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Homework Help: Existance of unique solutions for differential equation

  1. Oct 7, 2006 #1
    Hi guys,

    x'=A(t)x(t)+B(x,t)u(t) (1)

    x, t belongs to R^n and R+ respectively. u(t) is a function of t.

    I am required to find the sufficient conditions for A(t), f(x), B(x,t) and u(t)
    for the (1) to have a existence of a unique solution.

    In the lecture notes, the prof lists 4 conditions to be sufficient conditions for the
    existence of a unique solution for x'=f(x,t):

    1. for a fixed t, f(x) is continous.
    2. There exists a set S contained in R+ containing at most a finite
    number of points per unit interval. S will depend on f(x,t) and will denote
    possible discontinuity points. With this set, for x is fixed, f(t) will be discontinuous only when t belongs to S.

    3. If x(t) is continous, then f(x,t) is piecewise continous in t

    4.satisifies a global Lipschitz condition.


    Thus my solution is as follows:

    1. for a fixed t: B(x,t) are continuous.

    2. for a fixed x, A(t), u(t), B(x,t) are discontinuous only in S

    3. A(t)x(t)+B(x,t)u(t) is picewise continuous in t.

    4. I derive the global Lipschitz condition:


    ||A(x1-x2)+U(B(x1)-B(x2))||<=|A|*||(x1-x2)||+|u|*||B(x1)-B(x2)||

    so if ||B(x1)-B(x2)||<=K(t)||x1-x2|| then ||A(x1-x2)+U(B(x1)-B(x2))||<=K1||x1-x2||

    with k1=max(|A|, |k*u|).

    So the fourth condition will be B(x,t) satisfies the Lipschitz condition.


    I am not sure if the conditions can be narrower? can you please verify my solution.

    Thanks
     
  2. jcsd
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