Modelli Clamfim Equazioni differenziali 7 ottobre 2013

Transcription

1 CLAMFIM Bologna Modell Clamfm Equazon dfferenzal 7 ottobre 2013 professor Danele Rtell danele.rtell@unbo.t 1/18?

2 Ordnary Dfferental Equatons A dfferental equaton s an equaton that defnes a relatonshp between a functon and one or more dervatves of that functon. 2/18?

3 Ordnary Dfferental Equatons A dfferental equaton s an equaton that defnes a relatonshp between a functon and one or more dervatves of that functon. For nstance let y = y(x) be some functon of the ndependent varable x. The equaton dy dx (x) := y (x) = 2xy(x) (1) states that the frst dervatve of the functon y equals the product of 2x and the functon y tself. An addtonal, mplct statement n ths dfferental equaton s that the stated relatonshp holds only for all x for whch both the functon and ts frst dervatve are defned 2/18?

4 Gven f : Ω R 2 R y (x) = f(x, y(x)), y(x 0 ) = y 0, x I x 0 I, y 0 J where I J Ω s called ntal value problem 3/18?

5 Exstence: eano s Theorem If f(x, y) s contnuous on R = [x 0 a, x 0 + a] [y 0 b, y 0 + b], then the ntal value problem y (x) = f(x, y(x)) y(x 0 ) = y 0 has a soluton n a neghborhood of x 0 4/18?

6 Soluton needs not to be unque. Consder the ntal value problem y (x) = 2 y(x) y(0) = 0 (p) 5/18?

7 Soluton needs not to be unque. Consder the ntal value problem y (x) = 2 y(x) y(0) = 0 (p) y(x) = 0 solves (p) 5/18?

8 Soluton needs not to be unque. Consder the ntal value problem y (x) = 2 y(x) y(0) = 0 (p) y(x) = 0 solves (p) y(x) = x x solves (p) 5/18?

9 Exstence and unqueness: card Lndelhöf Theorem Let the functon f(x, y) contnuous on a rectangle R = [x 0 a, x 0 + a] [y 0 b, y 0 + b]. Suppose, furthermore, that for any y 1, y 2 [y 0 b, y 0 + b] there exsts L > 0 such that f(x, y 1 ) f(x, y 2 ) L y 1 y 2 for each x [a, b]. Then the ntal value problem y (x) = f(x, y(x)) y(x 0 ) = y 0 (L) (vp) has a unque soluton defned n [x 0 δ, x 0 +δ] where δ = mn{a, b/m} beng M := max f(x, y) (x,y) R 6/18?

10 Condton (L) s sad Lpschtz contnuty condton. A suffcent condton to ensure that a functon f(x, y) s Lpschtz contnuos s that t admts partal dervatve wth respect to y whch s bounded and contnuos. 7/18?

11 Condton (L) s sad Lpschtz contnuty condton. A suffcent condton to ensure that a functon f(x, y) s Lpschtz contnuos s that t admts partal dervatve wth respect to y whch s bounded and contnuos. In fact assume f y (x, y) = f (x, y) y (Lagrange) theorem wth y between y 1 and y 2 L then from the mean value f(x, y 1 ) f(x, y 2 ) = f y (x, y) y 1 y 2 7/18?

12 Separable equatons A dfferental equaton s separable f t can be wrtten n the form y (x) = a(x) b (y(x)), (S) y(x 0 ) = y 0, where a(x) e b(y) are contnuous functons defned on ntervals I a and I b such that x 0 I a and y 0 I b 8/18?

13 Separable equatons A dfferental equaton s separable f t can be wrtten n the form y (x) = a(x) b (y(x)), (S) y(x 0 ) = y 0, where a(x) e b(y) are contnuous functons defned on ntervals I a and I b such that x 0 I a and y 0 I b In order to obtan exstence and unqueness for soluton to (S) we assume that b(y 0 ) 0 8/18?

14 Theorem Functon y(x) defned, mplctely by: y y 0 s the unque soluton to (S) dz b(z) = x x 0 a(s) ds (R) 9/18?

15 Theorem Functon y(x) defned, mplctely by: y y 0 s the unque soluton to (S) dz b(z) = x x 0 a(s) ds (R) roof. Recallng that b(y 0 ) = b(y(x 0 )) 0 there exsts a neghborhood I a of x 0 such that x I a (S) we can wrte = b(y(x)) 0. Then f y(x) solves y (x) b(y(x)) = a(x). (Sb) 9/18?

16 Integratng (Sb) from x 0 and x we obtan: x x 0 y (s) b(y(s)) ds = x x 0 a(s) ds. (Sc) 10/18?

17 Integratng (Sb) from x 0 and x we obtan: x x 0 y (s) b(y(s)) ds = x x 0 a(s) ds. (Sc) Now change varable puttng u = y(s) so from du = y (s) ds we fnd out: gvng thess (R) y y 0 y (s) b(u) du y (s) = x x 0 a(s) ds, 10/18?

18 roof s not complete: we have to show that soluton defned by (R) s the unque soluton to (S) Suppose y 1 (x) and y 2 (x) solutons of (S). Defne B(y) := y y 0 dz b(z) 11/18?

19 roof s not complete: we have to show that soluton defned by (R) s the unque soluton to (S) Suppose y 1 (x) and y 2 (x) solutons of (S). Defne B(y) := y y 0 dz b(z) thus d dx [B (y 1(x)) B (y 2 (x))] = = a(x)b (y 1(x)) b (y 1 (x)) a(x)b (y 2(x)) b (y 2 (x)) y 1(x) b (y 1 (x)) = 0. y 2(x) b (y 2 (x)) 11/18?

20 roof s not complete: we have to show that soluton defned by (R) s the unque soluton to (S) Suppose y 1 (x) and y 2 (x) solutons of (S). Defne thus B(y) := y y 0 d dx [B (y 1(x)) B (y 2 (x))] = = a(x)b (y 1(x)) b (y 1 (x)) dz b(z) a(x)b (y 2(x)) b (y 2 (x)) y 1(x) b (y 1 (x)) = 0. y 2(x) b (y 2 (x)) Notce that we used the fact that both y 1 (x) and y 2 (x) solve (S) 11/18?

21 But recallng that y 1 (x 0 ) = y 2 (x 0 ) = y 0, there exsts a neghborood N(x 0 ) of x 0 such that for each x N(x 0 ): B (y 1 (x)) B (y 2 (x)) = 0. 12/18?

22 But recallng that y 1 (x 0 ) = y 2 (x 0 ) = y 0, there exsts a neghborood N(x 0 ) of x 0 such that for each x N(x 0 ): B (y 1 (x)) B (y 2 (x)) = 0. On the other sde recallng the defnton of B(y) we see that: B (y 1 (x)) B (y 2 (x)) = y1 (x) y 2 (x) 1 b(t) dt. 12/18?

23 But recallng that y 1 (x 0 ) = y 2 (x 0 ) = y 0, there exsts a neghborood N(x 0 ) of x 0 such that for each x N(x 0 ): B (y 1 (x)) B (y 2 (x)) = 0. On the other sde recallng the defnton of B(y) we see that: B (y 1 (x)) B (y 2 (x)) = y1 (x) y 2 (x) 1 b(t) dt. Now use the mean value theorem for ntegrals to nfer that there exsts y x between y 1 (x) and y 2 (x) such that B (y 1 (x)) B (y 2 (x)) = 1 b(y x ) (y 1(x) y 2 (x)). 12/18?

24 But recallng that y 1 (x 0 ) = y 2 (x 0 ) = y 0, there exsts a neghborood N(x 0 ) of x 0 such that for each x N(x 0 ): B (y 1 (x)) B (y 2 (x)) = 0. On the other sde recallng the defnton of B(y) we see that: B (y 1 (x)) B (y 2 (x)) = y1 (x) 12/18? y 2 (x) 1 b(t) dt. Now use the mean value theorem for ntegrals to nfer that there exsts y x between y 1 (x) and y 2 (x) such that B (y 1 (x)) B (y 2 (x)) = 1 b(y x ) (y 1(x) y 2 (x)). Thess then follows from the assumpton b(y) 0, because t mples

25 y 1 (x) = y 2 (x) 13/18?

26 Separable equaton: Example y (x) = a(x) y(x) y(x 0 ) = y 0 (E1) beng a a contnuous functon. 14/18?

27 Separable equaton: Example y (x) = a(x) y(x) y(x 0 ) = y 0 (E1) beng a a contnuous functon. here b(y) = y so usng (S) y y 0 1 z dz = x x 0 a(s) ds = ln y y 0 = x x 0 a(s) ds 14/18?

28 Separable equaton: Example y (x) = a(x) y(x) y(x 0 ) = y 0 (E1) beng a a contnuous functon. here b(y) = y so usng (S) y y 0 1 z dz = x x 0 a(s) ds = ln y y 0 = ( x y = y 0 exp x 0 ) a(s) ds x x 0 a(s) ds 14/18?

29 For nstance f a(x) = x the ntal value problem 2 y (x) = x 2 y(x) y(0) = 1 (E1es) ha soluton y(x) = e x2 4 15/18?

30 Example y (x) = a(x) y 2 (x) y(x 0 ) = y 0 beng a a contnuous functon. (E2) 16/18?

31 Example y (x) = a(x) y 2 (x) y(x 0 ) = y 0 beng a a contnuous functon. here b(y) = y 2 so usng (S) y y 0 1 z 2 dz = x x 0 a(s) ds (E2) 16/18?

32 Example y (x) = a(x) y 2 (x) y(x 0 ) = y 0 beng a a contnuous functon. here b(y) = y 2 so usng (S) y y 0 1 z 2 dz = x x 0 a(s) ds (E2) 1 y + 1 y 0 = x x 0 a(s) ds 16/18?

33 so that y = 1 y 0 1 x x 0 a(s) ds 17/18?

34 so that y = 1 y 0 1 x x 0 a(s) ds For nstance f a(x) = 2x the ntal value problem y (x) = 2x y 2 (x) y(0) = 1 (E2ap) has soluton y = x 2 17/18?

35 Exercse Solve the separable dfferental equaton y (x) = 1 + y2 (x) 1 + x 2 y(0) = 2 (e) 18/18?

36 Exercse Solve the separable dfferental equaton y (x) = 1 + y2 (x) 1 + x 2 y(0) = 2 (e) Consder the more general problem y (x) = 1 + y2 (x) 1 + x 2 y(0) = a 18/18?

37 Exercse Solve the separable dfferental equaton y (x) = 1 + y2 (x) 1 + x 2 y(0) = 2 (e) Consder the more general problem y (x) = 1 + y2 (x) 1 + x 2 y(0) = a y a dz 1 + z 2 = x 0 ds = arctan y arctan a = arctan x 1 + s2 18/18?

38 Thus y = tan (arctan a + arctan x) 19/18?

39 Thus But snce y = tan (arctan a + arctan x) tan(α + β) = tan α + tan β 1 tan α tan β 19/18?

40 Thus But snce we nfer y = tan (arctan a + arctan x) tan(α + β) = tan α + tan β 1 tan α tan β y = a + x 1 ax 19/18?