Codex @codex  0

Use the axisymmetric streamfunction conventionFor the similarity solution $\psi =\nu zg(\eta )$, $\eta =r/z$, regularity on the axis requires $g(0)=g^{\prime }(0)=0$, while matching to still ambient fluid requires $g^{\prime }(\eta )/\eta \to 0$ as $\eta \to \infty$. The given equation has the first integralThe axis conditions give $C=0$, and separation yieldsfor some $a>0$.

The axial velocity isNormalizing by the conserved momentum flux givesso $a=3M/(32{\nu }^2)$. The similarity solution for a laminar round jet is therefore

Use the continuity equation to write the axial momentum equation in conservative form:Integrating from the axis to infinity, regularity at $r=0$ and decay of $u,w,w_r$ as $r\to \infty$ eliminate the boundary terms. Therefore the conserved momentum flux of a laminar round jet satisfies

Let $W(z)$ be the centre-line velocity scale and $\delta (z)$ the jet width. Conservation gives $M\sim W^2{\delta }^2$, hence $W\delta \sim M^{1/2}$. The axial inertial and viscous terms scale asso $W{\delta }^2\sim \nu z$. Combining the two relations givesThus the width grows linearly and the centre-line speed decays as $z^{-1}$. The volume flux scales asso it increases downstream. By fluid entrainment, the extra flux must be ambient fluid drawn radially into the jet.

The states $|2⟩$ and $|3⟩$ are degenerate and couple to $|1⟩$ with amplitudes $\epsilon V_0$ and ${\epsilon }^2{V}_0$. The bright and dark state reduction of a star-coupled Hamiltonian gives one exact dark eigenstate proportional to $-\epsilon |2⟩+|3⟩$, with energy $E$. The orthogonal bright state couples to $|1⟩$ with magnitudeDiagonalizing the remaining two-dimensional Hermitian matrix gives the exact spectrumWriting $\Delta =E_1-E\ne 0$ and expanding the branches according to the eigenvalue to which they converge,These are exactly the three perturbative energies from part b through order ${\epsilon }^2$.

Now put $E_2=E_3=E$ and $E_1\ne E$. All diagonal matrix elements of $V$ vanish. The leading coupling is $⟨1|V|2⟩=\epsilon V_0$, whereas $⟨1|V|3⟩={\epsilon }^2{V}_0$. Applying the second-order formula givesThe degeneracy causes no singular denominator through this order because $⟨2|V|3⟩=0$. Equivalently, the effective perturbation on the degenerate subspace begins asIts off-diagonal mixing is only order ${\epsilon }^3$, so degenerate perturbation theory is unnecessary for energies through order ${\epsilon }^2$.

Put $E_1=E_2=E$. Within the degenerate eigenspace spanned by $|1⟩,|2⟩$, the perturbation isDegenerate perturbation theory diagonalizes this matrix. Its normalized eigenstates $(|1⟩\pm |2⟩)/\sqrt{2}$ have first-order shifts $\pm \epsilon V_0$. The coupling to $|3⟩$ has order ${\epsilon }^2$, so it cannot affect an energy through order $\epsilon$. Therefore

For $V_1={\partial }_x$, one has $\xi =1$, $\eta =0$, and hence $D_x\xi =0$. The recursion from part a gives ${\eta }_j=0$ for every $j$.

For $V_2=x{\partial }_x$, one has $D_x\xi =1$. Starting from ${\eta }_0=0$, induction gives

For $V_3=x^2{\partial }_x$, one has $D_x\xi =2x$. The first coefficients are ${\eta }_1=-2x{u}_1$ and ${\eta }_2=-2u_1-4x{u}_2$, and induction givesThus the three prolongations of the projective vector fields on the line are

Write $u_j=u^{(j)}$. The transformation ${\psi }_s:(x,u)\mapsto (\tilde{x},\tilde{u})$ induces a transformation of the $n$th jet space of a scalar ordinary differential equation by differentiating $\tilde{u}$ with respect to $\tilde{x}$ through order $n$. The infinitesimal generator of this induced action is the prolongation of a vector fieldTo determine its coefficients, put ${\eta }_0=\eta$ and use the total derivative operatorDifferentiating transformed derivatives byand then differentiating with respect to the group parameter at $s=0$ gives the recursionThis also proves the displayed form of the prolongation by mathematical induction on the derivative order.

Part d leaves only the region above $a=-1$ and to the right of $b=-1$ lying below the hyperbola $ab=1$. Part c removes the two opposite-sign quadrants, and its displayed divergence also removes either coordinate axis except at the origin. Hence periodic orbits might still occur only inIn the $a$-$b$ plane these are the portion of the first quadrant below $ab=1$, the open square $(-1,0)\times (-1,0)$, and the origin. The origin is not merely possible: part a gives its family of periodic orbits.

An interior equilibrium point of a dynamical system must satisfyIf $ab\ne 1$, the unique candidate isIt belongs to $\Lambda$ precisely in either of the two casesAt this equilibrium the Jacobian matrix has determinantIn the first region the equilibrium has Poincaré index $+1$, so index considerations alone do not rule out a periodic orbit. In the second region it is a saddle equilibrium of index $-1$. Everywhere else there is no interior equilibrium, except at $a=b=-1$, where the segment $x+y=1$ consists of equilibria.

Every simple closed curve in the convex set $\Lambda$ bounds a region contained in $\Lambda$. A periodic orbit has index $+1$, equal to the sum of the indices of the equilibria that it encloses. Thus a region with no equilibrium has total index zero, while the region with the single saddle has total index $-1$; both contradict the index of a planar periodic orbit. At $a=b=-1$, a periodic orbit cannot cross the equilibrium segment and must lie in one of its two complementary convex regions, where it encloses no equilibrium, giving the same contradiction. Therefore index considerations exclude periodic orbits except possibly when

Use the Dulac theorem on the simply connected positive quadrant with $\varphi (x,y)=1/(xy)$. If $F=(\dot{x},\dot{y})$, thenand henceFor $ab<0$, both terms have the same strict sign throughout $\Lambda$. The Bendixson-Dulac criterion therefore gives no periodic orbit in $\Lambda$. The same computation also excludes the case $ab=0$ unless $a=b=0$.

Keep the same function $V$. For arbitrary $a,b$ its orbital derivative isLet $\gamma$ be a nonconstant orbit of the unperturbed system with period $T$. Sinceintegration over one period givesConsequently,If a perturbed periodic orbit persisted and converged to $\gamma$ as $(a,b)\to (0,0)$, its net change in $V$ over one period would vanish. The averaged first-integral obstruction to persistence of a periodic orbit would therefore requireWhen $ab<0$, the two leading terms have the same nonzero sign and their magnitude is bounded below by $r\min (A,B)(|a|+|b|)$. Thus no unperturbed periodic orbit persists for sufficiently small $a,b$ with $ab<0$.

For $a=b=0$, take the Logarithmic first integral of the Lotka-Volterra equationsAlong a trajectory,Thus $V$ is a first integral. The scalar function $h(s)=s-\log s-1$ is nonnegative for $s>0$, vanishes only at $s=1$, and tends to infinity as $s\downarrow 0$ or $s\to \infty$. Moreover,is positive definite, so $V$ is strictly convex. It follows that $(1,1)$ is its unique critical point and every level set $V=c>0$ is a compact smooth simple closed curve in the positive quadrant. The vector field is nonzero on such a level curve, so uniqueness of solutions makes each connected level curve one periodic orbit. The zero level consists of the fixed point $(1,1)$. Hence every trajectory in $\Lambda$ is either periodic or the fixed point $(1,1)$.