# DSolve not satisfying such kind of initial conditions

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:

now how can we derive the correct solution in MMA?

• 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. – xzczd Aug 22 '19 at 12:37
• @xzczd I think this equation is not complex and the bcs are very clear, we should solve this equation without any difficults in MMA? – ABCDEMMM Aug 22 '19 at 12:39
• 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 – xzczd Aug 22 '19 at 12:45
• Two(not three ) boundary conditions should be enough I think! – Ulrich Neumann Aug 22 '19 at 12:57
• two Bcs works only for x-> 0 ->>>>... – ABCDEMMM Aug 22 '19 at 13:10

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!