Uniqueness of solutions for a differential equation on a manifold I have the following situation:
$M$ is a smooth manifold. Let $A_t$ be a smooth family of real functions on $M$,that is: $A:I \times M \to \mathbb{R}$ is smooth. (In particular, for each $t \in I ,\, A_t=A(t,\cdot) \in C^\infty (M)$).
Let $V\in \Gamma(TM)$. I would like to show the following equation has only one solution:
$\frac{\partial}{\partial t}A=V\cdot A_t \, , A_0=0$.   
How do I prove this? Note the equation is $\mathbb{R}$-linear in $A$. I guess there exist a simple uniquness theorem for linear PDE's which I am missing?
Does anything changes if I assume the manifold has a (non-empty) boundary?

Note:  $V\cdot A_t$ means we first fix a time $t \in I$ and then differentiate in the usual way the function $A_t \in C^\infty (M)$ along the vector field $V$. Thus, for each $t \, , (V\cdot A_t) \in C^\infty (M)$, so we can view it as a function $I \times M \to \mathbb{R}$.
$\frac{\partial}{\partial t}A$ is of the same type. (so the equation is meaningful).
Actually, another way to see these two differentiations, is to consider both $V,\frac{\partial}{\partial t}$ as vector fields of the product manifold $I \times M$. Then $A \in C^\infty(I \times M)$ and the differentiation is again the usual one w.r.t the considered vector fields.

(Of course the trivial solution is $A \equiv 0$).
 A: Since you want to speak about derivatives in $C ^\infty (M)$, you have to endow this with a topology, and the usual requirement of topological completeness will force you to restrict yourself to only a subspace of it.
Let us introduce the notation: $V ^n f  = \underbrace {V (V ( \dots (V} _{n \ \text{times}} f \underbrace {) \dots ))} _{n \ \text{times}}$ (i.e. the derivative along $V$ of the derivative along $V$ of ... of the derivative along $V$ of $f$), with $V ^0 f = f$.
Note also that your equation can be viewed as an ordinary differential equation: $\dfrac {\Bbb d A} {\Bbb d t} (t) = V (A(t))$, with $V$ viewed as a linear operator from functions to functions.
1) You may choose to work with a Banach space structure, since these are very well studied and understood. In this case the natural space to work with seems to be $X = \{ f \in C ^\infty (M) \mid \| f \| < \infty \}$, where the norm is $\| f \| = \sup \limits _{n \in \Bbb N, x \in M} |(V^n f) (x)|$.
Note that $V : X \to X$ is bounded, because
$$\| V \| = \sup \limits _{\| f \| = 1} \| Vf \| = \sup \limits _{\| f \| = 1} \sup \limits _{n \in \Bbb N} \| V^n (Vf) \|_\infty = \\ \sup \limits _{\| f \| = 1} \sup \limits _{n \ge 1} \| V^nf \|_\infty \le \sup \limits _{\| f \| = 1} \sup \limits _{n \ge 0} \| V^nf \|_\infty = \sup \limits _{\| f \| = 1} \| f \| = 1 ,$$
hence it is continuous. Under these assumptions, theorem 5.5 from chapter 5 of "Analysis Tools with Applications" (lecture notes by Bruce K. Driver from the UCSD) guarantees that your Cauchy problem has exactly one solution in $X$.
2) If you find the above Banach space to be too small for your needs, you may consider working in a larger Fréchet space, for the price of slightly increased technicality. Let $M = \bigcup \limits _{n \in \Bbb N} U_n$ be an exhaustion with open subsets (for instance, consider a compact exhaustion; the interiors of these compacts will form an exhaustion with open subsets). Let $p_k :C ^\infty (U_k) \to [0, \infty)$ be given by $p_k (f) = \sup \limits _{n \in \Bbb N, x \in U_k} |(V^n f) (x)|$ and $X_n = \{ f \big| _{U_n} \mid f \in C ^\infty (M), \ p_n (f) < \infty \}$. Note that $p_n$ is a norm on $X_n$, that the range of $V_n = V \big| _{X_n} : X_n \to X$ is $X_n$ and that $V_n$ is continuous, so that $(X_n, p_n)$ is a Banach space. Note that these spaces form a projective system, the projective limit of which is $C ^\infty (M)$. (In these considerations it is necessary to use the trivial remark that $V \big| _U ( f \big| _U) = (VF) \big| _U$.)
You can now use theorem 2.2 from "On a Type of Linear Differential Equations in Fréchet Spaces" by George N. Galanis, that guarantees that your Cauchy problem has exactly one solution.
No matter what, I don't think that anything meaningful can be said about the equation viewed in $C ^\infty (M)$.
A: $\newcommand{\M}{M}$
$\newcommand{\til}{\tilde}$
$\newcommand{\ep}{\epsilon}$
$\newcommand{\Ga}{\Gamma}$
$\newcommand{\brk}[1]{\left(#1\right)} $
$\newcommand{\R}{\mathbb{R}}$
$\newcommand{\pd}[2]{\frac{\partial#1}{\partial#2}}$
Here is a proof, using local existence-and-uniqueness result on Linear First-Order Cauchy problems, which works on all $C^{\infty}(M)$ without any topological restrictions:    
We first prove the theorem for compact manifolds $\M$.
We are going to use the following lemma (A proof can be found in John M.Lee's book Introduction to smooth manifolds - Theorem 9.51 page 240):
Let $\M$ be a smooth manifold. Suppose we are given an embedded hypersurface $S \subseteq M$, a smooth vector field $Y \in \Ga(TM)$ that is nowhere tangent to $S$, and functions $b,f \in C^{ \infty }(\M)$ and $\phi \in C^{\infty}(S)$. Then for some neighbourhood $U$ of $S$ in $M$, there exists a unique solution $u \in C^{\infty}(U)$ to the following (noncharacteristic) Cauchy problem:
$$  Yu+bu=f,$$
  $$  u|_S = \phi$$

The proof (for the compact case):
Define $\til M = I \times \M  $
Note that $S=\{0\} \times \M \subseteq \til M$ is an embedded hypersurface. (It's an inverse image $\pi^{-1}\brk{\{0\}}$ where $\pi$ is the projection $I \times \M \to I$).
Look at the vector field $Y=V-\pd{}{t} \in \Ga\brk{T \til \M}$. 
It's clearly nowhere tangent to $S$ (since $T_{\brk{0,p}}S=0 \oplus T_p\M$ and $Y(t,p)=(-1,V)$). Choose $b,f \in C^\infty\brk{\til \M} \, , \, b,f=0  \, \, \text{and} \, \, \phi \in C^\infty\brk{S} \, , \, \phi =0$. 
Then the above lemma implies that there exists a neighbourhood $U$ of $S=\{0\} \times \M $ in $\til \M = I \times \M$ where there exists a unique solution $u \in C^\infty\brk{U}$ to the equation: $$(*) Y \cdot u = 0,u|_S = 0$$. 
$\forall p \in \M , (0,p) \in S \Rightarrow  (0,p) \in U \Rightarrow$ there exists an open set $\til U_p \subseteq U$ which contains $(0,p)$. Hence, there exists $\ep_p \in \R \, , \, U_p \subseteq \M$ ($U_p$ open in $\M$)  such that $(-\ep_p,\ep_p) \times U_p \subseteq \til U_p$. $\{U_p|p \in \M\}$ form an open cover of $\M$, hence (by compactness of $M$) there is a finite subcover $U_{p_1},\dots ,  U_{p_n}$. 
Define $\ep = min\{\ep_{p_i}|i=1,\dots,n \}$. It follows immediately that $(-\ep,\ep) \times \M \subseteq U$. 
Remember we already have a global solution (zero), and we now have established that it's the unique solution on $(-\ep,\ep) \times \M $.
Now define the following set:
 $X = \{t \in I| \text{there exists a unique solution for the equation in  } (-t,t) \times \M \}$.
Look at $s= \text{sup} X$. We claim $\text{sup} X \in X$. Since $X$ is closed downward, i.e :  $x \in X \Rightarrow \brk{0 \le x' < x \Rightarrow x' \in X}$ it follows that $[0,s) \subseteq X$. 
It's easy to see that $u=0$ must be the unique solution for $(*)$ on $(-s,s) \times \M$. (If there were two differnet solutions, they would differ already at some $s'<s$ which contradicts $[0,s) \subseteq X$). Hence, any solution must be zero on $(-s,s) \times \M$, so by definition of $X$ , $s \in X$. 
By continuity, any solution must be zero on $[-s,s] \times \M$.  So, if we assume, that $s <1$, we can advance the uniqueness further, thus obtaining a contradiction.
In more detail, the same uniqueness result  which took us from $0$ to $\ep$ will take as from $s \to s+\ep'$ (and also for $-s$).  We will choose: $S = \{s\} \times \M$, and demand the solution to be zero on $S$ (this our initial conditions constraint this time) , and the same theorem will ensure us again uniqueness on some neighbourhood of $S$ in $\til \M = I \times \M$ ,so we can advance further: We already had uniqueness on $[-s,s]$, and now we have uniqueness on $(s-\ep',s+\ep')$, $(-s-\ep',-s + \ep') $ so together we have uniqueness on $(-(s+\ep'),s+\ep')$ so $s+\ep' \in X$ which contradicts the definition of $s$ as the supremum. 
We showed there $s=1 \in X$, so $f=0$ is the unique solution $u$ for $(*)$ defined on $\til \M = [-1,1] \times \M$ as required.

Proof of the non-compact case: 
I think it's enough to prove uniqueness locally (since we already have global existence). 
Now we can just take a coordinate ball around each point, and take a compact subset which contains an open subset. For each compact subset of this form we apply our previous argument. Now we 'glue' all these uniqueness results together, to obtain global uniqueness.
