sequence of solutions of a de

Question

Let $f:\mathbb R \rightarrow \mathbb R$ be continous and bounded and $u_n: [0,1] \rightarrow \mathbb R$ a sequence of solutions of the differential equation $y'=f(y)$. Show: If $u_n(0)$ is bounded, there exists a subsequence of $u_n$ which converges uniformly to a solution $u$ of the differential equation $y'=f(y)$. My idea: Because $u_n(0)$ is bounded in $\mathbb R$, there exists a convergent subsequence $u_{n_k}(0)$. How can i show, that a subsequence of $u_n$ converges uniformly?

Answer 1

As $f$ is bounded ($|f(x)| \le b$ on $\mathbb{R}$, say) the mean value theorem shows that each $u_n$ is Lipschitz-continuous with the same Lipschitz constant $b$. Thus $S:=\{u_n:n \in \mathbb{N}\}$ is an equicontinuous set in $C([0,1])$, and as $\{u_n(0):n \in \mathbb{N}\}$ is bounded in $\mathbb{R}$ the set $S$ is bounded in $C([0,1])$. To see this, just note that $$ |u_n(t)| \le |u_n(t)-u_n(0)| + |u_n(0)|\le bt+ |u_n(0)| \le b+ |u_n(0)| $$ for $t \in [0,1]$. Now, from Arcela-Ascoli's theorem we get that there is a subsequence $(u_{n_k})$ from $(u_{n})$ which is uniformly convergent on $[0,1]$ to a function $v$, say. Now $(f(u_{n_k}))$ is uniformly convergent to $f(v)$. Next, from $$ u_{n_k}(t)=u_{n_k}(0)+\int_0^t f(u_{n_k}(\tau)) d\tau \quad (t \in [0,1]) $$ we get (as $k \to \infty$) $$ v(t)=v(0)+\int_0^t f(v(\tau)) d\tau \quad (t \in [0,1]). $$ The integrand $f(v)$ is continuous, hence $v$ is differentiable and $v'(t)=f(v(t))$ on $[0,1]$.

Remark: Under the assumtions on $f$ IVPs are not uniquely solvable, in general.