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Let $c=\cos (\alpha {\xi }_3)$, $s=\sin (\alpha {\xi }_3)$ and $r^2=x_1^2+x_2^2=({\xi }_1^2+{\xi }_2^2)/\lambda$. Differentiating the deformation givesFor the neo-Hookean solid, $W_{,A_{i\alpha }}=\mu A_{i\alpha }$. The constrained nominal stress tensor is therefore $N_{\alpha i}=\mu A_{i\alpha }+q(A^{-1}{)}_{\alpha i}$. The multiplier cannot be set arbitrarily to a constant; equilibrium and the traction-free curved surface determine it.

Push the nominal stress to the Cauchy stress tensor: since $J=1$, $\sigma =AN=\mu A{A}^T+qI$. In the current cylindrical orthonormal basis, the left Cauchy-Green deformation tensor has entriesThus ${\sigma }_{rr}=\mu /\lambda +q$, ${\sigma }_{\vartheta \vartheta }=\mu /\lambda +\mu {\alpha }^2{r}^2+q$, and ${\sigma }_{\vartheta 3}=\mu \alpha \lambda r$. Static radial equilibrium without body force givesThe current outer radius is $a=R/\sqrt{\lambda }$. Setting ${\sigma }_{rr}(a)=0$ integrates the reaction field:The other curved-surface traction components vanish because ${\sigma }_{r\vartheta }={\sigma }_{r3}=0$.

On the reference end ${\xi }_3=H$, the outward reference normal is $e_3$, and the nominal traction has components $t_i=N_{3i}$. The third row of $A^{-1}$ is $(0,0,{\lambda }^{-1})$. Hence the nominal end traction of a twisted neo-Hookean cylinder isThe current $x_1,x_2$ here are evaluated at ${\xi }_3=H$. These components are per unit reference area, not per unit current area.

For the moment, use current lever arms and reference-area traction. Its axial component isThe axial traction gives no axial moment. This torque also agrees with the reference-energy derivative at fixed axial stretch: the twist-dependent energy per reference length is $\mu {\alpha }^2\pi R^4/(4\lambda )$. It is important that $\alpha$ is twist per reference length, not twist per current length.

Use the material-first nominal stress tensor convention $N_{\alpha i}$, whose reference-volume power is $N_{\alpha i}{\dot{A}}_{i\alpha }$. At a regular constraint point, $F_{,A}\ne 0$, all admissible deformation rates satisfy $F_{,A_{i\alpha }}{\dot{A}}_{i\alpha }=0$. Equality of mechanical power and the derivative of the strain energy density givesfor every vector in this eight-dimensional tangent hyperplane. Its orthogonal complement is the span of $F_{,A}$. Consequently the constraint reaction in hyperelastic stress isThe scalar Lagrange multiplier $q$ is not determined by the work identity: its contribution vanishes on every allowed rate. Regularity matters. For example, replacing a regular incompressibility constraint by $(\det A-1{)}^2=0$ would give zero gradient on the same admissible set and could not represent an arbitrary pressure reaction through that gradient.

For incompressibility use the regular function $F=J-1$, with $J=\det A$. Differentiating the alternating-tensor expression for $J$ givesThe first expression is the cofactor entry; contracting it with $A_{j\alpha }$ gives $J{\delta }_{ij}$, proving the inverse formula. On $J=1$ this yieldsWith this plus-sign convention $q$ is the negative of the usual pressure multiplier.

For an equibiaxially stretched sheet, incompressibility requires the principal stretches $(\lambda ,\lambda ,{\lambda }^{-2})$. The printed sheet expression's ${\lambda }^{-1/2}$ is inconsistent with this constraint and must be ${\lambda }^{-2}$; the balloon expression later on the same PDF page has the correct exponent. Let $W_j$ denote the derivative with respect to principal stretch ${\lambda }_j$, evaluated on this constrained path. With no applied thickness-direction traction,Isotropy gives $W_1=W_2$ at equal in-plane stretches. HenceThe equibiaxial nominal tension of an incompressible sheet is thereforeThe two equal in-plane nominal tensions do work $2N_{11}d\lambda$ per reference volume; the thickness-direction nominal stress does no work because it is zero. That is exactly the constrained energy differential, explaining energy conservation. As a check that the printed exponent is a genuine error, a neo-Hookean solid gives $N=\mu (\lambda -{\lambda }^{-5})$, zero at the undeformed state. The derivative with the erroneous exponent would instead give $3\mu /4$ at that state.

For the thin spherical balloon, use the leading reference shell volume $4\pi a_0^2{h}_0$. Its two tangential stretches are $\lambda =a/a_0$, while its thickness stretch is ${\lambda }^{-2}$ and its current thickness is $h=h_0{\lambda }^{-2}$. Define $\hat{W}(\lambda )=W(\lambda ,\lambda ,{\lambda }^{-2})$. Its total elastic energy at leading thin-shell order isDuring a quasistatic increment $da$, the enclosed volume changes by $d\mathcal{V}=4\pi a^{2}da$, and $d\lambda =da/a_0$. Equating energy change to work of the internal excess pressure givesThis is the inflation pressure of a thin incompressible elastic balloon. The approximation neglects through-thickness variation and curvature corrections beyond leading order, and requires $h/a\ll 1$. It also agrees with membrane force balance: $p=2h{\sigma }_t/a$, ${\sigma }_t=\lambda N$, and $N={\hat{W}}^{\prime }/2$. Here pressure is force per current area, while $N$ is force per reference area; confusing them would lose a factor of $\lambda$.

For odd $p$, write $e_p(t)=e^{2\pi it/p}$ and use the discrete paraboloidHere surface integration means $p^{-2}{\sum }_P$, whereas ambient norms below use unnormalized counting measure on ${\mathbb{F}}_p^3$. With the negative-phase transform convention,When $x_3=0$, additive-orthogonality of characters makes this one at $x=0$ and zero at all other points of that plane. When $x_3\ne 0$, completing the two squares givesThe Quadratic Gauss sum identity is $G(t{)}^2=\chi (-1)p$, where $\chi$ is the Legendre symbol. For completeness, counting the preimages of a square gives $G(t)=\chi (t)G(1)$ for nonzero $t$. Also $|G(1){|}^2={\sum }_{u,v}{e}_p(u^2-v^2)=p$, by the invertible change $(u,v)\mapsto (u-v,u+v)$ and orthogonality of characters. Since $\overline{G(1)}=G(-1)=\chi (-1)G(1)$, the asserted square identity follows. Thus for $p\equiv 3(\mathrm{mod}4)$,Using positive phase instead changes the sign inside the character, not the magnitude or any norm estimate.

Define $Eg(x)=p^{-2}{\sum }_{\xi \in P}g(\xi )e_p(x\cdot \xi )$, and define $R^{\ast }(2\to q)$ as the smallest constant forThe adjoint relative to these measures is $E^{\ast }h(\xi )={\sum }_{x}h(x)e_p(-x\cdot \xi )$. Consequently $E{E}^{\ast }h=h\ast \hat{\sigma }(-\cdot )$, where convolution is the unnormalized ambient sum. Put $K=\hat{\sigma }(-\cdot )-1_{\{0\}}$. Its supremum is at most $p^{-1}$, so convolution by $K$ has ${\ell }^1\to {\ell }^{\infty }$ norm at most $p^{-1}$.

Under the unnormalized ambient Fourier transform on a finite group, the convolution multiplier of $\hat{\sigma }(-\cdot )$ is $p^3\sigma =p{1}_P$, up to the harmless reflection convention. Thus the multiplier of $K$ is $p{1}_P-1$, with maximum absolute value at most $p$. The Plancherel theorem gives ${\ell }^2\to {\ell }^2$ norm at most $p$. The Riesz-Thorin interpolation theorem between these two bounds gives ${\ell }^{4/3}\to {\ell }^4$ norm at most one for convolution by $K$. Convolution by $1_{\{0\}}$ is the identity, whose ${\ell }^{4/3}\to {\ell }^4$ norm is at most one for counting measure. Hence $\Vert E{E}^{\ast }h{\Vert }_4\le 2\Vert h{\Vert }_{4/3}$.

By Lp duality and Hölder's inequality, $\Vert E^{\ast }h{\Vert }_{L^2(\sigma )}^2=⟨E{E}^{\ast }h,h⟩\le 2\Vert h{\Vert }_{4/3}^2$. Taking adjoints proves the finite-field paraboloid fourth-moment extension estimateIn fact this argument works for both residue classes of odd primes.

If $p\equiv 1(\mathrm{mod}4)$, choose $i\in {\mathbb{F}}_p$ with $i^2=-1$. The complete line $\ell =\{(t,it,0):t\in {\mathbb{F}}_p\}$ lies in $P$. For $g=1_{\ell }$ the normalized surface $L^2$ norm is $p^{-1/2}$, and orthogonality of characters gives $Eg(x)=p^{-1}{1}_{\{x_1+i{x}_2=0\}}$. That annihilator plane has $p^2$ points, so the isotropic-line obstruction to finite-field restriction givesThis tends to infinity along these primes for every $q<4$, as required.

Use $\widehat{f\sigma }(x)={\int }_{S^1}f(\omega )e^{-2\pi ix\cdot \omega }d\sigma (\omega )$, with arclength measure. We first bound the transform for constant density. By rotational symmetry, $\hat{\sigma }(x)={\int }_0^{2\pi }{e}^{-2\pi ir\cos \theta }d\theta$, $r=|x|$. For $r\ge 1$, remove neighborhoods of width $h=r^{-1/2}$ about the stationary points $0,\pi$. Their contribution is $O(h)$. On each remaining interval integrate by parts using the phase derivative $2\pi r\sin \theta$. The boundary terms are $O((rh{)}^{-1})$, and the integral of the derivative of its reciprocal is bounded bySplitting at the stationary points gives the same bound on the other half of the circle. Together with the trivial estimate for bounded $r$, this proves $|\hat{\sigma }(x)|\lesssim (1+|x|{)}^{-1/2}$.

For merely measurable $f$ there is no reason that its amplitude permits this integration by parts. Instead use Gaussian positivity for even extension moments. SetExpand the fourth power and integrate first in $x$. The Fourier transform of a Gaussian givesAll exchanges of integrals are justified by bounded amplitudes, finite surface measure and the integrable Gaussian. The kernel is nonnegative and $|f|\le 1$ almost everywhere. Taking the absolute value of the integral therefore gives $0\le I_R(f)\le I_R(1)$. The preceding scalar decay estimate and polar coordinates yieldFor the last bound split at $1$ and $R$; the tail becomes an integrable $e^{-\pi s^2}/s$ integral after $r=Rs$. Since the Gaussian is at least $e^{-\pi }$ on the radius-$R$ ball, the local fourth-moment restriction estimate for the circle follows:Every constant here is independent of $f$ and $R$.

For the requested connection to Besicovitch sets, let $\delta$ be small and dilate the unit segments of a bounded planar Kakeya set by ${\delta }^{-2}$. Their $\delta$-neighborhoods become direction-separated rectangles of length ${\delta }^{-2}$ and width ${\delta }^{-1}$. A smooth circle cap of angular width $c\delta$ has a Fourier extension operator of magnitude at least $c^{\prime }\delta$ on such a rectangle: after removing the constant phase, the tangential phase variation is $O(c)$ and the normal variation is $O(c^2)$. A frequency modulation translates the rectangle to any desired location. Choose separated caps for the rectangle directions, and multiply their modulated densities by independent Rademacher random variables. Because the caps are disjoint, the input remains bounded by one.

Apply the local fourth-moment estimate in a ball of radius $C_E{\delta }^{-2}$ containing the rectangles, then take expectation over the Rademacher random variables. The fourth moment dominates the square of the sum of squared packet magnitudes, givingThere are $M\asymp {\delta }^{-1}$ rectangles, each of area $\asymp {\delta }^{-3}$, so the integral of their sum is $\asymp {\delta }^{-4}$. Cauchy-Schwarz forces their union area to be at least $c_E{\delta }^{-4}/\log (1/\delta )$. Scaling back gives $|E_{C\delta }|{\gtrsim }_{E}1/\log (1/\delta )$, the planar Kakeya neighborhood lower bound. The covering-number argument in Question 1 then gives both Minkowski dimensions equal to two.

A finite-field Kakeya set, also called a finite-field Besicovitch set, is a subset containing a complete affine line in a vector space $a+{\mathbb{F}}_{p}v$ in every nonzero direction $v$, with scalar multiples counted as the same direction. The line's translate may depend on the direction.

For odd $p$, take the union of the lines $y=mx-m^2/4$, $m\in {\mathbb{F}}_p$. Completing the square gives $x^2-y=(m/2-x{)}^2$. Thus that union is exactly $K_0=\{(x,y):x^2-y\text{ is a square, including zero}\}$. For each fixed $x$ there are $(p+1)/2$ such $y$, so $|K_0|=p(p+1)/2$. It contains every nonvertical slope. Add the vertical line $x=0$, whose overlap with $K_0$ has $(p+1)/2$ points. The tangent-line finite-field Kakeya construction consequently hasThe prime two is irrelevant to this asymptotic statement and can be handled by taking its whole plane.

For the high-dimensional lower bound, we give a direct polynomial method in combinatorics proof, which yields a stronger estimate. Suppose a finite-field Kakeya set $K\subseteq {\mathbb{F}}_p^n$ has fewer than $\left(\genfrac{}{}{0ex}{}{p+n-1}{n}\right)$ points. That binomial coefficient is the dimension of a bounded-total-degree polynomial space of polynomials of total degree at most $p-1$. Evaluation at the points of $K$ imposes fewer linear conditions than this dimension. By the rank-nullity theorem, there is a nonzero polynomial $P$ of total degree $d\le p-1$ vanishing on $K$.

For every nonzero $v$ choose the line $a_v+tv\subseteq K$. The polynomial $P(a_v+tv)$ in $t$ has degree at most $d<p$ and vanishes at all $p$ field elements, so it vanishes identically. Its coefficient of $t^d$ is the leading homogeneous polynomial part $P_d(v)$. Thus $P_d(v)=0$ for every nonzero $v$. The case $d=0$ is already impossible for a nonzero constant vanishing on a nonempty set; for $d>0$, the absence of a constant term in this homogeneous polynomial also gives $P_d(0)=0$.

A polynomial of degree less than $p$ in each variable cannot vanish at every point of ${\mathbb{F}}_p^n$ unless it is zero. To see this, induct on $n$: fix the first $n-1$ variables, apply the one-variable root bound in the last variable, then apply the induction hypothesis to each resulting coefficient polynomial. This forces $P_d=0$, contradicting the choice of the leading part. We have proved the finite-field Kakeya polynomial boundAt $n=15$ this implies the required bound with an explicit absolute constant:The argument is valid in every prime characteristic and does not invoke the lower bound as an unproved theorem.

For a nonempty bounded set, let $N_{\delta }(E)$ be the least number of radius-$\delta$ balls covering $E$. The lower and upper Minkowski dimensions are respectivelyUsing comparable cubes or ball diameters gives the same limits. For an unbounded set one must specify a local or bounded-piece convention; these are the usual bounded-set definitions used in the Kakeya set problem.

For the digit set, fixing the first $k$ digits gives exactly $3^k$ cylinders in intervals of length $9^{-k}$. They cover the set with at most $3^k$ balls of comparable radius. Their base-nine prefix integers are separated by at least two, so these containing intervals have gaps at least $9^{-k}$. Choosing one point in each cylinder shows that a radius-$9^{-k}$ ball meets at most a bounded number of chosen points. Hence $N_{9^{-k}}(Q)\asymp 3^k$. For $9^{-(k+1)}<\delta \le 9^{-k}$, monotonicity sandwiches the covering number between constant multiples of $3^k$ and $3^{k+1}$. This proves the box-counting dimension of separated digit sets formula here:A planar Besicovitch set is a bounded set of zero Lebesgue measure containing a unit line segment in every unoriented direction. The following proof in fact applies to every bounded planar Kakeya set, whether its area is zero or positive.

Choose $M\asymp {\delta }^{-1}$ directions equally spaced in an angular interval of length, say, $\pi /2$, and one segment of the set in each direction. Let $T_j$ be its rectangle of length one and width comparable to $\delta$, lying in the $C\delta$ neighborhood $E_{C\delta }$. Put $F={\sum }_j{1}_{T_j}$. Then $\int F\asymp M\delta \asymp 1$. Two such Kakeya tubes at angle $\alpha$ have overlap at most $C{\delta }^2/\alpha$, since the intersection of their supporting strips is a parallelogram of that area. The trivial bound $C\delta$ applies for nearly parallel tubes. Thus, uniformly over their translations,Sum these intersections. Each row contributes at most $C\delta \log (2/\delta )$, so $\int F^2\lesssim \log (2/\delta )$. The Cauchy-Schwarz inequality on the union now gives the planar Kakeya neighborhood lower boundA cover of $E$ by $N_{\delta }(E)$ radius-$\delta$ balls covers $E_{C\delta }$ by their fixed-factor enlargements. Hence $N_{\delta }(E)\gtrsim {\delta }^{-2}/\log (2/\delta )$. Boundedness gives the matching upper exponent $N_{\delta }(E){\lesssim }_E{\delta }^{-2}$. Taking the lower and upper limits provesThe logarithmic neighborhood loss is compatible with zero area; it still forces full dimension.