Let's say that we have the following first order differential equation:
$$\frac{d \rho(t)}{d t}=F(\rho(t))$$
with some given initial condition $\rho(0)$.
I am interested in the steady-state solution $\rho_{ss}$, and to find it we set $\left.\frac{d\rho(t)}{d t}\right|_{t_{ss}}=0$ and solve the algebraic equation
$$F(\rho_{ss})=0$$
However, how in this situation do we account for the initial condition $\rho(0)=\rho_0$, without solving the full differential equation? After solving the algebraic equation I get infinetely many solutions (actually two distinct solutions, but their superposition is also a solution).
P.s. I don't know if this helps, but in my case there is a conserved quantity $Q$, which is the same for $\rho(0)$ and $\rho_{ss}$.
Best Answer
Solving the equation $\quad F(\rho)=0\quad $ gives $\rho_{ss}$, but doesn't gives the function $\rho(t)$. So we have to find the function $\rho(t)$ in which $\rho_0$ is a parameter.
$$\frac{d \rho(t)}{d t}=F(\rho(t))\quad\to\quad \frac{d\rho}{F(\rho)}=dt$$ $$\int \frac{d\rho}{F(\rho)}= t+c$$ $F(x)$ is a known function.
Suppose that it is possible to find an antiderivative for $\frac{1}{F(x)}$ , namely $G(x)$ , which then is a known function : $$G(\rho)=t+c$$ $$\rho=G^{-1}(t+c)$$ where $\quad G^{-1}\quad$ is the inverse function of $G$.
\begin{cases} \text{Initial point :} \quad G(\rho(0)) =0+c=c \quad\to\quad \rho= G^{-1}\left(t+G(\rho(0)) \right)\\ \text{Steady state point :} \quad \rho_{ss}=G^{-1}(t_{ss}+c) \end{cases} This leads to the relationship between $\rho_{ss}$ and $\rho_0=\rho(0)$ : $$\rho_{ss}=G^{-1}\left(t_{ss}+G(\rho_0)\right)$$ where $G$ and $G^{-1}$ are known functions derived from the given function $F$.
$t_{ss}$ is a root of the equation $\quad F\left(\rho(t)\right)= F\left( G^{-1}\left(t+G(\rho_0)\right) \right)=0\quad$ where all functions are known.