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Suppose the incident axisymmetric flow is an incompressible flow and an irrotational flow, with Stokes streamfunction $\Psi$ regular inside $r<a$. The displayed transformation adds an impermeable sphere at $r=a$ without new finite exterior singularities. Writing $s=a^2/r$ and $G=(r/a)\Psi (s,\theta )$, direct differentiation showsHence the Stokes streamfunction equation for irrotational flow is preserved. At $r=a$, $\tilde{\Psi }=0$, making the radial velocity zero. Exterior points sample only the regular interior under inversion. If $\Psi$ vanishes at the origin and is $O(r)$ there with smooth regular derivatives, the induced velocity decays at infinity. This is an inversion construction for an axisymmetric streamfunction, distinct from a scalar Kelvin transform.

For conforming piecewise affine elements on uniform squares split along their positive diagonals, a lower-triangle element matrix is $\frac{1}{2}[[1,-1,0],[-1,2,-1],[0,-1,1]]$ in consecutive vertex order. Finite element assembly gives diagonal entry 4 and four axis-neighbor entries $-1$. The diagonal mesh edges contribute zero coupling because the corresponding local shape-function gradients are perpendicular. Strong Dirichlet elimination removes known-neighbor columns and adjusts the load.

Consider $n$ players choosing a safe action worth $h$ or a joint action whose total reward $d$ is shared equally among its $s$ participants if $s\ge k$, and is zero otherwise. Assume $0<h<d$ and $2\le k\le n$. The safe-action profile is a Nash equilibrium. For $0<s<n$, a profile is a Nash equilibrium exactly under the displayed conditions: a joint participant must prefer $d/s$ to $h$, and a safe participant must prefer $h$ to $d/(s+1)$. An unsuccessful participant would strictly prefer the safe action. The all-joint endpoint is an equilibrium exactly when $d/n\ge h$.

Writing $r=d/h$, positive equilibrium counts lie at $s\le r\le s+1$ with $s\ge k$, or at $s=n$ when $r\ge n$. There are at most two. Exactly one exists when $r\ge k$ and $r\notin \{k+1,\dots ,n\}$. An integer $r$ in that excluded set permits both $r-1$ and $r$. Strict positivity matters: when $h=0<d$, the positive equilibrium counts instead include every $1\le s\le k-2$ as well as $n$.

Historical formulation name; composition is identified by pharmacopoeial edition.

Historical formulation name; composition is identified by pharmacopoeial edition.

Historical formulation name; composition is identified by pharmacopoeial edition.

A single tree rotation moving a node above its parent when that parent is the root. A left child uses a right tree rotation, and a right child uses a left tree rotation.

When a node and its parent are both left children, first apply a right tree rotation to the grandparent, moving the parent upward, and then another right tree rotation moving the selected node upward. Use the left-right mirror image when both are right children. The order of these two tree rotations distinguishes the operation from repeatedly rotating only the selected node.

When a node and its parent occupy opposite left/right directions, rotate the selected node above its parent and then above its grandparent. These two tree rotations move the selected node two levels upward while preserving the binary search tree ordering.

Every node of a splay tree with at least four distinct keys can become a leaf of a rooted tree using only successful splaying operations. At an extreme key, access that key and then its sole adjacent stored key; the consecutive keys in a binary search tree property makes the extreme key a leaf. For an interior key $x$ with two larger stored keys, take consecutive $a<x<b<c$ and access $c,a,b$; the mirror construction handles two smaller keys.

For a quick shape proof, use $k(L,R)$ for a node and its two subtrees. After accessing $c,a$, the right subtree of $a$ can only be $c(b(x,0),D)$, $c(x(0,b),D)$, $b(x,c(0,D))$, or $x(0,c(b,D))$, with outside subtrees $A$ below $a$ and $D$ above $c$. This follows by inspecting the final zig step, zig-zig step or zig-zag step that moves $a$ above the former root $c$. In each case, accessing $b$ yields $b(a(A,x),c(0,D))$, where $x$ is a leaf and $0$ denotes an empty subtree. These rotations prove the bound for arbitrary outside subtrees.

In dimensions at least two a function in the Sobolev space $H^1$ can have finite norm while $\int w^2|\nabla w{|}^2=\infty$. Localize the following radial examples with a smooth cutoff function in an interior ball. For $n\ge 3$, set $w(r)=r^{-\alpha }$, $\alpha =(n-2)/3$. Its gradient energy has radial exponent $n-3-2\alpha >-1$, but its weighted gradient energy has exponent $n-3-4\alpha <-1$. For $n=2$, take $w(r)=(\log (1/r){)}^{1/3}$ near zero. Substituting $s=\log (1/r)$ makes the two gradient integrals proportional to ${\int }^{\infty }{s}^{-4/3}ds$ and ${\int }^{\infty }{s}^{-2/3}ds$, respectively. The first converges and the second diverges. In dimension one the Sobolev embedding theorem gives $H^1\subset L^{\infty }$, preventing this failure. These examples show why a nonnegative coefficient $w^2$ need not even define a bounded gradient form when only $w\in H^1$ is assumed.

An initial layer is a short transient needed when the reduced outer solution cannot satisfy the original initial condition. On an inner time scale, terms multiplied by a small parameter remain comparable to the retained dynamics. For $ϵc_t=1-(1+\beta )c$, $c(0)=0$, the inner solution $c=(1-e^{-(1+\beta )t/ϵ})/(1+\beta )$ matches the outer value $1/(1+\beta )$ while retaining the initial datum.

For a fast-slow system $ϵ\dot{x}=f(x,y)$, $\dot{y}=g(x,y)$, the formal critical manifold is $f=0$. Where the fast linearization has eigenvalues with strictly negative real parts, trajectories rapidly approach a nearby attracting slow manifold and then evolve on the slow time. Near folds or loss of this spectral condition, simply imposing $f=0$ may fail.

For a fast-slow system $ϵ\dot{x}=F(x,y)$, $\dot{y}=G(x,y)$ with $0<ϵ\ll 1$, the rescaled time $\tau =t/ϵ$ gives $x_{\tau }=F$, $y_{\tau }=ϵG$. Away from zeros of $F$, an order-one change of $x$ takes order-one fast time but only order-$ϵ$ original time, while $y$ changes little. Setting $ϵ=0$ in these rescaled equations freezes $y$ and produces the leading fast subsystem. This limit differs from the slow reduction $F=0$ in the original time; matching the two describes a relaxation oscillation.

For $\ddot{x}+k(x^2-1)\dot{x}+x=\lambda$, with fixed $|\lambda |<1$ and $k\to \infty$, the Liénard coordinate $y=\dot{x}/k+x^3/3-x$ has fast horizontal jumps and slow motion on $y=x^3/3-x$. The slow time is $t/k$. Integrating $(x^2-1)dx/dT=\lambda -x$ along $[-2,-1]$ and $[2,1]$ gives the displayed period. The two jumps take negligible time at this scale. For fixed $\lambda >1$, the right attracting branch contains the stable equilibrium and the large relaxation cycle disappears.

An integral equation with a kernel singular on the integration contour needs a prescribed limiting interpretation such as a Cauchy principal value. For the Cauchy kernel, the Sokhotski–Plemelj theorem converts a linear scalar equation into a Riemann-Hilbert problem for its two analytic boundary values. Normalization, endpoint regularity and the index of that jump determine homogeneous freedom and solvability.

For an open cover $\mathcal{U}$ of a space $X$, let $C_{\ast }^{\mathcal{U}}(X)$ be the subgroup generated by singular simplices whose images lie in a single member of the cover. It is a chain complex, and its inclusion into the full singular chain complex induces an isomorphism on homology.

For a finite chain, enough iterations of barycentric subdivision make every simplex small: pull the cover back to each compact standard simplex and use its Lebesgue number. The subdivision operator commutes with the boundary and is related to the identity by the standard subdivision chain homotopy, which keeps each produced simplex inside its original image. Therefore any cycle becomes homologous to a small cycle, proving surjectivity.

If a small cycle $z$ bounds a full chain $b$, choose $N$ so that ${\mathrm{Sd}}^{N}b$ is small. Write ${\mathrm{Sd}}^N-1=\partial Q+Q\partial$. Since the homotopy preserves simplex images, $Qz$ is also small. Then $z=\partial ({\mathrm{Sd}}^{N}b-Qz)$ is a small boundary, proving injectivity. This is the chain-level locality input in the proof of the Mayer–Vietoris sequence.

Initialize source distance to zero and other distances to infinity, then process vertices in topological order, relaxing their outgoing edges once. Every predecessor on a path to a vertex appears earlier, so induction through that ordering proves that its distance is final when processed. Negative weights are allowed because there are no directed cycles. With adjacency lists, topological sorting and all edge relaxations take $O(|V|+|E|)$ time; predecessor pointers reconstruct paths.

An extension of simple type assignment that assigns a distinguished atomic type variable $\star$ to every lambda term in the empty typing context. The recorded assumptions therefore need not contain all free variables of the subject. This is a rule admitting a typing judgment, rather than an implicit conversion to every simple type or a general subtyping rule.

For $\gamma >0$, direct multiplication gives $|FN{|}^2-|N{|}^2=\gamma N_2(2N_1+\gamma N_2)$. Thus length-preserving line directions form the union of two planes in direction space: $N_2=0$ and $2N_1+\gamma N_2=0$. Their intersection is the third coordinate axis. This concerns segment length rather than pointwise fixed lines; a segment can move and turn while keeping its length.

The two nontrivial principal stretches are $v_{\pm }=(\sqrt{4+{\gamma }^2}\pm \gamma )/2$, with $v_{+}{v}_{-}=1$. The maximizing reference direction is proportional to $(1,v_{+},0)$. Consequently it tends to $(1,1,0)$ as $\gamma$ tends to zero from above. At zero itself all stretches equal one, so every direction maximizes length; the limit selects a direction from a degenerate endpoint. This follows from the largest eigenvector of the right Cauchy-Green deformation tensor.