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Math Help - Predict where I.V.P has a solution by using theorem(s)

  1. #1
    Member Mollier's Avatar
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    Predict where I.V.P has a solution by using theorem(s)

    Hi,

    the problem:

    Use each of the three theorems in this section to predict where this problem has a solution, and then solve the problem explicitly, comparing the theory to the fact:

    x'=x^2, x(0)=1.

    As stated in the problem, there are three theorems in my book dealing with existence and uniqueness of I.V.Ps.
    The first one is a pure existence theorem and says;

    If f is continuous in a rectangle R centered at (t_0,x_0), say

    R = \{(t,x): |t-t_0|\leq \alpha, |x-x_0|\leq \beta\}

    then the initial-value problem has a solution x(t) for |t-t_0|\leq min(\alpha,\beta/M), where M is the maximum of |f(t,x)| in the rectangle R.

    Attempt:
    For this problem, the rectangle is;

    R = \{(t,x): |t|\leq \alpha, |x-1|\leq \beta\}.

    Hence f(t,x)=x^2 is bounded by, (\beta+1)^2=M.

    Since f is continuous, the I.V.P has a solution for |t|\leq min(\alpha,\beta/(\beta+1)^2).

    Now, \beta/(\beta+1)^2 has a maximum when \beta=1, so,

    |t|\leq min(\alpha,1/4).

    Assuming \alpha\geq 1/4, we have that

    |t| \leq 1/4.

    --------------------------------------------------------------------------------------

    The explicit solutions is,

    x(t) = \frac{1}{1-t},

    which is not defined if t\neq 0.
    I can't say my prediction is very good...

    Any suggestions are welcome, thanks.
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    I've looked through your process and all is correct. However, why do you say the following?


    Quote Originally Posted by Mollier View Post
    The explicit solutions is, x(t) = \frac{1}{1-t}, which is not defined if t\neq 0.

    The function x(t)=1/(1-t) is not defined at t=1 but it is defined in |t-0|\leq 1/4 and satifies the I.V.P.
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  3. #3
    Member Mollier's Avatar
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    Quote Originally Posted by FernandoRevilla View Post
    I've looked through your process and all is correct. However, why do you say the following?





    The function x(t)=1/(1-t) is not defined at t=1 but it is defined in |t-0|\leq 1/4 and satifies the I.V.P.
    Hi,

    thank you for taking the time to look through my attempt.

    I was hoping that the theorem would give me a stronger indication that something is wrong when t=1. I do not know why I would expect such a thing though.

    Again, thanks.
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