ODE classification. February 7, Nasser M. Abbasi. compiled on Wednesday February 07, 2018 at 11:18 PM

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1 ODE classification Nasser M. Abbasi February 7, 2018 compiled on Wednesday February 07, 2018 at 11:18 PM 1

2 2 first order b(x)y + c(x)y = f(x) Integrating factor or separable (see detailed flow chart for 1D on page 3 if needs more information about 1D) non-homogeneous non constant coefficients Euler ODE ax 2 y + bxy + cy = f(x) homogeneous ax 2 y + bxy + cy = f(x) ax 2 y + bxy + cy = 0 solve the homogeneous first and find y 1,y 2. Can only use variation of parameters since not constant coefficients First find Wronskina W and then =y 1 u 1 + y 2 u 2 y2 f(x) u 1 = aw y1 f(x) u 2 = aw one root, repeated y 1 =x r y 2 =x r ln x Note: y 2 is found using reduction of order method. Let y h = Ax r, plug into ODE and find charaterestic equation ar(r 1) + br + c = 0 and find its roots r 1, r 2 Two distinct real roots r 1, r 2 y 1 =x r 1 y 2 =x r 2 Two complex conjugate roots α ± iβ y 1 =x (α+iβ) y 1 =x (α iβ) For complex roots, it can be simplified as follows. y h =x α ( c 1 x iβ + c 2 x iβ) ( ) =x α c 1 e ln xiβ ln x iβ + c 2 e ( =x α c 1 e iβ ln x iβ ln + c 2 e x) =x α (C 1 cos(β ln x) + C 2 sin(β ln x)) second order a(x)y + b(x)y + c(x)y = f(x) NO Use variation of parameters First find Wronskina W and then non-homogeneous ay + by + cy = f(x) solve the homogeneous first and find y 1,y 2. Does f(x) contain only exponential, trig, constants or polynomials? constant coefficients YES Use Undetermined coefficients (guess) Guess form based on f(x) from lookup table ay + by + cy = f(x) one root, repeated y 1 =e rx y 2 =xe rx homogeneous ay + by + cy = 0 Let y h = Ae rx, plug into ODE and find charaterestic equation ar 2 + br + c = 0 and find its roots r 1, r 2 Two distinct real roots r 1, r 2 y 1 =e r 1x y 2 =e r 2x =y 1 u 1 + y 2 u 2 Note: y 2 is found using reduction of or- 1 = y2 f(x) u aw y der method. 1 or y 2 present in y1 f(x) u 2 = f(x)? aw NO YES y Multiply guess h by extra x second order with one solution given Reduction of Order If the second order ODE is given along with one solution y 1 and asked to find general solution, then the homogenous solution can be found by assuming y h = v(x)y 1 (x) and plugging this back into the given ODE and solving for v(x). Two complex conjugate roots α ± iβ y 1 =e αx cos(βx) y 2 =e αx sin(βx) =e αx (c 1 cos(βx) + c 2 sin(βx)) found y h General solution y = y h +. If initial conditions are given, now we find c 1, c 2. Plugin guess found back into original ODE and determine the unknown coefficients in General solution y = y h +. If initial conditions are given, now we find c 1, c 2.

3 system of first order ODE s nonhomogeneous system. (A constant) x = Ax + G(t) First solve the homogeneous part to find x 1 and x 2 homogeneous system. (A constant) x = Ax Variation of parameters Undetermined coefficients Find eigenvalues λ by solving A λi = 0 3 Find fundamental matrix Φ = [ x 1 x 1 ] x p = Φ Φ 1 G(t) dt YES Adjust x p by multiplying by t, see example May 4, 2017 for illustration. Only works if G(t) contains sums and products of exponentials and/or polynomials of degree 0 and A is constant. Everything else, use Variation of parameters. Write G(t) = g 1 (t) + g 2 (t) guess x p (t) based on form of g 1 (t), g 2 (t) similar to scalar case but using vectors for constants. So end up with something like x p (t) = a + be t (see example May 4, 2017) is x 1 or x 2 in x p? NO Plug x p (t) into x p (t) = Ax p (t) + g 1 (t) + g 2 (t). Balance terms and solve for vectors a and b x 1 λ 1 is complete (rare case) Second eigenvector v2 is found by solving from same eigenvalue (A λ 1 I) v 2 = 0 Find second L.I. solution x 2 = v 2 e λ 1t λ is real and repeated Find first eigenvector v 1 by solving (A λ 1 I) v 1 = 0 Find first L.I. solution x 1 = v 1 e λ 1t Is the eigenvalue λ 1 complete? i.e. can we find the second v 2 from same eigenvalue λ 1? (in a 2 2 system, this will only happen when the two first order equations are decoupled). Hence, most of the time, the eigenvalue will not be complete. x 2 build homogeneous solution x h = c 1 x 1 + c 2 x 2 λ 1 is defective (common case) Solve for v 2 from (A λ 1 I) v 2 = v 1 Where v 1 was found in above step. This step requires Gaussian elimination. Find second L.I. solution x 2 = (tv 1 + v 2 ) e λ 1t x 2 λ 1, λ 2 are real and distinct Find first eigenvector v 1 by solving (A λ 1 I) v 1 = 0 Find second eigenvector v 2 by solving (A λ 2 I) v 2 = 0 build homogeneous solution x 1 x 1 = v 1 e λ 1t x 2 = v 2 e λ 2t x h = c 1 x 1 + c 2 x 2 λ 1 = α + iβ, λ 2 = α iβ are complex conjugates Find first eigenvector v 1 by solving (A λ 1 I) v 1 = 0 Find second eigenvector v 2 by solving (A λ 2 I) v 2 = 0 build homogeneous solution x 1 = v 1 e (α+iβ)t x 2 = v 2 e (α iβ)t x h = c 1 x 1 + c 2 x 2 convert to real basis x 1 = Re{x 1 } x 2 = Im{x 1 } x h = c 3 x 1 + c 4 x 2 x p x h x = x h + x p p8 sysem.ipe may 5, Nasser M. Abbasi

4 ode first order y = f(x, y, y 2,... ) linear y = f(x, y) non-linear separable if ODE has the form y = P (x)q(y). Example y = xy, then it is separable. dy = P (x) Q(y) dy Q(y) = P (x) and solve for y integrating factor if ODE has this form y + P (x)y = Q(x) Example y + xy = 2x, then Integrating factor µ = e P (x), hence d(µy) = µq(x) µy = µq(x) + c ODE is exact or can be made exact. Write ODE in form M(x, y) + N(x, y)dy = 0 M y = N x? Example 2xy + (x 2 + cos y) dy = 0 Set up these two equations du = M du dy = N From first equation find U = M + f(y) where f(y) is some arbitrary function of y only. Using second equation now solve for f(y). The final solution is U = c where c is constant. no Example (2y 2 x y) + xdy = 0. Find integrating factor µ A = M y N x N is A function of x alone? no µ = e A Multiply the original ODE by the found integrating factor µ And follow same method M = µm N = µn B = N x M y M is B function of y alone? µ = e B dy We come here if ODE is not separable, not exact, and can not be made exact. Example is y = ye y x +y x. With ODE in form dy = f(x, y), replace y in RHS with y = ux where u = u(x). If RHS now only has u in it, then ODE is homegeneous. Apply transformation y = ux it will become separable. no R = N x M y xm yn is R function of t = xy alone? µ = e R dt no Try other methods. see above Not linear Canonical form y = f 0 + f 1 y + f 2 y 2 + f 3 y 3 f 0 = 0, f 3 = 0 Bernoulli. Example y = xy + y 2 f 3 = 0 Riccati. Example y = 1 + xy + y 2 f 3 0 Abel first kind. Example y = 1+xy+y 2 +xy 3 start by divinding by y 2. Then let u be the coefficient of x term left on the right. Example using the above. y y 2 = x y Let u = 1 y or y = 1 u, hence y = u u 2, subtitute back to obtain u + xu = 1 which is solved using integrating factor. Hence y is now found. use substitution y = + 1/u know particular solution? No use substitution y = u uf2 this leads to second order ODE which can be solved To do. Few Abel ODE can be solved exactly. Nasser M. Abbasi August 20, d1.ipe 4

5 1 examples 1.1 second order, constant coeff second order, constant coeff. homogeneous second order, constant coeff. homogeneous, one root repeated y 2y + 1 = 0 Let y = Ae rx and plug into the above and simplify, we obtain the charaterstic equation r 2 2r + 1 = 0 (r 1) 2 = 0 r = 1 Repated root. Hence the two L.I. basis solutions are And the homogeneous solution is y 1 = e x y 2 = xe x y h = c 1 y 1 + c 2 y 2 = c 1 e x + c 2 e x second order, constant coeff. homogeneous, two real distinct roots y + y 2y = 0 Let y = Ae rx and plug into the above and simplify, we obtain the charaterstic equation r 2 + r 2 = 0 (r 1) (r + 2) = 0 r 1 = 1 r 2 = 2 Hence the L.I. basis solutions are And the homogeneous solution is y 1 = e x y 2 = e 2x y h = c 1 y 1 + c 2 y 2 = c 1 e x + c 2 e 2x second order, constant coeff. homogeneous, two complex conjugate roots y 6y + 13y = 0 Let y = Ae rx and plug into the above and simplify, we obtain the charaterstic equation r 2 6r + 13 = 0 Whose roots are r 1 = 3 + 2i r 2 = 3 2i 5

6 Hence the L.I. basis solutions are y 1 = e (3+2i)x y 2 = e (3 2i)x the homogeneous solution is y h = c 1 y 1 + c 2 y 2 = c 1 e (3+2i)x + c 2 e (3 2i)x This can be converted to real basis using Euler relation which results in y h = C 1 e 3x cos 2x + C 2 e 3 sin 2x = e 3x (C 1 cos 2x + C 2 sin 2x) 6