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Problem: Let $\theta:\mathbb{R}^2 \rightarrow \mathbb{R}$ be defined by $\theta(p,t) = p + t^3$. Prove that it is not the flow of any vector field on $\mathbb{R}$.

Attempt: I am not quite sure what approach to take here, I can get $\frac{\partial}{\partial t}(f (\theta(p,t))) = \frac{\partial}{\partial x}f(\theta(p,t)) \frac{\partial}{\partial t}(\theta(p,t))$, so $\frac{\partial}{\partial t} \theta(p,t) = 3t^2 \frac{\partial}{\partial x}|_{\theta(p,t)}$. Why is this not a vector field then? Any help is appreciated.

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A flow must satisfy the property $$ \theta(p, s + t) = \theta(\theta(p, s), t) \quad \text{ for all } p, s, t \in \mathbb{R}. $$ In your case this translates into $$ p + (s + t)^3 = (p + s^3 ) + t^3. $$

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