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The small-slope excess length is $\int (w^{\prime }{)}^2/2$, so the first term is the work needed against uniform tension. The second is the potential energy of the downward distributed load. Their sum is total potential energy subject to the obstacle constraint.

Expand the difference exactly:The first two integrals together are nonnegative by part (b), and the square is nonnegative. ThusFor $T>0$ this also proves uniqueness: equality for two minimizers forces identical derivatives, and the fixed endpoints remove their additive constant. The convex admissible set and strictly convex energy make the variational inequality a global minimization condition.

Let $w$ be the difference of two sufficiently regular solutions of the wave equation with the prescribed data. By linearity, $w$ has zero initial displacement, zero initial velocity, and homogeneous Dirichlet boundary conditions. Define its wave energy byFor $\alpha =0$, differentiation under the integral sign and integration by parts yieldThe boundary term vanishes because $w=0$ at both fixed endpoints implies $w_t=0$ there. The zero initial data give $E(0)=0$, so $w_t=w_x=0$ for all times. Thus $w$ is spatially constant, and its zero boundary values give $\boxed{w=0}$. This proves uniqueness for the undamped wave equation.

Choose the positive weightsThese are the volumetric heat capacities. The heat equations and integration by parts giveThe exterior boundary terms vanish because the temperatures there are zero. At the thermal interface, the remaining term isby continuity of temperature and heat flux. Hence the weighted energy estimate isBoth weights are positive, so $I\ge 0$. Zero initial temperatures give $I(0)=0$, forcing $I(t)=0$ for all $t>0$. Each temperature therefore vanishes throughout its slab. Applying this argument to a difference also proves uniqueness for arbitrary common initial data in this thermal interface problem.

Substitution of $y=x^p$ into the homogeneous Euler-Cauchy equation gives $(p(p-1)+p-1)x^{p-2}=(p^2-1)x^{p-2}$, so $\boxed{p=1,-1}$. Use the solution $u(x)=x$ that vanishes at zero and $v(x)=1/x$ that decays at infinity. Their Wronskian is $u{v}^{\prime }-u^{\prime }v=-2/x$. The half-line Green function for the first cylindrical harmonic is thereforeIt is continuous at $x=s$, its first derivative has jump $1/2-(-1/2)=1$, and it solves the homogeneous equation elsewhere. Thus $L_{x}G=\delta (x-s)$ for integration with ordinary measure $ds$. The boundary conditions hold for each fixed $s>0$.

Integrating against the forcing givesSince the two integrals are $2-(x^2+2x+2)e^{-x}$ and $e^{-x}$, respectively,A Taylor expansion gives $y=-x/2+O(x^2)$ near zero, while $y\sim -1/x$ at infinity. This verifies the two endpoint limits; direct differentiation verifies $Ly=e^{-x}$ on $x>0$. The differential expression itself is singular at zero, so the endpoint condition is imposed by continuity. No nonzero combination of $x$ and $1/x$ satisfies both homogeneous endpoint conditions, proving uniqueness.

The only nonzero angular Fourier mode is $\cos (2\sigma )$. Its radial equation is $T^{\prime \prime }-4T=0$. A solution vanishing at $\tau =a$ is proportional to $\sinh [2(\tau -a)]$; making its value one at $b$ fixes the constant. The Dirichlet problem between bipolar coordinate circles therefore givesIts second derivatives in $\sigma$ and $\tau$ cancel, and the endpoint values are respectively zero and $\cos (2\sigma )$. The difference of any two solutions would be a harmonic function with zero Dirichlet boundary conditions, so the maximum principle proves uniqueness.

Geometric wording. In the printed bipolar coordinates, the circle radius $1/\sinh \tau$ decreases with $\tau$. For $0<a<b$, the $b$ circle is contained inside the $a$ circle. Thus the source’s “inner” and “outer” labels are interchanged geometrically. The solution above follows the explicit coordinate boundary values, which unambiguously specify the problem.

Write $Y={\sum }_{n\ge 0}{b}_n{y}_n$ in the complete Sturm-Liouville eigenfunction expansion. To determine $b_m$, multiply the differential equation by $y_m$ and integrate. The Sturm-Liouville operator has the symmetry identitybecause both functions obey the homogeneous Dirichlet boundary conditions. Consequentlyso that $({\lambda }_m-\alpha )b_m=a_m$. The stipulated exclusion of every eigenvalue makes all denominators nonzero, and givesThis is the resolvent operator applied to $f$ in the weighted space. The same projection shows uniqueness: a homogeneous solution would have $({\lambda }_m-\alpha )b_m=0$ for every $m$, hence would vanish by completeness. For the regular problem the formula gives the solution in the operator domain when $f$ is square-integrable.

Let $d=b-a$. In the mixed Dirichlet-Neumann problem in an elliptic annulus, the angular Fourier modes separate. The outer data excite the $\cos \nu$ mode, while the inner derivative excites $\cos 2\nu$.

For the first mode, a radial factor with zero derivative at $\mu =a$ is $\cosh (\mu -a)$. Normalizing its outer value gives $F_1(\mu )=\cosh (\mu -a)/\cosh d$. For the second mode, a radial factor vanishing at $\mu =b$ is $\sinh [2(b-\mu )]$. Its derivative at $a$ must be $-1$, fixing the coefficient to $1/[2\cosh (2d)]$. ThereforeEach term satisfies Laplace's equation; at $\mu =b$ the second term vanishes and the first is $\cos \nu$. At $\mu =a$, differentiating gives $0-\cos 2\nu$, as required. All other modes, including the constant mode, vanish by the homogeneous mixed conditions. Alternatively, integration by parts applied to the difference of two solutions gives a zero integral of $|\nabla \Phi {|}^2$, proving uniqueness.

Boundary convention. The calculation follows the explicitly printed coordinate derivative ${\partial }_{\mu }\Phi$. In elliptic coordinates, the scale factor is $h=\sqrt{{\sinh }^2\mu +{\sin }^2\nu }$ and the outward unit normal at the inner boundary points toward decreasing $\mu$. A physical normal derivative there would be $-h^{-1}{\partial }_{\mu }\Phi$. Thus the phrase “normal derivative” in the source must be understood in its displayed coordinate sense; imposing the same numerical data on a unit-normal derivative would be a different problem.

Write the equation as $(L+\beta )y=-\cos x/x$ with the positive Sturm-Liouville operator from part (a). A Sturm-Liouville eigenfunction expansion therefore gives $a_n=-f_n/(4n^2+\beta )$, where weighted orthogonal projection givesHere the product-to-sum formulas reduce the integral to half the sum of the integrals of $\sin ((2n+1)x)$ and $\sin ((2n-1)x)$. ConsequentlyAt $x=0$, use the continuous limits of the terms. The coefficients are $O(n^{-3})$, so the resulting endpoint series converges. Completeness is the ordinary Fourier sine series completeness after $v=xy$. For an independent check, solving $v^{\prime \prime }-\beta v=\cos x$ with zero endpoint values givesIts numerator vanishes at zero, and it has the same orthogonal projection coefficients. Positivity of $L+\beta$ proves uniqueness.

Set $u=\cos \theta$. The boundary data simplify to $u+2u^2=\frac{2}{3}{P}_0(u)+P_1(u)+\frac{4}{3}{P}_2(u)$. The axisymmetric harmonic extension from a sphere keeps only decaying radial powers outside and regular powers inside. HenceandEach Legendre polynomial coefficient agrees at $r=a$. The exterior solution tends to zero, and the interior solution is finite at $r=0$. The maximum principle for harmonic functions proves uniqueness in these respective classes.

Take the trace first: the antisymmetric tensor has zero diagonal and $X$ is traceless, so $T_{ii}=3t$. Taking symmetric and antisymmetric parts then gives the scalar, symmetric-traceless and axial tensor decomposition:The Kronecker delta satisfies ${\delta }_{ii}=3$, which verifies $X_{ii}=0$. Substitution verifies the original decomposition and also proves its uniqueness.

Write $w=u-v$ and $r=f-g$. Subtraction of the two boundary value problems gives $-w^{\prime \prime }+w+u^3-v^3=r$, with $w(0)=w(1)=0$. Multiply by $w$ and apply integration by parts:Consequently the required energy identity isThe nonlinear term is nonnegative pointwise, sinceThis is the monotonicity energy estimate for a semilinear Dirichlet problem for the monotone function $s\mapsto s^3$. Apply Young's inequality for products, $rw\le (r^2+w^2)/2$, inside the last integral and discard the nonnegative nonlinear term. Moving $\frac{1}{2}{\int }_0^1{w}^{2}dx$ to the left givesIn particular, equal forcing gives $w=0$, proving uniqueness and a quantitative continuous dependence on forcing estimate for this boundary value problem.

Let $f_i=f(ih)$ and $g_i=g(ih)$, and define the scaled forward difference operator, backward difference operator and second-order central difference byThe two finite difference methods are the nonlinear systemswith the prescribed Dirichlet boundary conditions $U_0=V_0=A$ and $U_N=V_N=B$. These equations determine the interior nodal values.

For clarity distinguish the interior and edge inner products and their norms:The bracketed norm includes the last edge $i=N$. Let $W=U-V$, so $W_0=W_N=0$, and put $q_i=D_x^{-}{W}_i$ for $1\le i\le N$. Direct telescoping gives discrete summation by parts:Subtract the two nonlinear systems, multiply each equation by $h{W}_i$ and sum. The resulting discrete energy identity isThe nonlinear summands are nonnegative by the factorization in (a). Using Young's inequality for products at every node givesThis also proves uniqueness of each discrete solution. Existence can be seen by minimizing, with the endpoints fixed, the strictly convex energy functionalIt is a coercive function of its finitely many interior values, so it attains a minimum. Its first derivatives are $h(-D_x^{+}{D}_x^{-}{W}_i+W_i+W_i^3-f_i)$; thus its minimizer solves the scheme. The same argument applies to $g$.

By the spectral theorem a symmetric positive-definite matrix $C$ has $C=Q\mathrm{diag}(c_i)Q^{\mathsf{T}}$, with $Q$ orthogonal and $c_i>0$. Define $S=Q\mathrm{diag}(\sqrt{c_i})Q^{\mathsf{T}}$. It is symmetric positive-definite and $S^2=C$. For uniqueness, any symmetric positive-definite square root $T$ commutes with $C=T^2$, hence preserves each eigenspace of $C$. On its $c$-eigenspace, $T$ is symmetric positive-definite and all its eigenvalues have square $c$, so all equal $\sqrt{c}$. Therefore $T=S$, including repeated eigenspaces.

For $\det F>0$, putThen $R^{\mathsf{T}}R=I$ and $\det R=\det F/\det U>0$, so $R\in SO(3)$; also $F=RU=VR$. Any other right polar decomposition has $F^{\mathsf{T}}F=U^2$, so uniqueness of the positive root fixes $U$ and then $R$. The left factor is consequently fixed too.

The principal stretches $v_i$ are the positive eigenvalues of $U$, equivalently the singular values of $F$. For a unit vector $N={\sum }_i{c}_i{N}_i$ in an orthonormal eigenbasis of $U$,Thus $\boxed{\underset{|N|=1}{\max }|FN|=v_{\max }}$, attained precisely by the unit vectors in the eigenspace of $U$ belonging to $v_{\max }$. With a simple largest eigenvalue these are its two opposite unit eigenvectors; degeneracy allows every unit combination within that eigenspace.

For the simple shear,A line with nonzero direction $N$ keeps its length exactly whenHence $\boxed{N_2=0\text{or}2N_1+\gamma N_2=0}$, with $N_3$ arbitrary. Every segment in the cube parallel to one of these directions preserves length; there is no restriction on its starting point other than remaining in the cube.

The two in-plane stretches are $v_{\pm }=(\sqrt{4+{\gamma }^2}\pm \gamma )/2$, and the third is one. A maximizing direction for $\gamma >0$ is $N\parallel (1,v_{+},0)$, since the $C$ eigenvalue equation gives $N_2/N_1=(v_{+}^2-1)/\gamma =v_{+}$. Therefore $\boxed{N\to \pm (1,1,0)/\sqrt{2}}$ as $\gamma \downarrow 0$. At exactly zero shear all directions tie; the limit refers to the selected maximizing direction for positive shear. These are the length-preserving directions and principal directions of simple shear.

Cauchy's stress hypothesis says that traction across a cut depends locally on its point, time and oriented normal, rather than on the global shape of the cut. Here all areas and normals are referential: $E$ is a fixed material region in reference coordinates, ${\rho }_R$ its mass density per reference volume, $v$ the current particle velocity, $b$ body force per unit mass, and $s_R$ force per unit reference area. The time derivative on the left is total material momentum rate; the two right terms are contact and body forces.

A shrinking pillbox gives $s_R(-N)=-s_R(N)$. In a shrinking tetrahedron the volume contributions scale as its linear size cubed, whereas face forces scale as size squared. Their leading balance expresses traction on an arbitrary face as the normal-component-weighted sum of tractions on three coordinate faces. Defining those three tractions as tensor columns gives $\boxed{s_R(N)=T_{R}N}$. Applying the divergence theorem to momentum balance and localizing over arbitrary $E$ then yieldsThe referential tensor is the first Piola-Kirchhoff stress and need not be symmetric; it should not be silently identified with the spatial Cauchy stress.

Incompressibility is $\boxed{\det F=1}$. For a homogeneous hyperelastic material the referential constitutive relation becomeswhere the undetermined scalar pressure $p$ enforces the constraint. Homogeneity removes explicit reference-position dependence from $W$, not the possible spatial variation of that pressure.

For the neo-Hookean energy, $DW=2\alpha F$. For the positive diagonal ansatz the nominal principal stresses are $t_i=2\alpha v_i-p/v_i$. Since $F$ is constant, equilibrium requires $\mathrm{cof}F\nabla p=0$, so $p$ is constant. Stress-free endfaces give $\boxed{p=2\alpha v_3^2>0}$. The side normals include $(0,-1,0)$ and $\pm (\sqrt{3}/2,1/2,0)$, so their prescribed tractions force $\boxed{t_1=t_2=T}$.

For fixed positive $p$, the scalar function $2\alpha v-p/v$ is strictly increasing on $v>0$. Thus equality of the two nominal stresses implies $v_1=v_2=v$. Incompressibility gives $v_3=v^{-2}$, and substitution givesThe load function has derivative $2\alpha (1+5v^{-6})>0$, tends to minus infinity as $v\downarrow 0$ and to infinity as $v\to \infty$. It therefore determines exactly one positive $v$ for every specified load, with $v\ge 1$ for outward tensile $T\ge 0$. The resulting constant stress satisfies equilibrium and every prescribed face traction, proving existence and uniqueness within the requested positive diagonal class. It does not claim uniqueness among all possible nonhomogeneous deformations. This is the equibiaxial dead loading of a neo-Hookean prism.

Put $n=x/R$ and $q=r(R)/R$. Differentiating $y=qx$ uses $\nabla R=n$ and $\nabla q=(r^{\prime }-q)n/R$. HenceIts radial eigenvalue is $r^{\prime }$ and its two tangential eigenvalues are $q$. They are positive by hypothesis, so they are also the principal stretches:For an isotropic stored energy the nominal stress has the same radial/tangential projectors,where symmetry of $\Phi$ gives equal tangential derivatives. Since $\mathrm{Div}(n\otimes n)=2n/R$ and the scalar coefficients depend only on $R$,Thus equilibrium is exactly $\boxed{d{\Phi }_1/dR=2({\Phi }_2-{\Phi }_1)/R}$.

For the particular energy define $\delta =r^{\prime }{q}^2=\det F>0$. Its two derivatives areAlso $q^{\prime }=(r^{\prime }-q)/R$. Differentiating the first formula and substituting the equilibrium equation givesThe last terms cancel. Strict positivity of $h^{\prime \prime }$ and $q$ yields $\boxed{{\delta }^{\prime }=0}$, so $r^2{r}^{\prime }=D{R}^2$ for one positive constant $D$. Integration gives $r^3=D{R}^3+C$.

The displacement data mean $r(1)={\lambda }_1$ and $r(2)=2{\lambda }_2$, not $r(2)={\lambda }_2$. Solving the two scalar equations givesThe given inequality makes $D>0$. Since $r^3$ increases from ${\lambda }_1^3>0$ to $8{\lambda }_2^3$, it is positive on the whole shell, and $r^{\prime }=D{R}^2/r^2>0$. Thus the formula is smooth and admissible and satisfies equilibrium. Conversely every admissible radial equilibrium had to have this constant determinant and these two integration constants, proving uniqueness. This is a constant-Jacobian radial equilibrium for a stretch-sum energy.

A deformation maps a material point in the reference body to its current position, $y:\Omega \to {\mathbb{R}}^3$, with deformation gradient $F=Dy$. A homogeneous deformation has constant gradient and is affine, $y(X)=FX+b$.

If $\det F>0$, the matrix $F^{T}F$ is symmetric positive definite. The matrix-square-root theorem says that any such matrix has a unique symmetric positive definite square root, obtained by taking positive square roots in its orthogonal spectral decomposition. SetThen $R^{T}R=U^{-1}{F}^{T}F{U}^{-1}=I$ and $\det R=\det F/\det U>0$, so $R\in \mathrm{SO}(3)$ and $F=RU$. If also $F=\tilde{R}\tilde{U}$, then $F^{T}F={\tilde{U}}^2$; uniqueness of the positive square root gives $\tilde{U}=U$ and then $\tilde{R}=R$. This proves existence and uniqueness of polar decomposition.

The principal stretches are the positive eigenvalues of $U$, equivalently the singular values of $F$ or square roots of the eigenvalues of $F^{T}F$.