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Separation of variables technique

  1. Dec 5, 2014 #1
    If $$u=\frac{1}{2} E^2$$ and $$v=\frac{1}{2}B^2$$

    and we have that $$\frac{\partial L}{\partial u} \frac{\partial L}{\partial v} = -1$$

    The author says: to obtain explicit solution of the above, one must resort to techniques such as separation of variables in particular coordinate systems. For example, if one supposes that the solution separates multiplicatively in (u,v) coordinates one obtains:

    $$L = ± \sqrt{\alpha - \beta E^2}\sqrt{\gamma - \delta B^2}$$ where $$\beta \gamma =1$$

    How was this obtained? I didn't get this method of integration?
  2. jcsd
  3. Dec 5, 2014 #2


    Staff: Mentor

    Sometimes mathematicians pull these out of a hat. The fact that it works is the important thing and to remember the trick.

    I saw one recently in deriving the integral with x^n*e^(-x) dx where the author differentiated the constant to manipulate the integral into an easily integrable form. It was quite cool and I had never seen its like before. I had to wrestle with it for awhile before I accepted it.


    see section 1.2 on Parametric Differentiation
  4. Dec 5, 2014 #3
    I do not know how this is related to the equation I have?
  5. Dec 5, 2014 #4


    User Avatar
    2017 Award

    Staff: Mentor

    It is another example of a mathematical trick.

    There does not have to be a clear way to find those approaches. You can try many of them until one works. Then you throw away the 100 failed attempts and take the one that worked, which looks like magic to others afterwards ("how did he come up with that formula?").
    With experience, you can save some time and find good approaches faster.
  6. Dec 9, 2014 #5


    User Avatar
    Science Advisor

    They like to make it look like they magically pulled it out of a hat. Actually it is the result of "trying one thing after another" until something finally works!
  7. Dec 9, 2014 #6


    Staff: Mentor

    Yes, I heard Gauss was famous for this. His proofs were so polished that no one could figure out how he arrived at the conclusion.

    A similar case could be made for Ramanujan where he derived so many conjectures based on an old British math book:

  8. Dec 9, 2014 #7
    This is a first order non-linear pde, it may look more familiar if you write it like

    [itex] \tfrac{ \partial z}{\partial u} \tfrac{ \partial z}{\partial v} = \tfrac{ \partial z}{\partial (\tfrac{1}{2}x^2)} \tfrac{ \partial z}{\partial (\tfrac{1}{2}y^2)} = (\tfrac{1}{x}\tfrac{ \partial z}{\partial x})(\tfrac{1}{y} \tfrac{ \partial z}{ \partial y}) = \tfrac{1}{xy}\tfrac{ \partial z}{\partial x} \tfrac{ \partial z}{ \partial y} = - 1 \rightarrow = \tfrac{ \partial z}{\partial x} \tfrac{ \partial z}{\partial y} = - xy[/itex]

    with z = L, E = x, B = y.

    Here are examples on how to deal with these in general

    In this case there is another method being used, that of assuming a separable solution from the get-go, but it can also be solved with those methods in the videos. It's most definitely not a trick it's a standard method of solution with a nice geometric interpretation. Let this be a lesson never to accept the word 'trick' again ;)

    You basically just write L = f(u)g(v) and plug it into (dL/du)(dL/dv) = - 1 to get g(v)[dg/dv] = - 1/f(u)[df/du] and since both sides of this are equal to two different functions depending on two different variables, both have to be equal to a constant, i.e.
    [itex]g(v)[dg/dv] = C \rightarrow gdg = Cdv \rightarrow (1/2)g^2 = D + Cv \rightarrow g = \sqrt{\gamma - 2\delta v}= \sqrt{\gamma - \delta B^2}[/itex]
    and similarly for the other side, exact same method which gives you the same form of solution so you get the full solution you were given.

    Elsgoltz - Partial Differential Equations & the Calculus of Variations has a good chapter on all this.
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