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Use the classical regularity setting $u\in C^2(D)\cap C(\overline{D})$, with $D$ bounded. Continuity on the compact set $\overline{D}$ guarantees existence of a minimum. The inequality $\Delta u=f\le 0$ means that $u$ is superharmonic function.

A strict inequality $\Delta u<0$ would immediately exclude an interior minimum, because its Hessian matrix there would be positive semidefinite and have nonnegative trace. To handle equality, introduce the quadratic perturbation proof of the minimum principle:Its minimum must lie on $\partial D$. Choose $R$ so that $x^2+y^2\le R^2$ on $\overline{D}$. For any $p\in D$,Letting $\epsilon \downarrow 0$ gives $u(p)\ge {\min }_{\partial D}u$. ThereforeThis asserts that a minimum is attained on the boundary, not that it cannot also be attained inside; a constant function attains it everywhere.

Assume a classical solution $u\in C^2(D)\cap C(\overline{D})$, as needed to interpret the boundary values. At an interior local minimum its Hessian is positive semidefinite, hence $\Delta u\ge 0$. But the equation gives $\Delta u=-e^{-x^2}<0$, a contradiction. Since $D$ is bounded and $u=0$ on its boundary, a nonpositive interior value would either furnish an interior negative minimum or itself be a zero minimum. Thus $\boxed{u>0\text{ in }D}$.

For the upper bound use the barrier $w(x,y)=(1-x^2-y^2)/4$, which satisfies $-\Delta w=1$. Set $z=w-u$. On the boundary $z=w\ge 0$, because the domain lies inside the unit disk, whileThe quadratic perturbation proof of the minimum principle gives $z\ge 0$. The strong minimum principle for elliptic operators makes this strict in the connected interior unless $z$ is constant. Constancy is impossible since $1-e^{-x^2}$ is not identically zero on any open domain. HenceThis last strong-principle step handles points with $x=0$, where the comparison forcing itself vanishes. Simple connectedness is not needed for the comparison argument.

With source point $\xi \in D$ and field point $X=(x,y)$, the Dirichlet Green function for the operator $-\Delta$ satisfiesClassically it is harmonic off $\xi$, with the two-dimensional normalization $G(X,\xi )=-(2\pi {)}^{-1}\log |X-\xi |+H(X,\xi )$, where $H$ is regular near the source. Equivalently $-{\lim }_{r\downarrow 0}{\int }_{|X-\xi |=r}{\partial }_{r}Gds=1$. This sign convention gives the representation $u(X)={\int }_{D}G(X,\xi )e^{-{\xi }_1^2}d\xi$.