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Lie Symmetries Question

  1. Sep 24, 2012 #1
    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..)
  2. jcsd
  3. Sep 25, 2012 #2
    In symmetry analysis, you arrive at first order linear PDE's, e.g. your equations 1a,2b. So some background of first order PDE's is necessary:

    Consider the following general first order PDE:
    [itex]F = P(x,y,z)z_{x}+Q(x,y,z)z_{y}-R(x,y,z)=0[/itex]

    The PDE simply says that the scalar product of [P,Q,R] and [itex][z_x,z_y,-1][/itex] vanishes. Because they vanish, these two vectors are perpendicular.

    We also have the relationship that [itex]dz=z_xdx +z_ydy[/itex] due to the chain rule.
    Now, the normal vector in a point B(x,y,z) on the solution curve z(x,y) is simply [itex][z_x,z_y,-1][/itex]. You might see this intuitively, but it is also given by the chain rule above.

    Now, because [itex][z_x,z_y,-1][/itex] is normal to the solution surface, and because this vector is perpendicular to [P,Q,R], therefore [P,Q,R] must be tangent to the solution surface z(x,y).

    Now, define a curve s on the solution z(x,y). We can interpret s as a coordinate axis projected onto the solution and we can express x,y,z in terms of s: x(s),y(s),z(s). We now know that the gradients of x(s),y(s),z(s) with respect to s (which is lying on the solution curve) have to obey:
    dx/ds = P
    dy/ds = Q
    dz/ds = R
    because these gradients are given by the direction vector (P,Q,R) above. You can rewrite this as:
    or: dx/P = dy/Q = dz/R

    You can also derive this more formally by differentiating your first order PDE with respect to s and determine when dx/ds etc will satisfy the pde F=0.

    Note that in [itex]Q =\eta -y' \xi[/itex], you satisfy the characteristic equations if Q=0 (y'=dy/dx). So, yes it is just the characteristic equation, slightly rewritten.

    Also, the first integral leads to an equation of the form [itex]\phi(x,y)=C[/itex]. It doesn't mean that the first integral itself is constant (except on the solution curve!), for example the first integral can be [itex]\phi(x,y)=x^2+y^2=C[/itex].

    Hope this helps. I found Hydon's book useful, but the classic is from Bluman and Kumei. Also, The book of Hans Stephani was quite enjoyable.
  4. Sep 26, 2012 #3
    Thank you for the reply bigfooted, but I'm afraid your response raises some more questions, but it does indeed answer and shed light on a lot! heh, let me try to progress the discussion:

    Looking at these things with a vector interpretation (which my book NEVER talked about *sigh*) does seem necessary, but how do we interpret the RHS of these equations as a vector? Here it could be easy, all of the components would be zero, but what about my equation 1b (there was a typo where I called it "2b"..) where the RHS is equal to 1?

    I don't think I quite follow how we know the part I have bolded.

    Okay, I think I follow this although we would need another curve (say 'r') to map a surface to this new curvilinear axes. I could see the relation of this to what I'm doing, but I think I need to think about both of our posts a lot longer to really "get it"..

    However, this part brings up another question. It seems all well and good to do this with our vector interpretation of the PDE, but only if all of the components on the RHS are zero. In my post, there was that pesky equation 1b that I'm on about that was equal to "1". How would I come up with such a clean set of 3 equations if I didn't have the RHS of the PDE equal to 0? This whole seem sounds very PDE specific. Other than that I do indeed see how dx/P = dy/Q = dz/R = ds comes about.

    Again I agree as long as it's homogeneous.

    Not sure about the links between my equation #3 and the characteristic equation still :/

    Ohhhh, they're referring to a "level curve". It was written out in plain mathematical language there for me, but I guess because I didn't see the words "level curve" or any description like you gave so I didn't see it. Thanks a lot for this, not sure why they chose to word it the way they did...

    Oh okay, so the classic is from Bluman and Kumei, I'll have to check it out!

    Thanks for trying to get to the bottom of this seemingly intractable problem with me! I feel like you've shown me most of the way and when I completely figure it out I'm just going to sit awestruck at how impossible it would have been to figure this out from the book :(
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