EDIT: I tried to prove the continuous dependence of the problem by somehow use a weak Maximum principle and posted a modified Version of this post on mathoverflow: https://mathoverflow.net/questions/332041/using-weak-maximum-principle-to-prove-continuous-dependence-of-the-boundary-data

Maybe someone here can help me? :/

I am currently looking at the following Dirichlet problem:

\begin{align} \begin{cases} \text{div} (\sigma |\nabla u|^{p-2} \nabla u) = f &\quad \text{in } \Omega\\ u = g &\quad \text{in } \partial\Omega \end{cases} \end{align}

so my initial conditions are the right side $f \in L^2(\Omega)$ and the boundary condition $g \in W^{1,p}(\Omega)$. I already proved that a unique solution $u \in W^{1,p}_g$ (Subspace of Sobolev spave $W^{1,p}$ which is $g$ on $\partial\Omega$). Furthermore I have an a priori estimate

\begin{align} ||u||_{W^{1,p}(\Omega)} \leq C \left ( ||f||^{\frac{1}{p-1}}_{L^2(\Omega)} + ||g||_{W^{1,p}(\Omega)} \right) \end{align}

But how do I prove "continous dependence of initial conditions"? How is this property even defined? Do exactly do I have to show? Intuitivly I thought of something like...

Let $u$ be the solution of the problem with the initial conditions $g$,$f$. Let $f_k \rightarrow f$, $g_k \rightarrow g$ and $u_k$ shall be the solution of the problem with initial conditions $f_k$, $g_k$. Then $\lim_{k \rightarrow \infty} ||u - u_k||_{W^{1,p}(\Omega)} \rightarrow 0$?

I would guess, that I need some terms like $||u - u_k||_{W^{1,p}(\Omega)} \leq ... ||f - f_k|| ... ||g - g_k||$?

But how do I get there? I am not very experienced with such estimations...

What I know about the solutions $u$ and $u_k$ is, that they are the minimizer of the strict convex functional

\begin{align} J(v) = \frac{1}{p}\int_{\Omega} \sigma |\nabla v|^p~dx - \int_{\Omega} fv~dx \end{align}

Therefore we have

\begin{align} J(u) \leq J(u - u_k) \end{align}


\begin{align} J_k(u_k) \leq J_k(u - u_k) = \frac{1}{p}\int_{\Omega} \sigma |\nabla (u - u_k)|^p~dx - \int_{\Omega} f_k (u - u_k)~dx \end{align}

but still I don't know how to show the continuous dependence... Maybe someone more experienced can help me?

EDIT: Ok, I think I got the continuous dependence on the "f":

Let $u_1$, $u_2$ be the solutions for $f_1$, $f_2$ on the right side of the equation. Since they have the same values on the boundary we have $u_1 - u_2 \in W^{1,p}_0$. Because they minimize the functional, for the frechet derivates we have

\begin{align} \int_{\Omega} \sigma|\nabla u_1|^{p-2} \nabla u_1 \nabla \phi~d x = \int_{\Omega} f_1 \phi ~dx\\ \int_{\Omega} \sigma|\nabla u_2|^{p-2} \nabla u_2 \nabla \phi~d x = \int_{\Omega} f_2 \phi ~dx \end{align}

This is true for all $\phi \in W^{1,p}_0$. There it is also true for $u_1 - u_2 \in W^{1,p}_0$:

\begin{align} \int_{\Omega} \sigma|\nabla u_1|^{p-2} \nabla u_1 \nabla (u_1 - u_2)~d x = \int_{\Omega} f_1 (u_1 - u_2) ~dx\\ \int_{\Omega} \sigma|\nabla u_2|^{p-2} \nabla u_2 \nabla (u_1 - u_2)~d x = \int_{\Omega} f_2 (u_1 - u_2) ~dx \end{align}

Combined we have

\begin{align} \int_{\Omega} (\sigma|\nabla u_1|^{p-2} \nabla u_1 - \sigma|\nabla u_2|^{p-2} \nabla u_2)\nabla (u_1 - u_2)~d x = \int_{\Omega} (f_1 - f_2) (u_1 - u_2) ~dx \end{align}

For the left side we have the estimation \begin{align} \int_{\Omega} \alpha |\nabla u_1 - \nabla u_2|^p ~dx \leq \int_{\Omega} (\sigma|\nabla u_1|^{p-2} \nabla u_1 - \sigma|\nabla u_2|^{p-2} \nabla u_2)\nabla (u_1 - u_2)~d x \end{align}

with an $\alpha > 0$ (this estimation is not obvious, but valid). For the right side we have

\begin{align} \alpha || u_1 - u_2 ||^p_{W^{1,p}_0}=\int_{\Omega} (f_1 - f_2) (u_1 - u_2) ~dx \leq ||f_1 - f_2||_{L^2}~||u_1 - u_2||_{W^{1,p}_0} \end{align}

Therefore \begin{align} || u_1 - u_2 ||_{W^{1,p}_0} \leq C ||f_1 - f_2||^{\frac{1}{p}}_{L^2} \end{align}

meaning that $u_1 \rightarrow u_2$ for $f_1 \rightarrow f_2$... Is this correct? And does anyone know how to handle the dependence of the boundary condition?

EDIT2: My idea for the continuous dependence of the boundary values: Let $g_k \rightarrow g$. Then $\lim_{k \rightarrow \infty} (u_k - u)|_{\partial\Omega} = 0$. So we have (the same equation as above but with $f_1 = f_2$ this time)

\begin{align} \lim_{k \rightarrow \infty} \int_{\Omega} (\sigma|\nabla u|^{p-2} \nabla u - \sigma|\nabla u_k|^{p-2} \nabla u_k)\nabla (u - u_k)~d x = 0 ~dx \end{align}

so \begin{align} \lim_{k \rightarrow \infty} \alpha || u_1 - u_2 ||^p_{W^{1,p}_0} = 0 \end{align}

At first sight it looks correct for me... or did I overlook something crucial?


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