Apply the divergence theorem to $\varphi \mathbf{c}$, using ii. Because the vector $\mathbf{c}$ is constant, it can be taken outside both integrals:This holds for every $\mathbf{c}$, in particular each of the three Cartesian basis vectors. Equality of the three components proves the gradient volume-to-boundary identityIt suffices that $\varphi$ be a continuously differentiable function on a neighbourhood of the closure of $V$.

In Cartesian coordinates, write the constant vector as $\mathbf{c}=(c_1,c_2,c_3)$. The product rule and constancy of its components giveThe last expression is the dot product with the gradient, so

For a bounded region $V\subset {\mathbb{R}}^3$ with piecewise smooth boundary, suppose $\mathbf{F}$ is a continuously differentiable vector field on an open neighbourhood of its closure. The divergence theorem, with outward unit normal vector $\mathbf{n}$, isThe surface integral measures total outward flux. Edges and isolated vertices have zero surface area and do not affect the integral.

Here a potential of a conservative vector field means a single-valued scalar field $\psi$ with $\nabla \psi =\mathbf{F}$. By the fundamental theorem for line integrals, its line integral around every closed curve must vanish. Here the angular circulation obstruction to a cylindrical potential lets us test this using a circle about the axis.

On $V_1$, the circle from i lies entirely in the region and has line integral $4\sqrt{2}$. Therefore there is no potential of a conservative vector field on $V_1$.

On $V_2$, choose the continuous angular branch $-\pi <\theta <\pi$. By a, the equation for a potential of a conservative vector field reduces to ${\psi }_r={\psi }_z=0$ and ${\psi }_{\theta }=\sqrt{2}\cos (\theta /2)$. ThusThe denominator is positive throughout $V_2$, including when $y=0$ and $x>0$, so this is a well-defined differentiable potential of a conservative vector field on the entire region.

For $V_3$, choose the particular circle $x=2\cos \theta$, $y=2\sin \theta$, $z=0$. Its only point on the removed half-plane's supporting cut is $(-2,0,0)$, where $(x+2{)}^2+z^2=0<1$; this point is retained by $V_3$. Thus the whole circle lies in $V_3$ and again has line integral $4\sqrt{2}$. Consequently there is no potential of a conservative vector field on $V_3$. The opening in the cut allows a closed curve to go around the excluded axis.

Finally $V_4\subset V_2$. Restricting the potential of a conservative vector field already constructed on $V_2$ givesThe extra excluded cylinder does not invalidate an existing potential of a conservative vector field. Thus being simply connected is a sufficient condition in the usual curl-free criterion, not a necessary condition for a particular vector field to have a potential of a conservative vector field.

In cylindrical coordinates, $(-y,x,0)=r{\mathbf{e}}_{\theta }$ and $x+r=r(1+\cos \theta )$. Hence the vector field simplifies toTake the orientation of $C$ to be increasing $\theta$, as in its given parametrization. On the circle, $d\mathbf{r}=a{\mathbf{e}}_{\theta }d\theta$, so its line integral isWe used $\sqrt{1+\cos \theta }=\sqrt{2}|\cos (\theta /2)|$; the cosine is nonnegative on this interval. In particular, the answer does not depend on $a$.

The fixed-point iteration stays inside $[0,7]$ because $G$ maps that interval to itself. A fixed point satisfies $\xi =\sqrt{2+\xi }$, so $\xi \ge 0$ andOnly $\xi =2$ is admissible; $-1$ fails the unsquared equation. With $q=1/(2\sqrt{2})$, the mean value theorem gives the error boundInduction yields $|x_k-2|\le q^k|x_0-2|\le 5q^k\to 0$, since $0<q<1$. Therefore, for every allowed starting value,This estimate proves convergence directly, without assuming in advance that the sequence has a limit.

Set $H(x)=g(x)-x$. This function is continuous on $[a,b]$. Since $g$ takes values in $[a,b]$,The intermediate value theorem therefore supplies $\xi \in [a,b]$ with $H(\xi )=0$, equivalently $g(\xi )=\xi$. This includes the possibility that one endpoint is already a fixed point.

For uniqueness, the mean value theorem gives, for any $x,y\in [a,b]$,If both $\xi$ and $\eta$ are fixed points, thenSince $\gamma <1$, this forces $|\xi -\eta |=0$. Thus there exists exactly one fixed point. When $a=b$, the conclusion follows immediately from $g(a)=a$; otherwise the argument above applies. The contraction mapping estimate also proves convergence of the associated fixed-point iteration, as used next.

For $0\le x\le 7$, we have $2\le 2+x\le 9$. Taking the nonnegative square root givesThe derivative issoIn particular, the mean value theorem makes $G$ a contraction mapping on this interval.

At any division step $u=qv+r$, an integer divides both $u,v$ exactly when it divides both $v,r$: one direction uses $r=u-qv$, and the other uses $u=qv+r$. HenceStarting with $r_0=a$, $r_1=b$, the Euclidean division producesThe nonzero remainders form a strictly decreasing sequence of positive integers. Such a sequence must terminate, since there are only finitely many positive integers below $b$. Suppose $r_N$ is the last nonzero remainder, so $r_{N+1}=0$. The unchanged greatest common divisor at each step givesThe final equality holds because the positive common divisors of $r_N$ and $0$ are precisely the positive divisors of $r_N$. This proves both termination and correctness of the Euclidean algorithm.

The divisions in the Euclidean algorithm areThe last nonzero remainder is $17$, soAs a check, $323=19\cdot 17$ and $187=11\cdot 17$; back-substitution also gives $17=-4\cdot 323+7\cdot 187$.

At ${\mathbf{y}}^{\ast }=(4/3,2)$, the first three dual left sides are $10/3$, $2/3$ and $14/3$, strictly above their bounds $1$, $-1$ and $-5$. Thus complementary slackness forces $x_1=x_2=x_3=0$. Both dual variables are positive, so both primal inequalities are equalities:Solving this linear system givesAll components are nonnegative, the two original upper bounds are met exactly, and the objective function equals the dual value. The weak duality bound therefore certifies optimality. The forced zero components and unique solution of the two remaining equalities also prove that the primal optimum is unique.

After removing the two redundant inequalities, the feasible region isIts lower boundary consists of three line segments or rays: $y_2=6-3y_1$ for $0\le y_1\le 4/3$, then $y_2=3-3y_1/4$ for $4/3\le y_1\le 16/7$, then $y_2=y_1-1$ for $y_1\ge 16/7$. The feasible region is everything on or above this boundary in the first quadrant.

The boundary extreme points areAlong its three successive pieces, the objective function $2y_1+y_2$ has respective slopes $-1$, $5/4$ and $3$ as $y_1$ increases. Thus it decreases to the middle extreme point and increases thereafter. For fixed $y_1$, moving above the boundary only increases the objective function. HenceAs an algebraic check of the graphical answer, the active inequalities givewith equality at the stated extreme point.

The inequality $3y_1+y_2\ge 6$, together with $y_2\ge 0$, impliesAlso $-y_1+y_2\ge -1$ impliesThus both inequalities are strictly slack throughout the feasible region. They are redundant: the remaining inequalities imply them, even with strict inequality. By complementary slackness, any primal optimum must therefore have $x_1=x_3=0$; this is the principle that strictly slack dual constraints force zero primal variables.

There are two primal inequalities, so there are two nonnegative dual variables. The five primal variables yield five dual inequalities:Each coefficient column of the primal linear program becomes one row of this dual linear program.

The complementary slackness conditions areTo see their role, decompose the duality gap asEvery summand is nonnegative for feasible solutions, so the gap is zero exactly when every displayed product vanishes. Zero gap implies optimality by weak duality. Conversely, strong duality states that primal and dual optima, when both are attained, have equal values, and hence satisfy these conditions.

For every primal feasible solution $\mathbf{x}$, weak duality givesThus no primal feasible solution improves on ${\mathbf{x}}^{\ast }$. Similarly, for every dual feasible solution $\mathbf{y}$,Therefore both candidates are optimal, with the same objective function value. Equality is a linear programming optimality certificate and does not require a separate existence argument.

For feasible solutions $\mathbf{x},\mathbf{y}$, nonnegativity givesThe first inequality uses $A^T\mathbf{y}\ge \mathbf{c}$ and $\mathbf{x}\ge 0$; the second uses $A\mathbf{x}\le \mathbf{b}$ and $\mathbf{y}\ge 0$. This is weak duality.

If $A$ has $m$ rows and $n$ columns, attach a nonnegative dual variable $y_i$ to each of the $m$ upper-bound constraints. The dual linear program isHere $A^T$ is the transpose of $A$, and all vector inequalities hold componentwise. Nonnegative dual variables preserve the direction of the original inequalities when taking weighted sums.

If $g{f}^{\prime \prime }(a)<{\omega }^2$, putThe linearized equation becomes $\ddot{ϵ}-{\lambda }^2ϵ=0$, soArbitrarily small initial perturbations can have $C_1\ne 0$ and grow exponentially. Hence the strict unstable equilibrium condition isFor the particular wire, this occurs precisely at $a=2j\pi b$, $j=1,2,\dots$; the alternating radii $a=(2j-1)\pi b$ satisfy the stable equilibrium condition.

If $g{f}^{\prime \prime }(a)>{\omega }^2$, defineThe linearized equation is $\ddot{ϵ}+{\nu }^2ϵ=0$, with bounded simple harmonic motionThus the strict stable equilibrium condition isThis is also a sufficient condition for local nonlinear Lyapunov stability. Indeed the equation for the bead on a uniformly rotating wire has the conserved quantityIts derivative is $\dot{r}$ times the governing equation moved to one side, so it vanishes. At $a$, the last two terms have zero first derivative and positive second derivative $g{f}^{\prime \prime }(a)-{\omega }^2$. Their strict local minimum traps sufficiently small perturbations. The motion is bounded rather than asymptotically returning to rest.