How to solve a system of parabolic PDEs and 1st order ODEs?

I'm trying to solve a system of parabolic ODEs and first-order ODEs. I understand that Matlab's pdepe solver cannot solve 1st-order differential equations. How do I solve this sytem? I also tried solving the DAE system using a mass matrix, but encountered the following error:

Error in ode15s (line 310) [y,yp,f0,dfdy,nFE,nPD,Jfac] = daeic12(odeFcn,odeArgs,t,ICtype,Mt,y,yp0,f0,...

A sample set of equations is given below.

[Torsten]

As it stands, you can use "pdepe" to solve your system.

For the equations for c3 and c4, you have to prescribe f=0 in the pde function and pl=pr=0, ql=qr=1 in the boundary function of "pdepe".

[Monisha S R]

Oh thanks Torsten. Let me try. I was of the view that only parabolic and elliptic equations can be solved.

[Torsten]

Problematic are 1st order PDEs, but your 0-order PDEs (i..e. ODEs) for c3 and c4 should work.

[Bruno Luong]

I just don't understand your system of equations, doe sthat mean

c1 satisfies eqt (1)
c2 satisfies eqt (1+2)
c3 satisfies eqt (1+3)
c4 satisfies eqt (1+4)

I can't see how the variables (c2-c4) that can meet two partial equations at the same time.

[Torsten]

As I understand it, c in the first equation is c1. So equation i gives ci.

[Bruno Luong]

Then the equations are actually independent? This is not a system of PDE strictly speaking but two of them (c1+c2) can be solved with pdede, the other two (c3+c4), be solved with pdepe or ode independently.

Seem odd as question.

[Torsten]

We don't know the source terms ...

[Monisha S R]

@Bruno First equation is for (edited now). The source terms contain the dependence on other concentrations. For example, where k is a rate constant and so on.

[Bruno Luong]

Thanks.

Then PDEPE might work. I still have a little doubt about c3/c4 which is purely time ODE and whereas it wil get into conflict with spatial boundary conditions.

[Monisha S R]

@Bruno Yeah, I guess you may be right. I don't know what boundary conditions to give for c3 and c4.

[Torsten]

c3 and c4 don't need boundary conditions.

Artificially setting pl=pr=0, ql=qr=1 in the boundary function of "pdepe" makes the equations tractable for the "pdepe" solver.

[Bruno Luong]

Does that mean those bc will be ignored by PDEPE since there is no spatial eqts for c3 and c4?