DAE Problem : cannot understand how to find the problem in my code

    This is code to solve 6 ode's and 12 algebraic equations. All of them are interdependent. How do I go about it? I keep getting errors but I am unable to solve it.     
**M file**
 *M file* function f=ncs1_dae(t,x)

          %Input parameters

          global alpha A AH AI AO AV ACS beta CPH CPI CPIN CPO CPV dB g hB hI hO hH hV Hmax k M MH min Pset qein R rhol rhov ri ro TIN TR vB lambda;

          input()

          %Variables

          TI=x(1);
........
..........

              =x(16);

          %f(1) to f(6) are ODEs

          f(1)=

          f(2)=

          f(3)=

          f(4)=

          f(5)=

          f(6)=

          %f(7) to f(18) are Algebraic Equations

          f(7)=

          f(8)=

          f(9)=

          f(10)=

          f(11)=

          f(12)=

          f(13)=

          f(14)=

          f(15)=

          f(16)=

          f(17)=

          f(18)=

          f=f';

          This is what I enter in the command window:

          >> x0=[298 298 298 298 1 0.3 0 373 0 0 0 0 0 0 0 0.0013 0.0041 0.0056];

          >> tspan=[0 1200];

          >> M=[eye(6,18);zeros(12,18)];

          >> options=odeset('Mass',M);

          >> [t1,x1]=ode15s(f,[0,1200],x0,options);

          The warning msg I get is : Warning: Failure at t=3.014342e-01. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (8.881784e-16) at time t. > In ode15s at 753 >>    if true
                % code
              end

[Walter Roberson]

What if the equations are inconsistent?

[Urvi]

Inconsistent as in? I have 18 variables and 18 equations. All of which are interdependent.

Answer 1

[Jan]

You can define an DAE such that there is no feasible point. If the mass matrix is singular, as in your case, the initial slope M(t0, y0) * y'(0) = f(t0, y0) must exist.

Imagine a DAE which describes a point sliding on a circulare wire. If you start the integration with a point apart from the wire, there is no valid first step of the integration, because it is impossible to keep the trajectory on the feasible path.

[Urvi]

Oh ok I get it now! Thanks! Will look into my equations.