# Thread: Continuous dependence on the initial point

1. ## Continuous dependence on the initial point

Dear MHF members,

I need the proof of the following result but I don't know where can I get it.

Theorem. Let $\displaystyle A\in\mathrm{C}([0,\infty),\mathbb{R}_{n}^{n})$, $\displaystyle F\in\mathrm{C}([0,\infty),\mathbb{R}^{n})$ and $\displaystyle X_{0}\in\mathbb{R}^{n}$.
Then the unique solution $\displaystyle \varphi(\cdot,s)$ of the initial value problem
$\displaystyle \begin{cases}X^{\prime}=A(t)X+F,\ r\geq t\geq s\geq0\\ X(s)=X_{0}\end{cases}$,
where $\displaystyle r>0$ is fixed, is continuous in $\displaystyle s$.

I would be very glad if you can help me in this direction.
Thanks.
bkarpuz

2. ## Re: Continuous dependence on the initial point

Okay, I got the proof. Please let me know if you think that it is wrong.
Proof. Set $\displaystyle Y(t):=\varphi(t,s_{2})-\varphi(t,s_{1})$ for $\displaystyle t\geq s_{2}\geq s_{1}\geq0$. Then, for all $\displaystyle t\geq s_{2}$, we have
$\displaystyle Y(t)=\bigg(X_{0}+\int_{s_{2}}^{t}A(u)\varphi(u,s_{ 2}) \mathrm{d}u\bigg)-\bigg(X_{0}+\int_{s_{1}}^{t}A(u)\varphi(u,s_{1}) \mathrm{d}u\bigg)$
......_$\displaystyle =\int_{s_{2}}^{t}A(u)Y(u) \mathrm{d}u-\int_{s_{1}}^{s_{2}}A(u)\varphi(u,s_{1})\mathrm{d} u,$
which yields
$\displaystyle \|Y(t)\|\leq\int_{s_{2}}^{t}\|A(u)\|\|Y(u)\| \mathrm{d}u+\int_{s_{1}}^{s_{2}}\|A(u)\|\|\varphi( u,s_{1})\| \mathrm{d}u$
......___$\displaystyle \leq\bigg(\int_{s_{1}}^{s_{2}}\|A(u)\|\|\varphi(u, s_{1})\| \mathrm{d}u\bigg)\exp\bigg\{\int_{s_{2}}^{t}\|A(u) \| \mathrm{d}u\bigg\},$
where we have applied the Gronwall's inequality in the last line.
This shows that picking $\displaystyle s_{2}$ sufficiently close to $\displaystyle s_{1}$, the right-hand side of the above inequality can be made sufficiently small. This proves continuity of $\displaystyle \varphi(\cdot,s)$ in $\displaystyle s$ follows.

3. ## Re: Continuous dependence on the initial point

I don't have the book with me, but this sort of thing is in Coddington and Levinson. Check that out.