# Stability of nonlinear ODE linearized on a periodic solution.

I'm reading a paper about the effects of pulsed immunotherapy in cancer treatment. They have system$$\begin{array}{ccc} \dot{E} & = & cT-\mu_{2}E+\frac{p_{1}EI_{L}}{g_{1}+I_{L}}+s_{1}\\ \dot{T} & = & r_{2}T(1-bT)-\frac{aET}{g_{2}+T}\\ \dot{I_{L}} & = & \frac{p_{2}ET}{g_{3}+T}-\mu_{3}I_{L}+s_{2} \end{array}.$$ Now with $s_2=0$ and $$s_1=s_1(t)=\sum_{n=0}^{\infty}\sum_{i=1}^{m}=d_i\delta(t-(n\tau+t_{i-1}))$$ this function acts as a impulse of size $d_i$ in every $t= t_i$. Then, the system have a periodic solution $(\tilde{E}(t),0,0)$, where $$\tilde{E}(n\tau+t)=\tilde{E}(t)=(\tilde{E}_{j-1}+d_j)\exp{(-\mu_2(t-t_{j-1}))} ,t\in[t_{j-1},t_j]$$ $$\tilde{E_0}=\frac{\sum_{i=1}^{m}d_ie^{\mu_2 t_{i-1}}}{e^{\mu_2 \tau}-1}.$$ Now in order to investigate the stability of $(\tilde{E_0},0,0)$ they linearize the system letting $E(t)=\tilde{E}(t)+\epsilon u$, $T(t)=\epsilon v$, and $I_L(t)=\epsilon w$ which gives the linear system $$\begin{bmatrix} u'\\ v'\\ w' \end{bmatrix} = \begin{bmatrix} -\mu_2&c &p_1 \tilde{E}(t) \\ 0&r_2 - \frac{a\tilde{E}(t)}{g2} & 0 \\ 0 & \frac{p_2\tilde{E}(t)}{g_3} & -\mu_3 \end{bmatrix} \begin{bmatrix} u\\ v\\ w \end{bmatrix}.$$

Now they state that if the matrix is stable in $v$ then ($\tilde{E_0},0,0)$ is a stable fixed point. Why is that? , why the stability only depends on $v$ and not in the eigenvalues of the matrix?. I really appreciate any help or bibliography you can provide .

For a time dependent matrix $A(t)$, the behaviour of the system $x'(t) = A(t) x(t)$ cannot be determined from the (time dependent) eigenvalues of the matrix. If $A(t)$ is periodic in $t$, however, you can invoke Floquet theory, see e.g. J.D. Meiss, Differential Dynamical Systems, SIAM, 2007.
Furthermore, it is worthwhile to note that the article defines the Poincaré map $F(E_0,T_0,I_{L0}) = (E(\tau),T(\tau),I_L(\tau))$, and claims that $(\tilde{E}_0,0,0)$ is a stable fixed point of that map if $v$ is stable, i.e. if the solution to $v' = (r_2 - \frac{a}{g_2} \tilde{E}(t))v$ converges to $0$.