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In any two-colouring of $K_7$, there is a monochromatic triangle because $f(2)=6$. Call its colour blue. If any edge from one of its vertices to an outside vertex is blue, the triangle and that edge give the required copy of $H$. Otherwise all edges from that triangle to the four outside vertices are yellow. If any edge among those four is yellow, it makes a yellow triangle with a vertex of the original triangle, and another incident yellow edge supplies the pendant edge. If none is yellow, the four outside vertices form a blue $K_4$, which also contains $H$. Hence $g(2)\le 7$.

For the lower bound, divide six vertices into two triples. Colour edges inside each triple blue and all cross-edges yellow. Each blue component is only a triangle, and the yellow graph is bipartite, so neither colour contains $H$. Therefore $\boxed{g(2)=7}$. Copies are not required to be induced: an additional edge in a monochromatic $K_4$ does not invalidate the copy of $H$.

For three colours, $f(3)=17$ supplies a triangle in $K_{17}$. If it has no pendant edge of the same colour, each of its vertices has fourteen edges to outside vertices in the other two colours. Choose one vertex $v$ and a colour occurring on at least seven of those edges, with endpoint set $S$. If an internal edge of $S$ has this colour, it creates a triangle with $v$ and a fourth neighbour supplies the pendant edge. Otherwise $S$ is internally two-coloured, and $g(2)=7$ supplies $H$. Thus $g(3)\le 17$. Conversely the definition $f(3)=17$ ensures a three-colouring of $K_{16}$ without any monochromatic triangle, and therefore without $H$. Hence $\boxed{g(3)=17}$.

With one edge colour, $f(1)=3$. Suppose the assertion is known for $k-1$ colours, and take $n=k(f(k-1)-1)+2$. At any vertex, among its $n-1$ incident edges some colour occurs on at least $f(k-1)$ edges. Let $S$ be their other endpoints. If an edge within $S$ has that colour, it completes a monochromatic triangle with the chosen vertex. Otherwise $S$ uses only $k-1$ colours internally, and the inductive hypothesis gives a monochromatic triangle there. ThusInduction now yieldsTogether with the base case, $\boxed{f(k)\le 3k!}$ for every positive $k$. This is a constructive multicolour Ramsey bound.

Choose the ellipsoidal Lyapunov function $V=x^2+2y^2$. Direct substitution of the printed vector field givesThe weighted Cauchy-Schwarz inequality gives $(x-y{)}^2\le \frac{3}{2}(x^2+2y^2)=3V/2$, soFor any initial point with $V(0)<2/3$, choose $c$ with $V(0)<c<2/3$. The compact ellipse $V\le c$ is positively invariant, since $\dot{V}<0$ on its boundary. Inside it,This establishes global forward existence for those trajectories and convergence to the origin, while the First Lyapunov theorem establishes stability. The origin is asymptotically stable and its basin contains $\boxed{x^2+2y^2<2/3}$. The PDF coefficient in the second equation is $x^{2}y/2$, with the entire linear term $-x$; placing the factor $1/2$ on the linear term would destroy this factorization.

The First Lyapunov theorem states that for a locally Lipschitz autonomous system $\dot{x}=f(x)$ with equilibrium zero, if a $C^1$ function $V$ is positive definite near zero and $\dot{V}=\nabla V\cdot f\le 0$, then the equilibrium is Lyapunov stable. To prove it, choose a small sphere $|x|=\epsilon$ inside that neighborhood. Compactness and positivity give $m={\min }_{|x|=\epsilon }V(x)>0$. Continuity at zero gives a smaller ball $|x|<\delta$ where $V<m$. Since $V$ cannot increase along a trajectory, a trajectory starting there cannot first cross the sphere where $V\ge m$. It stays within the sphere for all future time, with continuation justified by confinement in a compact subset of the vector field's domain. This proves stability for every sufficiently small $\epsilon$.

The LaSalle invariance principle states that on a compact positively invariant set on which $\dot{V}\le 0$, every trajectory approaches the largest invariant subset of $\{\dot{V}=0\}$. If that subset is just the equilibrium, the trapping and stability argument proves asymptotic stability.

For example take $\dot{x}=y$, $\dot{y}=-x-y$ and $V=(x^2+y^2)/2$. Then $\dot{V}=-y^2$, which is not strictly negative on the punctured plane. Nevertheless its sublevel discs are compact and invariant, and a trajectory confined to $y=0$ must have $\dot{y}=-x=0$, hence $x=0$. LaSalle gives asymptotic stability despite the non-strict orbital derivative of this particular energy function.

The printed phrase about nonexistence of a strict Lyapunov function must be understood as a limitation of the chosen candidate. In this example a strict function actually exists:More generally the converse Lyapunov theorem supplies local strict functions for asymptotically stable sufficiently smooth systems. LaSalle avoids having to find such a function; it does not show that none exists.

Factor $n={\prod }_j{q}_j^{e_j}$. Since it is not a square, one exponent, say $e_1$, is odd. Choose a nonzero quadratic nonresidue $a$ modulo $q_1$; one exists because the squaring map on the nonzero residues is two-to-one and hence has only $(q_1-1)/2$ images. The Chinese remainder theorem supplies a positive odd integer $m>1$ satisfyingThen $\gcd (m,n)=1$ and $(m/n)=(-1{)}^{e_1}=-1$. The Jacobi reciprocity law proved in part (i), together with $m\equiv 1(\mathrm{mod}4)$, gives $(n/m)=-1$. Factoring $m={\prod }_{\ell }{p}_{\ell }^{r_{\ell }}$ givesAll $p_{\ell }$ are odd and none divides $n$. At least one factor must be negative, yielding an odd prime number $\boxed{p\text{ with }(n/p)=-1}$. No theorem about primes in arithmetic progressions is needed.

For odd coprime positive integers $a={\prod }_i{p}_i^{e_i}$ and $b={\prod }_j{q}_j^{f_j}$, define the Jacobi symbol by multiplying Legendre symbols over the prime factorization of the denominator. Multiplicativity in the numerator givesBy quadratic reciprocity for the Legendre symbol, this is $(-1{)}^{UV}$, where $U={\sum }_i{e}_i(p_i-1)/2$ and $V={\sum }_j{f}_j(q_j-1)/2$. For odd $x,y$, $(xy-1)/2\equiv (x-1)/2+(y-1)/2(\mathrm{mod}2)$, since $(x-1)(y-1)/2$ is even. Thus $U\equiv (a-1)/2$ and $V\equiv (b-1)/2(\mathrm{mod}2)$. We obtainIf $a,b$ are not coprime, both symbols are zero and the equivalent identity $(a/b)=(-1{)}^{(a-1)(b-1)/4}(b/a)$ still holds. The convention $(a/1)=1$ handles denominator one.

Put $a=2/\sqrt{3}$. On $[-a,a]$, the extrema of $|x(1-x^2)|$ occur at $\pm 1/\sqrt{3}$ or the endpoints. The corresponding values are $2/(3\sqrt{3})$, including the endpoint value $a(a^2-1)$. ConsequentlyThus $[-a,a]$ is a trapping region andIt contains a neighbourhood of the origin. Sign changes are possible because $f(x)$ has opposite sign to $x$ for $1<|x|<a$.

On $(0,1)$ the map is positive. Its maximum occurs at $x=1/\sqrt{3}$ and equals $2\lambda /(3\sqrt{3})$. Thus if $\lambda <3\sqrt{3}/2$, then $f((0,1))\subset (0,1)$. Since $f$ is odd, also $f((-1,0))\subset (-1,0)$. Induction givesThe case $x_0=0$ is constant. These intervals are positively invariant sets for the discrete dynamical system.

For $\lambda >1$, the multiplier at the origin exceeds one, so it is unstable. The other fixed points areUsing fixed point stability for an iteration, $|3-2\lambda |<1$ exactly when $1<\lambda <2$, and $|3-2\lambda |>1$ when $\lambda >2$. HenceThe endpoint $\lambda =2$ has multiplier $-1$ and requires nonlinear analysis, which is not part of the asserted open intervals.

For the iteration of a map $f(x)=\lambda x(1-x^2)$, a fixed point satisfies $x[1-\lambda +\lambda x^2]=0$. If $0<\lambda <1$, the bracket is strictly positive, so the sole fixed point is $0$. Its multiplier of a periodic orbit of an iteration is $f^{\prime }(0)=\lambda$, whose modulus is below one. Thus $\boxed{x=0\text{ is locally asymptotically stable}}$ by fixed point stability for an iteration.

The leapfrog finite-difference scheme for the diffusion equation uses centered time and space differences:A Fourier mode with amplification factor $z$ givesWriting $b=4\mu {\sin }^2(\theta /2)$, the roots are $z_{\pm }=-b\pm \sqrt{1+b^2}$. For any $\mu >0$ and any nonzero spatial frequency, $b>0$ andThusAlthough the other root decays, the growing parasitic root is excited by generic starting data or perturbations, so it cannot be discarded in a stability test.

For a Fourier stability analysis, substitute $u_j^n=g_n{e}^{ij\theta }$. The spatial second difference multiplies by $-4{\sin }^2(\theta /2)$. Set $a=4\mu {\sin }^2(\theta /2)\ge 0$. The mode satisfiesFor $a>0$ the two roots are real of opposite signs. Because $P(0)<0$ and $P(1)=a>0$, the positive root lies strictly between zero and one. The negative root is at least $-1$ exactly when $P(-1)=2-2a\ge 0$, namely $a\le 1$. At $a=0$ the roots are $1,0$; at $a=1$ they are $1/2,-1$, all simple. Thus the root condition for a multistep method for a stable two-step scheme holds exactly for $0\le a\le 1$. The discriminant $1-a+(9/4)a^2$ is bounded away from zero on that interval, so the mode powers are uniformly bounded, including the endpoints.

The maximum value of $a$ over all frequencies is $4\mu$, attained at $\theta =\pi$. ThereforeFor $\mu >1/4$, the highest-frequency mode has a negative amplification root of modulus greater than one and grows exponentially in the step number.

Put $\tau =\Delta t$ and $h=\Delta x$, with $h^2=\tau /\mu$ for fixed $\mu >0$. Substitute a smooth exact solution of the heat equation into the Adams-Bashforth method. Define its one-step defect $R$ as the left side minus the right side of the unscaled update. The centered finite difference satisfiesThe Taylor expansion in time givesSince $u_t=u_{xx}$ and $u_{tt}=u_{xxt}$, the first two terms cancel, leavingThis is the requested local error of the unscaled update. If “local truncation error” denotes the defect divided by $\Delta t$, it is instead $O(\Delta t)$ under fixed $\mu$: the centered spatial error is then $O(h^2)=O(\Delta t)$. The scheme is second order in time at fixed spatial discretization, which is a separate order statement.

A putative simultaneous assignment of classical values ${\tilde{s}}_x^{(i)},{\tilde{s}}_y^{(i)}\in \{\pm a\}$ reproducing the first three certain quantum outcomes must obey, almost surely,Multiply the three equalities. Each $y$ value occurs twice and its square is $a^2$, leaving $a^6{\tilde{s}}_x^{(1)}{\tilde{s}}_x^{(2)}{\tilde{s}}_x^{(3)}=a^9$. Thuscontrary to the quantum prediction $-a^3$.

This is the GHZ theorem against predetermined local measurement values. Experimentally prepare many copies of the three-spin state and choose independently among the four measurement settings on separated subsystems. The three mixed correlations force the classical prediction for the fourth, whereas quantum mechanics predicts the opposite sign. Since $s_x,s_y$ on the same subsystem do not commute, the four settings are measured on separate copies; the classical contradiction relies precisely on the common assignment of values across settings.

Put $a=\hslash /2$. Operators on distinct tensor product factors commute. Acting with ${\sigma }_x\otimes {\sigma }_y\otimes {\sigma }_y$ on $|\uparrow \uparrow \uparrow ⟩$ gives $-|\downarrow \downarrow \downarrow ⟩$, while acting on $|\downarrow \downarrow \downarrow ⟩$ gives $-|\uparrow \uparrow \uparrow ⟩$. Therefore their difference is an eigenvector with eigenvalue $+1$. The same computation applies to the permutations $yxy$ and $yyx$. The product $xxx$ flips the two basis tensors without a phase and so sends their difference to its negative.

Restoring the three factors of $a$, the GHZ state hasThese are certain measurement outcomes for the four product observables in this state.

In the spin-$z$ orthonormal basis, $|\uparrow ⟩=(1,0{)}^T$ and $|\downarrow ⟩=(0,1{)}^T$. The spin angular momentum operators are $s_i=(\hslash /2){\sigma }_i$. The Pauli matrices obey $[{\sigma }_i,{\sigma }_j]=2i{ϵ}_{ijk}{\sigma }_k$, soThe second identity is the correct $s(s+1){\hslash }^2$ eigenvalue for $s=1/2$. Matrix multiplication givesThus $s_x$ and $s_y$ flip the spin-$z$ basis states, with different complex phases.

If $U,V$ depend only on $x=\rho +\tau$, then ${\partial }_{\rho }={\partial }_{\tau }={\partial }_x$ on them. Put $A=U-V$ and $B=U+V$. The curvature equation givesBy cyclicity of matrix trace,This is an Isospectral Lax equation expressed as a Lax equation, and it proves the assertion for every positive integer $p$.

For the given matrices and $\varphi =\varphi (x)$,Using $(a\cdot \sigma {)}^2=(a\cdot a)I$ givesSince $\lambda$ is fixed, constancy of this trace yields the first integralDifferentiating confirms $dC/dx={\varphi }^{\prime }({\varphi }^{\prime \prime }-\sin \varphi )=0$ for the reduced equation.

Differentiating the two equations for a fundamental matrix of solutions givesEquality for all independent initial vectors is equivalent to the zero-curvature conditionA single possibly zero solution vector alone would not imply a matrix identity; compatibility is understood for the complete linear system.

Write $q={\varphi }_{\rho }$, $s=\sin \varphi$, $c=\cos \varphi$ and use Pauli matrices:The matrix products givewhereas $V_{\rho }=-(iqc/(4\lambda )){\sigma }_2+(iqs/(4\lambda )){\sigma }_3$. Thus these last two components cancel, and the remaining curvature is $(i/2)({\varphi }_{\rho \tau }-\sin \varphi ){\sigma }_1$. Thereforethe Sine-Gordon equation in light-cone coordinates.

Parameterize the initial line by $(s,2)$ and use characteristic time $\tau$. The characteristic equations for a transport equation integrate toHere $dU/d\tau =U^2$ and $U(0)=h$. Since $\tau =\frac{1}{4}\log (x_2/2)$, the solution isDirect differentiation gives $u_{x_1}=0$, $u_{x_2}=u^2/(4x_2)$, verifying the equation and the initial values. The map $(s,\tau )\mapsto (x_1,x_2)$ has determinant $8e^{6\tau }\ne 0$, so every point of this strip is reached exactly once.

At the upper boundary $u\to +\infty$, a finite-time blow-up of an ordinary differential equation. At the lower boundary $u\to 0$, but $u_{x_2}=u^2/(4x_2)$ diverges as $x_2\downarrow 0$, so no $C^1$ continuation through that boundary is possible. Thus the displayed strip is the maximal connected domain containing the initial line on which this classical Cauchy solution exists.

For the Cauchy problem for a partial differential equation, suppose $a,b,c$ are $C^1$ near the lifted initial data, $\Gamma$ has a regular $C^1$ parametrization $(\xi (s),\eta (s))$, and the initial function $u_0(s)$ is $C^1$. The local existence and uniqueness theorem requires the Noncharacteristic Cauchy data conditionat the starting point. Then there is a neighbourhood of that point carrying a unique $C^1$ classical solution attaining the initial values.

The method of characteristics proves this by solvingwith initial values $(\xi (s),\eta (s),u_0(s))$ at $\tau =0$. Local existence for these ordinary differential equations gives a $C^1$ map. The noncharacteristic determinant is, up to sign, the Jacobian determinant of $(s,\tau )\mapsto (x_1,x_2)$ initially. The inverse function theorem makes this map locally invertible, defining $u(x_1,x_2)=U(s,\tau )$ and proving uniqueness along characteristics.