# Differential-Algebraic equations

equ1 = a[z]^2 E (wx [z] wy[z] βx[z] βy[z])/
Sqrt[(wx[z]^2 +
2 βx[z]^2) ((wy[z]^2 + 2 βy[z]^2))] (βx[
z]^2/(wx[z]^2 + 2 βx[z]^2) + βy[
z]^2/(wy[z]^2 + 2 βy[z]^2)) - ν/
2 (α[z] E^2 βy[z])/βx[z]^3 - ν/
2 (α[z] E^2 βx[z])/βy[z]^3 -
q α[z] βx[z] βy[z] E^2    ;

equ2 = (3 ν)/4 (α[z] βy[z] E)/βx[z]^4 - ν/
4 (α[z] E)/βy[z]^3 - q/2 α[z] βy[z] E +
a[z]^2 ((
3 βx[z]^2 βy[z] wx[z]^3 wy[
z])/((wx[z]^2 + 2 βx[z]^2 )^(5/2) Sqrt[
wy[z]^2 + 2 (βy[z]^2) ]) + ( βy[z]^3  wx[z]^3 wy[
z])/((wx[z]^2 + 2 βx[z]^2 )^(3/
2) (wy[z]^2 + 2 βy[z]^2 )^(3/2)));

equ3 = (3 ν)/4 (α[z] βx[z] E)/βy[z]^4 - ν/
4 (α[z] E)/βx[z]^3 -
q/2 α[z]  βx[z] E +
a[z]^2 ((
3 βy[z]^2 βx[z] wy[z]^3 wx[
z])/((wy[z]^2 + 2 βy[z]^2 )^(5/2) Sqrt[
wx[z]^2 + 2 (βx[z]^2) ]) + ( βx[z]^3  wy[z]^3 wx[
z])/((wx[z]^2 + 2 βx[z]^2 )^(3/
2) (wy[z]^2 + 2 βy[z]^2 )^(3/2)));

eqns := ({
{Derivative[1][a][z] == -2 a[z] (Dx cx[z] + Dy cy[z])},
{Derivative[1][wx][z] == 4 Dx cx[z] wx[z]},
{Derivative[1][wy][z] == 4 Dy cy[z] wy[z]},
{Derivative[1][cx][z] == -4 Dx cx[z]^2 + (4 Dx)/wx[z]^4 - (
4 E α[z] βx[z] βy[
z] (3 wy[z]^2 βx[
z]^2 + (wx[z]^2 + 8 βx[z]^2) βy[
z]^2))/(π (wx[z]^2 + 2 βx[z]^2)^(
5/2) (wy[z]^2 + 2 βy[z]^2)^(3/2))},
{Derivative[1][cy][z] ==
4 Dy (-cy[z]^2 + 1/wy[z]^4) - (
4 E α[z] βx[z] βy[
z] (3 wx[z]^2 βy[z]^2 + βx[
z]^2 (wy[z]^2 + 8 βy[z]^2)))/(π (wx[z]^2 +
2 βx[z]^2)^(3/2) (wy[z]^2 + 2 βy[z]^2)^(5/2))},
{Derivative[1][σ][z] == -(2./wx[z]^2) + (
2 E wx[z]^2 α[z] βx[z]^3 βy[
z])/(π (wx[z]^2 + 2 βx[z]^2)^(5/2) Sqrt[
wy[z]^2 +
2 βy[
z]^2]) - (2 (-2 E wx[z]^2 wy[z]^6 α[z] βx[
z]^3 βy[z] -
3 E wy[z]^6 α[z] βx[z]^5 βy[z] -
3 E wx[z]^4 wy[z]^4 α[z] βx[z] βy[
z]^3 - 18 E wx[z]^2 wy[z]^4 α[z] βx[
z]^3 βy[z]^3 -
20 E wy[z]^4 α[z] βx[z]^5 βy[z]^3 -
3 E wx[z]^4 wy[z]^2 α[z] βx[z] βy[
z]^5 - 16 E wx[z]^2 wy[z]^2 α[z] βx[
z]^3 βy[z]^5 -
16 E wy[z]^2 α[z] βx[z]^5 βy[z]^5 +
1.5707963267948966 (wx[z]^2 + 2 βx[z]^2)^(
5/2) (wy[z]^2 + 2 βy[z]^2)^(5/2)))/(π wy[
z]^2 (wx[z]^2 + 2 βx[z]^2)^(
5/2) (wy[z]^2 + 2 βy[z]^2)^(5/2))},
{equ1 == 0},
{equ2 == 0},
{equ2 == 0}
})

ics = {a[0] == 1, wx[0] == 4.5, wy[0] == 4.5,  cx[0] == 0.0000,
cy[0] == 0.0000, σ[0] == 0.0000000, βx[0] ==
5.17, βy[0] == 5.36, α[0] == 0.4};

Dx := 1.0; Dy := 0.5; q := 2; ν := 200;

torusSol =
NDSolve[{eqns, ics}, vars, {z, 0, 1}, MaxSteps -> ∞,
Method -> {"IndexReduction" -> {True,
"ConstraintMethod" -> "DummyDerivatives"}}]


Could i get help?

NDSolve::ivres: NDSolve has computed initial values that give a zero residual for the differential-algebraic system, but some components are different from those specified. If you need them to be satisfied, giving initial conditions for all dependent variables and their derivatives is recommended.

NDSolve::ndcf: Repeated convergence test failure at z == 0.; unable to continue. >>]2

• Welcome to Mathematica.SE! To get help here, please make sure that all relevant information is contained within the question. Please review the asking guide and how to create a minimal example, then edit the question accordingly. – Szabolcs Sep 4 '17 at 20:37
• I got this message: NDSolve::ivres: NDSolve has computed initial values that give a zero residual for the differential-algebraic system, but some components are different from those specified. If you need them to be satisfied, giving initial conditions for all dependent variables and their derivatives is recommended. >> NDSolve::ndcf: Repeated convergence test failure at z == 0.; unable to continue. >> – Khaled Sep 4 '17 at 20:40
• 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. You may find this this meta Q&A helpful – Michael E2 Sep 4 '17 at 20:50
• Also, make sure that the example you post is complete (can be tried out by other people without needing additional definitions) and try to make it as short as possible while still showing the problem. Check the links in my first comment. – Szabolcs Sep 4 '17 at 21:07
• Judging from the error message, it may be that your initial conditions violate one or more of the algebraic equations. That is, maybe $\operatorname{eqn1} \ne 0$ at $z=0$. When you post your actual code we will be able to test such ideas and, hopefully, suggest ways to make progress. – LouisB Sep 4 '17 at 21:50