**Definitions**

Suppose $A$ is a commutative algebra over $\mathbb{R}$ with unity. $\mathbb{R}$-linear map $\xi\colon A\to A$ is a derivation of $A$ iff $\xi(ab)=a\xi(b)+\xi(a)b$ for any $a,b\in A$. If $\gamma\colon \mathbb{R}\to A$, $a\in A$, then we say, that $a=\frac{\partial}{\partial t}| _ {t=\tau} \gamma(t)$ iff $h(a) =\frac{\partial}{\partial t}|_{t=\tau} h(\gamma(t))$ for any $\mathbb{R}$-linear map $h\colon A\to\mathbb{R}$.
Suppose $\xi$ is a derivation of $A$. Then $\Phi\colon A\times\mathbb{R}\to A$ is it's flow iff $\Phi(a,0)=a$ for any $a\in A$ and

$$\frac{\partial}{\partial t} \Phi(a,t) = \xi \Phi(a,t).\tag{1}$$

**The question**

1. I like algebras $A$, such that any derivation of $A$ possesses a flow. Is there any simple sufficient condition for them?

2. Is there any simple condition for an algebra and it's derivation, from which it follows, that this derivation possesses a flow?

**Examples**

1. Algebra $C^\infty(M)$ of smooth functions on a closed manifold $M$ --- yes (if I haven't made a mistake), any derivation possesses a flow. This, I believe, can be checked using Picard-Lindelof theorem.

1'. Algebra $C^\infty(M)$ of smooth functions on a non-compact manifold without boundary --- no (see example 2), but a derivation possesses a flow if it preserves some function $H\in A$, such that for any $c\in\mathbb{R}$ subspace $\{x\mid H(x) < c \}$ of the topological space $M$ is compact.

2. Algebra $C^\infty((0,1))$ of smooth functions on an interval --- no, $\frac{\partial}{\partial x}$ does not possess a flow.

3. Algebra $C^\infty([0,1])$ of smooth functions on a segment --- yes, any derivation possesses a flow, but it's not always unique (for example, for $\frac{\partial}{\partial x}$ it is not). In order to prove this, one can consider an embedding of $[0,1]$ to some closed manifold $N$ and prolong any function from $[0,1]$ to $N$. Then use example 1.

4. Algebra $C^\infty(\mathbb{R})$ --- no, because it is isomorphic to the algebra from example 2.

5. Algebra $\mathbb{R}[x]$ --- no. In order to prove this one can consider derivation $\xi=x^2\frac{\partial}{\partial x}$ and manually solve equation (1) for $a=x$. Any solution locally should be of the form $\frac{x}{1+xt}$. It is not in $\mathbb{R}[x]$.

6. Algebra $\mathbb{R}[x,y]/(x^2+y^2-1)$ --- no. In order to prove this take $\xi = y (x\frac{\partial}{\partial y} - y \frac{\partial}{\partial x})=\sin(\varphi)\frac{\partial}{\partial \varphi}$ and solve equation (1) manually (locally) in polar coordinates (take, for example, $a=y$). Check that the answer is not a polynomial.

7.(from Greg Muller's answer) Algebra $A$ is finite-dimensional --- yes, every derivation possesses the flow.

This question was already posted here on math.stackexchange.com, but it has received no answers even with a bounty. Any help is appreciated, both in the theme of question and in improving its wording.