How to properly apply the Lie Series

Question

I am trying to solve this problem from Symmetry Methods for Differential Equations A Beginner's Guide (Peter E. Hydon).

Use the Lie Series $$F(\hat{x},\hat{y})=\sum_{j=0}^{\infty}\frac{\varepsilon^j}{j!}X^j F(x,y)$$

where $$X = \xi(x,y)\partial_x +\eta(x,y)\partial_y$$

to verify, that $$\hat{x}=\exp{(\varepsilon X)}x,$$ $$\hat{y}=\exp{(\varepsilon X)}y$$

holds for

$$a) X = x\partial_x-y\partial_y$$ $$b) X = x^2\partial_x+xy\partial_y$$ $$c) X = -y\partial_x+x\partial_y$$

I don't even know where to start ... . A step by step solution for one of the $X$ would be nice :).

Answer 1

A relevant reference is found here:

• Exponential of a function times derivative

It is advised to absorb this one-dimensional theory first, before proceeding to 2-D.

$c)\; X = -y\, \partial_x+x\, \partial_y$

Disclaimer. In our (LaTeX) notes we have $f$ instead of $F$ , $(x_1,y_1)$ instead of $(\hat{x},\hat{y})$ , $\theta$ instead of $\varepsilon$ , $k$ instead of $j$ , and more. I didn't replace notations because my eyes are bad and it is expected that the danger of making mistakes is greater than the advantage of being consistent with the question.

An example of a Continuous Transformation in two dimensions is a Rotation over an angle $\theta$: $$ \left\{ \begin{array}{c} x_1 = \cos(\theta) . x - \sin(\theta) . y \\ y_1 = \sin(\theta) . x + \cos(\theta) . y \end{array} \right. $$ It might be asked how rotation of the coordinate system works out for a function of these variables. With other words, how the following function would be expanded as a Taylor series expansion around the original $f(x,y)$: $$ f_1(x,y) = f(x_1,y_1) = f(\,\cos(\theta).x - \sin(\theta).y\, , \, \sin(\theta).x + \cos(\theta).y\, ) $$ Define other (polar) variables $(r,\phi)$ as: $$ x = r.\cos(\phi) \quad \mbox{and} \quad y = r.\sin(\phi) $$ Giving for the transformed variables: $$ x_1 = r.\cos(\phi).\cos(\theta) - r.\sin(\phi).\sin(\theta) = r.\cos(\phi+\theta) \\ y_1 = r.\cos(\phi).\sin(\theta) + r.\sin(\phi).\cos(\theta) = r.\sin(\phi+\theta) $$ We see that $\phi$ is a proper canonical variable. Another function $g(\phi)$ is defined with this canonical variable as the independent one: $$ g(\phi) = f(\,r.\cos(\phi)\, ,\,r.\sin(\phi)\,) = f(x,y) $$ Now rotating $f(x,y)$ over an angle $\theta$ corresponds with a translation of $g(\phi)$ over a distance $\theta$. Therefore $g(\phi+\theta)$ can be developed into a Taylor series around the point of departure: $$ g(\phi+\theta) = g(\phi) + \theta.\frac{dg(\phi)}{d\phi} + \frac{1}{2} \theta^2.\frac{d^2g}{d\phi^2} + ... $$ Working back to the original variables $(x,y)$ with a well known chain rule for partial derivatives: $$ \frac{dg}{d\phi} = \frac{\partial g}{\partial x}\frac{dx}{d\phi} + \frac{\partial g}{\partial y}\frac{dy}{d\phi} $$ Where: $$ \frac{dx}{d\phi} = - r.\sin(\phi) = - y \quad \mbox{and} \quad \frac{dy}{d\phi} = + r.\cos(\phi) = + x \quad \Longrightarrow \\ \frac{dg}{d\phi} = \frac{\partial g}{\partial x}.(-y) + \frac{\partial g}{\partial y}.(+x) \quad \Longrightarrow \quad \frac{d}{d\phi} = x.\frac{\partial}{\partial y} - y.\frac{\partial}{\partial x} $$ Herewith we find that $X = (x.\frac{\partial}{\partial y} - y.\frac{\partial}{\partial x})$ is the infinitesimal operator for Plane Rotations.
It is equal to differentiation with respect to the canonical variable, as expected. The end-result is: $$ f_1(x,y) = \sum_{k=0}^{\infty} \frac{1}{k!} \left[ \theta \left(x.\frac{\partial}{\partial y} - y.\frac{\partial}{\partial x}\right) \right]^k f(x,y) = e^{ \theta (x\, \partial / \partial y - y\, \partial / \partial x) } f(x,y) $$ This is true for any function $f(x,y)$. In particular, the independent variables themselves can be conceived as such functions. Which means that: $$ x_1 = e^{ \theta (x\, \partial / \partial y - y\, \partial / \partial x) } x \quad \mbox{and} \quad y_1 = e^{ \theta (x\, \partial / \partial y - y\, \partial / \partial x) } y $$ It is easily demonstrated that: $$ (x\frac{\partial}{\partial y} - y\frac{\partial}{\partial x}) x = - y \quad \mbox{and} \quad (x\frac{\partial}{\partial y} - y\frac{\partial}{\partial x}) y = x $$ Herewith we find: $$ \sum_{k=0}^{\infty} \left[ \theta (x.\frac{\partial}{\partial y} - y.\frac{\partial}{\partial x}) \right]^k x = 1 - \theta.y - \frac{1}{2} \theta^2.x + \frac{1}{3!} \theta^3.y + \frac{1}{4!} \theta^4.x + ... \\ = \cos(\theta).x - \sin(\theta).y = x_1 $$ Likewise we find: $$ \sum_{k=0}^{\infty} \left[ \theta (x.\frac{\partial}{\partial y} - y.\frac{\partial}{\partial x} ) \right]^k y = 1 + \theta.x - \frac{1}{2} \theta^2.y - \frac{1}{3!} \theta^3.x + \frac{1}{4!} \theta^4.y + ... \\ = \sin(\theta).x + \cos(\theta).y = y_1 $$ Thus, indeed, the formulas for a far-form-infinitesimal rotation over an finite angle $\theta$ can be reconstructed from the expansions.

$a)\; X = x\, \partial_x-y\, \partial_y$

Read the 1-D reference. We have the following results there: $$ e^{ln(\lambda) \,x \frac{d}{dx}} f(x) = f(\lambda\,x) $$ Where $\lambda$ is a positive scaling factor. We also have: $$ e^{-ln(\lambda) \,x \frac{d}{dx}} f(x) = e^{ln(1/\lambda) \,x \frac{d}{dx}} f(x) = f(x/\lambda) $$ These results translate to 2-D in the following manner: $$ e^{ln(\lambda) \,x \frac{\partial}{\partial x}} f(x,y) = f(\lambda\,x,y) \\ e^{-ln(\lambda) \,y \frac{\partial}{\partial y}} f(x,y) = f(x,y/\lambda) $$ The two exponents are commutative, so we can write, with $\;\ln(\lambda)=\mu\;\Longrightarrow\;\lambda=e^\mu=\exp(\mu)$ : $$ e^{\mu(x\, \partial_x - y\, \partial_y)}\; f(x,y) = f\left(e^\mu x,e^{-\mu} y\right) $$ In particular, with $\;X = x \frac{\partial}{\partial x} - y \frac{\partial}{\partial y}$ : $$ \exp(\mu X) x = exp(\mu) x \quad \mbox{and} \quad \exp(\mu X) y = \exp(-\mu) y $$

$b)\; X = x^2\,\partial_x+xy\,\partial_y$

As for this case, I don't see how we can say more than, in the OP's notation: $$ F(\hat{x},\hat{y})=\sum_{j=0}^{\infty}\frac{\varepsilon^j}{j!}X^j F(x,y) = \exp{(\varepsilon X)} F(x,y) $$ Then it follows that, for special functions $F(x,y)=x$ or $F(x,y)=y$ : $$ \hat{x}=\exp{(\varepsilon X)}x \\ \hat{y}=\exp{(\varepsilon X)}y $$ Please don't tell me that's all you want ..

Answer 2

An alternative and much more effective approach is proposed in this second answer. References:

• Exponential of a function times derivative
• How to derive these Lie Series formulas

In the first reference, the following formula is proved: $$ x_1(t) = e^{t\,g(x)\frac{d}{dx}} x \quad \Longleftrightarrow \quad \dot{x}_1(t) = g(x_1(t)) \quad \mbox{with} \quad x = x_1(0) $$ But I've found in my old notes that there is a far more general result. For any one $(t)$ parameter Lie group the following theorem holds: $$ {\bf x}_1(t) = e^{t X} {\bf x} = e^{t\,{\bf g(x)}\cdot\nabla} {\bf x} \quad \Longleftrightarrow \quad \dot{{\bf x}}_1(t) = {\bf g}({\bf x}_1(t)) \quad \mbox{with} \quad {\bf x} = {\bf x}_1(0) $$ Meaning that the problem of finding the Lie series for the independent variables ${\bf x}$ can be reduced to solving a system of ordinary differential equations. In the two dimensional case: $$ X = {\bf g(x)}\cdot \nabla = \xi(x,y)\,\partial_x +\eta(x,y)\,\partial_y $$ So we only have to solve the ODE system: $$ \left\{\begin{matrix}\dot{x}_1 = \xi(x_1,y_1) \\ \dot{y}_1 = \eta(x_1,y_1)\end{matrix}\right. $$ With boundary conditions, always the same: $$ \left\{\begin{matrix} x_1(0) = x \\ y_1(0) = y\end{matrix}\right. $$ Now let's do it for the operators at hand.

$a)\; X = x\, \partial_x-y\, \partial_y$

Accompanying ODE system: $$ \left\{\begin{matrix}\dot{x}_1 = x_1 \\ \dot{y}_1 = -y_1\end{matrix}\right. $$ Together with the boundary conditions giving a solution as has been found in the first answer: $$ \left\{\begin{matrix} x_1 = x\,e^t \\ y_1 = y\,e^{-t} \end{matrix}\right. $$

$b)\; X = x^2\,\partial_x+xy\,\partial_y$

Accompanying ODE system: $$ \left\{\begin{matrix}\dot{x}_1 = x_1^2 \\ \dot{y}_1 = x_1 y_1\end{matrix}\right. $$ Solve the first equation and substitute the solution into the second one. Solve again and apply the boundary conditions: $$ \left\{\begin{matrix} x_1 = {\Large \frac{x}{1-t\,x}} \\ y_1 = {\Large \frac{y}{1-t\,x}} \end{matrix}\right. $$ This solution deserves some attention because it is not in the first answer.

$c)\; X = -y\, \partial_x+x\, \partial_y$

Accompanying ODE system: $$ \left\{\begin{matrix}\dot{x}_1 = -y_1 \\ \dot{y}_1 = x_1\end{matrix}\right. $$ Two separate equations for $x_1$ and $y_1$ can be found from this: $$ \left\{\begin{matrix}\ddot{x}_1 + x_1 = 0 \\ \ddot{y}_1 + y_1 = 0 \end{matrix}\right. \quad \Longrightarrow \quad \left\{\begin{matrix}x_1 = A\cos(t) + B\sin(t) \\ y_1 = C\cos(t) + D\sin(t) \end{matrix}\right. $$ Employing the original ODE : $\;\dot{y}_1=x_1$ gives $\;-C = B\;$ and $\;D = A\;$ .
At last, apply the boundary conditions: $$ \left\{\begin{matrix}x_1 = x\cos(t) - y \sin(t) \\ y_1 = x\sin(t) + y \cos(t)\end{matrix}\right. $$ As has been found in the first answer too.