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Set $P=10x^4-2x{y}^3$ and $Q=-3x^2{y}^2$. Their partial derivatives satisfy $P_y=-6x{y}^2=Q_x$ throughout ${\mathbb{R}}^2$. The vector field is smooth on the entire simply connected plane, so zero curl proves path independence.

Choose the broken straight path $(0,0)\to (2,0)\to (2,1)$. Along its horizontal segment $y=0$ and $dy=0$, givingAlong its vertical segment $x=2$ and $dx=0$, givingTherefore the line integral is

The decreasing production term represents negative feedback: low red-cell abundance stimulates production, whereas high abundance suppresses it. The linear loss term models removal at constant per-cell rate. SetThen $u^{\prime }=f(u)-\delta u$, with $f(u)=1/(1+u^n)$. On the physical half-line $u\ge 0$, $f$ decreases from one to zero while $\delta u$ increases from zero to infinity. Their unique intersection $u_{\ast }>0$ is the steady state. Its linear derivative is $f^{\prime }(u_{\ast })-\delta <0$, so it is asymptotically stable.

For small $\delta$, the intersection lies at large $u$, where $f\sim u^{-n}$; for large $\delta$ it lies at small $u$, where $f\sim 1$. HenceImplicit differentiation of $f(u_{\ast })=\delta u_{\ast }$ gives $d{u}_{\ast }/d\delta =u_{\ast }/[f^{\prime }(u_{\ast })-\delta ]<0$. The plot joins these two decreasing asymptotic regimes.

The delay $t_0$ represents the time between feedback sensing and the appearance of newly produced mature cells. In dimensionless time it is $d=F_0{t}_0/\theta$, so the equation becomesIt requires a nonnegative initial history on $[-d,0]$, not just one initial value. Its constant equilibrium is unchanged. Put $a=-f^{\prime }(u_{\ast })=n{u}_{\ast }^{n-1}/(1+u_{\ast }^n{)}^2>0$. A small perturbation obeys $v^{\prime }=-\delta v-av(\tau -d)$, whose characteristic equation isThere is no real root with $\sigma \ge 0$, because all three terms would have positive sum; zero is never a root either. Starting from the stable zero-delay case, loss of linear stability can therefore occur only when a complex conjugate pair crosses the imaginary axis at a nonzero frequency, the linear signature of a Hopf bifurcation.

For completeness, substituting $\sigma =i\omega$ gives $\delta +a\cos (\omega d)=0$ and $\omega =a\sin (\omega d)$. Such a crossing requires $a>\delta$, withThe first crossing is transversal: $\mathrm{Re}(d\sigma /dd)={\omega }^2/[(1+\delta d{)}^2+{\omega }^2{d}^2]>0$. If $a\le \delta$, the magnitude relation rules out roots in the closed right half-plane for every delay. A nondegenerate nonlinear Hopf conclusion additionally uses the usual bifurcation hypotheses. This is delayed negative feedback in red-cell production.

Only the upper equilibrium can have zero trace with positive determinant. Setting $s_{+}=\lambda$ givesAt this point the determinant is ${\lambda }^2(2\lambda -1)>0$, so the angular frequency is $\lambda \sqrt{2\lambda -1}$. Moreover $d{s}_{+}/d\mu =1/\sqrt{1+4\mu }>0$, so the trace crosses zero transversely as $\mu$ increases. The stipulated supercritical Hopf bifurcation creates a stable periodic orbit on the side $\mu >{\mu }_H$ where the upper equilibrium becomes unstable.

The steady-state curves are the vertical lines $\mu =-1/4$ and $\mu =0$ in the $(\mu ,\lambda )$ plane. The Hopf parabola has its endpoint atThere the Jacobian has trace and determinant zero but is not the zero matrix: it has a nilpotent Jordan block. This is the codimension-two Takens-Bogdanov point. The intersection of the Hopf curve with $\mu =0$ at $\lambda =1$ concerns a different equilibrium from the transcritical crossing at the origin; it is not another double-zero equilibrium.

An efficient subdivision surface interrogation interface needs the mesh topology and patch adjacency, boundary and trimming information, patch-local parameter charts and their transition maps, and local restriction/refinement with the complete influencing control net. Numerical enquiries should supply limit positions $S$, first derivatives $S_u,S_v$, and second derivatives where defined; these determine normal vectors and curvatures at regular points. It must also return certified patch bounding volumes and convergent geometric/derivative error bounds. Near an extraordinary subdivision vertex, use the local subdivision matrix and limit/tangent evaluation appropriate to that scheme; second derivatives need not exist merely because a tangent plane does. Finite refined mesh vertices are not exact limit positions.

Represent the cutting plane as $n\cdot x=d$, with $n$ normalized. On each chart setThe desired plane section of a subdivision surface is the image of the zero set $g=0$. For a bounding volume $B$, compute a certified interval $[g_{\min },g_{\max }]$ for the plane functional over $B$. If that interval is strictly positive or strictly negative, reject the patch. With positive partition-of-unity weights, the minimum and maximum of $n\cdot p_i-d$ over all influencing control points already bound its limit patch. For masks without this convex hull property, the interface must provide a different valid enclosure.

An adaptive algorithm that isolates components before tracing is:

This is exhaustive to a chosen tolerance when certified subdivision isolates finitely many regular transverse branches and enclosures converge. If $\Vert \nabla g\Vert \ge \gamma >0$ locally and $\Vert DS\Vert \le M$, a residual tolerance $|g|\le \delta$ corresponds locally to a geometric correction bounded by $M\delta /\gamma$, using the gradient flow toward the level set. Bounds on second derivatives similarly control curve-segment chord error bound. As $\gamma$ approaches zero, a fixed residual tolerance no longer gives a fixed geometric error, so the step and stopping rules must reflect conditioning.

Seeding must not rely just on corner signs. For example $S(u,v)=(u,v,u^2+v^2-r^2)$ cut by $z=0$ has a closed circle wholly inside a patch whose corner values are all positive. Certified range tests plus subdivision retain it; a corner-only marching rule can miss it. Tangencies also change the nature of the output: $S=(u,v,u^2+v^2)$ meets $z=0$ in one isolated point, while a coplanar patch contributes an area rather than a curve. Thus a general plane section is not necessarily a union of regular curves; the algorithm should classify these cases instead of silently fabricating or deleting a branch.

For efficiency, cache subdivision stencils, patch extraction, limit-evaluation coefficients and bounds; reuse child data instead of reconstructing the entire refined mesh. Use local influencing neighborhoods, with the halo dictated by mask support, and keep adjacency/boundary topology separate from changing geometric coordinates. Plane-functional bounds can be computed by refining scalar signed-distance controls, which is cheaper than refining all three coordinates. Combine coarse inexpensive boxes with tighter local convex hull or interval arithmetic tests only where needed. Adapt steps to curvature and transversality, reuse branch seeds, and use regular spline evaluation away from extraordinary points. Stable edge ownership, outward-rounded bounds, scale-aware tolerances, and explicit degeneracy flags are correctness requirements as well as performance issues. The result is a limit-surface intersection computation; merely intersecting a finitely refined control polyhedron with the plane supplies only an uncontrolled approximation.

The vector $\mathbf{r}-(\mathbf{r}\cdot \mathbf{u})\mathbf{u}$ is the perpendicular component of $\mathbf{r}$ relative to the axis $\mathbb{R}\mathbf{u}$, so its length is the distance from that axis. Hence the surface is an infinite circular cylinder of radius $\beta$ about the line through the origin parallel to $\mathbf{u}$. Here $\beta$ is the radius and $\mathbf{u}$ fixes the axis direction; reversing $\mathbf{u}$ leaves the same circular cylinder.

If ${\mathbf{v}}_1,{\mathbf{v}}_2$ are an orthonormal basis of the plane perpendicular to $\mathbf{u}$, a parameterization isThis verifies that all heights and all angular positions, but no interior points, belong to the surface.

Set ${\mathbf{b}}_0=\mathbf{c}+h\mathbf{n}$, where $h=p-\mathbf{n}\cdot \mathbf{c}$. Then $\mathbf{n}\cdot {\mathbf{b}}_0=p$, so ${\mathbf{b}}_0$ lies in the plane. For any position $\mathbf{r}$ in that plane, $\mathbf{r}-{\mathbf{b}}_0$ is perpendicular to ${\mathbf{b}}_0-\mathbf{c}=h\mathbf{n}$. The Pythagorean theorem therefore givesOn the sphere the left side is $a^2$. Thus the intersection is the circle in the plane centred at ${\mathbf{b}}_0$, with squared radius $a^2-h^2$. Its centre must therefore be $\mathbf{b}={\mathbf{b}}_0$, andIn terms of the radii, $\lambda =\pm \sqrt{a^2-{\rho }^2}$, with the sign fixed by $\lambda =p-\mathbf{n}\cdot \mathbf{c}$. In the central-plane case $\rho =a$, both signs give zero. This is the center and radius of a sphere-plane intersection formula with its orientation made explicit.

Substitute the signed displacement $\lambda =p-\mathbf{n}\cdot \mathbf{c}$ into the Pythagorean theorem relation from part (a):A genuine circle requires $|p-\mathbf{n}\cdot \mathbf{c}|<a$. Equality gives a tangent plane and a single point, interpreted as radius zero; a larger distance gives no intersection. Under the stipulated circle intersection, the radicand is positive. Its centre is $\mathbf{c}+(p-\mathbf{n}\cdot \mathbf{c})\mathbf{n}$.

The sphere-plane intersection is governed by the orthogonal projection of the sphere's centre onto the plane. Since $\mathbf{n}$ has unit length, let $h=p-\mathbf{n}\cdot \mathbf{c}$, the signed distance from $\mathbf{c}$ to the plane along $\mathbf{n}$. The following parts determine the centre and radius and retain this sign; the radii alone cannot distinguish the two possible sides of the oriented plane.

Let $P\mathbf{x}=(\mathbf{n}\cdot \mathbf{x})\mathbf{n}$ be the axial orthogonal projection and $N\mathbf{x}=\mathbf{n}\times \mathbf{x}$. Then $NP=PN=0$, while the vector triple-product identity gives $N^2=P-I$. The map is $aP+a(I-P)+bN$. Its inverse is $P/a$ on the axis. On the perpendicular plane,Thus the inverse of an axis-angle linear map isor, in the requested vector form,The denominators are nonzero because $a,b>0$. Multiplication on each invariant component verifies the matrix inverse without choosing a coordinate frame.

Write $\mathbf{x}=s\mathbf{n}+\mathbf{v}$, with $\mathbf{v}\cdot \mathbf{n}=0$. The linear map sends $s\mathbf{n}$ to $as\mathbf{n}$ and sends $\mathbf{v}$ to $a\mathbf{v}+b(\mathbf{n}\times \mathbf{v})$. On the perpendicular plane, $\mathbf{v}$ and $\mathbf{n}\times \mathbf{v}$ are perpendicular vectors of equal length. ThusThe expression in parentheses is a rotation in three dimensions by $\vartheta$ about the oriented $\mathbf{n}$ axis, with the positive direction fixed by the cross product. Therefore the map scales the axial component by $a$, scales the perpendicular component by $\sqrt{a^2+b^2}$, and rotates the perpendicular plane by $\arctan (b/a)$. Since $a,b>0$, the angle lies between zero and $\pi /2$. This is an axis-angle decomposition, with different axial and transverse scale factors.

The principal argument is continuous and single-valued on $\mathbb{C}\setminus (-\infty ,0]$. The given loop stays in this slit plane, and its initial and final values have the same principal argument. The definition therefore gives $\boxed{w(\gamma ;0)=0}$.

At $p\in D$, each of the given ambient rotations fixes $p$, rotates its tangent plane and preserves the local surface. Naturality of the shape operator under that rotation makes $S_p$ commute with every planar rotation. A symmetric real two-by-two operator commuting with all rotations is scalar, so every point of $D$ is an umbilical point. The trace-free part of the shape operator is continuous. Density therefore makes it zero everywhere on $S$.

Locally write $S=\kappa I$. The Codazzi equation gives $(X\kappa )Y=(Y\kappa )X$ for tangent fields $X,Y$; choosing independent fields proves $d\kappa =0$. Thus the surface is locally a plane if $\kappa =0$, or has constant spherical centersince its differential is $I-S/\kappa =0$. A change of normal reverses both $N$ and $\kappa$, leaving this center unchanged, so these local descriptions patch across the connected surface. In the flat case the whole surface lies in one plane, where a regular surface without boundary is open; a nonempty such set cannot also be compact and closed. Hence the flat case is excluded. In the other case $S$ lies on one sphere of radius $1/|\kappa |$. Its inclusion in that sphere is locally open, since both have dimension two, and its compactness makes the image closed. Connectedness of the sphere forces the image to be the entire sphere. Therefore $S$ is a sphere.

The proposed stream function gives $u=Cx{H}^{-2}{\varphi }^{\prime }$ and $v=-C{H}^{-1}\varphi$. Since $\eta =y/H$, ${\partial }_t\eta =-\eta H^{\prime }/H$, the vorticity equation terms areFor $H=\sqrt{2Ct}$, $H{H}^{\prime }=C$. Divide by $C^{2}x/H^5$ (or compare coefficients at $x=0$ by continuity) to obtainThe planes move normally with velocities $v(\pm H)=\pm H^{\prime }=\pm C/H$ and have zero tangential velocity. No-slip boundary conditions therefore requireThese ensure both normal impermeability relative to the moving planes and tangential no slip.

Put $K=k\pi$, $s_k=(-1{)}^k$ and $\varphi =s_k\sin (K\eta )/K-\eta$. Its derivatives are ${\varphi }^{\prime }=s_k\cos (K\eta )-1$, ${\varphi }^{\prime \prime }=-s_{k}K\sin (K\eta )$, ${\varphi }^{\prime \prime \prime }=-s_k{K}^2\cos (K\eta )$ and ${\varphi }^{\prime \prime \prime \prime }=s_k{K}^3\sin (K\eta )$. The sine vanishes at both walls and $s_k\cos (K)=1$, verifying all four boundary conditions. In the ODE, the sine-cosine products from $(\eta +\varphi ){\varphi }^{\prime \prime \prime }$ and $-{\varphi }^{\prime }{\varphi }^{\prime \prime }$ cancel, leavingThus each displayed profile is an exact solution when $R=k^2{\pi }^2/4$.

For large $k$, $\varphi$ stays within $1/(k\pi )$ of $-\eta$, but the tangential velocity factor ${\varphi }^{\prime }$ continues to oscillate between $-2$ and $0$, and its shear grows like $k$. The sketch shows both effects; a nearly straight stream-function profile does not mean nearly uniform velocity. Such fine oscillatory branches occur at special discrete large Reynolds numbers and require specially prepared flow. Their many inflection points and strong shear permit instability, so generic disturbances are unlikely to preserve them. This is a physical plausibility discussion, not a proved stability classification.

For a point $\mathbf{x}$ on the plane, the Cauchy-Schwarz inequality gives $|d|=|\mathbf{x}\cdot \mathbf{n}|\le |\mathbf{x}|$, because $\mathbf{n}$ is a unit vector. Equality is achieved at $\mathbf{x}=d\mathbf{n}$. Thus the distance from the origin is $|d|$.

Let $\delta =\mathbf{p}\cdot \mathbf{n}-d$ and let $\mathbf{q}=\mathbf{p}-\delta \mathbf{n}$ be the perpendicular projection of the sphere's center onto the plane. Every point of the plane has form $\mathbf{x}=\mathbf{q}+\mathbf{u}$ with $\mathbf{u}\cdot \mathbf{n}=0$. The Pythagorean theorem givesIf $|\delta |=r$, the sphere equation therefore forces $\mathbf{u}=0$. There is exactly one contact point, $\mathbf{q}$; it lies on both surfaces.

Write $\tau =\mathbf{a}\cdot (\mathbf{b}\times \mathbf{c})>0$. The nonzero scalar triple product proves that $\mathbf{a},\mathbf{b},\mathbf{c}$ are linearly independent, so they form a positively oriented basis of ${\mathbb{R}}^3$ and define a nondegenerate tetrahedron.

For the three faces through the origin, choose the inward unit vectorsTheir signs are correct because each has positive inner product with the opposite vertex vector. Write the center as $\mathbf{p}=\alpha \mathbf{a}+\beta \mathbf{b}+\gamma \mathbf{c}$. It lies on the interior side of every face. Tangency of the interior sphere therefore means signed distance $r$, not $-r$, from each of those three planes. Taking inner products givesSolving proves the center formulaThis is the sphere center from three tetrahedron face distances; the interior hypothesis selects these signs rather than an exterior center.

Finally set $\mathbf{N}=\mathbf{a}\times \mathbf{b}+\mathbf{b}\times \mathbf{c}+\mathbf{c}\times \mathbf{a}$. The scalar triple product identities give $\mathbf{N}\cdot \mathbf{a}=\mathbf{N}\cdot \mathbf{b}=\mathbf{N}\cdot \mathbf{c}=\tau$. In particular $\mathbf{N}\ne 0$, and the plane through the three vertices isThe second formula follows by normalizing $\mathbf{N}$ to a unit vector and applying the first distance result.

An eigenvalue can be complex even though $P$ is real, so use the complex Hermitian inner product. For a nonzero complex eigenvector $\mathbf{v}$, the real orthogonal matrix also satisfies $P^{\ast }P=P^{T}P=I$. Henceand $\boxed{|\lambda |=1}$. Taking complex conjugates in $P\mathbf{v}=\lambda \mathbf{v}$ gives $P\overline{\mathbf{v}}=\overline{\lambda }\overline{\mathbf{v}}$ because $P$ is real. The conjugate vector is nonzero, so $\boxed{{\lambda }^{\ast }=\overline{\lambda }\text{ is also an eigenvalue}}$.

For the unheaded geometrical conclusion, the third eigenvalue of $Q$ must be real: nonreal eigenvalues of a real matrix occur in conjugate pairs. Its modulus is one, so it is $1$ or $-1$. Exact algebraic multiplicity two for $1$ excludes a third $1$, leaving $-1$. By the supplied diagonalisability fact, the $1$-eigenspace is a two-dimensional plane and the $-1$-eigenspace is a line. They are orthogonal: preservation of the inner product between a fixed vector and a negated vector gives $\mathbf{u}\cdot \mathbf{v}=-\mathbf{u}\cdot \mathbf{v}$. Therefore $Q$ is the orthogonal reflection in its fixed plane. It fixes the two tangential directions and reverses the normal direction; with unit normal $\mathbf{n}$, $Q=I-2\mathbf{n}{\mathbf{n}}^T$.

Take points $p=(3,0,4)$ and $q=(-2,3,3)$ with directions $u=(1,3,-1)$ and $v=(0,1,-1)$. A common normal is $u\times v=(-2,1,1)$, of length $\sqrt{6}$. The planes through the two lines with this common normal have separationThis is a lower bound for the distance between skew lines. It is attained at $s=-1$ and $t=-4$: the two points are $(2,-3,5)$ and $(-2,-1,7)$, whose difference $(4,-2,-2)$ is perpendicular to both directions and has length $2\sqrt{6}$. Thus the shortest distance is $\boxed{2\sqrt{6}}$.

The plane through $a_i$ has the constant normal coordinate $r\cdot \hat{n}=a_i\cdot \hat{n}$. Since $\hat{n}$ is a unit vector, the separation of these normal coordinates is a length. Every displacement joining the planes has this fixed normal component, so its length is at least its absolute value; a displacement parallel to $\hat{n}$ attains that bound. Hence the distance between parallel planes isThe absolute value, present in the PDF, is necessary because distance is nonnegative.

The first coordinate of the vector triple product isThe other two coordinates follow by cyclically permuting the indices, proving the identity.

A point on the line has the form $r=b+sm$. Substituting in the plane equation gives $s(m\cdot n)=(a-b)\cdot n$. When $m\cdot n\ne 0$ the parameter, and therefore the intersection point, is unique. The vector triple product gives $n\times (b\times m)=(n\cdot m)b-(n\cdot b)m$, so the unique intersection isFor the parallel case $m\cdot n=0$, the line is entirely in the plane if $(b-a)\cdot n=0$, and otherwise has no intersection. As usual, a genuine line and plane require nonzero direction and normal vectors. This is the line-plane intersection criterion.

The two displacement vectors $\mathbf{b}-\mathbf{a}$ and $\mathbf{c}-\mathbf{a}$ lie in the required plane. Because the three points are non-collinear, their cross productis nonzero and is a normal vector to that plane. A point with position vector $\mathbf{r}$ lies in the plane exactly whenExpanding the cross product givesThe terms $\mathbf{a}\cdot (\mathbf{a}\times \mathbf{b})$ and $\mathbf{a}\cdot (\mathbf{c}\times \mathbf{a})$ vanish, so the equation of a plane through three points becomesNon-collinearity is important: it ensures this is a genuine plane equation with nonzero normal, rather than the vacuous equality $0=0$.

For the specified coordinates,Their sum is $(3,3,6)$, with Euclidean norm $3\sqrt{6}$. Consequently either orientation of the unit normal is valid:As a check, the scalar triple product is $12$, so the plane equation simplifies to $x+y+2z=4$, which all three supplied points satisfy. The normal direction is $(1,1,2)$.

The chosen unit normal points toward the plane from the origin, so the plane equation isTo locate the circle centre, project $\mathbf{p}$ onto this plane. PutThen $\mathbf{n}\cdot \mathbf{h}=d$, so $\mathbf{h}$ lies in the plane. For any other point $\mathbf{r}$ of the plane, $\mathbf{r}-\mathbf{h}$ is perpendicular to $\mathbf{n}$. The Pythagorean theorem therefore givesRestricting the sphere equation to the plane consequently gives a circle centred at $\mathbf{h}$. Thus $\mathbf{m}=\mathbf{h}$, proving that the displacement of its centre is purely normal:The same Pythagorean theorem gives $q^2={\rho }^2+{\gamma }^2$, and henceThe sign of $\gamma$ is fixed by which side of the plane contains $\mathbf{p}$; $q$ and $\rho$ alone determine only its magnitude. Substituting the signed offset yields the sphere-plane intersection radiusA genuine circle requires $|d-\mathbf{n}\cdot \mathbf{p}|<q$; equality gives tangency and $\rho =0$, while a larger distance gives no intersection. The signed distance from a point to a plane in the direction $\mathbf{n}$ is $\mathbf{n}\cdot \mathbf{p}-d=-\gamma$; the ordinary distance from a point to a plane is its absolute value.