So, I know that there is this theorem (taken from here):
For Laplace's equation $\Delta u = 0$ on a domain $D$ and $u=f$ on $\partial D$ (and some regularity conditions on $D$), we have $$ u(x) = \mathbb E_x\left[ f(B_{\tau_D}) \right], $$ where $(B_t)_t$ is Brownian motion and $\tau_D = \inf\{ t\geq 0: B_t \notin D \}$.
My problem is that I need to show that $$ \Delta \mathbb E_x[ \tau_{B_r(y)} ] = -2 \text{ on $\mathbb T_2\setminus B_r(y)$}, $$ where $\mathbb T_2$ is a two-dimensional torus, $r>0$ und $y\in\mathbb T_2$. Here however, $\tau_{B_r(y)}=\inf\{ t\geq 0: (B_t) \in B_r(y) \}$ i.e. a hitting time and not an exit time.
Thus, I believe you have to turn the statement around meaning $$ \Delta E_x[\tau_{\mathbb T_2 \setminus B_r(x)}] = -2 $$ to get an exit time like in the theorem at the top.
Can I apply the theorem above to solve my problem? What do I need to change to get the value $-2$ instead of $0$. Also, how do I find the function $f$?
