Convergence of ODE with uniform $L^\infty \cap L^1$ bound on nonlinearity

Question

Consider the IVP $$ \left\{ \begin{aligned} \frac{d}{dt} \Phi_n(t,x) &= f_n(\Phi_n(t,x)) && \forall t \in \mathbf{R}_+ \\ \Phi_n(0,x) &= x && \forall x \in \mathbf{R} \end{aligned}\right. $$ for $n \in \mathbf N$.

Suppose $f_n:\mathbf R \to \mathbf R$ is Lipschitz for every $n \in \mathbf N$ and satisfies the bound $\|f_n\|_{L^1(\mathbf R)} + \|f_n\|_{L^\infty(\mathbf R)} < 1 $ (but not a uniform bound on the derivative $f'_n$). Then there exists $\overline f$ and a subsequence (say $f_{n_k}$) such that $f_{n_k} \overset{\ast}{\rightharpoonup} \overline f $ in $L^\infty$ (or weakly in $L^p$ for $1 < p < \infty$).

Is it true that $\Phi_{n_k}$ converges (possibly up to extracting another subsequence) in some sense to a Filippov solution $\Phi$ of $$ \left\{ \begin{aligned} \frac{d}{dt} \Phi(t,x) &= \bar f(\Phi(t,x)) && \forall t \in \mathbf{R}_+ \\ \Phi(0,x) &= x && \forall x \in \mathbf{R} \end{aligned} \right.\quad? $$

Answer 1

No. Consider $f_n(x)=(1/2)\cos^2 nx$. This converges weakly to $f=1/4$ (in $L^2(0,1)$, say). However, the solution $x(t)=(1/n)\arctan nt/2$ of $x'=f_n(x)$, $x(0)=0$, converges uniformly to $0$ rather than the solution $x(t)=t/4$ of $x'=f(x)$, $x(0)=0$.

In fact, we don't even need to solve this explicitly. It's clear from uniqueness that $x'=f_n(x)$ can't get past the first zero $x=\pi/(2n)$ of $f_n$.