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Math Help - Existence and Uniqueness

  1. #1
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    Existence and Uniqueness

    Determine the existence and uniqueness of solutions for each of the following. Indicate the interval (for t) where there exists a unique solution (if possible). DO NOT SOLVE THE DIFF EQ, JUST DETERMINE WHETHER A UNIQUE SOLUTION EXISTS AND EXPLAIN HOW YOU KNOW. State the results that you use to get to your conclusions on each one.

    1.) \displaystyle \frac{dy}{dt}=3y^{2/3}, y(0)=1

    2.) \displaystyle (t-1)y^{(4)}+(t+1)y''+\tan t y=0, y(0)=y'(0)=y''(0)=y'''(0)=0

    3.) \displaystyle y''+25y=0, y(0)=0, y(\pi)=0



    *** Well, I'm not sure, but I think the 2nd is unique.. not sure how to determine without solving.
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  2. #2
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    Quote Originally Posted by Ideasman View Post
    Determine the existence and uniqueness of solutions for each of the following. Indicate the interval (for t) where there exists a unique solution (if possible). DO NOT SOLVE THE DIFF EQ, JUST DETERMINE WHETHER A UNIQUE SOLUTION EXISTS AND EXPLAIN HOW YOU KNOW. State the results that you use to get to your conclusions on each one.

    1.) \displaystyle \frac{dy}{dt}=3y^{2/3}, y(0)=1

    2.) \displaystyle (t-1)y^{(4)}+(t+1)y''+\tan t y=0, y(0)=y'(0)=y''(0)=y'''(0)=0

    3.) \displaystyle y''+25y=0, y(0)=0, y(\pi)=0

    .
    You see the existence/uniqueness theorem only applies to linear differencial equations. If you look at #2 it is linear. Well, not exactly we need to divide by (t-1) but when we do that we need to be sure that this is not zero, i.e. t\not = 1. And we require that \tan t to be defined. In that case we are are on the interval (-\pi/2,1) then we are guarenntted to have a unique solution.

    If you look at equation #3 this is a linear equation BUT it is not an initival value problem. So the theorem does not apply.*


    *)This is actually a Sturm-Louiville boundary value problem which in fact has no solutions.
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  3. #3
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    Quote Originally Posted by ThePerfectHacker View Post
    You see the existence/uniqueness theorem only applies to linear differencial equations. If you look at #2 it is linear. Well, not exactly we need to divide by (t-1) but when we do that we need to be sure that this is not zero, i.e. t\not = 1. And we require that \tan t to be defined. In that case we are are on the interval (-\pi/2,1) then we are guarenntted to have a unique solution.

    If you look at equation #3 this is a linear equation BUT it is not an initival value problem. So the theorem does not apply.*


    *)This is actually a Sturm-Louiville boundary value problem which in fact has no solutions.
    Okay, I'm confused.

    For #1, this is non-linear, and therefore it has no unique solution.

    For #2, first off I am assuming its tan(t)*y instead of tan(ty) (my teacher should really learn to use parentheses!). Now, I don't see why we divide by (t - 1). It's linear without dividing by (t - 1). And how can you be so sure that t does not equal 1? Also, we know tan(t) is continuous between -Pi/2 and Pi/2. So, according to the unique/existence theorem, if the interval contains 0, it has to have a unique solution..but on this interval? I don't know what you mean by (-Pi/2,1).

    I understand your #3, and I'm impressed you knew WHAT type of boundary problem it was!
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  4. #4
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    Quote Originally Posted by Ideasman View Post
    For #1, this is non-linear, and therefore it has no unique solution.
    We cannot say that. My point is that the existence/uniqueness theorem was only proven for linear differencial equations. When we have non-linear differencial equations we have no idea whether it works or not without solving it.
    For #2, first off I am assuming its tan(t)*y instead of tan(ty) (my teacher should really learn to use parentheses!). Now, I don't see why we divide by (t - 1). It's linear without dividing by (t - 1). And how can you be so sure that t does not equal 1? Also, we know tan(t) is continuous between -Pi/2 and Pi/2. So, according to the unique/existence theorem, if the interval contains 0, it has to have a unique solution..but on this interval? I don't know what you mean by (-Pi/2,1).
    The theorem says that given,
    y^{(n)}+f_{n-1}(x)y^{(n-1)}+...+f_1(x)y'+f_0(x)y = g(x)
    With y(x_0)=y_0 , y'(x_0)=y'_0 , ... , y^{(n)} (x_0)=y^{(n)} _0.
    Where f_{n-1}(x),...,f_0(x),g(x) are continous functions on some open interval (a,b) and where x_0\in (a,b) then there exists exactly one solution to that differencial equation.

    So given (after division),
    y^{(4)} + \frac{t+1}{t-1} y'' + \tan t \ y = 0 \mbox{ and }0=y(0)=y'(0)=y''(0)=y'''(0)=y''''(0).
    We must work on the interval containing 0 because if you look at the theorem the initial point for the equation is contained in the interval. But that is not it. We also require that \frac{t+1}{t-1} and \tan t be continous. Now, \tan t is not continous at certain points (graph it). But if it needs to contain zero so the only open interval (largest interval) where \tan t is continous is on (-\pi/2,\pi/2). But we also require that (t+1)/(t-1) be continous, so we need to exclude the point t=1. This point t=1 is contained in (-\pi/2, \pi/2) we we need to omit it. This gives us that on (-pi/2,1) these coefficient functions are continous. So there is a unique solution.
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  5. #5
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    Quote Originally Posted by ThePerfectHacker View Post
    We cannot say that. My point is that the existence/uniqueness theorem was only proven for linear differencial equations. When we have non-linear differencial equations we have no idea whether it works or not without solving it.

    The theorem says that given,
    y^{(n)}+f_{n-1}(x)y^{(n-1)}+...+f_1(x)y'+f_0(x)y = g(x)
    With y(x_0)=y_0 , y'(x_0)=y'_0 , ... , y^{(n)} (x_0)=y^{(n)} _0.
    Where f_{n-1}(x),...,f_0(x),g(x) are continous functions on some open interval (a,b) and where x_0\in (a,b) then there exists exactly one solution to that differencial equation.

    So given (after division),
    y^{(4)} + \frac{t+1}{t-1} y'' + \tan t \ y = 0 \mbox{ and }0=y(0)=y'(0)=y''(0)=y'''(0)=y''''(0).
    We must work on the interval containing 0 because if you look at the theorem the initial point for the equation is contained in the interval. But that is not it. We also require that \frac{t+1}{t-1} and \tan t be continous. Now, \tan t is not continous at certain points (graph it). But if it needs to contain zero so the only open interval (largest interval) where \tan t is continous is on (-\pi/2,\pi/2). But we also require that (t+1)/(t-1) be continous, so we need to exclude the point t=1. This point t=1 is contained in (-\pi/2, \pi/2) we we need to omit it. This gives us that on (-pi/2,1) these coefficient functions are continous. So there is a unique solution.
    y^{(4)} + \frac{t+1}{t-1} y'' + \tan t \ y = 0 \mbox{ and }0=y(0)=y'(0)=y''(0)=y'''(0)=y''''(0)

    Um, why didn't you divide tan(t)y by (t-1) and how did you decide y''''(0) = 0
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