= Solution
At an <endemic equilibrium>, $I^*>0$. The equation $\dot I=0$ then gives
$$
\boxed{S^*=\frac{d+\delta}{\beta}}.
$$
Since also $S^*>0$, the equation $\dot S=0$ gives
$$
\boxed{\beta I^*+d=be^{-aS^*}},
\qquad
I^*=\frac{be^{-aS^*}-d}{\beta}.
$$
This infective population is positive precisely when
$$
S^*<\frac1a\log\frac bd=N^*,
$$
which is exactly the assumed instability condition for the disease-free equilibrium. Thus the <endemic equilibrium of the susceptible-infective model with exponential birth> exists.
At this equilibrium the relations above reduce the <Jacobian matrix> to
$$
J(S^*,I^*)=
\begin{pmatrix}
-aS^*be^{-aS^*}&-\beta S^*\\
\beta I^*&0
\end{pmatrix}.
$$
Hence
$$
\operatorname{tr}J=-aS^*be^{-aS^*}<0,
\qquad
\det J=\beta^2S^*I^*>0.
$$
By the <trace-determinant stability criterion>, both eigenvalues have negative real part. The <linearization stability theorem> therefore proves that the endemic equilibrium is locally <asymptotic stability>[asymptotically stable].
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