If rewrite your system as

eq0 := [diff(s(x, t), t) = -a*i(x, t)*s(x, t), diff(i(x, t), t) = a*i(x, t)*s(x, t)-b*i(x, t)+c*(diff(i(x, t), x))];

then it is equivalent to the following splited system

sys:=[i(x,t) = -diff(s(x,t),t)/a/s(x,t),S(x)+ (s(x,t)^2*a-s(x,t)*b*ln(s(x,t))-diff(s(x,t),t)+diff(s(x,t),x)*c)/c/s(x,t)];

where S(x) is an arbitrary function.

I think that the presence of ln(s(x,t)) item lead to main problems in (numeric) solving process.

I'd advise you to try numerics in the cases when exact solutions exist, e.g.

si_ans1:= {i(x,t) = m*(RootOf(Int(1/(_f*(-_f*a+b*ln(_f))),_f = `` .. _Z)*m-Int(1/(_f*(-_f*a+b*ln(_f))),_f = `` .. _Z)*c+t*m+x+_C1*m)*a-b*ln(RootOf(Int(1/(_f*(-_f*a+b*ln(_f))),_f = `` .. _Z)*m-Int(1/(_f*(-_f*a+b*ln(_f))),_f = `` .. _Z)*c+t*m+x+_C1*m)))/(-m+c)/a, s(x,t) = RootOf(-Int((-m+c)/_f/m/(-_f*a+b*ln(_f)),_f = `` .. _Z)*m+t*m+x+_C1*m)};

or

si_ans2:=[i(x,t) = -D(_F1)((x+t*c)/c)*(-c*k-RootOf(x-Int(1/(-c*k-_a*a+b*ln(_a))/_a*c,_a = `` .. _Z)+_F1((x+t*c)/c))*a+b*ln(RootOf(x-Int(1/(-c*k-_a*a+b*ln(_a))/_a*c,_a = `` .. _Z)+_F1((x+t*c)/c))))/c/a,s(x,t) = RootOf(x-Int(1/(-c*k-_a*a+b*ln(_a))/_a*c,_a = `` .. _Z)+_F1((x+t*c)/c))]; 

Your PDEs system can be solved exactly with help of Maple by the following scheme ( I leave out  bulky outputs).

with(PDETools);

sys:=[diff(v(s,n),n) + diff(u(s,n),s) +diff(xi(s,n),s) + A*n*diff(c(s),s) = 0,

A1*diff(xi(s,n),n) + diff(v(s,n),s) -c(s)+A2*v(s,n) + A3* c(s) = 0,

diff(u(s,n),s) + 2*A2* u(s,n)=A2*(xi(s,n) + A*n*c(s)) -A1*diff(xi(s,n),s)-A2*n*c(s)];

Here, as I understand,  c(s) is a parameter.

From the beginning we need to split the system

cas:=[casesplit(sys,[xi,u,v])];

As a result we obtain system of 4 equations, which is equivalent to initial one. First of all we have to solve the last equation

pp := diff(v(s,n),`$`(s,3)) = (diff(c(s),`$`(s,2))*A1-3*A1*A2*A*diff(c(s),s)-A1*diff(v(s,n),`$`(n,2),s)-2*A2*A1*diff(v(s,n),`$`(n,2))-A1*A*diff(c(s),`$`(s,2))+A1*A2*diff(c(s),s)-A3*diff(c(s),`$`(s,2))*A1-A2*diff(v(s,n),`$`(s,2))*A1+4*A2*diff(v(s,n),`$`(s,2))-diff(c(s),`$`(s,2))+3*A2*A3*diff(c(s),s)+3*A2^2*diff(v(s,n),s)-3*A2*diff(c(s),s)+A3*diff(c(s),`$`(s,2)))/(-1+A1);

for unknown function v(s,n). It is linear non-homogeneous third-order PDE (so, generally speaking, we need  three initial conditions !). One of exact solutions of this PDE can be obtained

an0:=subs({_c[2]=0,_C1=0,_C2=0,_C3=0,_C4=0,_C5=0

},pdsolve(pp,v, INTEGRATE, build));

Then we will seek the solution in the familiar form

an_W := v(s,n) = w(s,n) + rhs(an0);

which lead to the homogeneous linear PDE with constant coefficients  for w(s,n)

ppp:=collect(numer(factor(pdetest(an_W,pp))),diff);

ppp := -3*A2^2*diff(w(s,n),s)+2*A2*A1*diff(w(s,n),`$`(n,2))+(A1*A2-4*A2)*diff(w(s,n),`$`(s,2))+A1*diff(w(s,n),`$`(n,2),s)+(-1+A1)*diff(w(s,n),`$`(s,3));

This equation can be solved by Fourier method or by separation of variables (with similar result). Suppose that

an_w:=w(s,n)=Int(Phi(s,omega)*exp(I*n*omega),omega=-infinity..infinity);

and from

pdetest(an_w,ppp);

after some manipulation (by hand) we find that Phi(s,omega) must fit the following ODE

WW := (-A1*omega^2-3*A2^2)*diff(Phi(s,omega),s)+(A1*A2-4*A2)*diff(Phi(s,omega),`$`(s,2))+(-1+A1)*diff(Phi(s,omega),`$`(s,3))-2*A2*A1*Phi(s,omega)*omega^2;

which can be solved

ans_w:=pdsolve(WW);

Substituting it to an_W we obtain the general solution for  v(s,n). Now to produce a complete solution we need to specify 3 arbitrary functions {_F1(omega), _F2(omega), _F3(omega)}  in above solution by standard way from general solution and the set of initial conditions .

Then substituting it to cas  we get the split system for u and xi, which can be easily solved too .   

It seems to me that your method of attack the DE problems is very close to the operator method. It has a long history. See, please

A.A.Agrachev, R.V.Gamkrelidze, The exponential representation of flows and the chronological calculus. Matem. sbornik,1978, v.107. (in Russian)

A.A.Agrachev, S.A.Vakhrameev, Chronological series and Cauchy-Kovalevska theorem.Itogi nauki. VINITI. Problemy geometrii,1981, v.12 (in Russian)
 

and (unfortunately without references on above papers)

http://arxiv.org/abs/math-ph/0409035

Yuri

What does Cauchy mean in this sense?

Here the Cauchy prolem has  standard meanins - roughly speaking when PDE (system) is supplemented with conditions on one point of one independent variable.

Yuri

Dear Jurjen,

I think that procedure  "Formal series solutions to non-linear DE (ODE or PDE) or systems of them (Cauchy problem)"

http://www.maplesoft.com/applications/app_center_view.aspx?AID=1906

may help you.

First of all try to solve the Cauchy poblem for your system with some Order for t (or with some Order for r). Then test it and solve (by series) resulting ODE (system) with respect to free parameters with your boundary conditions. If your model is good enough, you obtain wholly satisfactory symbolic series solution (depending on used Order).

 Yuri