Lie Theory is definitely not my field, but I hope my question will be however meaningful.
I'm studying an article in which the vector field (in the plane) $X_1 = \partial_x$ and $X_2 = |x|^\alpha \partial_y$ is considered, with $\alpha >0 $. Then two curves are defined, which are supposed to connect the points $(\psi(y),y)$ and $(\psi(y-h),y-h)$, where $\psi$ is a $\mathcal{C}^1$ function defined on a certain open interval of $\mathbb{R}$. The first curve is defined by $$\gamma_1 (t) := \exp(t(X_1 - b X_2))(\psi(y),y)$$and the author says that this last expression is equal to $$\left(\psi(y) + t,y-b\int_0^t |\psi(y) + \tau|^\alpha \, d \tau \right)$$where $b = \min \{1,1/L\}$ and $L:=\sup_{|y|<\delta} | \psi' (y)|$. How did he get this last expression? I'm not familiar at all with exponentials of vector fields...
Thank you in advance!
It means you look at the linear combination of the two vector fields, which is the vector field $$X = X_1 - bX_2 = \partial_x - b|x|^{\alpha} \, \partial_y = \frac{\partial}{\partial x} - b|x|^{\alpha} \, \frac{\partial}{\partial y}.$$ Finding the vector field's exponent means you are finding the phase flow of the vector fields $$(x,y) \mapsto \Phi^t(x,y) = \Big(\Phi^t_1(x,y), \, \Phi^t_2(x,y)\Big) $$ i.e. it describes the curves tangent to the vector field, i.e. satisfies the condition $$\frac{d}{dt} \Phi^t(x,y) = X{\big(\Phi^t(x,y)\big)} \, , \, \,\,\, \Phi^0(x,y) = (x,y)$$ The phase flow $\Phi^t$, also called the exponent of the vector field $\exp(tX) = \Phi^t$ is obtained by solving the system of differential equations \begin{align} \frac{dx}{dt} &= 1 \\ \frac{dy}{dt} &= -b |x|^{\alpha} \end{align} with an arbitrary initial condition $x(0) = x_0$ and $y(0) = y_0$. This initial value problem is easy to solve: \begin{align} x(t) &= x_0 + t\\ y(t) &= y_0 + \int_{0}^t |x_0 + \tau|^{\alpha} \, d\tau \end{align} With initial conditions $x_0 = \psi(y)$ and $y_0 = y$ you get \begin{align} x(t) &= \psi(y) + t\\ y(t) &= y + \int_{0}^t |\psi(y) + \tau|^{\alpha} \, d\tau \end{align} exactly what you have. The choices of $b$ and $L$ have to do with something else, related to the specific problem you are reading about.