The kinematic relation is unchanged: $\alpha =1$ before contact and $\alpha =\lambda /(1+ct)$ afterwards. Therefore the nonlinear constitutive equation givesThe first term in the post-contact expression is $O(t^{-2})$. Hence its asymptotic expansion isIn this model the compressive axial stress diverges in magnitude, because the $-{\alpha }^{-1}$ term penalizes vanishing elastic stretch ratio without bound.

Write $\lambda =L_1/L_0>1$. The one-dimensional multiplicative decomposition of the deformation gradient reads $\ell /L_0=\alpha \gamma$. Before contact, $\alpha =1$; afterwards, the imposed fixed length givesSubstituting in the linear constitutive equation yieldsHere negative axial stress means compression. The two branches agree at $t_c$ because $1+c{t}_c=\lambda$. At late times,Thus the law permits arbitrarily severe elastic shortening at a finite compressive stress. It is a small-strain approximation, appropriate near $\alpha =1$, and is not a credible material law as $\alpha \to 0$. The one-dimensional constraint rules out buckling in this particular model; the finite limiting stress is a defect of extrapolating the linear constitutive equation, even with that constraint retained.

Before contact with both walls, the elastic rod can grow freely, so its elastic stretch ratio is $\alpha =1$ and its length is its stress-free grown length,The first contact occurs when $\ell (t_c)=L_1$. For the intended growth case $c>0$,The condition $L_1>L_0$ ensures $t_c>0$. If $c=0$, contact never occurs; negative $c$ describes shrinkage rather than the stated continuing growth.

Here a prime denotes $d/d\sigma$. It is useful to retain terms through $O({ϵ}^2)$ in the normal mechanical equilibrium equation, even though only first-order shape corrections are sought. The undeformed elastic membrane is tensionless, so a first-order change of curvature contributes to normal force only at second order.

The geometric equations written in reference arc length areTheir first-order Taylor expansions giveand henceThe principal stretches and the first-order constitutive equation areBecause $t_s=t_{\phi }=t$ exactly, the tangential mechanical equilibrium equation gives ${\partial }_{s}t=0$. Thus both $t_1$ and $t_2$ are spatial constants: $t_1^{\prime }=t_2^{\prime }=0$.

Expand the geometric factors in the normal balance:ConsequentlyMatching the first-order coefficient gives the requested relationSince the prescribed pressure has no second-order term, and ${\theta }_1^{\prime }=-r_1^{\prime \prime }$, its second-order coefficient givesA closed boundary value problem for the three requested unknowns is thereforeThe first equation has order two, the second order one, and the last order one, so the four boundary conditions match the total order. The unknown constant $t_2$ is selected by the axial endpoint constraint, rather than by an additional specified traction.

For $p\ne 0$, an equivalent formulation eliminates $z_1$ and displays that constraint as an integral:After solving this system, recover $z_1$ by integration from $z_1(0)=0$. No second derivatives of the unspecified function $f$ are required: they enter the next-order shape corrections through the second-order constitutive equation, whereas the displayed normal balance already determines the leading shape. The case $p=0$ has the undeformed solution $r_1=z_1=t_2=0$.

Take $s=\sigma$ up to an irrelevant additive constant, $r=a$ constant, $z=z_0-\sigma$, and $\theta =\pi /2$. These satisfy the geometric relations and give the principal stretchesThe constitutive equations therefore produce constant membrane tensionsThe tangential mechanical equilibrium equations hold automatically, while the normal equation reduces toEvery positive root of this algebraic equation provides a cylindrical solution with no axial stretch ratio change. There is no further constraint on $t_s$ in the interior equations when end boundary conditions are discarded. In particular, a closed-end pressure balance would impose additional axial loading data that are absent here. Existence of a positive root for an arbitrary specified $P$ depends on the given constitutive equation.

The first term is the bending energy of rod 1, whose preferred bending strain of a rod is zero. The second is the bending energy of rod 2 relative to its preferred intrinsic curvature of an elastic filament $\hat{u}$. The third is its stretching energy, with stretching stiffness $k{B}_2$ and stress-free stretch ratio ${\alpha }^{+}=1$. Each term is an energy per unit reference arc length before integration.

Substitute $u^{+}=u$ and ${\alpha }^{+}-1=-au$. The resulting quadratic function of $u$ has densityThere are no derivatives of $u$ or imposed end constraints in this minimum potential energy principle in elasticity. Sincethe unique pointwise minimizer is also the constant global minimizer:For example, completing the square giveswhich also proves uniqueness, up to equality almost everywhere, without merely assuming a uniform curvature.

With $k$, $a$ and $\hat{u}$ fixed, the two limits areIn the first limit, rod 1 dominates the bending energy and suppresses the bend. In the second, rod 1 offers negligible bending resistance, but following rod 2's preferred bending strain of a rod still shortens rod 2 through the bonded geometry. The competition between its bending energy and stretching energy leaves $u_{\ast }<\hat{u}$ when $\hat{u}>0$ and $k{a}^2>0$. Only when that stretching penalty also vanishes does $u_{\ast }\to \hat{u}$.

These formulae describe an admissible bonded elastic rod provided $1-a{u}_{\ast }>0$. If parameters instead force $a{u}_{\ast }\ge 1$, the unconstrained quadratic minimizer is outside the kinematic regime of the model; the given energy does not itself enforce positive stretch ratio.

Because the two elastic rods each have thickness $a$, their centrelines are separated by $a$ along the common director ${\mathbf{d}}_1$. The bonded geometry isDifferentiate with respect to reference arc length. The first rod is an inextensible rod, and the printed director kinematics of a rod giveFor the physically admissible orientation-preserving configuration $1-au>0$, the rods have the same unit tangent ${\mathbf{d}}_3^{+}={\mathbf{d}}_3$ and the same planar material frame of a rod. ThereforeIn particular, the second rod is shortened on the inside of a bend with $u>0$.

The bending strain of a rod is defined here by differentiation with respect to reference arc length. Since the reference labels match,The distinction from spatial curvature matters: the second centreline hasThus $u/(1-au)$ is its signed spatial curvature, rather than the $u^{+}$ appearing in the given elastic energy.

Zero applied end force and the force balance ${\mathbf{n}}^{\prime }=0$ imply $\mathbf{n}(S)=0$ everywhere. The moment balance then reduces to ${\mathbf{m}}^{\prime }=0$.

Choose an untwisted straight configuration: all three directors are constant, ${\mathbf{r}}^{\prime }={\mathbf{d}}_3$, and $\mathbf{u}=0$. The constitutive equation giveswhich is constant in space and therefore satisfies the moment balance. The required end bending moment is consequentlyIts sign opposes the positive intrinsic curvature of an elastic filament. If the initial position and frame from part (ii) are retained, this solution is simply $\mathbf{r}(S)=S{\mathbf{e}}_3$ with ${\mathbf{d}}_i={\mathbf{e}}_i$; the clamp supplies the opposite reaction moment at the other end.

The second moment of area about an axis weights each element of cross-sectional area by its squared perpendicular distance from that axis. The ellipse extends farther along ${\mathbf{d}}_2$ than along ${\mathbf{d}}_1$, so its material lies, on balance, farther from the ${\mathbf{d}}_1$ axis. ThereforeWith zero intrinsic curvature of an elastic filament, the respective bending moments required by the constitutive equation are $E{I}_1\kappa {\mathbf{d}}_1$ and $E{I}_2\kappa {\mathbf{d}}_2$. Thus their magnitudes satisfyfor $\kappa \ne 0$: bending about the minor axis is harder. Here $E$ is Young's modulus, and $E{I}_i$ is the corresponding bending stiffness.

The right-handed material frame of a rod has ${\mathbf{d}}_2\times {\mathbf{d}}_1=-{\mathbf{d}}_3$ and ${\mathbf{d}}_2\times {\mathbf{d}}_3={\mathbf{d}}_1$. Hence the director kinematics of a rod reduce toSolving these constant-coefficient ordinary differential equations with the given initial frame givesThe inextensible rod has ${\mathbf{r}}^{\prime }={\mathbf{d}}_3$. An integration using $\mathbf{r}(0)=0$ yields, for $\hat{u}\ne 0$,Indeed,so the centreline is a circular arc of radius $1/|\hat{u}|$ in the plane perpendicular to ${\mathbf{e}}_2$. The continuous $\hat{u}\to 0$ limit is ${\mathbf{d}}_1={\mathbf{e}}_1$, ${\mathbf{d}}_3={\mathbf{e}}_3$, and $\mathbf{r}=S{\mathbf{e}}_3$.

The material frame of a rod obeys the director kinematics of a rodIn the stress-free reference configuration, the constitutive equation gives $u_1=0$, $u_2=\hat{u}$, and $u_3=0$, since all three stiffnesses are positive. Thus $\mathbf{u}=\hat{u}{\mathbf{d}}_2$. Its cross product with ${\mathbf{d}}_2$ vanishes, soConsequently this director is a fixed vector in space, even though the other two directors rotate along the elastic rod.

Over the complex numbers, the fundamental theorem of algebra makes every irreducible polynomial factor $f_j(x)$ equal to $x-{\lambda }_j$. Part (b)(iv) giveswith distinct eigenvalues. The primary decomposition theorem for linear maps consequently expresses $V$ as the direct sum of the actual eigenspaces $E_{{\lambda }_j}=\ker (T-{\lambda }_{j}I)$, with no higher generalized eigenspaces needed. Part (ii) makes these summands mutually orthogonal.

Apply the Gram-Schmidt process to a basis within each eigenspace. Its linear combinations stay within that eigenspace, so all resulting vectors are still eigenvectors. Their union is orthonormal by part (ii) and spans $V$ by the primary decomposition theorem for linear maps. Thusfor an orthonormal basis $\mathcal{B}$ of eigenvectors. This proves the finite-dimensional complex spectral theorem for normal operators by the requested decomposition argument.

Let $u\in \ker (T-\lambda I)$ and $w\in \ker (T-\mu I)$, with $\lambda \ne \mu$. By part (i) and the adjoint operator identity,The last step uses conjugate-linearity in the second argument of the inner product. Therefore $(\lambda -\mu )⟨u,w⟩=0$, proving the orthogonality of the two eigenspaces:

Take $N=T-\lambda I$. It is a polynomial in the normal operator $T$, so part (b)(iii) makes $N$ a normal operator. Also $N^{\ast }=T^{\ast }-\bar{\lambda }I$. If $Tv=\lambda v$, part (b)(i), applied to $N$, givesThe norm vanishes only at the zero vector, soThis is the adjoint eigenvector identity for a normal matrix expressed independently of coordinates.

Suppose some exponent $q_j\ge 2$. DefinePart (iii) makes $N$ a normal operator. For every $v\in V$, put $w=h(T)v$. The minimal polynomial identity givesPart (ii), applied to $N$, forces $Nw=0$. As this holds for every $v$,This contradicts minimality: $m_T/f_j$ is a monic polynomial of smaller degree annihilating $T$. Therefore the minimal polynomial of a normal operator is a square-free polynomial over either of the given fields:This argument does not require the irreducible polynomials to be linear, so it also covers the real case.

For a polynomial $f(x)={\sum }_{j=0}^d{a}_j{x}^j$, the defining identity and uniqueness of the adjoint operator giveIndeed, the adjoint operator of a sum is the sum of the adjoint operators, the adjoint of $aT$ is $\bar{a}{T}^{\ast }$, and the adjoint of a product $AB$ is $B^{\ast }{A}^{\ast }$. Since the normal operator $T$ commutes with $T^{\ast }$, each power of $T$ commutes with each power of $T^{\ast }$. Multiplying the two finite sums therefore givesHence polynomials of a normal operator are normal, including constant polynomials.

Let $w=Tv$. If $T^{2}v=0$, then $Tw=0$, and part (i) gives $\Vert T^{\ast }w\Vert =\Vert Tw\Vert =0$. Thus $T^{\ast }w=0$ andThe positive definiteness of the inner product gives $w=0$. Since the reverse kernel inclusion always holds, we obtainThis is kernel stabilization for a normal operator.

For the normal operator $T$, the adjoint operator identity and $T^{\ast }T=T{T}^{\ast }$ giveFor the last equality, $⟨T{T}^{\ast }v,v⟩=⟨T^{\ast }v,T^{\ast }v⟩$ and complex conjugation allow reversal of the two arguments. Taking the nonnegative square roots provesIn particular the two kernels coincide.

If $R$ and $S$ both satisfy the defining identity of an adjoint operator, thenfor every $u,w\in V$. Choosing $u=(R-S)w$ gives $\Vert (R-S)w{\Vert }^2=0$. The positive definiteness of the inner product forces $(R-S)w=0$ for every $w$, proving

An adjoint operator of $T:V\to V$ is a linear map $T^{\ast }:V\to V$ satisfyingThis identity transfers $T$ from the first input of the inner product to the second.