Solvability of $u_t=\nabla_{u_\theta} u_\theta -\frac{1}{2}\nabla R +\frac{1}{2\tau}u_\theta + 2 Ric(u_\theta,\cdot)$ I ask a question in math.stackexchange,  but nobody answer it. So, I ask here.
In Cao and Zhu's paper about Ricci flow, the $\mathcal{L}$-geodesic is defined as picture below.In fact, $\mathrm{Ric}(X,\cdot)$ should be $g^{-1}\mathrm{Ric}(X,\cdot)$.



I want to study the heat flow of it, similar to the heat flow of ordinary geodesic. Let $u(t,\tau)$ be a family of curves, $t\ge0$ and for any $t_0\ge 0$, $u(t_0,\cdot):S^1\rightarrow M$ is a curve on $M$. $(M,g)$ is a compact Riemannian manifold. 
$$
u_t=\nabla_{u_\tau} u_\tau -\frac{1}{2}\nabla R +\frac{1}{2\tau}u_\tau + 2 \mathrm{Ric}(u_\tau,\cdot)
$$
In fact, I want to construct some convexity energy to prove the existence of solution. But fail. And fail for simple situation
$$
u_t=\nabla_{u_\tau} u_\tau +\nabla R
$$
I feel the method of energy is not suitable for it , because energy function is hard to construct . What method is suitable for this question ?
 A: *

*In Ottarsson's paper that you cited the curves are by definition closed, that is, they are images of $\mathbb{S}^1$. In Cao and Zhu, the $\mathcal{L}$-geodesics must be "open" in the sense that they are images of $[\tau_1,\tau_2]$. So if you want to write down any equation you absolutely have to write down boundary conditions. I assume you want something like Dirichlet where the end points are fixed. 

*The heat flow given in Ottarsson's paper is the gradient flow relative to the energy functional for immersions of $\mathbb{S}^1$ to $M$. If you want to define an analogous object you should have the gradient flow relative to the $\mathcal{L}$-length, and it seems to me that your definition is missing a factor of $\sqrt{\tau}$ on the right hand side of the flow equation. 

*In terms of the $\mathcal{L}$-length, you are using $\tau$ as the parameter for your curve, so all your $\theta$s should really be $\tau$s. 

*In terms of the basic energy: you are working with the gradient flow of an energy functional. Just use that as the energy. 
