1

How can I solve this equation:

$0 = (x+k)e^{-(x+k)^2}+(x-k)e^{-(x-k)^2}$

To find x in terms of k?

giorgiomugnaini
  • 1,456
Kenshin
  • 2,230
  • 1
    You can simplify it down to $x = k\tanh(2kx)$. There is no analytical solution in elementary functions to this equation. – Winther Apr 04 '15 at 08:46
  • ...apart from $x=0$ obviously! – Winther Apr 04 '15 at 08:57
  • @Winther, yes there should be 3 solutions for x. One is x = 0 and the other two depend on the value of k. – Kenshin Apr 04 '15 at 09:00
  • If $|k| < 1/\sqrt{2}$ there is only one solution. – Winther Apr 04 '15 at 09:02
  • @Winther, yes I was able to figure that one out. – Kenshin Apr 04 '15 at 09:04
  • Are you concerned by numerical solutions ? – Claude Leibovici Apr 04 '15 at 09:10
  • @ClaudeLeibovici, I was more interested in whether there was a way to manipulate to find x, rather than the solutions themselves. Claude are you able to help on this queston: http://math.stackexchange.com/questions/1219657/for-what-k-values-are-there-more-than-one-solution-to-the-following-equation – Kenshin Apr 04 '15 at 09:12
  • @Winther, can you help me with this similar quesiton: http://math.stackexchange.com/questions/1219657/for-what-k-values-are-there-more-than-one-solution-to-the-following-equation – Kenshin Apr 04 '15 at 09:12
  • 1
    Following Winther's first comment, I think that it could be an idea to change variable $2kx=y$ so the equation becomes $$\frac{y}{2k^2}=\tanh(y)$$ and this is just the intersection of a line going through origin with the $\tanh(y)$ function. – Claude Leibovici Apr 04 '15 at 09:18
  • @Mew That problem is even worse. Again no elementary analytical solution possible and yhe equation can hardly be simplified at all $x = \frac{k \sinh[k x]}{\cosh[k x] - e^{k^2}/2}$ is the 'simplest' expression I can find. Study it numerically is my suggestion. – Winther Apr 04 '15 at 09:39
  • If you are interested to analytical solutions, in the paper: "An Exact Solution of a Molecular Field Equation in the Theory of Ferromagnetism" C. E. Siewert and C. J. Essig, ( PDF: http://www4.ncsu.edu/~ces/pdfversions/56.pdf ) the equation

    $y=\tanh(ay+b)$ is solved by an integral representation. Could it help you?

     – giorgiomugnaini Apr 04 '15 at 09:46

2 Answers2

3

The only analytical way to solve this is with the generalization of Lambert's Omega function:

Your equation is:

$(x+k)e^{-x^2-2xk-k^2}+(x-k)e^{-x^2+2xk-k^2}=0$

Multiplying both sides for $e^{x^2-2xk+k^2}$ we got:

$(x+k)e^{-4kx}+x-k=0$

Dividing both sides for $(x-k)$ we have:

$e^{-4kx}\frac {x+k} {x-k} =-1$

Whose solutions are (in terms of generalized Lambert's function):

$x=-k-\frac 1 {4k}W_{e^{-4k^2}}(-8k^2e^{-4k^2})$

I don't know if we can do any simplification here (notice that the pedix of the GLF is the same as the the term which multiplies $-8k^2$ in the argument of the function and that $-8k^2$ is double the exponent we just talked about). Hope this will help ;)

AlienRem
  • 4,130
  • 1
  • 27
  • 38
1

This answer can be considered an expansion of the @Renato Faraone answer; we start from point reached by Renato :

$$e^{-4kx}\frac {x+k} {x-k} =-1$$

we can rewrite:

$$x=k-e^{-4kx}(x+k)$$

Appliying Lagrange Inversion: $$x=k+\sum_{n=1}\frac{(-1)^{n}D^{n-1}\left(e^{-4nkx}(x+k)^n\right)|_{x=k}}{n!}$$ Remembering the Rodrigues formula of the Laguerre associated polynomials (http://en.wikipedia.org/wiki/Laguerre_polynomials):

$$L_n^{(\alpha)}(x) = {x^{-\alpha} e^x \over n!}{d^n \over dx^n} \left(e^{-x} x^{n+\alpha}\right)$$

we can write the general term of Lagrange expansion:

$$L_{n-1}^{(1)}(4nk(x+k)) = {e^{4knx} \over (x+k)(n-1)!}{d^{n-1} \over dx^{n-1}} \left(e^{-4knx} {(x+k)}^{n}\right)$$

So the Lagrange expansion becomes: $$x=k\left(1+2\sum_{n=1}^\infty\frac{(-1)^nL_{n-1}^{(1)}(8nk)e^{-4nk^2x}}{n}\right)$$

This is nothing else but the Lagrange expansion of the above seen generalization of Lambert W function.

giorgiomugnaini
  • 1,456