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.