I try to solve the exercise 0.1 given here https://webspace.science.uu.nl/~kolk0101/Books/Analysis/implicitapplic.pdf. I have issues proving part (ii) the last formula $$D_2 \psi(0) = 1/g(0)$$ I will write $D_2 \psi(0)$ as $\partial_t \psi(t)$ since $t$ is the second variable and is one dimensional.
So by the implicit function theorem and choosing $x$ as the to be solved variable $x= \psi(y,t)$ it holds
$$\partial_{t}\psi(y,t) = -\partial_x F(\psi(y,t),y,t)^{-1}\partial_t F(\psi(y,t),y,t)$$ $$ \Rightarrow \partial_t \psi(0,0) = \underbrace{-g(0)^{-1}}_{-\partial_x F(\psi(y,t),y,t)^{-1} |y=t=0}\partial_t(\psi(0,t)g(\psi(0,t))-t) = ... $$
Where we used
$f(x,0) = xg(x)$ and $F(x,y,t) = f(x,y) -t $ $$... = -g(0)^{-1} (\partial_t \psi(0,0)g(\psi(0,0)) + \underbrace{\psi(0,0)}_{=0} g'(\psi(0,0))\partial_t \psi(0,0) -1 ) = -g(0)^{-1} (\partial_t\psi(0,0)g(0)-1)$$ where we used $\psi(0,0) = 0$ Hence $$\partial_t \psi (0,0) = -g(0)^{-1} (\partial_t \psi(0,0) g(0) -1) = -\partial_t\psi(0,0) + \frac{1}{g(0)} \Leftrightarrow \partial_t \psi(0,0) = \frac{1}{2g(0)}$$ which is not the desired formula. Does anyone spot the problem ?
Thank you so much
The problem is in your interpretation of $\partial_t F(\psi(y,t),y,t).$ You must not differentiate (w.r.t. $t$) $(y,t)\mapsto F(\psi(y,t),y,t)$ at the point $(y,t)$ but $(x,y,t)\mapsto F(x,y,t)$ (i.e. $F$) at the point $(\psi(y,t),y,t).$
Why so? to avoid such an ambiguity, let us denote by $D_1F,D_2F,D_3F$ the differentials of $F(x,y,t)$ w.r.t. $x,y,t.$ Differentiating the equation $$F(\psi(y,t),y,t)=0$$ w.r.t. $t$ gives: $$D_1F(\psi(y,t),y,t)\partial_t\psi(y,t)+D_3F(\psi(y,t),y,t)=0,$$ hence $$D_1F(0)\partial_t\psi(0)+D_3F(0)=0,$$ i.e. $$g(0)\partial_t\psi(0)-1=0.$$