How do we check whether the solution is continuouly dependent on parameters?

Let $\Omega$ be a domain with smooth boundary. Say $f$ and $h$ are smooth. Assume that for each $\theta\in (0, 1]$, the equation $\Delta u-f(\theta, u)=h$ with Neumann boundary condition has a unique classical solution $u(x, \theta)$. How do we check that $u$ is continously dependent on $\theta$? And what happens if $\theta\rightarrow 0$?

For example. for $\theta\in(0, 1]$, let $u$ be the solution of $\Delta u-\theta u= h$ subject to Neumann boundary condition. Is $u$ continuously dependent on $\theta$? It looks like $\lim_{\theta\rightarrow 0} u$ does not exist if $\int_\Omega h\neq0$; and the limit will exist if $\int_\Omega h =0$. Am I right? So in general, if $\theta=0$ the solution to $\Delta u-f(0, u)=h$ exists, then the limit exists?

Can anyone give me some hint? Thanks....

How do we check whether the solution is continuouly dependent on parameters?

Let $\Omega$ be a domain with smooth boundary. Say $f$ and $h$ are smooth. Assume that for each $\theta\in (0, 1]$, the equation $\Delta u-f(\theta, u)=h$ with Neumann boundary condition has a unique classical solution $u(x, \theta)$. How do we check that $u$ is continously dependent on $\theta$? And what happens if $\theta\rightarrow 0$?

For example. for $\theta\in(0, 1]$, let $u$ be the solution of $\Delta u-\theta u= h(x,\theta)$ subject to Neumann boundary condition. Is $u$ continuously dependent on $\theta$? If we use the implicit function theory, we need that: $\Delta v-\theta v=g$ has a unique solutoin for all $g\in C^\alpha(\overline\Omega)$. However, this is not true if $\theta=0$. How can I overcome this to show $u(x, \theta)$ is continuously differentible in $\theta$?.

Can anyone give me some hint? Thanks....