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Looking for a closed-form solution using Solve[ ]

sebastiano piccolroaz, EPFL
Posted 3 years ago

I was wondering if a closed form solution does exist for the following equation:

a^2 Exp[x/a] + x^2/2 (1 - b) - a^2 - a x Exp[x/a] = 0

where a and b are Real numbers. I would like to obtain a solution of the form x=f(a,b). I tried to use the following command:

Solve[a^2*Exp[x/a] + x^2/2*(1 - b) - a^2 - a*x*Exp[x/a] == 0, x, 
 Assumptions -> Element[a, Reals], Assumptions -> Element[b, Reals]]

but I obtained the following message: Solve: This system cannot be solved with the methods available to Solve.

I was wondering whether I could use the Lambert W function to solve this equation or if there is any way to obtain an approximated solution of this equation. I am looking for a solution for x<0, and I know that a>0, if this can make any difference. Thank you.

Edited after the 1st reply: The last sentence is probably superfluous: I am interested in the general closed-form solution of this equation. Then, I will use the part of solution when x<0 (which happens when 0<b<1).
POSTED BY: sebastiano piccolroaz
4 Replies
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It does not appear to have solutions for x<0. Just a cursory numerical exploration gives:

ans = Table[
  NSolve[a^2*Exp[x/a] + x^2/2*(1 - b) - a^2 - a*x*Exp[x/a] == 0 , x, 
   Reals], {a, 1, 10}, {b, -5, 5}]

All solutions are zero and a positive number for x. I would make sure the equation is correct before "diving into" solving it.

Regards

Neil
POSTED BY: Neil Singer

Thank you for the quick reply. As I expect based on the physical reasoning that stays behind the derivation of this equation and looking at its plot, the solutions of this equation should be:

  • x1=x2=0 when b>=1

  • x1=0 and x2<0 when 0<b<1

  • x1=0 and x2>0 when b<0

So, one solution is always x=0. I'm mainly interested to derive the solution when 0<b<1. For example, when I solve it with NSolve and assuming a = 0.3 and b=0.5, the solutions are x=0 and x=-0.327474 as confirmed by the plot. I guess that the cursory numerical exploration suggested above did not provide negative solutions because the table command did not explore values 0<b<1.
POSTED BY: sebastiano piccolroaz

Based on that, you can plot the answer. I put a Quiet[] in there because sometimes the solution has issues but you can get the idea by doing something like this:

Manipulate[
 ListLinePlot[
  Quiet[ans = 
    Table[x /. 
      FindRoot[
       a^2*Exp[x/a] + x^2/2*(1 - b) - a^2 - a*x*Exp[x/a] == 
        0, {x, -1, -100, -0.001}], {b, 0.01, 0.9, 0.01}]]], {{a, 0.3},
   0.1, 1}]

Giving a plot that looks like this for any value of a (only the y axis seems to change with a). This suggests you can easily do an excellent interpolation or an approximate curve fit.

enter image description here
POSTED BY: Neil Singer

Thank you for the suggestion.
POSTED BY: sebastiano piccolroaz
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