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Suppose I have the following PDE on $[0,1]^n$ $$\mathcal{L}u = -\nabla \cdot \left(a(x, r)\nabla u\right) = f(x,r), \qquad x\in [0,1]^n,$$ with periodic boundary conditions and $\int f(x) dx =0$ . Suppose that $a_{ij} \in C^\infty([0,1]^n \times \mathbb{R})$, and that for fixed $r \in \mathbb{R}$, $$ 0 < \lambda(r)|z|^2 \leq z\cdot a(x,r)z \leq \Lambda(r)|z|^2, \quad z \in \mathbb{R}^n,$$ for some smooth functions $\lambda(\cdot)$ and $\Lambda(\cdot)$.

Thus for fixed $r$ we can apply Lax-Milgram to give us the existence of a unique solution $u$ with $\int_{[0,1]^n} u(x) \,dx = 0$. Now, it is straightforward to show that $$ ||u(\cdot, r)||^2_{H^1} \leq \frac{C}{\lambda(r)}||f(\cdot, r)||^2_{L^2},$$ where $C$ depends only on $n$.

One can similarly derive $L^{\infty}$ bounds for $u$, however, I have not been able to see how the constants in these bounds depend on $r$. I am interested to see how these bounds grow with varying $r$. So my question is the following:

It is possible to derive similar upper bounds on $||u||_{L^{\infty}}$ as a function of $r$, where the upper bounds can be expressed explicitly? Of course, the bounds needn't be tight.

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It seems to me what you are asking is the same as to use elliptic regularity theory (there are several flavors that work with coefficients as smooth as you have provided) to provide a modulus of continuity for your function. Since your function $u$ has average 0, a continuous version of $u$ must be 0 at some point, and thus is bounded in magnitude by the modulus of continuity. If you pay attention to how that modulus changes with respect to the ellipticity of the equation (" $\lambda(r)$ ") and the norm of your source term $f$, then you have what you desire. –  Ray Yang Feb 3 '13 at 0:39
    
One rough idea is to take whatever estimates we have for $\Delta v = f$ and "reduce to this case" with the rescalings $\tilde{u}(x) = u(Ax)$ for some affine transformation $A$. Of course, when $a$ depends on $x$ this won't necessarily help the ellipticity at all points (it only works for constant coefficients) but it might give some insights. Otherwise, just look at the proofs and track explicitly the dependence on $\lambda,\Lambda$ at each step. –  Connor Mooney Sep 18 '13 at 3:07

1 Answer 1

Is the general $n$-dimensional case important?

With $n\leq3$, it can be done with off the shelf estimates. The following estimate can be proved using the De Giorgi-Nash-Moser method: If $a_{i,j} \xi_i \xi_j \geq 1$ (and in $L^\infty$) then the solution in $H^1_0(B_1)$ of $$ -\textrm{div}(a \nabla u) + u \leq f + div(g) $$ with $f\in L^p$ and $g \in L^{2p}$ and $p>n/2$ satisfies $$ \| u\|_{L^\infty(B_1)} \leq C(n,p) ( \|f\|_{L^p(B_1)} + \|g\|_{L^{2p}(B_1)}) $$
To go from your case to this one,

  1. Change $a \to a/\lambda(r)$ and $f \to f/\lambda(r)$

  2. Add +u to both sides.

At this point, you would are almost done since your right hand-side is $L^2$ and $2>3/2$. Your solution is not in $H^1_0(B_1)$ so you add a cut-off:

  1. you consider instead $u\times \chi$ with an explicit smooth $\chi$ of your choice supported in $B_1$ equal to $1$ in $B_{1/2}$.

This adds to the right-hand side two terms (I think), $$ 2 a \nabla \chi \cdot \nabla u + \textrm{div}(a \nabla \chi ) u $$ and these terms are still in $L^2$ with bounds depending on the $C^1$ norm of $a$ for example, since you already have an $H^1$ bound.

And you are done, covering your box with a few boxes (4 maybe), the result is explicit: $$ \| u\|_{L^\infty} \leq C(n) \frac{1+\sqrt{\Lambda(r)}+\|\nabla a\|_{L^{\infty}}}{\sqrt{\lambda(r)}} \|f\|_{L^2} $$ (you can lower the $L^\infty$ bound to a $L^3$ if it helps) If you want to use the same idea for bigger $n$, you need to derive $W^{n,p}$ estimates for $u$ and use that instead (replacing $f$ by div$(a \nabla u)$!). But that becomes just as bad as tracking constants in De Giorgi-Nash-Moser directly.

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