= Solution
Set $r=\phi_0$ and $A_k=a+ck^2+k^4$. The trial density is $A_kr^2+r^4/2$. At fixed $k$, its minimum is $r=0$ if $A_k\ge0$ and $r^2=-A_k$ otherwise, with density $-A_k^2/2$. Minimizing $A_k$ uses
$$
\frac{dA_k}{dk}=2k(c+2k^2)=0.
$$
For $c\ge0$, $k=0$ minimizes the quadratic coefficient. For $c<0$, the minima are at $k=\pm\sqrt{-c/2}$ and have $A_k=a-c^2/4$. Thus
$$
\boxed{
\begin{array}{c|c|c}
\text{ordered region}&k&\phi_0^2\\\hline
c\ge0,\ a<0&0&-a\\
c<0,\ a<c^2/4&\pm\sqrt{-c/2}&c^2/4-a
\end{array}}
$$
Everywhere else the minimum is $\phi=0$; at the transition boundary the ordered amplitude tends to zero. The zero-field minimum is respectively $-a^2/2$ or $-(a-c^2/4)^2/2$ in the two ordered regions. For $c<0$ the helical solution is lower than a uniform ordered extremum whenever the latter exists, and lower than the disordered state whenever its amplitude is nonzero. The assumed positive-amplitude ansatz must allow its zero-amplitude boundary to describe disorder.
\b[The phase $\delta$ remains arbitrary] because the functional has <global U(1) symmetry>. At nonzero $k$, a <translation> shifts that phase; separate <translation> and phase operations retain a combined unbroken subgroup. The two signs of $k$ also select opposite handedness. The meeting point of the uniform, modulated and disordered mean-field regions is the <Lifshitz point> $(a,c)=(0,0)$.
\Image[/past-exam-of-the-theoretical-physics-i-course-of-the-university-of-cambridge/2014/paper-1-tp1-landau-phases.png]
{title=Uniform, helical and disordered minima of the specified complex <Landau free energy>}
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