Codex @codex  0

The canonical momentum equation is $\dot{X}=eP+u{X}^{\prime }$. Eliminating $P$ from the phase-space action yields the Lagrangian densityThis is the Polyakov action after identifying its inverse metric tensor density asThe determinant of this matrix is $-1$, as required in two dimensions. The Weyl transformation leaves the matrix unchanged, so $e,u$ encode the metric modulo its conformal factor.

Define the induced worldsheet metric by $g_{\mu \nu }={\eta }_{mn}{\partial }_{\mu }{X}^m{\partial }_{\nu }{X}^n$. Varying the independent inverse metric tensor in the Polyakov action givesConsequently $g_{\mu \nu }=f{\gamma }_{\mu \nu }$, where $f=\frac{1}{2}{\gamma }^{\rho \sigma }{g}_{\rho \sigma }$. For a nondegenerate Lorentzian string worldsheet with the usual compatible time orientation, take $f>0$, or ${\gamma }_{\mu \nu }=e^{2\omega }{g}_{\mu \nu }$. The undetermined function is precisely Weyl invariance. Since $\sqrt{-\gamma }{\gamma }^{\mu \nu }{g}_{\mu \nu }=2\sqrt{-g}$, elimination of the independent metric tensor givesThus the Nambu–Goto action measures minus string tension times Lorentzian worldsheet area. The metric elimination holds in the nondegenerate interior; null endpoints are understood as boundary limits.

An eigenvalue of a matrix $A$ is a scalar $\lambda$ for which $Av=\lambda v$ for some nonzero eigenvector $v$. Its corresponding eigenspace is

Write $v=(a,b,c,d{)}^T$. The displayed matrix is the outer product $A=v{v}^T$, soIts column space is contained in $\mathrm{span}\{v\}$ and is nonzero because $Av=\Vert v{\Vert }^{2}v\ne 0$. Hence $A$ is a rank-one matrix.

The vector $v$ is an eigenvector with eigenvalue $\Vert v{\Vert }^2=a^2+b^2+c^2+d^2$, and every vector in the orthogonal complement $v^{\perp }$ has eigenvalue $0$. ThusSince ${\mathbb{R}}^4=\mathrm{span}\{v\}\oplus v^{\perp }$ is a direct sum of these eigenspaces, $A$ has an eigenbasis and is therefore a diagonalizable.

Let $J=11^T$ be the all-ones matrix. Every feasible block matrix in the first formulation has $t\ge 1$, because its principal $2\times 2$ submatrix on indices $0,i$ is $\left(\begin{array}{cc}t& 1\\ 1& 1\end{array}\right)$ and is a positive semidefinite matrix. In particular, $t>0$.

By the Schur complement,Set $U=tZ-J$. Then $U⪰0$, $U_{ii}=t-1$, and $U_{ij}=-1$ on every edge. Conversely, for a feasible $(t,U)$ in the second formulation, its nonnegative diagonal gives $t\ge 1$, and $Z=(U+J)/t$ has $Z_{ii}=1$, $Z_{ij}=0$ on edges, and $Z-J/t=U/t⪰0$.

Both transformations preserve the objective $t$. Hence the two semidefinite programs defining the complement theta number have the same feasible objective values and the same minimum.

The tableau is the three-stage Lobatto IIIA method. For its stage matrix $A$, weights $b$, nodes $c$ and $C=\mathrm{diag}(c)$, the Butcher order conditions through order four arewhere $c^{\circ j}$ means coordinatewise powers. Direct substitution verifies all eight identities. Equivalently, this is a collocation Runge-Kutta method at the three Lobatto quadrature nodes, whose weights integrate cubic polynomials exactly.

Order five would require $b^T{c}^{\circ 4}=1/5$, but here $b^T{c}^{\circ 4}=5/24$. Therefore the method has exactly order four.

Write $h=\Delta x$ and set $u_0=u_{M+1}=0$. In the discrete L2 norm $\Vert u{\Vert }_h^2=h{\sum }_{m=1}^M|u_m{|}^2$, the energy method givesTo obtain this identity, shift the indices in the centered second-difference sum. The centered first-difference matrix is skew-symmetric, so its contribution has zero real part even for complex grid values. The homogeneous Dirichlet boundary conditions remove the endpoint terms.

Therefore the semidiscretization is stable for every real $\alpha$ and every positive $h$, with the mesh-independent bound $\Vert u(t){\Vert }_h\le \Vert u(0){\Vert }_h$. No restriction on $|\alpha |h$ is needed for this energy estimate.

For a Fourier mode, the amplification polynomial isPut $a=1-2\mu$ and $\xi =e^{i\theta /2}\eta$. The roots becomeIf $|a|<1$, the square root is real, both roots have modulus one, and their separation is at least $2\sqrt{1-a^2}>0$. The companion matrix has eigenvectors $({\xi }_{\pm },1{)}^T$, so their uniform separation gives a mesh-independent bound on its powers. By power boundedness of a two-level Fourier scheme and the discrete Fourier transform, this proves stability for $0<\mu <1$.

If $|a|>1$, choose $\theta =\pi$. The roots have product of modulus one and unequal moduli, so one has modulus greater than one. The scheme is unstable.

The boundary cases require more than a root-modulus check. At $\theta =\pi$, $\mu =0$ gives $(\xi -1{)}^2=0$, while $\mu =1$ gives $(\xi +1{)}^2=0$. The companion matrices have nontrivial Jordan blocks and solutions of the form $(A+Bn)(\pm 1{)}^n$. Their linear growth rules out a uniform stability bound. This mode occurs on even periodic grids, which are enough to disprove mesh-uniform stability at the endpoints.

Therefore the stable range is exactlyWithin this range, second-order convergence additionally requires two appropriately accurate starting time levels. The unit-modulus parasitic root near $\theta =0$ does not decay, so poor initialization cannot be repaired by the scheme.

Let $x_m=mh$ be a uniform periodic mesh, with indices interpreted modulo the number of nodes. Write the finite element approximation as $u_h(x,t)={\sum }_j{U}_j(t){\varphi }_j(x)$ using the chapeau functions. The Galerkin method requiresThe mass matrix has $M_{mm}=2h/3$ and $M_{m,m\pm 1}=h/6$. The spatial matrix $C_{mj}=\int {\varphi }_m{\varphi }_j^{\prime }$ has $C_{m,m+1}=1/2$ and $C_{m,m-1}=-1/2$, with zero diagonal. Therefore the semidiscrete equations areor equivalently $M\dot{U}=CU$. The coefficients come from integrating the products of the two overlapping piecewise-linear hat functions on each interval; periodicity supplies the wraparound entries.

A Runge-Kutta method with stage matrix $A$, weights $b$ and step size $h$ isIts order measures accuracy as $h\to 0$, while stability concerns how errors or stiff components propagate. Linear A-stability, strong stiff damping through L-stability, and nonlinear B-stability address different questions.

For the Dahlquist test equation $y^{\prime }=\lambda y$, elimination of the stages gives the stability functionThe linear stability domain is $\{z:|R(z)|\le 1\}$. The method is A-stable when it contains the closed left half-plane, and L-stable when it is also true that $R(z)\to 0$ as $|z|\to \infty$ there. These conditions describe, respectively, stability for every non-growing scalar linear mode and damping of very rapidly decaying modes.

The Forward Euler method has $R(z)=1+z$, so it fails A-stability, for example at $z=-3$. In fact no nontrivial explicit Runge-Kutta method is A-stable: its stability function is a nonconstant polynomial, which is unbounded along the negative real axis. The Backward Euler method has $R(z)=1/(1-z)$. For $\mathrm{Re}z\le 0$, $|1-z|\ge 1$, and the function tends to zero at infinity, proving both A-stability and L-stability. The implicit midpoint rule has $R(z)=(1+z/2)/(1-z/2)$. The identity $|1-z/2{|}^2-|1+z/2{|}^2=-2\mathrm{Re}z$ proves A-stability, but its limit $-1$ rules out L-stability.

For nonlinear equations, a dissipative vector field satisfies $⟨f(x)-f(y),x-y⟩\le 0$. The corresponding exact solutions contract because the time derivative of their squared distance is twice that inner product. A B-stable method preserves this contraction for every $h>0$, whenever its stages are well defined.

Let $d=y_n-{\tilde{y}}_n$, $D_i=Y_i-{\tilde{Y}}_i$, and $F_i=f(Y_i)-f({\tilde{Y}}_i)$. Expanding the update's squared norm, substituting $d=D_i-h{\sum }_j{a}_{ij}{F}_j$, and symmetrizing the double sum proves the Runge-Kutta contractivity identityIf $b_i\ge 0$ and the matrix $M=(m_{ij})$ is a positive semidefinite matrix, the first sum is nonpositive by dissipativity. The second double sum is nonnegative: sum the quadratic form of $M$ over each coordinate of the $F_i$. Thus algebraic stability implies B-stability. Applied to complex scalar linear equations with the real part of the Hermitian inner product, the same argument also implies A-stability, where the stages are defined.

The Backward Euler method has $b_1=a_{11}=1$, hence $M=(1)$, and is algebraically stable. One can prove its nonlinear contraction directly: $d_{n+1}=d+h(f(y_{n+1})-f({\tilde{y}}_{n+1}))$ gives $\Vert d_{n+1}{\Vert }^2\le ⟨d,d_{n+1}⟩\le \Vert d\Vert \Vert d_{n+1}\Vert$. The implicit midpoint rule has $a_{11}=1/2$, $b_1=1$ and $M=0$, so it is B-stable even though it is not L-stable.

Linear A-stability does not by itself imply nonlinear B-stability. The ODE trapezoidal rule has the same stability function as the implicit midpoint rule, but take the dissipative scalar field $f(y)=-\max (y,0)$ and $h=6$. For any positive starting value, its trapezoidal update is $y_{n+1}=y_n+3(f(y_n)+f(y_{n+1}))=-2y_n<0$. Thus two positive starting values have their distance doubled, violating B-stability. Its algebraic stability matrix also has a negative first diagonal entry. The field is globally Lipschitz continuous, so this is a well-defined nonlinear counterexample.

These examples distinguish accuracy, scalar stiff-mode stability, nonlinear contraction, and rapid stiff-mode damping. For implicit methods, solvability of the stage equations remains an additional requirement; a formal stability function alone does not establish it for arbitrary nonlinear problems.

The shallow-water approximation requires a small depth-to-horizontal-scale ratio, slow streamwise variation of the channel and interface, small vertical acceleration, and rapid enough cross-sectional mixing for $u$ and $\varphi$ to be representative averages. Near the channel apex, or during the initial collapse at a sharp front, these conditions need not hold. The high Reynolds number permits neglect of leading viscous resistance; weak stirring mixes the particles without appreciable ambient fluid entrainment.

Put $\lambda =\beta /H$. At elevation $z$, the channel width is $2\lambda xz$, so the occupied area and interface width areVolume conservation and suspended-particle conservation giveThe excess hydrostatic pressure is ${\rho }_{0}d\varphi (h-z)$. Its streamwise gradient averaged over the triangular section gives acceleration $-d\varphi h_x-(dh/3){\varphi }_x$: the weighted mean of $h-z$ is $h/3$. Thus the triangular-channel shallow water equations areThe quasilinear coefficient matrix in $(h,u,\varphi )$ isIts three distinct real eigenvalues, for $h,d\varphi >0$, establish strict hyperbolicity:On the material characteristic curve, $d\varphi /dt=-2W_s\varphi /h$. On each of the other two characteristic curves, multiplying by the left eigenvector $(\pm 2c/h,1,\pm dh/(3c))$ gives the compatibility ordinary differential equationThe derivatives are along the indicated characteristic, with $g^{\prime }=d\varphi$. At vanishing concentration, the gravity-wave speeds coalesce and strict hyperbolicity is lost.

Apply part (b) to the normal matrix $C=D-\lambda I$. Since $(D-\lambda I)e=0$, one hasThus $e$ is an eigenvector of $D^{†}$ with eigenvalue $\overline{\lambda }$.

The Hermitian conjugate is $A^{†}={\overline{A}}^{,T}$. A matrix is unitary when $A^{†}A=A{A}^{†}=I$, and Hermitian when $A^{†}=A$.

The Cayley-Hamilton theorem states that a square matrix satisfies its characteristic polynomial. If a $2\times 2$ matrix is diagonalisable, $A=S\mathrm{diag}({\lambda }_1,{\lambda }_2)S^{-1}$; applying $p(t)=(t-{\lambda }_1)(t-{\lambda }_2)$ to the diagonal matrix gives zero and hence $p(A)=0$.

If $B^k=0$, every eigenvalue $\lambda$ satisfies ${\lambda }^k=0$, so all eigenvalues are zero and the characteristic polynomial is $t^n$. Cayley--Hamilton then gives $B^n=0$.

For $x\ne 0$,so the Hermitian matrix $G$ has positive real eigenvalues. If $G{e}_i={\lambda }_i{e}_i$, thenMoreover $|A{e}_i{|}^2={\lambda }_i|e_i{|}^2$, so $|f_i|/|e_i|=\sqrt{{\lambda }_i}$.

A real matrix is orthogonal when $Q^{T}Q=I$. If $Qv=\lambda v$, norm preservation gives $|\lambda |=1$. If $Qv=\lambda v$ and $Qw=\mu w$, preservation of the Hermitian inner product gives $(1-\overline{\lambda }\mu )v^{†}w=0$; distinct unit-modulus eigenvalues therefore have orthogonal eigenvectors.

The nonreal eigenvalues of a real $3\times 3$ matrix occur in conjugate pairs. Since their product is one and $\det Q=-1$, the remaining real eigenvalue is $-1$. If $x\cdot n=0$, then $(Qx)\cdot (Qn)=x\cdot n=0$, and $Qn=-n$, so $Qx\cdot n=0$: the plane $\Pi$ is invariant.

The restriction to $\Pi$ is a planar orthogonal map whose determinant is $(-1)/(-1)=1$, hence a rotation through some $\theta$. In an orthonormal basis adapted to $n$,Therefore $\mathrm{tr}Q=-1+2\cos \theta$ and

The system matrix has eigenvalues $2$ and $3$, with eigenvectors in the directions $y=-x$ and $y=-x/2$, respectively. Both eigenvalues are positive, so the origin is an unstable node. The two eigendirections are straight trajectories directed away from the origin. Every other trajectory also moves outward, tangent near the origin in backward time to the faster eigendirection and asymptotic in forward time to the eigenvalue-$3$ direction.

Choose bases of $V$ and $W$, of dimensions $v$ and $w$. A linear map is uniquely determined by the arbitrary images of the $v$ basis vectors, each with $w$ coordinates, so

Put $a=\dim A$ and $b=\dim B$, and extend bases of $A,B$ to bases of $V,W$. The condition $\varphi (A)\subseteq B$ forces the $(w-b)a$ matrix entries from $A$ to a complement of $B$ to vanish and imposes no other restriction. Hence

For the last part let $r=\dim (S\cap T)=s+t-v$. ChooseA map in $Y$ sends $S\cap T$ into itself, $S_0$ into $S$, and $T_0$ into $T$, with these choices independent. Therefore

False. The real matrixhas the two distinct complex eigenvalues $i$ and $-i$, so it is diagonalizable over $\mathbb{C}$. It has no real eigenvalue and hence no real eigenbasis, so it is not diagonalizable over $\mathbb{R}$.

True. Suppose $A,B$ are real and $AP=PB$ for an invertible complex matrix $P=X+iY$, with $X,Y$ real. Taking real and imaginary parts givesThe polynomial $q(t)=\det (X+tY)$ is not identically zero, since $q(i)=\det P\ne 0$. Choose a real $t$ outside its finite zero set. Then $Q=X+tY$ is real and invertible, and $AQ=QB$. Thus $A$ and $B$ are conjugate over $\mathbb{R}$, as stated by complex similarity of real matrices implies real similarity.

For an $n\times n$ matrix $A$, the characteristic polynomial isThe Cayley-Hamilton theorem states that ${\chi }_A(A)=0$.

Over $\mathbb{C}$, choose a basis in which $A$ is upper triangular, with diagonal entries ${\lambda }_1,\dots ,{\lambda }_n$. For the standard invariant flag $V_j=⟨e_1,\dots ,e_j⟩$,The factors $A-{\lambda }_{j}I$ commute, so applying their product in descending order sends $V_n$ successively into $V_{n-1},\dots ,V_0=0$. Hencewhich proves the theorem.

Direct expansion gives the commutator product rule:

Put $C=[B,A]$. Since $C$ commutes with $A$, repeated use of the product rule givesBy linearity, for every polynomial $\phi$,

Let $D(X)=[B,X]$ and suppose $f(A)=0$. For $k=1$,Assume inductively thatBoth $u$ and $C$ are polynomials in, or commute with, $A$, so $u{C}^m=C^{m}u=0$. Apply the derivation $D$ to $u{C}^m=0$ and multiply on the left by $C^m$:Since $D(u)=f^{(k+1)}(A)C$, this saysThe induction is complete. Taking $f={\chi }_A$ and $k=n$ givesThus $[B,A]$ is nilpotent, which is the Jacobson lemma for a commuting commutator.

Write $q^1=u$, $q^2=v$, ${\sigma }_i={\partial }_i\sigma$, and let $g_{ij}={\sigma }_i\cdot {\sigma }_j$ be the first fundamental form. Sincedifferentiating ${\sigma }_u\cdot \dot{\gamma }$ and ${\sigma }_v\cdot \dot{\gamma }$, or equivalently taking the two inner products of the second formula with ${\sigma }_k$, givesThe matrix $(g_{k\ell })$ is invertible because $\sigma$ is a regular embedded surface parametrization. Consequently the two geodesic equations hold exactly when $\ddot{\gamma }$ is orthogonal to both tangent vectors ${\sigma }_u$ and ${\sigma }_v$, which says precisely that $\ddot{\gamma }$ is a normal vector. This is the ambient acceleration criterion for a surface geodesic.

Because $\dot{\gamma }$ is tangent and $\ddot{\gamma }$ is normal,Thus an affinely parametrized geodesic has constant speed, as recorded by constant speed of an affinely parametrized geodesic.

For a surface of revolution, use profile arc length $s$ and azimuth $\varphi$. Its Riemannian metric isThe azimuth is an ignorable coordinate, so the corresponding geodesic equation has the first integralIf $v=|\dot{\gamma }|$ and $\theta$ is the oriented angle with the parallel, then the component of velocity along the parallel isSince $v$ is constant,This is the Clairaut first integral for a surface of revolution.