Science:Math Exam Resources/Courses/MATH215/April 2014/Question 01 (c)

MATH215 April 2014

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Question 01 (c)
Consider ${\frac {dy}{dt}}=y(1-y)(2-y)$. For what possible initial conditions $y(0)$ does $\lim _{t\to \infty }y(t)=1$?

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
You do not need to solve the equation in order to answer this question.

Checking a solution serves two purposes: helping you if, after having used the hint, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. The key is to realize that the question is asking for qualitative information which can be inferred by sketching trajectories for the ODE. In particular, it is not necessary to solve this ODE, which would require much more work. We first observe that the ODE has equilibrium (fixed) points at y = 0, y = 1, and y = 2. This means that any trajectories starting at 0, 1, or 2, will remain at that constant value for all time. In particular, we see that the initial condition y(0) = 1 satisfies $\lim _{t\to \infty }y(t)=1$. To sketch trajectories, we record the following signs of ${\frac {dy}{dt}}$ in the four regions $(-\infty ,0)\cup (0,1)\cup (1,2)\cup (2,\infty )$ separated by the fixed points: $y<0$ $y=0$ $0<y<1$ $y=1$ $1<y<2$ $y=2$ $y>2$ ${\frac {dy}{dt}}<0$ ${\frac {dy}{dt}}=0$ ${\frac {dy}{dt}}>0$ ${\frac {dy}{dt}}=0$ ${\frac {dy}{dt}}<0$ ${\frac {dy}{dt}}=0$ ${\frac {dy}{dt}}>0$ From this, we draw the following conclusions: Trajectories starting at y(0) < 0 will monotonically decrease without bound. In this case, $\lim _{t\to \infty }y(t)=-\infty$. Trajectories starting at 0 < y(0) < 1 will increase monotonically towards the critical point 1. In this case, $\lim _{t\to \infty }y(t)=1$. Trajectories starting at 1 < y(0) < 2 will decrease monotonically towards the critical point 1. In this case, $\lim _{t\to \infty }y(t)=1$. Trajectories starting at y(0) > 2 will increase monotonically without bound. In this case, $\lim _{t\to \infty }y(t)=+\infty$. Therefore, the only initial conditions for which $\lim _{t\to \infty }y(t)=1$ are $0<y(0)<2$.

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