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Math Help - Existence and Uniqueness Theorem

  1. #1
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    Post Existence and Uniqueness Theorem

    Greetings I have a question about the basic existence and uniqueness theorem for nth order homogeneous linear differential equations on a certain interval.If f(x_0)=0,f'(x_0)=0,...,f^{(n-1)}(x_0) where x, then  f(x)=0 on the whole interval. I did not understand the logic behind this, can someone explain it to me? I would be grateful. I think this is also valid for first order differential equations if  f(x_0)=0 then  f(x)=0 if I am not mistaken?

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    Last edited by JohnDoe; October 17th 2011 at 01:44 PM. Reason: Latex
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    Re: Existence and Uniqueness Theorem

    For the linear nth order homogeneous differential equation on an interval, you know that if the coefficients are continuous functions, and the coefficient of the highest-order derivative is nonzero on the interval, then there exists a unique solution to the initial value problem. Does f(x) = 0 solve the initial value problem? What could you say then?
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  3. #3
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    Re: Existence and Uniqueness Theorem

    Thanks for your reply.The attached pdf is from Shepley, Ross Differential Equations 3rd Edition and the theorem I did not understand is theorem 4.1 and especially its corollary. Can you shed some light on that ?
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  4. #4
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    Re: Existence and Uniqueness Theorem

    What exactly is your confusion? The theorem and it corollary state that the nth order linear differential equation, with continuous coefficients in some neighborhood of x_0, has unique solution to the "initial value" problem- that is finding a function, f, that satisfies the given differential equation and has specified values for the function and its derivatives, up to the n-1 derivative, at x_0. Specifically, the "theorem" says there is a solution and the corollary says there is only one solution.
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  5. #5
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    Re: Existence and Uniqueness Theorem

    I did not understand the corollory in the special case f(x_0)=0 ... f^{n-1}(x_0)=0. Why does it imply that f(x)=0 on the whole interval?
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    Re: Existence and Uniqueness Theorem

    First answer this question: is f(x) = 0 a solution of the initial value problem?
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  7. #7
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    Re: Existence and Uniqueness Theorem

    F is assumed to be the solution of the nth order homogeneous differential equation. The corollary states that if f(x_0)=0 and x_0 belongs to aforementioned certain interval I, then f(x)=0 for every x that is a member of this interval. How is this possible ?
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    Re: Existence and Uniqueness Theorem

    Quote Originally Posted by JohnDoe View Post
    F is assumed to be the solution of the nth order homogeneous differential equation. The corollary states that if f(x_0)=0 and x_0 belongs to aforementioned certain interval I, then f(x)=0 for every x that is a member of this interval. How is this possible ?
    You don't have to assume anything. Tell me this:

    1. Does f(x) = 0 satisfy the differential equation?
    2. Does it satisfy the initial conditions?
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    Re: Existence and Uniqueness Theorem

    Quote Originally Posted by Ackbeet View Post
    You don't have to assume anything. Tell me this:

    1. Does f(x) = 0 satisfy the differential equation?
    2. Does it satisfy the initial conditions?
    It satisfies the differential equation and the initial conditions.
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    Re: Existence and Uniqueness Theorem

    Quote Originally Posted by JohnDoe View Post
    It satisfies the differential equation and the initial conditions.
    Right. So that tells you that f(x) = 0 is a solution. The question is, is it the only solution? Now, you tell me: does the existence and uniqueness theorem apply to this problem?
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  11. #11
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    Re: Existence and Uniqueness Theorem

    Yes it does apply then I think I answered my own question. I think it is the trivial solution that is unique if I am not mistaken ?
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    Re: Existence and Uniqueness Theorem

    Quote Originally Posted by JohnDoe View Post
    Yes it does apply then I think I answered my own question.
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    Re: Existence and Uniqueness Theorem

    Quote Originally Posted by JohnDoe View Post
    Yes it does apply then I think I answered my own question. I think it is the trivial solution that is unique if I am not mistaken ?
    Your wording implies that you think this solution is unique because it is the trivial solution. In fact, for the conditions given on the coefficients, there exist a unique solution with f(x_0), f'(x_0), ..., f^{n-1}(x_0) any given numbers. Since the trival function, f(x)= 0, satisfies both the differential equation and the conditions f(x_0)= 0, f'(x_0)= 0, ..., f^{n-1}(x_0)= 0, it is the only solution to that problem.

    By the way, this is not true for "boundary value problems". The functions f(x)= C sin(x) satisfies the differential equation \frac{d^2f}{dt^2}= -f and the conditions y(0)= 0, y(\pi)= 0 for any value of C.
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  14. #14
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    Re: Existence and Uniqueness Theorem

    I concluded that this is only valid for an initial value problem and understood what is told in the book many thanks again for your help.
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