \sqrt{2k}\vert\Omega\vert^{\frac{1}{2}}e^{-\frac{1}{2}} \int_0^T \vert\vert u(., t)\vert\vert _{L^2(\overline{\Omega})}dt$  
$\displaystyle \left\vert \int_{\Omega} \int_0^T \frac{\partial u}{\partial t} (x, t) \phi(x, t)dtdx \right\vert$ $\displaystyle \leq$ $\displaystyle 2C\int_0^T\vert\vert u(., t)\vert\vert _{H^1_0{(\overline{\Omega}})}dt$  

entonces:
$\displaystyle \left\vert \int_{\Omega} u(x, T)\phi(x, T) dx\right\vert - \left\vert \int_{\Omega} u(x, 0)\phi(x, 0)) dx \right\vert$ $\displaystyle \leq$ $\displaystyle 2C\int_0^T\vert\vert u(., t)\vert\vert _{H^1_0{(\overline{\Omega}})}dt$  
  $\displaystyle +$ $\displaystyle \sqrt{2k}\vert\Omega\vert^{\frac{1}{2}}e^{-\frac{1}{2}} \int_0^T \vert\vert u(., t)\vert\vert _{L^2(\overline{\Omega})}dt$  

equivale a:
$\displaystyle \left\vert \int_{\Omega} u(x, T)\phi(x, T) dx\right\vert$ $\displaystyle \leq$ $\displaystyle \left\vert \int_{\Omega} u(x, 0)\phi(x, 0)) dx \right\vert + 2C\int_0^T\vert\vert u(., t)\vert\vert _{H^1_0{(\overline{\Omega}})}dt$  
  $\displaystyle +$ $\displaystyle \sqrt{2k}\vert\Omega\vert^{\frac{1}{2}}e^{-\frac{1}{2}} \int_0^T \vert\vert u(., t)\vert\vert _{L^2(\overline{\Omega})}dt$  

por el principio del máximo, la solución es no negativa:
$\displaystyle \int_{\Omega} u(x, T)\phi(x, T) dx$ $\displaystyle \leq$ $\displaystyle \int_{\Omega} \vert u_0(x)\vert\vert\phi(x, 0)\vert dx + 2C\int_0^T\vert\vert u(., t)\vert\vert _{H^1_0{(\overline{\Omega}})}dt$  
  $\displaystyle +$ $\displaystyle \sqrt{2k}\vert\Omega\vert^{\frac{1}{2}}e^{-\frac{1}{2}} \int_0^T \vert\vert u(., t)\vert\vert _{H^1_0(\overline{\Omega})}dt$  
  $\displaystyle \leq$ $\displaystyle \int_{\Omega} \vert u_0(x)\vert\vert\phi(x, 0)\vert dx + (2C + \s...
...rac{1}{2}}) \int_0^T \vert\vert u(., t)\vert\vert _{H^1_0(\overline{\Omega})}dt$  

luego:
$\displaystyle \int_{\Omega} u(x, T)\phi(x, T) dx$ $\displaystyle \leq$ $\displaystyle \int_{\Omega} \vert\vert u_0\vert\vert _{L^{\infty}(\Omega)} dx +...
...rac{1}{2}}) \int_0^T \vert\vert u(., t)\vert\vert _{H^1_0(\overline{\Omega})}dt$  
  $\displaystyle \leq$ $\displaystyle \vert\Omega\vert.\vert\vert u_0\vert\vert _{L^{\infty}(\Omega)} +...
...rac{1}{2}}) \int_0^T \vert\vert u(., t)\vert\vert _{H^1_0(\overline{\Omega})}dt$  

haciendo $D = (2C + \sqrt{2k}\vert\Omega\vert^{\frac{1}{2}}e^{-\frac{1}{2}})$ se tiene que:
$\displaystyle \int_{\Omega} u(x, T)\phi(x, T) dx$ $\displaystyle \leq$ $\displaystyle \vert\Omega\vert.\vert\vert u_0\vert\vert _{L^{\infty}(\Omega)} + D \int_0^T \vert\vert u(., t)\vert\vert _{H^1_0(\overline{\Omega})}dt$  

Por tanto, si para $T > 0$, se tiene que $\displaystyle \int_{\Omega} u(x,t)\phi(0, T)dx \to +\infty$, cuando $t \to T$ entonces $\displaystyle \int_0^T\vert\vert u(., t)\vert\vert _{H_0^1(\overline{\Omega})}dt \to +\infty$, de esto se tiene que: $\vert\vert u(., t)\vert\vert _{H_0^1(\overline{\Omega})}dt \to +\infty$ y en consecuencia $\vert u(x, t)\vert \to +\infty$. Con esto queda demostrado $(II)$.

Para demostrar $(III)$ se parte de la última desigualdad obtenida:

$\displaystyle \int_{\Omega} u(x, T)\phi(x, T) dx$ $\displaystyle \leq$ $\displaystyle \vert\Omega\vert.\vert\vert u_0\vert\vert _{L^{\infty}(\Omega)} + D \int_0^T \vert\vert u(., t)\vert\vert _{H^1_0(\overline{\Omega})}dt$  

pero, por ser $T$ tiempo de explosión, se cumple que:
$\displaystyle \int_{\Omega} u(x, T)\phi(x, T) dx > \int_{\Omega}u(x,t) \phi(x, t) dx,\;\;\;\forall t \in [0, T)$      

entonces:
$\displaystyle \int_{\Omega} u(x,t) \phi(x, t)dx$ $\displaystyle <$ $\displaystyle \vert\Omega\vert.\vert\vert u_0\vert\vert _{L^{\infty}(\Omega)} + D \int_0^T \vert\vert u(., t)\vert\vert _{H^1_0(\overline{\Omega})}dt$  

Se conoce que:
$\displaystyle \vert\vert u(., t)\vert\vert _{H^1_{0}(\overline {\Omega})} \geq ...
...\vert u(., t)\vert\vert _{L^2(\overline{\Omega})}, \;\;\; \forall t \in [0, T],$      

y además: $u(x, t) < +\infty$ para $t \in [0, T)$. entonces para $T_0 < T$, existe $\delta_1=\delta_1(T_0) > 0$ tal que:

$\displaystyle \vert\vert u(., t)\vert\vert _{H^1_{0}(\overline {\Omega})} \leq ...
...rt u(., t)\vert\vert _{L^2(\overline{\Omega})}, \;\; \forall \; t \in [0, T_0],$    

y como $\Omega$ es acotado se cumple que:
$\displaystyle \vert\vert u(., t)\vert\vert _{L^2(\overline{\Omega})}$ $\displaystyle \leq$ $\displaystyle \vert\vert u(., t)\vert\vert _{L^{\infty}(\overline{\Omega})}, \;\; \forall \; t \in [0, T_0],$  

en consecuencia:
$\displaystyle \vert\vert u(., t)\vert\vert _{H^1_{0}(\overline {\Omega})}$ $\displaystyle \leq$ $\displaystyle \delta_1\vert\vert u(., t)\vert\vert _{L^{\infty}(\overline{\Omega})}, \;\; \forall \; t \in [0, T_0],$  

Por otro lado:
$\displaystyle \int_{\Omega} u(x,t) \phi(x, t)dx$ $\displaystyle \leq$ $\displaystyle \int_{\Omega}\vert u(x,t)\vert. \vert\phi(x, t)\vert dx$  
  $\displaystyle \leq$ $\displaystyle \int_{\Omega}\vert\vert u(., t)\vert\vert _{L^{\infty}(\overline{\Omega})}dx$  
  $\displaystyle =$ $\displaystyle \vert\Omega\vert.\vert\vert u(., t)\vert\vert _{L^{\infty}(\overline{\Omega})}$  

y sea $\delta > 0$ tal que:
$\displaystyle \int_{\Omega} u(x,t) \phi(x, t)dx$ $\displaystyle \geq$ $\displaystyle \delta\vert\Omega\vert.\vert\vert u(., t)\vert\vert _{L^{\infty}(\overline{\Omega})}, \;\; \forall \; t \in [0, T),$  
  $\displaystyle \geq$ $\displaystyle \vert\Omega\vert.\vert\vert u\vert\vert _{L^{\infty}((0, T); L^{\infty}(\overline{\Omega}))}$  

Ahora, regresando a la desigualdad:
$\displaystyle \int_{\Omega} u(x,t) \phi(x, t)dx$ $\displaystyle <$ $\displaystyle \vert\Omega\vert.\vert\vert u_0\vert\vert _{L^{\infty}(\Omega)} + D \int_0^T \vert\vert u(., t)\vert\vert _{H^1_0(\overline{\Omega})}dt$  

o de forma equivalente:
$\displaystyle \vert\Omega\vert.\vert\vert u\vert\vert _{L^{\infty}((0, T); L^{\infty}(\overline{\Omega}))}$ $\displaystyle \leq$ $\displaystyle \vert\Omega\vert.\vert\vert u_0\vert\vert _{L^{\infty}(\Omega)} +...
...int_0^T \delta_1\vert\vert u(., t)\vert\vert _{L^{\infty}(\overline{\Omega})}dt$  
  $\displaystyle \leq$ $\displaystyle \vert\Omega\vert.\vert\vert u_0\vert\vert _{L^{\infty}(\Omega)} +...
...a_1\sqrt{T}\vert\vert u\vert\vert _{L^2((0, T); L^{\infty}(\overline{\Omega}))}$  

pero como: $\vert\vert u\vert\vert _{L^2((0, T); L^{2}(\overline{\Omega}))} \leq \vert\vert u\vert\vert _{L^{\infty}((0, T); L^{\infty}(\overline{\Omega}))}$ entonces:
$\displaystyle \vert\Omega\vert.\vert\vert u\vert\vert _{L^{\infty}((0, T); L^{\...
...a_1\sqrt{T}\vert\vert u\vert\vert _{L^2((0, T); L^{\infty}(\overline{\Omega}))}$ $\displaystyle \leq$ $\displaystyle \vert\Omega\vert.\vert\vert u_0\vert\vert _{L^{\infty}(\Omega)}$  
$\displaystyle (\vert\Omega\vert - D\delta_1\sqrt{T})\vert\vert u\vert\vert _{L^2((0, T); L^{\infty}(\overline{\Omega}))}$ $\displaystyle \leq$ $\displaystyle \vert\Omega\vert.\vert\vert u_0\vert\vert _{L^{\infty}(\Omega)}$  

donde se obtiene que:
$\displaystyle \vert\vert u\vert\vert _{L^2((0, T); L^{\infty}(\overline{\Omega}))}$ $\displaystyle \leq$ $\displaystyle \frac{\vert\Omega\vert}{\vert\Omega\vert - D\delta_1\sqrt{T}}.\vert\vert u_0\vert\vert _{L^{\infty}(\Omega)}$  

para todo $t \in [0, T_0]$. Y para que la desigualdad tenga sentido se debe cumplir que: $\vert\Omega\vert - D\delta_1\sqrt{T} > 0$, es decir:
$\displaystyle T < \left( \frac{\vert\Omega\vert}{D\delta_1}\right)^2. \;\Box$      

Ahora, para problemas particulares de la forma (2), se tiene a continuación
algunos resultados que proporcionan cotas para la solución:

Teorema 7   Sea el problema % latex2html id marker 6280
$ (\ref{eq:i_2})$. Entonces:
$(a)$
Si $f(x, t, u(x, t)) = g(x, t)h(u(x, t))$, con $g$ continua y $\int_{0}^{T} \vert g(x, s)\vert ds \leq M$, con $M \geq 0$, $x \in \overline \Omega$ y $h$ una función globalmente Lipschitziana satisfaciendo $h(-u) = -h(u)$. Entonces existe $C = C(M)$ tal que:

$\displaystyle \vert u(x, t)\vert \leq C\vert\vert u_0\vert\vert _{L^{\infty}(\overline{\Omega})}, \hspace{0.75 cm} (x, t) \in \overline{\Omega} \times [0, T],$    

donde $T$ es el tiempo máximo de existencia de la solución.

$(b)$
Si $f(x, t, u(x, t)) = g(x, t) + h(u(x, t))$, con $g$ continua y ${\small\int_{0}^{T} \vert g(x, s)\vert ds \leq M}$, con $M \geq 0$, $x \in \overline \Omega$ y $h$ una función globalmente Lipschitziana satisfaciendo $h(-u) = -h(u)$. Entonces:

$\displaystyle \vert u(x, t)\vert \leq e^T(\vert\vert u_0\vert\vert _{L^{\infty}...
...ne{\Omega})} + M), \hspace{0.75 cm} (x, t) \in \overline{\Omega} \times [0, T],$    

donde $T$ es el tiempo máximo de existencia de la solución.

Prueba:$~$
La formulación abstracta del problema es:

\begin{displaymath}\begin{cases}
u_t(t) = Au(t) + f(t, u(t)), \hspace{0.5 cm}t > 0,\\
u(0) = u_0 \in L^{\infty}(\Omega). &
\end{cases}\end{displaymath}    

Por el resultado de existencia y unicidad, existe $T > 0$ tal que $u \in L^{\infty}([0, T), L^{\infty}(\Omega))$ y además $u$ es solución de la ecuación:

$\displaystyle u(t) = T(t)u_0 + \int_{0}^{t}T(t - s)f(s, u(s)) ds$    

Luego:
$\displaystyle \vert\vert u(t)\vert\vert _{L^{\infty}(\Omega)}$ $\displaystyle =$ $\displaystyle \vert\vert T(t)u_0 + \int_{0}^{t}T(t - s)f(s, u(s)) ds\vert\vert _{L^{\infty}(\Omega)}$  
  $\displaystyle \leq$ $\displaystyle \vert\vert T(t)u_0\vert\vert _{L^{\infty}(\Omega)} + \int_{0}^{t}\vert\vert T(t - s)f(s, u(s))\vert\vert _{L^{\infty}(\Omega)}ds$  
  $\displaystyle \leq$ $\displaystyle \vert\vert u_0\vert\vert _{L^{\infty}(\Omega)} + \int_{0}^{t} \vert\vert f(s, u(s))\vert\vert _{L^{\infty}(\Omega)}ds$  

es decir:

$\displaystyle \vert\vert u(t)\vert\vert _{L^{\infty}(\Omega)} \leq \vert\vert u...
...\Omega)} + \int_{0}^{t} \vert\vert f(s, u(s))\vert\vert _{L^{\infty}(\Omega)}ds$ (14)

reemplazando $f(s, u(s)) = g(s).h(u(s))$ en (14) se tiene:
$\displaystyle \vert\vert u(t)\vert\vert _{L^{\infty}(\Omega)}$ $\displaystyle \leq$ $\displaystyle \vert\vert u_0\vert\vert _{L^{\infty}(\Omega)} + \int_{0}^{t} \ve...
...vert _{L^{\infty}(\Omega)}.\vert\vert h(u(s))\vert\vert _{L^{\infty}(\Omega)}ds$  

y aplicando la desigualdad de Gronwall se obtiene:
$\displaystyle \vert\vert u(t)\vert\vert _{L^{\infty}(\Omega)}$ $\displaystyle \leq$ $\displaystyle \vert\vert u_0\vert\vert _{L^{\infty}(\Omega)} \displaystyle e^{\int_{0}^{t} \vert\vert g(s)\vert\vert _{L^{\infty}(\Omega)}ds}$  
  $\displaystyle \leq$ $\displaystyle \vert\vert u_0\vert\vert _{L^{\infty}(\Omega)} \displaystyle e^{M}$  
  $\displaystyle =$ $\displaystyle C\vert\vert u_0\vert\vert _{L^{\infty}(\Omega)}$  

y con esto se obtiene (a). Para obtener (b) se reemplaza $f(s, u(s)) = g(s) + h(u(s))$ en (14):
$\displaystyle \vert\vert u(t)\vert\vert _{L^{\infty}(\Omega)}$ $\displaystyle \leq$ $\displaystyle \vert\vert u_0\vert\vert _{L^{\infty}(\Omega)} + \int_{0}^{t} \ve...
...rt _{L^{\infty}(\Omega)} + \vert\vert h(u(s))\vert\vert _{L^{\infty}(\Omega)}ds$  
  $\displaystyle \leq$ $\displaystyle \vert\vert u_0\vert\vert _{L^{\infty}(\Omega)} + \int_{0}^{t} \ve...
...}(\Omega)}ds + \int_{0}^{t}\vert\vert h(u(s))\vert\vert _{L^{\infty}(\Omega)}ds$  
  $\displaystyle \leq$ $\displaystyle \vert\vert u_0\vert\vert _{L^{\infty}(\Omega)} + M + \int_{0}^{t}\vert\vert h(u(s))\vert\vert _{L^{\infty}(\Omega)}ds$  

y por la desigualdad de Gronwall se obtiene:
$\displaystyle \vert\vert u(t)\vert\vert _{L^{\infty}(\Omega)}$ $\displaystyle \leq$ $\displaystyle (\vert\vert u_0\vert\vert _{L^{\infty}(\Omega)} + M) \displaystyle e^{\int_{0}^{t}ds}$  
  $\displaystyle \leq$ $\displaystyle (\vert\vert u_0\vert\vert _{L^{\infty}(\Omega)} + M) \displaystyle e^{T}$  
  $\displaystyle =$ $\displaystyle e^{T}(\vert\vert u_0\vert\vert _{L^{\infty}(\Omega)} + M),\;\;0 < t < T.\;\Box$