What about using
?
Hello,
So I have the differential system :
with some initial conditions, and where .
(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 is Lipschitz continuous in its second variable,
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So far... :
I've written , using the absolute value norm. ( )
Let
Basically, I have to prove that there exists a constant K such that :
I arrived at :
But now, I'm completely stuck... I don't see how to deal with
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I'm currently looking at the mean value theorem for vector-valued functions
But my intuition tells me it's not the right way...
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Then there is another question...
Show that if and are both positive, then and 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 !
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 ( is continuously differentiable), or you perform a more explicit proof using Danny Arrigo's formula.
Note that is a solution (and similarly with and ). Using the local unicity (granted by Cauchy-Lipschitz), you can see that the graphs of the solutions can't intersect. Therefore,...~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Then there is another question...
Show that if and are both positive, then and 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 ?
(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(edit: I've just had a look at the wikipedia, and Cauchy-Lipschitz theorem is just another name for Picard-Lindelöf theorem)
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 failedEither you use the mean value theorem on a compact ( is continuously differentiable)
Also, how can we know that f is continuously differentiable ? As we know nothing of u and v
I would have to find a majoration for and , but I still don't seeor you perform a more explicit proof using Danny Arrigo's formula.
I thought on it... But I must be missing the key point, as I don't understand how to conclude oONote that is a solution (and similarly with and ). Using the local unicity (granted by Cauchy-Lipschitz), you can see that the graphs of the solutions can't intersect. Therefore,...
Also, is it possible to do it without finding particular solutions ?
Thanks
I think you understood something wrong: is a very simple function, it is from (or just if you forget about ) to . In this function, is just a name for a variable, it is not connected with the solution at all. That's why it is obvious that, since each component of is a polynomial, it is . The MVT gives for any where 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 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 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 for some (and is anything), unicity gives that for all (and coincides as well with the particular solution). But : contradiction!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 ?
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 )
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 where the differential equation is defined?
If so, can this result be extended to ?
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 , or we can only prove it on for some , then repeatedly applying P-L theorem, we might have the unique solutions on , , , , where , then this result can not be extended to either the whole interval or the halfline .
So how local existence and uniqueness can imply the global one?
I just wonder to learn what you think.
Many thanks.
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 (if two solutions coincide at a point, they coincide on a small neighbourhood).
Suppose and are two solutions, on intervals and such that and .
Let . By local uniqueness, not only belongs to but every point in an open interval does as well. Thus we can prove that is open : for every , in the same way, apply the local uniqueness at . However, and are continuous, hence it is obvious that is closed. Since is connected, we deduce that : for every , . This is what I call "global uniqueness", and this is what Moo needed.
In particular, we can extend this solution to simply by taking if and if . (This is continuously differentiable since and coincide in )
This is not global existence however, which is usually much more complicated, or false, and indeed does not follow from local existence.
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By the way,
To Moo: Congratulations for you 18.5!