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.


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?

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    $\begingroup$ Similar question: math.stackexchange.com/questions/123441/… $\endgroup$ – Chee Han May 19 '16 at 9:10
  • $\begingroup$ Ah, thank you for the link. So basically for the steady states of $v(t), u(t),$ and $w(t),$ we want $$\lim_{s \to 0} s V(s), \lim_{s \to 0} s U(s), \lim_{s \to 0} s W(s).$$Is that right? So to find expressions for the Laplace transform of each of these, we must solve the three equations in my post above for three unknowns? Are my Laplace transform expressions above correct? $\endgroup$ – Merin May 19 '16 at 9:45
  • $\begingroup$ Yes, but you do not set the all three time derivatives equal to zero; instead you have to take Laplace transform of the full coupled ODEs, solve for $V(s), U(s), W(s)$. Then you evaluate the limit as stated above. $\endgroup$ – Chee Han May 19 '16 at 9:48
  • $\begingroup$ Why do you solve that with Laplace ? Setting the LHS equal to zero gives an easy linear system. $\endgroup$ – Yves Daoust May 19 '16 at 10:15
  • $\begingroup$ What do you mean by taking the Laplace transform of "the full coupled ODEs"? We need to set them to zero in order to be able to solve them simultaneously. For example in (1) and (3) I wrote both $U(s)$ and $W(s)$ in terms of $V(s),$ then by using (2) I found:$$V(s)=\frac{-\Gamma}{s\left( \frac{\Gamma}{2}+ \frac{2\Delta^2}{\Gamma}+ \frac{\Omega}{\Gamma} \right)}.$$Then we need to evaluate $$\lim_{s \to 0} s \frac{-\Gamma}{s\left( \frac{\Gamma}{2}+ \frac{2\Delta^2}{\Gamma}+ \frac{\Omega}{\Gamma} \right)}$$But I am not sure why the limit doesn't look right. $\endgroup$ – Merin May 19 '16 at 10:15

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.

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  • $\begingroup$ You should have $\lim_{s \to 0}$ after the second equation in the last line before the break. Also, it can happen that the Laplace transform of $x$ doesn't even exist for all $Re(s)>0$, that depends on the eigenvalues of $A$ and possibly on $x_0$. Without noticing this one would find that every linear system with constant coefficients approaches a steady state, which is obviously nonsense. $\endgroup$ – Ian May 19 '16 at 13:25
  • $\begingroup$ @Ian: right, I fixed the typo. Indeed all of this only makes sense for a stable system. (Anyway, a system exactly in the steady state remains in the steady state, even unstable.) $\endgroup$ – Yves Daoust May 19 '16 at 13:54

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