Mathematical Biology (steady states)

[ra_forever8]

non-dimensionalisation equation:
\begin{equation} \frac {du}{d\tau}=\frac{\overline{\lambda}_{1} u}{u+1} -\overline{r}_{ab}uv -\overline{d}u
\end{equation}
where $\overline{\lambda}_{1}= \frac {\lambda_{1}}{\lambda_{2} K_{1}}$
Another non-dimensionalisation equations
\begin{equation} \frac {dv}{d\tau}=(1-v) -\overline{r}_{ba}uv
\end{equation}
THE REAL QUESTION IS: calculate the steady state $(u_4,v_4)$ and $(u_5,v_5)$? Discuss the occurrence of these steady-states in respect of any relationships between the non-dimensional. Note you are not required to determine the stability of these two states.

=>
I have already calculated the steady-states of $(u_4,v_4)$ and $(u_5,v_5)$ which are
$(u_4,v_4)$ =($ \frac{\overline{\lambda}_{1} - \overline{d} -\overline{r}_{ab}}{\overline{r}_{ab} + \overline{d}} ,1)$
$(u_5,v_5)$ =($ \frac{\overline{\lambda}_{1} - \overline{d} -\overline{r}_{ab}}{\overline{r}_{ab} + \overline{d}}$ ,$ 1 - \overline{r}_{ba} \frac{\overline{\lambda}_{1} - \overline{d} -\overline{r}_{ab}}{\overline{r}_{ab} + \overline{d}}$ )

I HAVE ALREADY CALCULATED $(u_1,v_1)$ and $(u_2,v_2)$ and $(u_3,v_3)$ but i an not interested to talk about their steady-state with respect to non-dimensional parameters.

here are the non-dimensional parameters which I have also determined:
$\overline{\lambda}_{1}= \frac {\lambda_{1}}{\lambda_{2} K_{1}}$
$\overline{d}= \frac{d}{\lambda_{2}}$
$\overline{r}_{ab} = \frac{\overline{r}_{ab} K_{2}} {\lambda_{2}}$
$\overline{r}_{ba} = \frac{\overline{r}_{ab} K_{1}} {\lambda_{2}}$

can anyone please please help me in discussion of these steady states $(u_4,v_4)$ and $(u_5,v_5)$ with the above non-dimensional parameters .

 

[jvicens]

Sure, let's discuss the steady states $(u_4, v_4)$ and $(u_5, v_5)$ in relation to the non-dimensional parameters.

First, let's look at the steady state $(u_4, v_4)$. This steady state occurs when the rate of change of $u$ and $v$ is equal to zero, meaning that the system has reached a stable equilibrium. Looking at the non-dimensional equation for $du/d\tau$, we can see that the first term on the right-hand side ($\overline{\lambda}_{1} u/(u+1)$) is proportional to $\overline{\lambda}_{1}$, which is determined by the ratio of the two rate constants ($\lambda_1$ and $\lambda_2$) and the equilibrium constant for the first reaction ($K_1$). This term represents the production of $u$ by the first reaction. The second term ($-\overline{r}_{ab}uv$) is proportional to both $\overline{r}_{ab}$ and $v$, and represents the consumption of $u$ by the second reaction. Finally, the third term ($-\overline{d}u$) is proportional to $\overline{d}$ and represents the degradation of $u$.

At the steady state $(u_4, v_4)$, the production of $u$ is balanced by the consumption and degradation, resulting in a constant value for $u$. This value is determined by the non-dimensional parameters $\overline{\lambda}_{1}$, $\overline{d}$, and $\overline{r}_{ab}$. Similarly, the steady state value of $v$ at $(u_4, v_4)$ is determined by the non-dimensional parameters $\overline{r}_{ba}$ and $\overline{r}_{ab}$.

Now let's look at the steady state $(u_5, v_5)$. This steady state occurs when the rate of change of $v$ is equal to zero, but the rate of change of $u$ is not. This means that $u$ is still changing, but $v$ has reached a stable equilibrium. Looking at the non-dimensional equation for $dv/d\tau$, we can see that the first term on the right-hand side ($1-v$) represents the production of $v$ by the first reaction. The second term ($