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How to linearize these terms in q[t] and p[t,y]

term1=(covd[Derivative[1][q][t] + Derivative[1, 0][p][t, y], t] + 
      2*(1 + 2*\[Alpha][t, r, \[Theta], \[Phi]])*(Derivative[0, 1][a][t, y]*  (4*(p[t, y] + q[t])*Derivative[0, 1][a][t, y] - 
                a[t, y]*Derivative[0, 1][p][t, y]) + 
      2*a[t, y]*(p[t, y] + q[t])* Derivative[0, 2][a][t, y]))/(2*a[t, y]^2*(p[t, y] + q[t])^3* (1 + 2*\[Alpha][t, r, \[Theta], \[Phi]]))


term2=(Derivative[0, 1][a][t,y]*(4*(p[t, y] + q[t])*Derivative[0, 1][a][t, y] - a[t, y]*Derivative[0, 1][p][t, y]) + 2*a[t, y]*(p[t, y] + q[t])*
        Derivative[0, 2][a][t, y])/(a[t, y]^2*(p[t, y] + q[t])^3)

the codes I use are:

Normal[Series[term1 /. {q -> (eps q[#] &), p -> (eps p[#] &)}, {eps, 0, 1}]]

The output is:

(2*(1 + 2*\[Alpha][t, r, \[Theta], \[Phi]])*(4*(p[t] + q[t])* Derivative[0, 1][a][t, y]^2 + 
             2*a[t, y]*(p[t] + q[t])*Derivative[0, 2][a][t, y]) +     (Derivative[1][p][t] + Derivative[1][q][t])* Derivative[1, 0][covd][0, t])/
     (2*eps^2*a[t, y]^2*(p[t] + q[t])^3*(1 + 
      2*\[Alpha][t, r, \[Theta], \[Phi]])) + 
   ((Derivative[1][p][t] + Derivative[1][q][t])^2*
    Derivative[2, 0][covd][0, t])/
     (4*eps*a[t, y]^2*(p[t] + q[t])^3*(1 + 
      2*\[Alpha][t, r, \[Theta], \[Phi]])) + 
   ((Derivative[1][p][t] + Derivative[1][q][t])^3*
    Derivative[3, 0][covd][0, t])/
     (12*a[t, y]^2*(p[t] + q[t])^3*(1 + 
      2*\[Alpha][t, r, \[Theta], \[Phi]])) + 
   (eps*(Derivative[1][p][t] + Derivative[1][q][t])^4*    Derivative[4, 0][covd][0, t])/(48*a[t, y]^2*(p[t] + q[t])^3*
        (1 + 2*\[Alpha][t, r, \[Theta], \[Phi]]))

So it did not linearized cause (q[t]+p[t,y])^3 still there

or:

Normal[Series[term2 /. {q -> Function[{t}, eps q[t]], p -> Function[{t, y}, eps p[t, y]]}, {eps, 0, 1}]] /. eps -> 1

The output returns the same function.

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  • 1
    $\begingroup$ Use the replacement term1 /. {q :> (eps q[##] &), p :> (eps p[##] &)}. Otherwise p[t, y] will be replaced by eps p[t]. $\endgroup$ Nov 26, 2023 at 16:57
  • $\begingroup$ At which point do you want to linearize this? Currently, you seem to trying that at $q = 0$ and $p = 0$ where the expression has a pole of order 3 (so the expression is not linearizable there). $\endgroup$ Nov 26, 2023 at 16:58
  • $\begingroup$ @HenrikSchumacher. The replacements works, thanks. q[t] and p[t] do not equal zero so the terms linearized and only (q[t]+p[t])^2 remains. $\endgroup$
    – Dr. phy
    Nov 26, 2023 at 17:18

1 Answer 1

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Maybe you want to do something like this?

DerivativeWithRespectToq = D[
   term1 /. {q[args__] :> (q[args] + eps u[args])},
   eps
   ] /. {eps -> 0, u[args___] :> 1}

DerivativeWithRespectTop = D[
   term1 /.  {p[args__] :> (p[args] + eps v[args])},
   eps
   ] /. {eps -> 0, v[args___] :> 1}
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