The Brouwer fixed-point theorem says that every continuous map $T:{\overline{B}}_R(0)\to {\overline{B}}_R(0)$ has a fixed point. To use the supplied outward vector field zero theorem, first extend the construction to all of ${\mathbb{R}}^n$. LetThe radial retraction onto a closed ball $P_R$ is continuous, so $g$ is a continuous vector field on the required whole space. On $|x|=R$, the Cauchy-Schwarz inequality givesThe supplied theorem gives $g(x_0)=0$ with $|x_0|\le R$. There $P_R(x_0)=x_0$, hence $\boxed{T(x_0)=x_0}$. This explicit extension avoids presuming that the original map $T$ was defined outside the ball.

The contraction mapping theorem states that a self-map $T$ of a nonempty complete metric space with a uniform contraction constant $q<1$ has a unique fixed point $p$. For every starting point, the iteration $x_{n+1}=T{x}_n$ converges to $p$, with $d(x_n,p)\le q^{n}d(x_1,x_0)/(1-q)$.

For the cosine fixed-point iteration, any real $x_0$ gives $x_1\in [-1,1]$. This closed interval is complete, and cosine maps it into $[\cos 1,1]\subseteq [-1,1]$. The mean value theorem giveswith $\sin (1)<1$. Apply the contraction mapping theorem to the tail starting at $x_1$. Therefore $\boxed{x_n\to p,p=\cos p}$ for every real initial value. Numerically $p\approx 0.739085$. The global real-line derivative bound is only one; entering the invariant interval supplies the strict uniform bound needed by the theorem.

Here “contractive” means strictly distance-decreasing: $\boxed{d(Tx,Ty)<d(x,y)\text{ whenever }x\ne y}$. “Strongly contractive” means a contraction mapping with one uniform constant $q\in [0,1)$ satisfying $\boxed{d(Tx,Ty)\le qd(x,y)\text{ for all }x,y\in M}$.

If distinct $p,q$ were both fixed points, strict decrease would give $d(p,q)=d(Tp,Tq)<d(p,q)$, a contradiction. Thus there is at most one fixed point. This proves uniqueness if a fixed point exists; the pointwise strict inequality alone does not assert existence.

Take the clamp at $s=0$ with $x(0)=y(0)=\theta (0)=0$. An end force without an applied end moment gives the free-end condition ${\theta }^{\prime }(1)=0$. With $F=0$, the Euler elastica equation is $EI{\theta }^{\prime \prime }+G\cos \theta =0$. For $|G|/(EI)\ll 1$, $\theta$ is small, so the leading equation isIntegrating with the two boundary conditions givesTo first order, $x^{\prime }=\cos \theta =1+O(G^2)$ and $y^{\prime }=\sin \theta =\theta +O(G^3)$. HenceEquivalently $y(x)=G{x}^2(3-x)/(6EI)$ at leading order, the familiar cantilever response. Because $G<0$, the deflection is downward. The free-end displacement field and slope are $y(1)=G/(3EI)$ and $\theta (1)=G/(2EI)$. The remainder notation holds with fixed positive $EI$ and nondimensional unit rod length.

Set $\ell =\sqrt{EI/F}$. The Euler elastica equation becomes ${\theta }^{\prime \prime }={\ell }^{-2}\sin \theta$. Multiplying by ${\theta }^{\prime }$ gives the first integralStraight asymptotic ends require ${\theta }^{\prime }\to 0$ and a horizontal tangent, so $C=F$. This is the separatrix of a simple pendulum with its angle shifted by $\pi$. A nonconstant solution moves from $\theta =0$ to $2\pi$, with the same physical tangent at both ends, taking infinite arc length to approach either limit. One explicit representative isUsing $\cos \theta =1-2{\mathrm{sech}}^2(s/\ell )$ and $\sin \theta =-2\mathrm{sech}(s/\ell )\tanh (s/\ell )$ and integrating $r^{\prime }=(\cos \theta ,\sin \theta )$ gives the tensile elastica loopThis solution is nontrivial, tends to horizontal straight tails, and has finite excess bending energy. It is also necessarily self-intersecting. Indeed $y$ is even and $x$ is odd. For $z>0$, let $f(z)=z-2\tanh z$. Near zero $f(z)<0$, while $f(2)>0$, so there is a root $z_{\ast }\in (0,2)$. At the two distinct parameters $s=\pm \ell z_{\ast }$, both points have $x=0$ and the same $y>0$. Numerically $z_{\ast }\simeq 1.91501$.

Any nonconstant smooth solution with the same straight asymptotic conditions has this first integral. Between consecutive horizontal orientations, separating variables gives the displayed solution up to translation, reflection, and reversal. It cannot pass through a straight equilibrium at a finite parameter, by uniqueness of the ordinary differential equation. Thus all such nontrivial branches have the same loop crossing. This is an ideal zero-thickness rod solution; a physical rod with self-contact excluded cannot realize the crossing without adding contact forces.

Put $P=-F>0$. With $G=0$, the Euler elastica equation is $EI{\theta }^{\prime \prime }+P\sin \theta =0$. Linearization about the straight configuration gives $EI{\theta }^{\prime \prime }+P\theta =0$. Under the boundary conditions explicitly stated, $\theta (0)=\theta (L)=0$, a nonzero solution requiresThus with only the stated boundary conditions, the first critical compressive force is $\boxed{F_c=-{\pi }^{2}EI/L^2}$. The corresponding linear transverse displacement field, with $y(0)=0$, isand $x(s)=s$ to first order. These are the successive small-amplitude buckling modes. The nonlinear branches retain the simple pendulum equation; if the angular amplitude is ${\theta }_m$ and $m=\sin ({\theta }_m/2)$, fitting $n$ half-periods into the rod gives $P=4EI{n}^{2}K(m{)}^2/L^2$, which tends to $P_n$ as $m\downarrow 0$. Here $K$ is the complete elliptic integral of the first kind, with $m$ its modulus, not its squared parameter.

Endpoint position compatibility for the Euler elastica adds a condition when fully fixed clamps have equal transverse height. If the word “clamped” also imposes $y(L)=y(0)$, then ${\int }_0^L{\theta }_{n}ds=0$. With $G=0$, this retains only even $n=2j$, yieldingThe same exclusion is exact on the nonlinear branches: integrating the elastica equation gives $P\int \sin \theta ds=-EI[{\theta }^{\prime }(L)-{\theta }^{\prime }(0)]$, and an odd number of half-periods reverses the nonzero endpoint slope. Hence odd $n$ changes the endpoint height. The conventional fixed-clamp first Euler buckling load is the second boxed value, whereas the source's explicitly listed boundary conditions alone give the first. Additional clamped modes with a nonzero transverse reaction require releasing the imposed $G=0$. Load-controlled buckling also permits axial shortening; an exactly inextensible rod whose endpoint separation is fixed equal to its full length has no room to buckle.

Let $s$ be arc length, so inextensibility makes $r^{\prime }=d_3$ a unit tangent vector. The curvature is $\kappa ={\theta }^{\prime }$, hence $m^{\prime }=EI{\theta }^{\prime \prime }{e}_z$ for uniform bending stiffness. Balance of linear momentum in static equilibrium gives a constant resultant $n=F{e}_x+G{e}_y$.

Taking the cross product directly,The scalar component of conservation of angular momentum therefore becomesTogether with $x^{\prime }=\cos \theta$ and $y^{\prime }=\sin \theta$, this Euler elastica equation determines the centerline once the appropriate boundary conditions are supplied.

Substituting both profiles into the shear traction condition givesFor the normal traction, use $M{b}_1^2=M{k}^2-\rho k^2{c}^2$ and $b_2^2+k^2=k^2(2-c^2/c_T^2)$. Removing the common nonzero factors yields the second equation. ThusA nonzero displacement field requires the determinant to vanish:Let $s=c^2/c_T^2$ and $\eta =c_T^2/c_L^2$. Cancelling $k^4$ gives the Rayleigh secular equationSquaring and excluding the zero root gives $s^3-8s^2+(24-16\eta )s-16(1-\eta )=0$; any candidate must still satisfy the unsquared equation and the decay inequalities. For stable isotropic elasticity $0<\eta <3/4$, this cubic has one root in $(0,1)$: its values at zero and one have opposite signs, and its strictly negative second derivative on this interval prevents a second crossing back to a positive final value. Both sides of the unsquared equation are positive there, so this root is valid.

The root $s=0$ of the squared equation is not a propagating surface wave: at $c=0$, the two roots coincide, and the null combination $A_2=-A_1$ in this representation produces zero displacement field. As a numerical example, $\lambda =\mu$ gives $\eta =1/3$ and $c\simeq 0.9194c_T$. The surface-wave speed is independent of $k$ in this homogeneous half-space model.

Write the in-plane displacement field as $(u,v)$. The infinitesimal strain tensor gives $2e_{XY}=u_Y+v_X$, $e_{YY}=v_Y$, and $\mathrm{tr}e=u_X+v_Y$. The two nontrivial zero-traction conditions becomeEquivalently for the complex amplitudes they are $U^{\prime }(0)+ikV(0)=0$ and $ik\lambda U(0)+(\lambda +2\mu )V^{\prime }(0)=0$. Because the constitutive equation is real and linear, vanishing of these complex amplitudes enforces the conditions at every position and time.

The material occupies $Y\ge 0$, so its outward unit normal at the surface is $n=-e_Y$. A free surface has zero traction, $\tau n=0$. ThereforeThe third condition is automatic for the two-dimensional displacement field with zero third component and no dependence on the third coordinate. The first two impose zero shear and normal stress at the free surface.

For the $b_1$ root the elastic wave polarization is parallel to the complex derivative vector $(ik,-b_1)$, givingFor the $b_2$ root the field has zero divergence, $ik{A}_2-b_2{B}_2=0$, givingThese relations also follow by inserting each decay root into either row of the matrix. The complete decaying displacement field has complex componentsmultiplied by $e^{ik(X-ct)}$, followed by the real part. The first constituent is irrotational and the second is divergence-free; both are needed to cancel the two surface tractions.

Use the complex amplitude and take its real part at the end; the equations are linear with real coefficients. Let $M=\lambda +2\mu$, $\omega =kc$, and $v=(A,B{)}^T$. The differential operators on this exponential function are ${\partial }_X=ik$, ${\partial }_Y=-b$, and ${\partial }_t=-i\omega$. Substitution givesThe determinant factors asConsequently choose the positive decay rootsThe Rayleigh wave decay exponents must both give strict decay into the half-space. A genuine nonzero Rayleigh wave combines both, as the free-surface equations below also show. Thus $b_1,b_2>0$ requires $\boxed{0<c<\min (c_T,c_L)}$. In a stable ordinary isotropic solid $c_T<c_L$, so it is slower even than the transverse wave. The roots depend on the candidate speed as well as the material constants and wavenumber; the speed is fixed by the boundary condition.

Divide the Navier-Cauchy equation by mass density. Since $\mu /\rho =c_T^2$ and $(\lambda +\mu )/\rho =c_L^2-c_T^2$, it becomesEquivalently, using the vector calculus identity $\Delta u=\nabla \mathrm{div}u-\nabla \times (\nabla \times u)$,The divergence and curl consequently propagate at $c_L$ and $c_T$, respectively.

Resolve each amplitude by the orthogonal projectionsThe acoustic tensor acts by $\lambda +2\mu$ on the longitudinal line and by $\mu$ on its perpendicular plane. ThusThe longitudinal component is a longitudinal wave, while the two independent transverse polarizations are transverse waves. Each projection obeys its own wave equation, and superposition adds the corresponding solutions.

A general planar disturbance is therefore a sum of the two types. For a single sinusoid with a common speed, both projections cannot be nonzero unless the two speeds coincide: projecting the eigenvalue equation gives $(\rho c^2-\lambda -2\mu )a_L=0$ and $(\rho c^2-\mu )a_T=0$. This distinguishes superposition of waves with different speeds from a mixed polarization traveling rigidly at one speed.

Substituting the sinusoidal plane wave into the Navier-Cauchy equation gives $\Delta u=-u$, $u_{tt}=-c^{2}u$, and $\nabla \mathrm{div}u=-k(k\cdot a)\sin (k\cdot X-ct)$ because $|k|=1$. Equality of the amplitudes is thereforeThis is an eigenvalue problem for the acoustic tensor $\mu I+(\lambda +\mu )k\otimes k$. Choosing a nonzero amplitude parallel to $k$ gives $\rho c^2=\lambda +2\mu$; choosing any nonzero amplitude perpendicular to $k$ gives $\rho c^2=\mu$. For positive mass density and stable Lamé parameters, both squared speeds are positive. These choices construct nonzero solutions explicitly, proving existence.

Integrate the radial equilibrium equation across the thickness, retaining the integral rather than evaluating its antiderivative. The traction conditions give $t_r(b)-t_r(a)=P$. Hence $a$ must satisfyAll endpoints and coefficients here depend only on $a$ and the specified material and reference geometry. This is the required implicit pressure–radius relation for inflation and extension of an incompressible tube with unit axial stretch. In particular $a=A$ gives zero integrand and $P=0$. Once an admissible root is selected, the radial stress follows by integrating from the outer face, $t_r(r)=-{\int }_r^b(d{t}_r/ds)ds$, and the remaining stress components follow from the constitutive equation.

The radial equilibrium equation gives $d{t}_r/dr=(t_{\theta }-t_r)/r$. Set $C=a^2-A^2$, so that $R^2=r^2-C$. Substitution of the neo-Hookean solid stress difference eliminates both $q$ and $t_{\theta }$:This is a first-order ordinary differential equation for the radial stress, parametrized by the unknown inner radius. For inflation $C>0$, its right side is positive, so $t_r$ increases from the negative inner pressure to zero at the outer surface. The denominator stays positive because $r^2-C=R^2\ge A^2>0$.

The multiplier enforcing elastic incompressibility contributes an arbitrary hydrostatic stress. Denote it by $q(r)$ to distinguish it from the applied pressure $P$. For the neo-Hookean solid, the Cauchy stress tensor is $\tau =-qI+\mu F{F}^T$. Thus its principal components areThe axial component is $t_z=-q+\mu$, despite the unit axial stretch. The quantity $q$ is a Lagrange multiplier determined together with equilibrium and the traction conditions; it is not obtained by differentiating the reduced stretch energy while ignoring its constraint.

In particular the stress difference eliminates this multiplier:

Use orthonormal cylindrical bases in the reference configuration and current configuration. The deformation gradient maps a radial increment $dR$ to $r^{\prime }dR$, an azimuthal arc $Rd\Theta$ to $rd\Theta$, and an axial increment $dZ$ to itself. ThereforeDifferentiating the radius relation gives $r^{\prime }=R/r$. Its three principal stretches areTheir product is one, as required by elastic incompressibility. The symbols called “strains” in the source are these stretch ratios; engineering strain would instead subtract one. The axial stretch is one because this is plane strain.

For the axisymmetric deformation map, take $\theta =\Theta$, $z=Z$, and $r=r(R)$. A reference cylindrical shell of unit axial length has volume $2\pi RdR$, and the corresponding current shell has volume $2\pi rdr$. Elastic incompressibility therefore gives $rdr=RdR$.

Integrating from the inner surface, where $r(A)=a$, gives $r^2-a^2=R^2-A^2$. The positive radius is consequentlyThe outer surface is not an additional independent unknown: $\boxed{b=\sqrt{B^2+a^2-A^2}}$, and hence $h=b-a$.