Linear Ordinary Differential Equations

Transcription

1 MTH.B402; Sect ) 2 Linear Ordinary Differential Equations Preliminaries: Matrix Norms. Denote by M n R) the set of n n matrix with real components, which can be identified the vector space R n2. In particular, the Euclidean norm of R n2 induces a norm 1.1) X E = tr t n XX) = i,j=1 x 2 ij on M n R). On the other hand, we let { } Xv 1.2) X M := sup ; v R n \ {0}, v where on the right-hand side denotes the Euclidean norm of R n. Lemma ) The map X X M is a norm of M n R). 2) For X, Y M n R), it holds that XY M X M Y M. 3) Let λ = λx) be the maximum eigenvalue of semi-positive definite symmetric matrix t XX. Then X M = λ holds. 4) 1/ n) X E X M X E. 5) The map M : M n R) R is continuous with respect to the Euclidean norm. 12. June, Revised 19. June 2018) Proof. Since Xv / v is invariant under scalar multiplications to v, we have X M = sup{ Xv ; v S n 1 }, where S n 1 is the unit sphere in R n. Here, the S n 1 x Ax R is a continuous function defined on a compact space, and so the map takes maximum. Thus, the right-hand side of 1.2) is welldefined. It is easy to verify that M satisfies the axiom of the norm. Since A := t XX is positive semi-definite the eigenvalues λ j j = 1,..., n) are non-negative real numbers. In particular, there exists an orthonormal basis [a j ] of R n satisfying Aa j = λ j a j j = 12,..., n). Let λ be the maximum eigenvalues of A, and write v = v 1 a v n a n. Then it holds that Xv, Xv = λ 1 v λ n v 2 n λ v, v, where, is the Euclidean inner product of R n. The equality of this inequality holds if and only if v is the λ-eigenvector, proving 3). Noticing the norm 1.1) is invariant under conjugations X t P XP P On)), we obtain X E = λ λ n by diagonalizing t XX by an orthogonal matrix P. Then we obtain 4). Hence two norms E and M induce the same topology M n R). In particular, we have 5). Preliminaries: Matrix-valued Functions. Lemma 1.2. Let X and Y be C -maps defined on a domain U R m into M n R). Then 1) XY ) = X Y + X Y,

2 ) MTH.B402; Sect. 1 2) 3) det X = tr X X ), and X 1 1 X = X X 1, where X is the cofactor matrix of X, and we assume in 3). Proposition 1.3. Assume two C matrix-valued functions Xt) and Ωt) satisfy 1.3) Then = Xt)Ωt), X ) = X ) det Xt) = det X 0 ) exp tr Ωτ) dτ holds. In particular, if X 0 GLn, R), 1 then Xt) GLn, R) for all t. Proof. By 2) of Lemma 1.2, we have d det Xt) = tr Xt) ) ) = tr Xt)Xt)Ωt) = tr det Xt)Ωt) ) = det Xt) tr Ωt). Here, we used the relation XX = X X = det X) id 2. Hence d ρt) 1 det Xt) ) = 0, where ρt) is the right-hand side of 1.4). 1 GLn, R) = {A M n R) ; det A 0}: the general linear group. 2 In this lecture, id denotes the identity matrix. MTH.B402; Sect ) 4 Proposition 1.4. Assume Ωt) in 1.3) is skew-symmetric for all t, that is, t Ω + Ω is identically O. If X 0 On) resp. X 0 SOn)) 3, Xt) On) resp. Xt) SOn)) for all t. Proof. By 1) in Lemma 1.2, d Xt X) = dx t ) dx t X + X = XΩ t X + X t Ω t X = XΩ + t Ω) t X = O. Hence X t X is constant, that is, if X 0 On), Xt) t Xt) = X ) t X ) = X 0 t X 0 = id. if X 0 On), proves the first case of the proposition. Since det A = ±1 when A On), the second case follows by continuity of det Xt). Preliminaries: Norms of Matrix-Valued functions. Let I = [a, b] be a closed interval, and denote by C 0 I, M n R)) the set of continuous functions X : I M n R). For any fixed number k, we define 1.5) X I,k := sup { e kt Xt) M ; t I } for X C 0 I, M n R)). When k = 0, I,0 is the uniform norm for continuous functions, which is complete. Similarly, one can prove the following in the same way: 3 On) = {A M n R) ; t AA = A t A = id}: the orthogonal group; SOn) = {A On) ; det A = 1}: the special orthogonal group.

3 ) MTH.B402; Sect. 1 Lemma 1.5. The map I,k : C 0 I, M n R)) is a complete norm. Linear Ordinary Differential Equations. We prove the fundamental theorem for linear ordinary differential equations. Proposition 1.6. Let Ωt) be a C -function valued in M n R) defined on an interval I. Then for each I, there exists the unique matrix-valued C -function Xt) = X t0,idt) such that 1.6) = Xt)Ωt), X ) = id. Proof. Uniqueness: Assume Xt) and Y t) satisfy 1.6). Then Y t) Xt) = = Y τ) X τ) ) dτ Y τ) Xτ) ) Ωτ) dτ holds. Hence for an arbitrary closed interval J I, ) Y t) Xt) M Y τ) Xτ) Ωτ) M dτ Y τ) Xτ) M Ωτ) M dτ = e kτ Y τ) Xτ) M e kτ Ωτ) M dτ MTH.B402; Sect ) 6 t Y X J,k sup Ω M e kτ dτ J sup = Y X J Ω M J,k e kt 1 e kt ) k holds for t J. Here, setting J = [, a] and k = 2 sup J Ω M, we have Y X J,k 1 2 Y X J,k, that is, Y X J,k = 0, proving Y t) = Xt) for t J. Similarly, on the interval J = [a, ], we can conclude Y = X on J setting k = 2 sup J Ω M. Since J and J are arbitrary, Y = X holds on I. Existence: Let J := [, a] I be a closed interval, and define a sequence {X j } of matrix-valued functions defined on I satisfying X 0 t) = id and 1.7) X j+1 t) = id + X j τ)ωτ) dτ j = 0, 1, 2,... ). Let k := 2 sup J Ω M. Then X j+1 t) X j t) M e kt X j X j 1 J,k sup J Ω M k X j τ) X j 1 τ) M Ωτ) M dτ = ekt 2 X j X j 1 J,k, and hence X j+1 X j J,k 1 2 X j X j 1 J,k, that is, {X j } is a Cauchy sequence with respect to J,k. Thus, by completeness

4 ) MTH.B402; Sect. 1 MTH.B402; Sect ) 8 Lemma 1.5), it converges to some X C 0 J, M n R)). By 1.7), the limit X satisfies X ) = id, Xt) = id + Xτ)Ωτ) dτ. Applying the fundamental theorem of calculus, we can see that X satisfies X t) = Xt)Ωt) = d/). Since J can be taken arbitrarily, existence of the solution on I {t } is proved. Existence of I {t } can be proved in the same way. So far, existence of a differentiable function Xt) satisfying 1.6) is obtained. Finally, we shall prove that X is of class C. Since X t) = Xt)Ωt), the derivative X of X is continuous. Hence X is of class C 1, and so is Xt)Ωt). Thus we have that X t) is of class C 1, and then X is of class C 2. Iterating this argument, we can prove that Xt) is of class C r for arbitrary r. Corollary 1.7. Let Ωt) be a matrix-valued C -function defined on an interval I. Then for each I and X 0 M n R), there exists the unique matrix-valued C -function Xt) = X t0,x 0 t) defined on I such that 1.8) = Xt)Ωt), X ) = X 0. In particular, X t0,x 0 t) is of class C in X 0 and t. Proof. We rewrite Xt) in Proposition 1.6 as Y t) = X t0,idt). Then the function 1.9) Xt) := X 0 Y t) = X 0 X t0,idt), is desired one. Conversely, assume Xt) satisfies the conclusion. Noticing Y t) is a regular matrix for all t because of Proposition 1.3, W t) := Xt)Y t) 1 satisfies Hence dw = dx Y 1 1 dy Xt)Y Y 1 = XΩY 1 XY 1 Y ΩY 1 = O. W t) = W ) = X )Y ) 1 = X 0. Hence the uniqueness is obtained. The final part is obvious by the expression 1.9). Proposition 1.8. Let Ωt) and Bt) be a matrix-valued C - functions defined on I. Then for each I and X 0 M n R), there exists the unique matrix-valued C -function defined on I satisfying 1.10) = Xt)Ωt) + Bt), X ) = X 0. Proof. Rewrite X in Proposition 1.6 as Y t) := X t0,idt). Then ) 1.11) Xt) = X 0 + Bτ)Y 1 τ) dτ Y t) satisfies 1.10). Conversely, if X satisfies 1.10), W := XY 1 satisfies X = W Y + W Y = W Y + W Y Ω, XΩ + B = W Y Ω + B,

5 ) MTH.B402; Sect. 1 MTH.B402; Sect ) 10 and then we have W = BY 1. Since W ) = X 0, to the parameter α. We set Z = Zt) as the unique solution of W = X 0 + Bτ)Y 1 τ) dτ. 1.14) dz = Z Ω + X Ω α j + B α j, Z0) = O. Thus we obtain 1.11). Theorem 1.9. Let I and U be an interval and a domain in R m, respectively, and let Ωt, α) and Bt, α) be matrix-valued C - functions defined on I U α = α 1,..., α m )). Then for each I, α U and X 0 M n R), there exists the unique matrixvalued C -function Xt) = X t0,x 0,αt) defined on I such that 1.12) Moreover, = Xt)Ωt, α) + Bt, α), X ) = X 0. I I M n R) U t,, X 0, α) X t0,x 0,αt) M n R) is C -map. Proof. Let Ωt, α) := Ωt +, α) and Bt, α) = Bt +, α), and let Xt) := Xt + ). Then 1.12) is equivalent to 1.13) d Xt) = Xt) Ωt, α) + Bt, α), X0) = X0, where α :=, α 1,..., α m ). There exists the unique solution Xt) = X id,x0, αt) of 1.13) for each α because of Proposition 1.8. So it is sufficient to show differentiability with respect Then it holds that Z = X/ α j Problem 1-1). In particular, by the proof of Proposition 1.8, it holds that Z = X t = Xτ) Ωτ, α) + Bτ, ) ) α) Y 1 τ)dτ Y t). α j α j α j 0 Here, Y t) is the unique matrix-valued C -function satisfying Y t) = Y t) Ωt, α), and Y 0) = id. Hence X is a C -function in t, α). Fundamental Theorem for Space Curves. As an application, we prove the fundamental theorem for space curves. A C -map γ : I R 3 defined on an interval I R into R 3 is said to be a regular curve if γ 0 holds on I. For a regular curve γt), there exists a parameter change t = ts) such that γs) := γts)) satisfies γ s) = 1. Such a parameter s is called the arc-length parameter. Let γs) be a regular curve in R 3 parametrized by the arclength satisfying γ s) 0 for all s. Then es) := γ s), ns) := γ s) γ, bs) := es) ns) s) forms a positively oriented orthonormal basis {e, n, b} of R 3 for each s. Regarding each vector as column vector, we have the

6 ) MTH.B402; Sect. 1 MTH.B402; Sect ) 12 matrix-valued function 1.15) Fs) := es), ns), bs)) SO3). in s, which is called the Frenet frame associated to the curve γ. Under the situation above, we set κs) := γ s) > 0, τs) := b s), ns), which is called the curvature and torsion, respectively, of γ. Using these quantities, the Frenet frame satisfies 0 κ 0 df 1.16) ds = FΩ, Ω = κ 0 τ. 0 τ 0 Proposition The curvature and the torsion are invariant under the transformation x Ax + b of R 3 A SO3), b R 3 ). Conversely, two curves γ 1 s), γ 2 s) parametrized by arc-length parameter have common curvature and torsion, there exist A SO3) and b R 3 such that γ 2 = Aγ 1 + b. Proof. Let κ, τ and F 1 be the curvature, torsion and the Frenet frame of γ 1, respectively. Then the Frenet frame of γ 2 = Aγ 1 +b A SO3), b R 3 ) is F 2 = AF 1. Hence both F 1 and F 2 satisfy 1.16), and then γ 1 and γ 2 have common curvature and torsion. Conversely, assume γ 1 andγ 2 have common curvature and torsion. Then the frenet frame F 1, F 2 both satisfy 1.16). Let F be the unique solution of 1.16) with F ) = id. Then by the proof of Corollary 1.7, we have F j t) = F j )Ft) j = 1, 2). In particular, since F j SO3), F 2 t) = AF 1 t) A := F 2 )F 1 ) 1 SO3)). Comparing the first column of these, γ 2s) = Aγ 1t) holds. Integrating this, the conclusion follows. Theorem 1.11 The fundamental theorem for space curves). For given C -functions κs) and τs) defined on I such that κs) > 0 on I. Then there exists a space curve γs) parametrized by arc-length whose curvature and torsion are κ and τ, respectively. Moreover, such a curve is unique up to transformation x Ax + b A SO3), b R 3 ) of R 3. Proof. We have already shown the uniqueness in Proposition We shall prove the existence: Let Ωs) be as in 1.16), and Fs) the solution of 1.16) with Fs 0 ) = id. Since Ω is skewsymmetric, Fs) SO3) by Proposition 1.4. Denoting the column vectors of F by e, n, b, and let γs) := s s 0 eσ) dσ. Then F is the frenet frame of γ, and κ, and τ are the curvature and torsion of γ, respectively Problem 1-2). Exercises 1-1 Verify that Z in 1.14) coincides with X/ α j. 1-2 Complete the proof of Theorem 1.11.