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Let $G: \mathbb{R}^2\longrightarrow\mathbb{R}^2$ be a smooth function. Let also $\Gamma:[0,1]\longrightarrow\mathbb{R}^2$ be a smooth curve.

I am working on the problem of finding a curve $\gamma$ such that $G\circ\gamma = \Gamma$.

I think that a possible method to find an approximation of $\gamma$ would be sampling points $(X_k,Y_k)$ on $\Gamma$, for instance $(X_k,Y_k)=\Gamma(f(k))$ for some suitable function $f$, work to find points $(x_k,y_k)$ such that $G(x_k,y_k)=(X_k,Y_k)$ and then find an interpolation curve that passes through thesee points. This could work when $G$ is injective but even then finding preimages is not an easy task.

Moreover, this strategy would be utterly cumbersome if $G$ is not injective.

My question is this: is there a known method of finding such a curve $\gamma$?

I am currently trying via variational calculus, setting \begin{equation} I[\gamma] := \int_0^1\|G(\gamma(t))-\Gamma(t)\|^2\ \text{dt} \end{equation} as the functional to minimize, but the resulting differential equations are incredibly difficult.

Note: even an effective hint for finding an approximation for $\gamma$ would be really helpful.

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    This is known as "path lifting". The existence is assured if $G$ is a discrete and open map (preimage of every point is discrete, image of every open set is open). You don't say anything about $G$ beyond "smooth" (smoothness is actually not as important as topological properties of $G$). If $G$ is a projection onto a line, then it's very smooth but we can't really lift paths through it. Also, it looks like you want a numerical method for doing this? – 2017-01-29
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    I was looking for numerical methods because I'm dealing with a particular instance of the problem that doesn't really look like it's going to have an explicit solution... – 2017-01-29

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