非线性泛函分析讲义.pdf_三一文库31doc.com

资源描述

《非线性泛函分析讲义.pdf》由会员分享，可在线阅读，更多相关《非线性泛函分析讲义.pdf（67页珍藏版）》请在三一文库上搜索。

1、5 2013 c 11 ? 18 F 1 ?1 2 3 4 5 6 1? g 2 1?5N?Vg?n 5f(nonlinear operator)5N?5m5?f“u 5m(X?n 0 3 0 ?x U |x x0| 0 ?xn w x0?f(xn) =f 3x0? 4?. / Corollary 2.1.8. ?U gBanach mX?k.f48 fU ?efY K3x0 U ?f(x0) = infxUf(x). 3Ne3?m.8?4. dIOX e?r5: 2.1.9. f : X R1r?2 lim|x|f(x) = +. Corollary 2.1.10. ?Xg?Banachmf

2、X r?efYK3x0 X?f(x0) = infxXf(x) 2.1.2Caratheodory N? ? RnLebesgue8H(x,u)3R?XJu(x)? KH(x,u(x)?N?f : u(x) H(x,u(x) CaratheodoryN ?. ?IuCaratheodoryN?5?K?u(x) H(x,u(x)?U3?mCz fuUk.Y? 2.1.11. ? RnLebesgue8H(x,u)3 R?H vCaratheodory (H1) A?k?x H(x,u) U?Y; (H2) zuH(x,u) x ?Lebesgue . n 2.1.12. ?H : R RvCar

3、atheodory u ?Lebesgue KH(x,u(x)?Lebesgue. yyy duu?3?un, ?un ua. e. u. | (H2 ) ?H(xun(x) ux2d(H1) ,H(x,un(x) H(x,u(x) xA?.l?H(xu(x) ux . / n 2.1.13. ?mes() 0 p1,p2 1 b Lp2(). o Caratheodory N?f : Lp1() Lp2() Y3. yyy Iyun Lp1 u, Kf(un) Lp2 f(u).k|(2.1.1)?u Lp1()f(u) Lp2(). ?un Lp1 u , Kunuu|n2.1.13? f

5、M 0,?|u|p1 0 Ub?3|h| = 1? 0 ?|t| 0dDf3x0:Y3 0 ?x U,|x x0| 0 ?u W 1,p T k |u| C|u|W1,p T (2.2.4) ?b? T 0 u(t)dt = 0 u L(0,T) K |u| C| u|L(2.2.5) p |u|Lu3L?. yyy eu W 1,p T |n3 (0,T)?u() = 1 T ?T 0 u(t)dt,u |u(t)| = “ ? ? ? ? ? u() + ? t u(s)ds ? ? ? ? ? |u()| + ? t 0 | u(s)|ds # 1 T ? T 0 |u(t)|dt +

6、T 1/q ? T 0 | u(s)|pds !1/p 1 T T 1/q ? T 0 |u(t)|pdt !1/p + T 1/q ? T 0 | u(s)|pds !1/p c ? T 0 |u(t)|pdt + ? T 0 | u(s)|pds !1/p ,?d|u(t)| = | ?t u(s)ds|?(./ 2.2.13. ?L : 0,T Rn Rnv: 1) A?k?tL(t,x,y)u(x,y)Y; 2) z(x,y),L(t,x,y)utLebesgue. b?3a C(R+,R+),b L(0,T,R+)9c Lq(0,T,R+),1 ?f). Ly(t,u(t), u(t

7、) = Lx(t,u(t), u(t)a. e. in 0,T u(0) u(T) = Ly(0,u(0), u(0) Ly(t,u(t), u(t) = 0 (2.2.8) AO?Ly(t,x,y) y .Cu(0) u(T) = u(0) u(T) = 0. yyy f(u+h)f(u) = ?T 0 L(t,u(t)+h(t), u(t)+h(t)L(t,u(t), u(t) dt = ?T 0 (t,)dt (t,) = L(t,u(t) + h(t), u(t) + h(t) L(t,u(t), u(t) = L(t,u(t) + sh(t), u(t) + sh(t) ? ? ?

8、? ? 1 0 = ? 1 0 1 dL(t,u(t) + sh(t), u(t) + sh(t) ds ds = ? 1 0 h Lx(t,u(t) + sh(t), u(t) + sh(t)h(t) +Ly(t,u(t) + sh(t), u(t) + sh(t)h(t) i ds |(t,)| ? 1 0 ? ? ?Lx(t,u(t) + sh(t), u(t) + sh(t)h(t) +Ly(t,u(t) + sh(t), u(t) + sh(t)h(t) ? ? ?ds ? 1 0 ? |Lx ? t,u(t) + sh(t), u(t) + sh(t) ? | |h(t)| +|L

9、y ? t,u(t) + sh(t), u(t) + sh(t) ? | |h(t)| ? ds ? 1 0 a(|u(t) + sh(t)|) ? b(t) + | u(t) + sh(t)|p ? |h(t)|ds + ? 1 0 a(|u(t) + sh(t)|) ? c(t) + | u(t) + sh(t)|p1 ? |h(t)|ds dn2.2.12?|h| c1|h|W1,p T , Pa0= max(,t)0,00,T|a(|u(t)+sh(t)|)|-(t) = a0 h c1|h|W1,p T ? b(t) + | u(t) + sh(t)|p ? + ? c(t) + |

10、 u(t) + sh(t)|p1 ? |h(t)| i ,K L1(0,T,Rn) (t,) (t)dLebesguen? Df(u)h = lim0 f(u + h) f(u) 1?5N?Vg?n14 = ? T 0 lim0 L(t,u(t) + h(t), u(t) + h(t) L(t,u(t), u(t) dt = ? T 0 h Lx(t,u(t), u(t)h(t) + Ly(t,u(t), u(t)h(t) i dt d |Df(u)h| = ? ? ? ? ? ? T 0 h Lx(t,u(t), u(t)h(t) + Ly(t,u(t), u(t)h(t) i dt ? ?

11、 ? ? ? ? T 0 ? |Lx(t,u(t), u(t)| |h(t)| + |Ly(t,u(t), u(t)| |h(t)| ? dt |h| ? T 0 a(u(t)(b(t) + | u(t)|p)dt + ? T 0 a(u(t) ?c(t) + | u(t)|p1? |h(t)|dt c|h|W1,p T Gateaux?f3Df(u) ? W 1,p T ? 2d(2.2.6)!(2.2.7)9n2.1.14?Df() : W 1. p T L1Lq,Df(u) = (Lx(,u, u),Ly(,u, u) YqL1 Lq, (W 1,p T )Y10l?Df() : W 1

12、,p T (W 1,p T )Y“l?f(u) Frechet dFrechet?f(Lx(,u, u),Ly(,u, u) L1 Lq?W 1,p T ?“ ?d f0(u)h = ? T 0 h Lx(t,u(t), u(t)h(t) + Ly(t,u(t), u(t)h(t) i dt = 0,h W 1,p T ? ? T 0 Lx(t,u(t), u(t)h(t) + Ly(t,u(t), u(t)h(t)|T 0 ? T 0 ? d dtLy(t,u(t), u(t) ? h(t) ? dt = 0 ? T 0 (Lx(t,u(t), u(t),h(t) ? T 0 ? d dtL

13、y(t,u(t), u(t),h(t) ? dt +Ly(T,u(T), u(T)h(T) Ly(0,u(0), u(0)h(0) = 0 k?h(0) = h(T) = 0K? ? T 0 Lx(t,u(t), u(t)h(t) ? T 0 ? d dtLy(t,u(t), u(t) ? h(t)dt = 0 10(f,g) L1 Lq,(f,g),u)= ?T 0 f(t)u(t)dt + ?T 0 g(t) u(t)dt|f|L1|u|+ |g|Lq| | u|Lp c(|f|L1+ |g|Lq|)|u|W1,p T ,l?(f,g) (W1,p T ),|(f,g)|(W1,p T )

14、 c(|f|L1+ |g|Lq|) = c|(f,g)|L1Lq. 1?5N?Vg?n15 ? T 0 ? Lx(t,u(t), u(t) d dtLy(t,u(t), u(t) ? h(t)dt = 0 l? d dtLy(t,u(t), u(t) = Lx(t,u(t), u(t) Ly(T,u(T), u(T)h(T) Ly(0,u(0), u(0)h(0) = 0(2.2.9) 2?h(T) = h(0) = c Rn (Ly(T,u(T), u(T) Ly(0,u(0), u(0)c = 0 u Ly(T,u(T), u(T) Ly(0,u(0), u(0) = 0 AO/23(2.

15、2.9)-h(t) = u(t)K? (Ly(T,u(T), u(T),u(T) u(0) = 0 u(T) u(0) = Ly(T,u(T), u(T) Ly(0,u(0), u(0) = 0 uDf(u)h = 0?du d dtLy(t,u(t), u(t) = Lx(t,u(t), u(t) u(T) u(0) = Ly(T,u(T), u(T) Ly(0,u(0), u(0) = 0 (2.2.10) Remark 2.2.14. e f(u) = infvW1, T f(v)(2.2.11) KDf(u)h = 0,h W 1,p T ,l?Df(u) = 0O?(2.2.9) C

16、 ?g. / 2.2.2 ? 2.2.15. ?X,Y,ZBanachmf : D(f) X Y Z.?y-g(x) = f(x,y),egkF-?G-? Kf u1Cx? F-?G-?fx(x,y) = g0(x). fy(x,y)?aq“ 1?5N?Vg?n16 n 2.2.16. (a)ef3(x,y):F-K F-?3 f0(x,y)(h,k) = fx(x,y)h + fy(x,y)k(2.2.12) (b) L5ef F-?fx,fy3(x,y)?S33(x,y)YKf?F-?3 (2.2.12). (c) f3(x,y)?SYF-?=?k F-?3(x,y)?SY“ yyy (a

17、) f(x+h,y +k) = f(x,y)+f0(x,y)(h,k)+o(|(h,k)|),p|(h,k)| = |h|+|k| .f(x + h,y) = f(x,y) + fx(x,y)h + o(|h|) = f(x,y) + f0(x,y)(h,0) + o(|h|), f(x. y + k) = f(x,y) + fy(x,y)k + o(|k|) = f(x,y) + f0(x,y)(0,y) + o(|k|), f0(x,y)(h,0) = fx(x,y)h,f0(x,y)(0,y) = fy(x,y)k, l?f0(x,y)(h,k) = f0(x,y)(h,0)+f0(x,

18、y)(0,k) = fx(x,y)h+ fy(x,y)k. (b) |f(x + h,y + k) f(x,y) fx(x,y)h fy(x,y)k| |f(x + h,y + k) f(x,y + k) fx(x,y + k)h| +|fx(x,y + k)h fx(x,y)h| +|f(x,y + k) f(x,y) fy(x,y)k| = | ? 1 0 fx(x + th,y + k)hdt ? 1 0 fx(x,y + k)hdt| +|fx(x,y + k)h fx(x,y)h| +| ? 1 0 fy(x,y + tk)k ? 1 0 fy(x,y)kdt| sup(0. 1)|

19、fx(x + h,y + k) fx(x,y + k)| |h| +|fx(x,y + k) fx(x,y)| |h| +sup(0,1)|fy(x,y + k) fy(x,y)| |y| = r(h,k)(|h| + |k|) (c) H.W./ 1?5N?Vg?n17 2.2.3p?Taylor ?f : U Y F?Kzx U,f0(x) L(X,Y ).d?ff0 lU?L(X,Y )?5N?.?XJf0F?Kf00(x) L(X,L(X,Y ). X de?/k?ff(k)lU? L(X,L(X),L(X,Y ),)?f. ?ymL(X,L(X,L(X,Y. ,), 0 ? ?(x1

20、,x2, ,xn) X|A(x1,x2, ,xn)| Mn i=1|xi|i. 4L(X1,.,Xn;Y )LlX?Y ?n-k.5f?NzA L(X1,.,Xn;Y ) A? |A| = sup|xi|1,i=1,n|A(x1, ,xn)|Y KL(X1,.,Xn;Y )5Dm?eY ?KL(X1,.,Xn;Y )? ?. n 2.2.18. L(X1,.,Xn;Y )?uL(X1,L(X2, ,L(Xn,Y ) . yyy H.W. 2.2.19. ?f : U Y F?.XJ?N?f0= f(1): U Y 3x0?F Kf3x0?2?F.Pf3x0?F?ff(2)(x0),f00(x0

21、)d2f(x0). /ef(k): U L(X,X, ,X;Y )3x0?F-? Kf3x0?k + 1?F ?.f(k)3x0?F?ff3x0?k1?fPf(k+1)(x0)dk+1f(x0). X Jf(k)3UY? Kf3U ?Y Pf Ck(U,Y ). |5K8BNye?n. n 2.2.20. ?f Ck(U,Y ) ,x0 U K k t1tk f(x0+ k X i=0 tihi) = f(k)(x0)(h1, ,hk),h1, ,hk X 1?5N?Vg?n18 2.2.21. fA L(X ,X;Y )n?XJ (1,2,n)? = (1, ,n)kA(x1, ,xn) =

22、 A(x1, ,xn).f?NPL(n) s (X,Y ) . n 2.2.22. ?f Ck(U,Y ) Kf(k)(x) L(k) s (X,Y ). yyy H.W. n 2.2.23. (Taylor) ?L = x0+ th|0 t 1 U,q?f Cn(U,Y ) . K f(x0+ h) = n1 X k=0 1 k!f (k)(x0)h(k) + Rn(x0,h) h(k)= (h, ,h)k?|“ Rn(x0,h) = ? 1 0 (1 t)n1 (n 1)! f(n)(x0+ th)h(n)dt = 1 n!f (n)(x0)h(n) + 0(|h|n) yyy y Y ,

23、-(t) = y(f(x0 + th),K0(t) = y(f0(x0+ th)h),00(t) = y(f00(x0+ th)h(2), , (n)(t) = y(f(n)(x0+ th)h(n), (n) C(0,1), ?Taylor? (1) = (0) + 0(0) + 1 2! 00(0) + + 1 (n 1)! (n1)(0) + ? 1 0 (1 t)n (n 1)! (n)(t)dt = y(f(x0+ h) = y n1 X k=0 1 k!f (k)(x0)h(k) + Rn(x0,h) ! dy?5?f(x0+ h) = Pn1 k=0 1 k!f (k)(x0)h(

24、k) + Rn(x0,h)“/ 2.2.24. fKu(x) = ? k(x,y,u(y)dy, R nk.48k(x,y,u),ku(x,y,u) kuu(x,y,u)3 RYKK : C() C()k?YFrechet? K0(u)h(x) = ? ku(x,y,u(y)h(y)dy K00(u)h1h2(x) = ? kuu(x,y,u(y)h1(y)h2(y)dy yyy1-Bh(x) = ? ku(x,y,u(y)h(y)dy, K|Bh| maxx ? |ku(x,y,u(y)|dy|h|, B L(C().qd K(u+h)(x)K(u)(x)Bh(x) = ? (k(x,y,u

27、 ?B C()k.=3c 0?| = maxx|(y)| c, B,P M = max(x,y,u)c,c|k(x,y,u)|, K|K()(x)| ? |k(x,y,(y)|dy Mmes, B,l ?KBk.“qdk : c,c RY? 0, 3 0,?|x1x2| . ?H : R RvCaratheodory v |h(x,u)| a + b|u|p,(x,u) R(2.3.7) a,b?.“CaratheodoryN?f : H1 0() (H 1 0() ,?1 p n+2 n2Y ?1 p 0 ? |v|2n n2 M|v|H1 0(),k |f(uk) f(u)| = sup|

29、anach:n“ d|F(x,) F(y,)| = |x y (f0 x(x0,0) 1 (f(x,) f(y,)| sup0 0,z B(0,),T(x) = F(x,) 3M = BX(x0,r)Banach:n=“ eyx()?Y5“ |x() x()| = |F(x(),) F(x(),)| |F(x(),) F(x(),)| + |F(x(),) F(x(),)| k|x() x()| + |F(x(),) F(x(),)| l?|x() x()| 1 1k|F(x(),) F(x(),)|, duF(x,)Y?x()Y“/ 1?5N?Vg?n23 nnn4.2 ?f Cm(U,Y

30、) ?N?f0 x(x0,0) : X Y ?Kf. K3r, 0 9CmN? x : B(0,) BX(x0,r)x(0) = x0?f(x(),) = 0, B(0,). yyy m = 1,df Cm(U,Y ) ?N?f0 x(x0,0) : X Y ?Kf,?r, f0 x(x(),)?K? 0 = f(x( + th), + th) f(x(),) = f0 x(x(),)(x( + th) x() + f 0 (x(),)th + o(t) l? x( + th) x() = (f0 x(x(),) 1 f0 (x(),)th + o(t) x()GateauxGateaux?f

31、(f0 x(x(),) 1 f0 (x(),)Y?l?x()Frechet Frechet?f(f0 x(x(),) 1 f0 (x(),)Y?ux C 1(B(0,),X). m = 2,x0()h = (f0 x(x(),) 1 f0 (x(),)h, =f 0 x(x(),)x 0()h = f0 (x(),)h,d 0 = f0 x(x(+tk),+tk)x 0(+tk)hf0 (x(+tk),+tk)hf 0 x(x(),)x 0()hf0 (x(),)h = f0 x(x( + tk), + tk) f 0 x(x(),)x 0( + tk)h + f0 x(x(),)x 0( +

32、 tk)h x0()h f0 (x( + tk), + tk)h f 0 (x(),)h 0 = f00 xx(x(),)x 0()tk + f00 x(x(),)tk + o(t)(x 0()h + O(t)+f0 x(x(),)x 0( + tk)h x0()h f00 x(x(),)x 0()tk + f00 (x(),)tkh + o(t) x0( + tk)h x0()h = (f0 x(x(),) 1 f00 xx(x(),)x 0()tk f00 (x(),)tkh + o(t) x0()GateauxGateaux?f(f0 x(x(),) 1 f00 xx(x(),)x 0(

33、) f00 (x(),) Y?,l?x0(x)FrechetFrechet?f(f0 x(x(),) 1 f00 xx(x(),)x 0() f00 (x(),) Y?ux C2(B(0,),X). ?myaq“ 2.4.2n92 3nXJ? = Y ,f(x,y) = f(x) y , f(x0) = y0 “ ?e ?n. nnn4.3 ?UX?m8x0 U,f Cm(U. Y ), m 1.ef0(x0) : X Y ?Kf K3x0?V y0?W,?N?f : V WCm? =f : V W? f,f1km?Y?N?. 1?5N?Vg?n24 ny“?:.kNC/2.p=u ?N?f0

34、(x0) 4?/. 4.1 ?A : X Y k.5f. PKerA = x X,Ax = 0,R(A) = Ax|x X.A? XJ3X?4fmX1?X = X1 KerA. A ?XJ 3Y ?4fmY2?Y = R(A) Y2.?AQ?q?AV?. nnn4.4 ?f Cm(X,Y ),m 1,x0 X,y0= f(x0). 2 ?N?f0(x0) : X Y V? ?. PX2= Kerf0(x0), Y1= R(f0(x0)X = X1X2,Y = Y1Y2. N? : XY2 Y X2 (x,y2) = (y2+ f(x),x2) x = x0+ x1+ x2,x1 X1,x2 X

35、2 y = y0+ y1+ y2,y1 Y1,y2 Y2 K3(x0,0)?V (x0,0) = (y0,0)?W ? : V WCm?. yyy dn4.3Iy0(x0,0) : X Y2 Y X2_1?“ k0(x0,0) = (IY2+ f0(x0),IX2)?0(x0,0)(x,y2) = (y2+ f0(x0)(x1+ x2),x2) = (0,0), Kx2= 0,y2+ f0(x0)x1= 0,dY= Y1 Y2, y2= 0,f0(x0)x1= 0,df0(x0)3X1?x1= 0, l?(x,y2) = 0. dBanach_fnI2y0(x0,0) ?=“(y,x2) Y

36、X2, y Y = Y1 Y2, y = y1+ y2,y1 Y1= R(f0(x0),y2 Y2, df0(x0) : X1 Y1V? 3x1 X1,?f0(x0)x1= y1,-x = x1+ x2 X,K0(x0,0)(x,y2) = (y,x2),l?0(x0,0) ?“/ 5554.1 XJ4P : Y X2: Y P(y,x2) = y LKfi : X X Y2,i(x) = (x,0)?Lif Kn4.4 ?vf(x) = P(ix). 4.1 ?f Cm(X,Y ),m 1x0 X,y0= f(x0)2 ?N?f0(x0)? Kerf0(x0) = 0. Y = R(f0(x

38、5.2)?). yyy (2.5.2)?du x(t) = x0+ ? t 0 (s,x(s)ds(2.5.4) PTx(t) = x0+ ?t 0 (s,x(s)ds,KK=yT k:“d|(t,x) (t,x0)| K|x x0| Kr,?maxt0,a|(t,x0)| N,-sup(s,x)0,aB(x0,r)|(s,x)| = N + Kr = N1 0? |(t,x)| a| + b|x|,(t,x) 0,) X(2.5.5) KK(2.5.2)?)x(t)?43m0,). yyy dn5.1?(2.5.2)?)30,3)“?x(t)?43m0,T)e yT = . ?T max0,m nKf?.:8A?”Sf= f(A)3Nk“ i?nkN-?A“X?N%?:n3Rn?/?5 ndi?n“ 1.1 ? Rnmf Ck(,Rm),k 1,x f?K:XJ5ff0(x) : Rn Rm?; xf?.:.y Rmf?. XJ3f?. :x ?f(x) = y; yf?K. ?m = 1 =xf : Rn R?.:?duf0(x) = 0“d 4:?. :“? 4:? kO?.:Q:?“ ?n li,

展开阅读全文