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I am, at the moment, teaching calculus to students whose majors are, for example, biology, biochemistry, chemistry and geology. The course book is Claudia Neuhauser's "Calculus for biology and medicine", third edition. The course focuses on mathematical modelling and there are also calculation, but very few proofs. My role is that of an øvingslærer; I make exercises and give roughly one fifth of the lectures, in addition to bureaucracy and communication with student assistants (teaching assistants) and students.

I gave a lecture on equilibria and their stability of differential equations; essentially, for an an equation $$ \frac{\text{d}y}{\text{d}t} = g(y) $$ we have that $\hat y$ is an equilibrium point if and only if $g(\hat y) = 0$, and such a point is stable if $g'(\hat y) < 0$ and unstable if $g'(\hat y) > 0$.

The book justifies the stability by linearizing the equation. I tried to explain this during the lecture, but I am far from convinced the explanation was satisfactory. In addition to a small calculation I sketched a graph and a tangent line and explained how this relates to the linearization.

Two questions:

  1. How to best explain linearization in this case and use it to explain the stability conditions?
  2. A very tempting alternative is to bypass the justification and use the time to consider a nice example and do some modelling. How to decide between the mathematical justification and an example, in the context of this type of course?
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Linearization is a technique which is useful for systems of equations, but there are easier (and more powerful) ways of approaching stability in the single variable case.

The major thing that you are getting out of $g(\hat{y})=0$ and the sign of $g'$ is that

  1. the function $g$ is changing sign around $\hat{y}$, and
  2. whether is it going from $+$ to $-$ or $-$ to $+$.

The direction of the sign change determines if the fixed point is stable or not. For example, if $g$ goes from $+$ to $-$, the fixed point is stable as the ODE moves points below $\hat{y}$ up and points above $\hat{y}$ down.

I say that analyzing the change in sign is more powerful than looking at the derivative because of examples such as $g(y)=-y^3$. Here you have a stable fixed point at $y=0$ despite $g'(0)=0$. Further, students can graph $g$ and get an immediate idea about stability.

So if you must use linearization for single variable 1st order ODEs, I suggest explaining it as a way to determine the type of sign change of $g$.

(Linearization gives more benefits for the higher dimensional case since there are other questions, such as "does the system oscillate?", which can be answered by computing eigenvalues.)

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    $\begingroup$ Yeah, I discussed the sign change aspect via a graph before mentioning linearization. $\endgroup$ – Tommi Brander Jan 24 at 19:21
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Tommi, you say

"A very tempting alternative is to bypass the justification and use the time to consider a nice example and do some modelling. How to decide between the mathematical justification and an example, in the context of this type of course?"


I agree with the temptation to eschew more proof(ish) activity. Instead work on problem solving. Here is the rationale for "how to decide". (1) Nothing in your entire question (not just the quote but all of it) says that the students were unhappy with the rationale/justification/proof. Instead you mention your OWN dissatisfaction. Furthermore, you also mention (2) that proofs are not a big part of the course and (3) the audience is softer natural scientists (biologists, etc.) So definitely not mathematicians, but not even engineers and physicists. ALL of these factors (1-3) argue against pursuing a better rationale.

Note to pre-empt deletion: I am answering a solid HALF of the question. This IS AN ANSWER. Furthermore, you could even say someone who only gives a linear approximation lesson is ignoring THIS HALF of the question. And if anything, you could say such an answer luxuriates in the mathematics itself rather than the teaching of it.

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  • $\begingroup$ 1. I was dissatisfied with the way I explained the justification via linear approximation. I am uncertain of how much the students understood of that part, and so do not wish to speculate. $\endgroup$ – Tommi Brander Jan 24 at 19:23
  • $\begingroup$ 2. This is definitely an answer. It would be improved by also tackling the other half, of course. $\endgroup$ – Tommi Brander Jan 24 at 19:23

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