= Solution
Let the thin transverse length be $Re^{-1/2}$ times the longitudinal scale and write $Y=Re^{1/2}y$, $V=Re^{1/2}v$. Continuity then remains $u_x+V_Y=0$. Transverse momentum gives $p_Y=0$ to leading order; the quiescent exterior has constant <pressure>, hence $p_x=0$. Axial viscous diffusion in $x$ is smaller by $Re^{-1}$, leaving
$$
\boxed{uu_x+Vu_Y=u_{YY},\qquad u_x+V_Y=0.}
$$
<Symmetry> about the jet centreline requires $V=0$ and $u_Y=0$ at $Y=0$. At large positive $Y$, require $u\to0$ and $u_Y\to0$. The transverse <velocity> can approach a finite nonzero scaled <entrainment> <velocity>; imposing $V\to0$ as well would rule out the jet. The physical transverse <velocity> remains smaller by $Re^{-1/2}$ and matches an entraining outer flow.
Using continuity, the momentum equation is $\partial_x(u^2)+\partial_Y(uV-u_Y)=0$. Integrating across the jet, the boundary terms vanish because $u,u_Y\to0$ while $V$ is finite. Thus
$$
\boxed{M=\int_{-\infty}^\infty u^2\,dY=\text{constant}.}
$$
Apart from the dimensional <density> and <velocity>/length factors, this is the conserved axial <momentum flux> supplied by the nozzle. <Entrainment> increases <volume flux> without adding axial momentum from the resting exterior.
For $u=x^\alpha f'(\eta)$ with $\eta=Y/x^\beta$, use the <stream function> $\psi=x^{\alpha+\beta}f(\eta)$ so that $u=\psi_Y$ and $V=-\psi_x$. Momentum conservation requires $2\alpha+\beta=0$. Inertia and diffusion scale as $x^{2\alpha-1}$ and $x^{\alpha-2\beta}$, so $\alpha+2\beta=1$. Therefore $\boxed{\alpha=-1/3,\quad\beta=2/3}$. In particular
$$
V=x^{-2/3}(2\eta f'-f)/3.
$$
Substitution reduces momentum to $3f^{(3)}+ff''+(f')^2=0$, with $f(0)=f''(0)=0$ and $f'(\infty)=0$. Integrating first gives $3f''+ff'=0$, the integration constant vanishing at the centreline. A second integration gives $3f'+f^2/2=C$. Parameterize its positive constant as $C=18A^2$ and solve the resulting <Riccati equation> with $f(0)=0$:
$$
\boxed{f=6A\tanh(A\eta),\qquad u=6A^2x^{-1/3}\operatorname{sech}^2(AYx^{-2/3}).}
$$
The momentum <integral> fixes its scale, since $\int_{-\infty}^\infty\operatorname{sech}^4z\,dz=4/3$:
$$
\boxed{M=48A^3,\qquad A=(M/48)^{1/3}.}
$$
The upper-half <entrainment> is $V\to-2Ax^{-2/3}$; the negative side has the opposite sign, drawing ambient fluid inward. This is the <Bickley jet>, valid in the slender far-downstream laminar regime, rather than in the immediate nozzle region.
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