Suppose that $\textbf{a}:\mathbb R^n \to \mathbb R^n$ is a smooth map. We have a condition on $\textbf{a}$: $\forall p\in \mathbb R^n,\,|\textbf{a}(p)|\le C\,(1+|p|)$.
Let $U$ be a bounded, open, connected subset in $\mathbb R^n$. Recall that $H^1(U)=W^{1,2}(U)$. Suppose $u,v\in H_0^1(U)$, which is the closure of $C_c^\infty(U)$ in $H^1(U)$.
Let $\lambda>0.$
Now I want to prove that $$\int_U \textbf{a}(\mathrm Du-\lambda \mathrm Dv)\cdot \mathrm Dv \,\mathrm dx \to \int_U \textbf{a}(\mathrm Du)\cdot \mathrm Dv \,\mathrm dx \qquad\text{ as } \lambda \to 0.$$
The main problem which I encounter is that $(\mathrm Du-\lambda \mathrm Dv)(x)$ and $\mathrm Du(x)$ may not be bounded in $U$.
If both of them are bounded, then I can prove the convergence by Holder inequality and the smoothness of $\textbf{a}$ easily because we can use the mean value theorem in $\mathbb R^n$ to dominate the term $\textbf{a}(\mathrm Du-\lambda \mathrm Dv) - \textbf{a}(\mathrm Du)$.
I also thought of the norm continuity of $L^2$. By the condition on $\textbf{a}$, we know $\textbf{a}(\mathrm Du)$ is in $L^2(U)$. So we know $\| \textbf{a}(\mathrm Du+h) -\textbf{a}(\mathrm Du)\|_{L^2} \to 0$ when $|h|\to 0.$ By Holder inequality, we are done.
Can we regard $(\lambda Dv)(x)$ as a constant vector $h$? If we can, then why? Thanks.
Any other method is also welcome. My purpose is to prove that $$\int_U \textbf{a}(\mathrm Du-\lambda \mathrm Dv)\cdot \mathrm Dv \,\mathrm dx \to \int_U \textbf{a}(\mathrm Du)\cdot \mathrm Dv \,\mathrm dx \qquad\text{ as } \lambda \to 0.$$
Thanks again.
Edit: Here is a possibly useless condition on $\textbf{a}$:
$\textbf{a}$ is a monotone operator, which means $$(\textbf{a}(p)- \textbf{a}(q))\cdot (p-q)\ge 0$$ for all $p,q\in \mathbb R^n.$
Let $f_\lambda := \mathbf a(\mathrm D u - \lambda\,\mathrm D v)\cdot\mathrm Dv$ and notice that for $\lambda\in[0,1)$, $$ |f_\lambda| ≤ C \,(1+|\mathrm D u - \lambda\,\mathrm D v|) \,|\mathrm D v| ≤ C \,g $$ where $g = |\mathrm D v|+|\mathrm D u|\,|\mathrm D v| +\,|\mathrm D v|^2$ is integrable on $U$ since by the Cauchy-Schwarz inequality $$ ∫_U g ≤ |U|^{1/2} \,\|\mathrm Dv\|_{L^2(U)} + \|\mathrm Du\|_{L^2(U)} \,\|\mathrm Dv\|_{L^2(U)} + \|\mathrm Dv\|_{L^2(U)}^2. $$ Therefore, by the Lebesgue dominated convergence theorem, $$ \int_U f_\lambda \underset{\lambda\to 0}{\longrightarrow} \int_U f_0 $$ which is what you wanted to prove.