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Writing the velocity as $(u,v)=(y,ax)$ givesso the flow is incompressible. With the conventiona stream function isThe streamlines are its level sets. For $a>0$, they are the hyperbolaswith separatrices $y=\pm \sqrt{a}x$ and a saddle at the origin. For $a<0$, they are concentric ellipsestraversed clockwise. This is the streamline classification of a planar linear saddle or centre.

The scalar vorticity isHence the flow is irrotational exactly whenThen $(u,v)=(y,x)=\nabla \varphi$, and a velocity potential is

The coefficient condition says precisely that every nonconstant monomial has positive powers of both variables. ThusIf $c+XYf$ and $d+XYg$ lie in $R$, so do their sum and their productso $R$ is a subring of $K[X,Y]$.

For $n\ge 1$, letThen $I_n\subseteq I_{n+1}$. The containment is strict because $X{Y}^{n+1}\notin I_n$: multiplying a generator $X{Y}^j$ by an element of $R$ produces either a scalar multiple of $X{Y}^j$, or terms divisible by $X^2$. It cannot produce the monomial $X{Y}^{n+1}$ when $j\le n$. Henceis a strictly increasing ideal chain. The constant-plus-ideal non-Noetherian subring therefore satisfies

The steady Euler equations for an inviscid fluid areTaking the scalar product with $u$ and usinggivesThusThe Bernoulli function $H$ is constant along each streamline: fluid particles in steady flow exchange pressure, kinetic, and potential energy without changing their total mechanical energy density.

Let $y(t)$ be the downward displacement of the free surface from its initial level, and let $U$ be the speed in the tube. Conservation of volume givesThe free surface and the outlet are both at atmospheric pressure, and their vertical separation is $H+h_0-y$. Applying the Bernoulli equation between them, while retaining the small free-surface speed, givesThereforeThe surface reaches the upper tube end when $y=h_0$. Hence the draining time of a uniform tank through a siphon is

Fix $y\in V$. The map $x\mapsto \theta (x,y)$ is a linear functional, so the finite-dimensional Riesz representation theorem gives a unique vector $\beta (y)$ such thatFor scalars $\lambda ,\mu$,Uniqueness givesso $\beta$ is linear.

For $\alpha \in \mathrm{End}(V)$, apply this result toIt produces a unique linear map ${\alpha }^{\ast }$ satisfyingwhich proves existence and uniqueness of the adjoint.

Differentiate the spatial equation with respect to $t$ and the temporal equation with respect to $x$. Substitution ofinto ${\psi }_{xt}-ik{\psi }_t$ cancels every term involving $\psi$, $u$, and $u_x$, leavingThe left side is $u_t$ by differentiating ${\psi }_x-ik\psi =u$, so consistency is exactly $u_t=i{u}_{xx}$. This is a scalar Lax pair formulation.

Taking the half-line Fourier transform and integrating twice by parts givesMultiplication by $e^{i{k}^{2}t}$ and integration in time yields

Replace $k$ by $-k$ and subtract. The unknown Neumann boundary value cancels:For $x>0$, Fourier inversion of the zero extension givesInverting the preceding identity therefore givesThusThis elimination is the global relation for the half-line free Schrodinger equation.

Lorentz invariance requires $A^{\mu }$ and $J^{\mu }$ to transform as four-vectors, $\chi$ as a scalar, $m$ and ${\mu }_0$ as scalars, and the integration domain and boundary conditions to respect the transformation. Then every contracted term and $d^{4}x$ are Lorentz invariant.

Under $A_{\mu }\mapsto A_{\mu }+{\partial }_{\mu }\chi$, the field strength is unchanged. The source term changes by a boundary term pluswhich vanishes for a conserved current and suitable boundary behaviour of $\chi$. The mass term is not gauge invariant. Thus the action has the usual local gauge invariance whenwith the gauge function or its boundary contribution suitably controlled. For $m>0$, the mass term breaks this gauge symmetry.

For the static point source, take $A=0$ and $\alpha =0$. The scalar equation becomesIts spatial Fourier transform satisfiesThe stated inverse transform therefore giveswhere $\lambda$ absorbs the transform normalization and source strength. This is the Yukawa potential: a nonzero vector-field mass screens the interaction beyond the range $m^{-1}$. As $m\to 0$, it tends to the long-range Coulomb potential $\lambda /r$. The same limit removes the mass term that broke gauge invariance in part (b).

For scalar $\theta$, Taylor's theorem about the MLE, where
${\ell }_n^{\prime }({\hat{\theta }}_n)=0$, gives some ${\tilde{\theta }}_n$ between ${\theta }_0$ and ${\hat{\theta }}_n$ such thatThe Wald statistic isUnder the null, consistency gives
${\hat{\theta }}_n,{\tilde{\theta }}_n\to {\theta }_0$ in probability. A uniform law of large numbers for the observed information on a neighborhood of ${\theta }_0$ givesCancellation of the common squared displacement and Slutsky's theorem prove the Wald and likelihood-ratio asymptotic equivalence

Suppose the spectrum were empty. The resolventwould then be an entire operator-valued function. For $|\lambda |>\Vert T\Vert$,because the series converges in operator norm and multiplication by $\lambda I-T$ telescopes to $I$. Hence $\Vert R(\lambda )\Vert \le (|\lambda |-\Vert T\Vert {)}^{-1}$.

For fixed $x,y\in {\ell }^2$, the scalar function $⟨R(\lambda )x,y⟩$ is entire, bounded outside a disc by the estimate and bounded inside by compactness. Liouville's theorem makes it constant, and its limit at infinity makes that constant zero. This for all $x,y$ would imply $R(\lambda )=0$, impossible for an inverse. Thus the spectrum is nonempty.

An eigenvalue $\lambda$ is a scalar for which $\ker (A-\lambda I)$ contains a nonzero vector; that kernel is its eigenspace. The characteristic polynomial has degree $n$, so the fundamental theorem of algebra supplies at least one complex root and hence an eigenvalue. Its eigenspace dimension can be any integer from $1$ to $n$.

The fixed points are $0$ and, when $a\ge -1$, $1\pm \sqrt{1+a}$. For a scalar autonomous equation, a simple fixed point is stable when $f^{\prime }<0$. Since $f^{\prime }(0)=a$ and $f^{\prime }(r)=2r(1-r)$ at either nonzero root, excluding the bifurcation values as requested,and

The disc model isand the upper half-plane model isThe Cayley mapmaps $D$ bijectively to $\mathfrak{h}$. Since $C^{\prime }(z)=2i/(1-z{)}^2$ and $\mathrm{Im}C(z)=(1-|z{|}^2)/|1-z{|}^2$, direct substitution gives $C^{\ast }{g}_{\mathfrak{h}}=g_D$.

Representing $C$ by $K=\left(\begin{array}{cc}i& i\\ -1& 1\end{array}\right)$, the disc isometry corresponding to $g=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ is represented, up to a nonzero scalar, by

The dual $X^{\ast }$ is the vector space of bounded linear maps $f:X\to \mathbb{F}$, with normIf $(f_n)$ is Cauchy in this norm, then $(f_n(x))$ is Cauchy in the complete scalar field for every $x$. Define $f(x)={\lim }_n{f}_n(x)$. Pointwise passage to the limit preserves linearity. Given $\epsilon >0$, choose $N$ such that $\Vert f_n-f_m\Vert <\epsilon$ for $m,n\ge N$; sending $m\to \infty$ givesThus $f$ is bounded and $\Vert f_n-f\Vert \le \epsilon$. Hence $X^{\ast }$ is Banach, even if $X$ was only normed.

Let $E_2=(e_1,e_2)$ andBecause $Q_k{R}_k$ is the QR factorization of $H^{k+1}$, $S_k$ is the span of the first two columns $q_1^{(k)},q_2^{(k)}$ of $Q_k$.

The first row of the tridiagonal eigenvalue equation isIn the generic case ${\lambda }_{n-1}\ne {\lambda }_n$, the two dominant coefficient vectors are independent, sinceHere $h_{12}\ne 0$, since otherwise the displayed eigenvalue equation and the nonzero $b_j$ would force all eigenvalues to equal $h_{11}$. The strict spectral gap below the dominant pair now impliesThis is two-dimensional subspace iteration. If the dominant eigenvalue is repeated, $H$ acts as a scalar on $W$; the same QR argument either captures all of $W$ or separates its missing direction from the strictly smaller eigenspaces, and the conclusion below is unchanged.

Since $q_2^{(k)}\in S_k$ and $q_3^{(k)}\perp S_k$,The space $W$ is invariant under $H$, soShifting the index does not affect the limit, and thereforeThus the unshifted QR algorithm asymptotically deflates a $2\times 2$ block associated with the equal-modulus dominant pair.

Isotropic means invariant under every proper rotation. The general ranks $0,1,2,3$ are respectively a scalar $a$, zero, $b{\delta }_{ij}$, and $c{\epsilon }_{ijk}$ (for full orthogonal invariance the last must also vanish).

The Möbius group consists of maps $z\mapsto (az+b)/(cz+d)$ with $ad-bc\ne 0$, including the usual action on $\infty$. Scalar matrices act trivially and are the whole kernel, so $M\cong G{L}_2(\mathbb{C})/({\mathbb{C}}^{\ast }I)$.

By Schur lemma, $z\in Z$ acts on irreducible $V$ by a scalar. Induced representation shows that $V$ occurs in ${\mathrm{Ind}}_Z^G\Theta$ iff that scalar character is $\Theta$, and then with multiplicity $\dim V$.

An elementary row operation is one of three operations: interchange two rows, multiply one row by a nonzero scalar, or add a scalar multiple of one row to a different row. Each operation is reversed by another operation of the same kind.

Performing such an operation on an $m\times n$ matrix $A$ is equivalent to left multiplication by the $m\times m$ elementary matrix obtained by applying that operation to the identity matrix. Each elementary matrix is invertible, with inverse corresponding to the reverse operation. Thus a sequence of operations givesFor square matrices, if $A$ were invertible, then $B=PA$ would have inverse $A^{-1}{P}^{-1}$. Hence if $B$ is singular, $A$ must be singular as well. In fact the two are invertible together, because $A=P^{-1}B$ gives the converse.

The rank of a linear map and nullity of a linear map areThe rank-nullity theorem states thatTo prove it, choose a basis $u_1,\dots ,u_k$ of the kernel of a linear map and use basis extension to obtain a basisof $V$. The vectors $T(v_1),\dots ,T(v_{n-k})$ span the image of a linear map: applying $T$ to an arbitrary linear combination of the displayed basis removes all the $u_i$ terms. They are also linearly independent. Indeed,so ${\sum }_j{a}_j{v}_j={\sum }_i{b}_i{u}_i$ for some scalars $b_i$. The linear independence of the extended basis forces every $a_j$ and $b_i$ to vanish. Consequently $r(T)=n-k$ and $n(T)=k$, proving the identity. The same argument covers an empty basis when one of these dimensions is zero.

An elementary row operation is one of three reversible changes: interchange two rows; multiply one row by a nonzero scalar; or add a scalar multiple of one row to a different row. Their inverses are, respectively, the same interchange, multiplication by the reciprocal, and addition of the negative scalar multiple.

Applied to an augmented matrix, these changes interchange linear equations, rescale an equation by a nonzero factor, or add a multiple of one equation to another. Every old solution satisfies the new equations. Applying the inverse elementary row operation gives the converse, so the solution set is unchanged at every step. Equivalently, all the operations together givewhere $E$ is an invertible matrix, a product of elementary matrices.