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The horizontally invariant magnetized shearing-sheet equations are linear in the horizontal fields, so perturbations about the equilibrium satisfy the same equations. The fixed surface boundary conditions require $\delta B_x=\delta B_y=0$ at $z=\pm H$. Choose a normal mode withwhere the vertical wavenumber is $k=m\pi /(2H)$, $m=1,2,\dots$. With the Alfvén frequencythe four amplitude equations becomeEliminating the velocities leavesA nonzero amplitude requires the determinant of this system to vanish, yielding the ideal magnetorotational dispersion relationEquivalently, with ${\kappa }_r^2=2(2-q){\Omega }^2$,As a quadratic equation for $s^2$, its discriminant is ${\kappa }_r^4+16{\Omega }^2{\omega }_a^2>0$. If $0<{\omega }_a^2<2q{\Omega }^2$, its constant term is negative, so one root $s^2$ is positive and there is an exponentially growing mode. If ${\omega }_a^2>2q{\Omega }^2$, both the constant term and the coefficient of $s^2$ are positive, giving two negative roots and only oscillatory modes. Equality is marginal.

The lowest allowed vertical wavenumber, $k=\pi /(2H)$, is the last to be stabilized as $|B_z|$ increases. Therefore the finite-thickness magnetorotational instability criterion isThere is also a vertically uniform velocity mode with zero magnetic perturbation; for the usual orbitally stable regime $q<2$, it is just stable epicyclic motion. For $q>2$, that uniform mode is already hydrodynamically unstable, independently of the magnetic criterion. At $q=2$ it is marginal. The criterion above concerns the magnetic modes.

Set the noise strengths to the model A fluctuation-dissipation relation and the fluctuation-dissipation relation for a conserved flux:The preceding logarithmic ratio becomesThe last two terms combine as ${\partial }_i({\mu }_j{W}_{ij})$. They integrate to zero under the periodic boundary conditions, by integration by parts. The remaining term is $F_2-F_1$ by the functional chain rule, henceAfter including the equilibrium initial weights, the forward and reversed histories have equal probability. Thus microscopic reversibility is satisfied by the two channels together, with each channel's own thermal noise strength.

Write the action density asUnder a compactly supported variation $\varphi \mapsto \varphi +ϵ\eta$,Integrating the first term by parts in time and the second in space, and discarding the boundary terms, givesThe principle of stationary action requires this to vanish for every $\eta$. Sincedivision by $-a^3$ yields the field Euler-Lagrange equationThis is the equation generated by the inflaton action in an expanding universe.

The solution $x+\lambda$ satisfies the left boundary condition, while $e^{-x}$ satisfies decay at infinity. The proposed expression for $x<\xi$ is therefore the correct left homogeneous solution. For $x>\xi$, write $G=C{e}^{-x}$. Continuity at $x=\xi$ givesHence the Green function isIndeed, its derivative has the required unit jump,which produces the Dirac delta distribution in $L[G]$.

Choose the Sobolev space $V$ encoding the homogeneous essential boundary conditions, setafter the appropriate integration by parts, and define the energy functionalIts first variation is $J^{\prime }(u)v=a(u,v)-\ell (v)$, so its stationary points are exactly the solutions of the weak formulationIf $a$ is bounded, symmetric, and coercive and $\ell \in V^{\prime }$, the Lax-Milgram theorem supplies a unique weak solution $u$. Moreover, for every $w\in V$,unless $w=0$. Thus $J$ is strictly convex, and $u$ is its unique global minimizer. This proves existence and uniqueness of the minimizer and of the weak solution simultaneously.

Integrating the ordinary differential equation four times gives a quartic polynomial. It is useful to use the linearity of the equation and split the solution into a gravity part and a force part:wheresatisfies the clamped and torque-free boundary conditions with $F=0$, whilesatisfies the homogeneous equation and contributes the endpoint force. Direct differentiation verifiesThusthe clamped-free beam under uniform load and endpoint force.

Write $s=\log r$, so the two circular boundaries are $s=0$ and $s=2$. The zero angular mode must interpolate between $-1$ and $1$, giving $s-1=\log r-1$.

For the $n$th cosine mode, the radial factor has equal value $A_n$ at both boundaries. The unique harmonic function with those data isHence the solution of the annular Dirichlet problem isAt $r=1$ and $r=e^2$, the hyperbolic cosine quotient equals one, so the boundary conditions are satisfied term by term.

Away from $x=\xi$, the Green function solves $G^{\prime \prime }-G=0$. The solution satisfying the left boundary condition is proportional to $\sinh x$, while the solution decaying at infinity is proportional to $e^{-x}$. Continuity at $\xi$ and the unit derivative jumpgiveor $G=-\sinh (\min \{x,\xi \})e^{-\max \{x,\xi \}}$, the dirichlet half-line Green function for d2 minus 1.

The required solution isEvaluation of the two elementary integrals, split at $\xi =x$, givesIndeed, direct differentiation gives $y^{\prime \prime }-y=e^{-2x}$, and both boundary conditions are immediate.

In the one-dimensional limit, $v=0$ and $u=u(x)$. The first component of the Navier-Cauchy equation reduces toFor a nondegenerate one-dimensional elastic material, $\lambda +2\mu \ne 0$, so $u^{\prime \prime }=0$ and $u(x)=Ax+B$. The boundary conditions $u(0)=0$ and $u(L)=\Delta$ give $B=0$ and $A=\Delta /L$. Therefore the uniform extension of a one-dimensional elastic body is

Let $m_j$ be the expected remaining time when the walk is at $j$. The first-step recurrence for gambler's ruin iswith $m_{-a}=m_b=0$. The quadraticsatisfies both the recurrence and the boundary conditions, and their finite linear system has a unique solution. Starting from zero therefore gives

Suppose ${\varphi }_1,{\varphi }_2$ solve the Dirichlet problem and put $u={\varphi }_1-{\varphi }_2$. Then ${\nabla }^{2}u=0$ in $V$ and $u=0$ on $S$. The divergence theorem applied to $u\nabla u$ gives Green's identityThus $\nabla u=0$, so $u$ is constant; its zero boundary value makes it zero. The Dirichlet solution is therefore unique.

For homogeneous Neumann data the same calculation again shows that $u$ is constant, but the boundary condition does not determine that constant. Hence, whenever a Neumann solution exists, adding any constant produces another solution. This is the standard uniqueness of Poisson equation distinction.

For a radial function in three dimensions, the radial Laplacian givesHere regularity at zero selectsA further integration giveswhose apparent singularity is removable, with limit $3+C$ at zero. Since the nonconstant term has value $4e-8$ at $r=1$, the boundary condition $\varphi (1)=3$ gives $C=11-4e$. Therefore

Continuity at $x=\xi$ is automatic in the proposed expression. Integrating the differential equation through $x=\xi$ gives the derivative jumpThusIn terms of the Wronskianone hasThe boundary conditions hold because $y_1^{\prime }(0)=0$ and $y_2^{\prime }(1)=0$.

The Abel identity gives $W^{\prime }=-\alpha W$. If $\alpha =0$, the Wronskian and hence $c$ are constant. The two branches of the displayed formula are then interchanged by $x\leftrightarrow \xi$, provingThis is the symmetry of the Neumann Green function for a second-order ordinary differential equation in the self-adjoint case.

For $G^{\prime \prime }-G=\delta (x-\xi )$, chooseTheir Wronskian isWriting $x_{<}=\min (x,\xi )$ and $x_{>}=\max (x,\xi )$ givesThe solution of the inhomogeneous problem is $y(x)={\int }_0^{1}G(x,\xi )\xi d\xi$. Equivalently, solving directly givesThe two Neumann conditions yield $B=1$ andTherefore

Let $x$ point downslope and $z$ point normally away from the plane, with $0\le z\le h$. Write the parallel velocity as $u(z)e_x$ and take $\tau >0$ to be the magnitude of the air's upslope stress. The steady equations arewith boundary conditionsThe free-surface condition makes $p_x=0$, and integration gives

The surface velocity, downslope shear exerted by the fluid on the plane, and volume flux per unit width areThus the inclined viscous film with opposing surface shear reverses in the three senses whenThe order of increasing required air stress is therefore

Let $z$ measure height above the sponge. Under the long-wave approximation, hydrostatic pressure gives $p_x=\rho g{H}_x$, and the horizontal balance for viscous fluid flow isThe no-slip boundary condition $u(0)=0$ and stress-free boundary condition $u_z(H)=0$ giveIntegrating this lubrication theory profile gives the horizontal volume flux per unit span

Local mass conservation includes the downward loss found in part i:Consequently the required nonlinear partial differential equation isIf the imposed inlet flux is $Q_0{t}^{\alpha }$ and the moving front is $x=x_N(t)$, sufficient boundary conditions areFor a current released onto a dry substrate one also takes the initial condition $H(x,0)=0$ away from the source. The front position is part of this moving-boundary problem.

For $\alpha =2$, part iii gives $a=1$, $b=2$, and $\beta =1$. PutSubstitution into the nonlinear partial differential equation gives the ordinary differential equationwith boundary conditions

To find the leading edge, set $y={\eta }_N-\eta$ and suppose $F\sim C{y}^p$. The two singular terms in the ordinary differential equation have orders $y^{4p-2}$ and $y^{p-1}$. Their exponents agree only when $p=1/3$. Their leading coefficients then satisfywhereas $F$ and the constant drainage $D$ are lower-order terms. HenceThe draining gravity current therefore has a one-third-power leading edge and its flux vanishes there.

For a ridge-fixed disturbance, $c=0$. If the Scorer parameter varies on a height scale much longer than the vertical wavelength, the local WKB approximation giveswhere $m^2=l^2-k^2>0$. The disturbance is vertically oscillatory there and can carry wave activity upward. Where $l^2<k^2$, $m$ is imaginary and the solution is vertically evanescent.

A rigid ridge supplies the lower boundary condition and an upper radiation or decay condition selects the physical solution. If $l^2$ decreases through $k^2$, the crossing is a turning level: the wave is reflected or decays above it and is trapped beneath it. Increasing $\overline{u}$ normally reduces $N^2/{\overline{u}}^2$, while decreasing $N^2$ reduces it directly; subject to the curvature term $-{\overline{u}}_{zz}/\overline{u}$, either change therefore promotes vertical trapping of an atmospheric gravity wave.

Lorentz invariance requires $A^{\mu }$ and $J^{\mu }$ to transform as four-vectors, $\chi$ as a scalar, $m$ and ${\mu }_0$ as scalars, and the integration domain and boundary conditions to respect the transformation. Then every contracted term and $d^{4}x$ are Lorentz invariant.

Under $A_{\mu }\mapsto A_{\mu }+{\partial }_{\mu }\chi$, the field strength is unchanged. The source term changes by a boundary term pluswhich vanishes for a conserved current and suitable boundary behaviour of $\chi$. The mass term is not gauge invariant. Thus the action has the usual local gauge invariance whenwith the gauge function or its boundary contribution suitably controlled. For $m>0$, the mass term breaks this gauge symmetry.

The boundary conditions are now $C(0,t)=0$ and $C_x(L,t)=0$. Their wave numbers areThe principal growth rate is $\mu -D{\pi }^2/(4L^2)$, so long-term growth occurs whenClosing one end reduces the critical growth rate by a factor of four because there is only one absorbing escape boundary.