By using Laplace transforms find the steady state values of w, u and v:$$\frac{du}{dt}=-\frac{\Gamma}{2}u+\Delta v,$$$$\frac{dv}{dt}=-\frac{\Gamma}{2}v – \Delta u +w \Omega,$$$$\frac{dw}{dt}=-\Gamma-\Gamma w – \Omega v.$$ Where $\Gamma, \Omega,$ and $\Delta$ are constants, and $v(t), u(t),$ and $w(t)$ are functions.
Attempt:
So these DEs are in fact a form of Maxwell–Bloch equations. To find the steady state, I set all the time derivatives equal to zero on the LHS. Then I took the Laplace transform of both sides:
$$L \{-\frac{\Gamma}{2}u+\Delta v\}= \frac{- \Gamma}{2} U(s)+\Delta V(s) =0 \tag{1}$$
$$L \{-\frac{\Gamma}{2}v – \Delta u +w \Omega \}= \frac{- \Gamma}{2} V(s)-\Delta U(s)+ \Omega W(s) =0 \tag{2}$$
$$L \{-\Gamma-\Gamma w – \Omega v \}= -\frac{\Gamma}{s} – \Gamma W(s) – \Omega V(s) \tag{3}=0.$$
So what would be the steady state values? How do we proceed from here?
And can we assume that $s=0$ lies in the region of convergence?
Best Answer
Using a vector notation, the system reads
$$\frac{d\mathbf x}{dt}=A\mathbf x+\mathbf b.$$
Take the Laplace transform, giving
$$s\mathbf X-\mathbf x_0=A\mathbf X+\frac{\mathbf b}s,$$ solved by
$$\mathbf X=(sI-A)^{-1}\left(\mathbf x_0+\frac{\mathbf b}s\right).$$
To get the response at infinity, multiply by $s$ ant take the limit at $0$,
$$\mathbf x_\infty=\lim_{s\to0}s\mathbf X=\lim_{s\to0}\,(sI-A)^{-1}\left(s\mathbf x_0+\mathbf b\right)=-A^{-1}\mathbf b.$$
This is obviously the solution of the steady-state system
$$0=A\mathbf x+\mathbf b.$$
Using the Laplace transform is overkill for this problem.