Another twist to the same problem: for the PDE
v2 = 0.0005 t; L = 6; sol =
NDSolve[{D[f[t, x], t] + I v[t, x] f[t, x] ==
D[f[t, x], {x, 2}] + f[t, x] (1 - f[t, x] Conjugate[f[t, x]]),
D[v[t, x], t] ==
D[v[t, x], {x, 2}] -
0.5 I (f[t, x] Conjugate[D[f[t, x], {x, 2}]] -
D[f[t, x], {x, 2}] Conjugate[f[t, x]]), f[0, x] == 1,
f[t, -L] == E^(-0.5 v2 I t), f[t, L] == E^(0.5 v2 I t),
v[t, -L] == v2/2, v[t, L] == -v2/2, v[0, x] == 0}, {f, v}, {x, -L,
L}, {t, 0, 5}][[1]]
Mathematica claims that there are non-numeric derivatives. However, if I rescale x, so that L becomes a power of 2, e.g., xp=(4/3)x,
v2 = 0.0005 t; L = 8; sol =
NDSolve[{D[f[t, xp], t] +
I v[t, xp] f[t, xp] == (16/9) D[f[t, xp], {xp, 2}] +
f[t, xp] (1 - f[t, xp] Conjugate[f[t, xp]]),
D[v[t, xp], t] == (16/9) D[v[t, xp], {xp, 2}] -
0.5 I (f[t,
xp] Conjugate[(16/9) D[f[t, xp], {xp, 2}]] - (16/9) D[
f[t, xp], {xp, 2}] Conjugate[f[t, xp]]), f[0, xp] == 1,
f[t, -L] == E^(-0.5 v2 I t), f[t, L] == E^(0.5 v2 I t),
v[t, -L] == v2/2, v[t, L] == -v2/2, v[0, xp] == 0}, {f,
v}, {xp, -L, L}, {t, 0, 5}][[1]]
then Mathematica runs.