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

1. Nov 14, 2016

### lep11

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. Nov 14, 2016

### Krylov

No, I don't think so. Read the question carefully:

Last edited by a moderator: Nov 14, 2016
3. Nov 14, 2016

### lep11

O.k. How would one proceed in this case? Should I construct a counterexample?

Last edited by a moderator: Nov 14, 2016
4. Nov 14, 2016

### Krylov

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
5. Nov 15, 2016

### lep11

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
6. Nov 16, 2016

### lep11

Could someone please verify if I got it right?