Systems of Linear Differential EquationsNovember 14, 2007NotationWe work over an open interval:I := (a, b) := {t ∈ R : a < t < b}.Fix a positive integer n, and consider the vector space offunctions:V := {x: I → Rn}.Thus an element of V has the form:x(t) =x1(t)x2(t)· · ·xn(t).Let:IC0n(I) be the set of of those x such that each xiiscontinuous,IC1n(I) be the set of those x such that each derivative x0iexists and is continuous.These are linear subspaces of V .We could also allow complex-valued functions.Normal form for linear system of differential equationsLetIA be an n × n matrix of continuous functions on I.Iy be an n × 1 matrix of continuous functions on I, that is,an element of C0n(I).We consider a system of the formx01= a11x1+ a12x2+ · · · + a1nxn+ y1x02= a21x1+ a22x2+ · · · + a2nxn+ y2· · ·x0n= an1x1+ an2x2+ · · · + annxn+ ynWe can write this very simply using matrix notation:x0= Ax + yequivalentlyx0− Ax = y.The existence and uniqueness theoremTheoremGiven A and y as above, then for any t0∈ I and any x0∈ Rn,there is a unique x ∈ C1n(I) such thatIx0− Ax = y, andIx(t0) = x0.Why does this make sense?Think of a walk in the park, with signposts everywhere.Linear algebra point of viewGive A as above, consider the mappingL: C1n(I) → C0n(I) x 7→ x0− Ax.That is, L(x) := x0− Ax. Then L is a linear transformation.Furthermore, for each t0∈ I, the evaluation mappingEt0: C1n(I) → Rnis also a linear transformation.RestatementTheoremIThe linear transformationL : C1n(I) → C0n(I)is surjective.IFor any t0∈ I, the mapEt0: Ker(L) → Rnis an isomorphism.CorollaryThe dimension of the linear subspace Ker (L) ⊆ C1n(I) is a linearsubspace of dimension n.CorollaryLet (x1, x2, . . . xn) be any linearly independent sequence inKer(L). ThenI(x1, x2, . . . xn) is a basis for Ker (L).IAny x ∈ Ker(L) can be written uniquelyx = c1x1+ c2x2+ · · · + cnxn,for some ci∈ R.IIf x0is any solution to L(x) = y, then every solution x canbe written uniquelyx = c1x1+ c2x2+ · · · + cnxn+ x0,for some ci∈ R.The WronskianIf (x1, . . . xn) is a sequence of elements of Ker(L), letW := detx11x12· · · x1nx21x22· · · x2n· · ·xn1xn2· · · xnnCorollaryThen the following conditions are equivalent:IFor some t0∈ I, W (t0) 6= 0.IFor some t0∈ I, the sequence of vectorsEt0(x1), Et0(x2), . . . , Et0(xn)in Rnis linearly independent.I(x1, . . . , xn) is a basis for Ker (L).IW (t) 6= 0 for all t ∈
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