I need to find/define a function $G(x_1(t),x_2(t)) : \mathbb{R} \times \mathbb{R} \to \mathbb{R}$ such that the following holds:
$$ \frac{\partial G(x_1(t),x_2(t))}{\partial x_1(t)} =u_1(t)$$
and
$$ \frac{\partial G(x_1(t),x_2(t))}{\partial x_2(t)} =u_2(t)$$
where $x_1(t) = f_1(u_1(t),u_2(t))$ and $x_2(t) = f_2(u_1(t),u_2(t))$
$$ f_1,f_2: \mathbb{R} \times \mathbb{R} \to \mathbb{R} $$
How can I go about doing this?
Edit: $f_1$ and $f_2$ are bounded and continuously differentiable functions.
Best Answer
You can't find $G$ uniquely, since your equation tells you only about its behavior near points that are hit by $(x_1,x_2)$.
What we can do is compute $\frac{d}{dt}G(x_1,x_2)$ by the chain rule:
$$ \frac{d}{dt}G(x_1,x_2) = u_1(t)\frac{d}{dt}x_1(t) + u_2(t)\frac{d}{dt}x_2(t) $$ where everything on the right is known. Arbitrarily setting $G(x_1(0),x_2(0))=0$ (any constant term can be added of course) we get $$ G(x_1(s),x_2(s)) = \int_0^s \left[u_1(t)\frac{d}{dt}x_1(t)+u_2(t)\frac{d}{dt}x_2(t) \right] \, dt $$
This gives you some values of $G$, and you'd better hope that if $(x_1(t),x_2(t))=(x_1(s),x_2(s))$ for any $t,s$, the computed values of $G$ match -- otherwise there's no solution. Similarly, you must have $u(t)=u(s)$, since otherwise you'll have conflicting demands on the partial derivatives of $G$ at that point.
Afterwards you need to choose neighboring values of $G$ such that the partial derivatives are right. This can be done if the curve described by $(x_1,x_2)$ is smooth enough, but not at all uniquely, of course. The simplest solution may be to use make $G(p)$ vary linearly on a short perpendicular to the main curve at $(x_1(t),x_2(t))$, with a slope chosen to make the partial derivatives come out right.