This is a question about extensions of functions, given certain tangential data.
Let $A:= \{x_1^2+x_2^2 \leq 1, x_3 \leq 0\} \subset \mathbb{R}^3.$ Let $a := \{x_3=0\} \subset \mathbb{R}^3.$
Suppose I have a smooth function $\Phi: A \to \mathbb{R}^3$ and a smooth function $\phi: a \to \mathbb{R}^3$ such that $\Phi|_{\{A \cap a\}}=\phi.$
Suppose finally that there exists a smooth bundle isomorphism $F: T\mathbb{R}^3|_a \to T\mathbb{R}^3|_{\phi(a)}$ such that $F = d\Phi$ on their common domain of definition (namely on $T\mathbb{R}^3|_{\{A \cap a\}}$), and $F= d\phi$ as maps from $Ta \to T(\phi(a)).$
Question: does there exists a smooth extension $\Psi: N(A \cup a) \to \mathbb{R}^3,$ where $N(A \cap a)$ is an arbitrary neighborhood of $A \cup a,$ such that $\Psi|_A=\Phi,$ $\Psi|_a= \phi,$ and $d\Psi|_{T\mathbb{R}^3|_a}=dF$?
Note: Whitney's extension theorem might be useful here, but a direct application of it seems to only give a $C^1$ extension.
Here is a sketch of a solution which was explained to me.
Identifying $\mathbb{R}^3$ with $T\mathbb{R}^3_p$ for any p, we construct a smooth family of linear maps $L_t:= F(t)\circ F(0)^{-1}: \mathbb{R}^3 \to \mathbb{R}^3.$
It suffices to extend $\Phi(x_1, x_2, 0): \{x_1^2+x_2^2 \leq \epsilon, x_3=0\} \to \mathbb{R}^3.$
For $t \leq 1,$ define the extension by $G(x_1, x_2, t):= (1/(1-t)) L_t\left(\Phi((1-t)x_1, (1-t)x_2)\right)+ \phi(t).$ For $t \geq 1,$ just take the extension to be $G(x_1, x_2, t):= (L_t\circ d\Phi_{(0,0,0)})(x_1, x_2, 0)+ \phi(t).$ This will be an embedding for small $x_1, x_2.$