Ordinary differential equations

Transcription

1 CP3 REVISION LECTURE ON Ordinary differential equations 1 First order linear equations First order nonlinear equations 3 Second order linear equations with constant coefficients 4 Systems of linear ordinary differential equations

2 BASIC CONCEPTS Every differential equation involves a differential operator. Order of a differential operator: order of highest derivative contained in it Linearity: differential operator L is linear if for any two functions f and g L(αf +βg) = αlf +βlg with α and β any constants. Linearity superposition principle: If f 1 and f are solutions of Lf = 0, then any linear combination αf 1 +βf is also solution.

3 Lf = 0 Lf = h(x) 0 homogeneous differential equation inhomogeneous differential equation General solution of linear inhomogeneous ODE Lf = h is sum of a particular solution f 0 (the particular integral, PI) and the general solution f 1 of the associated homogeneous equation (the complementary function, CF): f = f 0 +f 1, i.e., GS = PI + CF. Initial conditions: n initial conditions needed to specify solution of linear ODE of order n

4 !"#$%&'#()#&*"+),#&)-.,/'+$& df General form : q( x) f h( x) dx + =. 4,$5&%'&$'*6)& 0+%)1#,/+1&,3%'#& Look for a function I(x) such that I(x) df dx +I(x)q(x)f!dIf dx =I(x)h(x)! I(x)=e x q(x')dx' Solution : f(x)= 1 I(x) x! x 0 I(x')h(x')dx'

5 June 006. Find the general solution of the differential equation 1 x dy dx y x = sinx June Solve the differential equation x(x+1) dy dx +y = x(x+1) e x September Solve the differential equation [Answ.: x(c cosx)] [Answ.: (c e x /)(x+1)/x] x dy dx +y = cosx [Answ.: (cosx+xsinx+c)/x ]

6 !"#$%&'#()#&*'*+"*),#&)-.,/'*$& 0+%1'.1&*'&)*)#,+&3)%1'(&4'#&$'+./'*&"$&,5,"+,6+)7&%1)#)&,#)&$)5)#,+&8,$)$&'4&& 91:$"8,++:&#)+)5,*%&*'*+"*),#&)-.,/'*$&;1"81&8,*&6)&$'+5)(&,*,+:/8,++:&<& =)9,#,6+)&)-.,/'*$& d y f( x) d x = g( y) ='+./'*&<& g ( y ) dy = f ( xdx )!! 0+3'$%&$)9,#,6+)&)-.,/'*$& >1,*)&5,#",6+)$&<&?'3')*)'.$&)-.,/'*$& >1,*)&5,#",6+)$&<& dy f( ax by) dx = + z = ax + by y =vx dy f( yx) dx = /. dz dx =a+bf(z) Separable dv dx =1 x (f(v)!v) Separable

7 !"#"$%&%"'()*'+),"-)."&(+/&+() dy x+ y+ 1 = dx x+ y+ Change variables : x =x'+a,y =y'+b dy' dx' =x'+y'+1+a+b x'+y'++a+b =x'+y', a=!3,b=1 Homogeneous x'+y' 01%)%-&"'334)%5'/6"&) dy n P( x) y Q( x) y, n 1 dx + =! Change variables : z =y 1!n dz dx +(1!n)P(x)z =(1!n)Q(x), First order linear

8 Exact equations : dy dx = φ/ x φ/ y for a given function φ(x,y) Then solution is determined by φ(x,y) = constant.

9 June Solve the differential equation dy dx = y(x+y) x homogeneous [Answ.: y = x/(1 c x)] June Find the general solution to the differential equation y dy dx = x 4x+3 September 009 separable 3. Solve the differential equation dy dx = x+y 1 x y [Answ.: y = c+x/ (3/8)ln(4x+3)] almost separable [Answ.: y (x+y) / = c]

10 1. Solve the differential equation dy dx +xy = xy Bernoulli [Answ.: y = 1/(1+ce x / )]. Solve the differential equation (y xcosy) dy dx = x+siny exact [Answ.: xsiny y +x / = c]

11 GEOMETRICAL INTERPRETATION OF SOLUTIONS General solution of a first-order ODE y = f(x,y) contains an arbitrary constant: y = (x,c) one curve in x,y plane for each value of c family of curves Example: y = x/y. separable equation y dy = x dx y / = x /+c i.e., x +y = constant : family of circles centered at origin y x Fig.1 Orthogonal family of curves : y = 1/f(x,y) Example : y = y/x y = cx

12 SUMMARY ON NONLINEAR FIRST-ORDER ODEs No general method of solution for 1st-order ODEs beyond linear case; rather, a variety of techniques that work on a case-by-case basis. Examples: i) Bring equation to separated-variables form, that is, y = α(x)/β(y); then equation can be integrated. Cases covered by this include y = ϕ(ax+by); y = ϕ(y/x). ii) Reduce to linear equation by transformation of variables. Examples of this include Bernoulli s equation. iii) Bring equation to exact-differential form, that is dy/dx = M(x,y)/N(x,y) such that M = φ/ x, N = φ/ y. Then solution determined from φ(x, y) = const.

13 SECOND-ORDER LINEAR ODEs f +p(x)f +q(x)f = h(x) General solution = PI + CF CF = c 1 u 1 +c u u 1 and u linearly independent solutions of the homogeneous equation nd-order linear ODEs with constant coefficients: a f +a 1 f +a 0 f = h(x) Complementary function CF by solving auxiliary equation Particular integral PI by trial function with functional form of the inhomogeneous term

14 !"#$%&'$(&"(')*%"+('",-+.$%'/*01'#$%0+%0'#$"3#*"%0' d f df Lf = a a 1 a0f h( x) dx + dx + =. d f 4$56)"5"%0+(7'8-%#.$%' Lf 0 =a 0 dx +a df 0 1 dx +a f = Try f 0 =e mx am + am+ a = m ±! a1 ± a1! 4aa 0 ", a 1 0 9:-;*)*+(7<'",-+.$%' a! 4 aa " +,0,! 4$56)"5"%0+(7'8-%#.$%' f 0 =A + e m + x +A! e m! x. =/$'#$%0+%0'$8'*%0">(+.$%'

15 !"#$%&'$(&"(')*%"+('",-+.$%'/*01'#$%0+%0'#$"3#*"%0' d f df Lf = a a 1 a0f h( x) dx + dx + =. d f 4$56)"5"%0+(7'8-%#.$%' Lf 0 =a 0 dx +a df 0 1 dx +a f = Try f 0 =e mx am + am+ a = m ±! a1 ± a1! 4aa 0 ", 9:-;*)*+(7<'",-+.$%' a a! 4 aa " +,0,! 1 0 4$56)"5"%0+(7'8-%#.$%' f 0 =Ae mx +Bxe mx ="6"+0"&'($$0'

16 !"#$%&'$(&"(')*%"+('",-+.$%'/*01'#$%0+%0'#$"3#*"%0' d f df Lf = a a 1 a0f h( x) dx + dx + =. 6$78)"7"%0+(9':-%#.$%' Lf 0 =a d f 0 dx +a 1 df 0 dx +a 0 f 0 =0. d f 4+(.#-)+('*%0"5(+)' Lf 1 =a 1 dx +a df 1 1 dx +a f =h. 0 1 General solution : f0 + f1

17 June Solve the differential equation 3 d y dx + dy dx 4y = 0 with initial conditions y(0) = 0 and dy/dx(0) = 7. September Find the general solution of the differential equation d y dt +7dy dt +1y = 3exp(t) [Answ.: 3(e x e 4x/3 )] [Answ.: c 1 e 4t +c e 3t +e t /10] June Find the general solution of the differential equation d y dx +3dy dx +y = e x [Answ.: c 1 e x +c e x +xe x ]

18 10. A mass m is suspended on a spring. Its displacement y as a function of time t is described by the equation September 009 d y dt + γdy dt + ω 0y = F 0 m eiωt, where γ, ω 0 and F 0 are constants. Initially no driving force is present (F 0 = 0). Determine the solution to the equation when γ < ω0, subject to the initial conditions y = y 0 and dy dt = 0 at t = 0. The driving force is now applied. Find the time dependence of y at sufficiently long times such that all transients have died away. Sketch as a function of ω the amplitude of y and its phase φ relative to the driving force. When ω = ω 0, show that the average power supplied by the driving force is P av = F 0 4mγ. CPSC 496 4

19 y +γy +ω 0y = (F 0 /m) e iωt. CF (F 0 = 0): Characteristic eq. λ +γλ+ω0 = 0 λ ± = γ ± γ ω0 γ < ω0 : λ ± = γ ±i ω0 γ }{{} ω So CF = e γt [Acosω t+bsinω t]. Initial conditions y(0) = y 0, y (0) = 0 A = y 0, Bω Aγ = 0 B = y 0 γ/ω Thus y(t) = y 0 e γt [cosω t+(γ/ω )sinω t].

20 PI (F switched on): Trial function y = Ce iωt C( ω +γiω +ω 0) = F 0 /m So C = F 0 /m ω 0 ω +γiω = C eiφ where C = F 0 /m (ω 0 ω ) +4γ ω (amplitude) tanφ = γω ω ω 0 (phase)

21 Power P supplied by the driving force F : P = F dy dt where F = F 0 e iωt, y(t) = C e i(ωt+φ) dy, dt = C iωei(ωt+φ). Using the explicit expressions determined previously for amplitude C and phase φ we get P = ωf0/m sin(ωt+φ)+sinφ (ω 0 ω ) +4γ ω where sinφ = tanφ 1+tan φ = γω (ω 0 ω ) +4γ ω Taking the average power P ( sin(ωt+φ) = 0) P = γω F 0/m (ω 0 ω ) +4γ ω Thus P = F 0 4mγ for ω = ω 0.

22 SYSTEMS OF ORDINARY DIFFERENTIAL EQUATIONS more than 1 unknown function: y 1 (x),y (x),...,y n (x) set of ODEs that couple y 1,...,y n physical applications: systems with more than 1 degree of freedom. dynamics couples differential equations for different variables. Example. System of first-order differential equations: y 1 = F 1 (x,y 1,y,...,y n ) y = F (x,y 1,y,...,y n ) y n = F n (x,y 1,y,...,y n )

23 Systems of linear ODEs with constant coefficients can be solved by a generalization of the method seen for single ODE: General solution = PI + CF Complementary function CF by solving system of auxiliary equations Particular integral PI from a set of trial functions with functional form as the inhomogeneous terms

24 June 007 The variables ψ(z) and φ(z) obey the simultaneous differential equations 3 dφ +5ψ = z dz 3 dψ +5φ = 0. dz Find the general solution for ψ. Differentiating nd equation wrt z gives 0 = 3 d ψ dφ +5 dz dz = 3 d ψ dz (z 5ψ) d ψ i.e. dz 5 9 ψ = 10 9 z Solve this nd-order linear ODE with constant coefficients:

25 CF: Auxiliary equation m 5/9 = 0 m ± = ±5/3. So CF = Ae 5z/3 +Be 5z/3. PI: Trial function ψ 0 = C 1 z +C 0 (5/9)C 1 z (5/9)C 0 = 10/9z C 1 = /5, C 0 = 0 PI = z/5. General solution for ψ : ψ(z) = CF+PI = A e 5z/3 +B e 5z/3 + 5 z.

26 June Consider the coupled differential equations du + au bv = f dt dv + av + bu = 0 dt where a,b and f are constants. i) Solve them for f = 0, subject to the boundary conditions u = 0 and v = v 0 when t = 0. [10] ii) Solve them for f 0, subject to the boundary conditions u = v = 0 when t = 0, and write down the steady state solutions. [10]

27 Using notation D = d/dt, (D +a)u bv = f bu+(d +a)v = 0 Apply D +a on 1st eq. and multiply nd eq. by b: (D+a) u b(d+a)v = (D+a)f b u+b(d+a)v = 0 Add the two equations: [(D +a) +b ]u = af which is nd-order ODE with constant coefficients u +au +(b +a )u = af.

28 General solution CF: Auxiliary equation λ +aλ+(b +a ) = 0 λ ± = a±ib. So CF = e at [Acosbt+Bsinbt]. PI: Trial function: constant u 0 = C C = af/(a +b ). General solution for u : u(t) = CF+PI = e at [Acosbt+Bsinbt]+af/(a +b ). General solution for v : v(t) = b 1 [(D+a)u f] = e at [ Asinbt+Bcosbt] bf/(a +b ).

29 i) f = 0,u(0) = 0 and v(0) = v 0 A = 0,B = v 0 Therefore u(t) = v 0 e at sinbt, v(t) = v 0 e at cosbt. ii) f 0,u(0) = v(0) = 0 A = af/(a +b ),B = bf/(a +b ) Therefore u(t) = f/(a +b )[e at ( acosbt+bsinbt)+a], v(t) = f/(a +b )[e at (asinbt+bcosbt) b]. Steady state solutions t : u = af/(a +b ), v = bf/(a +b ).