# Calculating Steady States: Exploring the "Two Cases

• MHB
• mt91
In summary, to calculate steady state solutions, set the derivatives of n and c to 0. There are two cases for solving these equations: when n is equal to 0 and when n is not equal to 0. In the first case, the solution is c=α/μ and n=0. In the second case, the solution is c=α/μ and n=(μ-αβ)/αβ, with the possibility for n to be positive or negative depending on the value of μ/αβ. The application of these solutions may depend on the importance of n being positive or negative.
mt91

I've got this question here, I know to calculate steady states you set dn/dT and dc/dT to 0 and then solve. However can anyone help me understand what it means by the "two cases" and how you go about this?

Yes, the "steady" in "steady state" means that it does not change. "Steady State" solutions for n and c are constant solutions so the derivatives are 0. The equations become:
$\frac{n}{n+1}- \beta nc=0$ and
$\alpha- \mu c= 0$.

Since the second equation involves only c, no n, I would solve it first:
$c= \frac{\alpha}{\mu}$.

Now put that into the first equation:
$\frac{n}{n+1}- \frac{\alpha \beta}{\mu}n= 0$
$\frac{n}{n+1}= \frac{\alpha \beta}{\mu}n$

$n= \frac{\alpha\beta}{\mu}n(n+ 1)$

Obviously n= 0 is a solution to that so one steady state solution is $c= \frac{\alpha}{\mu}$, $n= 0$.

If n is not 0 we can divide by it to get $1= \frac{\alpha\beta}{\mu}(n+ 1)$.
Then $n+ 1= \frac{\mu}{\alpha\beta}$,
$n= \frac{\mu}{\alpha\beta}- 1= \frac{\mu- \alpha\beta}{\alpha\beta}$.

So if n is not 0 then $c= \frac{\alpha}{\mu}$, $n= \frac{\mu}{\alpha\beta}- 1$ is a steady state solution.

But notice the "$\frac{\mu}{\alpha\beta}- 1$". If $\frac{\mu}{\alpha\beta}= 1$ then we are back to the "n= 0" solution, above. If $\frac{\mu}{\alpha\beta}> 1$ the steady state solution for n is positive, if $\frac{\mu}{\alpha\beta}< 1$ the steady state solution for n is negative.

Was this the entire question or was there some application in which n being positive or negative would be important?

## 1. What are the two cases in calculating steady states?

The two cases in calculating steady states are the case of no external inputs and the case of external inputs. In the first case, there are no external factors affecting the system, while in the second case, there are external inputs that can influence the system.

## 2. How do you calculate the steady state in the case of no external inputs?

In the case of no external inputs, the steady state can be calculated by setting the derivative of the system's variables to zero and solving for the values of the variables. This means that the system is in equilibrium and there is no change in the variables over time.

## 3. What is the role of external inputs in calculating steady states?

External inputs can affect the steady state of a system by introducing changes to the variables. These changes can either increase or decrease the values of the variables, thus altering the steady state of the system. External inputs can also cause the system to move away from equilibrium and into a new steady state.

## 4. Can a system have multiple steady states?

Yes, a system can have multiple steady states. This can occur when there are multiple solutions to the equations that describe the system. In this case, the system can exist in different steady states depending on the initial conditions or external inputs.

## 5. How do you determine which steady state a system will reach?

The steady state that a system will reach depends on the initial conditions and external inputs. If the system is in equilibrium, it will remain in that steady state. However, if there are external inputs, the system may reach a new steady state. The specific steady state that the system will reach can be determined by solving the equations that describe the system and considering the initial conditions and external inputs.

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