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Let $u_0 \in L^\infty(\Omega)$ and $f \in L^\infty((0,T)\times\Omega)$ and consider $$u_t - \Delta u = f$$ $$u(0) = u_0$$ with some boundary conditions. Consider the weak form

$$\int_{\Omega}u_t\varphi + \int_{\Omega}\nabla u \nabla \varphi = \int_{\Omega}f\varphi $$

How can I show that $u \in {L^\infty((0,T)\times \Omega)}$ or $u \in {L^\infty(0,T;L^\infty(\Omega))}$, without using classical approach (just a weak approach)?

If the classical theory cannot be avoid then I don't mind using it. I would have thought we should test with $(u(t) - \Vert u_0 \Vert_{L^\infty})^+$ but don't know what do with the term involving $f$.

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Consider some $v(t) = \alpha t + \beta$. Then $v(t)$ will satisfy $$ v_t - \Delta v = \alpha,\\ v(0) = \beta. $$ Take now $\alpha$ such that $\alpha \geq f$ in $(0, T)\times \Omega$ and $\beta$ such that $\beta \geq u(0)$. (it is possible due to the boundedness of $f$ and $u_0$).

Therefore, the Weak maximum principle for $w = v - u$ will give you that $w \geq 0$. Hence, $u \leq v$, but $v$ is bounded (for $T < +\infty$). Thus, $u$ is also bounded.

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    $\begingroup$ However, I realised that this approach is classical ( $\endgroup$
    – Voliar
    May 10 '14 at 9:26
  • $\begingroup$ Thanks. Can we not just do what you suggest in the weak formulation, and it still works if I am right? I don't think we need classical solutions??? $\endgroup$
    – riem
    May 10 '14 at 11:40
  • $\begingroup$ @riem Yes, you are right. The weak maximum principle can be proved for weak solutions using the way you suggested. $\endgroup$
    – Voliar
    May 10 '14 at 19:30
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    $\begingroup$ @Voliar Oh thanks. I have a manifold without boundary so nothing for me to worry about :) $\endgroup$
    – LapLace
    Jun 30 '14 at 16:52
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    $\begingroup$ @S.Cho Unfortunately, I don't know a way how to avoid the usage of maximum or comparison principles here... $\endgroup$
    – Voliar
    Dec 25 '18 at 12:23

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