Q:

Behavior of Non-Hyperbolic Equilibria?

So I'm working on a differential equation problem concerning epidemics - we're using the Kermack-McKendrick model.  I've reached a point where I need to sketch phase portraits near my equilibria, however, I'm a little lost with my results.  After evaluating my equilibria via the jacobian matrix (once I've linearized) I end up with eigenvalues of $\lambda=0$ and $\lambda=-a$, ($a>0$)
So since I have a real $\lambda=0$ I have a non-hyperbolic equilibira correct?  Does this negate my linearization? Is this not a robust case?  When I look at the trace-determinant criteria the only value that I can interpret would be when $-a=0$ at which point I think I have a star point but I'm not entirely sure... I could use some direction as to best sketch a phase portrait here... 
If it helps, the system I am modeling is:
$$S'=-rIS$$
$$I'=rIS-aI$$

A:

Every point on the axis $I=0$ is a fixed point and every solution starting from $(S_0,I_0)$ in the nonnegative quadrant converges to some point $(S_\infty,0)$. The only case when $S_\infty=0$ is if $S_0=0$. 
To see this, note that $S'+I'=-aI$ hence $S'=-rSI=bS(-aI)=bS(S'+I')$ with $b=r/a$. 
If $S\ne0$, this is solved by $S'S^{-1}=b(S'+I')$, that is, $\log|S|=b(S+I)+$Constant, or, $$S=S_0\,\mathrm e^{-b(S_0+I_0)}\,\mathrm e^{b(S+I)}.$$ In particular, for every positive $S_0$ and $I_0$, $S_\infty$ is the unique solution in the interval $(0,S_0)$ of the identity $$S_\infty\,\mathrm e^{-bS_\infty}=S_0\,\mathrm e^{-b(S_0+I_0)}.$$
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