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For bounded linear operators between Hilbert spaces with $\Vert Bu\Vert \ge \beta \Vert u\Vert$ and $\beta >0$, the operator $K^{\ast }K+\alpha B^{\ast }B$ is coercive and invertible for $\alpha >0$. The reconstruction uniquely minimizes $\Vert Ku-f{\Vert }^2/2+\alpha \Vert Bu{\Vert }^2/2$. Testing its normal equation for coercive Tikhonov regularization by $u$ gives $\Vert R_{\alpha }\Vert \le 1/(2\beta \sqrt{\alpha })$. If $K$ is injective, exact-data consistency and $\alpha \to 0$, $\delta /\sqrt{\alpha }\to 0$ give norm convergence to $K^{†}f$.

For a bounded linear operator between Hilbert spaces, the spectral regularization method for quadratic Tikhonov regularization has scalar coefficient $s/(s^2+\alpha )$. Since $(s-\sqrt{\alpha }{)}^2\ge 0$, this coefficient is at most $1/(2\sqrt{\alpha })$. The spectral theorem for normal operators transfers this scalar bound to the operator norm, providing the noise term in the noise-bias decomposition for linear regularization.

The reconstruction operator in Tikhonov regularization has scalar gain $\sigma /({\sigma }^2+\alpha )$. Its maximum over $\sigma \ge 0$ is $1/(2\sqrt{\alpha })$, so a data perturbation of norm at most $\delta$ changes the solution by at most $\delta /(2\sqrt{\alpha })$. A convergent noisy-data parameter choice satisfies $\alpha \to 0$ and $\delta /\sqrt{\alpha }\to 0$. Fixed-parameter stability is distinct from stability of the unregularized inverse.

For a singular system of a compact operator with $A{v}_i={\sigma }_i{u}_i$, Tikhonov regularization gives $x_{\alpha }={\sum }_i{g}_{\alpha }({\sigma }_i)(y,u_i)v_i$, where $g_{\alpha }(\sigma )=\sigma /({\sigma }^2+\alpha )$ and the inner product is linear in the first argument. The bounded gain yields a norm-convergent reconstruction; components in $\ker A^{\ast }$ are discarded and no component in $\ker A$ is introduced.

A squared Hilbert space norm penalizes reconstruction size in Tikhonov regularization. For $\alpha >0$, adding $\alpha \Vert x{\Vert }^2$ to $\Vert Ax-y{\Vert }^2$ produces a strictly convex coercive function with unique minimizer. Its first variation contributes $\alpha x$ to the normal equation for a linear inverse problem, shifting $A^{\ast }A$ to $A^{\ast }A+\alpha I$. It selects a minimum-norm representative and controls amplification of noisy small-singular-value components.

For a point-mass companion at separation $a$, subtract the constant and linear terms of its Newtonian gravitational potential in a frame accelerating with the target's center. The leading residual isIt is degree two in spherical harmonics. In a circular orbit with angular frequency ${\omega }_o$, its time-dependent $m=\pm 2$ components force at $2{\omega }_o$ in a frame with no spin; there is also a static $m=0$ component. This is the leading approximation in $r/a$, and the tidal amplitude may be absorbed into a harmonic-dependent constant.

Differential gravitational acceleration, multiplied by the affected mass, across a spatially extended body. In the linear approximation the acceleration difference is $T\Delta \mathbf{r}$, where $T$ is the tidal tensor. Stellar tidal forces limit the domain of a circumplanetary orbit; sufficiently strong ones can produce tidal disruption.

To isolate a tidal force, subtract the gravitational acceleration shared by the body as a whole. Let a reference test particle follow $\ddot{\mathbf{R}}=\mathbf{g}(\mathbf{R})$, and let another have position $\mathbf{R}+\mathbf{x}$. In a translating accelerated reference frame with parallel, nonrotating axes,for a source point mass, when $|\mathbf{x}|/R\ll 1$. Here $\mu =GM$ is its standard gravitational parameter and the last term uses Big O notation. This Taylor expansion removes the common inward pull and keeps the spatial variation. The near side has an excess pull toward the source; the far side has a deficit of inward pull, which is an outward relative acceleration. Both sides can stretch apart while both still accelerate toward the source.^[ref]

For a material element of mass $m$, the residual external force in this translating accelerated reference frame is $m[\mathbf{g}(\mathbf{R}+\mathbf{x})-\mathbf{g}(\mathbf{R})]$. Actual deformation also depends on elasticity, pressure and self-gravity. A rigid body approximation suppresses deformation but can still have internal forces that resist the tidal force. The point-mass tidal stretching and squeezing pattern describes freely falling, noninteracting test particles, rather than promising the same shape for every solid, star or ocean. A rotating reference frame introduces centrifugal acceleration and Coriolis acceleration, which must be included separately.

Disruption of a self-gravitating body when differential gravitational acceleration across it exceeds its binding acceleration. For a body of radius $R$, mass $m$ at distance $r$ from a much more massive body $M$, the competing scales are $GMR/r^3$ and $Gm/R^2$; their equality gives $r\sim R(M/m{)}^{1/3}$, up to geometry and internal-structure factors. The tidal tensor gives the directional differential acceleration. A tidal disruption event near a black hole is a specific astronomical application.

Subtracting the constant and uniform-acceleration terms from the point-mass relative potential gives ${\psi }_{\mathrm{tid}}=G{m}_p[3(\mathbf{r}\cdot {\hat{\mathbf{r}}}_p{)}^2-r^2]/(2r_p^3)$, with force convention $\nabla \psi$. Its tidal tensor is $G{m}_p(3{\hat{\mathbf{r}}}_p{\hat{\mathbf{r}}}_p^T-I)/r_p^3$. This is the leading term for target size small compared with separation.

Tidal dissipation converts mechanical energy associated with a time-varying tidal deformation into heat. It can reduce orbital eccentricity, drive synchronous rotation, and contribute deep planetary heating. A circular orbit with exact synchronous rotation, perpendicular spin axis and no additional forcing has a static tidal deformation in the body's frame and need not sustain substantial tidal heating.

For $f(n)=3^{\omega (n)}$,If $p^a\Vert n$, its contribution to $(f\ast \Lambda )(n)$ is $f(n)(a-1+1/3)\log p$. Since $3(a-1+1/3)\ge a$, summing over the prime factors proves the bound. Here $\ast$ is Dirichlet convolution and $\Lambda$ is the Von Mangoldt function.

The Euler product and Mertens second theorem give ${\sum }_{n\le D}{3}^{\omega (n)}/n\ll {\log }^{3}D$. The log-weighted convolution bound for three to the prime omega, with the Chebyshev estimate $\psi (y)\ll y$, then gives ${\sum }_{\sqrt{D}\le n\le D}{3}^{\omega (n)}\ll D{\log }^{2}D$. A direct bound for the full range follows from $3^{\omega (n)}\le {\tau }_3(n)$, where ${\tau }_3(n)=\#\{(a,b,c):abc=n\}$:

For the three-to-one spatially forced amplitude equation with forcing $e\exp [3i(\Omega T+\delta )]$, put $A=B\exp [i(\Omega T+\delta )]$. The real phase constant $\delta$ belongs inside the factor of $i$. If $c>0$ and the phase has been chosen so that $e>0$, the scaling $B=(e/c)C$, $\mathcal{T}=(e^2/c)T$ gives the displayed canonical amplitude equation, with $\mu =c\tilde{\mu }/e^2$ and $\omega =c\Omega /e^2$. A negative cubic saturation coefficient cannot give the same cubic sign under a forward-time normalization; zero forcing or zero cubic coefficient requires another scaling.

The canonical three-to-one spatially forced amplitude equation has nonzero equilibrium points $C=R{e}^{i\theta }$ satisfying $\cos 3\theta =(R^2-\mu )/R$ and $\sin 3\theta =-\omega /R$. Thus $(R^2-\mu {)}^2+{\omega }^2=R^2$. Each positive amplitude has three phases separated by $2\pi /3$: two amplitude branches normally mean six complex equilibrium points. For $\mu >{\omega }^2-1/4$ the larger branch is asymptotically stable and the smaller consists of saddle equilibria, except that its zero-amplitude root at $(\mu ,\omega )=(0,0)$ is not a nonzero equilibrium point. Equality gives a saddle-node bifurcation. This phase locking breaks continuous translation symmetry down to threefold symmetry.

For small positive forcing-frame frequency $\omega$, use $\mu =\hat{\mu }{\omega }^2$, $C=\omega z$ and $\tau =\omega T$ in the three-to-one spatially forced amplitude equation. The leading system is $z_{\tau }+iz={\overline{z}}^2$. Writing $z=x+iy$ gives the Hamiltonian system $(x_{\tau },y_{\tau })=(H_y/2,-H_x/2)$ and the displayed first integral. The origin is a center equilibrium; three saddle equilibria at $(0,1)$ and $(\pm \sqrt{3}/2,-1/2)$ lie on $H=1/3$. The factorization $H-1/3=(1+2y)[x^2-(y-1{)}^2/3]$ reveals a triangular heteroclinic cycle, containing closed periodic orbits for every $0<H<1/3$.

Every triple of subspaces is a direct sum of one-dimensional membership blocks and two-dimensional three-line blocks. Split off the triple intersection, then the pairwise intersections using projections killing the third subspace. Next split complements to $U_i\cap (U_j+U_{\ell })$, using projections killing the other two subspaces. The remaining subspaces are pairwise disjoint and each lies in the sum of the other two. Their sum is $U_1\oplus U_2$, with $U_3$ the graph of an isomorphism $U_1\to U_2$. A graph basis splits this into two-dimensional blocks. A complement to the total span supplies zero-membership blocks. This elementary argument classifies the three-subspace quiver.

For a compact oriented three-manifold, the integral relative class $u$ has norm the minimum of ${\chi }_{-}(F)={\sum }_j\max (0,-\chi (F_j))$ over properly embedded oriented topological surfaces representing $u$. Homogeneity and continuity extend it to real relative homology. It is a seminorm, since spheres, disks, annuli and tori have zero cost.

In the orientable case an excellent compact three-manifold is an irreducible three-manifold and a boundary-irreducible three-manifold, is not a ball, contains a two-sided properly embedded incompressible surface, and has every properly embedded incompressible zero-Euler characteristic surface a boundary-parallel surface. In particular it has no essential annulus or torus. These properties give lower bounds on the Thurston norm of classes with essential boundary.

An orientable three-manifold is irreducible if every embedded two-sphere bounds a three-ball. This makes innermost-disk simplifications of embedded topological surfaces possible without changing the ambient three-manifold.

A compact three-manifold is boundary-irreducible when there is no embedded disk with interior in the manifold and boundary an essential curve in its boundary. Thus its boundary has no compressing disk.