3.有限时间收敛梯度流外文翻译资料

3. Finite-time convergent gradient flows

Here, we formally introduce the normalized and signed gradient flows of a differentiable function, and characterize their convergence properties. We build on Section 2 to identify conditions for finite-time convergence. Let

Both equations have discontinuous right-hand sides. Hence, we understand their solutions in the Filippov sense. Note that the trajectories of (6a) and of = -grad(f)(x) describe the same

paths.

Lemma 6. The Filippov set-valued maps associated with the discontinuous vector fields (6a) and (6b) are

K[](x)

=co{} ,

K[sgn(grad(f))](x)

={

注：K[](x)=grad(f)(x)/

如果

The proof of this result follows from the definition (4) of the operator K and the particular forms of (6a) and (6b).

For a differentiable function f, let Critical(f)= {x | grad(f)(x)= 0} denote the set of its critical points, and let Hess(f )(x) denote its Hessian matrix at x. The next result establishes the general asymptotic properties of the flows in (6).

Proposition 7. Let f : →R be a differentiable function.Let S, with S compact and strongly invariant for (6a) (resp., for (6b)). Then each solution of Eq. (6a) (resp.Eq. (6b)) starting from asymptotically converges to Critical(f ).

Proof. For Eq. (6a), if grad(f)(x) 0, then

= {}

= {}.

If, instead, grad(f )(x)=0, then = {0}.Therefore, we deduce =- for all x..

Consequently, = Critical(f) is closed, and LaSalle Invariance Principle (cf. Theorem 3) implies that each solution of (6a) starting from x0 asymptotically converges to

the largest weakly invariant set M contained in Critical(f )cap;S,which is Critical(f ) cap; S itself.

For Eq. (6b), we have a if and only if there exists visin;K[](x) such that a = v·grad(f)(x). From Lemma 6, we deduce that a = sgn()· ... sgn()·=. Therefore, we deduce

.

Consequently, = Critical(f) is closed, and LaSalle Invariance Principle implies that each solution of (6b) starting from x0 asymptotically converges to the largest weakly invariant set M contained in Critical(f ) cap; S, which is Critical(f ) cap; S itself. 'Let us now discuss the finite-time convergence properties of the vector fields (6). Note that Proposition 4 cannot be applied.Indeed,

max=-，

max，

and both =0 and

=0, for any neighborhood U of Critical(f ) cap; S in S. Hence, the hypotheses of Proposition 4 are not verified by either (6a) or (6b).Under additional conditions, one can establish stronger convergence properties of (6). We show this next.

Theorem 8. Let f : →R be a second-order differentiable function. Let isin;Ssub; , with S compact and strongly invariant for (6a) (resp., for (6b)). Assume there exists a neighborhood

V of Critical(f )cap;S in S where either one of the following conditions hold:

(i) for all x isin; V , Hess(f )(x) is positive definite; or

(ii) for all xisin;V(Critical(f)cap;S), Hess(f )(x) is positive

semidefinite, the multiplicity of the eigenvalue 0 is constant,and grad(f )(x) is orthogonal to the eigenspace of Hess(f )(x) corresponding to 0.Then each solution of (6a) (resp. (6b)) starting fromconverges in finite time to a critical point of f. Furthermore, if V =S, then the convergence time of the solutions of (6a) (resp.(6b)) starting from x0 is upper bounded by

(resp﹒) ,

where '=.

Proof. Our strategy is to show that the hypotheses of Theorem 5 are verified by both vector fields. From Proposition 7,we know that each solution of (6a) (resp. (6b)) starting from converges to Critical(f ). Let us take an open set U sub; S such that Critical(f)Ssub;Usub;sub;V. Since S is compact,U is also compact. By continuity, under either assumption (i) or assumption (ii), the function '(Hess(f))：→R，x，→(Hess(f)(x)), reaches its minimum on , i.e., there

exists 'gt; 0 such that ' xisin;，(Hess(f)(x)).Moreover, from (1), we have for all uisin; ,

For (6a), recall from the proof of Proposition 7, that the function x is single-valued, locally Lipschitz and regular, and hypothesis (i) in Theorem 5 is satisfied. Additionally, = Critical(f). Let us take x Critical(f), and let us compute

. Noting

!

we deduce

=

Let xisin;(Critical(f)cap;S). Under either assumption (i) or (ii) in the theorem, !.Then, using (7) in Eq. (8), we conclude

=

for xisin;(Critical(f)cap;S). Hence, hypothesis (ii) in Theorem 5 is also verified, and we deduce that the set Critical(f ) is reached in finite time, which in particular implies that the limit of any solution of Eq. (6a) starting from x0 isin; S exists and is reached in finite time.

For (6b), recall from Proposition 7, that the function xisin;→is single-valued,locally Lipschitz and regular, and hypothesis (i) in Theorem 5

is satisfied. Additionally, = Critical(f). Let us take x Critical(f), and compute . By definition, aisin; if and only if there exists visin;K[sgn(grad(f))](x) such that a = a = v·,for anyisin;()(x). Note that

()(x)

={ , for some with

=sgn() if and

if , for i isin; {1, . . . , d}}.

In particular, Hess(f)(x) visin;()(x). Then a=. Let us now decompose v as v=, where and . Because visin;K[sgn(grad(f))](x), we deduce v·grad(f)(x)= . Let xisin;(Critical(f)cap;S).Under either assumption (i) or (ii),

= v·grad(f)(x)

=

.

Using for any uisin; , we deduce from this e

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