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Math Help - [SOLVED] Question - Stability of First ODE

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    [SOLVED] Question - Stability of First ODE

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

    The equation is stable if  Re(\lambda) \leq 0. What does Re() mean?
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    MHF Contributor chisigma's Avatar
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    The DE...

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

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

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

    The system is said to be stable if ...

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

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



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  3. #3
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    Quote Originally Posted by Paperwings View Post
    Consider the IVP
     y'=\lambda y and  y(0)=y_0=1

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

    If \lambda= a+ bi, then Re(\lambda)= a. 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".
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    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by HallsofIvy View Post
    The real part of the complex number \lambda
    For this simple equation, you can solve explicitely. dy/y= \lambda dt so [tex]ln(y)= \lambda t+ C[tex] and y= C'e^{\lambda t} where C'= e^C

    If \lambda= a+ bi, then Re(\lambda)= a. 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 \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 \lambda<0...



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    Quote Originally Posted by chisigma View Post
    Two minor observations...

    a) dinamic systems are usually characterized in term of real time functions, so that a complex \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 \lambda<0...



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    I think that most use the word asymptotely stable for {\it Re} \lambda < 0
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    MHF Contributor chisigma's Avatar
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    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...



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