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For point masses joined by unstretched central springs obeying Hooke's law, let $n_{ij}$ be the equilibrium unit vector along an edge and $u_i$ the small displacement of its endpoint. The leading extension is $n_{ij}\cdot (u_j-u_i)$, so its quadratic elastic potential energy is $k_{ij}[n_{ij}\cdot (u_j-u_i){]}^2/2$. Consequently the block stiffness matrix has off-diagonal blocks $K_{ij}=-k_{ij}{n}_{ij}{n}_{ij}^T$ and diagonal blocks $K_{ii}={\sum }_j{k}_{ij}{n}_{ij}{n}_{ij}^T$, with absent edges contributing zero. It is symmetric and positive semidefinite. The normal modes solve $Kq={\omega }^{2}Mq$, where $M$ is the mass matrix. Rigid translations and infinitesimal rotations are rigid-motion zero modes of a spring framework. A prestressed spring can add transverse quadratic terms, so the unstretched assumption matters.

A small arc of a rotating thin hoop needs inward force per length $\rho A{\omega }^{2}r$. Its circumferential tension supplies this through curvature, giving tension $\rho A{\omega }^2{r}^2$ and tensile stress $\rho {\omega }^2{r}^2$. Hooke's law converts that stress to circumferential strain, which equals fractional radius change. Small strain, a slender cross-section and uniaxial linear elasticity are required; the cross-sectional area cancels at leading order.

A finite-dimensional linear operator is diagonalizable over a specified field when it has an eigenbasis. Equivalently, its matrix $A$ admitswhere $P$ is an invertible matrix whose columns are eigenvectors, and $D$ is a diagonal matrix of the corresponding eigenvalues. This change of basis reveals a linear map as independent one-dimensional scalings. The matrix may look coupled in the original coordinates, but each coordinate in the eigenbasis evolves on its own.

For the symmetric matrix $A=\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)$ computed in eigenvector, takeThen $AP=PD$, hence $P^{-1}AP=D$. The orthonormal basis version is $Q=P/\sqrt{2}$, an orthogonal matrix satisfying $Q^{T}AQ=D$. Every real symmetric matrix has an orthonormal eigenbasis by the spectral theorem; a general diagonalizable matrix need not have orthogonal eigenvectors.

The immediate payoff is repeated application: adjacent factors $P^{-1}P$ cancel, giving $A^n=P{D}^n{P}^{-1}$. In this example,The component in $E_3$ grows while the component in $E_1$ stays fixed. After division by its Euclidean norm, this particular sequence approaches $(1,1{)}^T/\sqrt{2}$. This is the mechanism behind the power method; an initial vector entirely in $E_1$ would stay there and never find the dominant eigenvalue $3$.

Two different obstructions deserve comparison. The planar rotation $R$ in eigenvector is diagonalizable over $\mathbb{C}$ but not over $\mathbb{R}$, because its eigenvalues are $i$ and $-i$. By contrast, the shear mappinghas only $\lambda =1$, of algebraic multiplicity two, and $E_1=\mathrm{span}\{(1,0{)}^T\}$, of geometric multiplicity one. There is no second independent eigenvector, even over $\mathbb{C}$. This is a defective matrix, already a two-dimensional Jordan block. Its powers areThus even though both eigenvalues are one, repeated application can produce growth. Eigenvalues alone miss the coupling recorded by a Jordan block. A repeated eigenvalue is harmless when it has enough independent eigenvectors, as $I$ demonstrates. More generally, a matrix is diagonalizable exactly when its characteristic polynomial splits over the chosen field and every eigenvalue has equal geometric multiplicity and algebraic multiplicity. The proof outline is to join bases of the distinct eigenspaces; their direct sum fills the space exactly under these conditions.

A physical motivation is the separation of normal modes. Consider two equal masses $m$ between three identical springs of stiffness $k$, with the outer ends fixed. By Hooke's law and Newton's second law, the displacement vector $q=(q_1,q_2{)}^T$ obeysThe eigenvectors $(1,1{)}^T$ and $(1,-1{)}^T$ have eigenvalues $1$ and $3$: the masses move together in the first normal mode and oppositely in the second. Writing $q=a(t)(1,1{)}^T+b(t)(1,-1{)}^T$ separates the coupled equations into two harmonic oscillators,Their angular frequencies are $\sqrt{k/m}$ and $\sqrt{3k/m}$; the eigenvalues give squared angular frequencies after multiplication by $k/m$. This is the beauty of diagonalization: finding the right directions turns interacting motions into independent patterns. For a physical treatment, see Yen-Jie Lee's "Coupled Oscillators, Normal Modes" (MIT 8.03SC, 2016), and the local article normal modes of two equal masses between three springs.^[ref]

Take two point masses joined by a massless spring of spring constant $k>0$ and equilibrium length $\ell >0$. Assume one-dimensional motion along the spring, no external force, no damping, and an extension $s=x_2-x_1-\ell$ satisfying Hooke's law; keep $x_2>x_1$. For a real spring this is a model within its elastic range. Its potential energy is $U=k{s}^2/2$, soHere $\omega$ is angular frequency and $\nu$ is frequency. Choose $|A|<\ell$ so the model bodies do not cross. The internal normal mode has displacements $u_1=-m_{2}s/M$ and $u_2=m_{1}s/M$ about equilibrium offsets from the centre of mass; its amplitude ratio is $|u_1|/|u_2|=m_2/m_1$. The other normal mode is collective translation with zero frequency and no spring extension. This free pair has different boundary conditions from normal modes of two equal masses between three springs.

For $k=10N/m$ and two $1\mathrm{kg}$ masses, $\omega =\sqrt{20}{\mathrm{s}}^{-1}\approx 4.47214{\mathrm{s}}^{-1}$ and $\nu \approx 0.711763\mathrm{Hz}$. Replacing the second mass by $100\mathrm{kg}$ gives $\omega =\sqrt{10.1}{\mathrm{s}}^{-1}\approx 3.17805{\mathrm{s}}^{-1}$ and $\nu \approx 0.505802\mathrm{Hz}$. Fixing the second endpoint instead would give $\sqrt{10}{\mathrm{s}}^{-1}$: the free heavy partner still moves. The free pair's angular frequency exceeds that fixed-end limit by the factor $\sqrt{1+1/q}$, about $1.004988$ at $q=100$.

Represent a dilute polymer by two beads connected by a spring obeying Hooke's law, with stiffness $H>0$ and bead drag coefficient $\zeta >0$. Neglect bead inertia and hydrodynamic interactions. For component-first velocity gradient $K_{ij}={\partial }_j{u}_i$, the configuration Fokker-Planck probability current isThe factor two comes from relative motion of both beads. At rest its normalized probability density is ${\psi }_{\mathrm{eq}}=(H/(2\pi k_{B}T){)}^{3/2}{e}^{-H{R}^2/(2k_{B}T)}$. The conformation tensor $C=⟨\mathbf{R}{\mathbf{R}}^T⟩$ has dimensions of length squared; $(H/(k_{B}T))C$ is its dimensionless version. The Hookean dumbbell in planar extension is a particular imposed flow in this model.

Under uniform compressive pressure $p$, isotropic Hooke's law gives each normal strain as $-p(1-2\nu )/E$. The volume strain is three times this, so the bulk modulus is $B=E/[3(1-2\nu )]$, where $E$ is Young's modulus and $\nu$ is Poisson's ratio. Positivity of $E$ and finite $B$ gives $\nu <1/2$. Positive shear modulus $G_s=E/[2(1+\nu )]$ adds the lower bound $\nu >-1$. Negative Poisson's ratio is allowed; the bulk relation alone does not supply the lower bound.

For the same massless spring obeying Hooke's law, the frequency is $\nu =(2\pi {)}^{-1}\sqrt{k/m}$. The displayed mass ratio follows. For small independent frequency errors with common fractional standard uncertainty $ϵ$, monomial uncertainty propagation gives ${\sigma }_R/R\simeq \sqrt{8}ϵ$. If $ϵ$ instead describes individual measurements and each sample mean uses $n$ independent trials, its standard error is smaller by $\sqrt{n}$, giving ${\sigma }_R/R\simeq \sqrt{8/n}ϵ$. Four times as many trials halve this statistical uncertainty; a common calibration error needs its covariance included.

Take $\mathbf{R}={\mathbf{r}}_2-{\mathbf{r}}_1$ and label each dumbbell once. Locally assume its probability density and number density vary negligibly over a connector length. The oriented connector crossing count follows geometrically: a connector crosses a small plane element precisely when its midpoint lies within a slab of normal thickness $|\mathbf{R}\cdot \mathbf{n}|$. For configurations in $d^{3}R$, the unsigned crossing number per unit area is thereforeThe expression without the absolute value is an oriented crossing count, or an unsigned count restricted to $\mathbf{R}\cdot \mathbf{n}\ge 0$. It cannot be an unrestricted nonnegative number: a connector $\mathbf{R}=-R\mathbf{n}$ gives a negative printed expression despite physically intersecting the plane.

To obtain traction, keep the force orientation as well. For $\mathbf{R}\cdot \mathbf{n}>0$, the spring force on the bead below the plane exerted by the bead above it is $H\mathbf{R}$. For the opposite orientation it is $-H\mathbf{R}$. Multiplying that force by the unsigned crossing count in either case givesIntegrating over all labelled configurations yieldsThere is no extra factor of two: each dumbbell was labelled and counted once, and the orientation-dependent force has already supplied the sign. This is the local continuum stress approximation; resolving appreciable spatial variation across a finite connector would require a nonlocal stress construction.

Let $p_T=n{k}_{B}T$, $C=⟨\mathbf{R}{\mathbf{R}}^T⟩$, and $M=I+\alpha \sigma$, so the single-bead mobility is $M/\zeta$. Assume an incompressible solvent, constant $n,H,T,\zeta$, negligible bead inertia, and a symmetric positive-definite $M$. The last requirement makes the resistance invertible and Brownian diffusion physically admissible. At a given position and time, $M$ depends on the moment $C$ but not on the individual configuration $\mathbf{R}$.

Use the source's gradient convention $K_{ij}={\partial }_i{u}_j$, and write $G=K^T$ for the component-first gradient. Then $\mathbf{u}({\mathbf{r}}_2)-\mathbf{u}({\mathbf{r}}_1)\simeq G\mathbf{R}$. The spring forces on the two beads are ${\mathbf{F}}_1=H\mathbf{R}$ and ${\mathbf{F}}_2=-H\mathbf{R}$. Their overdamped probability currents areHere the Brownian term written as $-k_{B}T\nabla \psi$ is a force weighted by probability density. Its corresponding conditional osmotic force is $-k_{B}T\nabla \log \psi$; treating the given term as an unweighted deterministic force per bead would not produce the stated Fokker–Planck equation.

With midpoint $\mathbf{x}=({\mathbf{r}}_1+{\mathbf{r}}_2)/2$, the derivatives are ${\nabla }_{r_1}=\frac{1}{2}{\nabla }_x-{\nabla }_R$ and ${\nabla }_{r_2}=\frac{1}{2}{\nabla }_x+{\nabla }_R$. Subtracting the bead currents gives the configuration currentThe midpoint current is $\mathbf{u}\psi$ after neglecting the specified spatial Brownian diffusion. Configuration probability conservation therefore givesSince $G\mathbf{R}$ has components $R_j{\partial }_j{u}_i$, this agrees with the source's $\mathbf{R}\cdot \nabla \mathbf{u}$. The factor two in both spring relaxation and configuration diffusivity comes from relative motion of both beads.

The second moment $C$ is the unnormalized conformation tensor. Now multiply by $R_i{R}_j$ and integrate over configuration space. Assume sufficiently decaying tails so every boundary term at infinite separation vanishes. The affine current gives $GC+C{G}^T$; spring drift gives $-(2H/\zeta )(MC+CM)$; two integrations by parts of diffusion give $(4k_{B}T/\zeta )M$. HenceBecause $\sigma =nHC-p_{T}I$, $M$ is a polynomial in $C$ and commutes with it. ThusThis is the Hookean dumbbell conformation equation with anisotropic mobility, an exact second-moment closure within the stated model; no Gaussian approximation to $\psi$ was needed.

The upper-convected derivative of the identity is $I^{\nabla }=-(G+G^T)$. Applying it to the polymer stress therefore givesConsequently the coefficients and the closed stress equation arewhere ${\sigma }^{\nabla }=D_t\sigma -\sigma K-K^T\sigma$ in the printed convention. Here $\lambda$ is a polymer relaxation time, unrelated to the inverse-length parameter of question c; $\mu$ is the polymer viscosity scale. The coefficient $\alpha$ has inverse-stress units. In the dimensional mobility parameter in the Giesekus model, the dimensionless factor is $b=\alpha p_T$, not the dimensional $\alpha$ printed in this problem.

The upper-convected rate removes the affine stretching and rotation of material connectors. If $\dot{F}=GF$, then a tensor transported purely as $C=F{C}_0{F}^T$ has $C^{\nabla }=0$. Under a time-dependent rigid rotation $Q$, the rate transforms as $Q{C}^{\nabla }{Q}^T$: this is material frame indifference, not an ordinary componentwise time derivative. The resulting equation is the Giesekus model in these dimensional conventions.

The isotropic subtraction is fixed by zero polymer extra stress at thermal equilibrium. With no flow and positive mobility, the stationary distribution isso $\boxed{{\sigma }_{\mathrm{eq}}=0}$. The constitutive equation correspondingly has a zero-stress quiescent state and a Newtonian low-rate limit $\sigma \simeq \mu (K+K^T)$. Without subtracting exactly $p_{T}I$, the spring contribution would retain an equilibrium isotropic stress and the equation would acquire corresponding reference terms. Because mobility itself depends on $\sigma$, that reference cannot simply be shifted while retaining the same model. The separate fluid pressure enforcing incompressibility is not included in this polymer extra stress.

Use the deterministic Hookean dumbbell model, with a zero-rest-length spring and negligible connector drag. For $\mathbf{R}={\mathbf{r}}_1-{\mathbf{r}}_2$, its elastic forces on the two beads areHere $H>0$ is the spring constant. No independent thermal forcing from Brownian motion or bead inertia is included. The unperturbed linear velocity field is $\mathbf{Lr}$, with $\mathrm{tr}\mathbf{L}=0$ by incompressibility.

The fluid velocity used in each bead's drag law is the imposed flow plus the disturbance from the other bead, with its own singular field omitted. At leading order in $a/R$, the evenness of the Oseen tensor, $\mathsf{G}(-\mathbf{R})=\mathsf{G}(\mathbf{R})$, givesThese are the pair hydrodynamic mobility matrix relations. Their signs follow from balancing each spring force with $-\zeta ({\dot{\mathbf{r}}}_i-{\mathbf{u}}_{\mathrm{ext},i})$, the fluid drag on that bead.

Adding the equations cancels both the direct spring slips and the cross-induced velocities, since ${\mathbf{F}}_1+{\mathbf{F}}_2=0$. For the midpoint ${\mathbf{r}}_c=({\mathbf{r}}_1+{\mathbf{r}}_2)/2$,The midpoint therefore follows the undisturbed linear velocity field. Subtracting the equations instead gives the hydrodynamic interaction of a Hookean dumbbell:The dimensionless interaction matrix isIts transverse and longitudinal eigenvalues are $3a/(4R)$ and $3a/(2R)$, respectively. Since the spring force is longitudinal, the separation equation can also be writtenThe relative correction to the free-draining spring relaxation rate is therefore $3a/(2R)$, explicitly proportional to $a/R$. Entrainment by the other bead reduces their relative approach for given opposing spring forces.

This is a far-field hydrodynamic interaction, valid for $R\gg a$; it is not a near-contact law and omits higher reflections and lubrication. The printed target also fixes the spring's rest length implicitly. For an ordinary spring obeying Hooke's law with nonzero rest length $R_0$, the elastic force would be $-H(R-R_0)\hat{\mathbf{R}}$, and every spring term above would acquire the factor $1-R_0/R$. The displayed result uses the zero-rest-length spring convention appropriate to the ideal Hookean dumbbell model.

Linearized magnetic flux freezing gives the Eulerian magnetic field perturbationThe background is uniform, and incompressibility gives $\nabla \cdot \mathbf{\xi }=0$. For a horizontal displacement proportional to $e^{ikz}$, $\delta \mathbf{B}=ik{B}_0\mathbf{\xi }$ and $\delta B_z=0$. Thus the first-order magnetic pressure perturbation vanishes, and the magnetic-tension response to an incompressible displacement isBending a field line produces a restoring force proportional to displacement, like a spring obeying Hooke's law. Per unit mass, its squared spring frequency is $k^2{v}_A^2$, where $v_A=B_0/\sqrt{{\mu }_0\rho }$ is the Alfvén speed. The equilibrium can be independent of $z$ while the perturbation varies with $z$; neglect of background vertical structure does not set this perturbation's wavenumber to zero.

Put $y=e^{-r/\xi }$. Differentiating the Morse potential givesThe finite-separation minimum has $y=B/(2A)$. ThusThe positive bond-dissociation energy is $D=B^2/(4A)=0.250\mathrm{eV}$, measured relative to the separated-atom value $V(\infty )=0$. This distinguishes the well's negative potential energy from the positive energy required to break the bond.

For the displacement $x=r-a$, the Taylor expansion isConsequentlyThe restoring force is $-\kappa x$ to leading order, identifying the bond's Hooke's law spring constant.

The in-plane equilibrium equations are ${\tau }_{xx,x}+{\tau }_{xy,y}=0$ and ${\tau }_{xy,x}+{\tau }_{yy,y}=0$. On a simply connected patch, each divergence-free row has a streamfunction: write ${\tau }_{xx}={\varphi }_y$, ${\tau }_{xy}=-{\varphi }_x=-{\psi }_y$, ${\tau }_{yy}={\psi }_x$. Symmetry gives ${\varphi }_x={\psi }_y$, so another potential exists with $A_y=\varphi$, $A_x=\psi$. HenceIf two such Airy stress functions give the same stress, their difference has zero Hessian and is therefore an arbitrary affine function $a+bx+cy$ on a connected domain. This is the full nonuniqueness for a fixed stress. The existence argument is local, or global on a simply connected domain; domains that are multiply connected regions require the associated force and moment period conditions for a single-valued global potential. Equilibrium alone should not silently erase that topological qualification.

For plane strain put $s=e_{xx}+e_{yy}$. The constitutive equation gives ${\tau }_{xx}+{\tau }_{yy}=2(\lambda +\mu )s$. ConsequentlyThe displacement Saint-Venant compatibility equations $e_{xx,yy}+e_{yy,xx}-2e_{xy,xy}=0$ then becomesFor stable elastic moduli the coefficient is nonzero, so $\boxed{{\Delta }^{2}A=0}$. If $D$ is bounded and smooth and $A={\partial }_{n}A=0$ on its boundary, integrating twice by parts gives $0={\int }_{D}A{\Delta }^{2}AdV={\int }_D(\Delta A{)}^{2}dV$. Thus $A$ is harmonic with zero boundary value and $\boxed{A=0}$ by uniqueness of the Dirichlet problem. The normal-derivative condition is what removes the extra fourth-order boundary terms.

For the pressurized annulus, substituting the radial constitutive equations into equilibrium gives $u^{\prime \prime }+u^{\prime }/r-u/r^2=0$, so $u=Cr+D/r$. The stresses areUsing gauge stresses relative to ambient, the boundaries are ${\tau }_{rr}(a)=-P$, ${\tau }_{rr}(b)=0$. ThereforeandThe maximum occurs at the bore. Once the chosen yield criterion is reached, Hooke's law cannot describe the total strain alone: write total strain as elastic plus plastic strain and supplement the elastic stress law with a yield condition, a plastic flow rule and any required work hardening law. A plastic region starts near the bore and must be matched to the remaining elastic region. The exact conversion between a principal-stress difference and the quoted yield stress depends on whether the material criterion is Tresca or von Mises. This is the elastic stress concentration in a pressurized thick cylinder.

For antiplane displacement $u=(0,0,w)$ the only nonzero strain components are $e_{xz}=w_x/2$ and $e_{yz}=w_y/2$. Commutation of mixed partial derivatives gives the necessary local compatibility conditionThe supplied stress is symmetric, independent of $z$ and trace-free. Its first two equilibrium equations are automatic; its third givesaway from the origin. Antiplane shear has zero trace, so isotropic Hooke's law reduces to $\boxed{e=\tau /(2\mu )}$. HenceLocally this integrates to $w=\arg (x+iy)+C$, represented on a suitable branch by $\arctan (y/x)+C$. The strain condition is satisfied at every point of the punctured plane, but the displacement cannot be made globally single-valued there:A gradient of a globally single-valued smooth displacement would instead integrate to zero around any closed curve. Local compatibility is therefore not sufficient on this multiply connected region.

A physical construction is a Volterra screw dislocation: cut the cylinder along a radial half-plane ending on its axis, shift one cut face relative to the other parallel to the axis, and rejoin. The axial mismatch is the Burgers vector. In the given normalization its magnitude is $2\pi$; with a dimensional mismatch $b$ one uses $w=b\arg (x+iy)/(2\pi )$. An intact pristine solid cannot reach this state by a smooth single-valued elastic deformation alone. The axis is a dislocation core where the continuum elastic field is singular and must be regularized by microscopic structure or plastic deformation. On the punctured domain the differential condition holds locally; globally its nonzero circulation detects the defect. In a distributional description, that incompatibility is concentrated on the axis. This is local strain compatibility versus screw-dislocation circulation.

Multiply equilibrium by the displacement and integrate. Integration by parts and the traction condition giveThe stress is symmetric, so it contracts only with the symmetric part of $\nabla \mathbf{u}$, namely $e$. Hooke's law then gives $T:e=\lambda (\mathrm{tr}e{)}^2+2\mu e:e=2W$. HenceThe identity is the linear-elastic work balance: twice the stored quadratic energy equals the body-force and boundary-traction work paired with the final displacement.

In the completely elastic exterior, radial elastic compatibility makes $q+2t$ constant. The far-field stress vanishes, so $q+2t=0$ and $t=-q/2$. Radial force balance then gives $q^{\prime }+3q/r=0$, whence $q=C/r^3$. The pressure boundary condition is $q(a)=-P$: positive applied pressure is compressive. ThereforeFor completeness, the corresponding radial displacement field is $u=P{a}^3/(4\mu r^2)$, which gives these components through Hooke's law and decays at infinity.

The Tresca yield criterion becomesFor $P\ge 0$ this is largest at the cavity wall and decreases strictly with radius when $P>0$. The first yield point is therefore at $r=a$, where $3P/4={\tau }_Y$. HenceThe normalization matters: ${\tau }_Y$ in this paper is a shear yield stress, rather than the uniaxial yield strength.

The end displacement relative to the initial unit axial length isBy Hooke's law, the total resisting force supplied by the spring is $F_3=-K\delta$. To convert this resultant into the axial Cauchy stress tensor, divide by the current end cross-sectional area. The transverse total stretches are both ${\alpha }^{-1/2}$, so that area isUniform traction over that face consequently givesEquate this to the expression from part (ii):The extra factor $\alpha$ on the left is essential: the spring law specifies a total force, whereas the material constitutive equation specifies force per current area. The reference end area is one, so the same spring resultant can also be read as a nominal end traction.

Hooke's law and Newton's second law give $m\ddot{x}=-kx$, so ${\omega }_0=\sqrt{k/m}$. The initial conditions select the simple harmonic motionThe kinetic energy and elastic potential energy exchange periodically while their sum $E_0=k{x}_0^2/2$ is constant.

Take laboratory displacements from the equilibrium positions with the support at $f=0$. The left spring's dynamic extension is $x_1-f(t)$ and the spring between the masses has dynamic extension $x_2-x_1$. Hooke's law and Newton's second law giveEquivalently,The natural length sets the equilibrium separations but disappears from these displacement equations. The support is externally prescribed, so no additional support equation is required.

Let $x_1,x_2$ be displacements from equilibrium, with positive directions chosen alike. Hooke's law and Newton's second law giveFor a normal mode $x_j=a_j\cos \Omega t$, the determinant condition isSolving the quadratic for ${\Omega }^2$ givesThe angular frequencies are the positive square roots. Positive masses and spring constants make the stiffness quadratic form $k_1{x}_1^2+k_2(x_2-x_1{)}^2+k_3{x}_2^2$ positive definite, ensuring both squared frequencies are positive. Distinct spring constants do not require equal masses or any symmetry approximation.