2
$\begingroup$

Why there exist a solution when Evans constructs a parabolic equation in proving the strong maximum principle theorem 11 in chapter 7.

The theorem 11 is under below .

Theorem 11 (Strong maximum principle ) . Assume $u\in C_1^2(U_T)\cap (\overline{U}_T))$ and $c\equiv 0 \text{ in} U_T$ Suppose also $U$ is connected.$\cdots$

Proof. 1. Assume $u_t+Lu\leq 0 \text{in} U_T$ and $u$ attains its maximum at some point $(x_0,t_0)\in U_T$ . Select a smooth , open set $W\subset\subset U$, with $x_0\in W$ . Let $v$ solve $$\left\{ \begin{align} v_t+Lv&=&0 \ \text{in} W_T \\ v&=&u \ \text{on} \ \Delta _T \end{align} \right. $$ where $\Delta _T$ denotes the parabolic boundary of $W_T$ $\cdots$

My doubts :why there is a solution $v$? Existence of weak solution can guarantee if the condiction $v=0$ on $\partial U×[0,1]$. Could someone give me some advice, thank you!

$\endgroup$

1 Answer 1

1
$\begingroup$

The essential point is what Evans means by "parabolic boundary." It is not understood as the set $\partial U \times [0,T]$ as you write. Instead, if you check how this is defined in the book you'll see that $$ W_T = W \times (0,T] \text{ and } \Delta_T = \overline{W_T} \backslash W_T = (W \times \{0\}) \cup (\partial W \times [0,T]). $$ The solvability theory for the PDE with data specified on $\Delta_T$ is exactly what Evans establishes in the material prior to Theorem 11 in Section 7.1, though there he specializes to the case that the data on $\partial W \times [0,T]$ is zero. To recover the general case, though, we can simply subtract $u$. In other words we first solve $$ \begin{cases} w_t + L w = -u_t - L u & \text{in } W_T \\ w =0 & \text{on } \Delta_T \end{cases} $$ using the theory in Section 7.1. We then define $v = w + u$, which then solves

$$ \begin{cases} v_t + L v = 0 & \text{in } W_T \\ v =u & \text{on } \Delta_T. \end{cases} $$

$\endgroup$
1
  • $\begingroup$ Good idea , thank you! $\endgroup$
    – Yuri Manin
    Feb 5, 2021 at 6:07

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .