Consider the system of differential equations:
$x'(t)=-\beta x(t)v(t)$
$y'(t)=\beta x(t)v(t)-\gamma y(t)$
$u'(t)=b-(\mu+\delta y(t))u(t)$
$v'(t)=\delta y(t)u(t)-\mu v(t)$
with the boundaries $x(0)>0, y(0)\geq 0$, $u(0)>0$ and $v(0)\geq 0$.
I want to prove that this implies that $u(t)>0, v(t)\geq 0, u(t)>0, v(t)\geq 0$ for all $t>0$, how can I do that?
Thanks in advance.
Let us start with the $x$-coordinate. The hyperplane $x = 0$ is invariant (that is, any solution starting on it remains there). If $x(0) > 0$ then $x(t) > 0$ for $t$ sufficiently close to zero. $x(\tau) = 0$ for some $\tau \ne 0$ is impossible, since then some solution would, for $t = \tau$, take value on the hyperplane $x = 0$ and for $t = 0$ take value outside the hyperplane. If $x(\theta) < 0$ for some $\theta \ne 0$ then, by the intermediate value theorem applied to the $x$-coordinate of the solution, $x(s)$ would be zero for some moment $s$ between $0$ and $\theta$, which has been just excluded.
Regarding the $u$-coordinate, again, $u(t) > 0$ for all $t > 0$ sufficiently close to zero. Suppose to the contrary that $u(\theta) = 0$ for some $\theta > 0$. Let $\tau$ be the first moment at which $u(\tau) = 0$. Of course, $\tau > 0$. But then $u(\tau) = b > 0$, hence $u(t) < 0$ for $t < \tau$ sufficiently close to $\tau$. Applying the intermediate value theorem as in the first paragraph we obtain the existence of $s \in (0, \tau)$ such that $u(s) = 0$, which contradicts the definition of $\tau$.
We have thus proved that $x(t) > 0$ and $u(t) > 0$ for all $t > 0$ for which the solution exists. Now, fix some solution, and notice that its $y$- and $v$-coordinates satisfy the following system of nonautonomous linear homogeneous ordinary differential equations $$ \tag{$*$} \begin{cases} y'(t) = - \, \gamma y(t) + f(t) v(t) \\ v'(t) = g(t) y(t) - \mu v(t), \end{cases} $$ where $f(t) := \beta x(t) > 0$ and $g(t) := \delta u(t) > 0$. When $y(0) = v(0) = 0$ then $y(t) = v(t) = 0$ for all $t > 0$ for which the solution exists. If $y(0) > 0$ or $v(0) > 0$ (or both) then there is $\varepsilon > 0$ such that $y(t) > 0$ and $v(t) > 0$ for $t \in (0, \varepsilon)$. Suppose to the contrary that $\tau \ge \varepsilon$ is the first moment at which one of the coordinates $y$ or $v$ (say $y$) takes value $0$ (why not both?). Then $y'(\tau) = f(\tau) v(\tau) > 0$, which is impossible.
We have that the required inequalities are satisfied for all $t > 0$ for which the solution is defined. Now we need to prove that the domain of the solution contains $[0, \infty)$. If not, the norm of the solution would blow up in finite time, say $T < \infty$. Look at which coordinates may blow up. The $x$-coordinate is positive and nonincreasing, and the $u$-coordinate satisfies $0 < u(t) \le u(0) + b t$. So, the blow-up of those coordinates is excluded. Then the coefficients $f(t)$ and $g(t)$ in ($*$) are bounded on $[0, T]$, so it follows from Grönwall's inequality that the $y$- and $v$-coordinates are bounded as well.