# Thread: [SOLVED] Question - Stability of First ODE

1. ## [SOLVED] Question - Stability of First ODE

Consider the IVP
$\displaystyle y'=\lambda y$ and $\displaystyle y(0)=y_0=1$

The equation is stable if $\displaystyle Re(\lambda) \leq 0$. What does Re() mean?

2. The DE...

$\displaystyle y^{'}=\lambda\cdot y$ , $\displaystyle y(0)=1$ (1)

... represents the 'free response' of a dynamic system and its solution is...

$\displaystyle y_{free}(t)= e^{\lambda\cdot t}$ (2)

The system is said to be stable if ...

$\displaystyle \lim_{t \rightarrow \infty} y_{free} (t)=0$ (3)

... and that means that it must be $\displaystyle \lambda<0$ ...

Merry Christmas from Italy

$\displaystyle \chi$ $\displaystyle \sigma$

3. Originally Posted by Paperwings
Consider the IVP
$\displaystyle y'=\lambda y$ and $\displaystyle y(0)=y_0=1$

The equation is stable if $\displaystyle Re(\lambda) \leq 0$. What does Re() mean?
The real part of the complex number $\displaystyle \lambda$
For this simple equation, you can solve explicitely. $\displaystyle dy/y= \lambda dt$ so [tex]ln(y)= \lambda t+ C[tex] and $\displaystyle y= C'e^{\lambda t}$ where $\displaystyle C'= e^C$

If $\displaystyle \lambda= a+ bi$, then $\displaystyle Re(\lambda)= a$. $\displaystyle y= C'e^{(a+bi)t}= C'e^{at} e^{bit}= C'e^{at}(cos(bt)+ i sin(bt))$.

The sine and cosine terms are, of course, periodic. If a> 0, that exponential causes y to get larger and larger. Not "stable".

If a< 0, that exponential causes y to go to 0. "Stable".

If a= 0, that exponential is 1 and we just get a periodic solution. "Stable".

4. Originally Posted by HallsofIvy
The real part of the complex number $\displaystyle \lambda$
For this simple equation, you can solve explicitely. $\displaystyle dy/y= \lambda dt$ so [tex]ln(y)= \lambda t+ C[tex] and $\displaystyle y= C'e^{\lambda t}$ where $\displaystyle C'= e^C$

If $\displaystyle \lambda= a+ bi$, then $\displaystyle Re(\lambda)= a$. $\displaystyle y= C'e^{(a+bi)t}= C'e^{at} e^{bit}= C'e^{at}(cos(bt)+ i sin(bt))$.

The sine and cosine terms are, of course, periodic. If a> 0, that exponential causes y to get larger and larger. Not "stable".

If a< 0, that exponential causes y to go to 0. "Stable".

If a= 0, that exponential is 1 and we just get a periodic solution. "Stable".
Two minor observations...

a) dinamic systems are usually characterized in term of real time functions, so that a complex $\displaystyle \lambda$, at least for a first order system, is quite unusual...

b) a dinamic system is considered 'stable' if its 'free response' tends to 0 if t tends to infinity, or, in more prosaic words, if it has 'no memory of the past'. On the basis of that for a stable first order dinamic system must be tightly $\displaystyle \lambda<0$...

Merry Christmas from Italy

$\displaystyle \chi$ $\displaystyle \sigma$

5. Originally Posted by chisigma
Two minor observations...

a) dinamic systems are usually characterized in term of real time functions, so that a complex $\displaystyle \lambda$, at least for a first order system, is quite unusual...

b) a dinamic system is considered 'stable' if its 'free response' tends to 0 if t tends to infinity, or, in more prosaic words, if it has 'no memory of the past'. On the basis of that for a stable first order dinamic system must be tightly $\displaystyle \lambda<0$...

Merry Christmas from Italy

$\displaystyle \chi$ $\displaystyle \sigma$
I think that most use the word asymptotely stable for $\displaystyle {\it Re} \lambda < 0$

6. In N.Balabanian, T.A. Bickart Electrical Network Theory, John Wiley and Sons, 1969 the stability for linear elecrical network is trated at the chapter 9.4. At pag. 670 we have the definitionm of stability and the fundamental 'theorem 1'...

At the end of pag. 671 we have another 'fundamental theorem': the 'theorem 2'...

The concept of asymptotic stability is trated in a succesive chapter which is dedicated to the non linear electrical network. The basic concept is however the same...

Merry Christmas from Italy

$\displaystyle \chi$ $\displaystyle \sigma$