Here the question.

Consider the initial value problem: \begin{equation} \begin{cases} \frac{dy}{dx}=f(x,y) \\y(x_0)=y_0, \end{cases} \end{equation} $f(x,y)$ is continuous. Suppose that $y=\phi(x;x_0,y_0)$ is the maximum solution of the initial value problem. Prove:$\phi(x;x_0,y_0)$ is right continuous for $y$, that is \begin{equation} \lim_{y_1\to y_0^+}\phi(x;x_0,y_1)=\phi(x;x_0,y_0) \end{equation} establishs on $|x-x_0|\leq \alpha$, $\alpha$ is constant.

Below is my idea.

Consider the equations \begin{equation} (*)_n=\begin{cases}\frac{dy}{dx}=f(x,y)+\frac{1}{n}, \\y(x_0)=y_0, \end{cases} \end{equation} write the sequence of solutions as $\{\phi_n(x;x_0,y_0)\}$.It has a uniformly convergent subsequence,let itself be uniformly convergent.

Assume that $\phi(x;x_0,y_0)$ is not right continuous for $y_0$, then $\exists \epsilon>0,\forall \delta>0,\exists y_{\delta}-y_0<\delta$, such that $$|\phi(x;x_0,y_{\delta})-\phi(x;x_0,y_0)|\geq \epsilon.$$

Let $\{\delta_n\}$ decreases to $0$,The corresponding $y_{\delta}$ is denoted by $y_n$. According to the uniform convergence, $\exists N>0,\forall n>N$, we have \begin{equation} |\phi_n(x;x_0,y_0)-\phi(x;x_0,y_0)|<\epsilon. \end{equation} Then we can get \begin{equation} |\phi(x;x_0,y_{n})-\phi(x;x_0,y_0)|\leq|\phi(x;x_0,y_{n})-\phi_n(x;x_0,y_{0})|+|\phi_n(x;x_0,y_{0})-\phi(x;x_0,y_0)|. \end{equation} The latter item can be controlled by $\epsilon$,but I don't know how to handle the previous item.Do you have a good idea? I really appreciate it!

  • $\begingroup$ If you have only continuity of $f$, solutions to IVP might not be unique and you should not get continuity with respect to initial condition. $\endgroup$ Mar 29, 2023 at 3:19
  • $\begingroup$ The problem is to prove the right continuity of the maximum solution.@ArcticChar $\endgroup$
    – Ychen
    Mar 29, 2023 at 3:27
  • $\begingroup$ Are you sure you have only continuity of $f$ as an assumption, but not Lipschitz? $\endgroup$ Mar 29, 2023 at 3:38
  • $\begingroup$ There seems to be a confusion here with right continuity...confusing one thing for another. There is some continuity of the maximal interval of definition... and it is a half continuity. It has to do with the maximal domain of definition being open. Maybe your question is about this. $\endgroup$
    – orangeskid
    Mar 29, 2023 at 3:44
  • 1
    $\begingroup$ No,it means that the function value of other solutions is smaller than it.@ArcticChar $\endgroup$
    – Ychen
    Mar 29, 2023 at 3:55

1 Answer 1


Now that $\phi(n)$ is uniformly convergent to the $\phi$, what we need to prove is that $\phi(n)$ is continuous, according to the later theorem: the continuity of the solution with respect to the initial value, we can get the $\phi(n)$ is continuous for y.[notice: even the later theorem: the continuity of the solution with respect to the initial value need the Lipschitz condition, you just need the first half of the proof.]

  • $\begingroup$ Which later theorem are you referring to? $\endgroup$ Mar 29, 2023 at 9:13

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