A remark about polynomials with specified local minima and no other critical points
aa r X i v : . [ m a t h . D S ] F e b A remark about polynomials with specified local minimaand no other critical points
Eduardo D. Sontag
This fact must surely be well-known, but it seems worth giving a simple and quite explicit proof:
Proposition 1.1
Take any finite subset X of R n , n ≥ . Then, there is a polynomial function P : R n → R whichhas local minima on the set X , and has no other critical points.A weaker version, constructing one particular subset X of R with the stated property, was shown in a beautifullittle note by an undergraduate, Ian Robertson, as part of an REU conducted by Alan Durfee [1]; references to thiswork, and the context from degree theory, can be found in [2]. Our construction generalizes Robertson’s.We were interested in this question because of the following consequence, obtained immediately by consideringthe gradient vector field g ( x ) = −∇ P ( x ) : Corollary 1.2
Take any finite subset X of R n , n ≥ . Then, there is a polynomial vector field g : R n → R n which has asymptotically stable equilibria on the set X , and has no other equilibria.Suppose that g is as in Corollary 1.2 and that X has more than one element. Let O x , x ∈ X , be the domains ofattraction of the points in X . The union of the sets O x cannot equal all of R n , since these are disjoint open sets;pick any ξ ∈ R n \ S x O x . The omega-limit set Ω + ( ξ ) cannot intersect X (since points of X are asymptoticallystable). Thus there are points that do not converge to any equilibria. (Alternatively, one could arrive at the sameconclusion appealing to topological degree arguments.) Thus, the following is also of interest (and much easierto prove). Proposition 1.3
Take any finite subset X of R n , n ≥ . Then, there are a finite subset X ′ ⊂ R n and a polynomialvector field g : R n → R n which has asymptotically stable equilibria on the set X , saddles on the set X ′ , andno other equilibria, and moreover: (1) every solution of ˙ x = g ( x ) converges to X S X ′ and (2) except for ameasure-zero set of initial conditions, every solution converges to an equilibrium in X . We prove Proposition 1.1 by first treating the case of a set X ⊂ R of the special form X × { } , and then using acoordinate change to reduce the general case to this one. The proof of Proposition 1.3, in contrast, only requiresthe coordinate change, plus a trivial construction. Lemma 2.1
Let α : R → R be a C function whose zeroes are all simple; that is, on the set X := { x | α ( x ) = 0 }
1t holds that α ′ ( x ) = 0 . Introduce the following function f : R → R : f ( x, y ) := (cid:16) α ( x ) − [ α ( x ) − α ′ ( x )] y (cid:17) − Z α ( x ) [ α ( x ) − α ′ ( x )] dx (where the last term denotes an arbitrary anti-derivative). Consider the set of critical points of f , C ( f ) := { ( x, y ) | f x ( x, y ) = f y ( x, y ) = 0 } . Then:1. C ( f ) = X × { } .2. At each ( x, y ) ∈ C ( f ) , the Hessian of f is positive definite.As a consequence, f has local minima at the points in X × { } , and no other critical points. Proof . For convenience, we introduce β ( x ) := α ( x ) − α ′ ( x ) , so that f ( x, y ) := (cid:0) α ( x ) − β ( x ) y (cid:1) − Z α ( x ) β ( x ) dx . We have: f x ( x, y ) = 2 (cid:0) α ( x ) − β ( x ) y (cid:1) ( α ′ ( x ) − β ( x ) β ′ ( x ) y ) − α ( x ) β ( x ) f y ( x, y ) = − (cid:0) α ( x ) − β ( x ) y (cid:1) β ( x ) . Clearly, α ( x ) = 0 and y = 0 imply f x ( x, y ) = f y ( x, y ) = 0 , so we must only prove the converse. Pick any ( x, y ) ∈ C ( f ) .From f y ( x, y ) = 0 , we have that one of these must hold: β ( x ) = 0 , (1) β ( x ) = 0 and α ( x ) − β ( x ) y = 0 . (2)If (1) holds, then f x ( x, y ) = 2 α ( x ) α ′ ( x ) implies that either α ( x ) = 0 or α ′ ( x ) = 0 . On the other hand,the assumption of simple zeroes, “ α ( x ) = 0 ⇒ α ′ ( x ) = 0 ” can also be written in contrapositive form as“ α ′ ( x ) = 0 ⇒ α ( x ) = 0 ,” from which we have: α ( x ) = 0 ⇒ β ( x ) = − α ′ ( x ) = 0 α ′ ( x ) = 0 ⇒ β ( x ) = α ( x ) = 0 . This rules out both α ( x ) = 0 and α ′ ( x ) = 0 when β ( x ) = 0 , and thus case (1) cannot hold.So case (2) holds, which means that y = α ( x ) /β ( x ) . On the other hand, when α ( x ) − β ( x ) y = 0 , we have that f x ( x, y ) = − α ( x ) β ( x ) , and since β ( x ) = 0 , it follows that α ( x ) = 0 , from which it also follows that y = 0 .We conclude that ( x, y ) ∈ X × { } , as desired.To prove positive definiteness of the Hessian on C ( f ) , we must show that on the set C ( f ) , both f yy ( x, y ) > and ∆( x, y ) > , where ∆ = f xx f yy − f xy . In general: f xx = 2 ( α ′ ( x ) − β ( x ) β ′ ( x ) y ) + 2 (cid:0) α ( x ) − β ( x ) y (cid:1) (cid:0) α ′′ ( x ) − β ( x ) β ′′ ( x ) + ( β ′ ( x )) ] y (cid:1) − α ′ ( x ) β ( x ) − α ( x ) β ′ ( x ) f yy = 2 β ( x ) f xy = f yx = − α ′ ( x ) − β ( x ) β ′ ( x ) y ) β ( x ) − (cid:0) α ( x ) − β ( x ) y (cid:1) β ( x ) β ′ ( x )
2o in particular, on the set C ( f ) , since α ( x ) = 0 and y = 0 : f xx = 2 ( α ′ ( x )) − α ′ ( x ) β ( x ) f yy = 2 β ( x ) f xy = f yx = − α ′ ( x )) β ( x ) . Since on this set β = − α ′ ( x ) : f xx = 3 [ α ′ ( x )] f yy = 2[ α ′ ( x )] f xy = f yx = − α ′ ( x )] and thus ∆( x, y ) = 2[ α ′ ( x )] . As α ′ ( x ) = 0 on the set C ( f ) , it follows that both f yy > and ∆ > , which completes the proof. Lemma 2.2
Let X be a finite subset of R n , n ≥ . Then, there exists a polynomial map F : R n → R n such that: F ( X ) ⊆ R × { } n − , (3) F has a polynomial inverse . (4) Proof . For each pair of points ξ = η in X , let V ξη = { p ∈ R n | p T ( ξ − η ) = 0 } . Pick a point p not belonging tothe union of the finitely many hyperplanes V ξη . Choose any invertible mapping T which has p T as its first row,and consider the change of variables z = T x . Let X = { x (1) , . . . , x ( k ) } and Z := T X = { z (1) , . . . , z ( k ) } , where z ( i ) = T x ( i ) . Since x p T x is one to one on the set X , the first coordinates of the z ( i ) ’s are all distinct, that is z ( i ) = ( z ( i )1 , . . . , z ( i ) n ) and z ( i )1 = z ( j )1 for each i = j . For each j = 2 , . . . , n , let p j be the Lagrange interpolation polynomial that gives: p j ( z ( i )1 ) = z ( i ) j , i = 1 , . . . , k and define Π : R n → R n : z z ... z n z z − p ( z ) ... z n − p n ( z ) . This is invertible (with a polynomial inverse obtained by using instead “ z j + p j ( z ) ” for each j > ). Moreover,by definition, on the set Z we have that the j th coordinate of P ( z ) , P ( z ( i ) ) j = z ( i ) j − p j ( z ( i )1 ) = 0 , j > . Inother words, P maps into R × { } n − . The proof is completed by picking the composition F = Π ◦ T . Let X be a finite subset of R n , n ≥ , pick F as in Lemma 2.2, and let X be such that F ( X ) = X × { } n − .We will construct a polynomial Q : R n → R whose only critical points are on the set X × { } n − , and theHessian is positive definite there. Then P = Q ◦ F will be as desired, because diffeomorphisms preserve criticalpoints and their signature. (To be explicit: as ∇ P ( x ) = ∇ Q ( F ( x )) · JF ( x ) and the Jacobian JF is everywhere3onsingular, critical points map to critical points. Furthermore, at a critical point, the Hessian H transforms as J T HJ , so positive definiteness is preserved.)Let α ( x ) := Q ki =1 ( x − a i ) , where X = { a , . . . , a k } . This α is as in Lemma 2.1; let f be as there. We define Q ( x , . . . , x n ) := f ( x , x ) + 12 X i> x i . From ∂Q/∂x i = 0 for i > , the critical points of Q have x i = 0 for all i > , so C ( Q ) = C ( f ) × { } n − = X × { } n − . Moreover, at these points, the Hessian of Q is obtained by appending an identity matrix to theHessian of f , and thus it is also positive definite, as required. We use the same change of variables, so that, up to a diffeomorphism, we can assume without loss of generalitythat X = X × { } n − . Let γ ( x ) := − k Y i =1 ( x − a i ) k − Y i =1 ( x − b i ) where X = { a < . . . < a k } and the b i ∈ ( a i , a i +1 ) are arbitrary. The scalar differential equation ˙ x = γ ( x ) hasstable equilibria at X , unstable at X ′ = { b , . . . , b k − } , and no other equilibria. We then define g ( x ) = ( γ ( x ) , − x , . . . , − x n ) . This satisfies the conclusions of the Lemma, with X = X × { } n − and X ′ = X ′ × { } n − . References [1] Ian Robertson. A polynomial with n f ( x, y ) . Topology37