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A circular elastic plate with bending stiffness $B$, constant excess pressure $p$ and clamped boundary conditions at radius $R$ obeys $B{\nabla }_r^{4}h=p$, where ${\nabla }_r^2=r^{-1}{\partial }_r(r{\partial }_r)$. Its regular solution is $h=p(R^2-r^2{)}^2/(64B)$. Consequently its enclosed volume is $V=\pi p{R}^6/(192B)$, and its edge curvature is $h_{rr}(R)=p{R}^2/(8B)=24V/(\pi R^4)$.

A normal mode of a uniformly bending elastic filament fixed in position and slope at both ends satisfies $W^{\prime \prime \prime \prime }=k^{4}W$. With $\beta =kL>0$, the allowed wavenumbers solve $\cos \beta \cosh \beta =1$. The first root is $4.730040745$ and ${\beta }_n=(n+\frac{1}{2})\pi +O(e^{-(n+1/2)\pi })$ for $n\ge 1$. The apparent root at zero is not an eigenfunction: the zero-eigenvalue cubic polynomial satisfying all four clamped boundary conditions is identically zero. This spectrum differs from that of a filament clamped only at one end.

For $L=D^4+q(x)$ where $q$ is a real continuous function on $[a,b]$, integration by parts givesThe potential terms cancel, and differentiating the boundary expression verifies the identity. Under clamped boundary conditions $u=u^{\prime }=v=v^{\prime }=0$ at both endpoints, this form vanishes. Conversely, requiring its vanishing against every clamped test function forces the adjoint function and its first derivative to vanish at both endpoints, because the second and third test-function derivatives are independent endpoint traces. Thus, on a regular finite interval with the usual fourth-order Sobolev domain, the clamped realization is self-adjoint. The equality of domains is part of this conclusion; formal self-adjointness of the expression alone does not supply it.

Use the usual inner product $⟨u,v⟩={\int }_0^1\overline{u}vdx$. Moving four derivatives by integration by parts leaves $L^{\ast }=D^4+1$ and gives the fourth-order boundary identityThus the differential expression is formally self-adjoint. If both functions satisfy the clamped boundary conditions, every displayed endpoint product vanishes, proving symmetry on the specified domain.

To check full self-adjointness, also identify the adjoint domain. Allow $v$ to run through the clamped domain. At either endpoint, $v^{\prime \prime }$ and $v^{\prime \prime \prime }$ can be chosen independently, for instance using a polynomial near that endpoint multiplied by a smooth cutoff function. The remaining boundary form is $[\overline{u}{v}^{\prime \prime \prime }-\overline{u^{\prime }}{v}^{\prime \prime }{]}_0^1$. It vanishes for every such $v$ precisely when $u=u^{\prime }=0$ at both endpoints. These are exactly the original boundary conditions. On the regular finite interval, the adjoint domain consequently equals $\{u\in H^4(0,1):u(0)=u^{\prime }(0)=u(1)=u^{\prime }(1)=0\}$, rather than a larger domain. Here $H^4$ is the Sobolev space of functions whose weak derivatives through order four are square-integrable.

The operator together with these boundary conditions is fully self-adjoint. As a further consistency check, integration by parts gives $⟨y,Ly⟩={\int }_0^1(|y^{\prime \prime }{|}^2+|y{|}^2)dx>0$ for every nonzero admissible $y$. The reusable calculation is the Lagrange identity for a fourth-order differential operator.

With zero tension and zero distributed load, a free interval satisfies $w^{\prime \prime \prime \prime }=0$. Before contact the entire elastic beam is free, so $w$ is a cubic polynomial. The four end boundary conditions determine it uniquely. Reflection $x\mapsto -x$ preserves those boundary conditions, so the unique solution is an even function and its odd terms vanish. Direct substitution givesFor $\alpha \ge 0$, its minimum occurs at $x=0$ and is $-\alpha L/2$. The constraint $w\ge -H$ is therefore satisfied exactly when $\alpha \le 2H/L$. The first contact is at the midpoint:At equality the normal contact force is still zero. The uniqueness used here can also be checked by integrating $v^{\prime \prime \prime \prime }v$ for the difference of two solutions with homogeneous clamped boundary conditions, which gives ${\int }_{-L}^L(v^{\prime \prime }{)}^{2}dx=0$ and hence $v=0$.

Put $B=E{I}_2$, the filament bending modulus in the easier bending direction; $I_2$ is the relevant second moment of area. Take filament tension as positive in extension. A cut at height $z$ supports the weight of the material above it, so its axial force is compressive:The small transverse displacement induces a transverse tension force density $(T{X}_z{)}_z$, while bending produces $-B{X}_{zzzz}$. Their balance givesThe same result follows from elastic energy and gravitational energy. To quadratic order, local inextensibility shortens the height of the material point labelled $z$ by $\frac{1}{2}{\int }_0^z{X}_z(s{)}^{2}ds$. Integrating its weight and interchanging the integrals gives the second-order potential energyTwo integration by parts steps in its first variation giveThe clamped boundary conditions fix displacement and slope at the base, and the unconstrained tip must have zero bending moment and transverse end force. Thus the complete four conditions areSince the filament tension vanishes at the tip, the last condition is $X_{zzz}(L)=0$. Integrate the ordinary differential equation once: $B{X}_{zzz}-T{X}_z$ is constant, and its tip value makes that constant zero. For the slope $u=X_z$, it follows thatThe other tip condition $u_{zz}(L)=0$ already follows from this equation for a regular solution; it is not a third independent condition on the second-order slope problem. Finally reconstruct $X(z)={\int }_0^{z}u(s)ds$ to impose the base position.

Set $x=L-z$, $\alpha =\lambda g/B$, and $\eta =(2/3)\sqrt{\alpha }{x}^{3/2}$. Then $d\eta /dx=3\eta /(2x)$ and $d^2\eta /d{x}^2=3\eta /(4x^2)$, so $u_{xx}+\alpha xu=0$ becomesSubstituting $u={\eta }^{1/3}F(\eta )$ givesThis is the Bessel differential equation of order $1/3$, whose two independent Bessel functions of the first kind giveThe tip is $x=\eta =0$. To decide which branch is admissible, retain both leading behaviours, not just the fact that $u$ stays finite:Since ${\eta }^2\propto x^3$ and ${\eta }^{2/3}\propto x$, the first branch has zero tip derivative, while the second has a nonzero tip derivative. The moment-free condition therefore sets $b=0$, and it does not require the tip slope itself to vanish. The base condition now requiresLet $j_{-1/3,1}$ be the first positive zero. The Bessel threshold for self-buckling of a vertical rod isTo identify this as the first loss of stability, write the quadratic energy in terms of the slope as $\frac{1}{2}[B\int u_z^{2}dz-\lambda g\int (L-z)u^{2}dz]$. On functions with $u(0)=0$, its critical coefficient is the minimum Rayleigh quotient $\int u_z^{2}dz/\int (L-z)u^{2}dz$; variation supplies the moment-free condition at the tip and the same Sturm-Liouville problem. The first zero gives the lowest eigenvalue, so the straight state is stable below this load, neutral at it, and unstable above it, with the easier $I_2$ direction buckling first.

Take $A>0$ and use primes for spatial derivatives. Two integration by parts operations in the first variation of the elastic filament's bending energy giveThus the Euler-Lagrange equation is $A{h}^{\prime \prime \prime \prime }=0$, and the fluctuation differential operator is $K=A{d}^4/d{x}^4$. In the L2 space inner product, its boundary form isThe conjugate endpoint trace pairs are $(h,h^{\prime \prime \prime })$ and $(h^{\prime },h^{\prime \prime })$. Requiring one member of each pair to vanish gives the four standard self-adjoint endpoint conditions for filament bending, applied at both ends:

Each pair annihilates the boundary form for all $u,v$ in the domain. Conversely, the remaining two endpoint traces can be chosen freely: requiring the boundary form to vanish against every such $u$ forces an adjoint-domain function $v$ to satisfy the same two conditions. This proves self-adjointness, rather than just formal symmetry, on the corresponding fourth-order Sobolev space domain.

The count four concerns these elementary homogeneous choices. Identical-end boundary conditions do not restrict all self-adjoint operators to these four possibilities. For example, $h=0$, $h^{\prime \prime }=b{h}^{\prime }$ at both ends, with any fixed real $b\ne 0$, also annihilates the boundary form: the remaining expression is $-\overline{u^{\prime }}b{v}^{\prime }+b\overline{u^{\prime }}{v}^{\prime }=0$. This Robin boundary condition supplies a continuous family beyond the four listed pairs.

For each of the four standard pairs, integration by parts givesConsequently the eigenvalues are nonnegative. For a positive eigenvalue ${\mu }_n=A{k}_n^4$, the eigenvalue equation is $W_n^{\prime \prime \prime \prime }=k_n^4{W}_n$. Its four characteristic roots are $\pm k_n,\pm i{k}_n$, soHere $C_j$ are coefficients, avoiding a collision with the filament bending modulus $A$. The regular finite-interval self-adjoint operator has compact resolvent; applying the spectral theorem for compact self-adjoint operators to a shifted inverse supplies a complete orthonormal basis of eigenfunctions.

For clamped boundary conditions at zero, a clamped--clamped bending mode can be writtenAt $L$, writing $\beta =kL$, the two remaining boundary conditions areThe determinant is $2(1-\cos \beta \cosh \beta )$. Hence the positive wavenumbers obeyEquivalently, intersect $\cos \beta$ with $\mathrm{sech}\beta$. Numerical root bracketing givesThe entire sequence has the useful large-$n$ descriptionIndeed, put $\beta =(n+\frac{1}{2})\pi +\delta$ in $\cos \beta =\mathrm{sech}\beta$ and use $\cos \beta =(-1{)}^{n+1}\delta +O({\delta }^3)$ and $\mathrm{sech}\beta =2e^{-\beta }+O(e^{-3\beta })$.

The apparent root $\beta =0$ in the determinant equation is spurious for the clamped problem: at zero eigenvalue, $W$ is a cubic polynomial, and its four clamped conditions force $W=0$. For other endpoint choices, zero-energy filament modes must be treated separately from the trigonometric formula. The free-free kernel consists of affine functions, the torqued-torqued kernel consists of constants, and the hinged-hinged kernel is trivial. Including those kernels is necessary for a complete eigenfunction expansion.

Choose real eigenfunctions with ${\int }_0^L{W}_n{W}_{m}dx={\delta }_{nm}$. For positive modes, self-adjointness and integration by parts diagonalize the energy:At temperature $T$, the canonical ensemble is a product of centered Gaussian distributions. The equipartition theorem, with Boltzmann constant $k_B$, givesThe resulting thermal covariance of an elastic filament isFor an unnormalized eigenfunction of squared L2 norm $N_n$, divide its summand by $N_n$. This normalization factor cannot be absorbed silently into the modal variance.

For clamped boundary conditions at both ends, the inverse of $d^4/d{x}^4$ has Green function, for $x\le y$,It is cubic on each side of $y$, satisfies the four clamped conditions, has continuous first two derivatives, and has unit jump in its third derivative. Thus ${\partial }_x^{4}G=\delta (x-y)$, and its eigenfunction expansion is the sum above. In particular,These finite variances require removal of every zero-energy filament mode. An unconstrained free-free elastic filament can translate and tilt at no energy cost; a torqued-torqued elastic filament can translate. Their unrestricted Boltzmann distributions are not normalizable, so the full displacement variance is undefined. Fix those rigid degrees of freedom before applying the positive-mode formula.

For one small transverse displacement, the elastic energy of the elastic filament iswhere $A$ is the filament bending modulus. Its first variation isThe clamped boundary conditions remove the left boundary terms. A force-free and torque-free tip permits independent $\delta h(L)$ and $\delta h^{\prime }(L)$, so the natural boundary conditions for a free endpoint are $h^{\prime \prime }(L)=h^{\prime \prime \prime }(L)=0$: zero bending moment and transverse shear.

There is an important distinction between an energy minimum and a normal mode. The unloaded higher-order Euler-Lagrange equation is $h^{\prime \prime \prime \prime }=0$, whose only solution with these four boundary conditions is $h=0$. A general fluctuating shape is a sum of modes, not one of the stated sinusoidal/hyperbolic functions. To obtain the clamped--free bending modes, extremize the bending Rayleigh quotient, or equivalently $U[W]-\frac{1}{2}A{k}^4\int W^{2}dx$ with a fixed norm. Its Euler-Lagrange equation isThis is the constrained variational characterization of bending modes. Its four characteristic roots are $\pm k,\pm ik$. Clamping gives $F=-D$ and $E=-B$, henceThe free-tip conditions reduce toIts determinant is $-2(1+\cos q\cosh q)$, so nontrivial modes requireNo zero mode exists, since a cubic satisfying the homogeneous clamp/free conditions is zero. The clamped-free bending spectrum starts with $q_1\in (\pi /2,2)$; bisection or Newton iteration gives $q_1\simeq 1.875104$, so $k_1\simeq 1.875104/L$. Choosing $D=N(\sin q_n+\sinh q_n)$ and $B=-N(\cos q_n+\cosh q_n)$ givesThe second boundary condition follows from the root equation, and $N$ is arbitrary until a normalization is chosen.

The bending operator with these boundary conditions is positive and self-adjoint. Twice integrating by parts givesSymmetry implies orthogonality when $n\ne m$. Expand $h(x)={\sum }_n{a}_n{W}_n(x)$ and use $I_n=L^{-1}{\int }_0^L{W}_n^{2}dx$. Thenby the equipartition theorem. The supplied endpoint identity therefore yields the thermal bending fluctuations of a clamped filament:To evaluate the sum, apply a tip force $f$ and minimize $U-fh(L)$. The modified free-end conditions are $h^{\prime \prime }(L)=0$ and $-A{h}^{\prime \prime \prime }(L)=f$, and $h^{\prime \prime \prime \prime }=0$ in the interior. Integration gives $h(x)=f{x}^2(3L-x)/(6A)$, so the tip-force compliance of a cantilever is $L^3/(3A)$. On the other hand, minimizing the modal energy minus $f{\sum }_n{a}_n{W}_n(L)$ gives $a_n=f{W}_n(L)/(AL{k}_n^4{I}_n)$ and henceComparison evaluates the fourth inverse-power sum of the cantilever spectrum without truncating the modes:The boxed variance is for the specified single transverse direction. An independent equilibrium check follows by differentiating the Gaussian partition function with respect to $f$: $\partial ⟨h(L)⟩/\partial f=⟨h(L{)}^2⟩/(k_{B}T)$ at zero load, reproducing the same result from the static compliance. Keeping only the first normal mode gives about $0.32356k_{B}T{L}^3/A$, roughly $97.1\%$ of the exact variance. With two independent equivalent transverse directions, their summed variance is twice the boxed result. The small-slope model requires $k_{B}TL/A\ll 1$, or $L$ small compared with the usual three-dimensional persistence length $A/(k_{B}T)$.

For an axisymmetric thin elastic plate, define ${\nabla }_r^2=r^{-1}{\partial }_r(r{\partial }_r)$. Uniform excess pressure and bending stiffness $B$ give $B{\nabla }_r^{4}h=p$. The roof is attached to undeformed rock at the front, giving clamped boundary conditions $h(R)=h_r(R)=0$, and regularity excludes singular displacement or curvature at $r=0$. A regular particular solution is $p{r}^4/(64B)$; adding a constant and a multiple of $r^2$ and enforcing the two edge conditions gives the axisymmetric clamped-plate deflectionIntegrating the magma intrusion volume,therefore yieldsHere the axisymmetric phrase requires the radial plate biharmonic operator; a Cartesian one-dimensional beam equation would not give the same pressure coefficient. Hydrostatic contributions are excluded as instructed.

For the early front, let $U_p=\dot{R}$ be the leading propagation speed. The fluid and fracture radii differ only within a local region of length $l_p\ll R$. In lubrication theory, the elastic pressure gradient near the front is of order $B{h}_f/l_p^5$, and the volume flux through the gap is of order $B{h}_f^4/(\mu l_p^5)$. In a frame following the fluid front that flux is of order $U_p{h}_f$. The viscous peeling of an elastic plate law is thereforeMatch the peeling region to the interior curvature: $h_f/l_p^2\sim {\kappa }_i$. The stated vapour-gap relation $h_f\simeq p_T{l}_p^4/(24B)$, with $p_T>0$ interpreted as the relevant pressure magnitude, then givesSubstitution gives the vapour-tip peeling of a magma intrusion speedWith constant supplied volume flux $Q$ and negligible initial intrusion volume, $V=Qt$. Integrating $R^{14}\dot{R}\sim B^{3/2}{Q}^{7/2}{t}^{7/2}/(\mu p_T^{1/2})$ gives $R^{15}\sim B^{3/2}{Q}^{7/2}{t}^{9/2}/(\mu p_T^{1/2})$, orOrder-one coefficients require the full local peeling profile. This early regime is a front-controlled asymptotic regime, not an assertion that a continuum peeling zone remains thin at arbitrarily small $t$.

When fracture toughness instead imposes fixed ${\kappa }_f>0$, equating it to the outer edge curvature gives the toughness-controlled magma intrusion lawIts pressure can be written in either useful formThe pressure decreases as the intrusion grows on this advancing branch.

If $V_{r0}$ is the reservoir volume before any magma was transferred to the initially negligible intrusion, volume conservation and negligible conduit storage give $V_r=V_{r0}-V$. With the supplied pressure-volume law and conduit conductance, the reservoir-fed magma intrusion obeysIn this law $E$ is an effective reservoir pressure-volume stiffness, with dimensions of pressure per volume. If the stated initial reservoir volume refers instead to the start of the late regime, when the intrusion already has volume $V_i$, replace $V_{r0}$ by the total $V_T=V_{r0}+V_i$ throughout. An independent value of $V_i$ would then be needed.

For an advancing solution, inflow ceases when the reservoir and intrusion pressures equalize. Its final volume is the larger positive root in $0<V<V_{r0}$ ofEquivalently, $y=\sqrt{V_{\infty }}$ solves $y^3-V_{r0}y+\mathcal{A}/E=0$. The larger root is essential: $\sqrt{V}(V_{r0}-V)$ rises from zero to a maximum $2V_{r0}^{3/2}/(3\sqrt{3})$ at $V=V_{r0}/3$ and then decreases. A nonempty interval of positive inflow exists only ifThere are then two positive equilibria $V_{-}<V_{r0}/3<V_{+}$. At equilibrium the derivative of the volume ordinary differential equation is $\beta E(V_{r0}-3V)/(2V)$, so $V_{-}$ is unstable and $V_{+}$ has asymptotic stability. An initial advancing intrusion with $V_{-}<V_i<V_{+}$ approaches $V_{\infty }=V_{+}$. The equality case has a double equilibrium and no positive-inflow interval; above the threshold there is no advancing equilibrium branch of this model.

The late formula $p\propto V^{-1/2}$ is singular at $V=0$ and cannot nucleate the intrusion: the earlier propagation regime must supply its finite seed. Nor does this fixed-edge-curvature propagation law prescribe how an irreversible fracture closes under negative inflow. The final-size answer is conditional on reaching the advancing late branch. In particular the reservoir is generally not emptied, and $\beta$ controls the approach rate rather than the equilibrium size.

With $q=p^{\prime }$ and $r=0$, the equation isMultiply by the real eigenfunction $y$ and integrate over $[a,b]$. The clamped boundary conditions $y=y^{\prime }=0$ at both endpoints eliminate all boundary terms, soSince $p(x)<0$, the right side is a sum of nonnegative terms. It cannot vanish for a nonzero eigenfunction: if $y^{\prime \prime }=y^{\prime }=0$, the boundary conditions force $y=0$. Thus

True. Two applications of Green's identities, or two integrations by parts, giveThe biharmonic equation makes the left side zero, while the two clamped boundary conditions $u={\partial }_{n}u=0$ annihilate the boundary integral. Consequentlyso $\Delta u=0$. Apply the first of Green's identities once more with $\varphi =\psi =u$: because $u=0$ on the boundary, it gives ${\int }_V|\nabla u{|}^2=0$. By path connectedness, $u$ is constant, and the zero boundary value implies $\boxed{u=0}$. This argument assumes the classical differentiability needed for the displayed integrations by parts.

The dispersion relation shows that an infinite rod has unstable long-wave modes. A finite rod excludes sufficiently long wavelengths, so stability depends on its length and the precise end conditions. The fourth spatial derivative requires two conditions at each end; fixing transverse positions alone does not specify whether slopes are constrained.

First take displacement-fixed ends with free rotation. Setting $\varphi (0)=\varphi (l)=0$ leaves slope variations free; the bending boundary term in the first variation then gives the corresponding natural boundary conditions on the slope, ${\varphi }_{xx}(0)={\varphi }_{xx}(l)=0$. The spatial normal modes are $\sin (n\pi x/l)$, andThe first mode destabilizes first. ConsequentlyAt equality, the first mode has zero restoring force and is marginal, with a static component and a possible linear-in-time component in the linear theory. Every integer $n<l\sqrt{\alpha /\beta }/\pi$ gives an exponentially growing mode.

If "fixed" instead means clamped, the clamped boundary condition is $\varphi ={\varphi }_x=0$ at both ends. The critical compression is thenFor example, the static critical shape $1-\cos (2\pi x/l)$ obeys the four clamp conditions and the zero-frequency equation at this load. To establish that it is the first threshold, write $v={\varphi }_x$: it has zero endpoints and zero mean, and periodic extension gives ${\int }_0^l{v}_x^{2}dx\ge (2\pi /l{)}^2{\int }_0^l{v}^{2}dx$ by the zero-mean periodic Poincaré inequality, with equality at $v\propto \sin (2\pi x/l)$. This proves the critical Rayleigh quotient. Both cases are instances of the compressed Euler-Bernoulli beam and its endpoint-dependent buckling threshold. The source does not specify the endpoint rotation, so both interpretations are stated explicitly.

For real coefficient functions, the fourth-order scalar operatoris formally self-adjoint exactly when $q=p^{\prime }$. Indeed, $(p{D}^2{)}^{\ast }=p{D}^2+2p^{\prime }D+p^{\prime \prime }$ and $(qD{)}^{\ast }=-qD-q^{\prime }$. Under clamped boundary conditions, its quadratic form is