1. some example and observations

$latex {(M^2,g)}&fg=000000$,$latex {g(t)=e^{2u(t)}g_0}&fg=000000$,

$latex \displaystyle \frac{\partial u}{\partial t}=e^{-2u}\tilde\Delta u+\frac{r}{2}-e^{-2u}K_0 &fg=000000$

$latex {(M^n,g_{ij}(t))}&fg=000000$

$latex \displaystyle \frac{\partial g_{ij}(t)}{\partial t}=-2Ric(g_{ij})&fg=000000$

The given “smooth” initial :
$latex {\exists }&fg=000000$ T small ,$latex {T>0}&fg=000000$,the solution exists on $latex {[0,T]}&fg=000000$

equation is possible system.

Deturk Trick

“Threshold type theorem”
Ricci flow:
Mean curvature flow:
HMF:
Calabi flow:
pf of observation 4:if threshold condition hold for $latex {[0,T]}&fg=000000$,then we can bound ang $latex {C^k}&fg=000000$ norm of solution.

“geometry”

2. smooth manifold with conical singularities

on surface we can define conical singularity.

Definition 1 (conical singularity) $latex {M^2,p_i}&fg=000000$,$latex {\beta_i}&fg=000000$,where $latex {\beta > -1}&fg=000000$,the angle of conical singularity $latex {p_i}&fg=000000$ is $latex {2\pi(1+\beta)}&fg=000000$.
iff conical background metric $latex {g_0}&fg=000000$: $latex {g_0}&fg=000000$ near p,$latex {g_0=r^{2\beta}(dr^2+r^2d\theta^2)}&fg=000000$,$latex {(r,\theta)}&fg=000000$ is the interpolation coordinate chart.

it is easy to chake the form $latex {g_0=r^{2\beta}(dr^2+r^2d\theta^2)}&fg=000000$ is independent with the coordinate chart ,so the definition is well defined.

3. rough line of proof

initial $latex {u_0}&fg=000000$,

$latex \displaystyle \frac{\partial u}{\partial t}=e^{-2u}\tilde\Delta u+\frac{r}{2}-e^{-2u}K_0&fg=000000$

Step 1:(Short time existence)
state and proof the “magic theorem”:
1.we need to explain what is smooth,to define a Banach space $latex {A}&fg=000000$,maybe $latex {W^{k,p},C^{k,p}}&fg=000000$ type.
2.to proof the “magic theorem” under the setting.maybe use shauder fix point theorem or contraction map theorem or else.
3.so the problem reduce to get this type estimate,
if $latex {\frac{\partial u_{i+1}}{\partial t}=e^{-2u_i}\tilde\Delta u_{i+1}+\frac{r}{2}-e^{-2u_i}K_0}&fg=000000$.
define $latex {T_{[0,T]}:A(\Omega \times [0,T]) \longrightarrow A(\Omega \times [0,T])}&fg=000000$. $latex {T_{[0,T]}}&fg=000000$ is continuous,$latex {Dom(T_{[0,T]})}&fg=000000$ is convex,$latex {Im(T_{[0,T]})}&fg=000000$ is a pre-compact set.for $latex {T_{[0,T]}}&fg=000000$ suffice small
the difficult is to set up the continuous of operator $latex {T_{[0,T]}}&fg=000000$
Schauder estimate tell us:
$latex \displaystyle ||u_{i+1}||_{C^{2,\alpha}} \leq ||u_{i+1}||_{L^{\infty}}+||\frac{r}{2}-e^{-2u_i}K_0||_{C^{\alpha}}&fg=000000$

this give us some useful information to construct space $latex {A}&fg=000000$ .
Step 2:(Threshold type theorem,long time existence)
Threshold as long as $latex {||u||_{L^{\infty}}}&fg=000000$ is bounded.
This type theorem is relate to the maximal internal in which solution existence is closed.
Basically is based on Alzalo-Ascoli theorem.
Step3:(More regularity) 1.the question does not existence for smooth manifold.(why)
2.singular space.
$latex {u_0\in }&fg=000000$ small space $latex {\Longrightarrow}&fg=000000$ $latex {u(t)\in}&fg=000000$ small space.
what is the optimal regularity?
the problem naturally come from both “pure PDE” and “application for geometry problem”.

conical Kachler Ricci flow[Chen.Wang]
Donaldson setting $latex {C^{2,\alpha,\beta}}&fg=000000$

4. More seriously treat with the problem

in 07 years,consider the problem
$latex \displaystyle \frac{\partial u}{\partial t}=\Delta u&fg=000000$

on $latex {M-\{p\}}&fg=000000$
$latex \displaystyle u|_{t=0}=f&fg=000000$

where f is a function with nice regularity.
Functional analysis:
$latex \displaystyle \Delta: C_c^{\infty}(M-\{p\}) \longrightarrow C_c^{\infty}(M-\{p\}) &fg=000000$

extension to:
$latex \displaystyle \Delta: L^{2}(M-\{p\}) \longrightarrow L^2(M-\{p\})&fg=000000$

which is a self-adjoint extension.and then use the theory of operator semi-group.the problem can be solved.
remark:the extension is not unique so the information we know for the solution is very little.and because the really true extension which is suit for our geometry setting is just one extension.so the treat of Functional analysis is not enough for us.
Elementary treat:
consider the simplest case,smooth manifold with only one singularity.

we set $latex {M_i}&fg=000000$ is the manifold cut off form $latex {M}&fg=000000$ with a boundary more and more near the singularity. consider the equation on each $latex {M_i}&fg=000000$,i.e.:
$latex \displaystyle \frac{\partial u}{\partial t}=\Delta u&fg=000000$

on $latex {M_i}&fg=000000$
$latex \displaystyle u|_{t=0}=f&fg=000000$

with boundary condition:
Drichlet condition
$latex \displaystyle u|_{\partial M_i=0}&fg=000000$

or Neumann condition
$latex \displaystyle \frac{\partial u}{\partial v}=0 &fg=000000$

we choose Neumann condition there and at last we will see the solution come from Dirichlet condition is the same with the solution come from Neumann condition.
Under the general setting this become:

$latex \displaystyle \frac{\partial u_k}{\partial t}=a(k,t)\Delta u_k+b(k,t)\partial^i u_k +c(x,t)&fg=000000$

$latex \displaystyle \frac{\partial u_k}{\partial v}|_{\partial M_k}=0&fg=000000$

when $latex {k \longrightarrow \infty }&fg=000000$, do we have $latex {u_k \longrightarrow u}&fg=000000$?
we need priori estimate: Schauder estimate for serious parabolic equation tell us:
for equation $latex {\frac{\partial u}{\partial t}=\Delta u}&fg=000000$ on $latex {M}&fg=000000$ with priori estimate $latex {||u||_{L^{\infty}}\leq C}&fg=000000$, we have:

$latex \displaystyle |\nabla^k u(p)|\leq \frac{C}{r^k} &fg=000000$

wher $latex {r}&fg=000000$ is the maximal such that geodesic ball $latex {B(r,p) \subset\subset M_i}&fg=000000$.
For general setting :

$latex \displaystyle \frac{\partial u_k}{\partial t}=\Delta u_k+f&fg=000000$

$latex \displaystyle u_k|_{t=0}=u_0&fg=000000$

$latex \displaystyle \frac{u_k}{\partial t}|_{\partial M_k}=0&fg=000000$

we know

$latex \displaystyle ||u(t)||_{C^0(M_k)}\leq ||u_0||_{C^0(M_k)}+t||f||_{C^0(M_k)}&fg=000000$

this is what Schauder estimate tell us.
1.the uniform estimate with k:
$latex {||u_k||_{***}\leq C}&fg=000000$ independent of k.(now we do not know what the norm $latex {||\cdot||_{***}}&fg=000000$ need to be)
we have $latex {C^0}&fg=000000$ estimate and the energy estimate as follows:
from maximal principle,easy to get $latex {C^0}&fg=000000$ norm estimate.

the point is the equation $latex {\frac{\partial }{\partial t}u_k= \Delta u_k +f}&fg=000000$ is strict parabolic so we have strong maximal principle and to construct suit bump function we can estimate $latex {C^0}&fg=000000$ norm of $latex {u_k}&fg=000000$.

from energy method we can estimate $latex {\int_{M_k} ||\nabla u_k||^2}&fg=000000$.

the point is:
$latex {\frac{\partial}{\partial t}\int_{M_k} |\nabla u_k|^2=2\int_{M_k} \nabla u_k \cdot \frac{\partial}{\partial t}(\nabla u_k) }&fg=000000$
$latex {=2\int_{M_k} \Delta u_k \cdot \frac{\partial}{\partial t} u_k }&fg=000000$
$latex {=-2\int_{M_k}(\frac{\partial}{\partial t} u_k -f)\cdot \frac{\partial}{\partial t}u_k}&fg=000000$
$latex {=-2[\int_{M_k}|\frac{\partial}{\partial t}u_k|^2-\int_{M_k} f\cdot \frac{\partial}{\partial t}u_k]}&fg=000000$
$latex {=-2\int_{M_k}|\frac{\partial}{\partial t}u_k|^2-\int_{M_k}f \cdot (\Delta u_k +f)}&fg=000000$
$latex {\leq 2\int_{M_k} \nabla u_k\cdot \nabla f}&fg=000000$
$latex {\leq\int_{M_k} |\nabla f|^2 +\int_{M_k} |\nabla u_k|^2}&fg=000000$.
so we get:

$latex \displaystyle \frac{\partial}{\partial t}\int_{M_k}|u_k|^2\leq \int_{M_k}|\nabla f|^2+\int_{M_k} |\nabla u_k|^2 &fg=000000$

so we can bounded $latex {\int_{M_k}|u_k|^2}&fg=000000$.
for the general case:the equation becomes:
$latex \displaystyle \frac{\partial u}{\partial t}=a(x,t)\Delta u+b(x,t) \partial^i u+c(x,t)&fg=000000$

$latex \displaystyle \frac{\partial u}{\partial v}|_{\partial M_k}=0&fg=000000$

$latex \displaystyle u(0)=u_0&fg=000000$

but there is a hide Dragon,we need the condition $latex {\frac{\partial u(0)}{\partial v}|_{M_k}=0}&fg=000000$.
otherwise we will get solution $latex {u \notin W^{1,2}(M_k)\cap C^{2}(M_k)}&fg=000000$.
but in this case we still have the two necessary estimate(esay to see the above argument still make sense).
in this case to prove the short time existence we need follow four claims is ture.
$latex \displaystyle ||u_{i+1,k}(t)||_C^0\longrightarrow ||u||_{i,k}{C^0} &fg=000000$

as $latex {t \longrightarrow 0}&fg=000000$
$latex \displaystyle \int_{M_k}|u_{i+1,k}|^2 \longrightarrow \int_{M_{k}}|u_{i,k}|^2 &fg=000000$

as $latex {t \longrightarrow 0}&fg=000000$
$latex \displaystyle ||u_{i,k+1}(t)||_C^0\longrightarrow ||u||_{i,k}{C^0} &fg=000000$

as $latex {t \longrightarrow 0}&fg=000000$
$latex \displaystyle \int_{M_{k+1}}|u_{i,k+1}|^2 \longrightarrow \int_{M_{k}}|u_{i,k}|^2 &fg=000000$

as $latex {t \longrightarrow 0}&fg=000000$
5. Construct the suitable Banach space

call the space construct follow the Mixed-Holder-Sobolev space for simply case,consider smooth manifold with only one conical singularity.
first cover the whole manifold by a open set have positive distance t=with the conical singularity and a countable group of set $latex {A_n=B(\frac{d}{2^n},p)-B(\frac{d}{2^{n+1}},p)}&fg=000000$,which is balls center at singularity $latex {p}&fg=000000$ with radius $latex {\frac{d}{2^n}}&fg=000000$.(where $latex {M=B(d,p)\cup U}&fg=000000$)
i.e. $latex {M-\{p\}=U \cup (\cup_{i=1}^{\infty}A_i)}&fg=000000$

Definition 2 ($latex {||\cdot||_{\varepsilon^{k,\alpha}(S)}}&fg=000000$)
$latex \displaystyle ||f||_{\varepsilon^{k,\alpha}(S)}=sup_{k=1,2,…,\infty}||f(2^{-k},\theta)||_{C^{k,\alpha}(B_1-B_{\frac{1}{2}})}+||f||_{C^{k,\alpha}(U)}&fg=000000$

easy to see the definition is independent with the cover and the local interpolation coordinate chart.
one thing is also trivial,is that we have the schauder estimate under the norm $latex {||\cdot||_{\varepsilon^{k,\alpha}(S)}}&fg=000000$.
that is
$latex \displaystyle \delta u=f&fg=000000$

on $latex {S-\{p\}}&fg=000000$. $latex {|u|<C_1}&fg=000000$ on $latex {S}&fg=000000$. then
$latex \displaystyle ||u||_{\varepsilon^{k+2,\alpha}}\leq C(||u||_{L^{\infty}}+||f||_{\varepsilon^{k,\alpha}})\leq C(C_1+||f||_{\varepsilon^{k,\alpha}})&fg=000000$

in fact we only need to add each inequality come from each open set of the cover by Schauder estimate to proof this.
on the other hand we need a suitable Sobolev type norm.
Definition 3 ($latex {|\cdot|_w}&fg=000000$)
$latex \displaystyle |u|_w=(\int_S |\tilde \nabla u|^2d \tilde V)^{\frac{1}{2}}&fg=000000$

Definition 4 ($latex {W^{k,\alpha}}&fg=000000$) the set of all f in $latex {\varepsilon^{k,\alpha}}&fg=000000$ with finite $latex {|f|_w}&fg=000000$,
$latex \displaystyle ||f||_W^{k,\alpha}=||f||_{\varepsilon^{k,\alpha}}+|f|_w&fg=000000$

in Banach space.
Assume norm $latex {C^{k,\alpha}(B\times [o,T])}&fg=000000$ on $latex {B\times [0,T]}&fg=000000$

Definition 5 ($latex {||\cdot||_{\rho^{l,\alpha,{0,T}}}}&fg=000000$]
$latex {f: S\times [0,T] \longrightarrow R }&fg=000000$

$latex \displaystyle ||f||_{\rho^{l,\alpha,[0,T]}}=sup_{k=0,1,2,…,\infty}||f(2^{-k}\rho,\theta,4^{-k}t)||_{C^{l,\alpha}((B_1-B_{\frac{1}{2}})\times [0,4^{-k}T])}+||f||_{C^{l,\alpha}(U\times[0,T])}&fg=000000$

\end) from the definition,easy to see
$latex \displaystyle \frac{\partial u}{\partial t}=\Delta u+f&fg=000000$

om $latex {M}&fg=000000$
$latex \displaystyle u|_{t=0}=u_0&fg=000000$

$latex {\Longrightarrow}&fg=000000$
$latex \displaystyle ||u||_{\rho^{l+2,\alpha,[0,T]}}\leq C(||u_0||_{\varepsilon^{l,\alpha}}+||f||_{\rho^{l,\alpha,[0,T]}}+||u||_{C^0(S\times [0,t])})&fg=000000$

easy from the classical schauder estimate.

Definition 6 ($latex {|f|_v}&fg=000000$) $latex {f:S\times [0,T] \longrightarrow R}&fg=000000$
$latex \displaystyle |f|^2_v=max_{t\in [0,T]}\int_S|\tilde \nabla f|^2d\tilde V +\int_0^T\int_M |\frac{\partial f}{\partial t}|^2 d\tilde Vd t&fg=000000$

Key point:

Definition 7 ($latex {\nu^{k,\alpha,{0,T}}}&fg=000000$] $latex {\nu^{k,\alpha,[0,T]}}&fg=000000$ is the set of $latex {f}&fg=000000$ in $latex {\rho^{l,\alpha,[0,T]}}&fg=000000$ with finite $latex {|f|_v}&fg=000000$
$latex \displaystyle ||\cdot||_{\nu^{k,\alpha,[0,T]}}=||\cdot||_{\rho^{l,\alpha,[0,T]}}+|\cdot|_v&fg=000000$

\end)
6. What is a solution of equation

trivial sense:
satisfied equation point-wise on $latex {S-\{p\}}&fg=000000$.
weak sense:
1.trivial case
2.$latex {|u|_v<+\infty}&fg=000000$