Continuous dependence on the initial point

**bkarpuz** · December 14th 2011, 08:41 PM

Dear MHF members,

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

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

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

**bkarpuz** · December 14th 2011, 10:29 PM

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

**Ackbeet** · December 15th 2011, 02:49 AM

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

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