Math 211A Homework. Edward Burkard. = tan (2x + z)


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1 Mth A Homework Ewr Burkr Eercises 5C Eercise 8 Show tht the utonomous system: 5 Plne Autonomous Systems = e sin 3y + sin cos + e z, y = sin ( + 3y, z = tn ( + z hs n unstble criticl point t = y = z = 0 Solution First let s verify tht (0, 0, 0 is inee criticl point of our system: (0, 0, 0 = e0 sin 0 + sin 0 cos 0 + e 0 = = 0, y (0, 0, 0 = sin (0 + 0 = 0, z (0, 0, 0 = tn (0 + 0 = 0 so (0, 0, 0 is inee criticl point of the system To voi pin n heches, write our system s = F ( To check the if this is n unstble criticl point, we merely nee to look t the liner pproimtion of our system ner (0, 0, 0 Ner (0, 0, 0 the liner pproimtion is: First we compute DF : DF = n t (0, 0, 0 the mtri is: so our liner pproimtion is = F (0, 0, 0 + DF (0,0,0 ( (0, 0, 0 = DF (0,0,0 e sin 3y + cos 3 e cos 3y e z cos ( + 3y 3 cos ( + 3y 0 sec ( + z 0 sec ( + z 3 DF (0,0,0 = = 3 0 = A 0 To check if (0, 0, 0 is n unstble criticl point, we merely nee to know if t lest one of the eigenvlues of A hs positive rel prt So let s fin them (I will spre you the gruesome clcultions: First, the Chrcteristic Polynomil of A: p A (λ = λ 3 6λ + 6λ + 3
2 We merely wnt to know if there is n eigenvlue with positive rel prt In fct, there is, n though I will not ctully compute it, I will prove its eistence Notice tht the chrcteristic polynomil is continuous function from R to R Notice tht p A ( = = 4 > 0 n p A (3 = = 6 < 0, so by the intermeite vlue theorem, p A must hve zero in the intervl (, 3 This is sufficient to sy tht A hs positive rel eigenvlue (in fct, ll three re rel: one is negtive n the other two positive, n hence tht (0, 0, 0 is n unstble criticl point of our system 6 Eercises 6A 6 Eistence n Uniqueness Theorems Eercise 8 Show tht the curves efine prmetriclly s solutions of the system: y = F = F y (6 z = F z re orthogonl to the surfces F (, y, z = c, c R Wht ifferentibility conition on F must be ssume to mke this system stisfy Lipschitz conition? Solution Let g(t be solution to (6 Then it s irection vector t ny time t is g (t Notice tht g (t = F for ll t Recll tht the grient of F is lwys perpeniculr to ny of its level sets, hence g (t is perpeniculr to F (, y, z = c for ny c R Thus ny solution to (6 is perpeniculr to the level sets of F In orer to stisfy Lipschitz conition in compct, conve region R, we nee F C (R, since then the system (6 is in C (R n hence we cn pply the Lemm in Section 6 6 Aitionl Eercises Eercise 8 Show tht, if X(t = ij (t is mtri whose columns re solutions of the homogeneous liner system X = A(tX, then ( t et X(t = et X(] ep kk (s Proof Assume tht both X n A re n n mtrices Borrowing the nottion bove, let A(t = ij (t Viewing X (t = A(tX(t, then the entries of X (t hve the form n ij(t = ik (t kj (t (6 Spring the pin of the computtion, let s tke the erivtive of the et X(t: n n n n n n et X(t = n n nn n n nn n n nn Focusing on the first eterminnt for moment, by eqution (6, we hve: k k k k k kn n n n nn k=
3 3 Notice tht if we subtrct times the secon row from the first row we lose the secon term in ech of the sums in the first row (recll tht ing multiples of one row of eterminnt to nother row oes not chnge the vlue of the eterminnt Similrly if we subtrct 3 times the thir row from the first row, we lose the thir term in ech of the sums in the first row, oing this for ech row leves us with only the k = term in ech sum, ie n n n n nn Now this is precisely et X(t Doing similr thing for ech of the other eterminnts in the sum for et X(t, we see tht: n et X(t = kk (t et X(t Let et X(t = y(t, n k= n kk (t = η(t; then the bove eqution reuces to: k= This is seprble eqution with solution: y (t = η(ty(t ( t y(t = y( ep η(s for some ( woul be etermine by n initil vlue Replcing y n η with their originl menings, we get the esire result: ( t n et X(t = et X(] ep kk (s k= Remrk My resoning for this problem ws bsiclly try to pt proof of the similr formul for seprble eqution (which I use for the eqution in y The proof of the formul use bove for the erivtive of the eterminnt of X(t is s follows: We cn write et X(t = δ(j,, j n j (t j (t njn (t, (j,,j n S n where S n is the permuttion group on n letters (the specific letters re {,, n} n δ(j,, j n is the prity of the permuttion Now tking the erivtive et X(t = δ(j,, j n j (t njn (t (j,,j n S n n = δ(j,, j n j (t k,jk (t kj k (t k+,jk+ (t njn (t (j,,j n S n k= n = δ(j,, j n j (t k,jk (t kj k (t k+,jk+ (t njn (t (j,,j n S n k= n = δ(j,, j n j (t k,jk (t kj k (t k+,jk+ (t njn (t k= (j,,j n S n which is precisely the foruml use bove
4 4 0 Eercises 0D 0 SturmLiouville Systems Eercise 7 Derive Theorem 3 from the Sturm Comprison Theorem of Ch by introucing the new epenent vribles t = P (s n t = P (s Theorem 3: Theorem Let P ( P ( > 0 n Q ( Q( in the DEs: ( P ( u + Q(u = 0, ( P ( u + Q (u = 0 Then, between ny two zeros of nontrivil solution u( of the first DE, there lies t lest one zero of every rel solution of the secon DE, ecept when u( cu ( This implies P P n Q Q, ecept possibly in intervls where Q Q 0 The Sturm Comprison Theorem of Ch : Theorem Let f( n g( be nontrivil solutions of the DEs u + p( = 0 n v + q(v = 0, respectively, where p( q( Then f( vnishes t lest once between ny two zeros of g(, unless p( q( n f is constnt multiple of g Proof Recll Given functions g n h where g is invertible, the eqution f g = h cn be solve for f Let P, P, Q, n Q be efine s in Theorem 3 Since t = P (s n t = re both incresing P (s functions of (becuse ( P n P re positive functions they re invertible, n hence we my fin functions q(t ( n q (t such tht q = P (Q( n q = P (Q ( Notice tht, by efinition, since P (s P (s P P n Q Q, it follows tht q q Consier the ifferentil equtions given by: v + q(tv = 0 (0 n v + q (tv = 0 (0 Let v(t n v (t be nontrivil solutions to (0 n (0 respectively, n efine the new functions u( n u ( by ( u( = v P (s n u ( = v ( P (s
5 Let s see wht hppens when we mke the substitution t = cttywompus (ie chin rule glore : u = u = = P ( = P ( = u = u P ( P ( u ] = P u + P ( P (s ] u = P u + P u ( = P P (u + P (] u ] P u = P ( + P u ( = P ( P ( u ] Thus substituting t s bove into (0, we en up with the eqution: ( ] ( ( v + q v = u P (s P (s P (s which simplifies to since P ( > 0 Thus we similrly hve = P ( in (0 Now let s compute bunch of ( + P (Q(u( P ( u ] + P (Q(u( = 0 P ( u ] + Q(u( = 0 (03 P ( u ] + Q (u ( = 0 (04 when plugging t = into (0 P (s Recll tht q (t q(t Using the Sturm Comprison Theorem of Chpter on (0 n (0 we hve the result tht between ny two zeros of v(t there is t lest one zero of v (t, unless q(t q (t, in which cse v(t cv (t for some c R (c 0 Assuming tht v(t cv (t, then since to ny nontrivil solutions v n v of (0 n (0 respectively, there is unique corresponing u n u s bove, it follows tht between ny two zeros of u( there is t lest one zero of u ( Now if v(t cv (t, then q(t q (t n it follows tht u( cu ( Assume we re on n intervl such tht Q Q 0 Then since q(t q (t we hve tht P (Q( = P (Q ( = P (Q(, n since we re on n intervl where Q 0 we cn ivie both sies by Q to ttin: P ( P (, 5 on tht intervl On the other hn, if Q Q 0 on some intervl, then it is not necessry tht P P Eercise 8 For ny solution of u +q(u = 0, q( < 0, show tht the prouct u(u ( is n incresing function Infer tht nontrivil solution cn hve t most one zero
6 6 Proof Let u( be nontrivil solution of the DE bove Since u is solution of secon orer ifferentil eqution it is t lest twice ifferentible This implies tht the prouct u(u ( is ifferentible (n hence continuous Now let s look t its erivtive: u(u (] = u (u ( + u(u ( = u (] u(q(u( = u (] q(u(] 0 Hence u(u ( is n incresing function Now ssume tht u is nontrivil solution n tht it hs two zeros, sy n b ( < b, thus the function u(u ( hs zeros t n b This mens tht, since u(u ( is continuous, u(u ( is ecresing on some subintervl of, b] or u(u ( 0 But if u(u ( is ecresing, this implies tht u(u (] is negtive somewhere, which contricts the bove clcultion Hence it must be tht u(u ( 0 But if this is true, then since u( 0, it must be tht u ( 0 But then u( constnt, however, the only constnt solution to this ifferentil eqution is the trivil one Therefore it is not possible tht u(u ( 0, n hence it must be tht u( oes not hve more thn one zero
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