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Theorem of existence and unicity.

  1. Mar 3, 2012 #1
    1. The problem statement, all variables and given/known data
    In the following case: x'(t) = log (3t (x(t) - 2)) does the theorem of existence and unicity guarantee a unique solution for the initial value problem x(3) = 5, justify your answer?


    2. Relevant equations
    x'(t) = log (3t (x(t) - 2))


    3. The attempt at a solution
    Ok what I would do is take the domain of f(x,t) and the domain of ∂f(x,t)/∂x and then find the common interval and if the initial value problem is within this interval then there is a unique solution. However there is x(t) and t in the function so I don't understand how to find the domain?
     
    Last edited: Mar 3, 2012
  2. jcsd
  3. Mar 3, 2012 #2

    HallsofIvy

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    For what x and t is that function, log(3t(x- 2), continuous and differentiable? What is the largest interval containing (3, 5) on which the function is continuous and differentiable?
     
  4. Mar 3, 2012 #3
    Yeh, how would I find that out? Thats what i'm confused about.

    EDIT: Would I just plug numbers into the x and t? Until I got an x and t that came up with 'Math error'? Because if thats the case then when t = 0 the function is discontinuous and when x = 2? Am I right?
    and then the partial derivative is 3t/(3t(x-2)) = 1/(x-2) in which case x cannot equal 2 and t is eqal to all real numbers? So the common intervals are x cannot = 2 and t cannot = 0, with intial values t_0 = 3 and x_0 = 5 which are within the common intervals, therefore there is a unique solution? On a side note, should i have taken the partial derivative with respect to t or x? Or do I take partial derivative with respect to both and then compare all 3 intervals.
     
    Last edited: Mar 3, 2012
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