For a planar, simply connected domain $U \subset \mathbb{R}^2$ let $\gamma_0\colon [0, 1]\longrightarrow U$, $\gamma_1\colon [0, 1]\longrightarrow U$ be two differentiable curves in $U$ with same endpoints, i.e. $\gamma_0(0) = \gamma_1(0)$, $\gamma_0(1) = \gamma_1(1)$. Assume $\gamma_0$ and $\gamma_1$ are each injective (not self-intersecting) on $(0, 1)$, i.e. apart from the end points and also do not intersect each other. Let $H \colon [0, 1]^2 \longrightarrow U$ be a homotopy with $H(t, j) = \gamma_j(t)$ for $j \in \{0, 1\}$. I will denote the second parameter, the "morph parameter", of $H$ as $\theta$. Assume that for all $\theta \in [0, 1]$, $H([0, 1], \theta)$ are disjoint sets, i.e. not any curves defined by $H$ intersect.
Let $L_{t_0, t_1}(\gamma)$ be the length functional \begin{equation} L_{t_0, t_1}(\gamma) \quad = \quad \int_{t_0}^{t_1} \; || \dot{\gamma}(t) || \; \; d t \end{equation}
My questions:
Does always a length-monotonic homotopy exist? Precisely: does always exist a $H$ like described above such that $L_{0, 1}(H(\cdot, \theta))$ is monotonic in $\theta$ and strictly monotonic if we assume $L_{0, 1}(\gamma_0) \neq L_{0, 1}(\gamma_1)$?
If this is asked too generally, would it help or make a difference if we assume the shorter curve, let's say $\gamma_0$ is shorter, is actually the shortest connection between $\gamma_0(0)$ and $\gamma_0(1)$? Would it help or make a difference if we further assume $\gamma_0(0)$ and $\gamma_0(1)$ are on the boundary $\partial U$ and the longer curve $\gamma_1$ is actually a boundary section connecting $\gamma_0(0)$ and $\gamma_0(1)$?
Is it even possible to choose $H$ such that $L_{0, 1}(H(\cdot, \theta))$ is linear in $\theta$? (Intuitively I think if monotonic is possible it should be possible to exchange an amount of length between nearby curves by some perturbation, maybe enough to linearize $L_{0, 1}(H(\cdot, \theta))$.)
I especially wonder if this is a known (maybe classical) result or if it somehow follows "obviously" or "easily" from some known broader theorem. Or is it maybe not true?
Some context:
I am interested in finding the "length-energy" minimal homotopy in the sense of $H$ minimizing
\begin{equation} \text{minimize} \quad \int_{0}^{1} \Big( \frac{\partial}{\partial \theta} \; L_{0, 1}(H(\cdot, \theta)) \Big)^2 \; \; d \theta \end{equation}
under the boundary condition $H(t, j) = \gamma_j(t)$ for given $\gamma_j$ and $j \in \{0, 1\}$. By Euler-Lagrange equations I can deduce that the ideal $L_{0, 1}(H^*(\cdot, \theta))$ should be linear in $\theta$ but I wonder if one can assume that such a homotopy always exists.
My thoughts so far:
In case that $\gamma_0$ and $\gamma_1$ enclose a convex domain, I guess one can show that property by analyzing the linear blend from $\gamma_0$ to $\gamma_1$, i.e. $\gamma_{\theta} = \theta \gamma_0 + (1-\theta) \gamma_1$. I guess this would not work where $\gamma_{\theta}$ moves through the shortest connection of $\gamma_0(0)$ and $\gamma_0(1)$ within the enclosed domain. This is why I mentioned above that it may be helpful to assume $\gamma_0$ is itself that shortest connection.
For the non-convex case I think Riemann's mapping theorem may be helpful. However, so far I wasn't able to conclude an answer from that theorem.
Further note that it is not true for non-planar surfaces: $\gamma_0$ and $\gamma_1$ might enclose a "mountain" and the homotopy would have to cross that mountain. Such a mountain could be made arbitrarily high, leading to arbitrarily long intermediate curves. I guess it may be true for planar multiply connected domains, given that $\gamma_0$ and $\gamma_1$ are homotopic at all.
Please also don't hesitate to give partial answers or hints if you don't know the "perfect" full answer! Thanks in advance!
The answer to your question is positive. Below is a bit sketchy proof, making arguments complete would require much more work (pretty, much, writing a research paper).
First of all, I will work with simply-connected closed polygonal domains $X$, rather than open simply connected planar sets (see remarks at the bottom of the answer): The proof is already complicated in this setting.
Then $X$ is a length space: The length of a curve in $X$ is the usual Euclidean length: $$ \ell(c)= \int_0^1 |c'(t)|dt. $$ Define the intrinsic distance $d$ between two points $p, q$ in $X$ as the infimum of lengths of paths in $X$ connecting $p$ and $q$. One verifies that $X$ is complete and every two points are connected by a length-minimizing curve: Such curves are called geodesic. I will always parameterize geodesics with the constant speed. One verifies that geodesics are piecewise-linear paths in the plane.
One then verifies that $(X,d)$ is a CAT(0) space: Geodesic triangles in $(X,d)$ are "thinner" than their comparison triangles in the Euclidean plane $E^2$.
You can read more about CAT(0) spaces in the book
M. Bridson, A. Haefliger, Metric Spaces of Non-Positive Curvature, Springer-Verlag, 1999.
For instance, in order to see that $(X,d)$ is a CAT(0) space triangulate the domain $X$ (using Euclidean triangles; you can even do so that the vertices of the triangulation are all on the boundary of $X$) and observe that the resulting simplicial complex satisfies the "link condition" (page 206 of the book): For every vertex $v$ of the triangulation, the length of each cycle in the link $L_v$, is $\ge 2\pi$. This condition is trivially satisfied since $L_v$ contains no cycles! From this, it follows that $(X,d)$ is locally CAT(0). Since $X$ is simply-connected, it follows that $(X,d)$ is CAT(0).
The definition of geodesics in metric geometry is different from the one you know in Riemannian geometry: They are required to be globally (rather than locally) length-minimizers. However, in CAT(0) setting, there is a local-to-global principle: Each local length-minimizer is a global length-minimizer.
The most important (for us) property of CAT(0) spaces is that the distance function $d$ is convex:
For any two geodesics $c_1(t), c_2(t)$ in $(X,d)$ the function $f(t)=d(c_1(t), c_2(t))$ is a convex function of one variable. (See page 176 of the book.)
From this, it follows that any two points $p, q$ in $(X,d)$ are connected by a unique (constant speed) geodesic $c_{pq}(t), t\in [0,1]$, and the latter depends continuously on $(p,q)\in X\times X$.
Now, Suppose that $\gamma_0(s), \gamma_1(s), s\in [0,1]$ are two piecewise-linear paths in $X$ satisfying $$ \gamma_0(0)=\gamma_1(0), \gamma_0(1)=\gamma_1(1). $$ (By an approximation argument, one can also handle general rectifiable paths.)
Their length (with respect to $d$) is the same as their Euclidean length. We construct the "geodesic" homotopy between these curves by the formula: $$ F(s,t)= c_{\gamma_0(s)\gamma_1(s)}(t), s\in [0,1], t\in [0,1]. $$
Lastly, convexity of the distance function will imply that the function $$ L(t)=Length (F(\cdot, t)) $$ is convex as a function of $t$.
What you are asking for, however, is monotonicity of length. I will describe the modification so that it is non-strictly monotonic. With a bit more work, one can get strict monotonicity.
For concreteness, assume that $L(1)=\ell(\gamma_1)\ge L(0)= \ell(\gamma_0)$. Let $L(t_0)$ denote the minimum of the function $L(t)$, $t_0\in [0,1]$. By convexity of $L(t)$, it is decreasing on $[0,t_0]$ and increasing on $[t_0, 1]$ (both non-strictly on some interval containing $t_0$). Let $t_1\in [t_0, 1]$ be maximal such that $L(t_1)=L(0)$. I will modify the homotopy $F(s,t)$ to a new homotopy $G(s,t)$ as follows:
For $t\in [t_1,1]$, set $G(s,t)=F(s,t)$, since the length $L(t)$ is strictly increasing on $[t_1,1]$.
For $t\in [0,t_1]$ I will define $G(s,t)$ so that the curve $$ c_t: s\mapsto G(s,t) $$ has length $L(0)$. Namely, let $\lambda(t)$ denote the difference $L(0)-L(t)$: It is $\ge 0$ since $t\in [0, t_1]$. Extend $c_t$ to $[1,1+\lambda(t)]$ so that:
(a) The extension curve $b_t$ has constant speed (except for finitely many points where the derivative does not exist).
(b) For $s\in [1, 1+ \frac{1}{2}\lambda(t)]$ the extension curve backtracks length $\lambda(t)/2$ along $c_t$.
(c) For $s\in [1+ \frac{1}{2}\lambda(t), 1+\lambda(t)]$ the extension curve goes back to $c_t(1)$ covering length $\frac{1}{2}\lambda(t)$ along the way.
The resulting curve $a_t(s), s\in [0, 1+\lambda(t)]$, now has length $L(0)$. The only problem is that it is defined not on the unit interval. But we can reparameterize it with constant speed to redefine it on the unit interval. The result is a map $G(s,t)$. One verifies that this map is still continuous. By the construction, each curve $s\mapsto G(s,t)$ has length $L(0)$ for $s\in [0,t_1]$.
Two more things:
One can eliminate the assumption that $X$ has polygonal boundary by an approximation argument (the, suiably defined, limit of a sequence of CAT(0) spaces is again CAT(0)).
If $L(0)<L(1)$, one can also modify the above proof to make it work in open domains in the plane. (Again, by an approximation argument.) However, I am not sure how to deal with the case $L(0)=L(1)$: Approximation arguments will no longer work.