Estimate for a weak solution to a PDE Let $f \in L^2(B_R(0))$ and let $u \in W^{1,2}(B_R(0))$ be a weak solution of the equation
$$Lu = - \sum_{i,j=1}^{n} D_i(a_{ij}D_ju)+ \sum_{i=1}^{n} b_i D_i u + cu =f.$$
There are constants $0 \le B,C < \infty$ so that
$$\sup_{B_R(0)} \sqrt{\sum_{i=1}^{n} |b_i|^2 } \le B  \text{ and }\sup_{B_R(0)}|c| \le C.$$
We also assume the uniform ellipticity condition
$$\exists 0 < \lambda \le \Lambda < \infty \ \forall x \in B_R(0) \forall \xi \in \mathbb{R}^n : \lambda |\xi|^2 \le \langle \xi,A\xi \rangle \le \Lambda |\xi|^2.$$
I want to show that then $$||Du||_{L^2(B_{R/2}(0))} \le K (||u||_{L^2(B_R(0))} + ||f||_{L^2(B_R(0))}),$$
where $K $ depends only on $B,C, \lambda , \Lambda $ and $R$.
The hint is to use the test function $\phi = u \eta^2$ where $\eta \in C_c^{1} (B_{3R/4}(0))$ with $\eta=1$ on $B_{R/2}(0)$ and $ 0 \le \eta \le 1$ with $|D\eta(x)| \le \frac{4}{R} $ for almost all $x \in B_R(0).$
Does anybody have an idea?
 A: I will prove a different estimate, but it should help you get started and give you a feeling of how to use your knowledge.
Notice that since $\eta\in C^1_0(B_R)$ and $u\in W^{1,2}(B_R)$, you have $\phi\in W^{1,2}_{B_R}$.
Therefore $\phi$ is a valid test function.
Note also that it follows from your ellipticity estimate that $a_{ij}X_iX_j\leq\Lambda|X||Y|$ for any vectors $X$ and $Y$.
If you use your test function $\phi$, you get
$$
\int_{B_R}a_{ij}\partial_iu\partial_j\phi+b_i(\partial_iu)\phi+cu\phi
=
\int_{B_R}f\phi
.
$$
A little calculation gives you $\nabla\phi=\eta^2\nabla u+2u\eta\nabla\eta$.
Therefore
$$
\int_{B_R}\eta^2a_{ij}\partial_iu\partial_ju+2a_{ij}u\eta\partial_iu\partial_j\eta+b_i(\partial_iu)u\eta^2+cu^2\eta^2
=
\int_{B_R}f\eta^2u
.
$$
You have $\eta=1$ on $B_{R/2}$ and $0\leq\eta^2\leq1$ everywhere, so
\begin{equation}
\begin{split}
\|\nabla u\|^2_{L^2(B_{R/2})}
&=
\int_{B_{R/2}}|\nabla u|^2
\\&=
\int_{B_{R/2}}\eta^2|\nabla u|^2
\\&\leq
\int_{B_R}\eta^2|\nabla u|^2
\\&\leq
\lambda^{-1}\int_{B_R}\eta^2 a_{ij}\partial_iu\partial_ju
\\&=
\lambda^{-1}\int_{B_R}[f\eta^2u-2a_{ij}u\eta\partial_iu\partial_j\eta-b_i(\partial_iu)u\eta^2-cu^2\eta^2]
\\&\leq
\lambda^{-1}\left[\int_{B_R}|f\eta^2u|+\int_{B_R}|2a_{ij}u\eta\partial_iu\partial_j\eta|+\int_{B_R}|b_i\partial_iu||u\eta^2|+\int_{B_R}|cu^2\eta^2|\right]
\\&\leq
\lambda^{-1}\left[\int_{B_R}|f||u|+2\Lambda\int_{B_R}|u||\nabla u||\nabla\eta|+B\int_{B_R}|\nabla u||u|+C\int_{B_R}|u^2|\right]
\\&\leq
\lambda^{-1}\left[\int_{B_R}|f||u|+\frac8R\Lambda\int_{B_R}|u||\nabla u|+B\int_{B_R}|\nabla u||u|+C\int_{B_R}|u^2|\right]
\\&\leq
\lambda^{-1}\left[\|f\|_{L^2(B_R)}\|u\|_{L^2(B_R)}+\left(\frac8R\Lambda+B\right)\|u\|_{L^2(B_R)}\|\nabla u\|_{L^2(B_R)}+C\|u\|_{L^2(B_R)}^2\right].
\end{split}
\end{equation}
Does this give you a feeling for playing with your problem?

Continued:
Suppose you have three positive quantities $x,y,z$ that satisfy
$$
x^2\leq A(yx+yz+y^2)
$$
for some constant $A>0$.
Using $yz\leq y^2+z^2$, you get
$$
x^2\leq A\max(yx,z^2+2y^2).
$$
If $yx>z^2+2y^2$, you get $x^2\leq Axy$ and so $x\leq Ay$.
If $yx\leq z^2+2y^2$, you get $x^2\leq A(z^2+2y^2)$.
Either way, you get $x^2\leq A'(z^2+y^2)$ for some constant $A'$.
You can write a formula for $A'$ in terms of $A$ if you wish.
