= Solution
{c}
With $u=-t+z$ and $v=t+z$, the <Minkowski metric> and the specified <null vector> become
$$
\eta=du\,dv+dx^2+dy^2,\qquad
\eta_{uv}=\eta_{vu}=\frac12,\qquad
n=2\partial_v,\qquad n^\flat=du.
$$
The <Minkowski wave operator in null coordinates> is $\Box=4\partial_u\partial_v+\partial_x^2+\partial_y^2$. All components of the original <metric perturbation> still depend only on $u$, since this coordinate change is linear. The result of part (a), differentiated once, reads
$$
2\tilde h''_{v\nu}=\frac12H''\delta_\nu{}^u.
$$
It follows that $\tilde h''_{vv}=\tilde h''_{vx}=\tilde h''_{vy}=0$ and $\tilde h''_{uv}=H''/4$. Taking the <trace> gives
$$
H''=4\tilde h''_{uv}+\tilde h''_{xx}+\tilde h''_{yy},
\qquad \tilde h''_{xx}+\tilde h''_{yy}=0.
$$
Expanding the scalar $\phi=\tilde h''_{AB}x^Ax^B/2$ in these <null coordinates> therefore yields
$$
\boxed{\phi=a(u)(x^2-y^2)+b(u)xy+c(u)v+d(u)x+e(u)y+f(u),}
$$
where the coefficients are explicitly
$$
\boxed{a=\frac12\tilde h''_{xx},\quad b=\tilde h''_{xy},\quad
c=\frac14uH'',\quad d=u\tilde h''_{ux},\quad
e=u\tilde h''_{uy},\quad f=\frac12u^2\tilde h''_{uu}.}
$$
The quadratic transverse term is a <harmonic quadratic plane-wave profile>. In particular $\partial_v\phi=c$, $\partial^\mu\overline{\hat h}_{\mu\nu}=2c\,n_\nu$ and $\Box\phi=4c'$, consistently reproducing the failure of wave gauge in part (b).
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