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Generalizing separation technique

  1. May 2, 2008 #1
    [tex]
    \left[\begin{array}{c}
    y'_1(x) \\ \vdots \\ y'_n(x) \\
    \end{array}\right]
    =
    \left[\begin{array}{ccc}
    f_{11}(y) & \cdots & f_{1n}(y) \\
    \vdots & & \vdots \\
    f_{n1}(y) & \cdots & f_{nn}(y) \\
    \end{array}\right]
    \left[\begin{array}{c}
    g_1(x) \\ \vdots \\ g_n(x) \\
    \end{array}\right]
    [/tex]

    If n=1, then this can be solved with the separation technique. Suppose n>1 and that [itex]f[/itex] is invertible. Could the separation technique be generalized to give some explicit formula for solution y(x)? I tried without success. Anyone dealt with problems like this ever?
     
  2. jcsd
  3. May 4, 2008 #2

    lurflurf

    User Avatar
    Homework Helper

    As written I cannot say I do not know what y on the right hand side means.

    If you mean, or have use for
    [tex]\left[\begin{array}{c}y'_1(x) \\ \vdots \\ y'_n(x) \\\end{array}\right]=\left[\begin{array}{ccc}f_{11}(x) & \cdots & f_{1n}(x) \\\vdots & & \vdots \\f_{n1}(x) & \cdots & f_{nn}(x) \\\end{array}\right]\left[\begin{array}{c}y_1(x) \\ \vdots \\ y_n(x) \\\end{array}\right][/tex]

    Then yes this is called the method of Peano-Baker.
    A good reference is Ince Ordinary differential equations.
    The idea is
    y'=Ay
    A and y being functions of x
    A a linear operator y a vector
    let I be integration say from 0 to x
    suppose when x=0 y=c
    y=c+IAy
    y'=Ac+AIAy
    y=c+IAc+IAIAy
    y=c+IAc+IAIAc+IAIAIAy
    leting the sum go to infinity we have a geometric series
    y=((IA)^0+(IA)^1+(IA)^2+...)c
    or
    y={[(IA)^0-(IA)^1]^-1}c
    we can prove this works.
    It can be used in principle, but is more usefule to prove existance-uniqueness since it is highly impractical.
    we may think of this as a generalization of
    y'=Ay,A'=0,y(0)=c->y=exp(Ax)c
    ie
    y'=Ay,y(0)=c->y={[(IA)^0-(IA)^1]^-1}c
    in both cases the "explicit" formula as written is not enough we often want to know more
     
  4. Aug 19, 2008 #3
    I see. The equation

    [tex]
    y'(x) = A(x)y(x)
    [/tex]

    is equivalent with

    [tex]
    y(x) = y(0) + \int\limits_0^x du\; A(u)y(u),
    [/tex]

    so the iteration attempt

    [tex]
    y_0(x) = y(0)
    [/tex]

    [tex]
    y_{n+1}(x) = y(0) + \int\limits_0^x du\; A(u)y_n(u)
    [/tex]

    seems natural. If the iterations converge towards the solution, it follows that the solution is given by the series

    [tex]
    y(x) \;=\; \Big(1 \;+\; \int\limits_0^x du\; A(u) \;+\; \int\limits_0^x du\; \int\limits_0^u du'\; A(u)A(u') \;+\; \int\limits_0^x du\; \int\limits_0^u du'\; \int\limits_0^{u'} du''\; A(u) A(u') A(u'') \;+\; \cdots\Big)y(0).
    [/tex]

    But I didn't quite get the geometric series part. From equation

    [tex]
    S = 1 + q + q^2 + q^3 + \cdots
    [/tex]

    follows

    [tex]
    qS = q \;+\; q^2 \;+\; q^3 \;+\; \cdots = S - 1\quad\implies\quad S = \frac{1}{1-q},
    [/tex]

    but how do you do the same with the iterated integrals? If we define the time evolution operator to be

    [tex]
    U(x) \;=\; 1 \;+\; \int\limits_0^x du\; A(u) \;+\; \int\limits_0^x du\;\int\limits_0^u du'\; A(u) A(u') \;+\; \cdots,
    [/tex]

    then the calculation

    [tex]
    \int\limits_0^{x'} dx\; A(x)U(x) \;=\; \int\limits_0^{x'} dx\; A(x) \;+\; \int\limits_0^{x'} dx\;\int\limits_0^x du\; A(x) A(u) \;+\; \cdots \;=\; U(x') \;-\; 1
    [/tex]

    doesn't lead anywhere.
     
    Last edited: Aug 19, 2008
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