Let us consider following=20 problem
=20 z-t^2=3D0    where     1<=3D t=20 <=3D2.

I assume = that
x=3D(x1,x2)=3D(z,t),
(z0,t0)=3D(z(1),1)=3D(1,1),
(z1,t1)=3D(z(2),2)=3D(4,2),
X =3D[X1, X2] =3D [1, 4] x [1, = 2];

I introduce following systems = of equations=20 and functions
(1-)    =20 z-t*t=3D0,  z-1=3D0,    (i.e. G1(-)(x)=3D[x1-x2^2, = x1-1] =20 )
(1+)    = z-t*t=3D0, =20 z-4=3D0,    (i.e. G1(+)(x)=3D[x1-x2^2, x1-4] = )

i.e. if G1(-)(x)=3D0 then = x1-x2^2=3D0 and=20 x1-1=3D0.

Interval extensions of = matrices of Jacobi.
DG1(-)(X)  =3D[[1,=20 -2*X2], [1, 0]] =3D [[1, [-4, -2]],    [1,=20 0]]
DG1(+)(X) =3D[[1,=20 -2*X2], [1, 0]] =3D [[1, [-4, -2]],    [1,=20 0]]

det(DG1(-)(X))=3Ddet(DG1(+)(X))=3D2*X2=3D[-4,=20 -2];

We see that both matrices are=20 regular.

Conclusions
1) Because all matrices of Jacobi are = regular
then the following = equations

G1(-)(x)=3D0,  G1(+)(x)=3D0

have unique solutions (z0,t0) or (z1,t1) in the box=20 X.

2) Number z=3Dz(t) is a unique solution
of equation F(z,t)=3D0.
(this is obvious in this example)

3) Function z=3Dz(t) is continuous for all t between = 1 and=20 2.
(this is also obvious in this = example)

From (1), (2) and (3) follow

z0=3Dmin{ z(t): 1<=3D t <=3D2 }
z1=3Dmax{ z(t): 1<=3D t <=3D2 = }

i.e. Function z=3Dz(t) satisfy the following=20 inequality
=20 z0<z(t)<z(t)
for all t between 1 and 2 and
=20 z0=3Dz(1)        = z1=3Dz(2).

If exist such t* (1<t*<2) that z(t*)<z0 (or = z(t)>z1) then
equation G1(-)=3D0 (or G1(+)=3D0) has more than one=20 solutions
and matrix DG1(-)(X) (or DG1(+)(X)) cannot be = regular=20
over the box X.

Andrzej=20 Pownuk

P.S.
I am very sorry for all the trouble this has caused=20 you.

-------------------------------------------------------
Chair= of=20 Theoretical Mechanics
Faculty of Civil Engineering
Silesian = Technical=20 University
ul. Krzywoustego 7
44-100 Gliwice, Poland
tel/fax = 0048 032=20 371542
E-mail: pownuk [at] zeus [dot] polsl.gliwice.pl=
URL:=20 http://zeus.polsl.gliwice.= pl/~pownuk/
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