Reference:- An Elementary Course in Partial Differential Equations, T. Amaranth.
How do I apply implicit function theorem here?
My attempt:-
I have gone through the generalized example of implicit function theorem from Principle of Mathematical Analysis by W. Rudin.
Here I understood that $x_1=a,x_2=b, y_1=x,y_2=y,y_3=p,y_4=q.$
My doubt:- I really unable to find $f_1$ and $f_2$ from the $1.2.2,1.2.3$ and $1.2.4$.
Suppose $$f_1(a,b,x,y,p,q)=p-F_x(x,y,a,b)$$ $$f_2(a,b,x,y,p,q)=q-F_y(x,y,a,b)$$
By Implicit function theorem, we have only $2\times2$ matrix $\begin{pmatrix}p_{a}-F_{xa}&q_{a}-F_{ya}\\ p_{b}-F_{xb}&q_{b}-F_{yb}\end{pmatrix}.$ I am not getting (1.2.5). I request you to find equations such that matrix in (1.2.5) has rank 2.
Let's try to rewrite this carefully. Define $$\Phi\colon\Bbb R^7\to\Bbb R^3$$ by $$\Phi(x,y,z,p,q,a,b) = \begin{bmatrix}z-F(x,y,a,b)\\p-F_x(x,y,a,b)\\q-F_y(x,y,a,b)\end{bmatrix}.$$ We want to consider the level set $\Phi = 0$. We want to consider any pair of these functions, not all three. The Implicit Function Theorem will guarantee that we can solve for $a,b$ as (smooth) functions of the remaining variables if the appropriate $2\times 2$ submatrix of the derivative matrix of $\Phi$ with respect to $a,b$ has rank $2$. Since $$\frac{\partial\Phi(x,y,z,p,q,a,b)}{\partial (a,b)} = -\begin{bmatrix} F_a & F_b \\ F_{xa} & F_{xb} \\ F_{ya} & F_{yb}\end{bmatrix},$$ their assertion is correct if this $3\times 2$ matrix indeed has rank $2$.