Modelli Clamfim Equazioni differenziali 22 settembre 2016
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1 CLAMFIM Bologna Modell Clamfm Equazon dfferenzal 22 settembre 2016 professor Danele Rtell danele.rtell@unbo.t 1/22?
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. 2/22?
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 2/22?
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 We say that dfferental equatons are studed by quanttatve or exact methods when they can be solved completely,.e. all the solutons are known and could be wrtten n closed form n terms of elementary functons or sometme specal functons (or nverses of these type of functons). 3/22?
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 4/22?
6 Soluton needs not to be unque. Consder the ntal value problem y (x) = 2 y(x) y(0) = 0 (p) 5/22?
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) 5/22?
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) 5/22?
9 Moreover for each par of real numbers α < 0 < β (x α) 2 f x < α ϕ α, β (x) = 0 f α x β (x β) 2 f x > β solves (p) 6/22?
10 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 7/22?
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. 8/22?
12 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 8/22?
13 The nterval of exstence of an IV s the largest tme nterval where the soluton s vald. For nstance the IV y = 1 y 2 y(0) = 0 s solved by y(x) = e2x 1 e 2x + 1 so the nterval of exstence s R 9/22?
14 The nterval of exstence of an IV s the largest tme nterval where the soluton s vald. For nstance the IV y = 1 y 2 y(0) = 0 s solved by y(x) = e2x 1 so the nterval of exstence s R e 2x + 1 whle the smlar IV y = 1 + y 2 y(0) = 0 s solved by y(x) = tan x so the nterval of exstence s ] π 2, π 2 9/22? [
15 Separable equatons A dfferental equaton s separable f t can be wrtten n the form { y (x) = a(x) b (y(x)), y(x 0 ) = y 0, (S) 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 10/22?
16 Separable equatons A dfferental equaton s separable f t can be wrtten n the form { y (x) = a(x) b (y(x)), y(x 0 ) = y 0, (S) 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 10/22?
17 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) 11/22?
18 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 Example Solve the ntal value problem y = 4x y y(2) = 1 (R) 11/22?
19 Separable equaton: Example y (x) = a(x) y(x) y(x 0 ) = y 0 (E1) beng a a contnuous functon. 12/22?
20 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 12/22?
21 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 12/22?
22 For nstance f a(x) = x the ntal value problem 2 y (x) = x 2 y(x) y(0) = 1 (E1es) has soluton y(x) = e x2 4 13/22?
23 Example y (x) = a(x) y 2 (x) y(x 0 ) = y 0 beng a a contnuous functon. (E2) 14/22?
24 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) 14/22?
25 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 14/22?
26 so that y = 1 y 0 1 x x 0 a(s) ds 15/22?
27 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 15/22?
28 Exercse Solve the separable dfferental equaton y (x) = 1 + y2 (x) 1 + x 2 y(0) = a 16/22?
29 Exercse Solve the separable dfferental equaton 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 16/22?
30 Thus y = tan (arctan a + arctan x) 17/22?
31 Thus But snce y = tan (arctan a + arctan x) tan(α + β) = tan α + tan β 1 tan α tan β 17/22?
32 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 17/22?
33 Homogeneous ( Equatons ) If f(αx, αy) = f(x, y), and x 0, y 0 R, x 0 0 such that f 1, y 0 x 0 y 0 x 0 usng the change of varable y(x) = x u(x) the dfferental equaton y (x) = f(x, y(x)) y(x 0 ) = y 0 18/22?
34 Homogeneous ( Equatons ) If f(αx, αy) = f(x, y), and x 0, y 0 R, x 0 0 such that f 1, y 0 x 0 y 0 x 0 usng the change of varable y(x) = x u(x) the dfferental equaton s transformed n... y (x) = f(x, y(x)) y(x 0 ) = y 0 18/22?
35 u (x) = u(x 0 ) = y 0 f(1, u(x)) u(x) x x 0 19/22?
36 Equazon omogenee y = y2 x 2 2xy y(1) = 1 20/22?
37 Equazon omogenee y = y2 x 2 2xy y(1) = 1 f(x, y) = y2 x 2 2xy 20/22?
38 Equazon omogenee y = y2 x 2 2xy y(1) = 1 f(x, y) = y2 x 2 2xy = f(αx, αy) = α2 y 2 α 2 x 2 2αxαy 20/22?
39 Equazon omogenee y = y2 x 2 2xy y(1) = 1 f(x, y) = y2 x 2 = f(αx, αy) = α2 y 2 α 2 x 2 2xy 2αxαy u = 1 + u2 y(x) = xu(x) = 2xu u(1) = 1 20/22?
40 Rsolvere l problema a valor nzal per l equazone omogenea y = y2 x 2 2xy y(1) = 2 21/22?
41 u 1 ( 2v ) 1 + v 2 dv = x 1 1 s ds 22/22?
42 u 1 ( 2v ) x 1 dv = 1 + v 2 1 s ds [ ln(1 + v 2 ) ] u 1 = ln x 22/22?
43 u 1 ( 2v ) x 1 dv = 1 + v 2 1 s ds [ ln(1 + v 2 ) ] u 1 = ln x ln 2 ln(1 + u 2 ) = ln x 22/22?
44 u 1 ( 2v ) x 1 dv = 1 + v 2 1 s ds [ ln(1 + v 2 ) ] u 1 = ln x ln 2 ln(1 + u 2 ) = ln x = 2 x = 1 + u2 22/22?
45 u 1 ( 2v ) x 1 dv = 1 + v 2 1 s ds [ ln(1 + v 2 ) ] u 1 = ln x ln 2 ln(1 + u 2 ) = ln x = 2 x = 1 + u2 Then soluton s 2 y(x) = x x 1 22/22?
46 u 1 ( 2v ) x 1 dv = 1 + v 2 1 s ds [ ln(1 + v 2 ) ] u 1 = ln x ln 2 ln(1 + u 2 ) = ln x = 2 x = 1 + u2 Then soluton s 2 y(x) = x x 1 Note that soluton s defned for 0 < x 2 and lm y(x) = 0 x 0 22/22?
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