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Homework Help: Tail of an Almost Linear System

  1. Aug 13, 2007 #1
    1. The problem statement, all variables and given/known data
    This isn't really homework help, I'm just trying to see if there is a "proper" way of doing this.


    [tex]\frac{dx}{dt} = x - y^2[/tex]

    [tex]\frac{dy}{dt} = x - 2y + x^2[/tex]

    show that the system is almost linear.

    2. Relevant equations

    3. The attempt at a solution

    [tex]\left(\begin{array}{c}x&y\end{array}\right)' = \left(\begin{array}{cc}1&0\\1&-2\end{array}\right) \left(\begin{array}{c}x&y\end{array}\right) + \left(\begin{array}{c}-y^2&x^2\end{array}\right)[/tex],

    where [tex]\left(\begin{array}{c}-y^2&x^2\end{array}\right)[/tex],
    is the tail.

    I was taught that the way you get the tail, is just separate the parts that are not just [tex]x[/tex] or [tex]y[/tex]. But there has to be a better, or I should say correct way of finding the tail. Just separating the equation seems too simple. And besides, I tried this method for:

    [tex]\frac{dx}{dt} = (1 + x) \sin{y}[/tex]

    [tex]\frac{dy}{dt} = 1 - x - \cos{y}[/tex]

    And it didn't work.
  2. jcsd
  3. Aug 13, 2007 #2


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    First, never despise a method because it is "too easy"!!

    Second, to "show that the system is almost linear" you need to show that the definition of "almost linear" is satisfied. What is that?

    Third, to handle non-polynomial problems, use Taylor's series.
    The Taylor's series for sin(y) is, as I am sure you know,
    [tex]\sum_{i=0}^\infty \frac{(-1)^i y^{2i+1}}{(2i+1)!}[/tex]
    and the Taylor's series for cos(y) is
    [tex]\sum_{i=0}^\infty \frac{(-1)^i y^{2i)}{(2i)!}[/tex]
    Use that to convert (1+x) sin(y) and 1- x- cos(y) to power series and separate the linear and non-linear parts just as you did above.
  4. Aug 13, 2007 #3
    1) Heh, I don't despise "easy ways", it's just that I don't like to rely on tricks, without understanding the basis for them.

    2) I know a few ways to show an ALS (almost linear system):

    If the tail is [tex]g(x)[/tex],
    then a system is ALS, if

    [tex]\frac{||g(x)||}{r} \rightarrow 0[/tex]


    [tex]r \rightarrow 0[/tex]

    then it is ALS.

    Or if each function is at least twice differentiable, it is an ALS.

    3) THIS is what I was looking for. I KNEW there had to be a REAL method to finding the tail.

    Thanks HallsofIvy!!!
  5. Aug 13, 2007 #4


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    Do you understand that this makes no sense at all since you have not said what "r" is with a function of x?

    [/quote]Or if each function is at least twice differentiable, it is an ALS.[/quote]
    Okay, now this makes it easy!

    Don't you just love Taylor's series? The man's birthday should be national holiday!
  6. Aug 14, 2007 #5
    I didn't write what r is because I know what it is. But anyway, here it is:

    [tex]r = ||x||[/tex]

  7. Aug 14, 2007 #6
    Another quick question. I found this I think on some site:

    [tex]F(x,y) = F(x_0,y_0) + F_x(x_0,y_0)(x - x_0) + F_y(x_0,y_0)(y - y_0) + \eta_1(x,y)[/tex]

    [tex]G(x,y) = G(x_0,y_0) + G_x(x_0,y_0)(x - x_0) + G_y(x_0,y_0)(y - y_0) + \eta_2(x,y)[/tex]

    [tex]\frac{d}{dt}\left(\begin{array}{c}x - x_0&y - y_0\end{array}\right) = \left(\begin{array}{cc}F_x(x_0,y_0)&F_y(x_0,y_0)\\G_x(x_0,y_0)&F_y(x_0,y_0)\end{array}\right)\left(\begin{array}{c}x - x_0&y - y_0\end{array}\right) + \left(\begin{array}{c}\eta_1(x,y)&\eta_2(x,y)\end{array}\right)[/tex]

    So for my problem, letting

    [tex]F(x,y) = \frac{dx}{dt} = x - y^2[/tex]


    [tex]G(x,y) = \frac{dy}{dt} = x - 2y + x^2[/tex]

    [tex]F_x = 1..............F_y = -2y[/tex]

    [tex]G_x = 1 + 2x..............G_y = -2[/tex]


    [tex]\frac{d}{dt}\left(\begin{array}{c}x - x_0&y - y_0\end{array}\right) = \left(\begin{array}{cc}1&-2y\\1 + 2x&-2\end{array}\right)\left(\begin{array}{c}x - x_0&y - y_0\end{array}\right) + \left(\begin{array}{c}\eta_1(x,y)&\eta_2(x,y)\end{array}\right)[/tex]

    What I can't figure out is how to find the tail...
  8. Aug 14, 2007 #7


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    You have an error here:
    [tex]\frac{d}{dt}\left(\begin{array}{c}x - x_0&y - y_0\end{array}\right) = \left(\begin{array}{cc}1&-2y\\1 + 2x&-2\end{array}\right)\left(\begin{array}{c}x - x_0&y - y_0\end{array}\right) + \left(\begin{array}{c}\eta_1(x,y)&\eta_2(x,y)\end{ array}\right)[/tex]
    You don't have the "derivative" matrix evaluated at [itex]x_0[/itex] and [itex]y_0[/itex].
    It should be
    [tex]\frac{d}{dt}\left(\begin{array}{c}x - x_0&y - y_0\end{array}\right) = \left(\begin{array}{cc}1&-2y_0\\1 + 2x_0&-2\end{array}\right)\left(\begin{array}{c}x - x_0&y - y_0\end{array}\right) + \left(\begin{array}{c}\eta_1(x,y)&\eta_2(x,y)\end{ array}\right)[/tex]

    Use the "two variable" tangent plane formula you have to determine the "linear part" of the F and G at [itex](x_0,y_0)[/itex]. The tail is just the difference between F and its tangent plane formula and G and its tangent plane formula. In other words, multiply out those 2 matrices
    [tex]\left(\begin{array}{cc}1&-2y_0\\1 + 2x_0&-2\end{array}\right)\left(\begin{array}{c}x - x_0&y - y_0\end{array}\right)[/tex]
    and subtract from
    [tex]\left(\begin{array}{c}x- y^2\\x-2y+ x^2\end{array}\right)[/itex]
    Last edited by a moderator: Aug 14, 2007
  9. Aug 14, 2007 #8
    Thank you so much! Eveything is crystal clear now!
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