Hii!!!

Originally Posted by

**mathmari**
$$w_t(x,t)-\Delta w(x,t), x \in \Omega, t>0$$

and

$$-w_t(x,t)-\Delta (-w(x,t)), x \in \Omega, t>0$$

I think that should be $w_t(x,t)-\Delta w(x,t) = 0$.

Originally Posted by

**mathmari**
Since $w_tâˆ’\Delta w \leq 0$ from the maximum principle for $w$ we have that

$$\max_{x \in \Omega, t \in [0, T]}w(x, t)=\max_{(\Omega \times \{0\})\cup (\partial{\Omega} \times [0, T])}w(x, t)=0$$

I've looked up the maximum principle and found for instance on :

Let $u = u(x), x = (x1, …, xn)$ be a $C^2$ function which satisfies the differential inequality

$$Lu = \sum_{ij} a_{ij}(x)\frac{\partial^2 u}{\partial x_i\partial x_j} +

\sum_i b_i\frac{\partial u}{\partial x_i} \geq 0$$

in an open domain $Î©$, where the symmetric matrix $a_{ij} = a_{ij}(x)$ is locally uniformly positive definite in $Î©$ and the coefficients $a_{ij}, b_i = b_i(x)$ are locally bounded. If $u$ takes a maximum value $M$ in $Î©$ then $u â‰¡ M$.

If we would use this version, then I think all conditions of the proposition should be covered.

Btw, which version of the maximum principle do you have?