Steady-State Stability and the Second Derivative

[The Head]

Hi. I have a question about steady-state stability and the second derivative test. I have been reading about it in a book on mathematical modeling, and the section concerns differential equations. I believe this forum is more appropriate than "General Math," but let me know if it is not.

From what I was reading, my understanding is that if the second derivative is positive, then at a point where the first derivative is zero, then the system is unstable. Similarly, if the second derivative is negative, it is stable. Can someone please provide me with a reason why this is the case (either mathematically, graphically). Why would a function such as f(x)=x^2 (whose second derivative is positive) be unstable at its steady-state points?

What confuses me is when I think about an example of energy in physics. Take the graph of f(x) = x^2. Moving slightly away from x = 0 (because f'(x)=0 at x=o), the system easily slides back down to the point of equilibrium. Whereas with f(x)= -x^2, a slight movement to either side causes the system to fall away from these points.

Thanks!

 

[gtoguy97]

The reason why a function such as f(x)=x^2 is unstable at its steady-state points is because its second derivative is positive. This means that the slope of the function is increasing, which implies that the function is becoming steeper and thus is more likely to be pushed away from its steady-state point. On the other hand, a function such as f(x)=−x^2 has a negative second derivative, which implies that it is getting less steep as you move away from the steady-state point and thus is more likely to remain at or near the point. Graphically, this can be seen by looking at the graph of the two functions. For the function f(x)=x^2, the graph is an upward-opening parabola, whereas for the function f(x)=−x^2, the graph is a downward-opening parabola.