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First derive the formal turbulent plane jet similarity equations, then check the printed edge conditions. Define a streamfunction by $U={\psi }_y$, $V=-{\psi }_x$, so mean incompressibility is automatic. Momentum conservation gives $U_c^2\delta \sim M_0$. With $l=C_1|y|$, the closure balance $U_c^2/x\sim C_1^2{U}_c^2/\delta$ gives $\delta \propto x$. Hence $U_c\propto M_0^{1/2}{x}^{-1/2}$ and the consistent self-similar ansatz isAn odd $f$ gives symmetric $U$ and antisymmetric $V$. Direct differentiation givesThe advection and Reynolds stress terms becomeThus the two required consistency equations, the first for $\eta >0$ and the second for a symmetric finite-width profile, areThe second is momentum normalization; it is the equation referenced as (2) in the PDF, although its display in part (d) has no printed number. Use an infinite upper limit for a decaying profile of infinite support. Symmetry sets $f(0)=0$. A finite centreline velocity and continuous zero centreline stress give the integrated first equationThese equations establish the similarity reduction. They also give a direct finite-edge stress condition for a mixing-length jet. At a finite edge ${\eta }_w>0$, $U=0$ requires $f^{\prime }({\eta }_w)=0$. For finite $f({\eta }_w)$, the integrated equation forces $f^{\prime \prime }({\eta }_w)=0$. A nontrivial decreasing velocity profile has nonzero shear somewhere inside, so zero edge shear cannot be its maximum. Allowing a one-sided nonzero edge shear instead produces nonzero edge stress and an unbalanced jump in the conservative momentum equation.

For a concrete diagnostic, take $f^{\prime }=1-{\eta }^2$ on $0\le \eta \le 1$, so $f=\eta -{\eta }^3/3$. It has zero edge velocity and its largest one-sided shear magnitude at the edge, matching that descriptive condition. Yet at $\eta =1$ the integrated equation has left-hand side zero and right-hand side $8C_1^2>0$: such a profile is not a solution of the stated momentum closure.

The centreline is also problematic for a smooth interpretation. If $f^{\prime }(0)=a>0$, then $f\sim a\eta$, and the decreasing branch of the integrated equation givesA continuous formal profile can have this cusp, but it is not a differentiable symmetric velocity field, and molecular diffusion cannot be neglected uniformly in that core. Therefore there is no nontrivial regular similarity solution satisfying all the printed edge requirements. The boxed consistency equations are valid; the additional coincidence of a zero-velocity edge with maximum shear is not. A consistent jet model must relax that edge definition and regularize the core or use a different mixing length.

For the homologous spherical flow $\mathbf{u}=q\mathbf{r}$, the ideal magnetohydrodynamic induction equation and $\nabla \cdot \mathbf{B}=0$ give the material derivativeLet $x=r/R(t)$ and use the self-similar ansatzPart (b) gives $q=3\dot{R}/(4R)$, and thereforeThe radial induction equation becomes $x{f}^{\prime }=6f$. The boundary value $f(1)=1$ gives $f=x^6$. The solenoidal vector field condition requireswhich also matches the tangential shock value in part (c). Hence the interior field is

Outside the shock, the uniform-field lines obey $r\sin \theta =\text{constant}$. Inside, the magnetic-field-line equation givessoA sketch therefore shows straight exterior lines refracting at the spherical shock into north-south symmetric curves that bow toward the equatorial interior before leaving through the opposite hemisphere. The field strength falls as $r^6$ toward the centre.