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Instruction needed

  1. Jan 17, 2008 #1

    Mentz114

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    Gold Member

    This is not homework, these equations are constraints I've encountered in some GR work.

    But I'm hopeless with differential equations and I have two I need to solve. Even after reading some instructive texts, I still can't do it.


    [tex]\frac{dB(x)}{dx} = \frac{B(x)}{x} - \frac{1}{x}[/tex]

    [tex]\frac{dB(x)}{dx} = \frac{B(x)}{x} - \frac{1}{x} + \frac{K}{B(x)}[/tex]

    K is a constant. I know the first one has a solution but the second one might not.

    M
     
  2. jcsd
  3. Jan 17, 2008 #2
    For the 1st one

    [tex]B'(x) = \frac{B(x)}{x} - \frac{1}{x}\Rightarrow \frac{B'(x)}{x}-\frac{B(x)}{x^2}=-\frac{1}{x^2}\Rightarrow (\frac{B(x)}{x})'=(\frac{1}{x})'\Rightarrow \frac{B(x)}{x}=\frac{1}{x}+c\Rightarrow B(x)=c\,x+1[/tex]

    For the 2nd one, let [tex]B(x)=K\,x\,f(x)[/tex], then

    [tex]f'(x)=\frac{-f(x)+1}{K\,x^2\,f(x)}\Rightarrow\frac{f(x)\,d\,f(x)}{-f(x)+1}=\frac{d\,x}{K\,x^2}\Rightarrow f(x)+\log(f(x)-1)=\frac{1}{K\,x}+c\Rightarrow \frac{B(x)}{K\,x}+\log(\frac{B(x)}{K\,x}-1)=\frac{1}{K\,x}+c[/tex]

    which is the general implicit solution and can be written with in terms of the product log function.
     
  4. Jan 18, 2008 #3

    Mentz114

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    Hi Rainbow,

    thanks for the help, much appreciated. It looks like witchcraft to me. Unfortunately I transposed the signs and your solution depends crucially on the sign. So I'm back to square one.

    Which serves me right for being a pratt.
    M
     
    Last edited: Jan 18, 2008
  5. Jan 19, 2008 #4
    I get a slightly different solution:

    [tex]\frac{B(x)}{Kx}+\ln\left|1-\frac{B(x)}{Kx}\right|=\frac{1}{Kx}+c[/tex]

    There seems to be something with the absolute value.
    Mmm, I use ln, sorry for that, it makes it easier for me to see what is meant.

    @Rainbow Child: What do you exactly mean by "product log function"
     
  6. Jan 19, 2008 #5
    I was working on the complex plane, that's why I didn't put the absolute value.

    The product log function is the multivalued function [tex]w(z)[/tex] defined by

    [tex]z=w(z)\,e^{w(z)}[/tex]

    Thus for the solution at hand

    [tex]\frac{B(x)}{K\,x}+\log(\frac{B(x)}{K\,x}-1)=\frac{1}{K\,x}+c[/tex]

    we have

    [tex](\frac{B(x)}{K\,x}-1)\,\exp(\frac{B(x)}{K\,x})=\exp(\frac{1}{K\,x}+c)\Rightarrow (\frac{B(x)}{K\,x}-1)\,\exp(\frac{B(x)}{K\,x}-1)=\exp(\frac{1}{K\,x}+c-1)\Rightarrow \frac{B(x)}{K\,x}-1=w\left(\exp(\frac{1}{K\,x}+c-1)\right)[/tex]

    yielding to

    [tex]B(x)=K\,x\,\left(1+w(C\,e^{\frac{1}{K\,x}})\right), \quad C=\exp(c-1)[/tex]
     
  7. Jan 19, 2008 #6
    I was under the assumption that the differential equation of the original post was to have a real valued function as solution. Therefore the use of ln instead of log. In case I assume that it is to be real, then I think that the solution I gave was correct. That leaves us with a minus sign difficulty or am I making a mistake?

    I never used the product log function (or Lambert W function, google). Always good to learn something new. I looked in the books of Erdelyi, Magnus, Oberhettinger, Tricomi and the one of Abramowitz, Stegun, but couldn't find it. It seems to be a rarely used one. Very nice.
     
  8. Jan 19, 2008 #7
    I wrote [tex]\log[/tex] because I am used with this notation for the natural logarithm. :smile: For the real domain your solution is the correct one.

    The product log function [tex]w(z)[/tex] often appears when you are dealing with GR, i.e. in maximally symmetric two-dimensional surfaces. But in GR we are always allowed to make a coordinate transformation, and get rid of the product log function. :cool:
     
  9. Jan 21, 2008 #8

    Mentz114

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    Thanks to both of you. I may have learnt enough to solve the correct equations.
     
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