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I'm currently reading a textbook on the application of Lie symmetries to differential equations (Symmetry Methods for Differential Equations: A Beginner's Guide Hydon, Peter. Cambridge University Press. 2000.) I'm somewhat at the beginning (pg. 22-25) where a method is being discussed to find canonical coordinates (coordinates that allow you to have some translational symmetry to and ODE, I.E. [itex](\hat{x}, \hat{y}) = (x, y + \epsilon)[/itex].) Let's assume that the coordinate system we are working in does not have a translational symmetry. Currently we're trying to find some coordinate system that does have such a symmetry (because ODEs are solvable via integration in the presence of said symmetries..) So we're trying to find some (r, s) = (r(x, y), s(x, y)) such that [itex](\hat{r}, \hat{s}) = (r(\hat{x}, \hat{y}), s(\hat{x}, \hat{y})) = (r, s + \epsilon)[/itex]. Anything variable with a "hat" on it is in the transformed coordinate system (or the destination of the mapping in the same coordinate system, whatever you want to call it...) However, this book appears to be clinically terse in "explanations". The section in the book I'm confused on is as follows:
In the new coordinates [itex](\hat{r}, \hat{s})[/itex] the tangent vector to an orbit traced out by the continuum of epsilon values has a tangent vector at the point (r, s) of (0, 1):[tex]\frac{d \hat{r}}{d \epsilon} = 0, \quad \frac{d \hat{s}}{d \epsilon} = 1[/tex]
These should both be evaluated at [itex]\epsilon[/itex] = 0, but I don't know how to do that in LaTeX.
By the chain rule ([itex]\hat{r}[/itex] is a function of 'x' and 'y' which are in turn functions of '[itex]\epsilon[/itex]'):[tex]EQN\#1a: \quad \xi{(x, y)} r_{x} + \eta{(x, y)} r_{y} = 0[/tex][tex]EQN\#2b: \quad \xi{(x, y)} s_{x} + \eta{(x, y)} s_{y} = 1[/tex]Canonical coordinates can be obtained from EQUATIONS #1 by using the method of characteristics.
Note that I believe [itex]\frac{dx}{{d}{\epsilon}} = \xi{(x, y)}[/itex] and [itex]\frac{dy}{{d}{\epsilon}} = \eta{(x, y)}[/itex]. I KNOW that [itex]r_{x} = \frac{{\partial}{r}}{{\partial}{x}}[/itex]. Thus EQN#1 is a chain rule.
The characteristic equations are:[tex]\frac{dx}{\xi{(x, y)}} = \frac{dy}{\eta{(x, y)}} = ds[/tex]A first integral of a given first-order ODE[tex]EQN\#2: \quad \frac{dy}{dx} = f(x, y)[/tex]is a non-constant function [itex]\phi(x, y)[/itex] whose value is constant on any solution y = y(x) of the ODE (Equation #2). Therefore:[tex]EQN\#3: \quad \phi_{x} + f(x, y) \phi_{y} = 0 \quad \phi_{y} \neq 0[/tex]The general solution of the ODE (Equation #2) is:[tex]\phi (x, y) = c[/tex]Suppose that [itex]\xi (x, y) \neq 0[/itex]. Comparing equations #1 and #3, we see that the invariant canonical coordinate 'r' is a first integral of[tex]EQN\#4: \quad \frac{dy}{dx} = \frac{\eta{(x, y)}}{\xi{(x, y)}}[/tex]So [itex]r = \phi (x, y)[/itex] is found by solving equation #4...
My question is: I'm VERY confused as to how the equation [tex]\frac{dx}{\xi{(x, y)}} = \frac{dy}{\eta{(x, y)}} = ds[/tex] comes up. Where did this thing come from? What's all this talk about [itex]\phi (x, y)[/itex]? I realize EQN#3 is the "characteristic equation", but I don't see how it takes the form that it does. Previously the book had said that the characteristic equation was something like:[tex]Q = \eta (x, y) - y' \xi (x, y)[/tex][tex]where \quad \frac{dy}{dx} = f(x, y)[/tex] Why is the general solution to the ODE simply [itex]\phi (x, y) = c[/itex] after the book said that the function wasn't constant... How in the world do equations #1 and #3 combine to get equation #4? I'm so utterly confused I had to make this entirely lengthy verbatim post because I can't break down this problem into a simple question.. I will say that I suspect my problems have something to do with the functional relationships of [itex]x, y, \hat{x}, \hat{y}, r, s, \hat{r}, and \hat{s}[/itex], because that's the added complication to this problem that a previous problem did not have where there was a translational symmetry in just straight 2D cartesian coordinates without having to transform to a canonical coordinate space.
I guess can anybody fill in the gaps on this explanation? I know I left things out, I'm hoping this is a standard/familiar enough problem that people can guide me the right direction even if I didn't quite have the ability to explain things correctly (after all I'm pretty confused about the whole thing..)
In the new coordinates [itex](\hat{r}, \hat{s})[/itex] the tangent vector to an orbit traced out by the continuum of epsilon values has a tangent vector at the point (r, s) of (0, 1):[tex]\frac{d \hat{r}}{d \epsilon} = 0, \quad \frac{d \hat{s}}{d \epsilon} = 1[/tex]
These should both be evaluated at [itex]\epsilon[/itex] = 0, but I don't know how to do that in LaTeX.
By the chain rule ([itex]\hat{r}[/itex] is a function of 'x' and 'y' which are in turn functions of '[itex]\epsilon[/itex]'):[tex]EQN\#1a: \quad \xi{(x, y)} r_{x} + \eta{(x, y)} r_{y} = 0[/tex][tex]EQN\#2b: \quad \xi{(x, y)} s_{x} + \eta{(x, y)} s_{y} = 1[/tex]Canonical coordinates can be obtained from EQUATIONS #1 by using the method of characteristics.
Note that I believe [itex]\frac{dx}{{d}{\epsilon}} = \xi{(x, y)}[/itex] and [itex]\frac{dy}{{d}{\epsilon}} = \eta{(x, y)}[/itex]. I KNOW that [itex]r_{x} = \frac{{\partial}{r}}{{\partial}{x}}[/itex]. Thus EQN#1 is a chain rule.
The characteristic equations are:[tex]\frac{dx}{\xi{(x, y)}} = \frac{dy}{\eta{(x, y)}} = ds[/tex]A first integral of a given first-order ODE[tex]EQN\#2: \quad \frac{dy}{dx} = f(x, y)[/tex]is a non-constant function [itex]\phi(x, y)[/itex] whose value is constant on any solution y = y(x) of the ODE (Equation #2). Therefore:[tex]EQN\#3: \quad \phi_{x} + f(x, y) \phi_{y} = 0 \quad \phi_{y} \neq 0[/tex]The general solution of the ODE (Equation #2) is:[tex]\phi (x, y) = c[/tex]Suppose that [itex]\xi (x, y) \neq 0[/itex]. Comparing equations #1 and #3, we see that the invariant canonical coordinate 'r' is a first integral of[tex]EQN\#4: \quad \frac{dy}{dx} = \frac{\eta{(x, y)}}{\xi{(x, y)}}[/tex]So [itex]r = \phi (x, y)[/itex] is found by solving equation #4...
My question is: I'm VERY confused as to how the equation [tex]\frac{dx}{\xi{(x, y)}} = \frac{dy}{\eta{(x, y)}} = ds[/tex] comes up. Where did this thing come from? What's all this talk about [itex]\phi (x, y)[/itex]? I realize EQN#3 is the "characteristic equation", but I don't see how it takes the form that it does. Previously the book had said that the characteristic equation was something like:[tex]Q = \eta (x, y) - y' \xi (x, y)[/tex][tex]where \quad \frac{dy}{dx} = f(x, y)[/tex] Why is the general solution to the ODE simply [itex]\phi (x, y) = c[/itex] after the book said that the function wasn't constant... How in the world do equations #1 and #3 combine to get equation #4? I'm so utterly confused I had to make this entirely lengthy verbatim post because I can't break down this problem into a simple question.. I will say that I suspect my problems have something to do with the functional relationships of [itex]x, y, \hat{x}, \hat{y}, r, s, \hat{r}, and \hat{s}[/itex], because that's the added complication to this problem that a previous problem did not have where there was a translational symmetry in just straight 2D cartesian coordinates without having to transform to a canonical coordinate space.
I guess can anybody fill in the gaps on this explanation? I know I left things out, I'm hoping this is a standard/familiar enough problem that people can guide me the right direction even if I didn't quite have the ability to explain things correctly (after all I'm pretty confused about the whole thing..)