How to solve model with different ODEs at initial conditions?

I am trying to solve a set of ODEs. The model has different equations for initial point,so I am using If condition in ODE function to use this specific equations to solve the model at initial point t0. The reason the model have different equations for the initial condition is that, the actual ODES are undefined at t0.

This is my function;

Lets say at from f=1 to 0.2.

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function dxdf = cocurrenty(f,x,r,qA, qB, qC, xFa, xFb)

if (f>0) && (f<1.0)

dxdf(4,1)= ((xFa-f*x(1))/(1-f)) - x(4);

dxdf(5,1) = ((xFb-f*x(2))/(1-f))- x(5);

else t=10e-15; yPa= x(4);

           for n=1:100               
                        yPa(n+1)=yPa(n)-((yPa(n)+(xFb*qB/qA)/(r*...
                           (qB/qA-1)+(xFa/yPa(n)))+((1-xFa-xFb)*qC/qA)/(r*...
                           (qC/qA-1)+(xFa/yPa(n))))-1)/((xFb*xFa*qB/qA)/yPa(n)^2/...
                           (r*(qB/qA-1)+(xFa/yPa(n)))^2+((1-xFa-xFb)*xFa*...
                           qC/qA)/yPa(n)^2/(r*(qC/qA-1)+(xFa/yPa(n)))^2+1);

                        if abs(yPa(n+1)-yPa(n))<t
                                y1=yPa(n);
                                y2=(xFb*qB/qA)/(r*(qB/qA-1)+(xFa/yPa(n)));
                                y3=1-y2-y1;
                                x3=1-xFa-xFb;

                                 if (1> y1) &&  (y1 > 0) && (1> y2) &&  (y2 > 0)
                                     break
                                 end
                        end                    
           end
dxdf(4,1) = y1-x(4);
dxdf(5,1) = y2-x(5);   
end

dxdf(1,1)= (qA*(x(1)-r*x(4)) - x(1) * (qA*(x(1)-r*x(4))+qB*(x(2)-r*x(5))+qC*((1-x(1)-x(2))-r*(1-x(4)-x(5)))))/... (f*(qA*(x(1)-r*x(4))+qB*(x(2)-r*x(5))+qC*((1-x(1)-x(2))-r*(1-x(4)-x(5)))));

dxdf(2,1)= (qB*(x(2)-r*x(5)) - x(2) * (qA*(x(1)-r*x(4))+qB*(x(2)-r*x(5))+qC*((1-x(1)-x(2))-r*(1-x(4)-x(5)))))/... (f*(qA*(x(1)-r*x(4))+qB*(x(2)-r*x(5))+qC*((1-x(1)-x(2))-r*(1-x(4)-x(5)))));

dxdf(3,1)= (-1/(qA*(x(1)-r*x(4))+qB*(x(2)-r*x(5))+qC*((1-x(1)-x(2))-r*(1-x(4)-x(5)))));

end

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Newtons Algorithm is used to find x(4) and x(5) at t0 (f=1). The problem is when I increase the value of r (a constant in ODEs) I got this warning.

''Warning: Failure at t=1.000000e+00. Unable to meet integration tolerances without reducing the step size below the smallest value allowed (3.552714e-15) at time t. > In ode15s at 669 In PressureOnemli at 75 ''

Is there another way of typing ODEs with different initial conditions? Any help would be much appreciated. Thank you. Merve

[Torsten]

Why don't you start with f=0.999999 ?

I mean: There can't be any effect on the solution if the values for x(4) and x(5) differ in only one single point, namely f=1.

Best wishes

Torsten.