Subsolution of Laplace equation

Let $$\Omega$$ be a bounded open subset of $$\mathbb{R}^2$$ and $$w(>0)\in H_{0}^1(\Omega)$$ satisfies the equation $$-\Delta w\leq e^w\text{ in }\Omega.$$ Let $$v(>0)\in H_{0}^1(\Omega)$$ satisfies the equation $$-\Delta v=e^v\text{ in }\Omega.$$ Then is it true that $$w\leq v$$ in $$\Omega$$?

I am trying to prove by observing that $$w$$ is a subsolution of the second equation, but unable to derive the result.

Let $$K$$ be a compact subset of $$\Omega$$ and we set $$f = \chi_K$$ (characteristic function of $$K$$) and $$w_0 := (-\Delta)^{-1}(f) \in H_0^1(\Omega)$$. Then, we should have $$w_0 \ge c$$ on $$K$$ for some $$c > 0$$. Now, we choose $$\lambda > 0$$ with $$\lambda < e^{\lambda \, c}$$. We set $$w := \lambda \, w_0$$. This yields $$-\Delta w = -\lambda \, \Delta w = \lambda \, f.$$ On $$K$$, we have $$\lambda \, f = \lambda \le e^{\lambda \,c } \le e^{\lambda \, w_0} = e^{w}.$$ On $$\Omega \setminus K$$, we have $$\lambda \, f = 0 \le 1 \le e^{w}.$$
Hence, $$w$$ solves your equation, but with $$\lambda \to \infty$$ (which is possible), it cannot satisfy $$w \le v$$ for a fixed $$v$$.
• Thank you very much for your answer. Suppose $w$ is a given subsolution as above , then can one always get such $v\in H_0^1(\Omega)\cap C^1(\Omega)$ such that $w\leq v$? Please help me. – Mathlover Dec 17 '18 at 10:34