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Mathematica Equation Solving Numerical Computation

Unknown error in NDSolve and other issues

Konstantin Nosov

Konstantin Nosov, V. N. Karazin Kharkov National University

Posted 10 years ago

With the help of NDSolve I try to solve a very simple BVP: $Av''(t)+Bv(t)=f(t),$ where A, B are 5x5 constant matrices, f(t) 5-dimentional vector with components composing of InterpolatingFunction(s). Each component of solution should satisfy: $v_i(0)=1,\ \ v_i(1)=0.$ The system is enough "good" for solving, matrices A,B are not singular, but I get strange error: Moreover, last error (NDSolve::bvaux) cannot be found anywhere! Any suggestions are welcomed. The code is below and A, B, Cv in attachment. NDSolve[{A[[1, 1]] v1''[t] + A[[1, 2]] v2''[t] + A[[1, 3]] v3''[t] + A[[1, 4]] v4''[t] + A[[1, 5]] v5''[t] + B[[1, 1]] v1[t] + B[[1, 2]] v2[t] + B[[1, 3]] v3[t] + B[[1, 4]] v4[t] + B[[1, 5]] v5[t] == Cv[[1]], A[[2, 1]] v1''[t] + A[[2, 2]] v2''[t] + A[[2, 3]] v3''[t] + A[[2, 4]] v4''[t] + A[[2, 5]] v5''[t] + B[[2, 1]] v1[t] + B[[2, 2]] v2[t] + B[[2, 3]] v3[t] + B[[2, 4]] v4[t] + B[[2, 5]] v5[t] == Cv[[2]], A[[3, 1]] v1''[t] + A[[3, 2]] v2''[t] + A[[3, 3]] v3''[t] + A[[3, 4]] v4''[t] + A[[3, 5]] v5''[t] + B[[3, 1]] v1[t] + B[[3, 2]] v2[t] + B[[3, 3]] v3[t] + B[[3, 4]] v4[t] + B[[3, 5]] v5[t] == Cv[[3]], A[[4, 1]] v1''[t] + A[[4, 2]] v2''[t] + A[[4, 3]] v3''[t] + A[[4, 4]] v4''[t] + A[[4, 5]] v5''[t] + B[[4, 1]] v1[t] + B[[4, 2]] v2[t] + B[[4, 3]] v3[t] + B[[4, 4]] v4[t] + B[[4, 5]] v5[t] == Cv[[4]], A[[5, 1]] v1''[t] + A[[5, 2]] v2''[t] + A[[5, 3]] v3''[t] + A[[5, 4]] v4''[t] + A[[5, 5]] v5''[t] + B[[5, 1]] v1[t] + B[[5, 2]] v2[t] + B[[5, 3]] v3[t] + B[[5, 4]] v4[t] + B[[5, 5]] v5[t] == Cv[[5]], v1[0] == 1, v1[1] == 0, v2[0] == 1, v2[1] == 0, v3[0] == 1, v3[1] == 0, v4[0] == 1, v4[1] == 0, v5[0] == 1, v5[1] == 0}, {v1[t], v2[t], v3[t], v4[t], v5[t]}, t, WorkingPrecision -> 50, Method -> {"Chasing", Method -> "StiffnessSwitching"}, MaxSteps -> Infinity] Attachments:

With the help of NDSolve I try to solve a very simple BVP:

$Av''(t)+Bv(t)=f(t),$

where A, B are 5x5 constant matrices, f(t) 5-dimentional vector with components composing of InterpolatingFunction(s). Each component of solution should satisfy:

$v_i(0)=1,\ \ v_i(1)=0.$

The system is enough "good" for solving, matrices A,B are not singular, but I get strange error:

enter image description here

Moreover, last error (NDSolve::bvaux) cannot be found anywhere!

Any suggestions are welcomed. The code is below and A, B, Cv in attachment.

NDSolve[{A[[1, 1]] v1''[t] + A[[1, 2]] v2''[t] + A[[1, 3]] v3''[t] + 
    A[[1, 4]] v4''[t] + A[[1, 5]] v5''[t] + B[[1, 1]] v1[t] + 
    B[[1, 2]] v2[t] + B[[1, 3]] v3[t] + B[[1, 4]] v4[t] + 
    B[[1, 5]] v5[t] == Cv[[1]], 
  A[[2, 1]] v1''[t] + A[[2, 2]] v2''[t] + A[[2, 3]] v3''[t] + 
    A[[2, 4]] v4''[t] + A[[2, 5]] v5''[t] + B[[2, 1]] v1[t] + 
    B[[2, 2]] v2[t] + B[[2, 3]] v3[t] + B[[2, 4]] v4[t] + 
    B[[2, 5]] v5[t] == Cv[[2]],
  A[[3, 1]] v1''[t] + A[[3, 2]] v2''[t] + A[[3, 3]] v3''[t] + 
    A[[3, 4]] v4''[t] + A[[3, 5]] v5''[t] + B[[3, 1]] v1[t] + 
    B[[3, 2]] v2[t] + B[[3, 3]] v3[t] + B[[3, 4]] v4[t] + 
    B[[3, 5]] v5[t] == Cv[[3]], 
  A[[4, 1]] v1''[t] + A[[4, 2]] v2''[t] + A[[4, 3]] v3''[t] + 
    A[[4, 4]] v4''[t] + A[[4, 5]] v5''[t] + B[[4, 1]] v1[t] + 
    B[[4, 2]] v2[t] + B[[4, 3]] v3[t] + B[[4, 4]] v4[t] + 
    B[[4, 5]] v5[t] == Cv[[4]],
  A[[5, 1]] v1''[t] + A[[5, 2]] v2''[t] + A[[5, 3]] v3''[t] + 
    A[[5, 4]] v4''[t] + A[[5, 5]] v5''[t] + B[[5, 1]] v1[t] + 
    B[[5, 2]] v2[t] + B[[5, 3]] v3[t] + B[[5, 4]] v4[t] + 
    B[[5, 5]] v5[t] == Cv[[5]], v1[0]  ==  1, v1[1] == 0, 
  v2[0]  == 1, v2[1] == 0, v3[0]  == 1, v3[1] == 0, v4[0]  == 1, 
  v4[1] == 0, v5[0]  == 1, v5[1] == 0}, {v1[t], v2[t], v3[t], v4[t], 
  v5[t]}, t, WorkingPrecision -> 50, 
 Method -> {"Chasing", Method -> "StiffnessSwitching"}, 
 MaxSteps -> Infinity]

POSTED BY: Konstantin Nosov

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Konstantin Nosov

Konstantin Nosov, V. N. Karazin Kharkov National University

Posted 10 years ago

Dear Kay. You can make sure, that Cv is 5-dimentional function-vector. Get["<your_path>:\\env.txt"] Plot[{Cv[[1]], Cv[[2]]}, {t, 0, 1}] This gives the following: Other components of Cv are also function-vectors.

POSTED BY: Konstantin Nosov

Kay Herbert

Kay Herbert, Chief scientist, Sloan Valve Co.

Posted 10 years ago

Not sure what Cv[[i]] is I thought it's a vector of time dependent functions? If so I would construct the equations this way: eqns = Join[ Table[Sum[A[i, j] v[j]''[t] + B[i, j] v[j][t], {j, 1, 4}] == cv[i][t], {i, 1, 4}], Table[v[j][0] == 0, {j, 1, 4}], Table[v[j][1] == 0, {j, 1, 4}]] and then NDSolve[eqns, {v[1][t], v[2][t], v[3][t], v[4][t]}, {t}] sorry for the double post

POSTED BY: Kay Herbert

Kay Herbert

Kay Herbert, Chief scientist, Sloan Valve Co.

Posted 10 years ago

Not sure what Cv[[i]] is I thought it's a vector of time dependent functions? If so I would construct the equations this way: In[7]:= eqns = Join[Table[ Sum[A[i, j] v[j]''[t] + B[i, j] v[j][t], {j, 1, 4}] == cv[i][t], {i, 1, 4}], Table[v[j][0] == 0, {j, 1, 4}], Table[v[j][1] == 0, {j, 1, 4}]] Out[7]= {B[1, 1] v[1][t] + B[1, 2] v[2][t] + B[1, 3] v[3][t] + B[1, 4] v[4][t] + A[1, 1] (v[1]^\[Prime]\[Prime])[t] + A[1, 2] (v[2]^\[Prime]\[Prime])[t] + A[1, 3] (v[3]^\[Prime]\[Prime])[t] + A[1, 4] (v[4]^\[Prime]\[Prime])[t] == cv[1][t], B[2, 1] v[1][t] + B[2, 2] v[2][t] + B[2, 3] v[3][t] + B[2, 4] v[4][t] + A[2, 1] (v[1]^\[Prime]\[Prime])[t] + A[2, 2] (v[2]^\[Prime]\[Prime])[t] + A[2, 3] (v[3]^\[Prime]\[Prime])[t] + A[2, 4] (v[4]^\[Prime]\[Prime])[t] == cv[2][t], B[3, 1] v[1][t] + B[3, 2] v[2][t] + B[3, 3] v[3][t] + B[3, 4] v[4][t] + A[3, 1] (v[1]^\[Prime]\[Prime])[t] + A[3, 2] (v[2]^\[Prime]\[Prime])[t] + A[3, 3] (v[3]^\[Prime]\[Prime])[t] + A[3, 4] (v[4]^\[Prime]\[Prime])[t] == cv[3][t], B[4, 1] v[1][t] + B[4, 2] v[2][t] + B[4, 3] v[3][t] + B[4, 4] v[4][t] + A[4, 1] (v[1]^\[Prime]\[Prime])[t] + A[4, 2] (v[2]^\[Prime]\[Prime])[t] + A[4, 3] (v[3]^\[Prime]\[Prime])[t] + A[4, 4] (v[4]^\[Prime]\[Prime])[t] == cv[4][t], v[1][0] == 0, v[2][0] == 0, v[3][0] == 0, v[4][0] == 0, v[1][1] == 0, v[2][1] == 0, v[3][1] == 0, v[4][1] == 0}

Not sure what Cv[[i]] is I thought it's a vector of time dependent functions? If so I would construct the equations this way:

In[7]:= eqns = 
 Join[Table[
   Sum[A[i, j] v[j]''[t] + B[i, j] v[j][t], {j, 1, 4}] == 
    cv[i][t], {i, 1, 4}], Table[v[j][0] == 0, {j, 1, 4}], 
  Table[v[j][1] == 0, {j, 1, 4}]]

Out[7]= {B[1, 1] v[1][t] + B[1, 2] v[2][t] + B[1, 3] v[3][t] + 
   B[1, 4] v[4][t] + A[1, 1] (v[1]^\[Prime]\[Prime])[t] + 
   A[1, 2] (v[2]^\[Prime]\[Prime])[t] + 
   A[1, 3] (v[3]^\[Prime]\[Prime])[t] + 
   A[1, 4] (v[4]^\[Prime]\[Prime])[t] == cv[1][t], 
 B[2, 1] v[1][t] + B[2, 2] v[2][t] + B[2, 3] v[3][t] + 
   B[2, 4] v[4][t] + A[2, 1] (v[1]^\[Prime]\[Prime])[t] + 
   A[2, 2] (v[2]^\[Prime]\[Prime])[t] + 
   A[2, 3] (v[3]^\[Prime]\[Prime])[t] + 
   A[2, 4] (v[4]^\[Prime]\[Prime])[t] == cv[2][t], 
 B[3, 1] v[1][t] + B[3, 2] v[2][t] + B[3, 3] v[3][t] + 
   B[3, 4] v[4][t] + A[3, 1] (v[1]^\[Prime]\[Prime])[t] + 
   A[3, 2] (v[2]^\[Prime]\[Prime])[t] + 
   A[3, 3] (v[3]^\[Prime]\[Prime])[t] + 
   A[3, 4] (v[4]^\[Prime]\[Prime])[t] == cv[3][t], 
 B[4, 1] v[1][t] + B[4, 2] v[2][t] + B[4, 3] v[3][t] + 
   B[4, 4] v[4][t] + A[4, 1] (v[1]^\[Prime]\[Prime])[t] + 
   A[4, 2] (v[2]^\[Prime]\[Prime])[t] + 
   A[4, 3] (v[3]^\[Prime]\[Prime])[t] + 
   A[4, 4] (v[4]^\[Prime]\[Prime])[t] == cv[4][t], v[1][0] == 0, 
 v[2][0] == 0, v[3][0] == 0, v[4][0] == 0, v[1][1] == 0, v[2][1] == 0,
  v[3][1] == 0, v[4][1] == 0}

POSTED BY: Kay Herbert

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