How to exponentiate the vector field $v$?

Question

i didn't succed to exponentiate these vector field $$\boxed{v=\sqrt{x}\sqrt{y}\frac{\partial}{\partial x}-\sqrt{x}\sqrt{y}\frac{\partial}{\partial y}+2b\sqrt{x}\sqrt{y}u\frac{\partial}{\partial u}}$$ where $x,y,t$ are independents variables and $u$ is an dependant variable . this vector field is one of the basis element of a Lie algebra of symmetry associated to a certain pde .the solution of the pde is of the form $u(x,y,t)$

Remark:

i succeded to exponentiate the vector field $$w=\sqrt{x}\sqrt{y}\frac{\partial}{\partial x}-\sqrt{x}\sqrt{y}\frac{\partial}{\partial y}$$ it gives $$\exp \left(\varepsilon w\right)(x,y, t, u)=(\tilde{x},\tilde{y},\tilde{t}, \tilde{u})=\left(\sqrt{x}\sqrt{y}\sin(\varepsilon)+(\frac{x-y}{2})\cos(\varepsilon)+(\frac{x+y}{2}),-\sqrt{x}\sqrt{y}\sin(\varepsilon)-(\frac{x-y}{2})\cos(\varepsilon)-(\frac{x+y}{2}),t,u\right) $$ where $exp : \mathfrak{g}\longrightarrow G$ and $\mathfrak{g}$ is the Lie algebra of $G$ .

to exponentiate a vector field of the form $\mathbf{v}=\displaystyle{\sum_{k=1}^{p}\xi_{k}(x,u)\frac{\partial}{\partial x_{k}}+\sum_{i=1}^{q}\phi_{i}(x,u)\frac{\partial}{\partial u_{i}}}$we solve the system of odes $ \displaystyle{\frac{d\tilde{x_{i}}}{d\varepsilon}=\xi_{i}(\tilde{x_{i}})\,\,,\,\, \frac{d\tilde{u}}{d\varepsilon}=\phi(\tilde{x},\tilde{t},\tilde{y})}$ with initial conditions $\tilde{x_{i}}(0)=x_{i}$ , $\tilde{u}(0)=u$

Or we can just develop this Lie series for the flow, given by $$ \exp (\varepsilon \mathbf{v}) \cdot x = x+\varepsilon \xi(x)+\frac{\varepsilon^2}{2} \mathbf{v}(\xi)(x)+\cdots=\sum_{k=0}^{\infty} \frac{\varepsilon^k}{k !} \mathbf{v}^k(x) $$ where $\xi=\left(\xi^1, \ldots, \xi^m\right), \mathbf{v}(\xi)=\left(\mathbf{v}\left(\xi^1\right), \ldots, \mathbf{v}\left(\xi^m\right)\right)$,

I tried to develop the serie but it's getting more and more complicated from order 3 i guess

Thanks for the help !:)

Answer 1

For readability, denote the initial condition by $$(x(0), y(0), t(0), u(0)) = (x_0, y_0, t_0, u_0),$$ so that $$(x(\tau), y(\tau), t(\tau), u(\tau)) = \exp(\tau V)(x_0, y_0, t_0, u_0) .$$

Hint Since the $\partial_x$ and $\partial_y$ components of the vector field $V$ depend only on $X, Y$, we may as well first solve for the integral curves $(x(\tau), y(\tau))$ of the projection of $V$ to the $xy$-plane, namely, $W$---that is, solve $$x' = \sqrt{x y}, \qquad y' = -\sqrt{x y}$$ ---then lift the solution curve to $xytu$-space.

We can simplify our computation some by observing that $x' + y' = 0$, hence $x + y = x_0 + y_0$, which reduces our system to a single, separable o.d.e. in $x(t)$: $$x' = \sqrt{x (x_0 + y_0 - x)} .$$ As in the edited question statement, solving gives \begin{align*} x(\tau) &= \frac{1}{2}\left[x_0 + y_0 + 2 \sqrt{x_0 y_0} \sin \tau + (x_0 - y_0) \cos \tau\right] \\ y(\tau) &= \frac{1}{2}\left[x_0 + y_0 - 2 \sqrt{x_0 y_0} \sin \tau - (x_0 - y_0) \cos \tau\right] . \end{align*} To compute $u(\tau)$ efficiently, we can compare coefficients of the vector find and observe that $$u' = 2 b u x'.$$

Integrating gives $$u(\tau) = u_0 \exp \left[2 b \left(x(\tau) - x_0\right)\right] = u_0 \exp \left[ b \left(2 \sqrt{x_0 y_0} \sin \tau + (x_0 - y_0) (\cos \tau - 1)\right) \right] .$$

Of course, since the vector field $V$ does not have a $\partial_t$-component, $t(\tau) = t_0$.