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     W\frac{d\hat{l_E}}{\hat{dt}} = -\hat{l_E}\hat{\rho_F} + \psi_L\hat{l_B} \quad (1)

    Let  W = \frac{1}{\epsilon} where  \epsilon << 1 .

    So then

    \frac{1}{\epsilon}\frac{d\hat{l_  E}}{\hat{dt}} = -\hat{l_E}\hat{\rho_F} + \psi_L\hat{l_B}

    \frac{d\hat{l_E}}{\hat{dt}} = -\epsilon\hat{l_E}\hat{\rho_F} + \epsilon\psi_L\hat{l_B}, \quad \quad \hat{l_E}(0) = 1

    We expand \hat{l_E}(t) in a pertubation series

     \hat{l_E}(t) = \hat{l_{E0}}(t) + \epsilon\hat{l_{E1}}(t) + \epsilon^2\hat{l_{E2}}(t) + ...

    Inserting this into (1)

     \hat{l_{E0}}' + \epsilon\hat{l_{E1}}' + \epsilon^2\hat{l_{E2}}' +  ... = -\epsilon(\hat{l_{E0}} + \epsilon\hat{l_{E1}} + \epsilon^2\hat{l_{E2}} + ... )\rho_F + \epsilon\psi_L\hat{l_B}

    Inserting into initial condition

    \hat{l_{E0}}(0) + \epsilon\hat{l_{E1}}(0) + \epsilon^2\hat{l_{E2}}(0) + ... = 1

    Now let  \epsilon \rightarrow 0

    Then \hat{l_{E0}}(0) = 1

    That is

     1 + \epsilon\hat{l_{E1}}(0) + \epsilon^2\hat{l_{E2}}(0) + ... = 1

    Subtracting 1 from both sides and diving the rest with \epsilon we get

    \hat{l_{E1}}(0) + \epsilon\hat{l_{E2}}(0) + \epsilon^2\hat{l_{E3}}(0) + ... =

    One more time let  \epsilon \rightarrow 0

    Then  \hat{l_{E1}}(0) = 0

    if we repeat the procedure we realise that also

     \hat{l_{E2}}(0) = 0, \quad  \hat{l_{E3}}(0), \quad ... = 0

    We now determine  \hat{l_{E0}}, \hat{l_{E1}}, \hat{l_{E2}} ...

    subsequently by solving the equations that we get by comparing terms containing the same power of \epsilon.

     \epsilon^0: \hat{l_{E0}}' = 0

     \epsilon^1: \hat{l_{E1}}' = \psi_L\hat{l_B} - \hat{l_{E0}}

     \epsilon^2: \hat{l_{E2}}' = -\hat{l_{E1}}


    I'm unsure what to do now, as for  \epsilon^0 I need to integrate  \rho_F

    however my equation for  \rho_F is:

     \sigma\frac{d\hat{\rho_F}}{d\hat  {t}} = \sigma\gamma_{rr}\hat{\rho_I} - \chi_0\hat{\rho_F} - m\hat{l_E}\hat{\rho_F} + m\psi_L\hat{l_B} - \frac{m\chi_L\hat{l_B}\hat{\rho_  F}}{1 - \hat{\rho_F}}

    i.e. really complicated, and coupled with other functions.

    Does anybody have any idea what I can do from here?

    Thanks in advance!
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    (Original post by sexyzebra)


    Inserting this into (1)

     \hat{l_{E0}}' + \epsilon\hat{l_{E1}}' + \epsilon^2\hat{l_{E2}}' +  ... = -\epsilon(\hat{l_{E0}} + \epsilon\hat{l_{E1}} + \epsilon^2\hat{l_{E2}} + ... )\rho_F + \epsilon\psi_L\hat{l_B}


    Thanks in advance!
    I may be being thick here but at leading order doesn't

     \hat{l_{E0}}' = 0 ?
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    (Original post by thebadgeroverlord)
    I may be being thick here but at leading order doesn't

     \hat{l_{E0}}' = 0 ?

    Your not being thick at all, I'm the thick one Thank you for pointing that out. I have edited that above to avoid confusion, however I now seem to have the same situation for  \epsilon^1

    since my equation for  \hat{l_B} is:

     \frac{d\hat{l_B}}{\hat{dt}} = \hat{\rho_F}\hat{l_E} - \psi_L\hat{l_B} - \chi_L\hat{l_B} and I'm not sure how to integrate this...
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    (Original post by sexyzebra)

     \epsilon^1: \hat{l_{E1}}' = \psi_L\hat{l_B} - \hat{l_{E0}}

     \epsilon^2: \hat{l_{E2}}' = -\hat{l_{E1}}


    I'm unsure what to do now, as for  \epsilon^0 [/lattex] I need to integrate [latex] \rho_F


    Thanks in advance!
    These are also wrong. Your  \hat{I_{Ei}} are missing  \rho_F .
    You also can now workout  \hat{I_{E0}} term.

    In terms of actually doing the intergartion I have no idea. You could try subbing the other two equations in and look for cancelling. They are hideous equations. If they gave you those equations then thats either evil or there is some other infomation or you have worked them out yourself check them (based on your ability to equate coefficeints of  \epsilon ).
 
 
 
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