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Q: Use Implicit Function Theorem to show that there exists a unique solution of the equation $x^{e^y} + y^{e^x} = 0$ in a neighborhood of the point $(0, 0)$.

I tried to satisfy three conditions of IFT.

Let $F(x,y)= x^{e^y} + y^{e^x}$ then,

1: $F(0,0)=0$

2: $\displaystyle{\frac{\partial F}{\partial y}= x^{e^y}e^y \log x + \frac{y^{e^x +1}}{e^x + 1}}\,$, I don't know whether it is continuous in the neighborhood of $(0,0)$

3: $\frac{\partial F}{\partial y}(0,0)\neq 0$, but I don't know how to find the value of $\frac{\partial F}{\partial y}$ at $(0,0)$ since $\log x$ is not defined at $x=0$.

Please help me.

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Hint: $x \log x \to 0$ as $x\to 0$ and $e^y \approx 1$ in a neighborhood of zero – Quickbeam2k1 Mar 9 '13 at 14:32
up vote 1 down vote accepted

For #$2$: the term $x^{e^y}e^y \log x$ is the same as $x^{e^y}\log(x^{e^y})$. This is the composition of $x\log x$ with $x^{e^{y}}$, so it suffices to show $x^{e^{y}}$ is continuous at $(0,0)$ and $x \log x$ is continuous at $x^{e^{y}}|_{(x,y) = (0,0)} = 0$, with the understanding that when $x = 0$ one interprets $x \log x$ to be $\lim_{x \rightarrow 0} x \log x = 0$.

For #$3$: Let $x \rightarrow 0$ above... you view $x^{e^y}e^y \log x = x^{e^y}\log(x^{e^y})$ and interpret its value at $(0,y)$ to be the limit as above as $x \rightarrow 0$... namely $0$.

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meaning that the IFT cannot be applied after all? – Delio Mugnolo Feb 5 at 7:11

If $F\in C^{1}(U;\mathbb{R}^m)$ and $J_yF(x_0, y_o)\neq 0$ then the Implicit Function Theorem tells you that there exists an open set $V\subset U$, with $(x_0, y_0)\in V$, and open set $W\subset \mathbb{R}^n$, with $x_0\in W$, and a $C^1$ mapping $G:W\rightarrow\mathbb{R}^m$ such that:

  1. $G(x_{0})=y_0$,
  2. $F(x, G(x))=z_0\quad(x\in W)$, and

  3. if $(x,y)\in V$ and $F(x, y)=z_0$, then $y=G(x)$.

  4. If $F\in C^k$ then $G\in C^k, \ (k=1, 2, \dots)$.

The function $G$ is implicitly defined near $x_0$ by the equation $F(x,y)=z_0$

This statement is from Evan's PDE book in the Appendix

Now let us apply the theorem to your problem.

We see that $x_0, y_0, z_o$ are all $0$ and that $n=m=1$. Certainly $F\in C^1(U;R)$ for some $U\subset \mathbb{R}^2$ because it is a composition (and addition) of such functions.

We have:

\begin{equation} \frac{\partial F}{\partial y}=x^{e^{y}}e^y\log x+ e^x\Rightarrow J_y(0,0)=1\neq 0. \end{equation}

So the hypothesis of the theorem are satisfied and therefore in some neigbourhood $W\subset\mathbb{R}$ of $x_0$ there is a differentiable map $G: W\rightarrow \mathbb{R}$ such that: \begin{equation} G(0)=0. \end{equation}

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