Hello,

So I have the differential system :

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

with some initial conditions, and where $\displaystyle \alpha>0$.

(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 $\displaystyle f(t,z)=f(t,(u,v))=(u(1-v) ~,~ \alpha v(u-1))$ is Lipschitz continuous in its second variable, $\displaystyle z\in\mathbb{R}^2$

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So far... :

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

Let $\displaystyle z_i=(u_i,v_i)$

Basically, I have to prove that there exists a constant K such that :

$\displaystyle \|f(t,z_1)-f(t,z_2)\|_1\leq K|z_1-z_2|=K (|u_1-u_2|+|v_1-v_2|)$

I arrived at :

$\displaystyle \|f(t,z_1)-f(t,z_2)\|_1\leq |u_1-u_2|+\alpha|v_1-v_2|+(1+\alpha)|u_1v_1-u_2v_2|$

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

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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 $\displaystyle u(0)=u_0$ and $\displaystyle v(0)=v_0$ are both positive, then $\displaystyle u(t)$ and $\displaystyle v(t)$ 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 !