Return to Question

Is there a way to show in a relatively simple manner that for a Lipschitz bounded, connected, open domain $\Omega\subset\mathbb{R}^N$ for any $f\in L^2(\Omega)$ the solution of the problem:

$$\begin{cases}\dfrac{\partial u}{\partial t}=\Delta u, & (t,x)\in (0,T)\times \Omega\\ \dfrac{\partial u}{\partial\nu}=0, & (t,x)\in (0,T)\times\partial\Omega \\ u(0,x)=f(x), & x\in\Omega\end{cases}$$

is bounded and $\Vert u(t,\cdot)\Vert_{L^{\infty}(\Omega)}\leq c\cdot t^{-N/4}\cdot \Vert f\Vert_{L^2(\Omega)}$ for any $t\in (0,T)$, for some $c$ depending only on $\Omega$.

Is there a way to show in a relatively simple manner that for a Lipschitz bounded, connected, open domain $\Omega\subset\mathbb{R}^N$ for any $f\in L^2(\Omega)$ the solution of the problem:

$$\begin{cases}\dfrac{\partial u}{\partial t}=\Delta u, & (t,x)\in (0,T)\times \Omega\\ \dfrac{\partial u}{\partial\nu}=0, & (t,x)\in (0,T)\times\partial\Omega \\ u(0,x)=f(x), & x\in\Omega\end{cases}$$

is bounded and $\Vert u(t,\cdot)\Vert_{L^{\infty}(\Omega)}\leq c\cdot t^{-N/4}\cdot \Vert f\Vert_{L^2(\Omega)}$ for any $t\in (0,T)$, for some $c$ depending only on $\Omega$?

Is there a way to show in a relatively simple manner that for a Lipschitz bounded, connected, open domain $\Omega\subset\mathbb{R}^N$ for any $f\in L^2(\Omega)$ the solution of the problem:

$$\begin{cases}\dfrac{\partial u}{\partial t}=\Delta u, & (t,x)\in (0,1)\times \Omega\\ \dfrac{\partial u}{\partial\nu}=0, & (t,x)\in (0,1)\times\partial\Omega \\ u(0,x)=f(x), & x\in\Omega\end{cases}$$

is bounded and $\Vert u(t,\cdot)\Vert_{L^{\infty}(\Omega)}\leq c\cdot t^{-N/4}\cdot \Vert f\Vert_{L^2(\Omega)}$ for any $t\in (0,1)$, for some $c$ depending only on $\Omega$.

Is there a way to show in a relatively simple manner that for a Lipschitz bounded, connected, open domain $\Omega\subset\mathbb{R}^N$ for any $f\in L^2(\Omega)$ the solution of the problem:

$$\begin{cases}\dfrac{\partial u}{\partial t}=\Delta u, & (t,x)\in (0,T)\times \Omega\\ \dfrac{\partial u}{\partial\nu}=0, & (t,x)\in (0,T)\times\partial\Omega \\ u(0,x)=f(x), & x\in\Omega\end{cases}$$

is bounded and $\Vert u(t,\cdot)\Vert_{L^{\infty}(\Omega)}\leq c\cdot t^{-N/4}\cdot \Vert f\Vert_{L^2(\Omega)}$ for any $t\in (0,T)$, for some $c$ depending only on $\Omega$.

Is there a way to show in a relatively simple manner that for a Lipschitz bounded, connected, open domain $\Omega\subset\mathbb{R}^N$ for any $f\in L^2(\Omega)$ the solution of the problem:

$$\begin{cases}\dfrac{\partial u}{\partial t}=\Delta u, & (t,x)\in (0,T)\times \Omega\\ \dfrac{\partial u}{\partial\nu}=0, & (t,x)\in (0,T)\times\partial\Omega \\ u(0,x)=f(x), & x\in\Omega\end{cases}$$

is bounded and $\Vert u(t,\cdot)\Vert_{L^{\infty}(\Omega)}\leq c\cdot t^{-N/4}\cdot \Vert f\Vert_{L^2(\Omega)}$ for any $t\in (0,T)$, for some $c$ depending only on $\Omega$.

Is there a way to show in a relatively simple manner that for a Lipschitz bounded, connected, open domain $\Omega\subset\mathbb{R}^N$ for any $f\in L^2(\Omega)$ the solution of the problem:

$$\begin{cases}\dfrac{\partial u}{\partial t}=\Delta u, & (t,x)\in (0,1)\times \Omega\\ \dfrac{\partial u}{\partial\nu}=0, & (t,x)\in (0,1)\times\partial\Omega \\ u(0,x)=f(x), & x\in\Omega\end{cases}$$

is bounded and $\Vert u(t,\cdot)\Vert_{L^{\infty}(\Omega)}\leq c\cdot t^{-N/4}\cdot \Vert f\Vert_{L^2(\Omega)}$ for any $t\in (0,1)$, for some $c$ depending only on $\Omega$.