This problem is in Spivak's Differential Geometry (Ch.9 #37), and he gives a sketch of a proof which I have been unable to finish.
So let's compute $\frac{dL(\overline{\alpha}(u))}{du}\mid_{u=0}$ where $L(\overline{\alpha}(u))=L_{0}^{t_{1}}(\gamma)+\int_{t_{1}}^{t_{0}+u}F(\,,\,)dt+\int_{t_{0}+u}^{1}F(\,,\,)dt$ , and of course $F(\alpha(u,t),\frac{\partial\alpha}{\partial t}(u,t))=\sqrt{\underset{i,j}{\sum}g_{ij}(\alpha)\frac{\partial\alpha^{i}}{\partial t}\frac{\partial\alpha^{j}}{\partial t}}$ (all the "$(u,t)$" omitted).
Well, $\frac{dL(\overline{\alpha}(u))}{du}=\frac{d}{du}\int_{t_{1}}^{t_{0}+u}F(\,,\,)dt+\frac{d}{du}\int_{t_{0}+u}^{1}F(\,,\,)dt$ . Now I apply the Leibniz integral rule, and the terms become
- $F(\alpha(u,t_{0}+u),\frac{\partial\alpha}{\partial t}(u,t_{0}+u))+\int_{t_{1}}^{t_{0}+u}\frac{\partial}{\partial u}F(\alpha(u,t),\frac{\partial\alpha}{\partial t}(u,t))dt$
and
- $\int_{t_{0}+u}^{1}\frac{\partial}{\partial u}F(\alpha(u,t),\frac{\partial\alpha}{\partial t}(u,t))dt-F(\alpha(u,t_{0}+u),\frac{\partial\alpha}{\partial t}(u,t_{0}+u))$
, respectively. Evaluating their sum at $u=0$ , we just get $\int_{t_{1}}^{1}\frac{\partial}{\partial u}F(\alpha(0,t),\frac{\partial\alpha}{\partial t}(0,t))dt$ .
Note that $\alpha$ is not exactly a variation on $\gamma$ since $\alpha(0,t)$ has a piece of $\gamma$ replaced by a geodesic (*). But anyway, $\alpha$ is a variation on the piecewise smooth curve $\alpha(0,t)$ , and the integral obtained yields the First Variation Formula for Length of $\alpha(0,t)\mid_{[t_{1},1]} .$ . For the moment ignore *, and assume this is the thing we want to show $\neq0$ . The integral term in the First Variation Formula will disappear, leaving
$\left\langle \frac{\partial\alpha}{\partial u}(0,t_{0}),\frac{\partial\alpha}{\partial t}(0,t_{0}^{+})-\frac{\partial\alpha}{\partial t}(0,t_{0}^{-})\right\rangle$ .
This is sort of like $\left\langle \frac{\partial\alpha}{\partial u}(0,t_{0}),\Delta_{t_{0}}\frac{d\gamma}{dt}\right\rangle$ , where we know $\Delta_{t_{0}}\frac{d\gamma}{dt}\neq0$ .
Assuming everything was correct so far, I have two questions:
1) Why (where) do we need $t_{1}$ to be sufficiently close to $t_{0}$, as hinted by Spivak ?
2) How can I conclude that the inner product term is not 0?
Some more thoughts...
Can we just define $\alpha(0,t)$ to be $\gamma(t)$; in other words, let the geodesic "short-cuts" begin only when u>0? Would this destroy the required $C^\infty$ properties of $\alpha$? The geodesics are parametrized arclength, while the piece of $\alpha$ they approach can have an arbitrary parametrization.
Another note: By moving $t_1$ toward $t_0$ we can change the length of $\frac{\partial\alpha}{\partial t}(0,t_{0}^{-})$, since the geodesic portion is parametrized by arclength. This may be useful when we need to show the vectors in the inner product are not orthogonal.