Let ( α, β be the eigenvector associated with the eigenvalue λ i

Transcription

1 ENGI Sabiliy Analysis (Linear) Page 4.5 Le ( α, be he eigenvecor associaed wih he eigenvalue λ i of he coefficien i i) marix A Le c, c be arbirary consans. a b c d Case of real, disinc, negaive eigenvalues (wih λ < λ < 0 ): Two linearly independen soluions are x, y e λ, e λ and x, y e λ λ α α, e ( ) ( ) ( ) ( ) The general soluion is ( x, y ) ( c e λ ) c e λ, c e λ λ α + α + c e ( ) ( ) One can see ha x y lim, 0, 0. All orbis herefore erminae a he criical poin a he origin. The sysem is asympoically sable. If boh arbirary consans are zero, hen we have he rivial soluion (x y 0 for all ). If one of he arbirary consans is zero (say c ), hen x, y cα e λ, ce λ y x α which is a sraigh line hrough he origin, of slope α. [The siuaion is similar if c is zero.] ( ) ( ) We herefore obain sraigh-line rajecories ending a he singular poin, when exacly one of he arbirary consans is zero.

2 ENGI Sabiliy Analysis (Linear) Page 4.6 If neiher arbirary consan is zero, hen λ λ ( λλ) ( λλ) y c e + ce c + ce ce + c x λ λ ( λλ) ( λλ) cα e + c α e cα + c α e cα e + c α Because λ < λ < 0, lim and lim y x y x ( λ λ ) + c e c lim ( λ λ ) α c e c α + α ( λ λ ) + ( λλ) α α + α c c e lim c c e All orbis herefore come in from infiniy parallel o he line y x. α All orbis share he same angen a he origin, y x. α We obain a sable node ha is also asympoically sable. [The case illusraed here is α, α 3,,, λ 5, λ 0, which is + 3 generaed from A 4.] Case of real, disinc, posiive eigenvalues (wih λ > λ > 0 ): The analysis leads o he same phase space, excep ha he arrows are reversed. The resul is an unsable node.

3 ENGI Sabiliy Analysis (Linear) Page 4.7 Case of real, disinc eigenvalues of opposie sign (wih λ < 0 < λ): The general soluion is ( x, y ) ( c e λ ) c e λ, c e λ λ α + α + c e x y x y ( ) ( ( ) ( )) λ < 0 < λ lim, and lim, do no exis (infinie), (wih he excepion of he orbi for c 0). All orbis (excep c 0) herefore move away from he criical poin a he origin. The sysem is unsable. If boh arbirary consans are zero, hen we have he rivial soluion (x y 0 for all ). If one of he arbirary consans is zero (say c ), hen λ λ x, y cα e, ce y x α which is a sraigh line hrough he origin, of slope α. [The siuaion is similar if c is zero.] ( ) ( ) We herefore obain sraigh-line rajecories when one of he arbirary consans is zero. One of hem (c 0) ends a he singular poin while he oher begins here. If neiher arbirary consan is zero, hen λ λ ( λλ) ( λλ) y c e + ce c + ce ce + c x λ λ ( λλ) ( λλ) cα e + c α e cα + cα e cα e + c α Because λ < 0 < λ, lim and lim y x y x ( λ λ ) + c e c lim ( λ λ ) α c e c α + α ( λ λ ) + ( λλ) α α + α c c e lim c c e

4 ENGI Sabiliy Analysis (Linear) Page 4.8 All orbis herefore share he same asympoes, y x (incoming) and α y x (ougoing). α We obain a saddle poin, which is an unsable criical poin. [The case illusraed here is α, α 3,,, λ + 5, λ 5, which is generaed from A ]

5 ENGI Sabiliy Analysis (Linear) Page 4.9 Case of real, equal, negaive eigenvalues ( λ λ < 0) and b c 0: The sysem is uncoupled: dx d ax dy d dy and equal eigenvalues now require a d λ. x, y λ λ c e, c e. ( ) ( ) The general soluion is λ 0 lim ( x, y ) (, ) and lim ( x, y ) ( 0 ) <,0. All orbis herefore erminae a he criical poin a he origin. The sysem is asympoically sable. If boh arbirary consans are zero, hen we have he rivial soluion (x y 0 for all ). y c c 0 x c and c 0, c 0 x 0 The orbis are sraigh lines ending a he criical poin a he origin. The criical poin is an asympoically sable sar-shaped node. Addiional Noe: The eigenvalues of any riangular marix are he diagonal enries of ha marix: a b The characerisic equaion of A 0 d de A λ I 0 is ( ) a λ b 0 d λ ( a λ)( d λ) 0 λ a o r d

6 ENGI Sabiliy Analysis (Linear) Page 4.0 Case of real, equal, negaive eigenvalues ( λ λ < 0) and b, c no boh zero: The characerisic equaion λ ( a d ) λ ( ad bc) ( ) ( ) ( ) + + has he discriminan a + d 4 ad bc a d + 4bc 0. a+ d The soluion of he characerisic equaion simplifies o λ. x, y c c e λ λ α + α, c + c e. ( ) The general soluion is ( ) ( ) ( ) λ 0 lim x, y, and lim x, y 0 ( ) ( ) 0 ( ) ( ) <,0. All orbis herefore erminae a he criical poin a he origin. The sysem is asympoically sable. If boh arbirary consans are zero, hen we have he rivial soluion (x y 0 for all ). y c + c If c 0, hen as ± x cα + cα α All orbis (excep for c 0) herefore come in from infiniy parallel o he line y x, which is also a angen a he origin. I can be shown ha α when c 0, so ha he rajecories for c 0 and c 0 are boh y x. α α α Neiher eigenvalue can be zero, oherwise (0, 0) is no he only criical poin (as shown on page 4.4).

7 ENGI Sabiliy Analysis (Linear) Page 4. Case of real, equal, posiive eigenvalues ( λ λ > 0 ) The analysis leads o he same phase planes as in he case of real equal negaive eigenvalues, bu he signs of he arrows are reversed and he resul is an unsable node. Case of complex conjugae pair of eigenvalues wih negaive real par The eigenvalues (roos of he characerisic equaion) are λ a + jb, λ a jb, ( a<0 ). The general soluion has he form x c( Acosb Asin b) c( Asin b Acosb) + + e y c( Bcosb Bsin b) c( Bsin b Bcosb) + + e Using he definiions ( ) + ( + ), B ( c B c B ) + ( c B + c B ) A ca ca ca ca ca + ca ca ca cb + cb cb cb cos α, sinα, cos, sin A A B B he general soluion can be wrien more compacly as a a x, y A e cos b + α, B e cos b + a a ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) a< 0 lim x, y, and lim x, y 0,0. π If x 0 hen b n n π If y 0 hen b+ + nπ n y Bcos( b+ ) x Acos( b+ α ) y is periodic, wih period π. x b The orbis spiral in o he origin. + α + π ( ) ( ) We have an asympoically sable spiral, also known as a sable focus.

8 ENGI Sabiliy Analysis (Linear) Page 4. Case of complex conjugae pair of eigenvalues wih posiive real par The analysis leads o he same phase planes as in he case of negaive real par, bu he signs of he arrows are reversed and he resul is an unsable focus. Case of complex conjugae pair of eigenvalues wih zero real par (pure imaginary) The eigenvalues (roos of he characerisic equaion) are λ jb, λ + jb. The general soluion has he compac form ( x, y ) ( Acos ( b+ α ), Bcos( b+ )) π If α 0 and, hen x y ( x, y ) ( Acos b, Bsinb) A + B so ha he orbis are ellipses, cenred on he criical poin a he origin. This is a sable cenre. Oher choices of α and also lead o concenric ses of ellipses, bu roaed wih respec o he coordinaes axes. Noe ha his is he only case of a sable criical poin ha is no asympoically sable.

9 ENGI Sabiliy Analysis (Linear) Page 4.3 Summary for he Linear Sysem dx dy ax + by, d d cx + dy, abcd,,, consans Characerisic equaion: λ a + d λ + ad bc 0 Discriminan ( ) ( ) ( ) ( ) ( ) ( ) D a+ d 4 ad bc a d + 4bc a b Roos of characerisic equaion ( eigenvalues of A ): c d Cases: λ ( ) a+ d ± D a + d D oher condiion λ Type of poin a + d < 0 D > 0 ad bc > 0 real, disinc Sable negaive node a + d < 0 D 0 b c 0 real, equal Sable negaive sar shape a + d < 0 D 0 b, c no boh 0 real, equal Sable negaive node a + d < 0 D < 0 complex Sable conjugae pair focus [spiral] a + d 0 D < 0 Pure Sable imaginary pair cenre a + d > 0 D > 0 ad bc > 0 real, disinc Unsable posiive node (any) D > 0 ad bc < 0 real, disinc Unsable opposie signs saddle poin a + d > 0 D 0 b c 0 real, equal Unsable posiive sar shape a + d > 0 D 0 b, c no boh 0 real, equal Unsable posiive node a + d > 0 D < 0 complex Unsable conjugae pair focus [spiral] Noe ha ad bc de A and ha a + d he race of he marix A. In brief, if he real pars of boh eigenvalues are negaive (or boh zero), hen he origin is sable. Oherwise i is unsable. [See also he example a "

10 ENGI Sabiliy Analysis (Linear) Page 4.4 Example Find he naure of he criical poin of he sysem dx dy 4x 3 y, 5x4y d d and find he general soluion. a b 4 3 The coefficien marix is A c d 5 4. race(a) a + d 4 + ( 4) 0 D (a d) + 4bc (4 + 4) + 4( 3)(5) > de A < D > 0 and ad bc < 0 λ are real wih opposie signs and he criical poin is a saddle poin (unsable). Solving he sysem: ( a+ d) ± D 0 ± 4 λ ± x, y cα e + c α e, c e + c e ( ) ( ) α where is he eigenvecor associaed wih he eigenvalue λ α and is he eigenvecor associaed wih he eigenvalue λ +. To find he eigenvecors, find non-zero soluions o he equaion a λ b α 0 c d λ 0 A λ : 4+ 3 α 5 3 α Any non-zero choice such ha 5α 3 0 will provide an eigenvecor. α 3 Selec. 5

11 ENGI Sabiliy Analysis (Linear) Page 4.5 Example (coninued) A λ +: 4 3 α 3 3 α Any non-zero choice such ha α 0 will provide an eigenvecor. α Selec. The general soluion is ( x, y ) ( 3 ce, 5ce ) ]. [I is simple o check ha (4x 3y, 5x 4y) is indeed equal o ( x, y) Also noe ha y 5ce y ( ) 5 y( ) lim ( c 0) and lim c 0 x 3ce x 3 + x so ha all orbis for which boh c and c are non-zero share he same asympoes, 3y 5x (which is he incoming orbi, when c 0) and y x (which is he ougoing orbi, when c 0). ( ) A few represenaive orbis and he wo asympoes are ploed in his phase space diagram:

12 ENGI Sabiliy Analysis (Linear) Page 4.6 Example Find he naure of he criical poin of he sysem dx dy x + y, d d x y and find he general soluion. a b The coefficien marix is A c d. race(a) a + d + 4 < 0. D (a d) + 4bc ( + ) > 0 λ are real, disinc and negaive and de A ad bc 4 3 > 0 he criical poin is a sable node. Solving he sysem: ( a+ d) ± D 4 ± 4 λ ± 3,, 3, 3 x y cα e + c α e c e + c e ( ) ( ) α where is he eigenvecor associaed wih he eigenvalue λ 3 α and is he eigenvecor associaed wih he eigenvalue λ. To find he eigenvecors, find non-zero soluions o he equaion a λ b α 0 c d λ 0 A λ 3: + 3 α α Any non-zero choice such ha α + 0 will provide an eigenvecor. α Selec. A λ : + α α Any non-zero choice such ha α + 0 will provide an eigenvecor. α Selec.

13 ENGI Sabiliy Analysis (Linear) Page 4.7 Example (coninued) The general soluion is ( ) ( 3, 3 x y ce, ce ) ]. [I is simple o check ha ( x + y, x y) is indeed equal o ( x, y) Also noe ha y 3 ce ce + c c x 3 ce ce + c c y y lim ( c 0) and lim c 0 x + x and ( ) ( ) ( 3 3 x y ce ce ) ( ) lim, lim, 0, 0 so ha all orbis for which boh c and c are non-zero come in from a direcion parallel o y x (which is he orbi when c 0) and share he same angen a he origin, y x (which is he orbi when c 0). A few represenaive orbis and he common angen are ploed in his phase space diagram: