= Solution
<Parallel transport> the auxiliary <null vector> $N$ and a screen basis along each generator of the <null geodesic congruence>. This keeps the <screen-space projector> fixed under the corresponding transported <derivative>. Use the curvature convention $[\nabla_c,\nabla_b]V_a=-R^d{}_{acb}V_d$.
With $B_{ab}=\nabla_bU_a$, differentiation of the affine <geodesic equation> gives
$$
U^c\nabla_cB_{ab}
=-(\nabla_bU^c)(\nabla_cU_a)-R^d{}_{acb}U^cU_d.
$$
Here the first term comes from commuting the <derivative> past $U^c$, and the second from the curvature commutator. The constraints $B_{ab}U^a=B_{ab}U^b=0$ follow respectively from the null normalization and affine <geodesic equation>. Projecting onto the screen therefore yields the optical <matrix> equation
$$
\frac{D\widehat B_{IJ}}{d\lambda}
=-\widehat B_{IK}\widehat B_{KJ}-\mathcal R_{IJ},\qquad
\mathcal R_{IJ}=R_{acbd}e_I^aU^ce_J^bU^d.
$$
The trace of the tidal <matrix> is $R_{ab}U^aU^b$; extra terms from replacing the screen trace by the <spacetime> trace vanish by curvature antisymmetry. Taking the trace gives
$$
\frac{d\theta}{d\lambda}=-\operatorname{tr}(\widehat B^2)-R_{ab}U^aU^b.
$$
Decompose the <optical tensor> as $\widehat B=\tfrac12\theta I+\widehat\sigma+\widehat\omega$. The <null shear> is symmetric and trace-free and the <null twist> is antisymmetric. Their cross traces vanish, and
$$
\operatorname{tr}(\widehat B^2)
=\frac12\theta^2+\widehat\sigma^{ab}\widehat\sigma_{ab}
-\widehat\omega^{ab}\widehat\omega_{ab}.
$$
The negative sign in the last term is the identity $\operatorname{tr}(\widehat\omega^2)=-\sum_{I,J}\widehat\omega_{IJ}^2$ on the positive-definite screen. Substitution proves the four-dimensional <Null Raychaudhuri equation>:
$$
\boxed{\frac{d\theta}{d\lambda}
=-\frac12\theta^2-\widehat\sigma^{ab}\widehat\sigma_{ab}
+\widehat\omega^{ab}\widehat\omega_{ab}-R_{ab}U^aU^b.}
$$
The coefficient is $1/(D-2)$ in $D$ <spacetime> dimensions; the factor $1/2$ here belongs to the four-dimensional congruence in this question, not the five-dimensional geometry of the next question.
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