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Generalized W Lambert function

  1. Jan 4, 2012 #1
    Hi everyone,
    I'm currently trying to solve this equation : x²[A+B.exp(x)]=1 for A and B real numbers, and x a complex (this comes from physics, so in my case, Re(x)>0)

    I know that x.exp(x)=a has a solution using Lambert function : x=W(a)
    I know that x².exp(x)=a may be recast to use the Lambert function, the solution being something like x=2W(sqrt(a)/2)

    But what about my equation ? I tried a lot of things to recast the equation and use the Lambert function, but nothing, so I'm asking to you guys ...

    Thanks
     
  2. jcsd
  3. Jan 5, 2012 #2
    The roots of the equation x²[A+B.exp(x)]=1 cannot be expressed as a combination of a finite number of elementary functions.
    As far as I know, up to now, there is no standard special function which could help to analitically solve it.
    Of course, it is possible that someone already defined a particular function in order to formally express the roots. Even if such a function was defined, it cannot be of general use, since the function is not implemented in mathematical softwares, nor common in math background.
    Nevertheless, the equation can be solved, thanks to numerical methods. I think that it is presently the usual way to solve these kind of problems when they are encountered in Physics.
     
  4. Jan 5, 2012 #3
    Ok thanks. I know there are other methods to solve that. I have started to find x0 solution of my equation with B=0 (this case has a particular physical signification. For example, x has to be real, not complex in this case). And then, to solve my equation with B≠0, I assume that x=x0+dx and obtain solution with dx complex. But this is just for small variations !!! I don't know I can do that for stronger perturbation dx.

    I guess I cannot do something else... first I thought that I could do the assumption B>>A to remove the A in my equation because we know solutions for x².B.exp(x)=1 .... however it seems not correct because the condition to do that is in fact A<<B.cos(Re(x)) that may be not true for some values of x...

    But thanks anyway
     
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