I need help with solving this difficult fluid dynamic expression. I have tried using rules of logs, symbolab algebra calculator and Wolfram Alpha calculator, and I have got no solution.

How would you solve the following expression for $x$? $$e^{\frac{x^2}{4vt}} = 1+\frac{x^2}{2vt}$$

When solving this numerically, the solution is: $x=2.2418\sqrt{vt}$

I want to know how you could solve the first expression to get the solution. So could someone please provide a step-by-step solution please?

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    $\begingroup$ If you move everything to one side and ask Wolfram Alpha, you get a rather nasty-looking result involving the Lambert W function: wolframalpha.com/input/… $\endgroup$ Jan 3 '19 at 2:05
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    $\begingroup$ The expression you get is also much nastier than the numeric solution, so I think solving for $x$ might not be the way to go, because of how nontrivial it is. $\endgroup$ Jan 3 '19 at 2:06
  • $\begingroup$ Would you have to use use the Newton-Raphson formula? $\endgroup$
    – Alan Glenn
    Jan 3 '19 at 2:30
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    $\begingroup$ x = 0 is a solution. $\endgroup$ Jan 3 '19 at 2:54
  • $\begingroup$ What are the parameter values for $v$,$t$? Looks like it could be a perturbation problem if $v t $ is large. $\endgroup$
    – user150203
    Jan 3 '19 at 3:54

$$e^{\frac{x^2}{4vt}} = 1+\frac{x^2}{2vt}$$ Let $y=\frac{x^2}{4vt}$ $$e^y=1+2y$$ $$e^{-y}=\frac{1}{1+2y}$$ $$(1+2y)e^{-y}=1$$ $$(\frac12+y)e^{-y}=\frac12$$ $$(-\frac12-y)e^{-y}=-\frac12$$ $$ (-\frac12-y)e^{-y}e^{-\frac12}=-\frac12 e^{-\frac12}$$ $$ (-\frac12-y)e^{-\frac12-y}=-\frac12 e^{-\frac12}=-\frac{1}{2\sqrt{e}}$$ $X=(-\frac12-y)$ $$Xe^X=-\frac{1}{2\sqrt{e}}$$ From the definition of the Lambert W function : http://mathworld.wolfram.com/LambertW-Function.html $$X=W\left(-\frac{1}{2\sqrt{e}}\right)$$ $$y=-\frac12-X=-\frac12-W\left(-\frac{1}{2\sqrt{e}}\right)$$ $$x=\sqrt{4vty}=\sqrt{-2-4W\left(-\frac{1}{2\sqrt{e}}\right)}\sqrt{vt}$$ The Lambert W(z) function is a multi valuated function in $-\frac{1}{e} <z<0$ , real $z$.

This is presently the case where $z=-\frac{1}{2\sqrt{e}}$ since $-\frac{1}{e} <-\frac{1}{2\sqrt{e}}<0$

First root :

$W_0\left(-\frac{1}{2\sqrt{e}}\right)=-\frac12 \quad;\quad {-2-4W_0\left(-\frac{1}{2\sqrt{e}}\right)}=-2-4(-1/2)=0 \quad;\quad x=0$

Second root :

$W_{-1}\left(-\frac{1}{2\sqrt{e}}\right)\simeq -1.756431... $


One can use series expansion to compute it approximately. See page 13 in https://fr.scribd.com/doc/34977341/Sophomore-s-Dream-Function . The convergence is very slow. See the numerical calculus in the below addition. In practice, it is certainly faster to solve directly $e^y=1+2y$ with Newton-Raphson or similar iterative method.

Finally, an approximate value is :

$$x\simeq\sqrt{-2-4(-1.756431)}\sqrt{vt}\simeq 2.2418128 \sqrt{vt}$$

With more digits : https://www.wolframalpha.com/input/?i=sqrt(-2-4+lambertw(-1,-1%2F(2+sqrt(e))))


Example of recursive numerical calculus of $W_{-1}(x)$ in the range $-\frac{1}{e}<x<0$

enter image description here


As mentioned above, clearly $x=0$ is a solution.

Also (Mathematica):

$$x = \pm \sqrt{2} \sqrt{-2 t v W_{-1}\left(-\frac{1}{2 \sqrt{e}}\right)-t v}$$

  • $\begingroup$ Sorry pardon my ignorance, which special function is $W_n(x)$? $\endgroup$
    – user150203
    Jan 3 '19 at 3:55
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    $\begingroup$ The Lambert W function or ProductLog. en.wikipedia.org/wiki/Lambert_W_function . mathworld.wolfram.com/ProductLogFunction.html $\endgroup$ Jan 3 '19 at 3:58
  • $\begingroup$ Cheers @David G. Stork $\endgroup$
    – user150203
    Jan 3 '19 at 3:59
  • $\begingroup$ How would you expand the Lamber W function to get $x=2.2418\sqrt{vt}$ @DavidG.Stork $\endgroup$
    – Alan Glenn
    Jan 3 '19 at 4:07
  • $\begingroup$ @AlanGlenn: I would merely evaluate it numerically, just as we do with innumerable other functions, such as $\sin$, $\log$, etc. $\endgroup$ Jan 3 '19 at 4:15

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