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The case when W has no linear terms in y.

Proposition (19981018). In the following cases, the conclusion of Problem 1 (the uniqueness of critical point of w in D) holds without additional conditions under the assumption of Problem 1.

  1. If W is a homogeneous polynomial.
  2. If W is a polynomial of total degree 4 or less (not necessarily homogeneous).

Remarks.

  1. For both cases in the Proposition, no term of y-degree 1 is allowed under the assumptions of the Problem 1.
  2. For W of homogeneous degree, the y dependence of W is in a single term under the assumptions of Problem 1. Hence this case is actually a special case of Proposition (19981226).
  3. W with terms of degree 4 or less are the examples (the only known class of examples with non-trivial parameters in the coefficients) of W with more than one terms of non-trivial dependence on y.

Put F(x,y)=(y/x2) (Wx2/Wy)(x,y). Note that if F(x,y)> 1 holds on D-{(x,0)|x>0}, then y/x2 is a Lyapnov function (on the same set) of a dynamical system (Xn,Yn) determined by grad W; namely, if (Xn+1,Yn+1)= (grad W)(Xn,Yn) and if (Xn,Yn) is in D-{(x,0)|x>0}, then Yn+1/Xn+12 <Yn/Xn2 holds. Therefore the condition F(x,y)> 1 serves as a sufficient condition for the absence of critical point of w in D - {(x,0)|x>0}. In this respect, we have the following result.

Proposition (19981226). Consider a W which satisfies the conditions in Problem 1. Assume furthermore that W has just one term with y degree greater than 1, and the remaining terms do not depend on y; namely, W(x,y)=f(x) + b xk yk' for a polynomial f of one variable with positive coefficients and constants b>0, k>=0, k'>=2. Denote by L (resp., H) the lowest (resp., highest) degree among the terms in f.

Then, if at least one of the following holds then the infimum of F(x,y) (defined above) in D is attained at the boundary (possibly at infinity) of D.

  1. k=0 or k'=2,
  2. L >= 4 beta/(k'+2) or H <= 4 beta/(k'+2) (where beta=k'+(k/2)),
  3. otherwise, f'(x0) x01-2 beta >= b k (k'+2)/(k'-2), where x=x0 is the unique positive root of x f''(x)/f'(x) = 4 beta/(k'+2)-1.
Remarks.
  1. Note that F(x,x2) > 1 under the assumptions of Problem 1. Hence, to have F(x,y) > 1 in D, one only has to check boundaries of D other than the parabolla y=x2.
  2. The sufficient conditions in the Proposition are irrelevant of the detailed properties of W in the interior of D. All the conditions are reduced to the properties of functions of (at most) one variable. If F(x,y) > 1 holds at the boundaries of D, then the remark above the Proposition implies that there are no critical point of w in the interior of D.


A proof of Proposition (19981226).
Note first that the assumptions for
W(x,y)=f(x) + b xk yk' (b>0, k>=0, k'>=2)
in Problem 1 implies that L>=3 (i.e., f is a polynomial with terms of degree no less than 3) and k+k'>=3.
Put R(x,z):=F(x,x2z)1/2
=z1/2 (Wx/Wy1/2)(x,x2z) =(bk')-1/2(f'(x) x-beta+1z1-k'/2 +bk xbetaz1+k'/2),
where we put beta=k'+(k/2).
If k=0 or k'=2 then R(x,z) is obviously monotone in z, so that F(x,y)=R(x,y/x2)2 takes its extremum at the boundary z=0 or z=1 of D. Henceforth we may assume k>1 and k'>2.
Put g(x) := x f''(x)/f'(x). If R(x,z) attains its extremum at (x,z), then it must hold that
R,x(x,z)=R,z(x,z)=0,
where R,x denotes the partial derivative of R in x. By explicit calculation, we have, as extremum condition,
g(x) = 4 beta/(k'+2) - 1,
b k zk' = (k'-2) f'(x) x1-2 beta/(k'+2).
If f is a monomial, f(x)=a xm (a>0, m>2), then g(x) = 4 beta/(k'+2) - 1 implies L=H=m=4 beta/(k'+2), in which case the extremum of R is taken on a curve extending to x to infinity, hence the infimum of F is attained at the boundary of D. Henceforth we may assume that f has more than 2 terms, and put
f(x)=aLxL + ... + aHxH (aL aH >0, H>L>=3).
By explicit calculation we find g'(x) >0, x>0 (A note for correlation inequality enthusiasts: Use Griffiths' inequality for ferromagnetic spin system of 1 degree of freedom!). Therefore, L-1=g(0+)< g(x) < g(infinity)=H-1, x>0. In particular, if L >= 4 beta/(k'+2) or H <= 4 beta/(k'+2), then g(x) = 4 beta/(k'+2) - 1 has no solution for positive x, hence infimum of F is not attained in the interior of D. Henceforth we may assume H > 4 beta/(k'+2) > L.
In this case, let x0 be the unique positive root of x f''(x)/f'(x) = g(x)= 4 beta/(k'+2)-1. Then the conditions for an extremum of R imply that an extremum must be of a form (x0,z0), where
b k z0k' = (k'-2) f'(x0) x01-2 beta/(k'+2).
Hence if the right hand side of this equality is no smaller than b k, then the extremum of R cannot be attained in the interior of D. All the cases listed in the Proposition is thus exhausted, and the proof is complete.
generalization of RG for rSAP on dSG; case when potential has no linear term in y Remarks.
  1. One does have to check F(x,y) > 1 at the boundaries, especially for large x. For example, W(x,y)=x3/3 + 4 x2y3 /3 satisfies the assumptions of the Proposition (19981226) because H < =4 beta/(k'+2), but the corresponding w=W(x,y)-(x2+y2)/2 has 3 critical points in D: (1,0), (0.939352,0.283324), (0.734382,0.46355) (the large blobs in the right figure; one of them is the minimum of w, and is supposed to be in the dark area). In fact, F(x,y) = (1+8y3/(3x))2/(4y) -> 1/(4y), x->infinity, so that if y > 1/4, then F(x,y) eventually is less than 1 when x is large, hence y/x2 is not a Lyapnov function in D.
  2. This page (and the Proposition) is an answer to Hiro Ochiai's suggestion:
    Date: Fri, 6 Nov 1998 15:27:56 +0900 (JST)
    From: Hiroyuki Ochiai
    Actually, your counterexample has only two terms.
    Then, we should clarify the meaning of this counterexample.
    ^^^ you ```````
    (The counterexample mentioned in Hiro's e-mail is the example introduced above. I first found the example as an counterexample against a Conjecture which I stated in Oct. 1998.)
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