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The Laplace transform of a function on $t\ge 0$ isfor values of $s$ for which the integral converges. Integration by parts gives the Laplace transform of a derivative

Write ${\hat{N}}_i(s)=\mathcal{L}\{N_i\}(s)$. Transforming the first two equations and using the initial data givesThe third equation then givesand henceA partial fraction decomposition therefore yields the sequential radioactive decay formula

Now let ${\lambda }_1={\lambda }_2=\lambda$ and put $\Delta ={\lambda }_3-\lambda >0$. The transformed contribution originating from the initial $N_1$ population becomesEquivalently, solve the middle equation first to obtainand use an integrating factor in the final equation:Evaluating the elementary integral gives

At distance $r$, isotropic luminosity $L$ produces radiative acceleration $a_{\mathrm{rad}}=\kappa L/(4\pi r^{2}c)$ in material of opacity $\kappa$. In fully ionized hydrogen, one Electron supplies a Thomson scattering cross-section ${\sigma }_T$ for approximately one proton mass of material, so $\kappa ={\sigma }_T/m_p$. Balancing radiation against the Newtonian gravitational field of the black hole gives the Eddington luminosityBelow this luminosity gravity can confine optically thin ionized hydrogen; above it, electron scattering alone accelerates the gas outward.

An electron--positron pair has mass $2m_e$ and total scattering cross-section $2{\sigma }_T$, so its opacity is ${\sigma }_T/m_e$. ThereforeThe limit is lower because the radiative force per unit inertial mass is larger in a pair plasma.

For a cloud with arbitrary constant opacity per unit mass,Suppose the luminosity jumps to a constant $L>L_{\mathrm{Edd}}$ while the cloud is at rest at $r_0$. With $\Gamma =L/L_{\mathrm{Edd}}$, its radial acceleration isMultiplication by $\dot{r}$ and integration from $r_0$ to infinity gives conservation of energyThis result assumes that $L$ and $\kappa$ remain constant and that the cloud stays optically thin.

Write $f_{\theta }(x)$ for the joint density and assume regularity permits differentiation under the integral. Sinceone differentiation givesDifferentiating the score itself,Taking expectation givesThus the information identity is

For $N=1$, writeThe given evolution of the discrete scattering data givesSolving the resulting two scalar linear equations for the components of ${\psi }_1$ and substituting in the reconstruction formula givesChoosing the additive multiple of $2\pi$ so that $u\to 0$ as $X\to +\infty$ and integrating in $X$ yields the One-soliton solution of the sine-Gordon equation in light-cone coordinatesIt depends only on $X+T$ exactly when the two positive coefficients in the exponent agree:The unique positive solution is $l=1/2$, and then

The transformations satisfy $g^s{g}^r=g^{s+r}$, $g^0$ is the identity map, and $(g^s{)}^{-1}=g^{-s}$, so they form a one-parameter group. Ifthen the chain rule givesThus $g^s$ is a Lie point symmetry. Applied to the one-soliton family, it replaces $l$ by $l^{\prime }=e^{-s}l$. Taking $e^s=2l$ gives $l^{\prime }=1/2$, so every member is transformed to the function $F(X+T)$ found above.

For the stated $N=2$ solution, set $x=X+T$ and $t=T-X$. When $l^2+m^2=1/4$, its two arguments reduce toAt fixed $x$, only $\sin (2mt)$ varies, so the Sine-Gordon breather has fundamental period

Finally put $r^2=l^2+m^2$. Under the same symmetry, the parameters becomeand hence $(l^{\prime }{)}^2+(m^{\prime }{)}^2=e^{-2s}{r}^2$. Choosing $e^s=2r$ makes this sum $1/4$, proving that every solution in the family is symmetry-equivalent to the normalized breather.

For a bounded function, the lower and upper integrals are the supremum of lower Darboux sums and infimum of upper Darboux sums. It is Riemann integrable when these agree.

Here $u(x)=1/x-\lfloor 1/x\rfloor$ for $x>0$, so $0\le u<1$. On $[\delta ,1]$ it has only finitely many discontinuities and is piecewise continuous, hence integrable. On $[0,\delta ]$ its upper-minus-lower contribution is at most $\delta$. Taking $\delta \downarrow 0$ proves integrability on $[0,1]$.

The characteristic polynomial of the homogeneous equation isso $y_h=C_0+(C_1+C_{2}x)e^{2x}$. For a particular integral put $y=e^{2x}v$. ThenTaking $v^{\prime \prime }=x/2-1/4$ gives $v=x^3/12-x^2/8$ after discarding homogeneous terms. Therefore

Expanding with the Levi-Civita symbol and the product rule givesThe Stokes theorem stateswhere the boundary orientation follows the right-hand rule about $n$.

Apply it to $F=m\times v$. On $S$, $m=n$ and $v\cdot n=0$. Substitution of the vector identity and contraction with $n$ leavesThe boundary integrand satisfies $(n\times v)\cdot dx=v\cdot (dx\times n)$, proving the formula. The vector $dx\times n$ is the outward co-normal in the tangent plane of $S$.

For the hemisphere, $n=e_r$ and $v=r\sin \theta e_{\theta }$. The projected divergence is $2\cos \theta$, so the left side isOn the positively oriented equator, $dx=R{e}_{\varphi }d\varphi$, hence $dx\times n=R{e}_{\theta }d\varphi$ and the boundary integral is also $2\pi R^2$.

The first isomorphism theorem says that for a homomorphism $\varphi :G\to H$,The map is well-defined because equal cosets differ by a kernel element; it is a surjective homomorphism onto the image, and injectivity follows because its kernel is the identity coset.

Here $G{L}_n(\mathbb{R})$ consists of invertible real $n\times n$ matrices and $S{L}_n(\mathbb{R})$ consists of those with determinant one. The latter is the kernel of the surjective determinant map to ${\mathbb{R}}^{\ast }$, so it is normal and

The given integral matrices form a group because products and inverses remain integral. The inverse formula shows that an integral inverse exists exactly when the determinant divides every cofactor; taking determinants shows more directly that $\det A\det A^{-1}=1$, hence $\det A=\pm 1$, and the adjugate formula proves the converse. The matrices $\left(\begin{array}{cc}1& m\\ 0& 1\end{array}\right)$ show infinitude.

Reduction modulo two is a homomorphism $G\to G{L}_2({\mathbb{F}}_2)$. Its kernel is exactly $H$, so $H$ is normal; its index is finite because the target has only six elements.

The residue theorem states that if a meromorphic function has finitely many poles inside a positively oriented simple closed contour and none on it, then its contour integral is $2\pi i$ times the sum of the enclosed residues.

Apply it toon a keyhole contour around the positive real axis. The outer and inner circles vanish as their radii tend to infinity and zero because $\alpha <1$ and $\alpha >-1$, respectively. On the upper bank the numerator tends to $x^{\alpha }$, while on the lower bank it tends to $e^{2\pi i\alpha }{x}^{\alpha }$ and the direction is reversed. Therefore the limiting contour integral is

The only enclosed pole is the double pole at $z=-1$. Since $L(-1)=i\pi$,The residue theorem now givesFor $\alpha \ne 0$, division andyieldAt $\alpha =0$, the integral is one, agreeing with the continuous limit. This is the positive-axis keyhole beta integral.

Because the parameter enters through $\lambda wy$,Use this as $f$ in part (a). The necessary orthogonality condition isWith $y=\epsilon y_0+{\epsilon }^2{y}_1$ and $\mu =O(1)$,ConsequentlyThe normalization makes the second integral one. Divide by ${\epsilon }^{m+1}$ to obtainThis is the leading nonlinear eigenvalue shift in a Sturm-Liouville problem.

For $\psi (x)=C{e}^{-\alpha x^2}$,The stationary Schrodinger equation becomesHence, choosing the positive root and a positive normalization constant,The state is even, so $⟨x⟩=⟨p⟩=0$. The Gaussian integrals giveand therefore $\Delta x\Delta p=\hslash /2$. These values are collected in gaussian eigenstate for a quadratic potential.

Finally, equality in the derivation of the uncertainty relation requires the centred vectors to be linearly dependent with a purely imaginary proportionality constant. Thus for some $s>0$,In position space this sayswhose normalizable solutions areThus every saturating state is Gaussian up to translation, a plane-wave factor, and an overall phase, as in the equality case of the Heisenberg uncertainty relation.

Suppose the additive group of the integral domain $R$ is isomorphic to ${\mathbb{Z}}^n$. Every ideal $I$ is then an additive subgroup of a finitely generated free abelian group, so it has finitely many additive generators $x_1,\dots ,x_m$. These also generate $I$ as an $R$-ideal: integer coefficients are coefficients from the canonical copy of $\mathbb{Z}$ in $R$. Thus every ideal is finitely generated, proving that $R$ is Noetherian by noetherianity from finite additive rank.

For an example that fails (i)--(iv), takeIt is an integral domain and has additive group ${\mathbb{Z}}^2$. The idealhas index two: modulo $I$, one has $2=0$ and $\sqrt{-5}=-1=1$, and the resulting quotient is ${\mathbb{F}}_2$. If $I=(a+b\sqrt{-5})$, multiplication by its generator would have determinant and absolute indexwhich has no integer solution. Hence $I$ is not principal, as detailed in nonprincipal ideal in the integers adjoined a square root of minus five. Condition (i), and therefore all four equivalent conditions, fails.

Take a variation $y+\epsilon \eta$, where the differentiable function $\eta$ obeysbecause both $y$ and its first derivative have fixed endpoint values. The first variation of the functional isApplying integration by parts once to the second term and twice to the third givesThe endpoint conditions on $\eta$ and ${\eta }^{\prime }$ make every boundary term zero. Since the remaining integral vanishes for every admissible variation, the fundamental lemma of the calculus of variations yields the higher-order Euler-Lagrange equation

Forthe parity condition is preserved by componentwise addition, negation, and multiplication, and $(1,1)$ is the identity. Thus $R$ is a ring.

It is not an integral domain, since $(2,0)$ and $(0,2)$ are nonzero elements of $R$ whose product is zero. It is also not a product of two nontrivial rings. Indeed, an idempotent in ${\mathbb{Z}}^2$ has each coordinate in $\{0,1\}$, and the parity condition leaves only $(0,0)$ and $(1,1)$. The proved equivalence then excludes a nontrivial product decomposition.

To deduce the final assertion, use the characterization that an integral domain is a unique factorization domain when it is atomic and every irreducible element is prime. Factoring the contents in $R$ and then the primitive parts by degree shows that $R[X]$ is atomic.

An irreducible constant of $R$ is prime in $R[X]$. Every irreducible polynomial of positive degree is, up to a constant unit, primitive. Part (i) makes it irreducible in $F[X]$; because $F[X]$ is a principal ideal domain, it is prime there, and part (ii) makes it prime in $R[X]$. Thus every irreducible of $R[X]$ is prime, andThis is the polynomial ring over a unique factorization domain theorem.