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Determine if the non-linear set of equations has unique solution

  1. Nov 14, 2016 #1
    1. The problem statement, all variables and given/known data
    Determine if the following set of equations has unique solution of the form ##g(z)=(x,y)## in the n-hood of the origin. $$\begin{cases} xyz+\sin(xyz)=0 \\ x+y+z=0 \end{cases}$$

    2. Relevant equations
    I assume I am supposed to use the implicit function theorem
    https://en.wikipedia.org/wiki/Implicit_function_theorem


    3. The attempt at a solution
    Let's consider function ##F:ℝ^3\rightarrow ℝ^2, F(x,y,z)=(xyz+\sin(xyz),x+y+z).## Now ##F(0,0,0)=(0,0)## and ##F\in{C^{1}}## but the Jacobian of ##F## at the origin is: ##
    J_f(0)=[Df(0)]=\begin{bmatrix}
    0 & 0 & 0 \\
    1 & 1 &1
    \end{bmatrix},## which implies that ##[Df(0)]## is not surjective, thus the implicit function theorem doesn't apply directly. I think the next step is to define another function such that we can apply the implicit function theorem to that and maybe find a solution which uniqueness is guaranteed by the theorem. Am I on the right track?

     
    Last edited: Nov 14, 2016
  2. jcsd
  3. Nov 14, 2016 #2

    Krylov

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    No, I don't think so. Read the question carefully:
     
    Last edited by a moderator: Nov 14, 2016
  4. Nov 14, 2016 #3
    O.k. How would one proceed in this case? Should I construct a counterexample?
     
    Last edited by a moderator: Nov 14, 2016
  5. Nov 14, 2016 #4

    Krylov

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    If the answer to the question in the problem were positive, there would exist ##\epsilon > 0## and a function ##g : (-\epsilon,+\epsilon) \times (-\epsilon,+\epsilon) \to \mathbb{R}## such that for all ##x,y \in (-\epsilon,+\epsilon)## the triple ##(x,y,z) = (x,y, g(x,y))## satisfies the system of equations.

    Please make sure you understand why this is.

    It follows from the system of equations that such a function cannot exist, no matter how small you choose ##\epsilon > 0## to be. It is up to you to explain why this is. (You do not need any theorems for this.)
     
    Last edited by a moderator: Nov 14, 2016
  6. Nov 15, 2016 #5
    I think I got the right intuition, but I am not sure if this counterexample is rigorious enough.

    If we consider the function ##F##, we notice that ##F(\epsilon,-\epsilon,0)=(0,0)## ##∀\epsilon>0##.
    Because we can choose ##\epsilon## to be arbitrarily small, ##\nexists{R>0}## such that n-hood of the origin ##B(0,R)## would contain a unique solution ##(x,y,z)## such that ##F(x,y,z)=0##. Thus the set of equations doesn't have a unique solution in any n-hood of the origin.
     
    Last edited: Nov 15, 2016
  7. Nov 16, 2016 #6
    Could someone please verify if I got it right?
     
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