2
$\begingroup$

Consider the non-linear differential operator

$$\mathfrak{L}: \ C^2((0,T)\times\mathbb{R}^2)\ni\varphi\equiv\varphi(t,x,y) \, \mapsto \, \partial_x^2\varphi + (\partial_x\varphi)^2.$$

For $U\subset\mathbb{R}^2$ open and bounded with smooth boundary $\Gamma:=\partial U$ and $T>0$, can you guarantee that the (smooth classical) solution to the Cauchy-problem

$$\left\{\begin{aligned}\label{lem:ContrastiveDiffusions:eq2} \partial_t\varphi \ &= \ \mathfrak{L}\varphi\quad &&\text{in } \ (0,T)\times U,\\ \partial_x\partial_y\varphi \ &= \ 0 \quad &&\text{on } \ \{T\}\times\Gamma, \end{aligned}\right.$$

with initial condition $\lim_{t\rightarrow 0+}\varphi = \delta(x-y)$, is unique?

(Given that the answer is not immediate, references to the literature are appreciated.)

$\endgroup$
3
  • 2
    $\begingroup$ Do you really want the boundary condition only at t=T? And I suppose you also want an initial condition? $\endgroup$ Commented Aug 4, 2020 at 17:01
  • 1
    $\begingroup$ Thanks @MichaelRenardy. The existing solution $\varphi$ that I found is a probability density (modelling the transition probability of a stochastic system to jump from state $x$ to state $y$ in $t>0$ units of time). Yes, I want the boundary at $t=T$ only. $\endgroup$
    – cts12
    Commented Aug 4, 2020 at 17:14
  • 1
    $\begingroup$ Writing $\phi=\log v$ leads to the heat equation $v_t=v_{xx}$ but the boundary condition looks more complicate. $\endgroup$ Commented Aug 10, 2020 at 8:32

0

You must log in to answer this question.