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Implicit Function Theorem

  1. Nov 23, 2008 #1
    1. The problem statement, all variables and given/known data
    I'm looking for someone to walk me through a problem, because I keep falling further and further behind ...

    Let f be a real-valued function on he strip [itex] R_2: a \leq x \leq b, -\infty < y < \infty [/itex]. Can the equation f(x,y) = 0 be solved for y as a function of x;
    f is continuous w/ continuous partial derivatives of first order, and
    [tex] 0 < m \leq \frac{\partial f}{\partial y} \leq M [/tex]
    show a unique function y = u(x) so that f(x,u(x)) = 0

    2. Relevant equations
    There was s hint to consider a continuous function w/ the sup norm and define the transformation [itex] T:C[a,b] \rightarrow C[a,b] [/itex] by
    [tex] (Tu)(x) = u(x) - \frac{2}{m + M}f(x,u(x)) [/tex]
    and show that T is a contraction.

    3. The attempt at a solution
    I'm just not getting it, but here's what I have:
    Apply the contraction mapping theorem:

    [tex] \parallel (Tu)(x) - (Tv)(x) \parallel = \parallel u(x) - \frac{2}{m + M}f(x,u(x)) - v(x) + \frac{2}{m + M}f(x,v(x)) \parallel [/tex]
    [tex] = \parallel u(x) - v(x) + \frac{2}{m + M}\left(f(x,v(x)) - f(x,u(x)) \right ) \parallel [/tex]
    If I apply the Mean Value theorem, I should be able to get
    [tex] = \parallel u(x) - v(x) - \frac{2}{m + M}f'(c)(v(x) - u(x)) \parallel [/tex]
    taking the sup norm, I would state
    [tex] \leq \parallel u(x) -v(x) + \frac{2M}{m + M}(u(x) - v(x)) \parallel_\infty [/tex]
    [tex] \leq \left(1+\frac{2M}{m + M} \right) \parallel u(x) - v(x) \parallel_\infty [/tex]
    [tex] \leq \frac{m+3M}{m+M} \parallel u(x) - v(x) \parallel_\infty [/tex]

    I know that's not correct (definitely not a contraction), but I don't know what I'm doing wrong. Please use small words, so I can understand. Thanks.
     
  2. jcsd
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