Let $\Omega$ be a bounded domain in $R^N$ and suppose $\gamma$ is some smooth bounded function in $\Omega$. Let $ -\Delta \phi_0 = \lambda_0 \phi_0 $ in $ \Omega$ with $ \phi_0=0$ on $ \partial \Omega$ denote the first eigenpair and we suppose $ \sup_\Omega \phi_0=1$. Let $ \phi_\epsilon>0$ denote a positive solution of $$-\Delta \phi_\epsilon + \epsilon \gamma \phi_\epsilon = \lambda_\epsilon \phi_\epsilon$$ in $ \Omega$ with zero BC and so $(\lambda_\epsilon, \phi_\epsilon)$ is the first eigenpair and lets normalize so $ \sup_\Omega \phi_\epsilon=1$. I can show that $ \phi_\epsilon \rightarrow \phi_0$ directly, but I would like some better asymptotics. I realize this is not a research level question.
So I define $(\psi_\epsilon, \lambda^\epsilon)$ via $ \lambda_\epsilon = \lambda_0 + \epsilon \phi_1 + \epsilon \psi_\epsilon$ and $ \lambda_\epsilon = \lambda_0 + \lambda_1 \epsilon + \epsilon \lambda^\epsilon$. So once we determine $ \phi_1, \lambda_1$ then $ \psi_\epsilon$ and $ \lambda^\epsilon$ are determined. Lets assume we can show that $ \lambda^\epsilon \rightarrow 0$. Assume $ \psi_\epsilon \rightarrow 0$ (at least for now) we can see that $ (\lambda_1, \phi_1)$ should satisfy $$ -\Delta \phi_1 - \lambda_0 \phi_1 = \lambda_1 \phi_0 - \gamma \phi_0,$$ and so we pick $ \lambda_1$ such that the right hand side is orthogonal to $ \phi_0$. Then we can find a $ \phi_1$ that solves this. Now we can pick $ \phi_1$ orthogonal to $ \phi_0$ or we can choose it another way (i will leave that open for now).
Now we write out the equation for $ \psi_\epsilon$. Here is where I get confused. If we knew $ \psi_\epsilon$ was orthogonal to $ \phi_0$ then we could try and show it must converge to zero uniformly; if not then we can probably find something in the kernel of $L$ where $ L(\zeta) = -\Delta \zeta - \lambda_0 \zeta$ which is nonzero and perpendicular to $ \phi_0$; a contradiction. But this is the part I just can't seem to see.
Anyhow comments would be appreciated.
