AMS 2000 Subject Classification: 49M15, 26A09, 47A60. Key words: convexity, Newton's method for convex monotone operators, approximation of the solution, ...
APPLICATIONS OF NEWTON’S METHOD FOR CONVEX MONOTONE OPERATORS ˘ A, ˘ ALINA OLTEANU JANINA MIHAELA MIHAIL and OCTAV OLTEANU In Section 1 we recall a result proved in [9] and used in [15], concerning the equation P (x) = 0, where P is a convex ”increasing” operator. Then we state a similar result for convex ”decreasing” operators (see also [1] and [14]). In Section 2 we apply some results of [15] to some concrete equations and to a linear differential equation. In Section 3 we consider a particular linear homogeneous differential equation in an infinite dimensional space of operators. AMS 2000 Subject Classification: 49M15, 26A09, 47A60. Key words: convexity, Newton’s method for convex monotone operators, approximation of the solution, linear differential equations. 1. GENERAL-TYPE RESULTS CONCERNING NEWTON’S METHOD FOR OPERATORS Let X be an order-complete vector lattice endowed with a norm || · ||, such that |x| ≤ |y| ⇒ ||x|| ≤ ||y||, x, y ∈ X. Assume also that
o
τ
xn → x ⇒ xn → x, where τ is the norm-topology (see [5] and [6]). Let Y be a normed vector space endowed with a linear order relation defined by a closed convex cone Y+ . Let a, b ∈ X, a < b. Denote [a, b] := {x ∈ X; a ≤ x ≤ b}. 1
Let P : [a, b] → Y be a convex operator such that P ∈ C 1 ([a, b], Y ). In these hypothesis, we have 1.1. Theorem (Theorem 2 from [9] and Theorem 2.2. from [15]). Assume in addition that for any x ∈ [a, b] there exists [P 0 (x)]−1 ∈ L(Y, X), such that [P 0 (x)]−1 ≥ 0 (i.e. [P 0 (x)]−1 (Y+ ) ⊂ X+ ) and a ≤ x ≤ b ⇒ P 0 (a) ≤ P 0 (x) ≤ P 0 (b). If P (a) < 0, P (b) > 0, then there exists a unique root x∗ ∈]a, b[ of the equation P (x) = 0, with x∗ defined by x∗ :=
^
xn = inf xn = lim xn , n
n∈N
n
where (xn )x≥0 is the sequence from Newton’s method x0 := b,
xn+1 := xn − [P 0 (xn )]−1 (P (xn )),
n ∈ N.
The convergence speed for xn → x∗ is controlled by the inequalities ||xk − x∗ || ≤ ||[P 0 (a)]−1 || · ||P (xk )||,
k ∈ N.
For the proof of this theorem see [9], p.26-27. For some of its applications, see [15] and [1]. 1.2. Corollary. Let P : [a, b] ⊂ R → R be a convex C 1 -function such that P 0 (x) > 0 ∀x ∈ [a, b]. If P (a) < 0 and P (b) > 0, then there exists a unique root x∗ ∈]a, b[ of the equation P (x) = 0, with x∗ defined by x∗ = inf xn , n
where x0 := b, xn+1 := xn −
P (xn ) , P 0 (xn )
n∈N
The convergence speed for xk → x∗ is controlled by |xk − x∗ | ≤
|P (xk )| , P 0 (a) 2
k ∈ N.
Note that in the preceding results P is ”increasing”. Now we state a similar result, but this time for convex ”decreasing” operators. 1.3. Theorem. Let X, Y, a, b be as in Theorem 1.1. Let P ∈ C 1 ([a, b], Y ) be a convex operator. Assume that for any x ∈ [a, b], there exists [P 0 (x)]−1 ∈ L(Y, X) such that −[P 0 (x)]−1 ≥ 0 (i.e. −[P 0 (x)]−1 (Y+ ) ⊂ X+ ). Assume also that a ≤ x ≤ b ⇒ [P 0 (x)]−1 ≥ [P 0 (b)]−1 . If P (a) > 0 and P (b) < 0, then there exists a unique root x∗ ∈]a, b[ of the equation P (x) = 0, x∗ being constructed in the following way x=
_
xn = sup xn = lim xn , n→∞ n∈N n∈N
where (xn )n is defined by x0 := a,
xn+1 := xn − [P 0 (xn )]−1 (P (xn )),
n ∈ N.
Moreover, the sequence (xn )n is increasing and the convergence speed for xk → x∗ is controlled by the inequalities ||xk − x∗ || ≤ ||[P 0 (b)]−1 || · ||P (xk )||,
k ∈ N.
1.4. Corollary. Let [a, b] ⊂ R and f ∈ C 1 ([a, b], R) be a strictly decreasing convex function such that P (a) > 0, P (b) < 0. Then the sequence defined by x0 := a,
xn+1 := xn −
(P (xn ) , P 0 (xn )
n ∈ N,
is strictly increasing and converges to the unique root x∗ ∈]a, b[ of the equation P (x) = 0. The convergence speed for xk → x∗ is given by x∗ − xk ≤
P (xk ) , −P 0 (b) 3
k ∈ N.
Theorem 1.3. (and its Corollary 1.4.) will be proved in [1]. For some applications of these general results see [14] and [1]. Note that all results in this section are valid under the assumption o
τ
xn → x ⇒ xn → x, where τ is the norm-topology on X. This assumption is accomplished (in particular) when X is finite dimensional. 2. APPLICATIONS We start by an application of Theorem 2.4 [15] p.79 to a differential equation. Let H be a complex or real Hilbert space. Denote by A(H) the real vector space of all self-adjoint operators acting on H. Let A ∈ A(H) be a fixed element. Let A1 := {U ∈ A(H); AU = U A} (1)
X := {U ∈ A1 ; U V = V U
∀V ∈ A1 }
(see [6], p.303 − 305) X+ := {U ∈ X; hU (h), hi ≥ 0 ∀h ∈ H}.
It is well known that X is an order-complete vector lattice and a commutative algebra of operators (see [6]). We recall the following result. 2.1. Theorem (Theorem 2.4 [15], p. 79). Let H, A, X, X+ be as above. Let B0 , B1 , . . . , Bn ∈ X+ , B0 > 0, Bn > 0 be such that B0
0 and Y (t) ∈]et I, I[
Proof. By Theorem 2.4 [15] U ∈ ]0, I[⊂ X such that
and its
for t < 0.
proof, there exists a unique
P (U ) := Bn U n + . . . + B1 U − B0 = 0 and ||I − U || ≤ ||B1−1 || · ||Bn + . . . + B2 + B1 − B0 ||. Put Y (t) := exp(tU ),
t ∈ R.
Then Y : R → X obviously satisfies the equalities Y (k) (t) = U k exp(tU ),
t ∈ R, k ∈ N.
These relations lead to Bn Y (n) (t) + . . . + B1 Y 0 (t) − B0 Y (t) = (Bn U n + . . . + B1 U − B0 ) exp(tU ) = 0. This (i) and (ii) are proved. On the other hand, since U ∈ ]0, I[, for t > 0 we have Y (t) := exp(tU ) ∈ ] exp(0), exp(tI)[ = ]I, et I[ 5
and for t < 0 we have Y (t) := exp(tU ) ∈ ] exp(tI), exp(0)[ = ]et I, I[ . Thus (iii) is proved too. Note that the hypothesis ”H is finite dimensional” is used to assume that the Newton’s sequence (which is convergent with respect to the order relation on X) converges in the norm topology on X too, so that we may apply Theorem 2.4 [15]. The proof is complete. 2 We next approximate the solution of an algebraic equation.¸ · 3 2.3. Proposition. There exists un unique solution x∗ ∈ , 2 of the 2 equation 2x3 − 4x2 + 1 = 0
(2) and we have (3)
2−
27 ≤ x∗ < 2 152
Proof. We rewrite the equation (2) as x−2 + 2x − 4 = 0, ·
x 6= 0.
¸
3 Let P : , 2 → R, P (x) := x−2 + 2x − 4. Then we have: 2 µ
−2 1 P (x) = 3 + 2 = 2 1 − 3 x x 0
¶
·
·
00
−4
P (x) = 6x
> 0,
¸
3 ∀x ∈ , 2 . 2 ·
Thus P is strictly increasing and strictly convex on µ ¶
P
3 2
µ ¶−2
=
3 2
P (2) =
¸
2 3 = 3 (x3 − 1) > 0 ∀x ∈ , 2 x 2
+2·
3 4 −4= −1 0. 4 4
6
¸
3 , 2 . We also have: 2
3 Hence P satisfies all the requirements of Corollary 1.2 for a = , b = 2, so 2 ¸ · 3 ∗ that there exists an unique x ∈ , 2 such that 2 P (x∗ ) := x∗−2 + 2x∗ − 4 = 0. Moreover, from Corollary 1.2 we infer that |x0 − x∗ | ≤
P (x0 ) µ ¶, 3 0 P 2
where x0 := b = 2. Thus we have got: 1 27 P (2) 1 27 µ ¶ = Ã 4 ! = · = . 2−x ≤ 3 3 8 19 152 2 0 P 2 1− 3 2 3 ∗
The conclusion follows and the proof is complete. 2 2.4. Remark. To approximate the solution of the equation 2x3 −4x2 +1 = 0, we may take P1 (x)¸ := ·2x3 − 4x2 + 1. We have: P10 (x) = 6x2 −¸ 8x = · 3 3 00 2x(3x − 4) > 0, ∀x ∈ , 2 , P1 (x) = 12x − 8 = 4(3x − 2) > 0, ∀x ∈ , 2 . 2 2 · ¸ 3 , 2 . We also have Thus P1 is striclty increasing and strictly convex on 2 µ ¶
P1
3 2
=2·
27 9 27 −4· +1= − 8 < 0. 8 4 4 P1 (2) = 1 > 0.
3 , b = 2) are 2 accomplished. Applying this corollary, we have for k = 0 (x0 = b = 2):
Thus all conditions upon P from Corollary 1.2 (where a =
0 < 2 − x∗ ≤
P1 (2) 1 2 µ ¶ = µ ¶ = , 3 3 3 3 P10 2· 3· −4 2 2 2
which implies 2−
2 ≤ x∗ < 2. 3 7
Comparing to the approximation 2−
27 ≤ x∗ < 2 152
of Proposition 2.3., we find that Proposition 2.3. gives a better approximation of x∗ . That is why we have considered the function P (x) := x−2 + 2x − 4 in the proof of Proposition 2.3, which leads to the approximation (3). We next consider the same equation, but in a space X of operators, defined by (1). 2.5. Theorem. Let A be a self-adjoint operator ¸ acting on the finite · 3 , 2 . Let X = X(A) be dimensional Hilbert space H, such that σ(A) ⊂ 2 the space defined by (1). Then there exists an unique operator U ∈X such that 2U 3 − 4U 2 + I = 0 and
¸
·
27 σ(U ) ⊂ 2 − ,2 152
(σ(V ) is the spectrum of·V ). ¸ 3 Proof. From σ(A) ⊂ , 2 we infer that 2 3 I ≤ A ≤ 2I 2 Let
·
in X.
¸
3 P : I, 2I → X, 2 P (U ) := U −2 + 2U − 4I.
P2 (U ) = U −1 is convex on · ¸ 3 {U ∈ X; σ(U ) ⊂]0, ∞[}, it follows that the operator P is convex on I, 2I . 2 On the other hand, we have Since
one
can
prove that the operator
[P 0 (U )](V ) = (−2U −3 + 2I)V = 2(I − U −3 )V 8
·
∀U ∈
¸
3 I, 2I , ∀V ∈ X, which yields 2
·
1 [P 0 (U )]−1 (V ) = (I − U −3 )−1 V ≥ 0 2
¸
·
¸
3 3 I, 2I , ∀V ∈ X+ (the relation (I − U −3 )−1 V ≥ 0, ∀U ∈ I, 2I , 2 2 µ ¶3 3 2 2 ∀V ∈ X+ is true since U ≥ I ⇒ U −1 ≤ I ⇒ U −3 ≤ I < I ⇒ 2 3 3 I − U −3 > 0 ⇒ (I − U 3 )−1 > 0 and (I − U 3 )−1 V ≥ 0 as a product of two positive permutable operators (X is commutative). On the other hand, we compute ∀U ∈
µ
P
3 I 2
¶
µ ¶2
=
2 3
P (2I) =
3 I + 2 · I − 4I = 2
µ
¶
4 − 1 I < 0, 9
1 1 I + 2 · 2I − 4I = I > 0, 2 2 4
3 so that all conditions of Theorem 1.1 are accomplished, where a := I, 2 ¸ · 3 b := 2I. By Theorem 1.1, there exists an unique U ∈ I, 2I , such that 2 P (U ) := U −2 + 2U − 4I = 0, or, equivalently, 2U 3 − 4U 2 + I = 0. (Since H is supposed to be finite dimensional so is X and hence in X we have o τ xn → x ⇒ xn → x in the norm topology τ , so that we may 1.1). To finish the · apply Theorem ¸ 27 ,2 . proof, we have to prove that σ(U ) ⊂ 2 − 152 To do this, we apply the last inequality of Theorem 1.1, namely ||xk − x∗ || ≤ ||[P 0 (a)]−1 || · ||P (xk )||, 3 to k = 0 (x0 = b = 2I, a = I, 2 °· µ ¶¸ ° ° −1 ° 3 1 ° ° 0 −1 ||[P (a)] || = ° P 0 I °= ° ° 2 2
9
k ∈ N,
°" ° ° µ ¶3 #−1 ° ° ° 2 ° I− °= I ° ° 3 ° °
µ
1 8 = 1− 2 27
¶−1
||I|| =
27 27 = ; 2 · 19 38
P (x0 ) = P (2I) = 2−2 I + 2 · 2I − 4I = 2−2 I). We obtain
° ° ° 27 ° 27 ° ||2I − U || ≤ ° I ° · ||2−2 I|| = . ° 38 152
The above relations lead to ¸
27 σ(ϕ(U )) = σ(2I − U ) ⊂ 0, 152
·
, ¸
·
27 where ϕ(t) := 2 − t, t ∈ R. These yield ϕ(σ(U )) ⊂ 0, , i.e. ϕ(t) := 152 ¸ · 27 27 2 − t ∈ 0, ∀t ∈ σ(U ), or, equivalently 2 − < t < 2 ∀t ∈ σ(U ). 152 152 The conclusion follows and the proof is complete. 2 2.6. Proposition. Let b > 0, c > 0 be such that b ch (1) < c < ch (1) + . 2 Let
µ ¶
P (x) := ch (x) +
b x − c, 2
x ∈ R.
Then there exists a unique root x∗ = x∗ (b, c) ∈]0, 1[ of the equation P (x) = 0, such that
·
¸
2 b 0 < 1 − x (b, c) < ch (1) + − c . b 2 ∗
In particular, we have lim x∗ (b, c) = 1.
b↓0 c↓ch(1)
Proof. We have P 0 (x) = sh(x) + 2b > 0, ∀x > 0, P 00 (x) = ch(x) > 0, ∀x ∈ R. In particular, P is strictly increasing and strictly convex on [0,1]. On the other hand, P (0) = 1 − c < 1 − ch (1) < 0 10
b − c > 0. 2 Thus all conditions of Corollary 1.2 are satisfied (for a = 0, b = 1), so that there exists an unique x∗ ∈]0, 1[ which satisfies P (1) = ch(1) +
µ ¶
P (x∗ ) = ch (x∗ ) + and |x0 − x∗ | ≤
P (x0 ) , P 0 (a)
b x∗ − c = 0, 2
where x0 = b = 1, a = 0.
Thus we have: b · ¸ ch(1) + − c P (1) 2 b 2 1−x ≤ 0 = = ch(1) + − c . b P (0) b 2 2 ∗
The conclusion follows and the proof is complete. 2 The operatorial version of Proposition 2.6. is stated in the following 2.7. Theorem. Let X be the space defined by (1). Let B ∈ X be such that σ(B) ⊂]0, ∞[. Let C ∈ X be such that µ ¶
(I
