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The biharmonic equation is ${\nabla }^{4}f=0$, where the biharmonic operator is the square of the Laplacian. In planar Stokes flow, taking the curl of the momentum equation and using a Cartesian streamfunction produces this equation. Its decaying spatial Fourier series modes have form $(A+By)e^{-|k|y}{e}^{ikx}$.

A local real scalar field with Lagrangian density $\frac{1}{2}[({\partial }_t\varphi {)}^2-{\kappa }^2({\nabla }^2\varphi {)}^2-m^2{\varphi }^2]$ has the equation of motion ${\partial }_t^2\varphi +{\kappa }^2{\nabla }^4\varphi +m^2\varphi =0$. Two spatial integrations by parts produce the positive biharmonic operator in this equation. Its Hamiltonian is a sum of positive squares and each nonzero-frequency spatial Fourier mode quantizes as an oscillator with the displayed dispersion relation. Higher spatial derivatives do not require higher time derivatives. The theory has a preferred time and does not possess Lorentz invariance.

For a sufficiently regular solution, put $w=-\Delta u$, with $\Delta$ the Laplacian. Then $-\Delta u-w=0$ and $u-\Delta w=u+{\Delta }^{2}u=f$. The zero Dirichlet boundary conditions on $u$ and $\Delta u$ become zero Dirichlet boundary conditions on $u$ and $w$. Conversely, eliminating $w$ from these two equations recovers the fourth-order equation with the biharmonic operator. For the weak solution we will construct, these interior identities hold as distributional identities; zero Sobolev traces are imposed on $u$ and $w$, without assuming classical fourth derivatives of $u$.

Let $V=H_0^1(\Omega )\times H_0^1(\Omega )$, the Cartesian product of two zero-boundary Sobolev spaces, with normThis is a Hilbert space. Applying integration by parts separately to the two equations gives the weak formulationEquivalently, for $U=(u,w)$ and $T=(v,z)$, define the bilinear form and linear functionalWe seek $U\in V$ with $b(U,T)=F(T)$ for every $T\in V$.

The skew-coupled elliptic system has a useful bounded bilinear form estimate with constant one. Indeed, write the integrand as the dot product of $(\nabla p,\nabla q,-q,p)$ and $(\nabla v,\nabla z,v,z)$. The Cauchy-Schwarz inequality givesFor equal arguments, the two zeroth-order coupling terms cancel. If $K$ is the constant in the supplied Poincaré inequality, thenThus $b$ is a coercive bilinear form, although it is not a symmetric bilinear form. The Lax-Milgram theorem applies and yieldsIn particular, $\Vert (u,w){\Vert }_V\le (1+K)\Vert f{\Vert }_0$. Testing with compactly supported smooth test functions identifies $-\Delta u=w$ and $-\Delta w=f-u$ as distributional identities. This gives the asserted relation to the original boundary value problem at the available regularity.

Write ${\omega }_h=\dot{\alpha }$, positive on opening. The no-slip boundary condition equates the fluid velocity with each plate's velocity:The upper condition is a moving-wall no-penetration boundary condition, not zero normal velocity in the laboratory frame. At the hinge the velocity tends to zero. With the arbitrary constant of the streamfunction chosen so that $\psi (r,0)=0$, these conditions becomeA changing angle cannot give a literally steady flow in fixed coordinates. We interpret the stated steadiness as the quasi-steady Stokes approximation at each instantaneous $\alpha$.

The printed expression is not biharmonic. To test it without reproducing the source formula, write its bracket as $G_p(\theta )$ and its streamfunction as ${\psi }_p=-\Omega r^2{G}_p/2$. For any angular function $G$, the Laplacian givesAt the concrete angle $\alpha =\pi /4$, differentiation of the printed bracket givesFor example, at the interior point $\theta =\pi /8$,The printed bracket does satisfy $G_p(0)=G_p^{\prime }(0)=G_p^{\prime }(\alpha )=0$ and $G_p(\alpha )=1$, so its boundary velocities would require $\Omega ={\omega }_h$. Satisfying these boundary conditions does not cure its nonzero biharmonic operator. The stated series is likewise not a general biharmonic basis: its $r^2\theta \sin 2\theta$ term already fails the displayed test. The separated polar solutions of the biharmonic equation contain additional radial powers; for the required $r^2$ dependence, the angular basis is $1,\theta ,\sin 2\theta ,\cos 2\theta$.

The corrected slowly opening Stokes wedge similarity solution follows by setting $\psi =-{\omega }_h{r}^{2}G(\theta )/2$. The biharmonic equation gives $G^{(4)}+4G^{\prime \prime }=0$. The two lower-wall conditions reduce its general solution toThe upper-wall condition $G^{\prime }(\alpha )=0$ fixes the ratio of $A$ and $B$, and $G(\alpha )=1$ fixes their scale. An expression that also remains finite at $\alpha =\pi /2$ isFor $0<\alpha <\pi$, its denominator is positive: the derivative of $\sin \alpha -\alpha \cos \alpha$ is $\alpha \sin \alpha >0$. Direct differentiation verifies all four boundary conditions and the biharmonic equation. The resulting velocity components are $v_r=-\dot{\alpha }r{G}^{\prime }/2$ and $v_{\theta }=\dot{\alpha }rG$.

For a fixed angle of order one, $U(r)\sim |\dot{\alpha }|r$, so the local Reynolds number is ${\mathrm{Re}}_r\sim |\dot{\alpha }|r^2/\nu$. One breakdown length is thereforeA second is the geometric length $l_2\sim \min (L,W)$, where $L$ is the distance to the radial ends and $W$ the spanwise width. At $r\sim l_2$, matching to the surrounding bath or three-dimensional end flow matters. Thus $r\ll \min (l_1,l_2)$ is required for this local two-dimensional Stokes flow, with angle-dependent factors in the inertial estimate. Ideal infinite plates remove the geometric cutoff but not the inertial one.

There is also a temporal restriction: the relevant viscous diffusion time must be short compared with the time $T_s$ over which the shape and its velocity field change appreciably. Using $r$ as a conservative diffusion length gives $r\ll \sqrt{\nu T_s}$. For order-one angles and constant $\dot{\alpha }$, $T_s\sim |\dot{\alpha }{|}^{-1}$ gives the same scale as $l_1$; narrow angles require using their actual gap and angular factors rather than extrapolating this order-one-angle estimate.

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.

Let $\Omega =(-1,1{)}^2$ and use the clamped energy space $V=H_0^2(\Omega )$, whose traces satisfy $u={\partial }_{n}u=0$ on $\partial \Omega$. Two applications of integration by parts giveIf equality holds, then $\Delta u=0$. The maximum principle for harmonic functions and the zero Dirichlet boundary condition imply $u=0$. The biharmonic operator is therefore positive definite. The same identity and conclusion hold for the simply supported conditions $u=\Delta u=0$; either standard interpretation of the paper's phrase “zero boundary conditions” gives the result.

In the dark, energy minimization gives $h_{xx}=-H_0$. Translation and rotation are zero-energy freedoms; one representative is $h_{-}(x)=-H_0{x}^2/2$. After illumination, choose the equivalent light-adapted equilibriumThen $u=h-h_{+}$ has homogeneous free-end conditions $u_{xx}=u_{xxx}=0$, and its initial valueis orthogonal to the two rigid zero modes $1$ and $x$.

Let $q_n>0$ solve the free--free biharmonic eigenvalue equationand chooseThese are orthogonal eigenfunctions of the biharmonic operator with ${\varphi }_n^{\prime \prime }={\varphi }_n^{\prime \prime \prime }=0$ at both ends. The shape iswhereThe omitted zero modes would only translate or rotate the whole filament.

For a Lagrangian density depending on $\varphi$, ${\partial }_t\varphi$, ${\partial }_i\varphi$ and ${\partial }_i{\partial }_j\varphi$, direct variation givesTake $\delta \varphi$ to have compact support, so every boundary term vanishes. One integration by parts for the first-derivative terms and two for the second-derivative term yield the Euler-Lagrange field equation with second derivativesThe plus sign on the last term is essential. For the action in a, it is particularly transparent to vary the Laplacian directly:Thus the classical equation of motion isHere ${\nabla }^4=({\nabla }^2{)}^2$ is the biharmonic operator. Substitution of the plane wave $\varphi =A{e}^{-i\omega t+i\mathbf{p}\cdot \mathbf{x}}$ givesA real classical solution combines complex-conjugate plane waves. Quantization assigns the positive branch to the excitation energy, giving the required dispersion relation. The negative-frequency branch supplies the creation operator part of the field, rather than negative excitation energy. Independently, the Hamiltonian in a gives $\dot{\varphi }=\pi$, $\dot{\pi }=-{\kappa }^2{\nabla }^4\varphi -m^2\varphi$, confirming the same sign.

Put $R=h_t-\nu {\nabla }^{2}h$, where the diffusion coefficient $\nu$ is constant. ThenOne temporal and two spatial integrations by parts givewith compactly supported variations. HenceHere ${\nabla }^4=({\nabla }^2{)}^2$ is the biharmonic operator. The cross terms cancel because $\nu$ is constant and spatial and temporal derivatives commute. Equivalently, $P^0=2R$ and $Q^{ij}=-2\nu R{\delta }^{ij}$ in part (b), producing $-2{\partial }_{t}R-2\nu {\nabla }^{2}R=0$.

This squared diffusion-residual action does not have only the first-order diffusion equation as its stationary equation. The zero-residual paths $h_t=\nu {\nabla }^{2}h$ have $S=0$ and are certainly solutions, but the variational equation is the higher-order one above. For a spatial Fourier mode with nonzero wavevector $\mathbf{k}$, it givesBoundary data or an additional physical selection principle are needed to retain only the decaying diffusion branch. The spatially uniform mode instead has $h_0(t)=A_0+B_{0}t$ in the stationary equation.

A skew-coupled elliptic system has zeroth-order couplings with opposite signs. For the displayed example with homogeneous Dirichlet boundary conditions, its weak formulation on $V=H_0^1(\Omega )\times H_0^1(\Omega )$ has the bilinear formThe coupling cancels for equal arguments, so $b((p,q),(p,q))=\Vert \nabla p{\Vert }_0^2+\Vert \nabla q{\Vert }_0^2$. On a domain satisfying the Poincaré inequality, this is a coercive bilinear form. With the product first-order Sobolev norm, the Cauchy-Schwarz inequality applied to $(\nabla p,\nabla q,-q,p)$ and $(\nabla v,\nabla z,v,z)$ gives $|b(P,T)|\le \Vert P{\Vert }_V\Vert T{\Vert }_V$. The Lax-Milgram theorem therefore applies even though the bilinear form is not symmetric. Eliminating $w=-\Delta u$ recovers ${\Delta }^{2}u+u=f$, involving the biharmonic operator.