Upper bounds for the solution of an elliptic PDE depending on a parameter.

Question

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.

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.