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Suppose first that a retraction $R:D\to C$ exists. The continuous map $p\mapsto -R(p)$ takes $D$ to itself, so the Brouwer fixed-point theorem gives $p=-R(p)$. Such $p$ lies on $C$, where $R(p)=p$, contradicting $p=-p$.

Conversely, suppose a continuous map $T:D\to D$ has no fixed point. For each $p$, follow the ray from $T(p)$ through $p$ to its last intersection with $C$. This defines a retraction. More explicitly, put $a=p-T(p)\ne 0$ and setThe positive-root formula is continuous, $t(p)\ge 1$, and $R(p)$ is on the unit circle. If $p\in C$, the last intersection is $p$ itself, so $R(p)=p$. Thus the nonexistence of a retraction rules out a fixed-point-free $T$, proving the equivalence.

The planar Brouwer fixed-point theorem states that every continuous map from a closed disc to itself has a fixed point.

Write $A=D+L+U$, with $D$ diagonal and $L,U$ the strict triangular parts. The Jacobi method updates all coordinates using the preceding iterate:For a symmetric positive definite matrix, $D$ has positive diagonal entries. In its splitting take $M=D$, $N=D-A$. The theorem requires $M^{\ast }+N=2D-A$ to be positive definite, which need not hold for every positive definite matrix.

For this tridiagonal matrix, let $S=\mathrm{diag}((-1{)}^1,\dots ,(-1{)}^n)$. Neighboring indices have opposite signs, so $SAS=2D-A$: the diagonal stays fixed and every nonzero off-diagonal entry changes sign. ThusThe Householder-John theorem now applies and proves Jacobi convergence for every initial vector. This is precisely why the tridiagonal structure matters.

The Householder-John theorem says that, for a splitting $A=M-N$ with $A$ Hermitian positive definite, positivity of $H=M^{\ast }+N=M^{\ast }+M-A$ implies $M$ is invertible and $\rho (M^{-1}N)<1$. Invertibility follows since $Mx=0$ for $x\ne 0$ would give $x^{\ast }Hx=-x^{\ast }Ax<0$.

For clarity, the spectral-radius conclusion follows from a useful energy identity. Put $T=M^{-1}N=I-M^{-1}A$. Direct multiplication givesIf $Tv=\lambda v$, thenThe eigenvalue one is impossible because $A$ is invertible. Both quadratic forms are positive, so $|\lambda |<1$. The splitting iteration $M{x}^{(k+1)}=N{x}^{(k)}+b$ propagates its error by $T$; spectral radius less than one therefore gives convergence from every initial vector.

The baffles create an induced pressure gradient even though no external gradient is applied. Away from them the same Couette-Poiseuille flow has flux per unit widthZero flux therefore gives $G=-6U/a^2$ andThe positive pressure gradient drives the return flow near the stationary lower plate, balancing the forward flow near the moving plate. The rectilinear approximation excludes the short end regions next to the baffles.

The steady rectilinear equation is $u^{\prime \prime }(y)=-G$, with no-slip conditions $u(0)=0$ and $u(a)=U$. Two integrations give the Couette-Poiseuille flowThe first term is driven by the moving wall and the second by the specified pressure gradient.

The substitution $u=e^{-x}$ changes the beta function toFor fixed $t>0$, its factor near zero expands asAway from zero the factor is bounded, so the extension from the finite-interval Watson lemma to this exponentially damped tail is immediate. ThereforeThe change $u\mapsto 1-u$ gives $B(s,t)=B(t,s)$, so, with $s>0$ fixed,These expansions hold with the other parameter fixed; the first does not claim uniformity when $t$ grows with $s$. The printed gamma-integral hint has inconsistent variable labels: the integration differential is $dx$ and the positive parameter is $y$.

Fix an integer $N\ge 0$ and writeFor each power, substitute $y=\lambda x$ and extend the upper limit to infinity; the discarded tail is exponentially small. Since $\alpha +j\beta >-1$, the gamma integral givesTo control the remainder, choose $\delta >0$ such that $|r_N(x)|\le ϵx^{\alpha +N\beta }$ for $0<x<\delta$. Its integral there is at most $ϵ\Gamma (\alpha +N\beta +1){\lambda }^{-\alpha -N\beta -1}$. On $[\delta ,b]$ the remainder is bounded and its integral is $O(e^{-\lambda \delta })$. Since $ϵ$ is arbitrary, we have the full Watson lemma expansionMore precisely, after the term $j=N$, the remainder is $o({\lambda }^{-\alpha -N\beta -1})$. This proof uses neither differentiability of $f$ nor convergence of its asymptotic series.

Repeated integration by parts would require derivatives of $f$ and finite endpoint values of those derivatives. The hypothesis gives only continuity away from zero, and even the leading behavior $x^{\alpha }$ can diverge at zero or have singular derivatives when $\alpha$ is nonintegral. Thus those integrations need not be defined. The appropriate method integrates the asymptotic powers themselves, rather than differentiating an insufficiently regular function.

Write $[u]=u_r-u_l$ and $[f]=u_r^2/2-u_l^2/2$. Splitting the test-function integral along $x=s(t)$ and integrating by parts gives a remaining interface contributionFor example the normal to the interface in $(x,t)$ coordinates is proportional to $(1,-s^{\prime })$, so the jump in the conserved spacetime flux $(u^2/2,u)$ has normal component $[f]-s^{\prime }[u]$. Since the traces of the test function are arbitrary, the interior weak equation holds if and only if $s^{\prime }[u]=[f]$. At a genuine jump $[u]\ne 0$, this isThis is the Rankine-Hugoniot condition. An initial-value weak formulation also requires the prescribed initial trace; the jump condition alone concerns the interior equation. The source's reference to “(1)” is to the displayed Burgers equation, not a different PDE.

Inside the fan $u=x/t$, one has $u_t=-x/t^2$ and $u{u}_x=x/t^2$, while the constant regions also satisfy the equation. Across $x=0$ and $x=t$, the two traces of both $u$ and its flux $u^2/2$ agree. Integrating by parts separately in the three regions therefore cancels all interface terms and givesfor every smooth compactly supported test function. The initial term is correct because $u(\cdot ,t)\to 1_{\{x>0\}}$ in local $L^1$: the difference is supported in $(0,t)$ and its integral is at most $t$. Thus the fan is a weak solution of the Burgers equation with the required initial data.

The characteristic equations for a transport equation are ${\dot{x}}_1=x_1$, ${\dot{x}}_2=2x_2$, $\dot{u}=5u$. Starting from $(\xi ,1)$ at parameter zero gives $(x_1,x_2)=(\xi e^s,e^{2s})$, $u=e^{5s}g(\xi )$. Eliminating $s,\xi$ givesEvery point of this upper half-plane lies on exactly one characteristic from the initial line, so it is the maximal characteristic domain determined by the data. Neither the line $x_2=0$ nor the lower half-plane is reached in finite characteristic time; extra data would be needed there.

For $g\in C^1$ this is a classical solution, directly verified by differentiation. The printed assumption is only continuity, which in general does not give a classical $C^1$ solution. Under that assumption the same formula is the unique continuous characteristic solution and solves the transport equation distributionally; smoothing $g$ locally and passing to limits verifies the latter interpretation. A classical interpretation needs the additional regularity just stated.

No. Take the complete metric space $X=\mathbb{R}$ with its usual metric, $E=\mathbb{Q}$, and $G=\mathbb{Q}\cup (1,2)$. This $G$ is uncountable and dense, whereasmisses, for example, the nonempty open interval $(3,4)$. Uncountability does not replace the open-dense hypothesis in the Baire category theorem.

The Baire category theorem says that a countable intersection of open dense sets in a nonempty complete metric space is dense. Write $E=\{e_1,e_2,\dots \}$. Each singleton is closed in a metric space and has empty interior because there are no isolated points. Therefore every $X\setminus \{e_j\}$ is open and dense. Applying the theorem also to $G$ yieldswhich is dense in $X$. Thus every nonempty open subset meets the required difference, not merely the original dense set $G$.

At a positive-rate state the jump chain moves to either other state with probability $1/2$, independently of the holding rate. At a zero-rate state there is no subsequent jump; extend the chain as permanently absorbed there, giving an identity row. This convention is needed to speak of all powers of its transition matrix in the absorbing cases.

When all rates are positive,Its unique equilibrium is uniform and its powers tend to the matrix with all rows $(1/3,1/3,1/3)$. This differs from the holding-time-weighted equilibrium of the continuous-time chain unless the rates coincide.

With one zero-rate state the unique jump-chain equilibrium is its point mass and the limit has all rows that point mass. With two zero-rate states the equilibria are all mixtures on them, and the limiting rows are exactly the absorbing rows and half-half transient row found in part (d).

When all rates are positive, finite irreducible convergence gives $P(t)\to 1\pi$, with $\pi$ as above. With one zero-rate state $z$, each visit to either transient state has probability $1/2$ of entering $z$ on the next jump. Survival for $n$ jumps therefore has probability $2^{-n}$, so absorption is almost sure andWith two zero-rate states $z_1,z_2$ and the sole positive-rate state $k$, absorption occurs at the first jump, with probability $1/2$ into each absorbing state. The limit rows are $e_{z_1}^T,e_{z_2}^T$ for those states and $(e_{z_1}^T+e_{z_2}^T)/2$ for row $k$. These limits depend on the initial state, as nonuniqueness requires.

The stationary equations simplify by setting $a_i={\pi }_i{\lambda }_i$. They read $2a_i={\sum }_{j\ne i}{a}_j$, so all three $a_i$ are equal. If all rates are positive, this gives the unique equilibriumIf any rate is zero, the common $a_i$ is zero, forcing zero equilibrium mass on every positive-rate state. Every zero-rate state is absorbing. With one such state $z$, the unique equilibrium is ${\delta }_z$; with two, all probability mixtures on those two states are equilibria, giving nonuniqueness. The excluded all-zero case would make every distribution an equilibrium.

For an irreducible, nonexplosive, positive recurrent continuous-time Markov chain on a countable state space, there is a unique equilibrium distribution $\pi$ andIn particular every finite irreducible chain has this convergence. Continuous time removes the discrete-time periodicity obstruction. Reducible finite chains instead converge according to absorption probabilities into their closed classes.

An equilibrium distribution of a continuous-time Markov chain is a probability row vector $\pi$ invariant under the transition semigroup: $\pi P(t)=\pi$ for every $t\ge 0$. Differentiating gives $\pi Q=0$, or ${\sum }_i{\pi }_i{q}_{ij}=0$ for each $j$, when the generator sums are defined. Reducible chains may have several closed classes supporting different equilibrium distributions; convex mixtures are then also equilibria. Irreducibility and positive recurrence give uniqueness for a conservative nonexplosive chain.

All three assertions are false. For the transversality assertion, use $f:\mathbb{R}\to \mathbb{R}$, $f(x)=x^3$, and $Z=\{0\}$. At the preimage point the derivative is zero, so $f$ is not transverse to $Z$, yet $f^{-1}(Z)=\{0\}$ is a codimension-one submanifold. Transversality is sufficient, not necessary.

For openness of regular values, the smooth function $f(x)=e^{-x}(2+3\sin x)$ gives a connected-domain counterexample. Every zero has $\sin x=-2/3$ and derivative $3e^{-x}\cos x\ne 0$, so zero is a regular value. Critical points solve $\cos x-\sin x=2/3$. Choose any such point $x_0$; its value is nonzero, since simultaneous vanishing would force $\sin x=-2/3$ and $\cos x=0$. The critical points $x_0+2\pi n$ have nonzero critical values $e^{-2\pi n}f(x_0)\to 0$. Thus no neighborhood of the regular value zero consists entirely of regular values. Properness would supply the missing compactness needed for the usual open-regular-value conclusion.

For the measure assertion, a constant map $f:\mathbb{R}\to \mathbb{R}$ has every point critical. Its critical-point set has positive, indeed infinite, measure. Sard theorem concerns the measure of the set of critical values in the target, not the measure of critical points in the domain.