eqs = {D[Ar[x, t], t] + 1/4*(Ar[x, t]^3*Ai[x, t]^2*Ar[x, t]) - D[Ar[x, t], {x, 2}] - 2*Ar[x, t] == 0, D[Ai[x, t], t] + 1/4*(Ar[x, t]^2*Ai[x, t] + Ai[x, t]^3) - D[Ai[x, t], {x, 2}] - 2*Ai[x, t] == 0}; 

But I don't know how to make an anti-periodic initial condition to be consistent with the boundary condition.

eqs = {D[Ar[x, t], t] + 1/4*(Ar[x, t]^3+Ai[x, t]^2*Ar[x, t]) - D[Ar[x, t], {x, 2}] - 2*Ar[x, t] == 0, D[Ai[x, t], t] + 1/4*(Ai[x, t]^3+Ar[x, t]^2*Ai[x, t]) - D[Ai[x, t], {x, 2}] - 2*Ai[x, t] == 0}; 

But I don't know how to make an anti-periodic initial condition (Ai[x, 0] = inianti[x]) to be consistent with the boundary condition.

Response to @user64494's comment:

Yes, I can split the real and imaginary parts by writing the 2nd term as

$(A^\ast A)A=\vert A\vert^2A=(A_r^2+A_i^2)(A_r+i A_i)=A_r^3+A_i^2A_r+i(A_r^2A_i+A_i^3)$

Then the equation can be split into

eqs = {D[Ar[x, t], t] + 1/4*(Ar[x, t]^3*Ai[x, t]^2*Ar[x, t]) - D[Ar[x, t], {x, 2}] - 2*Ar[x, t] == 0, D[Ai[x, t], t] + 1/4*(Ar[x, t]^2*Ai[x, t] + Ai[x, t]^3) - D[Ai[x, t], {x, 2}] - 2*Ai[x, t] == 0}; 

But I don't know how to make an anti-periodic initial condition to be consistent with the boundary condition.

ibcs = {Ar[-L, t] == Ar[L, t], Ai[-L, t] == -Ai[L, t], Ar[x, 0] == ini[x], Ai[x, 0] = inianti[x]};