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Suppose a compact connected component $C$ of $\{x:V(x)\le c\}$ contains an equilibrium $x_{\ast }$ and $\dot{V}<0$ throughout $C\setminus \{x_{\ast }\}$. Then $C$ is an invariant sublevel set, and every trajectory in $C$ converges to $x_{\ast }$ by the LaSalle invariance principle. Thus the interior of $C$ lies in the basin of attraction of $x_{\ast }$. The largest such component supplies the strongest basin estimate available from that Lyapunov function alone.

This Lotka-Volterra predator-prey model includes self-limitation of the prey through the term $-b{x}^2$, with positive constants $a,b,c,d,e$. A positive coexistence equilibrium point exists when $a>be/d$, at $x_{\ast }=e/d$, $y_{\ast }=(a-b{x}_{\ast })/c$. The Lyapunov functionsatisfies $\dot{V}=-bd(x-x_{\ast }{)}^2$: its predator-prey cross terms cancel. Its sublevel sets are compact in the positive quadrant. The only invariant subset of $\{\dot{V}=0\}$ is the coexistence equilibrium point, because staying on $x=x_{\ast }$ requires $y=y_{\ast }$. The LaSalle invariance principle therefore proves convergence to coexistence for every positive initial state.

Introduce $\omega =\dot{\theta }$. The pendulum with quadratic damping is the autonomous planar systemIts equilibrium points are $(k\pi ,0)$. Since the derivative of $\omega |\omega |$ is zero at zero, the linearization of a dynamical system at each equilibrium point isAt an odd multiple of $\pi$ its eigenvalues are $\pm 1$, so it is a saddle equilibrium; the stable and unstable tangents are $\omega =-(\theta -k\pi )$ and $\omega =\theta -k\pi$. At an even multiple the eigenvalues are $\pm i$. This linearization is a centre and does not decide the nonlinear stability. In particular it would be incorrect to infer a linear damped focus from this Jacobian.

For the full system the mechanical energy is a Lyapunov function near an even equilibrium point:A sufficiently small sublevel set is compact and contains only that equilibrium point. The only invariant motion with $\dot{H}=0$ has $\omega =0$ and $\sin \theta =0$, hence is the equilibrium point itself. The LaSalle invariance principle proves asymptotic stability. Near it the trajectories turn clockwise and spiral inward, as the near-linear angular motion persists while energy decreases. Thus even equilibria are nonhyperbolic asymptotically stable spiral points; odd equilibria remain saddles.

For explicit trajectories put $W={\omega }^2$ on a branch where $\omega \ne 0$. The chain rule gives $dW/d\theta =2\dot{\omega }$, so the two signs yield the first-order linear differential equations $W^{\prime }+2cW=-2\sin \theta$ for $\omega >0$, and $W^{\prime }-2cW=-2\sin \theta$ for $\omega <0$. Their integrating factors giveOnly portions with a nonnegative right-hand side are physical. At a turning point the constants of the two branches are matched using $\omega =0$; neither expression is one global trajectory through both signs. When $c=0$ they reduce to energy contours, with closed orbits around the even equilibrium points. For every $c>0$, the quadratic damping destroys those closed orbits and makes the centres attracting spirals. Its effect is higher order in amplitude, so decay is weaker near rest than with linear damping. The saddle equilibria remain unstable, though their separatrices change shape.

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.

In congestion control, resource $j$ observes only its own aggregate load $y_j={\sum }_{r:j\in r}{x}_r$ and generates the nonnegative feedback price ${\mu }_j=p_j(y_j)$. User $r$ receives the sum $M_r={\sum }_{j\in r}{\mu }_j$ of prices along its route. Its desired increase $w_r$ competes with the congestion charge $x_r{M}_r$, while ${\kappa }_r$ sets the adjustment speed. This is local information: a resource needs the rates passing through it, and a user needs the feedback along its own route, rather than the complete network state.

There is a hypothesis to make explicit before claiming global convergence. If “increasing” means nondecreasing, the printed assumptions allow $p_j\equiv 0$, in which case ${\dot{x}}_r={\kappa }_r{w}_r>0$ and there is no equilibrium point. The intended theorem holds when every route contains at least one resource whose price is not identically zero, with nonempty routes and nonnegative, continuous, nondecreasing prices. Strictly increasing prices satisfy this condition automatically. We prove the theorem under this necessary qualification; without it the zero-price example disproves the literal claim.

Let $P_j(y)={\int }_0^y{p}_j(u)du$. Each $P_j$ is a nonnegative convex function. For positive rates define the primal congestion potentialThe weighted logarithmic utility is a strictly concave function, and subtracting the resource costs preserves strict concavity. Thus $U$ has at most one maximizer and at most one critical point on the positive orthant.

It also attains its maximum there. A nonzero nondecreasing price is bounded below by some positive constant at all sufficiently large loads; hence its primitive satisfies $P_j(y)\ge c_{j}y-K_j$ with $c_j>0$. Choose $c_j=0$ at zero-price resources. The route coverage assumption gives $c={\min }_r{\sum }_{j\in r}{c}_j>0$, soThis bounds all superlevel sets of $U$. Within a bounded set, approaching $x_r=0$ makes $U\to -\infty$. Consequently every nonempty superlevel set is a compact set contained in the positive orthant. A maximizing sequence stays in one such set, giving a unique interior maximizer $x^{\ast }$. Its critical-point equations areThe rate dynamics are the gradient flow with diagonal mobility ${\dot{x}}_r={\kappa }_r{x}_r{g}_r(x)$. Along any positive trajectory,with equality only at $x^{\ast }$. Thus $V(x)=U(x^{\ast })-U(x)$ is a strict Lyapunov function. The initial superlevel set traps the trajectory away from both infinity and the boundary, ensuring global continuation. Positive initial rates remain positive: variation of constants in ${\dot{x}}_r+{\kappa }_r{M}_r(t)x_r={\kappa }_r{w}_r$ makes this explicit. Starting with finite nonnegative rates also gives positive rates at every later positive time.

For completeness, mere continuity of $p_j$ suffices for convergence; one need not silently assume differentiable prices. On the trapping compact set, both the vector field and $D(x)={\sum }_r{\kappa }_r{x}_r{g}_r(x{)}^2$ are continuous and bounded. The trajectory satisfies a uniform Lipschitz condition in time, so $D(x(t))$ is uniformly continuous. Also ${\int }_0^{\infty }D(x(t))dt<\infty$, since $U$ increases to a finite limit. The decay of a nonnegative uniformly continuous integrable function applies: otherwise disjoint intervals of a fixed positive width would each contribute a fixed positive amount. Every accumulation point therefore has $D=0$ and is $x^{\ast }$. Compactness then proves the global convergenceThis is the dissipation argument behind the LaSalle invariance principle. If uniqueness of trajectories is desired under continuous prices, put $z_r=2\sqrt{x_r/{\kappa }_r}$. The transformed equation is $\dot{z}={\nabla }_{z}U(\kappa z^2/4)$. The transformed potential is concave: it is a sum of positive multiples of $\log z_r$, minus convex nondecreasing $P_j$ composed with nonnegative quadratic sums. Its gradient is monotone decreasing, so the squared distance between two solutions cannot increase. This proves uniqueness for positive initial data without a locally Lipschitz function hypothesis on $p_j$.

For the alternative price dynamics, restrict first to positive prices on resources used by at least one route, and discard unused resources. Define the dual congestion potentialIts domain has ${\mu }_j\ge 0$ and $M_r>0$ for every route. Strictly increasing $q_j$ make each $Q_j$ strictly convex, and $-\log M_r$ is convex. Thus $H$ is a strictly convex function. Each $q_j$ is eventually bounded below by a positive constant, so its integral grows at least linearly, dominating the negative logarithms as $\Vert \mu \Vert \to \infty$. If a route price tends to zero in a bounded set, $H\to \infty$. Therefore $H$ attains a unique minimum on its domain including the admissible boundary faces.

That minimum has every used-resource price positive. At a face ${\mu }_j=0$, all route prices are still positive and the inward derivative isHence the minimum cannot lie on that face. At the unique interior minimum,Since the price dynamics are ${\dot{\mu }}_j=-{\kappa }_j{\mu }_j{\partial }_{j}H$, their positive equilibrium points are exactly these critical points. There is one equilibrium in the positive price domain. An unused resource instead has its only nonnegative equilibrium at zero.

The positive-domain qualification matters here even with strictly increasing $q_j$. If zero prices are admitted, the multiplicative factor ${\mu }_j$ permits boundary equilibria of multiplicative price dynamics. Take one route using two resources, $w=1$, and $q_1(a)=q_2(a)=a$. Both $({\mu }_1,{\mu }_2)=(1,0)$ and $(0,1)$ are equilibria, with $x=1$, as is the positive equilibrium $(1/\sqrt{2},1/\sqrt{2})$, with $x=1/\sqrt{2}$. Thus the literal assertion of uniqueness on the whole nonnegative orthant is false. A boundary equilibrium need not satisfy load equals supplied capacity at its zero-price resources.

The relationship between the two intended equilibria is inverse resource response. The first system has ${\mu }_j=p_j(y_j)$; the positive equilibrium of the second has $y_j=q_j({\mu }_j)$. If $q_j=p_j^{-1}$ on the relevant ranges, these are the same resource equations, and both systems also impose $x_r{M}_r=w_r$. They therefore have the same rates and prices. For strictly increasing continuous $p_j$ with $p_j(0)=0$ and unbounded range, the ordinary inverses satisfy all the stated conditions on $q_j$. With arbitrary independently chosen $p_j$ and $q_j$, the equilibria need not agree. Under inverse response, $P_j$ and $Q_j$ are convex conjugates on the nonnegative half-line, and $H$ supplies the Lagrangian dual problem for maximizing the weighted logarithmic utility minus resource costs, up to constants independent of $\mu$.

Use the complete system in the PDF: the TeX omits $\dot{y}=y(4x-1)/8$. Both coordinate axes are invariant, and the equilibrium points are $(0,0)$, $(1,0)$ and $(1/4,3/4)$. The Jacobian matrix isAt $(0,0)$ its eigenvalues are $1,-1/8$, and at $(1,0)$ they are $-1,3/8$: both points are saddle equilibria. At the coexistence point,Thus the interior point is a stable spiral. The nontrivial nullclines are $x+y=1$ and $x=1/4$: $x$ increases below the first and decreases above it, while $y$ increases to the right of the second and decreases to the left. The rotation near coexistence is counterclockwise.

A local linearization by itself does not prove that every interior solution has the same limit. For this logistic predator-prey model, let $x_{\ast }=1/4,y_{\ast }=3/4$ and use the Lyapunov functionThe inequality $u-1-\log u\ge 0$ makes $V$ nonnegative, with equality only at coexistence. Its sublevel sets are compact inside the positive quadrant, because $V$ diverges both at the axes and at infinity. Direct differentiation cancels the predator-prey cross terms and givesOn $\{\dot{V}=0\}$ one has $x=x_{\ast }$, and remaining there requires $\dot{x}=-x_{\ast }(y-y_{\ast })=0$. The largest invariant subset is therefore the coexistence point. The LaSalle invariance principle provesOn $x=0$, $y$ decays as $e^{-t/8}$ toward the origin. On $y=0$, the logistic differential equation takes every $x(0)>0$ to one. The origin itself stays fixed. These boundary trajectories are the exceptions to the interior convergence statement.

The direct First Lyapunov theorem says that a continuously differentiable positive-definite function $V$ near an equilibrium, with $\dot{V}\le 0$, proves stability. LaSalle invariance principle says that in a compact positively invariant set, trajectories approach the largest invariant subset of $\{\dot{V}=0\}$. Under the terminology in which “first theorem” means the linearization theorem, eigenvalues with negative real part imply asymptotic stability, while zero-real-part eigenvalues leave the test inconclusive. Here the linearization has eigenvalues $0,-k$, so the direct criterion is needed.

Write $y=\dot{x}$ and chooseNear zero, $V=y^2/2+x^4/4+O(x^6)$ is positive definite. Along solutions,The component containing the origin of a sufficiently small closed sublevel set is compact, lies in $|x|<\pi$ and is positively invariant. The set $\dot{V}=0$ has $y=0$; to remain there also requires ${\sin }^{3}x=0$, so its largest invariant subset is just the origin. The two theorems establish the origin is asymptotically stable.

Write $z=x+y$. When ${\mu }_1={\mu }_2=0$ the system gives $\dot{z}=y-z^3$ and $\dot{y}=-z^3$. Choosingmakes $V$ positive definite and radially unbounded, while its orbital derivative isThe set $\{\dot{V}=0\}$ is $z=0$. A trajectory remaining there must also have $\dot{z}=y=0$, so its largest invariant subset is the origin. The LaSalle invariance principle therefore proves that the origin is globally asymptotically stable. In the phase plane, $\dot{x}=0$ on $y=0$ and $\dot{y}=0$ on $x+y=0$; every nonstationary trajectory crosses the nested level curves of $V$ inward and tends to the origin.

At an equilibrium point, $y=0$ andHence the origin always exists, while $x=\pm \sqrt{{\mu }_1}$ exist for ${\mu }_1>0$. At $(x_{\ast },0)$ the Jacobian matrix isA stationary bifurcation occurs when a real eigenvalue passes through zero, hence onA Hopf bifurcation requires zero trace and positive determinant. At the origin this givesand on either nonzero branch it givesThus the Hopf locus consists of the negative ${\mu }_1$-axis and the ray ${\mu }_2=3{\mu }_1$ in the first quadrant; the stationary locus is the ${\mu }_2$-axis.

Now set ${\mu }_2=-1$, write $\mu ={\mu }_1$, and append $\dot{\mu }=0$. At $(x,y,\mu )=(0,0,0)$ the extended centre manifold for a parameter is tangent to the $(x,\mu )$-plane, so put $y=h(x,\mu )$. Its centre-manifold invariance equation isUnder the ordering $\mu =O(x^2)$, solving through cubic order givesThis is the pitchfork bifurcation normal form: $x=0$ is stable for $\mu <0$ and unstable for $\mu >0$, while the two stable branches $x=\pm \sqrt{\mu }$ emerge for $\mu >0$. The bifurcation is therefore a supercritical pitchfork bifurcation.

Let ${\varphi }_t(x_0)$ denote the flow map generated by $\dot{x}=f(x)$.

is nonpositive on $E$.

The First Lyapunov theorem says that the existence of such a function makes the origin Lyapunov stable. To prove it, choose a sufficiently small closed ball ${\overline{B}}_{\epsilon }\subset E$. By compactness and positive definiteness,By continuity, some $\delta >0$ satisfies $V(x)<\alpha$ whenever $\Vert x\Vert <\delta$. Along a trajectory beginning in $B_{\delta }$, $V$ cannot increase. Such a trajectory therefore cannot reach the sphere $\Vert x\Vert =\epsilon$, where $V\ge \alpha$. Hence it remains in $B_{\epsilon }$ for every $t\ge 0$, which is Lyapunov stability.

The LaSalle invariance principle says that if a trajectory remains in a compact positively invariant set on which $\dot{V}\le 0$, then it approaches the largest invariant subset of $\{x:\dot{V}(x)=0\}$.

To obtain a basin estimate from a Lyapunov sublevel set, take the largest sublevel component around the origin that remains inside the strict-decrease rectangle. On its boundary $|x|=\sqrt{2}$ or $|y|=\pi /2$, the minimum value of $V$ is $1$. Hence the maximal open estimate isEvery smaller closed sublevel set $V\le c<1$ is compact and positively invariant, and $\dot{V}<0$ there except at the origin. The LaSalle invariance principle therefore implies convergence to the origin. Taking the union over $c<1$ shows that ${\mathcal{D}}_V$ lies in the domain of stability. No larger sublevel component can be certified from this $V$ alone, because every level $c>1$ contains points $(0,y)$ just beyond $|y|=\pi /2$ where $\dot{V}>0$.

Let $x_{\ast }$ be an equilibrium point of a dynamical system. A Lyapunov function on a neighbourhood of $x_{\ast }$ is a continuously differentiable function $V$ such thatand whose derivative along trajectories satisfies

The First Lyapunov theorem says that the existence of such a positive-definite $V$ proves Lyapunov stability of $x_{\ast }$. If $\dot{V}<0$ away from $x_{\ast }$, the Second Lyapunov theorem strengthens this to asymptotic stability.

LaSalle invariance principle says that if a trajectory remains in a compact positively invariant set $\Omega$ on which $\dot{V}\le 0$, then it approaches the largest invariant subset ofIn particular, if that largest invariant subset consists only of $x_{\ast }$, every trajectory in $\Omega$ approaches $x_{\ast }$.

SetThe stated identity givesHence, along the system,Near the origin, $E$ is positive definite because $U(x)>0$ for $x\ne 0$. The First Lyapunov theorem therefore gives stability.

Choose a sufficiently small compact invariant sublevel setOn $\{\dot{E}=0\}$ we have $y=0$. A trajectory can remain there only ifwhich occurs at $x=0,\pm 1$. The choice $c<1/4=U(\pm 1)$ excludes the two latter points, so the largest invariant subset of $\{\dot{E}=0\}\cap {\Omega }_c$ is the origin. LaSalle invariance principle now gives convergence to the origin. Thus, for every $k>0$,This is an instance of damped mechanical energy as a Lyapunov function.

For $\dot{x}=f(x)$, a fixed point $x^{\ast }$ satisfies $f(x^{\ast })=0$. It is Lyapunov stable when, for every $\epsilon >0$, there is $\delta >0$ such that

A Lyapunov function on a neighbourhood of $x^{\ast }$ is a continuously differentiable function $V$ such thatThe First Lyapunov theorem states that such a function makes $x^{\ast }$ Lyapunov stable. If $\dot{V}<0$ away from $x^{\ast }$, then $x^{\ast }$ is locally asymptotically stable.

To prove stability, choose a closed ball contained in the domain of $V$. On its boundary sphere $|x-x^{\ast }|=\epsilon$, compactness and positive definiteness giveContinuity at $x^{\ast }$ supplies $\delta <\epsilon$ such that $|x-x^{\ast }|<\delta$ implies $V(x)<m$. Along a trajectory, $V$ cannot increase. Such a trajectory therefore cannot first reach the boundary sphere, where its value would be at least $m$. This proves Lyapunov stability.

If $\dot{V}<0$ away from $x^{\ast }$, take a sufficiently small compact sublevel set of $V$. The LaSalle invariance principle says that every trajectory in it approaches the largest invariant subset of $\{\dot{V}=0\}$, which is just $\{x^{\ast }\}$. Hence the equilibrium is also asymptotically stable.

Put $y=\dot{x}$. The first-order system isAt an equilibrium point of a dynamical system, $y=0$ and $x(1-x^2{)}^2=0$, so the three fixed points are

DefineThen $U^{\prime }(x)=2x(1-x^2{)}^2=-F(x)$ andMoreover,so $E$ is positive definite at the origin. It is therefore a damped mechanical energy as a Lyapunov function, and the First Lyapunov theorem proves that the origin is Lyapunov stable.

Since $U(\pm 1)=1/3$, choose a compact energy sublevel $\{E\le c\}$ with $0<c<1/3$. In this set, $\dot{E}=0$ means $y=0$. A trajectory can remain in $y=0$ only when $F(x)=0$, and the chosen sublevel excludes $x=\pm 1$. Thus the largest invariant subset of $\{\dot{E}=0\}$ is the origin. The LaSalle invariance principle proves that the origin is asymptotically stable.

Finally, $E$ is radially unbounded because its leading terms are $x^6/3+y^2/2$. Every forward trajectory is therefore bounded. LaSalle's principle puts its omega-limit set insideAn omega-limit set of a bounded continuous trajectory is nonempty and connected. A connected subset of this three-point set is a singleton, so every trajectory has precisely one of the three fixed points as its omega-limit set.

Take the regular radial solution with $u(0)=1$. In the transformed variables it satisfiesso $J(s)\to 0$. Since $s_c>1$, the Lyapunov function immediately becomes negative. The trajectory cannot reach $v=0$, where $J={\dot{v}}^2\ge 0$, and $J<0$ confines it below the positive zero of $f$. Hence it remains positive and bounded for all $s$.

The identity $\dot{J}=-4(s_c-1){\dot{v}}^2$ and the LaSalle invariance principle force the omega-limit set to consist of equilibria. The negative limiting energy excludes $v=0$, leavingConsequentlyso $\alpha =2/(p-1)$. At infinity, $|u^{\prime }(r)|\sim a{c}_{\infty }{r}^{-a-1}$, andbecause $s_c>1$. Thus $\nabla u\notin L^2({\mathbb{R}}^d)$.