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

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 at 0:39