Math 334 Fall 2011 Homework 11 Solutions
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1 Dec. 2, 2 Mah 334 Fall 2 Homework Soluions Basic Problem. Transform he following iniial value problem ino an iniial value problem for a sysem: u + p()u + q() u g(), u() u, u () v. () Soluion. Le v u. Then v u and he equaion becomes and he iniial value becomes The sysem we are looking for is hen wih iniial values v + p()v + q() u g() (2) u()u, v() v. (3) v q() u p() v + g() (4) u v (5) u()u, v() v. (6) Inermediae Problem 2. Express he soluion of he following iniial value problem in erms of a convoluion inegral: Soluion. Firs ransform he equaion: y +4 y + 4 y g(); y() 2, y () 3. (7) L{y s 2 Y s y() y ()s 2 Y 2 s + 3; (8) L{y sy y() s Y 2 (9) Denoing L{g G(s), we have he ransformed equaion as So Now ake inverses: { L G(s) s s+ 4 { L L { 2 s + 5 s s+ 4 So he final answer is (s 2 +4 s +4) Y G(s)+2s+5. () G(s) Y s s s + 5 s s+4. (). We use he convoluion heorem: { G(s) L s s + 4 (s + 2) 2 { L + 2 (s + 2) 2 (s + 2) L {G (e 2 ) g e 2( τ) ( τ) g(τ) dτ. (2) e e 2. y e 2( τ) ( τ) g(τ)dτ +e 2 ( +2). (3) Problem 3. Express he soluion of he following iniial value problem in erms of a convoluion inegral: y (4) y g(); y() y () y () y (). (4) Soluion. Taking ransform of he equaion we obain Therefore (s 4 ) Y G(s) Y G(s) s 4. (5) { y() L g. s 4 (6)
2 2 Mah 334 Fall 2 Homework Soluions We compue { L { s 4 2 L s 2 { s L s s + 2 s [e e 2 sin ]. (7) So he answer is y() 4 Problem 4. Find all eigenvalues and eigenvecors for ) ; a) A ( 2 2 b) A Soluion. a) We have Solving So eigenvalues are 3,. [e ( τ) e ( τ) 2 sin ( τ)] g(τ) dτ. (8) ( ) 2 λ de (A λi) de λ 2 λ λ+3. (9) Eigenvecors corresponding o 3: We solve λ λ + 3 λ 3, λ 2. (2) (A ( 3)I) x (2) which becomes ( )( ) ( ) ( ) x x a. (22) Eigenvecors corresponding o : We solve ( )( ) x b) We have de (A λ I) de 3 λ λ λ Now we solve Observe: λ is a roo. Facorize: Now solve: ( x (3 λ) ( λ) (3 λ) ( λ) (3 λ) 2 2 (3 λ) ) ( ) a. (23) λ λ 2 9 λ λ λ 2 +4 λ λ λ λ+8. (24) λ 3 +6 λ λ +8. (25) λ λ λ + 8(λ + ) ( λ λ + 8). (26) λ λ + 8 λ 2 8, λ 3. (27) So in fac we have wo eigenvalues: λ λ 2, λ 3 8. Nex we find eigenvecors corresponding o. We need o solve x. (28) Noe ha he soluions are given by x,, saisfying 2 x (29)
3 3 In oher words he eigenvecors are all vecors saisfying his equaion. To ge an explici formula for eigenvecors, we wrie x x 2 x 2 x 2 + x3 2. (3) There are no resricion on x,. Therefore he eigenvecors corresponding o is given by a 2 +b 2. (3) Remark. Keep in mind ha for an eigenvalue, is eigenvecors are no several single vecors, bu a collecion of infiniely many vecors. As a consequence, here are more han one way o represen hem. For example, in he above we have shown ha eigenvecors corresponding o can be represened as a 2 +b 2. (32) wih a, b arbirary consans. The same se of vecors can also be wrin as a 2 +b. (33) To see ha hey indeed represen he same se of vecors, we check:. The former includes he laer: Tha is any vecor in he form of he laer can be represened by he former. 2 ( ) 2, ( ) (34) 2. The laer includes he former: 2 ( ) 2, 2 ( ) 2 +. (35) Now we urn o he eigenvalue 8. We need o solve x. (36) We use Gaussian eliminaion: (Simplify he 2nd row) (Swich s and 2nd row) (firs row 5 add o 2nd; ( 4) add o 3rd)
4 4 Mah 334 Fall 2 Homework Soluions So he sysem for x,, is equivalen o x 4 + (37) 2 + (38) Represen x, by : x (39) 2. (4) This gives x x3 2 /2. (4) So he eigenvecors corresponding o 8 are where a is an arbirary number. a /2 (42) Advanced Problem 5. Prove he basic properies of convoluion: f g g f; f (g + g 2 ) f g + f g 2 ; (f g) h f (g h); f f. Proof. f g g f. Recall definiion: Now do he change of variable: and he inegral becomes f( τ) g(τ)dτ We have f (g + g 2 ) f g + f g 2. Use definiion: f g f( τ) g(τ) dτ. (43) f( τ) [g (τ) + g 2 (τ)] dτ (f g) h τ dτ d (44) f( ) g( )( d ) g( ) f( ) d g f. (45) (f g)( τ) h(τ) dτ τ [ τ f( τ) g (τ) dτ + f( τ s)g(s) ds ] h(τ) dτ f( τ) g 2 (τ) dτ (46) f( τ s) g(s)h(τ) ds dτ. (47) As we would like o pair g and h ogeher, we have o wrie f as f( ). So inroduce s + τ in he inner inegral Thus ds d. Then we have [ ] τ [ f( τ s)g(s) ds h(τ) dτ f( ) g( τ) d ]h(τ) dτ τ f( ) g( τ) h(τ)d dτ. (48) τ
5 5 Now we swich he order of he inegraion. The domain of he inegraion is <τ < <. So runs from o while τ from o. Therefore [ ] f( ) g( τ) h(τ) d dτ f( ) g( τ) h(τ) dτ d τ f( ) [ f( ) (g h)( ) d ] g( τ) h(τ) dτ d f (g h). (49) This one is rivial: f f( τ) dτ. (5) Noe ha, all he above can be easily proved by he propery L{f g L{f L{g. However, implici in ha approach is he assumpion ha L {L{f f whose proof is acually no easy. Challenge Problem 6. Derive he formula L {e as F(s) f( a) u( a) using convoluion. Proof. We have L {e as F(s) L {e as L {F(s) δ( a) f() f( τ) δ(τ a) d f( a) u( a). The las sep follows from he following observaion: When < a, τ a < and herefore in he inegral δ( a). Problem 7. Recall ha we can wrie any single linear homogeneous equaion of order n ino a s order sysem consising of n equaions. Show ha he Wronskian of he laer is he same as he Wronskian of he former. Proof. Le he n-h order equaion be I can be wrien ino a sysem of n firs order equaions hrough seing x y, y,, y (n ), x x, P() y + p () y (n ) + + p n () y. (5) The Wronskian for he n-h order equaion reads: y y n y y n de ẋ P() x (52) y (n ) p n () p n () p n 2 () p () y n (n ) (53) (54)
6 6 Mah 334 Fall 2 Homework Soluions which becomes he Wronskian for he sysem afer idenifying y i x (i) y i. (55) y i (n ) Problem 8. Le W be he Wronskian of n soluions x (),, x o he sysem ẋ p () x + + p n () (56) ẋ n p n () x + + p nn (). (57) Prove ha dw d (p () + + p nn ()) W. (58) Proof. From properies of deerminans we have ( ) d de () x x () d de () ẋ () () ẋ + + de x () x () ẋ n ẋ n (59) Here we have used he following propery: The derivaive of a deerminan is he sum of n deerminans, each obained by puing derivaive on one single row (or one single column). This can be proved by using he ulimae definiion of deerminans: de (M) (sign of σ) m σ() m nσ. (6) σ All permuaions of {,,n or hrough definiion of derivaive (he lim δ one) and use he following propery of deerminans: de a + b a n + b n de a a n + de b b n (6) Now we have ẋ () p ()x () + p 2 () () + ; ; x p () x + + p n (). (62) Subsiuing ino he firs deerminan and use he propery de a + b a n + b n de a we have de () ẋ () () ẋ () p () x de () () a n + de b b n (63) () p () p 2 () p 2 () + de x () () +Terms similar o he 2nd one. (64) Now using he following propery: If a marix has one row a muliple of anoher, hen he deerminan is, we see ha only he firs one is no. Bu he firs one is simply () p () x p () x n () () de x x 2 p x () de p ()W. (65) () ()
7 7 Dealing wih he res similarly, we reach dw d (p () + + p nn ()) W. (66) Remark. I s ineresing ha if we pu derivaive on each column and wrie ( ) d de () x x () d de ( ẋ () x ) + + de ( x () ẋ ) (67) and hen use ẋ () P() x () and so on, we seem o ge suck. The philosophical reason for his difference beween he row-by-row approach and column-by-column approach seems o be ha, when doing he row-by-row approach we are using he fac ha x (),,x are all soluions in each deerminan, while when in he column-by-column approach, in each deerminan in he righ hand side, we only ake advanage of one x (i) being a soluion.
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