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Differentiation gives ${\mathbf{r}}^{\prime }=e^t(\sqrt{2},-\sin t-\cos t,\cos t-\sin t)$. The squared speed is $4e^{2t}$, so the arc length isThe speed never vanishes, so this is a regular curve throughout its parameter range.

The Taylor series definition says that $f\in C^{\infty }(\mathbb{R})$ and, for every $a$, there is $r>0$ such thatThe equivalent factorial derivative criterion for real analyticity says that, for every $a$, there are a neighborhood $I$ of $a$ and constants $C,A>0$ such thatThe uniformity over $I$ matters: bounds only at $a$ do not exclude a flat function.

Assume the factorial derivative criterion for real analyticity. The Taylor theorem with Lagrange remainder giveswhen the segment from $a$ to $a+h$ is contained in $I$. For sufficiently small $|h|<A^{-1}$ the Taylor remainder tends to zero, proving the Taylor series definition.

Conversely, write the convergent power series at $a$ as $\sum b_k{h}^k$. Choose $\rho$ strictly inside its radius of convergence; then $|b_k|\le M{\rho }^{-k}$ for some $M$. Termwise differentiation on $|h|\le \rho /2$ givesHere the sum is $(1-1/2{)}^{-n-1}$, obtained by differentiating the geometric series. This is the required locally uniform bound. The two definitions of a real analytic function are equivalent.

Write $p_r=(A^T\mu {)}_r$ for a route congestion price and $y_j=(Ax{)}_j$ for resource demand. A route with weighted logarithmic utility $w_r\log x_r$ chooses $x_r=w_r/p_r$, since its marginal utility equals its route congestion price. Each resource raises its price when $y_j>C_j$ and lowers it when $y_j<C_j$; ${\kappa }_j$ controls the speed. This is a multiplicative resource-price dynamics model of supply-demand feedback.

For a well-defined finite network, assume positive capacities, nonempty routes, and strictly positive initial resource congestion prices. The relevant rank condition is full row rank, $\mathrm{rank}A=|J|$. Zero initial prices are invariant under the printed differential equation, so the unrestricted phrase “all trajectories” needs a qualification; a concrete counterexample is given below.

Define $g(\mu )=Ax(\mu )-C$. On the domain $\mu \ge 0$ with $p_r>0$, direct differentiation givesThus $V$ is a concave function, and it is a strictly concave function when $A$ has full row rank. It has a maximum: as $\Vert \mu {\Vert }_1\to \infty$, its negative linear capacity term dominates its logarithmic terms, while on bounded sets $V\to -\infty$ if any route congestion price approaches zero. A maximizing price vector ${\mu }^{\ast }$ therefore exists with every $p_r^{\ast }>0$, though some individual resource congestion prices may be zero. The Karush-Kuhn-Tucker conditions areThey express feasibility and complementary slackness: positively priced resources are fully used, and an underused resource has zero price. Full row rank gives uniqueness of this maximizing price vector.

To prove convergence, rather than merely monotonicity of $V$, use the relative-entropy Lyapunov function for resource priceswhere a summand with ${\mu }_j^{\ast }=0$ means ${\mu }_j/{\kappa }_j$. Every summand is nonnegative and vanishes only at its reference price. Along a positive-price trajectory,The last equality follows by substituting $x_r=w_r/p_r$, and complementary slackness removes ${\mu }^{\ast T}g({\mu }^{\ast })$.

A sublevel set of $L$ bounds every price from above and bounds every coordinate with ${\mu }_j^{\ast }>0$ away from zero. Each route contains at least one such coordinate, because $p_r^{\ast }>0$. All route congestion prices consequently remain bounded away from zero, so the vector field is regular on the resulting compact set. Positive coordinates cannot hit zero in finite time, since their equations have the form ${\dot{\mu }}_j={\mu }_j$ times a bounded function. These bounds also give global existence.

The LaSalle invariance principle now applies. Equality in $\dot{L}\le 0$ requires $A^T\mu =A^T{\mu }^{\ast }$ and ${\mu }_j{g}_j({\mu }^{\ast })=0$. Then $x=x^{\ast }$ and $g=g^{\ast }$, so every such point is an equilibrium satisfying the same Karush-Kuhn-Tucker conditions. If $A$ has full row rank, $A^T$ is injective, making this set the singleton ${\mu }^{\ast }$. Thusfor every strictly positive initial price vector. This proves full-row-rank convergence of multiplicative resource prices.

If $A$ is row-rank deficient, all maximizing price vectors still have the same route-price vector: otherwise strict concavity of ${\sum }_r{w}_r\log p_r$ makes their midpoint improve $V$. Consequently the limiting rates are unique, while individual prices may not be. Positive-price trajectories actually converge to a single point in that possibly non-singleton equilibrium set. To see this, choose any subsequential limit $\bar{\mu }$ in it, and form the same $L$ with reference $\bar{\mu }$. It is nonincreasing, and along the chosen subsequence it tends to zero. Hence it tends to zero on the whole trajectory, forcing $\mu (t)\to \bar{\mu }$.

For example, two identical unit-capacity resources serving one route of weight one have $A=(1,1{)}^T$. With ${\kappa }_1={\kappa }_2=1$, $p={\mu }_1+{\mu }_2$ obeys $\dot{p}=1-p$, while ${\mu }_1/{\mu }_2$ is constant. The equilibrium prices can be any nonnegative pair with sum one, and positive initial values select the limiting split; the route rate always tends to one. Rank deficiency can, but need not, cause price nonuniqueness. This example also shows why full column rank alone is insufficient.

Finally, the zero-price qualification is essential even for an invertible matrix. TakeThe vector $\mu =(2,0)$ has positive route congestion prices and rates $(1/2,1/2)$. Resource one is balanced, but resource two is overloaded; its zero price keeps its derivative zero. This is a boundary equilibrium of multiplicative price dynamics, distinct from the optimizing equilibriumTherefore full row rank gives the unique optimizing equilibrium and convergence from strictly positive prices; it does not give uniqueness of every equilibrium on the nonnegative orthant.

The Hopf boundary point lemma requires an interior sphere condition, strictness of the boundary maximum, and quantitative control of the uniformly elliptic operator near the contact point. Assume that $y\in \partial \Omega$, that $B_R(z)\subset \Omega$ and $\overline{B_R(z)}\setminus \{y\}\subset \Omega$, and that $|y-z|=R$. After shrinking the tangent open ball, we may suppose that the coefficients are bounded there and that, for constants $0<\lambda \le \Lambda$ and $B<\infty$,The coefficient matrix may be replaced by its symmetric matrix part because the Hessian matrix is symmetric. Assume also that $u(x)<u(y)$ throughout the tangent open ball. In the usual statement, one assumes $u(x)<u(y)$ throughout $\Omega$; when $\Omega$ is connected, the strong maximum principle for elliptic operators supplies this strictness from a nonconstant solution attaining its global maximum at $y$.

Put $\mathbf{n}=(y-z)/R$, the outward unit normal of the tangent open ball. With only the given continuity at $y$, the precise conclusion isIf the outward normal derivative exists, it is strictly positive. In particular, if $u$ is continuously differentiable up to $y$ and the domain has this outward unit normal, then ${\partial }_{\mathbf{n}}u(y)>0$. The stated $C^0$ boundary regularity alone does not assert existence of a normal derivative.

For the proof, use the annulus $A=\{R/2<|x-z|<R\}$ and the barrier for the Dirichlet problemWriting $q=x-z$, direct differentiation givesChoose $k>2(n\Lambda +BR)/(\lambda R^2)$, so that $Lv>0$ on $A$. On the inner sphere, compactness and the strict maximum give $m={\min }_{|x-z|=R/2}(u(y)-u(x))>0$. Choose $\epsilon >0$ with $\epsilon {\max }_{|x-z|=R/2}v\le m$. On the outer sphere $v=0$ and $u-u(y)\le 0$, including at $y$ by continuity. Consequently $w=u-u(y)+\epsilon v$ has $Lw\ge 0$ and nonpositive boundary values. The allowed comparison principle, or the weak maximum principle for elliptic operators, yields $w\le 0$ in $A$.

Along the inward radius this givesThis proves the Hopf boundary point lemma. There is no sign restriction on $u(y)$ here, because the operator has no zeroth-order term.

Assume the particle masses $m_i$ are fixed and their total mass $M={\sum }_i{m}_i$ is positive. The centre of mass and the relative position vectors areTheir weighted sum is zero, so differentiation gives ${\sum }_i{m}_i{\dot{\mathbf{y}}}_i=0$. Expanding the kinetic energy with ${\dot{\mathbf{x}}}_i=\dot{\mathbf{R}}+{\dot{\mathbf{y}}}_i$ makes the mixed term vanish:This kinetic-energy decomposition about the centre of mass separates translation from internal motion without assuming any particular force law.

For the rigid body, its relative velocity is ${\dot{\mathbf{y}}}_i=\omega \mathbf{n}\times {\mathbf{y}}_i$. The cross product identity for a unit vector gives $|\mathbf{n}\times {\mathbf{y}}_i{|}^2=|{\mathbf{y}}_i{|}^2-(\mathbf{n}\cdot {\mathbf{y}}_i{)}^2$, the squared perpendicular distance to the rotation axis. Hence the moment of inertia of the particle is $I_i=m_i[|{\mathbf{y}}_i{|}^2-(\mathbf{n}\cdot {\mathbf{y}}_i{)}^2]$, and summing the rotational kinetic energy yields

Take $P_0>0$, real $\alpha$, and the real branch $T\ge T_0$, with physical temperature $T>0$. In hydrostatic equilibrium the pressure decreases with height $z$:Thus an atmospheric thermal inversion means $dT/dz>0$, equivalently $dT/dP<0$. For $\alpha \ne 0$, the logarithm of the given relation yieldsThe sign restriction must be retained after squaring. In particular, the inverted branch has $0<P\le P_0$, whereas the non-inverted branch has $P\ge P_0$. By differentiation, on the interior of either branch,Consequently,At $T=T_0$, the inverse temperature gradient vanishes at the branch endpoint, rather than producing a finite isothermal interval. If $\alpha =0$, the relation becomes $P=P_0$ independently of $T$; it is not a nontrivial vertical atmospheric pressure-temperature profile under hydrostatic equilibrium. These distinctions prevent the two sides of the squared parabola from being interpreted as one permissible atmosphere.

A Runge-Kutta method advances $y^{\prime }=f(t,y)$ through stagesThe Butcher tableau records $A,b,c$, normally with $c=Ae$. Linear absolute stability tests the amplification of a linear decay mode; nonlinear B-stability compares distances between two solutions of a dissipative vector field. Both describe propagation of perturbations, but they impose different conditions.

For the Dahlquist test equation, elimination of stages givesThe linear stability domain consists of $z$ for which the step is defined and $|R(z)|\le 1$. Poles or singular stage systems must be excluded, even if an output formula has a formal cancellation. For a scalar test equation, the proof is $y_n=R(z{)}^n{y}_0$: the powers are bounded exactly when $|R(z)|\le 1$. For $y^{\prime }=Ly$, the update is $R(hL)$. If $L$ is a normal matrix, unitary diagonalization of a normal matrix makes the operator norm equal to the largest scalar amplification modulus. For a diagonalizable matrix, powers are bounded by the condition number of the eigenvector matrix times the maximal scalar power. These constants must remain uniform for mesh-dependent systems. A defective unit-modulus eigenvalue can instead produce a growing Jordan block; the scalar spectral test alone does not control a general matrix.

An A-stable method accepts the entire closed left half-plane. The Explicit Euler method has $R(z)=1+z$ and a disk of stability $|1+z|\le 1$, so it is not A-stable. No nonconstant polynomial stability function can be A-stable, since it is unbounded on the negative real axis; this excludes all consistent explicit Runge-Kutta methods. The Backward Euler method has $R(z)=1/(1-z)$, and $|1-z|\ge 1$ when $\mathrm{Re}z\le 0$, proving A-stability. It is also L-stable, because $R(z)\to 0$ for large left-half-plane $z$.

The implicit midpoint rule and the trapezoidal rule have the same scalar stability function,The identity $|1-z/2{|}^2-|1+z/2{|}^2=-2\mathrm{Re}z$ proves A-stability. On the imaginary axis their amplification has unit modulus, and along the negative real axis it tends to $-1$. Thus neither is L-stable; very stiff decay can persist as alternating numerical values. In contrast, the two-stage Radau IIA method has $R(z)=(1+z/3)/(1-2z/3+z^2/6)$, with the explicit nonnegative modulus-gap proof in question 1, and is L-stable. This illustrates why L-stability matters beyond A-stability for a stiff differential equation.

For nonlinear problems, assume a real or complex Euclidean inner product and the dissipativity hypothesisfor all relevant states at each common time. Exact solutions satisfyA B-stable Runge-Kutta method reproduces this nonexpansion of distances whenever its stages exist. Put $d=y_n-{\tilde{y}}_n$, $D_i=Y_i-{\tilde{Y}}_i$, and $G_i=f(t_n+c_{i}h,Y_i)-f(t_n+c_{i}h,{\tilde{Y}}_i)$. Then $D_i=d+h{\sum }_j{a}_{ij}{G}_j$. Expansion of the output squared norm, followed by substituting $d=D_i-h{\sum }_j{a}_{ij}{G}_j$ in its linear terms, proves the Runge-Kutta contractivity identityIf $b_i\ge 0$ and $M=(m_{ij})$ is a positive semidefinite matrix, the first sum is nonpositive by dissipativity, and the second is nonnegative: expand each coordinate of the stage vectors to express it as a sum of nonnegative quadratic forms. These are exactly the conditions for algebraic stability of a Runge-Kutta method. Hence algebraic stability implies B-stability. Applying B-stability to the linear dissipative problem $y^{\prime }=\lambda y$, or its two-dimensional real form, also proves A-stability whenever stages are well defined.

The Backward Euler method has $b=A=1$, hence $M=1$; the implicit midpoint rule has $b=1$, $A=1/2$, hence $M=0$. Both are algebraically stable and B-stable. For the two-stage Radau IIA method,a positive semidefinite matrix, while both weights are positive. It therefore combines third-order accuracy, B-stability, and L-stability. The distinction between stability and solvability is essential: for Lipschitz continuous $f$, the stage map is a contraction when $hL{\max }_i{\sum }_j|a_{ij}|<1$, so the contraction mapping theorem proves small-step unique solvability. The preceding algebraic stability proof is an estimate for existing stages; it is not a theorem asserting stage existence for arbitrary step sizes and domains.

Linear A-stability is insufficient for nonlinear B-stability. The trapezoidal rule fails B-stability, even though its scalar stability function is identical to that of the B-stable implicit midpoint rule. Its Runge-Kutta method coefficients areFailure of this sufficient algebraic stability test alone would not prove failure of B-stability. An actual counterexample does: take $f(y)=-y^3$ and $h=2$. Dissipativity follows fromThe unique step map $T$ obeys $T+T^3=y-y^3$, because the left-hand side is strictly increasing and onto. At $y=1$, $T=0$, and implicit differentiation givesNearby initial states expand in distance. This proves a true nonlinear failure rather than merely failure of a coefficient criterion.

Finally, stability of a numerical method on finite intervals connects these contractivity properties to accuracy. If a one-step map is nonexpansive and its exact one-step defect has norm at most $C{h}^{p+1}$, then the error satisfies $\Vert e_{n+1}\Vert \le \Vert e_n\Vert +C{h}^{p+1}$, hence $\Vert e_n\Vert \le \Vert e_0\Vert +CT{h}^p$ for $nh\le T$. A Lipschitz step bound $1+Kh$ gives the same order using the discrete Gronwall inequality. For a stiff differential equation, the value of B-stability is that the propagation estimate need not grow with a large negative dissipative rate. The accuracy constant still needs appropriate smoothness and uniform derivative bounds; stability alone does not establish a mesh-uniform error order.

Termwise differentiation of the matrix exponential gives $A^{\prime }=LA$ and $A(0)=I$. Thus $x(t)=A(t)x(0)$ satisfies $x^{\prime }=Lx$ and the prescribed initial condition. Uniqueness for a linear differential equation makes it the solution.

At fixed incidence angle, differentiation of the grating equation gives $d\cos \beta d\beta =md\lambda$. Near the camera axis, the focal-plane displacement is $dX\simeq f_{\mathrm{cam}}d\beta$, so the local grating dispersion isIt has units of distance per unit wavelength. If the camera axis is at ${\beta }_0$ and the flat focal-plane coordinate is retained exactly as $X=f_{\mathrm{cam}}\tan (\beta -{\beta }_0)$, thenThe boxed expression is the local value at the optical axis.

The probability generating function isTermwise differentiation givesThe coefficients are nonnegative, and $\mathbb{P}(X\ge 2)>0$, so both derivatives are positive and nondecreasing on $(0,1]$.

A field is irrotational when $\nabla \times F=0$. In ${\mathbb{R}}^3$ definewhere $\gamma$ is any path from $x_0$ to $x$. The difference between two choices is the circulation around a closed loop. By the Stokes theorem it is the flux of $\nabla \times F$ through a spanning surface and hence zero. Thus $V$ is well-defined, $V(x_0)=0$, and differentiation gives $F=-\nabla V$.

If $\nabla \times A_0=B$, every solution isfor an arbitrary differentiable scalar $\chi$, since the difference is curl-free. A necessary condition is $\nabla \cdot B=0$, because the divergence of a curl vanishes.

Differentiation of $x=\rho \cos \varphi$, $y=\rho \sin \varphi$ givesThe two radial face areas are ${\rho }_0\Delta \varphi \Delta z$ and $({\rho }_0+\Delta \rho )\Delta \varphi \Delta z$. Each constant-$\varphi$ face has area $\Delta \rho \Delta z$, and each constant-$z$ face has areaThe volume is this last expression times $\Delta z$.

Only the radial faces contribute to the flux of $F(\rho )e_{\rho }$. Dividing their difference by the volume and taking $\Delta \rho \to 0$ givesat $\rho ={\rho }_0$.

Termwise differentiation inside the disc givesConsequentlySince $f(1)=0$, the constant is one, so $e^{f(z)}=z$. Thus $f$ is an analytic branch of the logarithm on $D(1,1)$ with the required value.

Given $a\in D$, write $a=|a|e^{i{\theta }_a}$ with $0<{\theta }_a<2\pi$. On $|z/a-1|<1$, defineThen $e^{{\ell }_a(z)}=z$ and $\mathrm{Im}{\ell }_a(a)={\theta }_a$. After shrinking the neighbourhood of $a$, continuity keeps its imaginary part in $(0,2\pi )$.

Integrating the ordinary differential equation four times gives a quartic polynomial. It is useful to use the linearity of the equation and split the solution into a gravity part and a force part:wheresatisfies the clamped and torque-free boundary conditions with $F=0$, whilesatisfies the homogeneous equation and contributes the endpoint force. Direct differentiation verifiesThusthe clamped-free beam under uniform load and endpoint force.

Away from $x=\xi$, the Green function solves $G^{\prime \prime }-G=0$. The solution satisfying the left boundary condition is proportional to $\sinh x$, while the solution decaying at infinity is proportional to $e^{-x}$. Continuity at $\xi$ and the unit derivative jumpgiveor $G=-\sinh (\min \{x,\xi \})e^{-\max \{x,\xi \}}$, the dirichlet half-line Green function for d2 minus 1.

The required solution isEvaluation of the two elementary integrals, split at $\xi =x$, givesIndeed, direct differentiation gives $y^{\prime \prime }-y=e^{-2x}$, and both boundary conditions are immediate.

For $f_{ϵ}(x)=({ϵ}^2+x^2{)}^{1/2}$, direct differentiation givesSince the quadratic variation of standard Brownian motion is $[B{]}_t=t$, the Itô formula gives

A central force conserves angular momentumPut $u=1/r$. Since $\dot{\theta }=\ell u^2/m$,where primes now denote differentiation with respect to $\theta$. Substitution into the radial equation gives the Binet equation

For $0<a<2$, the beta-integral evaluation givesDifferentiation under the integral sign near $a=1$ is justified by domination at zero and infinity. ThereforeWriting $a=1+t$ givesSincethe logarithmic moments of the Cauchy kernel yieldThe same expansion gives $I^{\prime \prime }(1)={\pi }^3/8$, agreeing with the supplied identity.

Put $W=\sqrt{1+h_x^2+h_y^2}$. An upward-pointing Gauss map isFor tangent vectors $v,w$, differentiation ofgivesThe Weingarten map is self-adjoint and $S=-dN$, so the right side is $-2I{I}^S(v,w)$. HenceAlternatively differentiating $⟨N,{\varphi }_i⟩=0$ givesThereforeThis is the fundamental forms of a graph surface.

Substitute $r=1/u$ and $\dot{r}=-h{u}^{\prime }$ into the conserved radial energy. Differentiation with respect to $\varphi$ givesAfter division by $h^2{u}^{\prime }$ and continuity through turning points,

Put $q=m/h^2$ and substituteKeeping the leading nontrivial powers of $m/h$ and terms through first order in $A$, the constant and oscillatory terms giveThereforeThe radial oscillation has angular period $2\pi /(1+\beta )$, so successive periapses advance byto leading order. This is the Schwarzschild perihelion precession.