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Put $h_0=1$ and $h_r=0$. The Markov property and first-step conditioning give $h_i={\sum }_j{P}_{ij}{h}_j$. Consider a finite history ending at state $i$ and having positive conditional probability given $A$. The chance of eventual absorption at zero after that history is $h_i$: the future depends only on the present state, and if zero has already been reached, absorption keeps the chain there. If the next state is $j$, this chance is $h_j$. Bayes' rule therefore givesThis depends on the history only through $i$, proving the conditional Markov property. The conditional transition matrix isIts row sums are one by harmonicity of $h$, and $Q_{ir}=0$. State $r$ is never visited under the conditioning and is naturally removed; any row assigned there on an enlarged formal state space is immaterial, since conditioning starting at $r$ is undefined. State zero stays absorbing. For a random initial law $\nu$, its conditional law is ${\nu }_i{h}_i/{\sum }_j{\nu }_j{h}_j$. This is the Doob h-transform with the absorption probability as harmonic function.

For the simple symmetric walk, $h_i=(r-i)/r$ solves the linear harmonic recurrence and its boundary values. ThusIn particular, the conditioned chain moves downward with probability one from $r-1$.

Let $\tau$ be the first hit of $\{0,r\}$ in the original walk and let $g_i={\mathbb{E}}_i[\tau 1_A]$. Its absorption time has finite expectation: from any transient state a run of at most $r$ downward steps hits zero with probability at least $2^{-r}$, giving a geometric tail bound by blocks. First-step conditioning then givesEquivalently, $g_{i+1}-2g_i+g_{i-1}=-2(r-i)/r$. The cubic polynomialhas exactly these second differences and boundary values. It is the unique solution, since the difference of any two solutions is a linear sequence vanishing at both endpoints. Conditional on $A$, $\tau$ is the first time in zero, so division by $h_i$ provesThis is the duration of gambler's ruin conditioned on a chosen boundary.

A steady localized electric current density satisfies $\nabla \cdot \mathbf{j}=0$, from charge conservation or the divergence of Ampère's equation. Choose an enclosing region whose boundary lies outside its support; extending the integration region does not change the source integrals. The divergence theorem applied componentwise givesTheir boundary fluxes vanish, proving the localized steady-current moment identitiesFor discontinuous but localized currents the same argument uses the distributional steady conservation law and an outer enclosing surface.

The magnetic vector potential is a vector field with $\mathbf{B}=\nabla \times \mathbf{A}$, consistent with $\nabla \cdot \mathbf{B}=0$. It is defined up to $\mathbf{A}\mapsto \mathbf{A}+\nabla \chi$. The steady Maxwell equation $\nabla \times \mathbf{B}={\mu }_0\mathbf{j}$ givesIn the Coulomb gauge $\nabla \cdot \mathbf{A}=0$, the equation is ${\nabla }^2\mathbf{A}=-{\mu }_0\mathbf{j}$, whose decaying Green-function solution is the one supplied in the source.

Let $r=|\mathbf{x}|$ and expand its kernel uniformly over the bounded source region:The first term integrates to zero. Put $T_{ki}=\int x_k^{\prime }{j}_i({\mathbf{x}}^{\prime })d{V}^{\prime }$, an antisymmetric tensor by the earlier identity. Define the magnetic dipole moment $m_{\ell }=\frac{1}{2}{ϵ}_{\ell ki}{T}_{ki}$, so $T_{ki}={ϵ}_{ki\ell }{m}_{\ell }$. ConsequentlyorThe displayed dipole term has size $r^{-2}$. If $\mathbf{m}=0$, it vanishes and a higher multipole determines the leading magnetic field instead.

For $r>0$, $\nabla \cdot (\mathbf{x}/r^3)=0$ andTherefore the far magnetic field is

For a stationary spherically symmetric state, set $u(r)=r\psi (r)$. The Schrödinger equation becomesA bound state must have $E<0$, while the well's lower bound gives $E>-U$ for a nonzero square-integrable eigenstate. Define $k=\sqrt{2m(U+E)}/\hslash$ and $\kappa =\sqrt{-2mE}/\hslash$. Regularity at the origin requires $u(0)=0$, so the interior cosine solution is excluded; square integrability at infinity excludes the growing exterior exponential. Matching $u$ and $u^{\prime }$ at $r=a$ therefore givesThe limit at the origin is the finite value $Ck$. Continuity of $u,u^{\prime }$ is equivalent here to continuity of $\psi ,{\psi }^{\prime }$ at the finite potential jump. If desired, the normalized state hasThe equation for $k$ implicitly determines the energy, with $k^2+{\kappa }^2=2mU/{\hslash }^2$.

Write $K=a\sqrt{2mU}/\hslash$ and $z=ka$. The matching condition is $z\cot z=-\sqrt{K^2-z^2}$. The first possible solution has $\pi /2<z<\pi$. If $K\le \pi /2$, every permitted interior $z<K$ has nonnegative cotangent and cannot match a negative exterior logarithmic derivative. If $K>\pi /2$, at $z\downarrow \pi /2$ the sum $z\cot z+\sqrt{K^2-z^2}$ is positive; it is negative either at $z\uparrow K$ when $K<\pi$, or at $z\uparrow \pi$ when $K\ge \pi$. Continuity supplies a first bound-state solution. Thus at least one spherically symmetric bound state exists exactly whenAt equality the limiting energy is zero and the exterior wave is proportional to $1/r$, whose norm diverges. Hence the inequality is strict. Deeper wells can support additional radial states; their negative-energy matching solutions have the same displayed form. This is the spherically symmetric bound state of a finite well.

Integrate the entire function $e^{i{z}^2}$ around the positively oriented sector boundary: the real segment $0\to R$, the arc $R{e}^{i\theta }$ for $0\le \theta \le \pi /4$, and the radial segment back to zero. Cauchy integral theorem givesThe last term follows from $z=r{e}^{i\pi /4}$ and $i{z}^2=-r^2$; its minus sign is the return orientation.

On the arc, $|e^{i{z}^2}|=e^{-R^2\sin (2\theta )}$. The supplied sine bound gives $\sin (2\theta )\ge 4\theta /\pi$ throughout this sector, soThe Gaussian integral then yields ${\int }_0^{\infty }{e}^{i{x}^2}dx=e^{i\pi /4}\sqrt{\pi }/2$. Taking real and imaginary parts proves both Fresnel integrals:The contour estimate proves existence of the improper oscillatory integrals as well as their values; absolute convergence is neither asserted nor required.

Differentiability at $p$ means there is a linear map $A=Df{|}_p$ such thatThis is the Fréchet derivative definition; the same linear approximation must work in every direction, not merely along coordinate axes.

The chain rule says that if $f$ is differentiable at $p$ and $g$ is differentiable at $f(p)$, then $g\circ f$ is differentiable at $p$ andFor its proof, put $A=Df{|}_p$, $B=Dg{|}_{f(p)}$, and $k=f(p+h)-f(p)=Ah+r(h)$. Since a linear map between these finite-dimensional normed spaces is bounded, $\Vert k\Vert =O(\Vert h\Vert )$. Differentiability of $g$ gives $g(f(p)+k)=g(f(p))+Bk+s(k)$ with $s(k)=o(\Vert k\Vert )$. ThereforeIf $k=0$, take $s(0)=0$; otherwise $\Vert s(k)\Vert /\Vert h\Vert =(\Vert s(k)\Vert /\Vert k\Vert )(\Vert k\Vert /\Vert h\Vert )\to 0$. This proves the required full linear approximation.

For the Euclidean norm, at every $x\ne 0$ the ordinary one-coordinate differentiation givesAt the origin its difference quotient in direction $e_i$ is $|t|/t$, whose right and left limits are $1$ and $-1$. Thus none of its coordinate partial derivatives exists at the origin.