Let $u$ be the weak solution on a smooth bounded domain $\Omega \subset \mathbb{R}^n$ (for $n \leq 3$) of $$u_t - \Delta u = f$$ $$u(0) = u_0$$ $$\partial_\nu u = 0 \quad\text{on $\partial\Omega$}$$ for $u_0 \in L^\infty(\Omega)$ (non-negative) and $f \in L^\infty(0,T;L^2(\Omega))$. We know that $$u(t,x) \leq C(\lVert u_0 \rVert_\infty + \lVert f \rVert_{L^\infty(0,T;L^2)})$$ is satisfied almost every $(x,t)$ by De Giorgi's method (eg. Theorem 4.2.2 in "Elliptic and Parabolic Equations" book by Wu, Yin and Wang).

If instead we had the equation $$u_t - \Delta u +\alpha u= f$$ for a constant $\alpha > 0$ (and all else is the same as above), can I expect some kind of decay on the $L^\infty$ estimate involving $\alpha$, eg. $$u(t,x) \leq \frac{C}{\alpha}(\lVert u_0 \rVert_\infty + \lVert f \rVert_{L^\infty(0,T;L^2)})?$$ It is essential for me that $f$ is only in $L^\infty(0,T;L^2)$.

Of course I tried using an integrating factor, but this gives a bad contribution on the term involving the $f$ norm. I wonder if it can be true.

  • $\begingroup$ my impression is that if you take $\Omega=(0,1)$, $u_0=0$ and $f(x,t)=e^{-\alpha t}$, then $u(x,t)=te^{-\alpha t}$ and the estimate is wrong $\endgroup$
    – poupy
    Apr 19, 2016 at 20:21
  • $\begingroup$ Maybe it's obvious, but could you define the norm you take on $f$ explicitely? $\endgroup$
    – Amir Sagiv
    Apr 19, 2016 at 21:29
  • $\begingroup$ @poupy you're right, I just meant some kind of decay not exactly what I wrote. $\endgroup$
    – TLE
    Apr 21, 2016 at 9:39
  • $\begingroup$ @AmirSagiv it is equal to $esssup_{t \in [0,T]}\lVert u(t) \rVert_{L^2(\Omega)}$ $\endgroup$
    – TLE
    Apr 21, 2016 at 9:39

1 Answer 1


The estimate as stated is clearly false since $u\to u_0$ as $t\to0$ hence $\|u\|_{L^\infty}\ge \|u_0\|_{L^\infty}$.

Anyway you can write the kernel explicitly and extract the information you need from it. Just define $v=e^{at}u$ so that $v_t-\Delta v=e^{at}(u_t-\Delta u+au)=e^{at}f$. Thus $$ u(t)=c t^{-n/2}\int e^{-\frac{|x-y|^2}{4t}-at}u_0(y)dy + \int_0^t \int (t-s)^{-n/2} e^{-\frac{|x-y|^2}{4s}-as}f(t-s,y)dy ds. $$


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