Linear Differential EquationsNovember 7, 2007NotationWe work over an open interval:I := (a, b) := {t ∈ R : a < t < b}.Let Ck(I) := {f : I → R : f, f0· · · f(k)exist and are continuous }.Note that Ck(I) is a linear subspace of the vector space of allfunctions from I to R. We could also consider complex-valuedfunctions, which form a vector space over the field of complexnumbers.The main existence and uniqueness theoremTheoremGiven p1, p2, . . . pn∈ C0(I) and f ∈ Cn(I), letL(f ) := f(n)+ p1f(n−1)+ · · · + pnf .Then:IL is a linear transformation: Cn(I) → C0(I).IGiven g ∈ C0(I), t0∈ I, and y0, y1, . . . yn−1∈ R, there existsa unique f ∈ Cn(I) such thatIL(f ) = g, andIf(i)(t0) = yifor i = 0, 1, . . . , n − 1.ConsequencesRecall that Ker(L) := {f : L(f ) = 0}. That is, Ker(L) is the set ofsolutions to the homgeneous equationf(n)+ p1f(n−1)+ · · · + pnf = 0.CorollaryKer(L) is a linear subspace of the vector space Cn(I).CorollaryFor each t0∈ I,Et0: Ker(L) → Rnf 7→f (t0), f0(t0), . . . , f(n−1)(t0)is an isomorphism.CorollaryThe operator L : Cn(I) → C0(I) is surjective, and its kernel is alinear subspace of Cn(I) of dimension n.CorollaryIf (f1, · · · fn) is a linearly independent sequence in the kernelKer(L) of L then (f1, . . . , fn) forms a basis for Ker (L).The WronskianCorollaryLet (f1, . . . , fn) be a sequence of elements of Ker(L), and letW (f1, . . . , fn) := detf1f2· · · fnf01f02· · · f0n· · ·f(n−1)1f(n−1)2· · · f(n−1)n.Then the following conditions are equivalent:IFor some t0∈ I, W (f1, . . . , fn)(t0) 6= 0.IFor some t0∈ I, the sequence of vectorsEt0(f1), Et0(f2), . . . , Et0(fn)in Rnis linearly independent.I(f1, . . . , fn) is a basis for Ker (L).IW (f1, . . . , fn)(t) 6= 0 for all t ∈ I.Abel’s theoremTheoremIf (f1, . . . , fn) is a sequence of elements of Ker(L) andW := W (f1, . . . , fn), then W0+ pW = 0. Hence W = ce−P,where c is some constant and P0= p1.Note that this proves again the fact that W is either always zeroor never zero.Constant CoefficientsIf p1, . . . , pnare constant and g = 0, a fundamental solution setto the equation can be found fairly easily. One method is to tryexponential solutions of the form ert.TheoremSuppose r is a root of the polynomialXn+ p1Xn−1+ · · · pn= 0.Then ertbelongs to the kernel of L.If r has multiplicity m, thenthe functionsert, tert, . . . , tm−1ertall belong the the nullspace, and form a linearly independentsequence.ExampleConsider the equation y000− 2y00+ y0= 0. The correspondingcharaceristic equation is r3− 2r2+ r = 0.This factors:r3− 2r+r = r (r2 − 2r + 1) = r (r − 1)2So a basis for the solution space should be (e0, et, tet).Let’s compute the Wronskian:W :=1 ettet0 etet+ tet0 et2et+ tet=1 ettet0 etet+ tet0 0 et= e2tNote that W0− 2W = 0.Another exampleRecall the damped spring equation of the formy00+ py0+ qy = 0.The characteristic equation is r2+ pr + q = 0, which has roots:r =−p ±pp2− 4q2For example, if p = 2 and q = 5, we getr = −1 ± 2i,which leads to solutions (e−tsin2t,
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