1

Good morning. I was trying to calculate soliton-type solutions (kinks) in Rajarman's book, particularly with a potential of the form $$ U(\phi) = \frac{1}{2} \phi^2(\phi^2-1)^2 $$ According to the book, to obtain this solution, one must solve the integral $$ x-x_0 = \int \frac{1}{\sqrt{2 U(\phi)}} ~d\phi = \frac{1}{\phi} + \frac{1}{2} \ln(1+\phi) - \frac{1}{2} \ln(1-\phi) $$ which I have done, however, I don't see how to solve the equation $$ x-x_0 = \frac{1}{\phi} + \frac{1}{2} \ln(1+\phi)- \frac{1}{2} \ln(1-\phi) $$ for $\phi$ and obtain the same result as the book: $$ \phi(x)= \pm\left(1+\mathrm{e}^{ \pm 2 x}\right)^{-1 / 2} $$ I would greatly appreciate it if someone could indicate how to solve said equation.

Gorga
  • 47
  • 1
    That seems to be more of an asymptotic approximation... – Cesareo Mar 03 '24 at 11:26
  • You can simplify it as $\tanh^{-1}(\phi)+\frac1\phi=x-x_0$ which almost certainly cannot be solved for with elementary functions. The equation actually solved was $\pm\frac12\ln(\frac1{\phi^2}-1)=x$ – Тyma Gaidash Mar 03 '24 at 13:58

0 Answers0