Let $\Omega=B_R(0) \in \mathbb{R}^n$ and $\lambda\in \mathbb{R}$. Consider the equation $$-\Delta u=e^{-\frac{\lambda}{u+1}} \text{ in } \Omega$$
$$u=0 \text{ on } \partial \Omega$$
Show that a weak solution $u$ is unique if $\lambda<0$ or $\lambda>0$ and sufficiently large or sufficiently small.
Let $u_1$ and $u_2$ satisfy the equation. tried to use the Mean Value Theorem and write $\Delta(u_1(x)-u_2(x))=-e^{\frac{-\lambda}{\xi(x)+1}}(\frac{\lambda}{u_1(x)+1}-\frac{\lambda}{u_2(x)+1})$ for some $\xi(x)\in (\min (\frac{\lambda}{u_1(x)+1}, \frac{\lambda}{u_2(x)+1}), \max (\frac{-\lambda}{u_1(x)+1}, \frac{-\lambda}{u_2(x)+1}))$. Testing with $u_1-u_2$ in the weak formulation we get using integration by parts $\int |\Delta(u_1-u_2)|^2 dx=\int e^{\frac{-\lambda}{\xi(x)+1}}(\frac{\lambda}{u_1(x)+1}-\frac{\lambda}{u_2(x)+1}) dx \geq$ but this just gave me $(u_1+1)^2\geq 0$ which is always true. How should I approach questions like this?
