MIT OpenCourseWarehttp://ocw.mit.edu 18.034 Honors Differential Equations Spring 2009 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.� � �LECTURE 12. SOLUTION BASES We present the results pertaining to the linear differential equation (12.1) L0y = y(n) + p1y(n−1) + + pn−1y� + pny· · · with constant coefficients. Some results we establish apply to equations with variable coefficients. Let p(λ) = λn + p1λ(n−1) + + pn−1λ + pn be the characteristic polynomial corresponding to · · · L0. By the fundamental theorem of algebra∗, p is factored into linear factors in the complex fields as p(λ) = (λ − λ1)k1 (λ − λ2)k2 (λ − λm)km ,· · · where λj are (complex) roots of p(λ) and kj � 1 are the multiplicity of λj . We recall from the last lecture that the functions tr e λj t , where r = 0, 1, , λj−1, and j = 1, 2, , m, · · · · · · are (complex) solutions of L0y = 0. Moreover, each pair of complex conjugate roots, λj = ¯ µ + iνj, λ = µj − iνj , with µj , νj ∈ R, gives real solutions treµj t cos νj t, treµj t sin νj t, where r = 0, 1, , kj−1, of L0y = 0.· · · Our goal is to show that all solutions of L0y = 0 are linear combinations of these n solutions. Namely, these solutions form a basis of solutions of L0y = 0. Linear independence. There are two notions of linear independence, depending on the scalar field. A set os n real or complex functions f1, , fn defined on an interval I is said linearly · · · independent over the real field if c1f1(t) + c2f2(t) + + cnfn(t) = 0 on I, cj ∈ R, implies cj = 0 for all j. · · · We may define, similarly, a set of real or complex functions to be linearly independent over the complex field. Lemma 12.1. A set of real-valued functions on an interval I is linearly independent over the real field if and only if it is linearly independent over the complex field. Next, we state the main result of this subsection. Lemma 12.2. Any set of functions of the form (12.2) frj (t) = tr e λj t , j = 1, 2, , n, · · · where the r are nonnegative integers and λj ∈ C, is linearly independent on any nonempty open interval, unless two or more of the functions are identical. Proof. Suppose that frj (t) are all distinct. Suppose that crj frj(t) = 0 on an open interval of t, but crj = 0 � for some r and j. Fix such a j, and choose R to be the largest r such that crj = 0� . We form a linear differential operator of constant coefficient p(D) = (D − λj )R (D − λi)ki+1 , i=j ∗It was first proved by Carl Friedrich Gauss. 1� � � �� �� �where �ki is the largest r such that treλit belongs to te set of functions in (12.2). It is obvious that p(D)( crj frj ) = p(D)(0) = 0. On the other hand, p(D)( crj frj ) = (D − λi)ki+1(D − λj )R( crj frj ) i=j =cRj (D − λi)ki+1(D − λj )R(tR e λj t) i=j =cRj (R!) (λj − λi)ki+1 e λit = 0� . i=j A contradiction then proves the theorem. � Corollary 12.3. The differential equation L0y = 0, where L0 is given in (12.1), has at least n linearly independent, real or complex solutions of the form treλt . The differential equation L0y = 0, where pj are real constants, has a set of n solutions of the form tr e λt , or tr eµt cos νt, tr eµt sin νt which is linearly independent over the real field in any nonempty interval. Solution bases. We now show that all solutions of the homogeneous equation L0y = 0, where L0 is given in (12.1) and pj are real, are linear combinations of the special solutions obtained in Corollary 12.3. To this end, we establish some results for the more general equation (12.3) Ly = y(n) + p1(t)y(n−1) + + pn−1(t)y� + pn(t)y = 0,· · · where pj (t) are real-valued continuous functions on an interval I. The solution set, denoted by N(L), of Ly = 0 is a collection of solutions of Ly = 0. Exercise. Show that N (L) forms a linear subspace of Cn(I). A basis of solutions of Ly = 0 is then defined as a basis of N(L), as a linear space. In other words, any solution of Ly = 0 is uniquely expressed as a linear combination of members in the basis of solutions. We associate to the differential equation Ly = 0, where L is given in (12.3), a transformation T : N(L) Rn , T y = (y(t0), y�(t0), , y n−1(t0)).→ · · · It is clear that T is linear. We show that T is one-to-one. That is, Ly = 0 has uniqueness. Lemma 12.4. (Uniqueness) If y is a real or complex solution of Ly = 0, where L is given in (12.3) and pj are real-valuead continuous in the closed interval I containing t0, and if y(t0) = y�(t0) = = y(n−1)(t0) = 0,· · · then y(t) = 0 for all t ∈ I. Proof. The proof is similar to that for the second-order equations. If y is a real solution, then let E(t) = y 2(t) + (y�(t))2 + + (y(n−1)(t))2 · · · dEand derive a differential inequality � KE for some constant K > 0. The detail is left as an dt exercise. If y is a complex solution, then its real and imaginary parts are both real solutions of Ly = 0 satisfying the initial conditions. This completes the proof. � Lecture 12 2 18.034 Spring 2009� Furthermore, if we show that T is onto, that is, for any n-vector (y0, y1, , yn−1) ∈ Rn, the · · · differential equation Ly = 0 has a solution satisfying the initial conditions y(t0) = y0, , y(n−1)(t0) = yn−1,· · · then it implies that T is an isomorphism. In particular, the dimension of N(L) is n. Thus, a basis of solutions of Ly = 0 consists of n functions. If the coefficients of L are real constants, that is, L = L0 where L0 is in (12.1), then Ly = 0 has a n linearly independent (real) solutions of the form tr e λt , tr eµt cos νt, tr eµt sin νt. In this case, these n solutions form a basis of solutions of Ly = 0. This result holds true for equations with real-variable coefficients. Exercise. If y1, , yn are n linearly independent solutions of Ly = 0, where L is given in (12.3) · · · and pj (t) are real-valued contiuous on an interval t0 ∈ I, then show that for given arbitrary real nnumbers y0, y1, , yn−1 there exist unique constant c1, , cn such …
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