# Conceptual understanding

1. Jun 15, 2005

The function y=0 is an equilibrium solution of the natural decay equation and the natural growth equation $$\frac{dy}{dt}=ky\\k>0$$.

An equilibrium solution is stable if solutions that begin sufficiently close to the equilibrium soltuion tend toward that equilibrium solution as $$t\rightarrow\infty$$

What does this mean?

2. Jun 16, 2005

### TenaliRaman

Lets consider
dy/dt = ky for all k's for now

Now,
Any constant solution to a differential equation is an equilibrium solution.
I am sure you would agree that y=0 is an equilibrium solution of the given equation.

Now the above equation (as stated by me above), has another solution,
y = e^(kt) ... (1)

So any **system** that adheres to this equation(1) satisfies the differential equation.
Now if a system has the k factor such that k<0
then as t->oo, y -> 0
So this system will follow the equilibrium solution after some period of time. Such systems are called stable systems.

Now if a system has the k factor such that k>0
then as t->oo, y->oo
So this system is unbounded and becomes completely unstable after some period of time.

I am sorry to have introduced the word system, which may be completely alien to you right now. You can think of a system as a machine that takes input and gives certain output and the equation y=e^(kt) as a characteristic equation that describes how much residual noise inputs exist internally in the machine.

Now if i give an input to this machine and if the machine has some residual noise inputs, then it will give me incorrect output.
i.e lets say the machine is supposed to calculate the square of a number. if i give input x to machine then it should ideally give me x^2 but if it has some residual noise, then it will give me (x+some_error)^2

Now you can see how stability and unstability can be realised here. If the above squaring system has the k factor < 0 , then as time progresses the error factor will tend to zero and the machine would give proper output. Thus this system will become stable.

However if it has k factor > 0, then as time progresses, the error factor will tend to infinity and at some point the output will not have any relation to input you give. Thus this system becomes unstable.

I hope this clears some of your doubts if not all.

-- AI

Last edited: Jun 16, 2005
3. Jun 16, 2005

### HallsofIvy

Staff Emeritus
With reference to the "decay" problem, the fact that 0 is a stable equilibrium means that if you don't have any radioactive material to start with you never will have- it won't suddenly appear. In fact, if you start with a small amount of radioactive material it will slowly disappear so, eventually, you wind up with none!

An equilibrium solution is one that doesn't change: dy/dt= 0.

Imagine holding a meter stick loosely by one end and letting it hang straight down. The only forces on it are your hand and gravity and they cancel- there is no force to change the position so that is an "equilibrium" position. If something bumps the meter stick very slightly, it will swing a little bit but eventually, due to friction, stop- again hanging straight down. That's not only an "equilibrium" position is it is a "stable equilibrium".

Now imagine balancing the meter stick upright on the palm of your hand. If you can get it exactly upright, your palm is giving it a force upward, gravity a force downward and the two cancel- again there is no motion. This is also an "equilibrium" position. But if your hand shakes a tiny amount, or if a breeze moves the meter stick slightly, there will be an unbalanced force and the meter stick will fall- not going back to the equilibrium position- this is an "unstable equilibrium".

4. Jun 16, 2005