Ritelli, Daniele

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Almae Matris Studiorum Campus

1/15 Pi?22333ML232CLAMFIM 2012-2013Modelli @ ClamfimLesson 8Marted`ı 20 Novembre 2012Riccati equation.professor Daniele Ritelliwww.unibo.it/faculty/daniele.ritelli2/15 Pi?22333ML232Riccati equationy′ = a(x) + b(x)y + c(x)y2 (R)2/15 Pi?22333ML232Riccati equationy′ = a(x) + b(x)y + c(x)y2 (R)Suppose that y1(x) solves (R).2/15 Pi?22333ML232Riccati equationy′ = a(x) + b(x)y + c(x)y2 (R)Suppose that y1(x) solves (R). Then assuming that the other solutionsof (R) have the form y(x) = y1(x) + u(x) being u(x) an unknownfunction to be determinated, then u(x) solves the associated Bernoulliequationu′ = (b(x) + 2c(x)y1(x))u+ c(x)u23/15 Pi?22333ML232Otherwise substitutiony(x) = y1(x) +1u(x)drives (R) into the linear equationu′ = −c(x)− (b(x) + 2c(x)y1(x))u4/15 Pi?22333ML232Example Knowing that y1(x) = 1 solves the Riccati equationy′ = −1 + xx+ y +1xy2 (r)show that the general solution to equation (r) isy(x) = 1 +x2exc+ (1− x)ex4/15 Pi?22333ML232Example Knowing that y1(x) = 1 solves the Riccati equationy′ = −1 + xx+ y +1xy2 (r)show that the general solution to equation (r) isy(x) = 1 +x2exc+ (1− x)exWe use the change of variable y(x) = 1 +1u(x)to obtain the linearequationu′ = −1x−(1 +2x)u5/15 Pi?22333ML232A(x) = −∫ (1 +2x)dx = −x− 2 lnx =⇒ eA(x) = e−xx25/15 Pi?22333ML232A(x) = −∫ (1 +2x)dx = −x− 2 lnx =⇒ eA(x) = e−xx2∫b(x)e−A(x)dx = −∫1xx2exdx = −∫xexdx = (1− x)ex5/15 Pi?22333ML232A(x) = −∫ (1 +2x)dx = −x− 2 lnx =⇒ eA(x) = e−xx2∫b(x)e−A(x)dx = −∫1xx2exdx = −∫xexdx = (1− x)exu(x) =e−xx2(c+ (1− x)ex) = ce−x + 1− xx2=c+ (1− x)exx2ex5/15 Pi?22333ML232A(x) = −∫ (1 +2x)dx = −x− 2 lnx =⇒ eA(x) = e−xx2∫b(x)e−A(x)dx = −∫1xx2exdx = −∫xexdx = (1− x)exu(x) =e−xx2(c+ (1− x)ex) = ce−x + 1− xx2=c+ (1− x)exx2exConclusion y = 1 +1u= 1 +x2exc+ (1− x)ex6/15 Pi?22333ML232ExampleUsing the fact that y1(x) = x solvesy′ = −x5 + yx+ x3y2 (E)find the general solution of (E)6/15 Pi?22333ML232ExampleUsing the fact that y1(x) = x solvesy′ = −x5 + yx+ x3y2 (E)find the general solution of (E)Putting y = x+ u we get1 + u′ = −x5 + x+ ux+ x3(x+ u)26/15 Pi?22333ML232ExampleUsing the fact that y1(x) = x solvesy′ = −x5 + yx+ x3y2 (E)find the general solution of (E)Putting y = x+ u we get1 + u′ = −x5 + x+ ux+ x3(x+ u)2Simplifying we get the Bernoulli equationu′ =2x5 + 1xu+ x3u2which can be solved with the change of variable v = u−17/15 Pi?22333ML232which gives− v′v2=2x5 + 1xv+x3v27/15 Pi?22333ML232which gives− v′v2=2x5 + 1xv+x3v2Multiplying by −v2 we get the linear differential equationv′ = −2x5 + 1xv − x3whose solution isv(x) =ce−2x55x− 12xbeing c an integration constant.7/15 Pi?22333ML232which gives− v′v2=2x5 + 1xv+x3v2Multiplying by −v2 we get the linear differential equationv′ = −2x5 + 1xv − x3whose solution isv(x) =ce−2x55x− 12xbeing c an integration constant.Therefore the general solution of the given Riccati equation isy =2x2ce−2x55 − 1+ x8/15 Pi?22333ML232Exercise Knowing that y1(x) = 1 is a solution ofy′(x) = −1− ex + exy(x) + y2(x) (?)show that the general solution to (?) isy(x) = 1 +e2x+exc− eex (ex − 1)Hint:∫e2x+exdx = eex(ex − 1)9/15 Pi?22333ML232Change of variabile in ordinary differential equationsGiveny′x = f(x, y(x)) (E)9/15 Pi?22333ML232Change of variabile in ordinary differential equationsGiveny′x = f(x, y(x)) (E)Consider the map (x, y) 7→ (ξ, η) where ξ = ξ(x, y) and η = η(x, y)are C1 anddetξx(x, y) ξy(x, y)ηx(x, y) ηy(x, y) 6= 09/15 Pi?22333ML232Change of variabile in ordinary differential equationsGiveny′x = f(x, y(x)) (E)Consider the map (x, y) 7→ (ξ, η) where ξ = ξ(x, y) and η = η(x, y)are C1 anddetξx(x, y) ξy(x, y)ηx(x, y) ηy(x, y) 6= 0Thus (E) is changed toDxη(x, y)Dxξ(x, y)= η′ξ =ηx + ηy y′ξx + ξy y′=ηx + f(x, y)ηyξx + f(x, y)ξy(T)10/15 Pi?22333ML232the last expression in (T) contains (x, y) to complete the coordinatechange we have to revert the map solving the systemξ(x, y) = ξη(x, y) = η =⇒x = xˆ(ξ, η)y = yˆ(ξ, η)and substituing in (T) the founded expressions for x and y11/15 Pi?22333ML232ExampleConsider the Riccati equationy′x = xy2 − 2yx− 1x3(R)with the change of variables (ξ, η) = (x2y, lnx) we findη′ξ =1x+(xy2 − 2yx− 1x3)× 02xy +(xy2 − 2yx− 1x3)x2=1xx3y2 − 1x=1x4y2 − 1 (C)12/15 Pi?22333ML232solve with respect to (x, y)ξ = x2yη = lnx =⇒x = eηy = ξe−2η12/15 Pi?22333ML232solve with respect to (x, y)ξ = x2yη = lnx =⇒x = eηy = ξe−2ηand then substituite in (C)η′ξ =1x4y2 − 1 =1e4ηξ2e−4η − 1 =1ξ2 − 1 =12(1ξ − 1 −1ξ + 1)(C1)12/15 Pi?22333ML232solve with respect to (x, y)ξ = x2yη = lnx =⇒x = eηy = ξe−2ηand then substituite in (C)η′ξ =1x4y2 − 1 =1e4ηξ2e−4η − 1 =1ξ2 − 1 =12(1ξ − 1 −1ξ + 1)(C1)Thus if ln c is an integration constant integrating (C1)η = ln c+12lnξ + 1ξ − 113/15 Pi?22333ML232eventually going back to the original variableslnx = ln c+12lnx2y + 1x2y − 1 = ln(c√x2y + 1x2y − 1)13/15 Pi?22333ML232eventually going back to the original variableslnx = ln c+12lnx2y + 1x2y − 1 = ln(c√x2y + 1x2y − 1)thereforex = c√x2y + 1x2y − 1 =⇒ y =c2 + x2x2(c2 − x2)14/15 Pi?22333ML23215/15 Pi?22333ML2320000599130 Colonnese Giusy 25 0000664942 Pagano Rossella 40000600534 Bignani Giulia 22 0000665721 Lenzi Francesca 30 +0000602324 Di Domenico Danila 23 0000666344 Rinaldi Elena 30 ++0000607428 Fontana Stefano 28 0000666534 Furlotti Luca 300000632288 Lazar Valeriu 27 0000666734 Valenti Valentina 270000633861 Donadel Riccardo 30 0000667189 Quici Laura 270000638471 Lalli Gianluca 27 0000668649 Paolucci Antonio 270000639500 Radovani Stivi 27 0000669193 Mariani Marco 260000640225 Baravelli Emanuela 18 0000672247 Ciabatti Francesco 300000644195 Marinelli Mariangela 26 0000673238 Rinaldi Francesco 280000645469 Fiocco Barbrara 24 0000673676 Brighi Alberto 270000645870 Finzi Marina 27 0000673783 Messina Dario 300000654365 Luna Costanza 27 0000675448 Ceccarelli Giacomo 300000661583 Di Girolamo Jonathan 27 ++ 0000676010 Lamuraglia Nicola 260000661663 Galato Gianluca 30 0000676306 Marconi Alberto 260000661808 Carloni Carlotta 30 0000676637 Caligiuri Giorgia 300000661814 Minghetti Riccardo 23 0000677441 Nanni Matteo 270000661924 Gentile Francesco 28 Faggioli Alessia 250000662032 Altini Luca 28 Graziosi Marco 300000662683 Manserra Valentina 30 Luisi Mattia 280000663017 Marzotto Mario 27 Metti Cristian 270000663084 Rivera Susana 4 Pentassuglia Gianni 270000664450 Salaroli Claudio 27 Rettino Soledania 27

Lezione 8, 20 novembre 2012

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Lezione 8, 20 novembre 2012

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