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I have the following code for solving differential equations:

DSolve[{si1''[x] - 6/(270000*0.054) (si1[x] + se1[x]) == 0, 
si1''[x] - se1''[x] == -(6/(7000*4)) se1[x], 
si2''[x] - 6/(270000*0.054) se2[x] == 0, 
si2''[x] - se2''[x] == -(6/(7000*4)) se2[x], si1'[0] == 0, 
si1'[0] == se1'[0], si2'[1000] == se2'[1000], 
si1'[1000 - a] - se1'[1000 - a] == si2'[1000 - a] - se2'[1000 - a], 
se1'[1000 - a] == se2'[1000 - a], si1[1000 - a] == si2[1000 - a], 
se1[1000 - a] == se2[1000 - a], si1[1000 - a] == 0.2}, {si1, si2, 
se1, se2}, x]

Then, I would like to evaluate the value of the parameter "a" such that se2[1000]=0.2. How can I solve this problem? Thanks

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Clear["Global`*"]

Rationalize equations to get exact solution and enable subsequent selection and control of desired precision.

eqns = {si1''[x] - 6/(270*54) (si1[x] + se1[x]) == 0, 
   si1''[x] - se1''[x] == -(6/(7000*4)) se1[x], 
   si2''[x] - 6/(270*54) se2[x] == 0, 
   si2''[x] - se2''[x] == -(6/(7000*4)) se2[x], si1'[0] == 0, 
   si1'[0] == se1'[0], si2'[1000] == se2'[1000], 
   si1'[1000 - a] - se1'[1000 - a] == si2'[1000 - a] - se2'[1000 - a], 
   se1'[1000 - a] == se2'[1000 - a], si1[1000 - a] == si2[1000 - a], 
   se1[1000 - a] == se2[1000 - a], si1[1000 - a] == 1/5};

sol = DSolve[eqns, {si1, si2, se1, se2}, x][[1]];

eqn2 = se2[1000] == 1/5 /. sol // Simplify // N[#, 20] &;

Use FindRoot[lhs == rhs, {x, x0, x1}] to search for a solution using x0 and x1 as the first two values of x, and avoid the use of derivatives.

sola = FindRoot[eqn2, {a, 5, 10}, WorkingPrecision -> 15]

{* {a -> 38.9567170435138} *)

Verifying that sola satisfies eqn2

se2[1000] == 1/5 /. sol /. sola

(* True *)

EDIT Plot requested in comment

f[x_] := Piecewise[
     {{se1[x], 0 < x < a}, {se2[x], a < x < 1000}}] /.
    sol /. sola;

Plot[f[x], {x, 0, 1000},
 PlotRange -> All,
 Exclusions -> {x == a /. sola},
 AxesLabel -> (Style[#, 12, Bold] &) /@ {"x", "f[x]"},
 ImageSize -> 500,
 Epilog -> Inset[
   Plot[f[x], {x, 800, 1000},
    Frame -> True, Axes -> False,
    PlotRange -> {-2, 0.25},
    FrameLabel -> (Style[#, 12, Bold] &) /@ {"x", "f[x]"}],
   Scaled[{0.6, 0.45}], Automatic, 800]]

enter image description here

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  • $\begingroup$ Thanks. My further question is: I would like to plot a piecewise function with se1 for 0<x<sola and se2 for sola<x<1000. $\endgroup$ – Gae P Feb 11 at 21:40
  • $\begingroup$ Thanks! I change the boundary conditions of the plot: se1 for 0<x<1000-a and se2 for 1000-a<x<1000. If I plot a similar piecewise function for si1 and si2 I have troubles for a which tends to 1000. $\endgroup$ – Gae P Feb 12 at 10:08
  • $\begingroup$ Rather than changing the question, ask a new question. Post the code you are using and explain what problems you are unable to resolve. $\endgroup$ – Bob Hanlon Feb 12 at 13:08

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