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After spending several hours exploring different options, I still couldn't not figure out how to solve the following differential equations. Any suggestions would be highly appreciated.

W[δ_]=Γ*1/√(π*ϵ0^2) Exp[-δ^2/ϵ0^2];
densityEqns = {D[ρ22[δ,t],t] ==-(γsp+Γ)*ρ22[δ,t]-(I*χ*ρ12[δ,t]-I*χ*ρ21[δ,t])+W[δ]*Integrate[ρ22[δ,t],δ],
D[ρ12[δ,t],t] ==-(γt-I*δ)*ρ12[δ,t]-I*χ*(2 ρ22[δ,t]-1),
D[ρ21[δ,t],t] ==-(γt+I*δ)*ρ21[δ,t]+I*χ*(2  ρ22[δ,t]-1),ρ22[δ,0]==0,ρ12[δ,0]==0,ρ21[δ,0]==0};
soln=DSolve[densityEqns,{ρ22[δ,t],ρ12[δ,t],ρ21[δ,t]},t]

As per @Dr.WolfgangHintze suggestions, I separated the variables and got the following equations, eqns={D[D[[Rho]22[[Delta],t],t]/W[[Delta]],[Delta]] ==-([Gamma]sp+[CapitalGamma])D[[Rho]22[[Delta],t]/W[[Delta]],[Delta]]-D[(I[Chi][Rho]12[[Delta],t]-I[Chi][Rho]21[[Delta],t])/W[[Delta]],[Delta]]+[Rho]22[[Delta],t], D[[Rho]12[[Delta],t],t] ==-([Gamma]t-I[Delta])[Rho]12[[Delta],t]-I[Chi](2 [Rho]22[[Delta],t]-1), D[[Rho]21[[Delta],t],t] ==-([Gamma]t+I[Delta])[Rho]21[[Delta],t]+I[Chi]*(2 [Rho]22[[Delta],t]-1),r22t[0]==0,r12t[0]==0,r21t[0]==0} Not sure how to solve this.

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  • $\begingroup$ @ crossingsymmetry: First of all, the system is a system of linear (all three functions rho appear in the first power) partial differential equations. Therefore you can separate variables, i.e. write f(x,t) = g(x)*h(t), and in the first equation get rid of the integral by differentiating partially with respect to x. Ok, no MMA specifics up to now. So please prepare your problem up to this point. Then we can discuss how to use MMA to proceed further. $\endgroup$ – Dr. Wolfgang Hintze Nov 10 '14 at 10:01
  • $\begingroup$ Thank you for the reply. I don't see how I can get rid of the integral by differentiating partially wrt delta. My equations have W[d]*Integrate[rho22[d,t],d], so the integration will never go away, am I missing something? $\endgroup$ – crossingsymmetry Nov 13 '14 at 4:18
  • $\begingroup$ In order to get rid of the integral divide the equation by W[d] before differentiating. $\endgroup$ – Dr. Wolfgang Hintze Nov 14 '14 at 8:47
  • $\begingroup$ @Dr.WolfgangHintze I am not sure how that helps. When I separate the variables and get rid of integral, I still have some complicate mess to solve. I have written the updated equations above. $\endgroup$ – crossingsymmetry Nov 15 '14 at 20:09
  • $\begingroup$ Could you please specify the limits of your integral over delta in the original version of the problem? $\endgroup$ – Dr. Wolfgang Hintze Nov 20 '14 at 8:31
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We can find the exact solutions to the equations as follows:
For the ease of reference let us first write down the equations one by one.

eq1 = D[\[Rho]22[\[Delta], t], 
  t] == -(\[Gamma]sp + \[CapitalGamma])*\[Rho]22[\[Delta], 
    t] - (I*\[Chi]*\[Rho]12[\[Delta], t] - 
    I*\[Chi]*\[Rho]21[\[Delta], t]) + 
  W[\[Delta]]*Integrate[\[Rho]22[\[Delta], t], \[Delta]]

eq2 = D[\[Rho]12[\[Delta], t], 
  t] == -(\[Gamma]t - I*\[Delta])*\[Rho]12[\[Delta], t] - 
  I*\[Chi]*(2 \[Rho]22[\[Delta], t] - 1)

eq3 = D[\[Rho]21[\[Delta], t], 
  t] == -(\[Gamma]t + I*\[Delta])*\[Rho]21[\[Delta], t] + 
  I*\[Chi]*(2 \[Rho]22[\[Delta], t] - 1)

The initial conditions for t = 0 are

{\[Rho]22[\[Delta], 0] == 0, \[Rho]12[\[Delta], 
  0] == 0, \[Rho]21[\[Delta], 0] == 0}

eq1 shows that at t=0 the derivative of \[Rho]22[\[Delta],t] is 0. Hence this quantity is = 0 always. Also the function w[x] is irrelevant.

This reduces the equations eq2 and eq3 to

eq2a = D[\[Rho]12[\[Delta], t], t] == 
-(\[Gamma]t - I*\[Delta])*\[Rho]12[\[Delta], t] + I*\[Chi]

eq3a = D[\[Rho]21[\[Delta], t], t] == 
-(\[Gamma]t + I*\[Delta])*\[Rho]21[\[Delta], t] - I*\[Chi]

Which are solved for constant \[Chi] as follows::

sol1a = 
 DSolve [{\[Rho]12[\[Delta], 0] == 0, 
    D[\[Rho]12[\[Delta], t], 
      t] == -(\[Gamma]t - I*\[Delta])*\[Rho]12[\[Delta], t] + 
      I*\[Chi]}, \[Rho]12[\[Delta], t], t] /. K[1] -> s

(* Out[7]= {{\[Rho]12[\[Delta], t] -> -((
    I (-1 + E^(t (-\[Gamma]t + I \[Delta]))) \[Chi])/(\[Gamma]t - 
     I \[Delta]))}} *)

sol2a = 
 DSolve [{\[Rho]21[\[Delta], 0] == 0, 
    D[\[Rho]21[\[Delta], t], 
      t] == -(\[Gamma]t + I*\[Delta])*\[Rho]21[\[Delta], t] - 
      I*\[Chi]}, \[Rho]21[\[Delta], t], t] /. K[1] -> s

(* Out[8]= {{\[Rho]21[\[Delta], t] -> (
   E^(-t (\[Gamma]t + I \[Delta])) (-1 + E^(
      t (\[Gamma]t + I \[Delta]))) \[Chi])/(I \[Gamma]t - \[Delta])}} *)

Hence you have the explicit Solutions of the equations with the given zero initial conditions and constant \[Chi].

Regards,
Wolfgang

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  • $\begingroup$ Thanks @Dr.WolfgangHintze for spending your time trying to help me. I am not sure that the assumption W[delta] is irrelevant is true. At t=0, D[[Rho]22[[Delta], 0], t]==0, that doesn't necessary mean that [Rho]22[[Delta], 0] == 0. A simple example, lets take y[x,t] = x*t. Now at t=0 derivative of y wrt t is 0. That doesn't give y[x,t]=0. I know the term W[delta] is relevant, but still have no clue how to solve the equations beside taking some perturbative solution, which unfortunately result in an "incomplete" physics. $\endgroup$ – crossingsymmetry Nov 25 '14 at 0:50
  • $\begingroup$ In case you are curious, the equations are taken from the following paper, see equation 6. imgur.com/5NlafaK $\endgroup$ – crossingsymmetry Nov 25 '14 at 1:07
  • $\begingroup$ 1) From the paper quoted I gather that the first equation is of the form d/dt r22[w,t] = i/h (V r12[w,t] - c.c.) - g r22[w,t] + f[w] Integrate[ r22[w',t], {w',-oo,oo}]. That is the integral leaves no dependence on frequency. 2) Now your initial condition r22[w,t=0] = 0 gives at t = 0 : d/dt r22[w,t] = i/h (V r12[w,t] - c.c.) but this is 0 because also r12[w,t=0] = 0. 3) your example contains an error: if y=x*t then dy/dt = x rather than 0. $\endgroup$ – Dr. Wolfgang Hintze Nov 25 '14 at 13:27

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