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Math Help - Continuous dependence on the initial point

  1. #1
    Senior Member bkarpuz's Avatar
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    Exclamation 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 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
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  2. #2
    Senior Member bkarpuz's Avatar
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    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 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.
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  3. #3
    A Plied Mathematician
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    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.
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