# Proving the solution of a Cauchy problem is continuous

Determine for $$I_{(t_0,x_0)}$$ the maximal domain that defines the solution $$\phi_{(t_0,x_0)}(.)$$ for the Cauchy problem:

$$\begin{cases} \dot{x}=f(x)g(t)\\ x(t_0)=x_0 \end{cases}$$ where $$f$$ is continuous and non-zero in the interval $$[a_1,a_2]$$ and $$g$$ is contious in the interval $$(t_1,t_2)$$.

Show that the set $$\mathscr{D}=\{(t,t_0,x_0):t_0,x_0\in(t_1,t_2)\times(a_1,a_2)\}$$ and $$t\in I(t_0,x_0)$$ is open and the function $$\varphi:D\to\mathbb{R}$$ given by $$\varphi_{(t,t_0,x_0)}=\varphi_{(t_0,x_0)}(t)$$ is continuous on $$D$$.

I have no idea on how to solve this exercise. I thought of using the Picard theorem: admitting the problem has a solution then $$f$$ would be Lipschitz contious which would imply $$\varphi$$ to be continuous. But I am not sure I could use the reverse implicarion of the theorem is valid.

Question:

How should I solve this problem?

First, a counterexample: for $$-a_1=a_2=1,g=1,f(x)=1+x^2$$, the maximal interval $$I_{(0,0)}$$ is $$[-\pi/4,\pi/4]$$, not open!
So I am assuming $$f$$ is actually defined on $$(a_1,a_2)$$. The “proof” below isn’t complete, some detail is omitted.
First, it is easy to show that if $$F$$ is a given antiderivative of $$1/f$$ (so $$F$$ monotonous), $$G$$ some antiderivative of $$g$$, $$x$$ is a solution of the differential equation iff $$x=F^{-1}(G(t)+c)$$ for some constant $$c$$.
As a consequence, $$(t,t_0,x_0) \in D$$ iff $$t_1 < t,t_0 and $$a_1 and there exists some $$c$$ such that $$F^{-1}(G(t_0)+c)=x_0$$ and $$G(t)+c$$ is in the domain of $$F^{-1}$$ (which we denote as $$(b_1,b_2)$$). Note that in this case, $$c=F(x_0)-G(t_0)$$.
Thus, $$(t,t_0,x_0) \in D$$ iff $$t,t_0 \in (t_1,t_2)$$, $$x_0 \in (x_1,x_2)$$ and $$G(t)-G(t_0)+F(x_0) \in (b_1,b_2)$$. This is clearly an “open condition”.
For the second part, just note from the above that $$\varphi_{(t_0,x_0)}(t)=F^{-1}(G(t)-G(t_0)+F(x_0))$$, hence the global flow is clearly continuous.