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Math Help - [SOLVED] Uniqueness of the solution

  1. #1
    Moo
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    [SOLVED] Uniqueness of the solution

    Hello,

    So I have the differential system :

    \begin{cases} \frac{du}{dt}=u(t)(1-v(t)) \\ \frac{dv}{dt}=\alpha v(t)(u(t)-1) \end{cases}

    with some initial conditions, and where \alpha>0.
    (It's taken from the Lotka-Volterra system)

    Now, I have to show that it has a unique solution. In order to prove this, we've been taught to use the Picard-Lindelöf theorem.

    So I have to prove that f(t,z)=f(t,(u,v))=(u(1-v) ~,~ \alpha v(u-1)) is Lipschitz continuous in its second variable, z\in\mathbb{R}^2

    ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
    So far... :

    I've written \|f(t,z_1)-f(t,z_2)\|, using the absolute value norm. ( \|(u,v)\|_1=|u|+|v|)

    Let z_i=(u_i,v_i)

    Basically, I have to prove that there exists a constant K such that :
    \|f(t,z_1)-f(t,z_2)\|_1\leq K|z_1-z_2|=K (|u_1-u_2|+|v_1-v_2|)

    I arrived at :
    \|f(t,z_1)-f(t,z_2)\|_1\leq |u_1-u_2|+\alpha|v_1-v_2|+(1+\alpha)|u_1v_1-u_2v_2|

    But now, I'm completely stuck... I don't see how to deal with |u_1v_1-u_2v_2|

    ---------------
    I'm currently looking at the mean value theorem for vector-valued functions
    But my intuition tells me it's not the right way...

    ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
    Then there is another question...
    Show that if u(0)=u_0 and v(0)=v_0 are both positive, then u(t) and v(t) are both positive.

    So how can I do this one ? oO Thought of finding the minimum (at (1,1) or something like that), but we wouldn't use the initial conditions ?



    Any help will be appreciated Thanks !
    Last edited by mr fantastic; May 31st 2009 at 03:43 AM. Reason: Changed post title (by request)
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  2. #2
    MHF Contributor
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    What about using

    <br />
|u_1v_1-u_2v_2| = |u_1 v_1 - u_1 v_2 + u_1 v_2 - u_2 v_2| \le |u_1| |v_1 - v_2| + |v_2| |u_1 - u_2| ?
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  3. #3
    Moo
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    Quote Originally Posted by danny arrigo View Post
    What about using

    <br />
|u_1v_1-u_2v_2| = |u_1 v_1 - u_1 v_2 + u_1 v_2 - u_2 v_2| \le |u_1| |v_1 - v_2| + |v_2| |u_1 - u_2| ?
    I thought about it.
    But the constant has to be independent of u,v.
    And if we take the sup... we don't know if u_1,u_2 are bounded ?

    Thanks for your answer
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  4. #4
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    Quote Originally Posted by Moo View Post
    So I have to prove that f(t,z)=f(t,(u,v))=(u(1-v) ~,~ \alpha v(u-1)) is Lipschitz continuous in its second variable, z\in\mathbb{R}^2
    (...)
    I'm currently looking at the mean value theorem for vector-valued functions
    But my intuition tells me it's not the right way...
    Your function is not globally Lipschitz, but it is locally Lipschitz, and that's enough for Cauchy-Lipschitz theorem (for the unicity, at least: local unicity implies global unicity).

    Either you use the mean value theorem on a compact ( f is continuously differentiable), or you perform a more explicit proof using Danny Arrigo's formula.

    ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
    Then there is another question...
    Show that if u(0)=u_0 and v(0)=v_0 are both positive, then u(t) and v(t) are both positive.

    So how can I do this one ? oO Thought of finding the minimum (at (1,1) or something like that), but we wouldn't use the initial conditions ?
    Note that u(t)=0, v(t)=e^{-\alpha t} is a solution (and similarly with v(t)=0 and u(t)=..). Using the local unicity (granted by Cauchy-Lipschitz), you can see that the graphs of the solutions can't intersect. Therefore,...

    (edit: I've just had a look at the wikipedia, and Cauchy-Lipschitz theorem is just another name for Picard-Lindelöf theorem)
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  5. #5
    Moo
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    (edit: I've just had a look at the wikipedia, and Cauchy-Lipschitz theorem is just another name for Picard-Lindelöf theorem)
    Yes, I picked the name by looking at the wikipedia ^^ But we learnt it as the Cauchy-Lipschitz theorem

    Quote Originally Posted by Laurent View Post
    Your function is not globally Lipschitz, but it is locally Lipschitz, and that's enough for Cauchy-Lipschitz theorem (for the unicity, at least: local unicity implies global unicity).
    How can one know that it's not globally Lipschitz ?

    Either you use the mean value theorem on a compact ( f is continuously differentiable)
    I tried to find what particular property gives the MVT on a compact, but I failed
    Also, how can we know that f is continuously differentiable ? As we know nothing of u and v

    or you perform a more explicit proof using Danny Arrigo's formula.
    I would have to find a majoration for |u_1| and |v_2|, but I still don't see

    Note that u(t)=0, v(t)=e^{-\alpha t} is a solution (and similarly with v(t)=0 and u(t)=..). Using the local unicity (granted by Cauchy-Lipschitz), you can see that the graphs of the solutions can't intersect. Therefore,...
    I thought on it... But I must be missing the key point, as I don't understand how to conclude oO
    Also, is it possible to do it without finding particular solutions ?

    Thanks
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  6. #6
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    Quote Originally Posted by Moo View Post
    How can one know that it's not globally Lipschitz ?

    I tried to find what particular property gives the MVT on a compact, but I failed
    Also, how can we know that f is continuously differentiable ? As we know nothing of u and v
    I think you understood something wrong: f is a very simple function, it is f(t,(u,v))=(u(1-v),\alpha v(u-1)) from \mathbb{R}^3 (or just \mathbb{R}^2 if you forget about t) to \mathbb{R}^2. In this function, u is just a name for a variable, it is not connected with the solution u(t) at all. That's why it is obvious that, since each component of f is a polynomial, it is \mathcal{C}^\infty. The MVT gives \|f(u,v)-f(0,0)\|\leq \|(u,v)-(0,0)\|\max_K \|df\| for any (u,v)\in K where K is a compact convex subset containing 0. By compacity and continuity, the maximum is finite. The end.

    I claimed it is not globally Lipschitz because for instance \frac{\|f(u,u)\|}{\|(u,u)\|}=\frac{\|u(1-u)|+|\alpha u(u-1)|}{2|u|} = \frac{1+|\alpha|}{2}|u-1| is not bounded. A globally Lipschitz function can not grow faster than linearly, and I noticed that along the diagonal this function grows quadratically.

    I thought on it... But I must be missing the key point, as I don't understand how to conclude oO
    Also, is it possible to do it without finding particular solutions ?
    I don't know, in this case the particular solutions are very simple. The important thing is that the axes are graphs of solutions. Then, if u(t)=0 for some t (and v(t) is anything), unicity gives that u(t)=0 for all t (and v(t) coincides as well with the particular solution). But u(0)=u_0>0: contradiction!
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  7. #7
    Moo
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    Okay, so finally, I've got a 18.5/20 mark on this lol
    Though, as I didn't see my paper after it has been corrected, I don't know if I did this part right...

    Thanks guys =)


    (for these well-aware moderators, we were not told it is forbidden to get help for this )
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  8. #8
    Senior Member bkarpuz's Avatar
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    As far as I know Cauchy-Lipschitz=Picard-Lindelöf, but the proof methods are different.
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    Senior Member bkarpuz's Avatar
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    Exclamation

    Quote Originally Posted by Laurent View Post
    Your function is not globally Lipschitz, but it is locally Lipschitz, and that's enough for Cauchy-Lipschitz theorem (for the unicity, at least: local unicity implies global unicity).

    Either you use the mean value theorem on a compact ( f is continuously differentiable), or you perform a more explicit proof using Danny Arrigo's formula.


    Note that u(t)=0, v(t)=e^{-\alpha t} is a solution (and similarly with v(t)=0 and u(t)=..). Using the local unicity (granted by Cauchy-Lipschitz), you can see that the graphs of the solutions can't intersect. Therefore,...

    (edit: I've just had a look at the wikipedia, and Cauchy-Lipschitz theorem is just another name for Picard-Lindelöf theorem)
    Laurent, I need some explanation about your answer if possible.
    How local uniqueness implies global uniqueness?
    And one more, what is meant by a global solution?
    Does it means that unique solution exists on the whole interval [a,b] where the differential equation is defined?
    If so, can this result be extended to [a,\infty)?

    The reason that I am asking the second and the third question is that, if we can not prove existence and uniqueness of a solution on [a,b], or we can only prove it on [a,a+\delta_{0}] for some b-a>\delta_{0}>0, then repeatedly applying P-L theorem, we might have the unique solutions on [a+\delta_{0},a+\delta_{0}+\delta_{1}], [a+\delta_{0}+\delta_{1},a+\delta_{0}+\delta_{1}+\d  elta_{2}], \cdots, [a+\delta_{0}+\delta_{1}+\cdots+\delta_{n},a+\delta  _{0}+\delta_{1}+\delta_{2}+\cdots+\delta_{n+1}], where \sum\nolimits_{n\in\mathbb{N}}\delta_{n}=c<b, then this result can not be extended to either the whole interval [a,b] or the halfline [a,\infty).
    So how local existence and uniqueness can imply the global one?
    I just wonder to learn what you think.
    Many thanks.
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  10. #10
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    Quote Originally Posted by bkarpuz View Post
    Laurent, I need some explanation about your answer if possible.
    How local uniqueness implies global uniqueness?
    There is a celebrated trick that helps going from local to global : a connectivity argument. Suppose we know local uniqueness for solutions of the equation y'(t)=f(t,y(t)) (if two solutions coincide at a point, they coincide on a small neighbourhood).

    Suppose y_1 and y_2 are two solutions, on intervals I_1 and I_2 such that t_0\in I_1\cap I_2 and y_1(t_0)=y_2(t_0).

    Let A=\{t\in I_1\cap I_2|y_1(t)=y_2(t)\}. By local uniqueness, not only t_0 belongs to A but every point in an open interval (t_0-\varepsilon,t_0+\varepsilon) does as well. Thus we can prove that A is open : for every t\in A, in the same way, apply the local uniqueness at t. However, y_1 and y_2 are continuous, hence it is obvious that A is closed. Since I_1\cap I_2 is connected, we deduce that A=I_1\cap I_2 : for every t\in I_1\cap I_2, y_1(t)=y_2(t). This is what I call "global uniqueness", and this is what Moo needed.

    In particular, we can extend this solution to I_1\cup I_2 simply by taking y(t)=y_1(t) if t\in I_1 and y(t)=y_2(t) if t\in I_2. (This is continuously differentiable since y_1 and y_2 coincide in I_1\cap I_2)

    This is not global existence however, which is usually much more complicated, or false, and indeed does not follow from local existence.


    -------

    By the way,
    To Moo: Congratulations for you 18.5!
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