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Since the order of the derivatives off on the right-hand side of (8.84), (8.84 ), and so forth, is one less than the orderon the left-hand side of the equality, we find by induction that f ∈ C (p) (Ix ; Iy ) ifF ∈ C (p) (U, R).Example 1 Let us return to relation (8.80) studied above, which defines a circlein R2 , and verify Proposition 1 on this example.In this caseF (x, y) = x 2 + y 2 − 1,and it is obvious that F ∈ C (∞) (R2 ; R).
Next,Fx (x, y) = 2x,Fy (x, y) = 2y,so that Fy (x, y) = 0 if y = 0. Thus, by Proposition 1, for any point (x0 , y0 ) of thiscircle different from the points (−1, 0) and (1, 0) there is a neighborhood such thatthe arc of the circle contained in that neighborhood √can be written in the form√ y=f (x). Direct computation confirms this, and f (x) = 1 − x 2 or f (x) = − 1 − x 2 .4868 Differential Calculus in Several VariablesNext, by Proposition 1,x0Fx (x0 , y0 )=− .Fy (x0 , y0 )y0f (x0 ) = −(8.87)Direct computation yieldsf (x) =⎧x⎨−√1−x⎩√x1−x 2if f (x) =√1 − x2,√if f (x) = − 1 − x 2 ,,2,which can be written as the single expressionf (x) = −xx=− ,f (x)yand computation with it leads to the same result,f (x0 ) = −x0,y0as computation from formula (8.87) obtained from Proposition 1.It is important to note that formula (8.84) or (8.87) makes it possible to computef (x) without even having an explicit expression for the relation y = f (x), if onlywe know that f (x0 ) = y0 .
The condition y0 = f (x0 ) must be prescribed, however,in order to distinguish the portion of the level curve F (x, y) = 0 that we intend todescribe in the form y = f (x).It is clear from the example of the circle that giving only the coordinate x0 doesnot determine an arc of the circle, and only after fixing y0 have we distinguishedone of the two possible arcs in this case.8.5.3 Transition to the Case of a Relation F (x 1 , . .
. , x m , y) = 0The following proposition is a simple generalization of Proposition 1 to the case ofa relation F (x 1 , . . . , x m , y) = 0.Proposition 2 If a function F : U → R defined in a neighborhood U ⊂ Rm+1 ofthe point (x0 , y0 ) = (x01 , . . . , x0m , y0 ) ∈ Rm+1 is such that10 F ∈ C (p) (U ; R), p ≥ 1,20 F (x0 , y0 ) = F (x01 , . . . , x0m , y0 ) = 0,1 , .
. . , x m , y ) = 0,30 Fy (x0 , y0 ) = Fy (x0,00then there exists an (m + 1)-dimensional interval I = Ixm × Iy1 , whereIxm = x = x 1 , . . . , x m ∈ Rm x i − x0i < α i , i = 1, . . . , m ,8.5 The Implicit Function Theorem487Iy1 = y ∈ R |y − y0 | < β ,which is a neighborhood of the point (x0 , y0 ) contained in U , and a function f ∈C (p) (Ixm ; Iy1 ) such that for any point (x, y) ∈ Ixm × Iy1F x1, . .
. , xm, y = 0 ⇔ y = f x1, . . . , xm ,(8.88)and the partial derivatives of the function y ∈ f (x 1 , . . . , x m ) at the points of Ix canbe computed from the formula*−1 ) *) ∂fFx x, f (x) .(x) = − Fy x, f (x)i∂x(8.89)Proof The proof of the existence of the interval I m+1 = Ixm × Iy1 and the existenceof the function y = f (x) = f (x 1 , . . . , x m ) and its continuity in Ixm is a verbatimrepetition of the corresponding part of the proof of Proposition 1, with only a singlechange, which reduces to the fact that the symbol x must now be interpreted as(x 1 , . .
. , x m ) and α as (α 1 , . . . , α m ).If we now fix all the variables in the functions F (x 1 , . . . , x m , y) and f (x 1 , . . . ,mx ) except x i and y, we have the hypotheses of Proposition 1, where now the role ofx is played by the variable x i . Formula (8.89) follows from this. It is clear from this∂fm 1(1) (I m ; I 1 ). Reasoningformula that ∂xi ∈ C(Ix ; Iy ) (i = 1, .
. . , m), that is, f ∈ Cxyas in the proof of Proposition 1, we establish by induction that f ∈ C (p) (Ixm ; Iy1 )when F ∈ C (p) (U ; R).Example 2 Assume that the function F : G → R is defined in a domain G ⊂Rm and belongs to the class C (1) (G; R); x0 = (x01 , . . . , x0m ) ∈ G and F (x0 ) =F (x01 , . .
. , x0m ) = 0. If x0 is not a critical point of F , then at least one of the par∂Ftial derivatives of F at x0 is nonzero. Suppose, for example, that ∂xm (x0 ) = 0.Then, by Proposition 2, in some neighborhood of x0 the subset of Rm definedby the equation F (x 1 , . . . , x m ) = 0 can be defined as the graph of a functionx m = f (x 1 , . . . , x m−1 ), defined in a neighborhood of the point (x01 , .
. . , x0m−1 ) ∈Rm−1 that is continuously differentiable in this neighborhood and such thatf (x01 , . . . , x0m−1 ) = x0m .Thus, in a neighborhood of a noncritical point x0 of F the equationF x1, . . . , xm = 0defines an (m − 1)-dimensional surface.In particular, in the case of R3 the equationF (x, y, z) = 0defines a two-dimensional surface in a neighborhood of a noncritical point(x0 , y0 , z0 ) satisfying the equation, which, when the condition ∂F∂z (x0 , y0 , z0 ) = 04888 Differential Calculus in Several Variablesholds, can be locally written in the formz = f (x, y).As we know, the equation of the plane tangent to the graph of this function at thepoint (x0 , y0 , z0 ) has the formz − z0 =∂f∂f(x0 , y0 )(x − x0 ) +(x0 , y0 )(y − y0 ).∂x∂yBut by formula (8.89)∂fF (x0 , y0 , z0 )(x0 , y0 ) = − x,∂xFz (x0 , y0 , z0 )Fy (x0 , y0 , z0 )∂f(x0 , y0 ) = − ,∂yFz (x0 , y0 , z0 )and therefore the equation of the tangent plane can be rewritten asFx (x0 , y0 , z0 )(x − x0 ) + Fy (x0 , y0 , z0 )(y − y0 ) + Fz (x0 , y0 , z0 )(z − z0 ) = 0,which is symmetric in the variables x, y, z.Similarly, in the general case we obtain the equationmFx i (x0 ) x i − x0i = 0i=1of the hyperplane in Rm tangent at the point x0 = (x01 , .
. . , x0m ) to the surface givenby the equation F (x 1 , . . . , x m ) = 0 (naturally, under the assumptions that F (x0 ) = 0and that x0 is a noncritical point of F ).It can be seen from these equations that, given the Euclidean structure on Rm ,one can assert that the vector∂F∂F(x0 )grad F (x0 ) =,...,∂x m∂x 1is orthogonal to the r-level surface F (x) = r of the function F at a correspondingpoint x0 ∈ Rm .For example, for the functionF (x, y, z) =x 2 y 2 z2++ ,a 2 b2 c2defined in R3 , the r-level is the empty set if r < 0, a single point if r = 0, and theellipsoidx 2 y 2 z2++=ra 2 b2 c28.5 The Implicit Function Theorem489if r > 0. If (x0 , y0 , z0 ) is a point on this ellipsoid, then by what has been proved, thevector2x0 2y0 2z0,,grad F (x0 , y0 , z0 ) =a 2 b2 c2is orthogonal to this ellipsoid at the point (x0 , y0 , z0 ), and the tangent plane to it atthis point has the equationx0 (x − x0 ) y0 (y − y0 ) z0 (z − z0 )++= 0,a2b2c2which, when we take account of the fact that the point (x0 , y0 , z0 ) lies on the ellipsoid, can be rewritten asx0 x y0 y z0 z+ 2 + 2 = r.a2bc8.5.4 The Implicit Function TheoremWe now turn to the general case of a system of equations⎧ 1 1F x , .
. . , x m , y 1 , . . . y n = 0,⎪⎪⎨...⎪⎪⎩ n 1F x , . . . , x m , y 1 , . . . , y n = 0,(8.90)which we shall solve with respect to y 1 , . . . , y n , that is, find a system of functionalrelations⎧ 1y = f 1 x1, . . . , xm ,⎪⎪⎨..(8.91).⎪⎪ 1⎩ nnmy = f x ,...,x ,locally equivalent to the system (8.90).For the sake of brevity, convenience in writing, and clarity of statement, let usagree that x = (x 1 , . . . , x m ), y = (y 1 , . . . , y n ).
We shall write the left-hand side ofthe system (8.90) as F (x, y), the system of equations (8.90) as F (x, y) = 0, and themapping (8.91) as y = f (x).Ify0 = y01 , . . . , y0n ,x0 = x01 , . . . , x0m ,α = α1, . . . , αm ,β = β 1, . . . , β n ,the notation |x −x0 | < α or |y −y0 | < β will mean that |x i −x0i | < α i (i = 1, . . . , m)jor |y j − y0 | < β j (j = 1, . .
. , n) respectively.490We next set8 Differential Calculus in Several Variables⎛∂f 1∂x l···...∂f 1∂x m···∂f n∂x m···...∂F 1∂x m∂F n∂x l···∂F n∂x m∂y 1···...∂F l∂y n···∂F n∂y n⎜ ..f (x) = ⎜⎝ .∂f n⎛∂x 1∂F 1∂x 1⎜ ..Fx (x, y) = ⎜⎝ .⎛ ∂F 1⎜ ..Fy (x, y) = ⎜⎝ .∂F n∂y l⎞.. ⎟. ⎟⎠ (x),(8.92)⎞.. ⎟. ⎟⎠ (x, y),(8.93)⎞.. ⎟. ⎟⎠ (x, y).(8.94)We remark that the matrix Fy (x, y) is square and hence invertible if and only ifits determinant is nonzero.
In the case n = 1, it reduces to a single element, and inthat case the invertibility of Fy (x, y) is equivalent to the condition that that singleelement is nonzero. As usual, we shall denote the matrix inverse to Fy (x, y) by[Fy (x, y)]−1 .We now state the main result of the present section.Theorem 1 (Implicit function theorem) If the mapping F : U → Rn defined in aneighborhood U of the point (x0 , y0 ) ∈ Rm+n is such that10 F ∈ C (p) (U ; Rn ), p ≥ 1,20 F (x0 , y0 ) = 0,30 Fy (x0 , y0 ) is an invertible matrix,then there exists an (m + n)-dimensional interval I = Ixm × Iyn ⊂ U , whereIyn = y ∈ Rn |y − y0 | < β ,Ixm = x ∈ Rm |x − x0 | < α ,and a mapping f ∈ C (p) (Ixm ; Iyn ) such thatF (x, y) = 0 ⇔ y = f (x),(8.95)for any point (x, y) ∈ Ixm × Iyn and*−1 ) *) Fx x, f (x) .f (x) = − Fy x, f (x)(8.96)Proof The proof of the theorem will rely on Proposition 2 and the elementary properties of determinants.
We shall break it into stages, reasoning by induction.For n = 1, the theorem is the same as Proposition 2 and is therefore true.Suppose the theorem is true for dimension n − 1. We shall show that it is thenvalid for dimension n.8.5 The Implicit Function Theorem491a) By hypothesis 30 , the determinant of the matrix (8.94) is nonzero at the point(x0 , y0 ) ∈ Rm+n and hence in some neighborhood of the point (x0 , y0 ). Consequently at least one element of the last row of this matrix is nonzero. Up to a changenin the notation, we may assume that the element ∂F∂y n is nonzero.b) Then applying Proposition 2 to the relationF n x 1 , . .
. , x m , y 1 , . . . , y n = 0,we find an interval I˜m+n = (I˜xm × I˜yn−1 ) × Iy1 ⊂ U and a function f˜ ∈ C (p) (I˜xm ×I˜yn−1 ; Iy1 ) such that n 1F x , . . . , x m , y 1 , . . . , y n = 0 in I˜m+n ⇔⇔ y n = f˜ x 1 , . . . , x m , y 1 , . . . , y n−1 , 1x , . . . , x m ∈ I˜xm , y 1 , . . . , y n−1 ∈ I˜yn−1 .yn(8.97)c) Substituting the resulting expression y n = f˜(x, y 1 , . . . , y n−1 ) for the variablein the first n − 1 equations of (8.90), we obtain n − 1 relations⎧ 1 1Φ x , . . . , x m , y 1 , . . .
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