I'm reading an article leading to the following system $$\left\lbrace \begin{array}{lll} y_a'(t) &= &-k_1y_a(t)+k_3y_b(t)y_c(t) \\ y_b'(t) &= & k_1y_a(t)-k_3y_b(t)y_c(t)-k_2[y_b(t)]^2 \\ y_c'(t) &= & k_2[y_b(t)]^2 \end{array} \right. $$ $$\left\lbrace \begin{array}{lll} y_a'(0) &= &1 \\ y_b'(0) &= &0 \\ y_c'(0) &= &0 \end{array} \right. $$ with $k_1,k_2,k_3 >0$. We can prove that for $t>0$, $y_a(t),y_b(t),y_c(t)\geq0$. And that the critical points are $(0,0,\gamma )$, $\gamma \geq 0$.
To know if $(0,0,\gamma )$ is stable we can first verify that it is the case for the linearised system $$\left\lbrace \begin{array}{lll} y_a'(t) &= &-k_1y_a(t)+k_3\gamma y_b(t) \\ y_b'(t) &= & k_1y_a(t)-k_3\gamma y_b(t) \\ y_c'(t) &= & 0 \end{array} \right. $$ The author is saying that it's trivially the case but I don't understand why. I've tried to calculate the eigenvalues to use Lyapunov's theorem, I found that $$P_A(X) = -X[X^2+X(k_1+k_3) +k_1k_3(1-\gamma )]$$ then I have to discuss $(1-\gamma)\leq 0$, etc.. and can't conclude. Plus the eigenvalue $\lambda_1 =0$ do not verify that $\Re (\lambda_1 )<0$.
Am I missing something? Should I try to calculate $\exp (tA)$, planning to use Duhamel formula?
Any help will be greatly appreciate.