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Math Help - Existence of solution

  1. #1
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    Existence of solution

    Thm: Suppose that both the function f(x,y) and its partial Derivative fy(x,y) are continuous on some rectangle R in the xy plane that contains the point (a,b) in its interior. Then, for some open interval I containing the point a, the initial value problem has one and only one solution that is defined on the interval I.

    I have to determine if the above Thm does or does not guarantee existence of a solution of:

    dy/dx = x lny ; y(1) = 1

    Can someone show how this is done? Thanks a lot!!
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  2. #2
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    Quote Originally Posted by jzellt View Post
    Thm: Suppose that both the function f(x,y) and its partial Derivative fy(x,y) are continuous on some rectangle R in the xy plane that contains the point (a,b) in its interior. Then, for some open interval I containing the point a, the initial value problem has one and only one solution that is defined on the interval I.

    I have to determine if the above Thm does or does not guarantee existence of a solution of:

    dy/dx = x lny ; y(1) = 1

    Can someone show how this is done? Thanks a lot!!
    This is a special case of the Picard–Lindelöf Theorem

    You just need to verify the hypothesis of the theorem...

    \displaystyle \frac{dy}{dx}=f(x,y)=x\ln(y) and

    \displaystyle \frac{\partial f}{\partial y}=\frac{x}{y}

    The question is can you find a rectangle around the point (1,1)

    Where both f(x,y) and f_y(x,y) are continuous?
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  3. #3
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    I guess I should've mentioned that Ive gotten this far. I don;t know what to do from here...
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  4. #4
    Behold, the power of SARDINES!
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    Ask yourself this question what is the domain of f(x,y) and what is the domain of f_y(x,y)?

    The domain for f(x,y) is
    \displaystyle \left\{ (x,y) \in \mathbb{R}^2\bigg| y > 0\right\}

    The domain for f_y(x,y) is
    \displaystyle \left\{ (x,y) \in \mathbb{R}^2\bigg| y \ne  0\right\}

    Now the question is can you find a rectangle around the point (1,1)

    That is in both of these domains? I bet you can.
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  5. #5
    MHF Contributor chisigma's Avatar
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    The proposed DE has something 'not fully clear' because if we change the roles of x and y it becomes...

    \displaystyle \frac{dx}{dy}= \frac{1}{x\ \ln y} , x(1)=1 (1)

    In that case the conditions of existence of a solution are not satisfied because f(x,y)= \frac{1}{x\ \ln y} has a singularity in y=1. That 'anomaly' is confirmed by the fact that is we try to solve (1) we obtain as general solution...

    \displaystyle \frac{x^{2}}{2} = Li(y) + c (2)

    ... and it is well known that the function Li(z) has a 'pole' in z=1... May be that all that requires further examination...

    Kind regards

    \chi \sigma
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  6. #6
    MHF Contributor chisigma's Avatar
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    The 'thriller' about the DE...

    y^{'} = x\ \ln y , y(1)=1 (1)

    ... has been 'solved': the solution is [simply...] y=1. The solution cannot be derived from the 'general solution' with an appropriate value of the 'arbitrary constant' and it is called 'singular solution'...

    Kind regards

    \chi \sigma
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