# Prove that differential equations are right continuous with respect to initial values.

Here the question.

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

Below is my idea.

Consider the equations $$$$(*)_n=\begin{cases}\frac{dy}{dx}=f(x,y)+\frac{1}{n}, \\y(x_0)=y_0, \end{cases}$$$$ write the sequence of solutions as $$\{\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 $$$$|\phi_n(x;x_0,y_0)-\phi(x;x_0,y_0)|<\epsilon.$$$$ Then we can get $$$$|\phi(x;x_0,y_{n})-\phi(x;x_0,y_0)|\leq|\phi(x;x_0,y_{n})-\phi_n(x;x_0,y_{0})|+|\phi_n(x;x_0,y_{0})-\phi(x;x_0,y_0)|.$$$$ The latter item can be controlled by $$\epsilon$$,but I don't know how to handle the previous item.Do you have a good idea? I really appreciate it!

• 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. Mar 29, 2023 at 3:19
• The problem is to prove the right continuity of the maximum solution.@ArcticChar Mar 29, 2023 at 3:27
• Are you sure you have only continuity of $f$ as an assumption, but not Lipschitz? Mar 29, 2023 at 3:38
• 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. Mar 29, 2023 at 3:44
• No,it means that the function value of other solutions is smaller than it.@ArcticChar Mar 29, 2023 at 3:55

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.]