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I'm trying to numerically solve a system of second order differentials equations that are connected only by boundary conditions. I can't figure out how to define the problem in mathematica.

The first equation:

$y_1''(x) = k_1*Sinh(y_1(x)) - c_1(x)$, is defined on $0<x\leq d$

The second equation:

$y_2''(x) = k_2*Sinh(y_2(x)) - c_2(x)$, is defined on $d<x<\infty$

Boundary conditions are:

$y_1'(0) = 0$

$y_2(\infty) = 0$

$y_1(d^-) = y_2(d^+)$

$\epsilon_1y_1'(x)|_{x=d^-} = \epsilon_2y_2'(x)|_{x=d^+}$

$k_1$, $k_2$, $\epsilon_1$, $\epsilon_2$, $c_1(x)$, $c_2(x)$ are given constants/functions. Can you please help to define and solve the system in mathematica ? I tried the following code, where put constant values for all parameters:

c1 = 1; 
c2 = 1; 
k1 = 1; 
d = 1; 
e1 = 1; 
e2 = 1; 
system = {Derivative[2][y1][x1] == k1*Sinh[y1[x1]] - c1, 
   Derivative[2][y2][x2] == k2*Sinh[y2[x2]] - c2, y2[Infinity] == 0, 
   Derivative[1][y1][0] == 0, y1[d] == y2[d], e1*Derivative[1][y1][d] == 
    e2*Derivative[1][y2][d]}
s1 = NDSolve[system, {y1[x1], y2[x2]}, {x1, 0, d}, {x2, d, Infinity}]

However I get

NDSolve::allort: "The dependent variable y1[x1] should depend either on the temporal variable alone or on all independent variables."

Can't understand what can be the problem.

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    $\begingroup$ See How to copy code from Mathematica so it looks good on this site and Hints and help on using the site efficiently and How do I format my posts using Markdown or HTML? and read the FAQ $\endgroup$ – user9660 Aug 15 '16 at 12:41
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    $\begingroup$ People here generally like users to post code as Mathematica code instead of images or TeX, so they can copy-paste it. It makes it convenient for them and more likely you will get someone to help you. $\endgroup$ – user9660 Aug 15 '16 at 12:42
  • $\begingroup$ I do not think this is well posed. You have basically only 3 boundary conditions, not 4. You need 4. The condition that $y_1[d]=y_2[d]$ does not add anything new. To find $y_2[d]$ you need to solve second ODE. But to solve second ODE you need $y_1[d]$ first, which requires solving the first ODE first. But to solve the first ode, you need $y_2[d]$, which requires solving the second ODE first. And so on. $\endgroup$ – Nasser Aug 15 '16 at 15:47
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    $\begingroup$ Please add a value for k2. $\endgroup$ – bbgodfrey May 13 '17 at 14:01