This DAE appears to be of index greater than 1 but det(M + lambda*dF/dx) isn't zero

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Marco 2024-6-11

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回答： Marco 2024-6-11

Hi everyone i'm trying to solve a DAE of the form M*y'(t) = F(y,t), with M a singular square matrix. Just for the context, my DAE comes from the equation of the dynamic response of a beam

when the space derivatives are transformed into finite difference.

F is defined as follow : F(y,t) = A.y(t) - b(t) with A a square matrix non singular and b a vector. A and M are defined bellow :

A = zeros(9,9);
a = EI/(m*h);
A(1,:) = [h 0 -1 0 0 0 0 0 0];
A(2,:) = [1 0 0 -1 0 0 0 0 0];
A(3,:) = [0 h 0 0 -1 0 0 0 0];
A(4,:) = [0 a 0 0 0 -a 0 0 0];
A(5,:) = [0 0 1 h 0 0 0 0 0];
A(6,:) = [0 0 0 1 h 0 0 0 0];
A(7,:) = [0 0 0 0 1 h 0 0 0];
A(8,:) = [0 0 0 0 0 0 1 0 0];
A(9,:) = [0 0 0 0 0 a 0 0 -a];
M = zeros(9,9);
M(8,2) = 1;
M(9,7) = 1;

The function for the DAE system is :

function F = Vect4(t,y,w,P,m,A)
ligne = length(A);
% Création de b
b = zeros(ligne,1);
b(4) = -P/m;
b(9) = -P/m;
% Calcul de la DAE
F = A*y - b;
end

w isn'nt used because i simplified the equation by removing the cosine term in b(t).

And finally the output :

n = 2; % Number of nodes
ligne = 4 + 5*(n-1); % Nombre de lignes (Nombre d'équations)
w = 2*pi * 5; % Pulsation
P = 100;
tspan = [0 2];
y0 = zeros(14,1);
opt = odeset('Mass',M);
[t,y] = ode23t(@(t,y) Vect4(t,y,w,P,m,A), tspan, y0,opt);

But here comes the problem, i get the classical error :

Error using daeic12 (line 78)

This DAE appears to be of index greater than 1.

Error in ode23t (line 282)

[y,yp,f0,dfdy,nFE,nPD,Jfac] = daeic12(odeFcn,odeArgs,t,ICtype,Mt,y,yp0,f0,...

Error in Script_Dyna_vrai (line 114)

[t,y] = ode23t(@(t,y) Vect4(t,y,w,P,m,A), tspan, y0,opt);

I don't really understand because firstly there are only first derivatives. Secondly, i belive that the index is greater than 1 when the matrix

(M +

) is singular, which is not the case here for every lambda. I tried using other ODE solvers like ODE15i for example but the problem persists.

If someone has any clues, i would apreciate. Thank you very much.

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Marco 2024-6-11

Ok, EI is equal to 5*10^5, m = 25*(0.12)^2/9.81, h = L/n with L = 2 and n = 2 so h = 1.

n is the number of nodes in the discretisation, but here i simplify the problem with only two nodes.

The PDE is :

With the boundary conditions :

at t = 0, every state variables are zeros.

When i do the finite difference :

, it gives for all nodes except the last one (so i = 0,1 if there is only two nodes) :

for i = 0,1 and where

design the j-eme derivative of V for the i-eme node

It results to the following matrix equation

with A and M defined previously,and x =

if n=2

are all equal to zero due to boundary conditions.

is equal to zero too because

= 0. Then these variabales which are null are not put into x(t) to get a non singular matrix.

Torsten 2024-6-11

在 MATLAB Online 中打开

This is your DAE system written out. There is no equation from which y(8) and y(9) could be deduced.

h*y(1) - y(3) = 0
y(1) - y(4) = 0
h*y(2) - y(5) = 0
a*y(2) - a*y(6) + P/m = 0
y(3) + h*y(4) = 0
y(4) + h*y(5) = 0
y(5) + h*y(6) = 0
dy2/dt = y(7)
dy7/dt = a*y(6) - a*y(9) + P/m

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Answer 1

Marco 2024-6-11

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2127496-this-dae-appears-to-be-of-index-greater-than-1-but-det-m-lambda-df-dx-isn-t-zero#answer_1470381

Thank you very much ! the error was quite "easy" but by looking again and again and again on the problem i lost my mind and got confused. Thank you infinitely.