What is the maximum yield for a steady state solution?

In summary, the conversation discusses the concept of a "steady state" solution to a differential equation and how to find the maximum value for a certain variable. The conversation also includes an example of how to solve for the maximum value using derivatives. The final answer for the maximum value is E*= 2/3.
  • #1
mt91
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I've got a question here which I'm really unsure what the wording is asking me to do, I've calculated (5), so worked out the steady states. However question 6 has really thrown me off with it's wording, any help would be appreciated.
 
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  • #2
A "steady state" solution to a differential equation is a constant solution. Since the derivatives of a constant are 0, for a "steady state" solution du/dt= 0. For this problem that means u(1- u)(1+ u)- Eu= 0. Factoring out "u", u[(1- u)(1+u)- E]= u[1- u^2- E]= 0. Either u= 0 or u^2= 1- E so the "steady state" solutions are u*(E)= 0 and u*(E)= sqr{1- E}. Is that what you got for (5)?

Problem (6) asks about the "yield" which is defined as Y= Eu*(E) where u* is a "steady state solution". Since the steady state solutions are u*(E)= 0 and u*(E)= sqrt(1- E), either Y= E(0)= 0 or Y= E sqrt(1- E)= sqr(E^2- E^3). The first is identically equal to 0 so cannot be maximized. To find the maximum of the second, set the derivative equal to 0.

Y= sqrt(E^2- E^3)= (E^2- E^3)^(1/2). Y'= (1/2)(E^2- E^3)^(-1/2)(2E- 3E^2)= 0. That is equivalent to 3E^2- 2E= E(3E- 2)= 0 so E= 0 or E= 2/3. Again, E= 0 cannot give a maximum (it gives a minimum) so E*= 2/3.
 
  • #3
A "steady state" solution to a differential equation is a constant solution. Since the derivatives of a constant are 0, for a "steady state" solution $\frac{du}{dt}= 0$. For this problem that means $u(1- u)(1+ u)- Eu= 0$. Factoring out "u", $u[(1- u)(1+u)- E]= u[1- u^2- E]= 0$. Either $u= 0$ or $u^2= 1- E$ so the "steady state" solutions are $u^*(E)= 0$ and $u^*(E)= \sqrt{1- E}$. Is that what you got for (5)?

Problem (6) asks about the "yield" which is defined as $Y= Eu^*(E)$ where u* is a "steady state solution". Since the steady state solutions are $u^*(E)= 0$ and $u^*(E)= \sqrt(1- E)$, either $Y= E(0)= 0$ or $Y= E \sqrt(1- E)= \sqrt(E^2- E^3)$. The first is identically equal to 0 so cannot be maximized. To find the maximum of the second, set the derivative equal to 0.

$Y= \sqrt(E^2- E^3)= (E^2- E^3)^{1/2}$. $Y'= (1/2)(E^2- E^3)^(-1/2)(2E- 3E^2)= 0$. That is equivalent to $3E^2- 2E= E(3E- 2)= 0$ so $E= 0$ or $E= 2/3$. Again, $E= 0$ cannot give a maximum (it gives a minimum) so $E^*= 2/3$.
 

1. What is the concept of steady state understanding?

Steady state understanding refers to the idea that a system or process is in a state of equilibrium, where the inputs and outputs are balanced and the system remains relatively constant over time.

2. How does steady state understanding apply to scientific research?

In scientific research, steady state understanding is used to study and analyze complex systems by observing their behavior over time and identifying patterns and relationships. It allows scientists to make predictions and understand how a system will respond to changes.

3. What are some examples of steady state understanding in scientific fields?

Examples of steady state understanding in scientific fields include studying the carbon cycle in ecology, analyzing the equilibrium of chemical reactions in chemistry, and observing the balance of forces in physics.

4. How is steady state understanding different from dynamic equilibrium?

While both steady state understanding and dynamic equilibrium refer to a state of balance, steady state understanding focuses on the overall behavior of a system over time, while dynamic equilibrium looks at the instantaneous balance between inputs and outputs.

5. What are the benefits of using steady state understanding in scientific research?

Steady state understanding allows scientists to better understand complex systems and their behavior, make predictions, and identify potential areas of change or instability. It also helps to simplify and model complex systems, making them easier to study and analyze.

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