AMATH 731: Applied Functional Analysis Fall Additional notes on Fréchet derivatives


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1 AMATH 731: Applied Functionl Anlysis Fll 214 Additionl notes on Fréchet derivtives (To ccompny Section 3.1 of the AMATH 731 Course Notes) Let X,Y be normed liner spces. The Fréchet derivtive of n opertor F : X Y is the bounded liner opertor DF() : X Y which stisfies the following reltion, F(+h) F() DF()h lim =. (1) h It is generliztion of the derivtive of function f : R R encountered in firstyer clculus nd the Jcobin of function f : R n R m studied in dvnced clculus. Indeed, for functions f : R R, the connection is cler if we go bck to the definition of f (): We my rewrite this reltion s f f(+h) f() () = lim. (2) h h f(+h) f() f ()h lim =. (3) h h The Fréchet derivtive of f is the sclr f (), which multiplies the sclr R s such, f () is liner opertor in R. For functions F : R n R m, the Fréchet derivtive DF() is the Jcobin of F, liner opertor which is represented by n m n mtrix, s written in Exmple 3.12 of the Course Notes nd reproduced below, DF() = F 1 F x 1 () 1 x n () F m F x 1 () m x n (). (4) Here, the rte of chnge of F : R m R n in the direction h R m is mesured t the point R m. In fct, the term, DF()h = DF()ĥ, (5) in Eq. (1) is, by definition, the directionl derivtive of F t. The Fréchet derivtive, s defined in Eq. (1) extends the bove concepts of the derivtive to opertors in generl normed spces, for exmple, infinitedimensionl function spces. This is of gret importnce to computtionl methods for solving nonliner opertor equtions. We consider few exmples below. In ll cses, it is best to employ the forml definition in Eq. (1). In the nlysis of n opertor F : X Y, the usul procedure is to exmine the difference F(+h) F(). All terms tht re liner in h (nd possibly its derivtives) will comprise the Fréchet derivtive. Higherorder terms in h (nd derivtives) will comprise reminder term, i.e. F(+h) F() = Lh+R(,h), (6) L is liner opertor. (It my be, for exmple, n integrl opertor or differentil opertor, or n expression involving both.) From Eq. (1), it then remins to show tht R(,h) lim =. (7) h 1
2 If this cn be done, then the liner opertor L is the Fréchet derivtive of F, i.e., L = DF() (8) Exmple 1: Let X = Y = C[,b] with norm nd let T : X X be the liner integrl opertor defined by (Tu)(x) = K(x,s) is continuous on [,b] [,b]. We first clculte T(u+h) T(u) for n rbitrry h X: [T(u+h) T(u)](x) = = = K(x, s)u(s) ds, (9) K(x,s)[u(s)+h(s)] ds K(x, s)u(s) ds K(x,s)[u(s)+h(s) u(s)] ds K(x, s)h(s) ds. (1) Note tht the finl term is liner opertor on h, which my not hve been unexpected fter ll, T is liner opertor. But let us go through the formlities. We my rerrnge the bove result to red [ ] 1 b T(u+h) T(u) K(x, s)h(s) ds =. (11) Since this eqution is true for ll h, it follows tht the reltion in Eq. (1) is stisfied. Therefore, the Fréchet derivtive is given by DT(u) = K(x, s)h(s) ds, (12) independent of u, i.e., the bounded liner opertor T itself. This illustrtes Proposition 3.8, p. 46 of the Course Notes: The Fréchet derivtive of bounded liner opertor L is L itself. Exmple 2: As before, let X = Y = C[,b] with norm. Now let T : X X be the nonliner integrl opertor defined by (Tu)(x) = u(x) K(x, s)u(s) ds, (13) K(x,s) is continuous on [,b] [,b]. Once gin, we clculte T(u+h) T(u) for n rbitrry h X: [T(u+h) T(u)](x) = [u(x)+h(x)] = u(x) K(x,s)h(s) ds+h(x) K(x,s)[u(s)+h(s)] ds u(x) K(x, s)u(s) ds K(x, s)u(s) ds + R(u, h)(x), (14) R(u,h)(x) = h(x) K(x, s)h(s) ds. (15) 2
3 Note tht the reminder term R(u,h) is nonliner in h. If / s h, then the first two terms in the lst line of (14) will define the Fréchet derivtive of T. We hve = mx h(x) K(x, s)h(s) ds M2. (16) x [,b] Thus, the Fréchet derivtive of T is given by M = (b ) mx K(x,s). (17) [,b] [,b] [DT(u)]h(x) = u(x) K(x,s)h(s) ds+h(x) K(x, s)u(s)ds. (18) Note tht it is liner opertor on h. It is lso bounded (why?). Exmple 3: Let X = C 1[,1] be the spce of ll C1 functions on [,1] which vnish t the endpoints, with norm [ 1/2 u = [u 2 +(u ) 2 ] dx]. (19) This norm is often clled the energy norm we shll study it lter in this course. Now consider the opertor K : X R defined by K(u) = [u 3 +(u ) 2 ] dx. (2) The opertor K is clled functionl it is relvlued mpping of the spce X. The gol is to compute the Fréchet derivtive of K. After little clcultion, we find tht K(u+h) K(u) = R(u,h) = [3u 2 h+2u h ] dx+r(u,h), (21) [3uh 2 +h 3 +(h ) 2 ] dx. (22) Note tht, once gin, the RHS of Eq. (21) hs been rrnged so tht the first term includes ll terms tht re liner in h, s the reminder, R(u,h), includes ll terms tht re nonliner in h. We suspect tht the first term represents the Fréchet derivtive, but in order to prove this we must show tht / s. This is, however, somewht complicted with the energy norm selected for this problem. In n effort to express in terms of, we try the following: = R(u,h) 3mx u(x) h 2 dx+mx h(x) [,1] [,1] Now note, from the definition of the energy norm in Eq. (19), tht h 2 dx+ (h ) 2 dx. (23) We use this result in Eq. (23): h 2 dx 2, (h ) 2 dx 2. (24) (3 u(x) + h(x) +1) 2. (25) 3
4 A simple rerrngement yields (3 u(x) + h(x) +1). (26) It is now tempting to let nd conclude tht the rtio on the left vnishes in this limit, but there is one compliction: Cn we gurntee tht h(x) is bounded, so tht the middle term on the right hnd side does not blow up? In fct, h(x) must be bounded, since it is C 1 function on [,b], i.e., there exists n M > such tht h(x) M. But for ech h, there is n M wht is necessry is to connect M with. This is mde possible with the following result. Lemm: If h C 1 [,1] nd h() =, then [ 1/2 = mx h(x) 2 (h ) dx] 2. (27) [,1] Proof: If h = on [,1], then the equlity is stisfied. We now consider the cse tht h does not vnish identiclly over [, 1]. From the Fundmentl Theorem of Clculus, x h(s)h (s) ds = 1 2 h(x)2 1 2 h()2 = 1 2 h(x)2. (28) Applying the CuchySchwrtz inequlity to the integrl on the left yields Thus [ 1 x ] 1/2 [ x 1/2 2 h(x)2 h(s) 2 ds (h (s)) ds] 2. (29) [ ] 1/2 [ ] 1/2 h(x) 2 2 h(s) 2 ds (h (s)) 2 ds [ ] 1/2 2mx (h (s)) 2 ds [,1] [ 1/2 = 2 (h (s)) ds] 2. Since this inequlity holds for ll x [,1], it follows tht [ 1/2 mx [,1] h(x)2 = 2 2 (h (s)) ds] 2. (3) Division on both sides by > yields the desired result. From the Lemm nd the second inequlity in (24), it follows tht Using this result in Eq. (26) yields h(x) 2. (31) (3 u(x) +2+1). (32) 4
5 It now follows tht s h. (33) Therefore, the Fréchet derivtive of the nonliner functionl K in Eq. (2) is given by [DK(u)]h = [3u 2 h+2u h ] dx. (34) 5
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