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I have a difficulty to solve this system equations:

Eqn1 := D[fθϕ[r, θ], r] == D[frϕ[r, θ], θ]
Eqn2 := D[fθt[r, θ], r] == D[frt[r, θ], θ]
bce = {ϕ1b[r, 0] == 1, ϕ1b[r, π] == -1, ϕv[r, 0] == 1, ϕv[r, π] == -1};
ice = {ϕ1b[M, θ] == Cos[θ], ϕv[M, θ] == Cos[θ]};

soldifeq = DSolve[Join[{Eqn1, Eqn2}, bce, ice], {ϕ1b[r, θ], ϕv[r, θ]}, {r, θ}]

With functions:

fθϕ[r_, θ_] := (4 M*Sin[θ])/Σ^3 (a*r*ℜ2*Sin[θ]*Sqrt[Δ]*κ3[r, θ])
frϕ[r_, θ_] := (4 M*Sin[θ])/Σ^3 (3 a (r^2 + a^2)*ℜ1*Cos[θ]*κ3[r, θ] + r (r^2 + a^2)*ℜ2*κ6[r, θ])
fθt[r_, θ_] := ((2 a*r*ℜ2*Sin[θ])/Sqrt[Δ]*κ4[r, θ] + (6 a^2*ℜ1*Cos[θ]*Sin[θ])/Sqrt[Δ]*κ5[r, θ])
frt[r_, θ_] := (4 M)/(Sqrt[Δ]*Σ^3) (-3 a^2*ℜ1*Cos[θ]*Sin[θ]*κ3[r, θ] + 6 a*ℜ1*Cos[θ]*Sqrt[Δ]*κ4[r, θ])

And other functions

Δ := r^2 - 2 M*r + a^2
Σ := r^2 + a^2*Cos[θ]^2
ℜ1 = r^2 - 1/3 a^2 Cos[θ]^2;
ℜ2 = r^2 - 3 a^2 Cos[θ]^2;

κ3[r_, θ_] := 2 ϕ1b[r, θ] Sqrt[-ϕv[r, θ]^2 + (Sqrt[(a^2 + r (-2 M + r)) (r^2 + a^2 Cos[θ]^2)] Subscript[c,ψ])/(2 (a^2 + r^2))]
κ4[r_, θ_] := (Sqrt[-ϕ1b[r, θ]^2 + ϕv[r, θ]^2] (4 (a^2 + r^2) ϕv[r, θ]^2 -Sqrt[(a^2 + r (-2 M + r)) (r^2 + a^2 Cos[θ]^2)] Subscript[c,ψ]))/(2 (a^2 + r^2) ϕv[r, θ])
κ5[r_, θ_] := -Sqrt[(((ϕ1b[r, θ] - ϕv[r, θ]) (ϕ1b[r, θ] + ϕv[r, θ]) (4 (a^2 + r^2) ϕv[r, θ]^2 - 2 Sqrt[(a^2 + r (-2 M + r)) (r^2 + a^2 Cos[θ]^2)] Subscript[c, ψ]))/(a^2 + r^2))]
κ6[r_, θ_] := 2ϕ1b[r, θ]* ϕv[r, θ] - (ϕ1b[r,θ] Sqrt[(a^2 - 2 M r + r^2) (r^2 + a^2 Cos[θ]^2)]Subscript[c, ψ])/(2 a^2* ϕv[r, θ] + 2 r^2* ϕv[r, θ])

I tried separating the variables, but without success. Could someone help me?

I need ϕ1b[r, θ] and ϕv[r, θ]. Consider: $a=1$, $M=1$, $c_ψ=1$

Anybody?

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  • 3
    $\begingroup$ And you need an analytical solution? Do you know that such a solution exists for equations of this kind? Would you be able to work with numerical solutions instead? $\endgroup$ – MarcoB May 30 '18 at 16:35
  • $\begingroup$ I need analytical solutions, based on "r" and "θ"... I only know that there is a solution of this type, only that I find myself with difficulty to solve. $\endgroup$ – will.al May 30 '18 at 16:41
  • $\begingroup$ You cannot have both ℜ2[r, θ] and ℜ2 in the system of equations. $\endgroup$ – bbgodfrey May 31 '18 at 1:19

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