This is the affine plane through $b,d,f$, provided $d-b$ and $f-b$ are independent. The scalars $\lambda ,\mu$ are affine coordinates in those two directions.

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This is the plane through $a$ perpendicular to the nonzero normal vector $n$.

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Let $L_h$ be the spatial matrix, so the forward Euler method isThe conservative use of shared edge coefficients makes $L_h$ a real symmetric matrix. In an interior row, put $s=\alpha +\beta +\gamma +\delta <4a_{\max }$. Its diagonal entry is $-s/h^2$, and the sum of the absolute off-diagonal entries is $s/h^2$. Boundary rows have no larger disk because homogeneous Dirichlet data remove some unknown neighbors.

The Gershgorin circle theorem and symmetry therefore place every eigenvalue of $L_h$ inIf $\mu =k/h^2$, each amplification eigenvalue is $1+k\lambda$. It lies in $[-1,1]$ wheneverorSince the amplification matrix is symmetric, bounding all its eigenvalues in modulus by one bounds its discrete Euclidean operator norm by one. This proves the stated stability condition and is the Gershgorin stability bound for a variable-coefficient diffusion stencil.

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Sample the diffusion coefficient at edge midpoints and writeCentered differences of the two fluxes give the conservative finite-difference stencilFor example, the $x$ part isTaylor expansion about $(ih,jh)$ shows that the odd powers cancel between the two face fluxes and that the result equals ${\partial }_x(a{u}_x)+O(h^2)$. The same calculation in $y$ gives total truncation error

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The fluid velocity at the piston equals its velocity,Using $c-c_0=(\gamma -1)u/2$ in part (ii) givesOver $0\le t\le 1$,ThusThe linear pressure oscillation has zero mean, while the quadratic compressibility produces a positive mean pressure, the finite-amplitude acoustic analogue of radiation pressure.

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The motion is isentropic before a shock forms, and the perfect-gas relations giveWriting $\delta =\delta c/c_0$ and applying the generalized binomial expansion,

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Before shock formation, the piston launches a right-moving simple wave into the undisturbed gas. The opposite invariant is fixed at its ambient value:Therefore the simple-wave relation is

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The covariant derivative adds one covariant index, so $\nabla S$ has type $(2,2)$. The outer product of two tensors adds their index counts. Hence

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The map takes one covector argument and returns two vector slots, so $S$ has two contravariant and one covariant index. Its tensor type isor type $(2,1)$, with components $S^{ab}{}_c$.

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Removing the closed unit ball leaveswhich is an open subset of ${\mathbb{R}}^3$. Every point therefore has an ordinary Euclidean coordinate neighborhood, soIt has no boundary; the unit sphere was removed together with the ball.

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The antipodal action fixes the origin. Away from the origin it acts freely, so the quotient is locally Euclidean there. A small punctured neighborhood of the image of the origin, however, has linkthe real projective plane, whereas the link of a point in a three-dimensional smooth manifold is $S^2$. Equivalently, the neighborhood is a cone on ${\mathbb{RP}}^2$ and cannot be a three-ball. Thus

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Direct contact permits no work extraction, so conservation of internal energy givesThe total entropy change isThe arithmetic-geometric mean inequality makes the last expression positive when $T_1\ne T_2$, with equality only when the bodies were already at the same temperature. This is the entropy production of irreversible heat flow.

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Let the common final temperature be $T_f$. A reversible isolated process has zero total entropy production, so part (i) givesHencefor both bodies. The decrease in their total internal energy is extracted as work by the reversible engine mediating the transfer.

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At fixed volume, a reversible infinitesimal heating supplies $\delta Q_{\mathrm{rev}}=C_{V}dT$. Since entropy satisfies $dS=\delta Q_{\mathrm{rev}}/T$,

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Here $b_n=\lambda 2^{-n}$, so detailed balance givesConsequentlyThe series converges for every $\rho >0$, so this defines the unique invariant distribution.

By the PASTA property, an arrival joins with probability ${\sum }_{n\ge 0}{\pi }_{n}p(n)$. In equilibrium the accepted-arrival rate must also equal the departure rate. Departures occur at rate $\mu$ precisely when the queue is nonempty, henceThe required joining probability is therefore

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With state-dependent admission, the effective birth rate in state $n$ iswhile the death rate is $\mu$ for $n\ge 1$. The stationary ratios for an M-M-1 queue with state-dependent admission are thereforeInduction gives ${\pi }_n={\pi }_0{\rho }^n/n!$, and normalization identifies a Poisson distribution of mean $\rho$:

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Statement (iii) can fail because a nontransitive ambient set can omit the elements that distinguish two of its members. TakeNeither element of $y$ has an element that also lies in $y$: the second contains $\{\varnothing \}$, which is omitted. Their predecessor sets for membership restricted to $y$ are therefore both empty, although the two elements are distinct. Hence this restricted membership relation, and any isomorphic relation, is not extensional.

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Statement (ii) is always true even when $y$ is not transitive. Given a nonempty subset $z\subseteq y$, the Axiom of foundation supplies $a\in z$ with $a\cap z=\varnothing$. Thus no member of $z$ is a member of $a$, so $a$ is minimal for membership restricted to $y$. An isomorphic relation $r$ is therefore well-founded.

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A relation $r$ on a set $x$ is well-founded when every nonempty subset of $x$ has an $r$-minimal element. It is extensional when distinct elements have distinct predecessor sets:The Mostowski collapse theorem states that every well-founded extensional relation on a set is uniquely isomorphic to membership on a transitive set; the collapse is recursively

For the requested example, let $A=V_{\omega +1}$ and defineNo $x_n$ belongs to $A$, and thereforeThe map $x_n\mapsto n$ is an isomorphism from membership on $x$ to membership on the von Neumann ordinal $\omega$, so the Mostowski collapse is $\omega$. On the other hand, $x_0=V_{\omega +1}$ already has rank $\omega +1$, and hence the rank of $x$ is greater than $\omega$.

For statement (i), suppose $(x,r)$ is isomorphic to $(y,\in )$ with $y$ transitive. Membership is well-founded by the axiom of foundation. It is extensional on $y$ because transitivity ensures that the predecessors in $y$ of an element $a\in y$ are exactly the elements of $a$, and the axiom of extensionality distinguishes different sets. Both properties are preserved by isomorphism. Thus (i) is always true.

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The annulus does not have the fixed-point property. The antipodal mapis a continuous self-map of the annulus and has no fixed point there.

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