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I am trying to solve this equation in MMA 11.3

$2 \epsilon s^{\prime \prime}+\frac{1-s}{2 \epsilon}=0$;

BCs: $s^{\prime}( \pm 1)=0$ and $s\left(0\right)=0$;

The analytical solution is expressed as:$s^{ \pm}(x)=1-\cosh \left(\frac{x}{2 \epsilon}\right) \pm \operatorname{coth}\left(\frac{1}{2 \epsilon}\right) \sinh \left(\frac{x}{2 \epsilon}\right)$.

However, the solution from MMA 11.3 test code is not corret:

Code

pf = DSolve[{s''[x]*2*\[Epsilon] + 0.5 (1 - s[x])/\[Epsilon] == 0, 
   s[0] == 0, s'[-1] == 0, s'[1] == 0}, s, x] 

if we validate the solution:

\[Epsilon] = 0.009
Plot[Evaluate[s[x] /. pf], {x, -1, 1}]

Output:enter image description here

now how can we derive the correct solution in MMA?

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  • $\begingroup$ 1. With "11.3.0 for Microsoft Windows (64-bit) (March 27, 2018)", I got {} as the output with DSolve::bvnul warning, have you Clear[s]? 2. You should solve with s[0] == 0, s'[-1] == 0 and s[0] == 0, s'[1] == 0 separately. $\endgroup$ – xzczd Aug 22 at 12:37
  • $\begingroup$ @xzczd I think this equation is not complex and the bcs are very clear, we should solve this equation without any difficults in MMA? $\endgroup$ – ABCDEMMM Aug 22 at 12:39
  • $\begingroup$ It's complicated enough in certain sense, because the solution actually isn't a classical one, it doesn't satisfy the ODE at $x=0$. Some debates can be found under this answer: mathematica.stackexchange.com/a/188918/1871 $\endgroup$ – xzczd Aug 22 at 12:45
  • $\begingroup$ Two(not three ) boundary conditions should be enough I think! $\endgroup$ – Ulrich Neumann Aug 22 at 12:57
  • $\begingroup$ two Bcs works only for x-> 0 ->>>>... $\endgroup$ – ABCDEMMM Aug 22 at 13:10
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Not an answer, only verification of the analytical solution:

positive sign:

{Limit[s[x], x -> 0], s'[-1], s'[1], s[x] == s[-x], s''[x]*2*\[Epsilon] + 1/2 (1 -s[x])/\[Epsilon] == 0 } /.
s -> Function[x,1 - Cosh[x/(2 \[Epsilon])] + Coth[x/(2 \[Epsilon])] Sinh[x/(2 \[Epsilon])]] // FullSimplify      
(*{1, 0, 0, True, True}*)

Solution fullfills the ode, is symmetrix s[x]==s[-x] and s[0]==1,s'[-1]==s[1]==0!

negative sign:

{Limit[s[x], x -> 0], s'[-1], s'[1], s[x] == s[-x], s''[x]*2*\[Epsilon] + 1/2 (1 -s[x])/\[Epsilon] == 0 } /.
s -> Function[x,1 - Cosh[x/(2 \[Epsilon])] -Coth[x/(2 \[Epsilon])] Sinh[x/(2 \[Epsilon])]] // FullSimplify      
(*{-1, Sinh[1/(2 \[Epsilon])]/\[Epsilon], -(Sinh[1/(2 \[Epsilon])]/\[Epsilon]), True, True}*)

Solution fullfills the ode, is symmetrix s[x]==s[-x] and s[0]==-1,s'[-1]==-s[1]!=0!

Obviously the analytical solutions and the boundary conditions don't match!

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