2
$\begingroup$

Let $u$ be defined on $\overline{U_T} = \overline{(U \times (0,T])}$ where $U$ is a bounded open set. Let $u$ solve

$$ u_t + b(x,t) \cdot \nabla u= \Delta u $$ where $b$ is an arbitrary, bounded vector field. Prove that u acheives its max on the parabolic boundary $\partial pU_T$

I was thinking of this proof, although I am unsure it's correct.

Assume that $u$ attains a maximum point $u(x_0,t_0) > \max_{\partial_pU_T} u$ with $(x_0,t_0) \in U_T$. Now, take $\epsilon$ small such that: $$ u(t_0,x_0) > u(t,x) + \epsilon T $$ Define $v(t,x)$ by: $$ v(t,x) = u(x,t) - \epsilon t $$ Now, it is clear that $v(t_0,x_0) \geq v(t,x)$ for any $(t,x) \in \partial_p U_T$ by construction. Now, let $(t_1,x_1) \in \overline{U_T}$ be the point such that: $$ v(t_1,x_1) = \max_{\overline{U_T}} v $$ This point exists by compactness. Moreover, this point cannot be in the parabolic boundary. Because this is an interior maximum, we have that $\Delta v(t_1,x_1) = \Delta u (t_1,x_1) \leq 0$, and moreover, that $v_t(t_1,x_1) = u_t(t_1,x_1) \geq 0 $ (this accounts for the case that $t_1 = T$). Hence, we have that: $$ v_t(t_1,x_1) - \Delta v(t_1,x_1) \geq 0 $$ And expanding in terms of $u$: $$ u_t(t_1,x_1) -\epsilon - \Delta u(t_1,x_1) \geq 0 $$ Now, we recall that: $$ u_t + b(x,t) \cdot \nabla u = \Delta u $$ Hence: $$ u_t(t_1,x_1) - \epsilon - u_t(t_1,x_1) -b(x_1,t_1) \cdot \nabla u(x_1,t_1) = -\epsilon - b(x_1,t_1) \cdot \nabla u(x_1,t_1)\geq 0 $$ But because $x_1,t_1$ is an interior maximum of $v$, and the gradient is taken only with respect to spatial variables (???) we have that $\nabla v(t_1,x_1) = 0$, and hence this implies $-\epsilon \geq 0 $, a contradiction.

This does not use some of the assumptions, which is why I think it may be invalid.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.