I have the set of nonlinear differential equations describing a system I modeled for my research (spread of epidemics or information for instance):

$$\begin{array}{rl} \dot{p}(t) &= \gamma r(t)-u(t)p(t)\\ \dot{q}(t) &= u(t)p(t)-\delta q(t)\\ \dot{r}(t) &= \delta q(t)-\gamma r(t)\end{array}$$

Since $\dot{p}+\dot{q}+\dot{r}=0$, I only consider the last two equations:

$$\begin{array}{rl} \dot{q}(t) &= u(t)(1-q(t)-r(t))-\delta q(t)\\ \dot{r}(t) &= \delta q(t)-\gamma r(t)\end{array}$$

with input $$u(t)\in \{{u}:u\textrm{ is Lebesgue integrable, } 0\leq u(t)\leq u_{\max}\}$$ and the set $$\Omega=\{(q,r)\vert 0 \leq {q},\,{r}\leq 1, {q}+{r}\leq 1\}$$

My goal is to show that $\Omega$ is positively invariant with respect to the nonlinear dynamical system expressed above, i.e., a solution starting from any initial condition $(q,r)\in\Omega$ remains confined to $\Omega$.

Edit: $p(t),\,q(t),$ and $r(t)$ represent probabilites that sum up to $1$: ${p}(t)+{q}(t)+{r}(t)=1$.

  • $\begingroup$ Excuse my ignorance, but what does it mean "$\Omega$ is positively invariant with respect to the dynamical system .." ? $\endgroup$ – Konstantinos Kanakoglou Aug 14 '17 at 17:22
  • $\begingroup$ @KonstantinosKanakoglou I believe it means that if the state is in $\Omega$ at $t_0$ then it stays in $\Omega$ for all $t \geq t_0$. $\endgroup$ – Rodrigo de Azevedo Aug 14 '17 at 17:26
  • $\begingroup$ Thanks, @RodrigodeAzevedo, correct :) I tried to include some background information. $\endgroup$ – Nickie Aug 14 '17 at 18:06
  • $\begingroup$ thank you for updating the post. It is certainly clear now. $\endgroup$ – Konstantinos Kanakoglou Aug 14 '17 at 18:11
  • $\begingroup$ This should be simple enough to prove if you draw $\Omega$. You need to show that the vector field is pointing "inwards" at the boundary $\partial\Omega$, for all possible values of control u. From equation for $\dot{q}$, it looks like the term containing $u$ vanishes on part of $\partial\Omega$. $\endgroup$ – Piyush Grover Aug 14 '17 at 18:17

Your Answer

By clicking "Post Your Answer", you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.