Giáo trình vi phân tích hàm nhiều biến nguyễn cao đạt

BO GIAO DUO VA DAO TAO TRUONG DAI HOC DAN LAP CUU LONG —oOo— NGUYEN CAO DAT - TRAN NGOC HAN VO PHUfdC HAU - DINH NGOC THANH VI TICK PHAN HAM NHIEU BIEN (Litu hanh noi bo) TKb'vi£N I 2002 Lditea Tritdng Dai Hoc Dan Lap Cilu Long nd Inc thnc hien nhiem vu mot trung tam dao tao cii nhan, ky sU chink quy cd chat luqng coo, mot trung tam gdn dao tao vdi nghien cilu ling dung khoa hoc ky thuqt tien tien, mot trung tam truyen ba khoa hoc viing Dong Bang Song Cdu Long Mot nhilng you to yeu la doi ngu gidng vien cua TrUdng vdi trinh do, kinh nghiem gidng day, uy tin khoa hoc cua ho cung vdi de citong, gi-do trinh dnoc soati thdo theo chitang trinh quy dinh cua Bq Gido Due vd Dao Tao An phdm la gido trinh ndm hoc 2001-2002 cua Bq mdn Loan - Tin hoc, Ban Khoa Hoc Co Ban, Tritdng Dai Hoc Dan Lap Cilu Long Dai Hoc Dan Lap Cilu Long in gido trinh de sinh vien nhd tritdng doc, nghien cilu HIEU TRUdNG Trifcrng Dai Hoc Dan Lap Guru Long GS NGUYfiN CONG BINH ( Chitdng KHONG GIAN IRn Trong chifOng nay, chung ta khao sat khong gian lRn, vdi n > 1, nhtf la mot tap hop cac diem va cung la khong gian cac vectcf, lam co sb d£ khao sat phep tinh vi tich phan ham nhieu bien Trtrdc het, khong gian IRn difOc gidi thieu tong quat cimg cac tap dftc biet cua nd Sau do, mat phAng IR2 va khong gian IR3 se dtfqc khao sat chi tiet voi cac khai niem can ban ve die d’ng va mdt KHONG GIAN lRn Xet tap h(?p IRn = 1R x x IR — {(a?i, xz x n) E IR, Vz — 1,2 7?} n I'An Moi phcin tii cua lRn diroc gpi la mot diem Hay ndi khac di, mdi di&m M cua lRn difOc lien ket vdi mot bo thu’ ttf (a?i, xz, ■■■, ^n) v& ta goi Xi la tga thii i cua M IRn, ky hieu M(zi, xz,zn) Diem cd toa la (0, 0,0) duoc goi la goc tga va thu’bng dtfOc ky hieu la O Khodng each giua hai diem ?l(ai, az, ,an) va Chtfcmg Khong gian IR7' ky hieu la AB, cho’ bd'i 1/2 AB y2(a2- - 6i)2 1=1 1.1 Mdt so tap ddc biet cua IRA 02, ■■■, a.n), B(6i, 62, •••> ^n) (i) Doan thdng V6i hai diem cua IRn, ta xac dinh dime mot du’bng thing, ky hieu At AB, vdi AtAB = {{ta1 + (l-t)b1,ta2 + (l-t)b2, ,tan + ('i t)bn) | t IR} va mot doan thing, ky hieu AB, vdi AB = {(tai d- (1 — t)6i, ta2 + (1 — t)&25 tan + (1 — t)^n) | < t < 1} Diem Mt (tai d- (1 — t)6i, ta2 d~ (1 — t)&2i •••» t^n d- (1 — t)&n), voi < t < 1, nim doan AB va di chuyen tif B (ling vdi t = 0) d£n A (ting vdi t = 1) Ngiroc lai, diim Af'(s6i + (1 - s)ai,s62 + (1 - s)a2,sbn d- (1 - s)an) vdi < s < cung nim doan AB nhifng di chuyen tir A (ting vdi s = 0) den B (tfrtg vdi s = 1) (ii) Mot o IRA la tich cua n khoang (bi chan) IR Neu tit ca cac khoang la khoang dong, tirong dug difOc goi la o dong Neu tat ci cac khoang la md, tirong dng du’oc goi la mA (Hi) Qua cdu - mat cau Ung vdi mot diem A(ai, ,an) va mot so thifc r > Ta goi qua cdu dong tam A ban kind r, ky hieu B'(A, r) la tap hop nhu’ng diim M(zi,xn) cho khoang each tif M den A nhd hon hay bing ban kinh r, nghia la, {M^xi,xn) | MA2 < r2} / M(zi, ,zn) ^2(xi - aj2 < r2 > I £=1 ) Chifcfng Khong gian lRn Ttrong tif, ta cd qua cau md B(A, r) va mat cau S(A, r) tam A ban kinh r, n B(A,r) M^}, ,xn) ^Xi-ai)2 xq hay yi(x') - 3/2(®) = neu x > xq De chti’ng minh z(z) = x < x0 ta dat t = z0 - x, (zx(Z), wi(i)) = (z(xq - t'),w(zo — 0)- Khi zi,wi thda he -W1 dtwi ftWj + (■hu’o’ng Nhap mon phifcfng trinh vi phan 210 Z1(0) = W](0) - 2(i’o) = w(z0) = 0- Ly luan tuvng tu’ nhif tren zY = t > nghia la z(x) = neu x < x0 Dinh ly da du’o'c chti’ng niinh □ Bay gicf ta xet phifcfng trinh y" 4- ay' + fry = Day la dang dSc biet cua phtfcmg trinh c? muc 7.6 Xet bai toan tren vdi dieu kien dau yM = a,y\xo) = b Dat A = a - 4fr Ta cd 7.7 Dinh ly i) Neu A > thi bdi loan tren cd nghiem nhat y = AeX1X 4- BeX2X vdi Aj,A2 la, hai nghiem cua A2 +oA + ,3 = vd A,B thoa he y(xo') = a,y'(xo) = b hay AeX1X° 4- BeX2X° AiXeAia:o X2Bex*X0 a b (ii) Neu A = thi bdi loan cd nghi€m nhat y = (A + Bx)e 211 Chifiyng Nhap mon phu’o’ng tr'mh vi phan vdz A, B thoa he t/(so) = a^y^xo) - b hay _ arp (A + Bso)e a ail b e~~ (Hi) Neu A < thi bai loan cd nghiem nhat y = Ae sino’s COSO’S + Be - a2 vd A, B thoa he y(x0) = a,y'(xo) - b hay vdi a) = asp _ aa:0 , I Ae 2“ cosa’so + ^e sinu’So < e~^S~ (A (-| coscusq - o’sinu’So) + [ 4-B (-^ sinivso + o’coscuso)) Lu’u y neu khong cd dieu kien dau t/C^o) = a b y'(xo) — b ta khong th< xac dinh cu the A,B Nghiem tru’dng hop nghiem goi la nghiem tong quat cua phttong trinh Vi du (i) Tim nghiem tong quat cua phuong trinh y" - 31/' + 2?/ = Ta co A=9 4.2 = Vay phtrong trinh A2 3A + = cd hai nghiem A = 1, Nghiem t6ng quat cua phuong trinh la y = Aex + Be2x 212 C'huxfng Nhap mon phuxmg trinh vi phan (iiI Tim nghiem cua phirang trinh y" 4- w2?/ = ()(u; > 0) 2/(0) = 0,/(0) = Giai Ta co A 4uj2, a = 0,(3 — a; Vay phu’ong trinh cd nghiem tong quat dang y = A cos 2u>2! 4- B sin 2ivjI a tim A, B d£ y(0) — 0, y'(O') = Ta cd theo bi£u thii’c cua y la y(0) = A Vay A = Ta suy Vay y'fO) y B sin ‘Iwx y' 2a)B cos 2