Put $M=\mathcal{N}(K{)}^{\perp }$ for a bounded linear operator between Hilbert spaces. Its restriction $K{|}_M:M\to \mathcal{R}(K)$ is bijective. Define the Moore–Penrose inverse of an operator byThe inverse vanishes on $\mathcal{R}(K{)}^{\perp }$. For each datum in this domain, $K^{†}f$ is the minimum-norm least-squares solution; all least-squares solutions are $K^{†}f+\mathcal{N}(K)$. On their natural domains,The equivalent continuity condition isIf the range is closed, the inverse restriction is bounded by the bounded inverse theorem. Conversely, if $\Vert K^{†}f\Vert \le C\Vert f\Vert$ on its domain and $K{u}_n$ converges, replace $u_n$ by $P_M{u}_n$. The inequality makes these preimages a Cauchy sequence, whose limit maps to the limiting datum; the range is therefore closed. Boundedness here uses the data norm, not a graph norm that would conceal instability.

Assume the vertical dynamical timescale of a disk is shorter than the thermal timescale of an accretion disk, which is itself shorter than the viscous timescale. This ordering is controlled in a thin disk with $\alpha \ll 1$: $t_{\mathrm{dyn}}\sim {\Omega }^{-1}$, $t_{\mathrm{th}}\sim (\alpha \Omega {)}^{-1}$ and $t_{\nu }\sim (r/H{)}^2{t}_{\mathrm{th}}$. Therefore compare heating and cooling at fixed surface density of a disk, radius, $\alpha$ and composition. Vertical hydrostatic equilibrium adjusts on the vertical dynamical timescale of a disk, giving $H\propto T^{1/2}$ and $\rho \propto T^{-1/2}$. Consequently the stipulated negative hydrogen ion opacity varies as $\kappa \propto T^{10-1/6}=T^{59/6}$. The rates per disk face then scale asThus heating increases while cooling decreases after a positive temperature perturbation. At equilibrium $Q^{+}=Q^{-}=Q_0$, a small change gives $\delta (Q^{+}-Q^{-})=(41/6)(Q_0/T)\delta T$, which drives the perturbation further from equilibrium for positive effective heat capacity. A negative temperature perturbation likewise increases the net cooling. This proves thermal instability from negative hydrogen ion opacity within the local closure; it is the temperature-slope test of thermal stability of an accretion disk. Holding volume density rather than surface density of a disk fixed would give a different cooling exponent, but still the same qualitative instability; the fixed-column comparison is the relevant thin-disk thermal test.

The radial null geodesics satisfyThe ingoing family has $v=\text{constant}$, with $r$ decreasing toward the future; its tangent $-{\partial }_r$ is affinely parametrized because ${\Gamma }^a{}_{rr}=0$. The outgoing family hasAway from the Schwarzschild event horizon, integration gives $v=2r_{\ast }+\text{constant}$. These are null geodesics up to reparametrization: a null direction in the two-dimensional radial geometry is automatically pregeodesic, and the angular geodesic equations are satisfied by constant angles. On the Schwarzschild event horizon the outgoing family instead has $r=2M$, with tangent proportional to ${\partial }_v$.

In a Finkelstein diagram, plot $T=v-r$ vertically and $r$ horizontally. Ingoing rays obey $T+r=\text{constant}$; outgoing rays obeyOutside the Schwarzschild event horizon, outgoing rays increase $r$; on it they remain at $r=2M$; inside it they decrease $r$ even though $v$ increases. Thus both radial future null directions point toward smaller $r$ inside the black hole. Every future timelike direction lies between these null directions, so it also moves toward smaller $r$ there.

Arrows show future propagation. The vertical red line is the Schwarzschild event horizon; the black boundary at $r=0$ is the Schwarzschild singularity.

Use the same method of multiple scales and its solvability condition in the method of multiple scales. The position-dependent damping function is odd, so the lemma that odd position-dependent damping has zero first-order amplitude drift applies. In its wave amplitude average, $\psi \mapsto \psi +\pi$ changes $\sin (A\cos \psi )$ to its negative while preserving ${\sin }^2\psi$, so the average is zero. Hence $A^{\prime }=0$; the wave phase average again gives ${\theta }^{\prime }=0$.

Equivalently, expand the position-dependent damping in its uniformly convergent power series for bounded $A$. Each term ${\cos }^{2m+1}\psi \sin \psi$ has only even sine Fourier harmonics, so none resonates with the unit-frequency oscillator. With the initial wave amplitude and wave phase unchanged, the leading uniform approximation isThere is no first-order slow drift. Effects at the next order accumulate only an $O(ϵ)$ correction on this time scale.

Write $M=\Vert u_0{\Vert }_{L^1}$. Since $|U(0,y)|\le M$, choosing $y=x-ct$ gives $g((x-y)/t)=g(c)=0$ and ${\max }_y{G}_{t,x}(y)\ge -M$. At the maximizing foot $x_0$, the inverse-flux quadratic bounds giveCombining the two estimates and dividing by $-k_{+}>0$ givesThe last step uses $u=h((x-x_0)/t)$ and the sign-independent bound on $h$. This is square-root decay before characteristic crossing; it is proved only for $0<t<t^{\ast }$. No continuation past characteristic crossing or global-time smoothness is assumed. If $M=0$, the initial function and the solution vanish.

The printed linear ordering is false for $z<c$. For example, $f(s)=-s^2/2$, $c=0$, $h(z)=-z$, $k_{-}=-3/4$, $k_{+}=-1/4$ satisfy the derivative hypothesis, but at $z=-1$ the printed lower bound would require $3/4\le 1/2$.

The correct inverse-flux quadratic bounds areIndeed $h(c)=0$ and $h(z)/2={\int }_c^z{h}^{\prime }(r)/2dr$; reversing the integration limits reverses the linear inequalities. Equivalently, for $z\ne c$, $k_{-}\le h(z)/(2(z-c))\le k_{+}$. Since $g(c)=0$ and $g^{\prime }=h$, integration once more gives the quadratic inequalities on both sides of $c$. A useful sign-independent consequence is $|h(z)|\le -2k_{-}|z-c|$.

Because $f^{\prime }$ is a decreasing bijection, its inverse $h$ exists and is continuous. For any $y$, choose $s=h((x-y)/t)$ in the preceding inequality. The line from $(0,y)$ then reaches $(t,x)$, giving $U(t,x)\ge U(0,y)+tg((x-y)/t)$.

Now let $x_0=X_t^{-1}(x)$ and $s=u_0(x_0)$. On this characteristic curve, $u=s$, so the tangent-line inequality is an equality throughout. This proves attainment andIn fact the maximizing foot is unique. The concave Legendre dual can be written $g(z)={\inf }_s(zs-f(s))$ and satisfies $g^{\prime }(z)=h(z)$: compare the minimizing values at $z$ and $z+\epsilon$ and use continuity of $h$. This avoids assuming differentiability of $h$, which strict concavity by itself does not guarantee. Differentiating the maximizing expression with respect to $y$ gives $u_0(y)-h((x-y)/t)=0$, hence $X_t(y)=x$ and $y=x_0$. The maximum representation for a concave conservation law therefore reproduces the characteristic solution of a scalar conservation law throughout the smooth lifespan.

The tangent-line inequality for a differentiable concave function is $f(r)\le f(s)+f^{\prime }(s)(r-s)$. Taking $r=u=U_x$ and using the Hamilton-Jacobi equation yieldsAlong $x(\tau )=x_0+\tau f^{\prime }(s)$, the chain rule identifies the left side with $dU(\tau ,x(\tau ))/d\tau$. Integrating givesStrict concavity makes equality possible exactly when $u=s$ along the line, which will select the maximizing characteristic curve.

The compact support just established and $f(0)=0$ permit differentiation of the antiderivative and integration of the scalar conservation law from $-\infty$:This is a Hamilton-Jacobi equation. There is no arbitrary function of time: the normalization at $-\infty$ and the zero exterior flux determine it.

The inverse function theorem applied to $X_t$ makes $u_0\circ X_t^{-1}$ smooth. Since $u_0(\pm A)=0$, $X_t(\pm A)=\pm A+ct$. Monotonicity of the characteristic flow map and the vanishing of $u_0$ outside $[-A,A]$ implyThus the solution is a smooth function with compact support at each such time. The fact that some interior characteristic speeds differ from $c$ causes no difficulty: they cannot cross the two exterior characteristic curves before $t^{\ast }$.

For the scalar conservation law, put $v_0(\xi )=f^{\prime }(u_0(\xi ))$ and $m={\min }_{\xi }{v}_0^{\prime }(\xi )={\min }_{\xi }{f}^{\prime \prime }(u_0(\xi ))u_0^{\prime }(\xi )$. The concave-flux characteristic lifespan and characteristic flow map areFor $t<t^{\ast }$, $X_t^{\prime }=1+t{v}_0^{\prime }>0$, and $X_t(\xi )=\xi +ct$ outside the support of $u_0$. Thus $X_t$ is a global smooth diffeomorphism and the formula is $u(t,x)=u_0(X_t^{-1}(x))$. The method of characteristics proves existence and uniqueness among smooth solutions.

If $m<0$, at a minimizer ${\xi }_{\ast }$ the numerator $u_0^{\prime }({\xi }_{\ast })$ is nonzero, since its product with $f^{\prime \prime }(u_0({\xi }_{\ast }))$ is negative. Thereforeblows up as $t\uparrow t^{\ast }$, showing that this is the maximal smooth lifespan. Strict concavity alone does not require $f^{\prime \prime }$ to be negative at every point; the argument uses no such extra hypothesis.

For the inflow transport boundary condition, the weak formulation isHere $\phi$ is a smooth compactly supported test function on the closed quadrant. Put $z=x-at$ and $\psi (t,z)=\phi (t,z+at)$. The transformed region is $t>\tau (z):=\max (0,-z/a)$. The formula in the preceding solution is $v(t,z)=b(z)$, with $b(z)=u_0(z)$ for $z\ge 0$ and $b(z)=f(-z/a)$ for $z<0$. Consequently the first integral equals $-\int b(z)\psi (\tau (z),z)dz$. Its positive-$z$ portion cancels the initial integral; the substitution $z=-at$ in its negative-$z$ portion cancels the inflow integral, including the factor $a$. This proves weak existence for arbitrary bounded data without corner compatibility.

For uniqueness, subtract two weak solutions. Interior test functions in these characteristic coordinates give ${\partial }_{t}v=0$ distributionally, so $v(t,z)=b(z)$ on each vertical ray. This follows first on rectangles compactly contained in $t>\tau (z)$ using tensor test functions, and then on the whole region by overlapping rectangles. In the full weak formulation the remaining lower-boundary term is $-\int b(z)\psi (\tau (z),z)dz=0$. Arbitrary smooth traces supported separately on $z>0$ and $z<0$ force $b=0$ there. The single point $z=0$ has zero measure. Thus the two-branch formula is the unique bounded weak solution.

A characteristic curve has $x-at$ constant. Backtracking reaches the initial line if $x\ge at$ and the inflow boundary otherwise. HenceValues on the dividing characteristic curve are immaterial for bounded weak solutions.

For a classical solution on the closed quadrant, choose $u_0,f\in C^1([0,\infty ))$ and impose the corner compatibility for constant-speed transportThese make the values and both first derivatives agree across $x=at$; each branch solves the transport equation. They are also necessary for a $C^1$ solution up to the initial and inflow boundaries.

Spatial $C^{\infty }$ regularity for every $t\ge 0$, up to $x=0$, holds precisely for smooth data with all matching jets:Indeed the spatial derivatives of order $n$ at $x=at$ are $u_0^{(n)}(0)$ and $(-1/a{)}^n{f}^{(n)}(0)$. Necessity of smooth $u_0$ follows at $t=0$; smoothness of $f$ on any finite interval follows by reading the boundary branch in a spatial slice at a larger time.

Absent in the minimal Standard Model. The single charged-lepton Yukawa matrix supplies both the fermion mass matrix and the Higgs coupling, so the same left/right rotations diagonalize both. There is no off-diagonal $h\tau \mu$ vertex. With massless neutrinos, the separate charged-lepton family numbers prevent this transition perturbatively. If neutrino masses and mixing are added, loop-induced charged-lepton flavour violation can occur but is extremely suppressed; a sizeable rate would require an additional source of flavour violation. Part (c) provides one such source.

The coupling is allowed, with an on-shell threshold condition. The Higgs field kinetic term producesand hence a tree-level Feynman vertex. A decay into two real W bosons requires $m_h>2m_W$ for nonzero phase space. Below that threshold the vertex still mediates $h\to W{W}^{\ast }$ or decays through two virtual bosons into fermions. Thus an allowed Lagrangian coupling does not by itself guarantee an on-shell two-body decay.

Allowed through second-order weak interactions, not a tree-level neutral-current vertex. This is the flavour-changing transition responsible for neutral kaon mixing, with change of strangeness by two. Box Feynman diagrams contain two W bosons and internal up-type quarks. Their vertices come from the weak charged current and contain CKM matrix factors. At low energy they generate an operator $({\bar{s}}_L{\gamma }_{\mu }{d}_L)({\bar{s}}_L{\gamma }^{\mu }{d}_L)$ plus its conjugate. The GIM mechanism cancels the flavour-independent part of the internal-quark sum; unequal quark masses leave a nonzero loop amplitude.

There are two diagrams: $t$-channel Z boson exchange and $s$-channel W boson exchange. The incoming electron and electron-antineutrino can annihilate into $W^{-}$, which then produces the outgoing pair. The charged-current propagator carries momentum $k+p$, hence invariant $s$. The solid-line arrows in the diagram indicate fermion-number flow; those on antineutrino legs run opposite their physical motion.

There are two diagrams: the $t$-channel Z boson exchange from the weak neutral current, and a crossed $u$-channel W boson exchange from the weak charged current. The charged-current vertices pair the incoming neutrino $k$ with the outgoing electron $p^{\prime }$ and the incoming electron $p$ with the outgoing neutrino $k^{\prime }$. The exchanged momentum has square $u=(k-p^{\prime }{)}^2$. With charge flow from the first vertex to the second, the exchanged boson is $W^{+}$. The corresponding panels show both topologies; no photon or Higgs diagram survives the stated neutrino assumptions.

The same affine transformation of the normal distribution givesIts variance stays one while its mean moves positively. This is the exact alternative distribution used to calculate statistical power. The square, if used instead, has a noncentral chi-squared distribution with one degree of freedom and noncentrality $n({\delta }^{\ast }{)}^2/{\sigma }^2$.

Every epimorphism in an abelian category is the coequalizer of its kernel pair. Its kernel-pair square is a pullback in a category along an epimorphism and therefore also a pushout. A map equalizing the two projections supplies the same map on both pushout legs and so factors uniquely through the original epimorphism. In particular, all such epimorphisms are regular epimorphisms.