0
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ClearAll
w1 = A1*Cos[\[Beta]*x] + A2*Sin[\[Beta]*x] + A3*Cosh[\[Beta]*x] + 
  A4*Sinh[\[Beta]*x]
w2 = B1*Cos[\[Beta]*x] + B2*Sin[\[Beta]*x] + B3*Cosh[\[Beta]*x] + 
  B4*Sinh[\[Beta]*x]
eq1 = FullSimplify[(w1 /. x -> 0)]
eq2 = FullSimplify[((D[w1, {x, 2}]) /. x -> 0)]
const = Solve[{eq1 == 0, eq2 == 0}, {A1, A3}]
w1 = w1 /. const
eq3 = FullSimplify[(w2 /. x -> 1)]
eq4 = FullSimplify[((D[w2, {x, 2}]) /. x -> 1)]
(*Compalability conditions*)
eq5 = (w1[[1]] /. x -> z) - (w2 /. x -> z)
eq6 = (D[w1[[1]], {x}] /. x -> z) - (D[w2, {x}] /. x -> z)
eq7 = (D[w1[[1]], {x, 2}] /. x -> z) - (D[w2, {x, 2}] /. x -> z)
eq8 = (D[w1[[1]], {x, 3}] /. x -> z) - (D[w2, {x, 3}] /. x -> z)
R = FullSimplify[
  Normal@CoefficientArrays[{eq3, eq4, eq5, eq6, eq7, eq8}, {A2, A4, 
      B1, B2, B3, B4}][[2]]]
MatrixForm[R]
MatrixRank[R]
b = {{0}, {0}, {0}, {0}, {0}, {0}, {0}, {1}}
MatrixForm[b]
(*Reduce[eq3\[Equal]0&&eq4\[Equal]0&&eq5==0&&eq6\[Equal]0&&eq7\[Equal]\
0&&eq8\[Equal]1,{A2,A4,B1,B2,B3,B4},Reals]*)
xx = FullSimplify[Inverse[R]]
Det[R];
(*LinearSolve[R,b]*)
Solve[{eq3 == 0, eq4 == 0, eq5 == 0, eq6 == 0, eq7 == 0, 
  eq8 == 1}, {A2, A4, B1, B2, B3, B4}, Integers, 
 MaxExtraConditions -> 2]

Now I am able to solve the system ax=b, but I am not getting the unique solution.How to simplify to get that?. I used matrix inverse to get the x but seems like its giving all vectors which satisfy the ax=b. And also I don't know why linear solve is not working on this problem.

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  • $\begingroup$ Your equations depend on {A2, A4, B1, B2, B3, B4} and {beta,z} (very nonlinear). Are you looking for a solution depending on {beta,z} ? $\endgroup$ – Ulrich Neumann Jan 6 '18 at 11:12
  • $\begingroup$ yes. I am looking for that solution only $\endgroup$ – Vijay Kumar S Jan 6 '18 at 11:38
  • $\begingroup$ I tried to run your code, it doesn't finish... In the last line you force the solution to be integer ? The Det[R]== 8 \[Beta]^8 Sin[\[Beta]] Sinh[\[Beta]] only depends on beta. For some beta the determinant equals zero , so how can you expect a unique solution??? $\endgroup$ – Ulrich Neumann Jan 6 '18 at 11:52
  • $\begingroup$ Sorry, I forget to comment the last line, i am looking for the solution for all betas which makes the det =0. A general solution in beta, but not for a particular value of beta. $\endgroup$ – Vijay Kumar S Jan 6 '18 at 12:10
  • $\begingroup$ Quite important to know... So you look for the solution of a singular system, which normally isn't unique! Just set beta==Pi k (*k: Integers*) . Perhaps you can get the minimal solution using PseudoInverse[] ! $\endgroup$ – Ulrich Neumann Jan 6 '18 at 12:27

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