# Thread: [SOLVED] Uniqueness of the solution

1. ## [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 !

2. What about using

$
|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|$
?

3. Originally Posted by danny arrigo

$
|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 ?

4. Originally Posted by Moo
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)

5. (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

Originally Posted by Laurent
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

6. Originally Posted by Moo
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!

7. 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 )

8. As far as I know Cauchy-Lipschitz=Picard-Lindelöf, but the proof methods are different.

9. Originally Posted by Laurent
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, 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.

10. Originally Posted by bkarpuz
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.

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