For any real positive number \(n\) show that the vector field \(X=x^{n} \partial_{x}\) is differentiable on the manifold \(\mathbb{R}^{+}\)consisting of the positive real line \([x \in \mathbb{R} \mid x>0]\). Why is this not true in general on the enture real line \(\mathrm{R}\) ? As done for the case \(n=2\) in Example 15,13, find the maximal one-parameter subgroup \(\sigma_{i}\) generated by this vector field at any point \(x>0\).

Short Answer

Expert verified
The vector field is differentiable on \(\mathbb{R}^{+}\) because \(x^n\) is continuously differentiable for any positive real number \(x\). It's not differentiable on the entire real line \(\mathbb{R}\) because the derivative at \(x=0\) doesn't exist if \(n<1\). The maximal one-parameter subgroups generated by this vector field at any \(x>0\) are \(((1-n)t + x_0^{1-n})^{1/(1 - n)}\) for \( n ≠ -1\) and \(x_0e^{t}\) for \( n=-1\).

Step by step solution

01

Prove the given vector field is differentiable

The given vector field is \(X=x^{n} \partial_{x}\). To show that it's differentiable on the manifold \(\mathbb{R}^{+}\), we need to show its smoothness at every point of the manifold. Since for any positive real number \(x>0\), \(x^n\) is also continuously differentiable, thus the vector field \(X=x^{n} \partial_{x}\) is smooth or differentiable on the manifold \(\mathbb{R}^{+}\).
02

Explain why it's not differentiable on the entire real line

The vector field \(X=x^{n} \partial_{x}\) is not smooth at \(x=0\), because the derivative doesn't exist at this point if \(n<1\) . This is why the vector field is not differentiable on the entire real line \(\mathbb{R}\).
03

Finding the maximal one-parameter subgroup

The maximal one-parameter subgroup generated by a vector field \(X\) at any point \(x_0\) is the integral curve of \(X\) with initial condition \(\sigma_i(0) = x_0\). We start by solving the ordinary differential equation related to \(X\) vector field. The ODE here is \(\frac{dx}{dt} = x^n\). When separating and integrating the variables we find that if \( n ≠ -1\), \(\sigma_i(t) = ((1-n)t + x_0^{1-n})^{1/(1 - n)}\) and when \( n=-1\), \(\sigma_i(t) = x_0e^{t}\). Hence, these are the maximal one-parameter subgroups for each condition.

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