4
$\begingroup$

Given:

1)a bounded domain $\Omega$ in $\mathbb R^n$ of class $\mathcal{C}^{\infty}$

2) the function $f\in L^{\infty}(\Omega)$ with $\int_{\Omega} f=0$

3)$g=(g_i,\ldots,g_n)\in \mathcal{C}^\alpha(\Omega)$

4) a scalar diffusion coefficient $d \in \mathcal{C} ^0(\Omega)$ s.t. $0<\lambda<d(x)<\Lambda<+\infty \quad \forall x\in \Omega$

there exists an estimate for the $\mathcal{C}^{1,\alpha}(\Omega)$-norm of $u$, weak solution of the elliptic equation

$\int_{\Omega}d\ \nabla u \cdot \nabla \Phi=\int_\Omega f \Phi +g \cdot \nabla \Phi \quad \forall \Phi \in H^1(\Omega)$

with Neumann boundary condition $d\nabla u \cdot n_{\partial \Omega}=0$ and with the condition $\int u=0$ (to fix the solution up to additive constant).

I found in theorem 8.33 and 8.34 in the book of Gilbarg and Trudinger "Elliptic Partial Differential Equations of Second Order" an estimate for the above equation but with Dirichlet boundary condition $u=\varphi\in \mathcal{C}^{1,\alpha}(\partial \Omega) \ on \ \partial \Omega$ of type

$\|u\|_{\mathcal{C}^{1,\alpha}(\Omega)}\leq C(d)(\|u\|_{C^0}+\|\varphi\|_{\mathcal{C}^{1,\alpha}}+\|f\|_{L^{\infty}}+\|g\|_{\mathcal{C}^{0,\alpha}})$

and assuming $d \in \mathcal{C}^{0,\alpha}$.

My questions now are:

1)"Is it possibile to have an estimate of type

$\|u\|_{\mathcal{C}^{1,\alpha}(\Omega)}\leq C(d)(\|f\|_{L\infty}+\|g\|_{\mathcal{C}^0,\alpha})$ having just $d\in \mathcal{C}^0(\Omega)$?

2) What can be say about the constant $C(d)$? Is it possible possible to say that is proportinal to $\frac{\Lambda}{\lambda}$?

$\endgroup$
  • $\begingroup$ I don't see why it should be proportional to $\Lambda/\lambda$ (certainly $\lambda/\Lambda$ would feel more natural). But that would be for an homogeneous problem. Here it critically depends on $\lambda$, but not so much on the upper bound . $\endgroup$ – username May 17 '15 at 21:58
  • $\begingroup$ As explained below, you need $C^{0,\alpha}$ for $C^{1,\alpha}$. To get $C^1$ you need a bit more than just $C^0$, Dini continuity is enough for example. $\endgroup$ – username May 17 '15 at 22:02
1
$\begingroup$

Continuity of $d$ does not guarantee that $\nabla u$ is Holder. In fact, the regularity of $d$ is the best regularity one can expect for $\nabla u$.

To see this, consider the one-dimensional divergence-form equation $$(du')' = f.$$ Integrating and dividing by $d$, we see that $u'$ is at best the same regularity as $d$. (This is also an easy way to see why we have one less derivative in estimates for divergence-form equations than in those for nondivergence equations).

This also shows that the constant $C(d)$ depends on the modulus of continuity for $d$. Indeed, even if $d$ is $C^{\alpha}$ it can oscillate wildly and stay close to $1$, and in the model problem $u' = 1/d$ would have the same behavior.

$\endgroup$
  • $\begingroup$ Best regularity in the sense of 'at most', yes. But precisely, C^0 does not work in dimension more than 1 to get C^1, I think. $\endgroup$ – username May 17 '15 at 21:54
  • $\begingroup$ Thanks for the answer and for the comments. They were very usefull $\endgroup$ – enrico May 25 '15 at 6:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.