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Math Help - Existence and Uniqueness Theorem

  1. #1
    MHF Contributor alexmahone's Avatar
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    Existence and Uniqueness Theorem

    1. Consider the differential equation \frac{dy}{dx}=\sqrt{x-y};\ y(2)=2.

    f(x, y)=\sqrt{x-y} and D_yf(x, y)=-\frac{1}{2\sqrt{x-y}}

    Neither f(x, y) nor D_yf(x, y) is continuous on some rectangle that contains (2, 2) in its interior. Neither existence nor uniqueness is guaranteed in any neighbourhood of x = 2.

    2. Consider the differential equation \frac{dy}{dx}=2\sqrt{y};\ y(0)=0.

    f(x, y)=2\sqrt{y} and D_yf(x, y)=\frac{1}{\sqrt{y}}

    Here also, neither f(x, y) nor D_yf(x, y) is continuous on some rectangle that contains (0, 0) in its interior. However, two different solutions exist: y_1(x)=x^2 and y_2(x)\equiv 0. Can someone please explain this anomaly?
    Last edited by alexmahone; September 3rd 2011 at 09:45 AM.
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    MHF Contributor chisigma's Avatar
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    Re: Existence and Uniqueness Theorem

    Quote Originally Posted by alexmahone View Post
    2. Consider the differential equation \frac{dy}{dx}=2\sqrt{y};\ y(0)=0.

    f(x, y)=2\sqrt{y} and D_yf(x, y)=\frac{1}{\sqrt{y}}

    Here also, neither f(x, y) nor D_yf(x, y) is continuous on some rectangle that contains (0, 0) in its interior. However, two different solutions exist: y_1(x)=x^2 and y_2(x)\equiv 0. Can someone please explain this anomaly?
    For the 'initial condition problem' ...

    y^{'}= 2\ \sqrt{y}\ ;\ y(0)=0 (1)

    ... the so called 'Lipschitz conditions' aren't satisfied so that existence and uniquess is not guaranted. Effectively observing (1) it is evident that y^{'} is function of the y alone, so that if \varphi(x) is solution of (1), then \varphi(x-\xi)\ ; \xi>0 is also solution of (1) and the solutions are infinite. In detail is...

    \varphi(x)=\begin{cases} x^{2}& x \ge 0\\0& x<0\end{cases} (2)

    Kind regards

    \chi \sigma
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  3. #3
    MHF Contributor alexmahone's Avatar
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    Re: Existence and Uniqueness Theorem

    Quote Originally Posted by chisigma View Post
    For the 'initial condition problem' ...

    y^{'}= 2\ \sqrt{y}\ ;\ y(0)=0 (1)

    ... the so called 'Lipschitz conditions' aren't satisfied so that existence and uniquess is not guaranted. Effectively observing (1) it is evident that y^{'} is function of the y alone, so that if \varphi(x) is solution of (1), then \varphi(x-\xi)\ ; \xi>0 is also solution of (1) and the solutions are infinite. In detail is...

    \varphi(x)=\begin{cases} x^{2}& x \ge 0\\0& x<0\end{cases} (2)

    Kind regards

    \chi \sigma
    I've only started learning differential equations and don't know about 'Lipschitz conditions'.

    Is it true that even if the Existence and Uniqueness Theorem doesn't guarantee existence, solution(s) may exist? Similarly is it true that even if the theorem doesn't guarantee uniqueness, there may be only one solution?
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    Re: Existence and Uniqueness Theorem

    What "anomaly" are you referring to? The existance and uniqueness theorem says that if certain hypotheses are satified then there exist a unique solution. It says NOTHING about what happens when those hypotheses are not satisfied. You can't be sure there is a unique solution but you can't be sure there isn't either. It is even possible to have a problem in which the hypotheses are not satisfied but there does exist a unique solution. That would not violate the theorem.
    Last edited by HallsofIvy; September 3rd 2011 at 10:26 AM.
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    MHF Contributor alexmahone's Avatar
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    Re: Existence and Uniqueness Theorem

    Quote Originally Posted by HallsofIvy View Post
    It is even possible to have a problem in which the hypotheses are satisfied but there does exist a unique solution.
    Surely, you mean "hypotheses are not satisfied"?
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    Re: Existence and Uniqueness Theorem

    Yes, I did. I just noticed that and editted.
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  7. #7
    MHF Contributor alexmahone's Avatar
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    Re: Existence and Uniqueness Theorem

    Quote Originally Posted by HallsofIvy View Post
    Yes, I did. I just noticed that and editted.
    Thanks.

    Another question: Does the first hypothesis [continuity of f(x, y)] guarantee existence and the second hypothesis [continuity of D_yf(x, y)] uniqueness? My textbook seems to imply this.
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  8. #8
    MHF Contributor chisigma's Avatar
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    Re: Existence and Uniqueness Theorem

    Quote Originally Posted by alexmahone View Post
    I've only started learning differential equations and don't know about 'Lipschitz conditions'.

    Is it true that even if the Existence and Uniqueness Theorem doesn't guarantee existence, solution(s) may exist? Similarly is it true that even if the theorem doesn't guarantee uniqueness, there may be only one solution?
    If You manage French language, here a good enough explanation...

    Théorème de Cauchy-Lipschitz - Wikipédia

    Kind regards

    \chi \sigma
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