1 周期線形系とモノドロミー行列 定義 1.1. K ∈ { R , C } \mathbb K\in\{\R,\C\} K ∈ { R , C } d ∈ N ≥ 1 d\in\NN d ∈ N ≥ 1 T > 0 T>0 T > 0 A ∈ L l o c 1 ( R ; K d × d ) A\in L^1_{\mathrm{loc}}(\R;\mathbb K^{d\times d}) A ∈ L loc 1 ( R ; K d × d ) t ∈ R t\in\R t ∈ R A ( t + T ) = A ( t ) A(t+T)=A(t) A ( t + T ) = A ( t ) x ′ = A ( t ) x \boldsymbol x'=A(t)\boldsymbol x x ′ = A ( t ) x T T T T-periodic linear system )
補題 1.2. K ∈ { R , C } \mathbb K\in\{\R,\C\} K ∈ { R , C } d ∈ N ≥ 1 d\in\NN d ∈ N ≥ 1 T > 0 T>0 T > 0 x ′ = A ( t ) x \boldsymbol x'=A(t)\boldsymbol x x ′ = A ( t ) x T T T Φ 0 \Phi_0 Φ 0 Φ 0 ( 0 ) = I \Phi_0(0)=I Φ 0 ( 0 ) = I R \R R M = Φ 0 ( T ) M=\Phi_0(T) M = Φ 0 ( T )
任意の基本行列Φ \Phi Φ t ↦ Φ ( t + T ) t\mapsto\Phi(t+T) t ↦ Φ ( t + T ) t ∈ R t\in\R t ∈ R Φ ( t + T ) = Φ ( t ) C Φ \Phi(t+T)=\Phi(t)C_\Phi Φ ( t + T ) = Φ ( t ) C Φ C Φ ∈ K d × d C_\Phi\in\mathbb K^{d\times d} C Φ ∈ K d × d C Φ = Φ ( 0 ) − 1 M Φ ( 0 ) C_\Phi=\Phi(0)^{-1}M\Phi(0) C Φ = Φ ( 0 ) − 1 M Φ ( 0 )
任意のs ∈ R s\in\R s ∈ R x ′ = A ( t ) x \boldsymbol x'=A(t)\boldsymbol x x ′ = A ( t ) x x : R → K d \boldsymbol x\colon\R\to\mathbb K^d x : R → K d x ( s + T ) = Φ 0 ( s ) M Φ 0 ( s ) − 1 x ( s ) \boldsymbol x(s+T)=\Phi_0(s)M\Phi_0(s)^{-1}\boldsymbol x(s) x ( s + T ) = Φ 0 ( s ) M Φ 0 ( s ) − 1 x ( s )
すべてのt ∈ R t\in\R t ∈ R n ∈ Z n\in\Z n ∈ Z Φ 0 ( t + n T ) = Φ 0 ( t ) M n \Phi_0(t+nT)=\Phi_0(t)M^n Φ 0 ( t + n T ) = Φ 0 ( t ) M n
証明. Φ \Phi Φ Ψ ( t ) = Φ ( t + T ) \Psi(t)=\Phi(t+T) Ψ ( t ) = Φ ( t + T ) Ψ \Psi Ψ Φ ′ ( s ) = A ( s ) Φ ( s ) \Phi'(s)=A(s)\Phi(s) Φ ′ ( s ) = A ( s ) Φ ( s ) s s s N 1 N_1 N 1 A ( t + T ) ≠ A ( t ) A(t+T)\ne A(t) A ( t + T ) = A ( t ) t t t N 2 N_2 N 2 t ∉ ( N 1 − T ) ∪ N 2 t\notin(N_1-T)\cup N_2 t ∈ / ( N 1 − T ) ∪ N 2
Ψ ′ ( t ) = Φ ′ ( t + T ) = A ( t + T ) Φ ( t + T ) = A ( t ) Ψ ( t ) \Psi'(t)=\Phi'(t+T)=A(t+T)\Phi(t+T)=A(t)\Psi(t) Ψ ′ ( t ) = Φ ′ ( t + T ) = A ( t + T ) Φ ( t + T ) = A ( t ) Ψ ( t ) である。det Ψ ( t ) = det Φ ( t + T ) ≠ 0 \det\Psi(t)=\det\Phi(t+T)\ne0 det Ψ ( t ) = det Φ ( t + T ) = 0 Ψ \Psi Ψ §E13.9 定理 2.2 によりΦ \Phi Φ S A ( R ) \mathcal S_A(\R) S A ( R ) Ψ \Psi Ψ S A ( R ) \mathcal S_A(\R) S A ( R ) Ψ = Φ C Φ \Psi=\Phi C_\Phi Ψ = Φ C Φ C Φ ∈ K d × d C_\Phi\in\mathbb K^{d\times d} C Φ ∈ K d × d Φ ( t ) \Phi(t) Φ ( t ) C Φ = Φ ( t ) − 1 Ψ ( t ) C_\Phi=\Phi(t)^{-1}\Psi(t) C Φ = Φ ( t ) − 1 Ψ ( t ) det C Φ = det Φ ( t ) − 1 det Ψ ( t ) ≠ 0 \det C_\Phi=\det\Phi(t)^{-1}\det\Psi(t)\ne0 det C Φ = det Φ ( t ) − 1 det Ψ ( t ) = 0
Φ 0 \Phi_0 Φ 0 t = 0 t=0 t = 0 C Φ 0 = Φ 0 ( T ) = M C_{\Phi_0}=\Phi_0(T)=M C Φ 0 = Φ 0 ( T ) = M t t t Φ 0 ( t + T ) = Φ 0 ( t ) M \Phi_0(t+T)=\Phi_0(t)M Φ 0 ( t + T ) = Φ 0 ( t ) M t ↦ Φ 0 ( t ) Φ ( 0 ) t\mapsto\Phi_0(t)\Phi(0) t ↦ Φ 0 ( t ) Φ ( 0 ) t = 0 t=0 t = 0 Φ \Phi Φ §E13.9 定理 2.2 の評価写像の単射性によりΦ ( t ) = Φ 0 ( t ) Φ ( 0 ) \Phi(t)=\Phi_0(t)\Phi(0) Φ ( t ) = Φ 0 ( t ) Φ ( 0 )
Φ ( t + T ) = Φ 0 ( t ) M Φ ( 0 ) = Φ ( t ) Φ ( 0 ) − 1 M Φ ( 0 ) \Phi(t+T)=\Phi_0(t)M\Phi(0)=\Phi(t)\Phi(0)^{-1}M\Phi(0) Φ ( t + T ) = Φ 0 ( t ) M Φ ( 0 ) = Φ ( t ) Φ ( 0 ) − 1 M Φ ( 0 ) であり、C Φ = Φ ( 0 ) − 1 M Φ ( 0 ) C_\Phi=\Phi(0)^{-1}M\Phi(0) C Φ = Φ ( 0 ) − 1 M Φ ( 0 ) 主張 (1) は示された。
主張 (2) を示す。x ( t ) \boldsymbol x(t) x ( t ) Φ 0 ( t ) x ( 0 ) \Phi_0(t)\boldsymbol x(0) Φ 0 ( t ) x ( 0 ) t = 0 t=0 t = 0 §E13.9 定理 2.2 によりx ( t ) = Φ 0 ( t ) x ( 0 ) \boldsymbol x(t)=\Phi_0(t)\boldsymbol x(0) x ( t ) = Φ 0 ( t ) x ( 0 ) Φ 0 ( s + T ) = Φ 0 ( s ) M \Phi_0(s+T)=\Phi_0(s)M Φ 0 ( s + T ) = Φ 0 ( s ) M x ( s + T ) = Φ 0 ( s ) M x ( 0 ) = Φ 0 ( s ) M Φ 0 ( s ) − 1 x ( s ) \boldsymbol x(s+T)=\Phi_0(s)M\boldsymbol x(0)=\Phi_0(s)M\Phi_0(s)^{-1}\boldsymbol x(s) x ( s + T ) = Φ 0 ( s ) M x ( 0 ) = Φ 0 ( s ) M Φ 0 ( s ) − 1 x ( s )
主張 (3) を示す。n ≥ 0 n\ge0 n ≥ 0 Φ 0 ( t + n T ) = Φ 0 ( t ) M n \Phi_0(t+nT)=\Phi_0(t)M^n Φ 0 ( t + n T ) = Φ 0 ( t ) M n Φ 0 ( t + ( n + 1 ) T ) = Φ 0 ( t + n T ) M = Φ 0 ( t ) M n + 1 \Phi_0(t+(n+1)T)=\Phi_0(t+nT)M=\Phi_0(t)M^{n+1} Φ 0 ( t + ( n + 1 ) T ) = Φ 0 ( t + n T ) M = Φ 0 ( t ) M n + 1 t − T t-T t − T Φ 0 ( s + T ) = Φ 0 ( s ) M \Phi_0(s+T)=\Phi_0(s)M Φ 0 ( s + T ) = Φ 0 ( s ) M Φ 0 ( t − T ) = Φ 0 ( t ) M − 1 \Phi_0(t-T)=\Phi_0(t)M^{-1} Φ 0 ( t − T ) = Φ 0 ( t ) M − 1 n ≤ 0 n\le0 n ≤ 0 ▨
定義 1.3. K ∈ { R , C } \mathbb K\in\{\R,\C\} K ∈ { R , C } d ∈ N ≥ 1 d\in\NN d ∈ N ≥ 1 T > 0 T>0 T > 0 x ′ = A ( t ) x \boldsymbol x'=A(t)\boldsymbol x x ′ = A ( t ) x T T T Φ 0 \Phi_0 Φ 0 Φ 0 ( 0 ) = I \Phi_0(0)=I Φ 0 ( 0 ) = I M = Φ 0 ( T ) M=\Phi_0(T) M = Φ 0 ( T ) モノドロミー行列 (monodromy matrix )M M M C \C C Floquet 乗数 (Floquet multiplier )ρ \rho ρ e μ T = ρ e^{\mu T}=\rho e μ T = ρ μ ∈ C \mu\in\C μ ∈ C Floquet 指数 (Floquet exponent )
定義 1.5. K ∈ { R , C } \mathbb K\in\{\R,\C\} K ∈ { R , C } d ∈ N ≥ 1 d\in\NN d ∈ N ≥ 1 T > 0 T>0 T > 0 x ′ = A ( t ) x \boldsymbol x'=A(t)\boldsymbol x x ′ = A ( t ) x T T T ρ ∈ C \rho\in\C ρ ∈ C x ′ = A ( t ) x \boldsymbol x'=A(t)\boldsymbol x x ′ = A ( t ) x x : R → C d \boldsymbol x\colon\R\to\C^d x : R → C d t ∈ R t\in\R t ∈ R x ( t + T ) = ρ x ( t ) \boldsymbol x(t+T)=\rho\boldsymbol x(t) x ( t + T ) = ρ x ( t ) ρ \rho ρ Floquet 解 (Floquet solution )
命題 1.6. K ∈ { R , C } \mathbb K\in\{\R,\C\} K ∈ { R , C } d ∈ N ≥ 1 d\in\NN d ∈ N ≥ 1 T > 0 T>0 T > 0 x ′ = A ( t ) x \boldsymbol x'=A(t)\boldsymbol x x ′ = A ( t ) x T T T M M M ρ ∈ C \rho\in\C ρ ∈ C x ′ = A ( t ) x \boldsymbol x'=A(t)\boldsymbol x x ′ = A ( t ) x x : R → C d \boldsymbol x\colon\R\to\C^d x : R → C d t ∈ R t\in\R t ∈ R x ( t + T ) = ρ x ( t ) \boldsymbol x(t+T)=\rho\boldsymbol x(t) x ( t + T ) = ρ x ( t ) F ρ \mathcal F_\rho F ρ
F ρ \mathcal F_\rho F ρ C \C C x ↦ x ( 0 ) \boldsymbol x\mapsto\boldsymbol x(0) x ↦ x ( 0 ) F ρ \mathcal F_\rho F ρ ker ( M − ρ I ) ⊆ C d \ker(M-\rho I)\subseteq\C^d ker ( M − ρ I ) ⊆ C d ρ \rho ρ ρ \rho ρ 零でないT T T 1 1 1 t t t x ( t + T ) = − x ( t ) \boldsymbol x(t+T)=-\boldsymbol x(t) x ( t + T ) = − x ( t ) − 1 -1 − 1 2 T 2T 2 T
K = R \mathbb K=\R K = R ρ ∈ R \rho\in\R ρ ∈ R F ρ \mathcal F_\rho F ρ R \R R x ↦ x ( 0 ) \boldsymbol x\mapsto\boldsymbol x(0) x ↦ x ( 0 ) F ρ \mathcal F_\rho F ρ ker ( M − ρ I ) ∩ R d \ker(M-\rho I)\cap\R^d ker ( M − ρ I ) ∩ R d F ρ \mathcal F_\rho F ρ R \R R dim C F ρ \dim_{\C}\mathcal F_\rho dim C F ρ 証明. Φ 0 \Phi_0 Φ 0 Φ 0 ( 0 ) = I \Phi_0(0)=I Φ 0 ( 0 ) = I K = R \mathbb K=\R K = R A A A L l o c 1 ( R ; C d × d ) L^1_{\mathrm{loc}}(\R;\C^{d\times d}) L loc 1 ( R ; C d × d ) Φ 0 \Phi_0 Φ 0 x : R → C d \boldsymbol x\colon\R\to\C^d x : R → C d §E13.9 定理 2.2 によりx ( t ) = Φ 0 ( t ) x ( 0 ) \boldsymbol x(t)=\Phi_0(t)\boldsymbol x(0) x ( t ) = Φ 0 ( t ) x ( 0 ) 補題 1.2 (1) により
x ( t + T ) − ρ x ( t ) = Φ 0 ( t ) ( M − ρ I ) x ( 0 ) \boldsymbol x(t+T)-\rho\boldsymbol x(t)=\Phi_0(t)\bigl(M-\rho I\bigr)\boldsymbol x(0) x ( t + T ) − ρ x ( t ) = Φ 0 ( t ) ( M − ρ I ) x ( 0 ) である。Φ 0 ( t ) \Phi_0(t) Φ 0 ( t ) x ∈ F ρ \boldsymbol x\in\mathcal F_\rho x ∈ F ρ x ( 0 ) ∈ ker ( M − ρ I ) \boldsymbol x(0)\in\ker(M-\rho I) x ( 0 ) ∈ ker ( M − ρ I ) F ρ \mathcal F_\rho F ρ x ↦ x ( 0 ) \boldsymbol x\mapsto\boldsymbol x(0) x ↦ x ( 0 ) ker ( M − ρ I ) \ker(M-\rho I) ker ( M − ρ I ) §E13.9 定理 2.2 の評価写像の同型を制限して主張 (1) を得る。
主張 (2) は主張 (1) のρ = 1 \rho=1 ρ = 1 ρ = − 1 \rho=-1 ρ = − 1 x ( t + T ) = − x ( t ) \boldsymbol x(t+T)=-\boldsymbol x(t) x ( t + T ) = − x ( t ) x ( t + 2 T ) = − x ( t + T ) = x ( t ) \boldsymbol x(t+2T)=-\boldsymbol x(t+T)=\boldsymbol x(t) x ( t + 2 T ) = − x ( t + T ) = x ( t )
主張 (3) を示す。Φ 0 \Phi_0 Φ 0 x = Φ 0 x ( 0 ) \boldsymbol x=\Phi_0\boldsymbol x(0) x = Φ 0 x ( 0 ) x ( 0 ) ∈ R d \boldsymbol x(0)\in\R^d x ( 0 ) ∈ R d M − ρ I M-\rho I M − ρ I R \R R C \C C dim R ( ker ( M − ρ I ) ∩ R d ) = d − rank ( M − ρ I ) = dim C ker ( M − ρ I ) \dim_{\R}\bigl(\ker(M-\rho I)\cap\R^d\bigr)=d-\operatorname{rank}(M-\rho I)=\dim_{\C}\ker(M-\rho I) dim R ( ker ( M − ρ I ) ∩ R d ) = d − rank ( M − ρ I ) = dim C ker ( M − ρ I ) ▨
命題 1.7. K ∈ { R , C } \mathbb K\in\{\R,\C\} K ∈ { R , C } d ∈ N ≥ 1 d\in\NN d ∈ N ≥ 1 T > 0 T>0 T > 0 x ′ = A ( t ) x \boldsymbol x'=A(t)\boldsymbol x x ′ = A ( t ) x T T T M M M ρ 1 , … , ρ d \rho_1,\ldots,\rho_d ρ 1 , … , ρ d
ρ 1 ⋯ ρ d = det M = exp ( ∫ 0 T tr A ( s ) d s ) \rho_1\cdots\rho_d=\det M=\exp\left(\int_0^T\operatorname{tr}A(s)\,ds\right) ρ 1 ⋯ ρ d = det M = exp ( ∫ 0 T tr A ( s ) d s ) である。
証明. M M M ∏ j = 1 d ( t − ρ j ) \prod_{j=1}^d(t-\rho_j) ∏ j = 1 d ( t − ρ j ) t = 0 t=0 t = 0 det M = ρ 1 ⋯ ρ d \det M=\rho_1\cdots\rho_d det M = ρ 1 ⋯ ρ d Φ 0 \Phi_0 Φ 0 Φ 0 ( 0 ) = I \Phi_0(0)=I Φ 0 ( 0 ) = I M = Φ 0 ( T ) M=\Phi_0(T) M = Φ 0 ( T ) §E13.9 注意 3.2 の等式によりdet M = exp ∫ 0 T tr A ( s ) d s \det M=\exp\int_0^T\operatorname{tr}A(s)\,ds det M = exp ∫ 0 T tr A ( s ) d s ▨
2 Floquet の標準形 定義 2.1. d ∈ N ≥ 1 d\in\NN d ∈ N ≥ 1 C ∈ C d × d C\in\C^{d\times d} C ∈ C d × d ρ ∈ C \rho\in\C ρ ∈ C C C C C C C ρ \rho ρ G ρ ( C ) = ker ( C − ρ I ) d G_\rho(C)=\ker(C-\rho I)^d G ρ ( C ) = ker ( C − ρ I ) d G ρ ( C ) = ker ( C − ρ I ) G_\rho(C)=\ker(C-\rho I) G ρ ( C ) = ker ( C − ρ I ) ρ \rho ρ C C C 半単純固有値 (semisimple eigenvalue )
系 2.2. d ∈ N ≥ 1 d\in\NN d ∈ N ≥ 1 B ∈ C d × d B\in\C^{d\times d} B ∈ C d × d ρ ∈ C ∖ { 0 } \rho\in\C\setminus\{0\} ρ ∈ C ∖ { 0 } B B B e β = ρ e^\beta=\rho e β = ρ E ρ E_\rho E ρ E ρ = ∅ E_\rho=\emptyset E ρ = ∅ E ρ E_\rho E ρ { 0 } \{\boldsymbol 0\} { 0 }
G ρ ( e B ) = ⨁ β ∈ E ρ G β ( B ) G_\rho(e^B)=\bigoplus_{\beta\in E_\rho}G_\beta(B) G ρ ( e B ) = ⨁ β ∈ E ρ G β ( B ) ker ( e B − ρ I ) = ⨁ β ∈ E ρ ker ( B − β I ) \ker(e^B-\rho I)=\bigoplus_{\beta\in E_\rho}\ker(B-\beta I) ker ( e B − ρ I ) = ⨁ β ∈ E ρ ker ( B − β I ) B B B β 1 , … , β d \beta_1,\ldots,\beta_d β 1 , … , β d e B e^B e B e β 1 , … , e β d e^{\beta_1},\ldots,e^{\beta_d} e β 1 , … , e β d ρ \rho ρ e B e^B e B ρ \rho ρ e B e^B e B β ∈ E ρ \beta\in E_\rho β ∈ E ρ B B B 証明. 主張 (1) を示す。M M M §E3.31 定理 4.2 によりe L = M e^L=M e L = M L ∈ C d × d L\in\C^{d\times d} L ∈ C d × d B = T − 1 L B=T^{-1}L B = T − 1 L e T B = M e^{TB}=M e T B = M P ( t ) = Φ 0 ( t ) e − t B P(t)=\Phi_0(t)e^{-tB} P ( t ) = Φ 0 ( t ) e − tB §E13.12 定理 2.2 (3) を− B -B − B I = R I=\R I = R t ↦ e − t B t\mapsto e^{-tB} t ↦ e − tB §E13.12 定理 2.1 によりe − t B e^{-tB} e − tB §E13.2 定理 3.1 (2) によりP P P P ( t ) P(t) P ( t ) P ( 0 ) = I P(0)=I P ( 0 ) = I 補題 1.2 (1) と§E13.12 定理 2.1 により
P ( t + T ) = Φ 0 ( t ) M e − T B e − t B = Φ 0 ( t ) e T B e − T B e − t B = P ( t ) P(t+T)=\Phi_0(t)Me^{-TB}e^{-tB}=\Phi_0(t)e^{TB}e^{-TB}e^{-tB}=P(t) P ( t + T ) = Φ 0 ( t ) M e − T B e − tB = Φ 0 ( t ) e T B e − T B e − tB = P ( t ) である。T B TB T B T β 1 , … , T β d T\beta_1,\ldots,T\beta_d T β 1 , … , T β d 系 2.2 (3) をT B TB T B M = e T B M=e^{TB} M = e T B e T β 1 , … , e T β d e^{T\beta_1},\ldots,e^{T\beta_d} e T β 1 , … , e T β d e T β j e^{T\beta_j} e T β j β j \beta_j β j
主張 (2) を示す。K = R \mathbb K=\R K = R Φ 0 \Phi_0 Φ 0 M M M §E3.31 補題 4.1 (2) によりe B 2 = M 2 e^{B_2}=M^2 e B 2 = M 2 B 2 ∈ R d × d B_2\in\R^{d\times d} B 2 ∈ R d × d B = ( 2 T ) − 1 B 2 B=(2T)^{-1}B_2 B = ( 2 T ) − 1 B 2 P ( t ) = Φ 0 ( t ) e − t B P(t)=\Phi_0(t)e^{-tB} P ( t ) = Φ 0 ( t ) e − tB e − t B e^{-tB} e − tB P P P P ( 0 ) = I P(0)=I P ( 0 ) = I 補題 1.2 (3) のn = 2 n=2 n = 2
P ( t + 2 T ) = Φ 0 ( t ) M 2 e − 2 T B e − t B = P ( t ) P(t+2T)=\Phi_0(t)M^2e^{-2TB}e^{-tB}=P(t) P ( t + 2 T ) = Φ 0 ( t ) M 2 e − 2 T B e − tB = P ( t ) である。
主張 (3) を示す。P ( t ) = Φ 0 ( t ) e − t B P(t)=\Phi_0(t)e^{-tB} P ( t ) = Φ 0 ( t ) e − tB §E13.12 定理 2.2 (3) を− B -B − B I = R I=\R I = R t ↦ e − t B t\mapsto e^{-tB} t ↦ e − tB §E13.12 定理 2.1 によりすべての点でd d t e − t B = − e − t B B \frac{d}{dt}e^{-tB}=-e^{-tB}B d t d e − tB = − e − tB B §E13.2 定理 3.1 (2) によりP P P
P ′ = A Φ 0 e − t B − Φ 0 e − t B B = A P − P B P'=A\Phi_0e^{-tB}-\Phi_0e^{-tB}B=AP-PB P ′ = A Φ 0 e − tB − Φ 0 e − tB B = A P − P B である。§E13.2 系 3.2 (2) によりP − 1 P^{-1} P − 1 x ∈ A C l o c ( R ; C d ) \boldsymbol x\in AC_{\mathrm{loc}}(\R;\C^d) x ∈ A C loc ( R ; C d ) z = P − 1 x \boldsymbol z=P^{-1}\boldsymbol x z = P − 1 x x = P z \boldsymbol x=P\boldsymbol z x = P z §E13.2 定理 3.1 (2) を適用すると、ほとんど至る所で
x ′ = P ′ z + P z ′ = A P z − P B z + P z ′ = A x + P ( z ′ − B z ) \boldsymbol x'=P'\boldsymbol z+P\boldsymbol z'=AP\boldsymbol z-PB\boldsymbol z+P\boldsymbol z'=A\boldsymbol x+P\bigl(\boldsymbol z'-B\boldsymbol z\bigr) x ′ = P ′ z + P z ′ = A P z − P B z + P z ′ = A x + P ( z ′ − B z ) である。P ( t ) P(t) P ( t ) x ′ = A x \boldsymbol x'=A\boldsymbol x x ′ = A x z ′ = B z \boldsymbol z'=B\boldsymbol z z ′ = B z ▨
3 乗数による安定性の判定 系 3.1. d ∈ N ≥ 1 d\in\NN d ∈ N ≥ 1 C d \C^d C d ∥ ⋅ ∥ \lVert\cdot\rVert ∥ ⋅ ∥ B ∈ C d × d B\in\C^{d\times d} B ∈ C d × d α \alpha α B B B
任意のγ > α \gamma>\alpha γ > α C ≥ 1 C\ge1 C ≥ 1 t ≥ 0 t\ge0 t ≥ 0 ∥ e t B ∥ ≤ C e γ t \lVert e^{tB}\rVert\le Ce^{\gamma t} ∥ e tB ∥ ≤ C e γ t
sup t ≥ 0 ∥ e t B ∥ < ∞ \sup_{t\ge0}\lVert e^{tB}\rVert<\infty sup t ≥ 0 ∥ e tB ∥ < ∞ B B B β \beta β ℜ β ≤ 0 \Re\beta\le0 ℜ β ≤ 0 ℜ β = 0 \Re\beta=0 ℜ β = 0 B B B β \beta β B B B v \boldsymbol v v t ∈ R t\in\R t ∈ R e t B v = e t β v e^{tB}\boldsymbol v=e^{t\beta}\boldsymbol v e tB v = e tβ v ∥ e t B v ∥ = e t ℜ β ∥ v ∥ \lVert e^{tB}\boldsymbol v\rVert=e^{t\Re\beta}\lVert\boldsymbol v\rVert ∥ e tB v ∥ = e t ℜ β ∥ v ∥ 定理 3.2. K ∈ { R , C } \mathbb K\in\{\R,\C\} K ∈ { R , C } d ∈ N ≥ 1 d\in\NN d ∈ N ≥ 1 T > 0 T>0 T > 0 C d \C^d C d ∥ ⋅ ∥ \lVert\cdot\rVert ∥ ⋅ ∥ x ′ = A ( t ) x \boldsymbol x'=A(t)\boldsymbol x x ′ = A ( t ) x T T T M M M ρ 1 , … , ρ d \rho_1,\ldots,\rho_d ρ 1 , … , ρ d x : R → C d \boldsymbol x\colon\R\to\C^d x : R → C d
すべての解がt → ∞ t\to\infty t → ∞ 0 \boldsymbol 0 0 j j j ∣ ρ j ∣ < 1 \lvert\rho_j\rvert<1 ∣ ρ j ∣ < 1 C ≥ 1 C\ge1 C ≥ 1 γ > 0 \gamma>0 γ > 0 x \boldsymbol x x s ≤ t s\le t s ≤ t s , t ∈ R s,t\in\R s , t ∈ R ∥ x ( t ) ∥ ≤ C e − γ ( t − s ) ∥ x ( s ) ∥ \lVert\boldsymbol x(t)\rVert\le Ce^{-\gamma(t-s)}\lVert\boldsymbol x(s)\rVert ∥ x ( t )∥ ≤ C e − γ ( t − s ) ∥ x ( s )∥
すべての解が[ 0 , ∞ ) [0,\infty) [ 0 , ∞ ) j j j ∣ ρ j ∣ ≤ 1 \lvert\rho_j\rvert\le1 ∣ ρ j ∣ ≤ 1 1 1 1 M M M
∣ ρ ∣ > 1 \lvert\rho\rvert>1 ∣ ρ ∣ > 1 ρ \rho ρ t ∈ R t\in\R t ∈ R n ∈ Z n\in\Z n ∈ Z ∥ x ( t + n T ) ∥ = ∣ ρ ∣ n ∥ x ( t ) ∥ \lVert\boldsymbol x(t+nT)\rVert=\lvert\rho\rvert^n\lVert\boldsymbol x(t)\rVert ∥ x ( t + n T )∥ = ∣ ρ ∣ n ∥ x ( t )∥ x ≠ 0 \boldsymbol x\ne\boldsymbol 0 x = 0 [ 0 , ∞ ) [0,\infty) [ 0 , ∞ ) K = R \mathbb K=\R K = R 主張 (1) と主張 (2) の同値は、解を実数値の解に限っても成り立つ。 証明. 定理 2.3 (1) によりe T B = M e^{TB}=M e T B = M B B B T T T P P P Φ 0 ( t ) = P ( t ) e t B \Phi_0(t)=P(t)e^{tB} Φ 0 ( t ) = P ( t ) e tB P P P P − 1 P^{-1} P − 1 T T T K P = sup t ∥ P ( t ) ∥ ⋅ sup t ∥ P ( t ) − 1 ∥ < ∞ K_P=\sup_t\lVert P(t)\rVert\cdot\sup_t\lVert P(t)^{-1}\rVert<\infty K P = sup t ∥ P ( t )∥ ⋅ sup t ∥ P ( t ) − 1 ∥ < ∞ §E13.9 定理 2.2 により各解はx ( t ) = Φ 0 ( t ) x ( 0 ) \boldsymbol x(t)=\Phi_0(t)\boldsymbol x(0) x ( t ) = Φ 0 ( t ) x ( 0 ) §E13.12 定理 2.1 により
x ( t ) = Φ 0 ( t ) Φ 0 ( s ) − 1 x ( s ) = P ( t ) e ( t − s ) B P ( s ) − 1 x ( s ) , ∥ x ( t ) ∥ ≤ K P ∥ e ( t − s ) B ∥ ∥ x ( s ) ∥ \boldsymbol x(t)=\Phi_0(t)\Phi_0(s)^{-1}\boldsymbol x(s)=P(t)e^{(t-s)B}P(s)^{-1}\boldsymbol x(s),\qquad
\lVert\boldsymbol x(t)\rVert\le K_P\bigl\lVert e^{(t-s)B}\bigr\rVert\lVert\boldsymbol x(s)\rVert x ( t ) = Φ 0 ( t ) Φ 0 ( s ) − 1 x ( s ) = P ( t ) e ( t − s ) B P ( s ) − 1 x ( s ) , ∥ x ( t )∥ ≤ K P e ( t − s ) B ∥ x ( s )∥ が成り立ち、e t B = P ( t ) − 1 Φ 0 ( t ) e^{tB}=P(t)^{-1}\Phi_0(t) e tB = P ( t ) − 1 Φ 0 ( t ) 定理 2.3 (1) により、B B B β \beta β ρ = e T β \rho=e^{T\beta} ρ = e T β ∣ ρ ∣ = e T ℜ β \lvert\rho\rvert=e^{T\Re\beta} ∣ ρ ∣ = e T ℜ β
主張 (1) を示す。すべての∣ ρ j ∣ < 1 \lvert\rho_j\rvert<1 ∣ ρ j ∣ < 1 B B B α \alpha α γ = − α / 2 \gamma=-\alpha/2 γ = − α /2 − γ > α -\gamma>\alpha − γ > α 系 3.1 (1) により∥ e t B ∥ ≤ C ′ e − γ t \lVert e^{tB}\rVert\le C'e^{-\gamma t} ∥ e tB ∥ ≤ C ′ e − γ t t ≥ 0 t\ge0 t ≥ 0 C ′ ≥ 1 C'\ge1 C ′ ≥ 1 C = K P C ′ C=K_PC' C = K P C ′ s = 0 s=0 s = 0 0 \boldsymbol 0 0 ∣ ρ ∣ ≥ 1 \lvert\rho\rvert\ge1 ∣ ρ ∣ ≥ 1 ρ \rho ρ 命題 1.6 (1) により乗数ρ \rho ρ x \boldsymbol x x ∥ x ( n T ) ∥ = ∣ ρ ∣ n ∥ x ( 0 ) ∥ ≥ ∥ x ( 0 ) ∥ > 0 \lVert\boldsymbol x(nT)\rVert=\lvert\rho\rvert^n\lVert\boldsymbol x(0)\rVert\ge\lVert\boldsymbol x(0)\rVert>0 ∥ x ( n T )∥ = ∣ ρ ∣ n ∥ x ( 0 )∥ ≥ ∥ x ( 0 )∥ > 0 x \boldsymbol x x 0 \boldsymbol 0 0
主張 (2) を示す。∣ ρ ∣ = 1 \lvert\rho\rvert=1 ∣ ρ ∣ = 1 ρ \rho ρ e T β = ρ e^{T\beta}=\rho e T β = ρ B B B β \beta β ℜ β = 0 \Re\beta=0 ℜ β = 0 ℜ β = 0 \Re\beta=0 ℜ β = 0 B B B β \beta β 1 1 1 e T β e^{T\beta} e T β G T β ( T B ) = G β ( B ) G_{T\beta}(TB)=G_\beta(B) G T β ( T B ) = G β ( B ) ker ( T B − T β I ) = ker ( B − β I ) \ker(TB-T\beta I)=\ker(B-\beta I) ker ( T B − T β I ) = ker ( B − β I ) 系 2.2 (4) をT B TB T B ρ \rho ρ M M M e T β = ρ e^{T\beta}=\rho e T β = ρ B B B β \beta β B B B 系 3.1 (2) のB B B sup t ≥ 0 ∥ e t B ∥ < ∞ \sup_{t\ge0}\lVert e^{tB}\rVert<\infty sup t ≥ 0 ∥ e tB ∥ < ∞ s = 0 s=0 s = 0 [ 0 , ∞ ) [0,\infty) [ 0 , ∞ ) [ 0 , ∞ ) [0,\infty) [ 0 , ∞ ) Φ 0 \Phi_0 Φ 0 e t B = P ( t ) − 1 Φ 0 ( t ) e^{tB}=P(t)^{-1}\Phi_0(t) e tB = P ( t ) − 1 Φ 0 ( t ) [ 0 , ∞ ) [0,\infty) [ 0 , ∞ ) 系 3.1 (2) によりB B B
主張 (3) を示す。命題 1.6 (1) により乗数ρ \rho ρ x \boldsymbol x x x ( t + T ) = ρ x ( t ) \boldsymbol x(t+T)=\rho\boldsymbol x(t) x ( t + T ) = ρ x ( t ) n n n x ( t + n T ) = ρ n x ( t ) \boldsymbol x(t+nT)=\rho^n\boldsymbol x(t) x ( t + n T ) = ρ n x ( t ) n ∈ Z n\in\Z n ∈ Z x ( 0 ) ≠ 0 \boldsymbol x(0)\ne\boldsymbol 0 x ( 0 ) = 0 ∥ x ( n T ) ∥ = ∣ ρ ∣ n ∥ x ( 0 ) ∥ → ∞ \lVert\boldsymbol x(nT)\rVert=\lvert\rho\rvert^n\lVert\boldsymbol x(0)\rVert\to\infty ∥ x ( n T )∥ = ∣ ρ ∣ n ∥ x ( 0 )∥ → ∞ n → ∞ n\to\infty n → ∞
主張 (4) を示す。A A A x \boldsymbol x x x ‾ \overline{\boldsymbol x} x ℜ x = ( x + x ‾ ) / 2 \Re\boldsymbol x=(\boldsymbol x+\overline{\boldsymbol x})/2 ℜ x = ( x + x ) /2 ℑ x \Im\boldsymbol x ℑ x 0 \boldsymbol 0 0 [ 0 , ∞ ) [0,\infty) [ 0 , ∞ ) x = ℜ x + i ℑ x \boldsymbol x=\Re\boldsymbol x+i\Im\boldsymbol x x = ℜ x + i ℑ x ▨
4 随伴系と周期系の Fredholm の交代定理 定義 4.1. K ∈ { R , C } \mathbb K\in\{\R,\C\} K ∈ { R , C } d ∈ N ≥ 1 d\in\NN d ∈ N ≥ 1 J ⊆ R J\subseteq\R J ⊆ R A ∈ L l o c 1 ( J ; K d × d ) A\in L^1_{\mathrm{loc}}(J;\mathbb K^{d\times d}) A ∈ L loc 1 ( J ; K d × d ) X ∗ X^* X ∗ K = R \mathbb K=\R K = R ψ ′ = − A ( t ) ∗ ψ \boldsymbol\psi'=-A(t)^*\boldsymbol\psi ψ ′ = − A ( t ) ∗ ψ x ′ = A ( t ) x \boldsymbol x'=A(t)\boldsymbol x x ′ = A ( t ) x 随伴系 (adjoint system )
補題 4.2. K ∈ { R , C } \mathbb K\in\{\R,\C\} K ∈ { R , C } d ∈ N ≥ 1 d\in\NN d ∈ N ≥ 1 J ⊆ R J\subseteq\R J ⊆ R A ∈ L l o c 1 ( J ; K d × d ) A\in L^1_{\mathrm{loc}}(J;\mathbb K^{d\times d}) A ∈ L loc 1 ( J ; K d × d ) b ∈ L l o c 1 ( J ; K d ) \boldsymbol b\in L^1_{\mathrm{loc}}(J;\mathbb K^d) b ∈ L loc 1 ( J ; K d )
x \boldsymbol x x x ′ = A ( t ) x + b ( t ) \boldsymbol x'=A(t)\boldsymbol x+\boldsymbol b(t) x ′ = A ( t ) x + b ( t ) J J J ψ \boldsymbol\psi ψ ψ ′ = − A ( t ) ∗ ψ \boldsymbol\psi'=-A(t)^*\boldsymbol\psi ψ ′ = − A ( t ) ∗ ψ J J J ψ ∗ x \boldsymbol\psi^*\boldsymbol x ψ ∗ x J J J ( ψ ∗ x ) ′ = ψ ∗ b (\boldsymbol\psi^*\boldsymbol x)'=\boldsymbol\psi^*\boldsymbol b ( ψ ∗ x ) ′ = ψ ∗ b Φ \Phi Φ x ′ = A ( t ) x \boldsymbol x'=A(t)\boldsymbol x x ′ = A ( t ) x ( Φ ∗ ) − 1 (\Phi^*)^{-1} ( Φ ∗ ) − 1 証明. 主張 (1) を示す。複素共役はK d \mathbb K^d K d ψ ‾ \overline{\boldsymbol\psi} ψ ψ ‾ ′ = ψ ′ ‾ \overline{\boldsymbol\psi}\,'=\overline{\boldsymbol\psi'} ψ ′ = ψ ′ ψ ∗ x = ψ ‾ T x \boldsymbol\psi^*\boldsymbol x=\overline{\boldsymbol\psi}^{\mathsf T}\boldsymbol x ψ ∗ x = ψ T x ( u , v ) ↦ u T v (\boldsymbol u,\boldsymbol v)\mapsto\boldsymbol u^{\mathsf T}\boldsymbol v ( u , v ) ↦ u T v §E13.2 定理 3.1 (2) を適用すると、ψ ∗ x \boldsymbol\psi^*\boldsymbol x ψ ∗ x
( ψ ∗ x ) ′ = ( ψ ′ ) ∗ x + ψ ∗ x ′ = − ψ ∗ A x + ψ ∗ ( A x + b ) = ψ ∗ b (\boldsymbol\psi^*\boldsymbol x)'=(\boldsymbol\psi')^*\boldsymbol x+\boldsymbol\psi^*\boldsymbol x'=-\boldsymbol\psi^*A\boldsymbol x+\boldsymbol\psi^*(A\boldsymbol x+\boldsymbol b)=\boldsymbol\psi^*\boldsymbol b ( ψ ∗ x ) ′ = ( ψ ′ ) ∗ x + ψ ∗ x ′ = − ψ ∗ A x + ψ ∗ ( A x + b ) = ψ ∗ b である。
主張 (2) を示す。§E13.2 系 3.2 (2) によりΦ − 1 \Phi^{-1} Φ − 1 ( Φ − 1 ) ′ = − Φ − 1 Φ ′ Φ − 1 = − Φ − 1 A (\Phi^{-1})'=-\Phi^{-1}\Phi'\Phi^{-1}=-\Phi^{-1}A ( Φ − 1 ) ′ = − Φ − 1 Φ ′ Φ − 1 = − Φ − 1 A Ψ = ( Φ ∗ ) − 1 = ( Φ − 1 ) ∗ \Psi=(\Phi^*)^{-1}=(\Phi^{-1})^* Ψ = ( Φ ∗ ) − 1 = ( Φ − 1 ) ∗ Ψ ′ = ( ( Φ − 1 ) ′ ) ∗ = − A ∗ Ψ \Psi'=\bigl((\Phi^{-1})'\bigr)^*=-A^*\Psi Ψ ′ = ( ( Φ − 1 ) ′ ) ∗ = − A ∗ Ψ det Ψ = det Φ ‾ − 1 ≠ 0 \det\Psi=\overline{\det\Phi}^{-1}\ne0 det Ψ = det Φ − 1 = 0 Ψ \Psi Ψ ▨
補題 4.3. K ∈ { R , C } \mathbb K\in\{\R,\C\} K ∈ { R , C } d ∈ N ≥ 1 d\in\NN d ∈ N ≥ 1 T > 0 T>0 T > 0 A ∈ L 1 ( [ 0 , T ] ; K d × d ) A\in L^1([0,T];\mathbb K^{d\times d}) A ∈ L 1 ([ 0 , T ] ; K d × d ) b ∈ L 1 ( [ 0 , T ] ; K d ) \boldsymbol b\in L^1([0,T];\mathbb K^d) b ∈ L 1 ([ 0 , T ] ; K d ) ρ ∈ K \rho\in\mathbb K ρ ∈ K Φ 0 \Phi_0 Φ 0 x ′ = A ( t ) x \boldsymbol x'=A(t)\boldsymbol x x ′ = A ( t ) x [ 0 , T ] [0,T] [ 0 , T ] Φ 0 ( 0 ) = I \Phi_0(0)=I Φ 0 ( 0 ) = I M = Φ 0 ( T ) M=\Phi_0(T) M = Φ 0 ( T ) c = ∫ 0 T Φ 0 ( s ) − 1 b ( s ) d s \boldsymbol c=\int_0^T\Phi_0(s)^{-1}\boldsymbol b(s)\,ds c = ∫ 0 T Φ 0 ( s ) − 1 b ( s ) d s ξ ∈ K d \boldsymbol\xi\in\mathbb K^d ξ ∈ K d
x ξ ( t ) = Φ 0 ( t ) [ ξ + ∫ 0 t Φ 0 ( s ) − 1 b ( s ) d s ] ( t ∈ [ 0 , T ] ) \boldsymbol x_{\boldsymbol\xi}(t)=\Phi_0(t)\left[\boldsymbol\xi+\int_0^t\Phi_0(s)^{-1}\boldsymbol b(s)\,ds\right]\qquad(t\in[0,T]) x ξ ( t ) = Φ 0 ( t ) [ ξ + ∫ 0 t Φ 0 ( s ) − 1 b ( s ) d s ] ( t ∈ [ 0 , T ]) とおく。このとき、ξ ↦ x ξ \boldsymbol\xi\mapsto\boldsymbol x_{\boldsymbol\xi} ξ ↦ x ξ K d \mathbb K^d K d x ′ = A ( t ) x + b ( t ) \boldsymbol x'=A(t)\boldsymbol x+\boldsymbol b(t) x ′ = A ( t ) x + b ( t ) x ∈ A C ( [ 0 , T ] ; K d ) \boldsymbol x\in AC([0,T];\mathbb K^d) x ∈ A C ([ 0 , T ] ; K d ) x ↦ x ( 0 ) \boldsymbol x\mapsto\boldsymbol x(0) x ↦ x ( 0 ) ( ρ I − M ) ξ = M c (\rho I-M)\boldsymbol\xi=M\boldsymbol c ( ρ I − M ) ξ = M c x ξ ( T ) = ρ x ξ ( 0 ) \boldsymbol x_{\boldsymbol\xi}(T)=\rho\,\boldsymbol x_{\boldsymbol\xi}(0) x ξ ( T ) = ρ x ξ ( 0 )
証明. §E13.11 定理 2.1 をI = [ 0 , T ] I=[0,T] I = [ 0 , T ] t 0 = 0 t_0=0 t 0 = 0 η = ξ \boldsymbol\eta=\boldsymbol\xi η = ξ x ξ \boldsymbol x_{\boldsymbol\xi} x ξ x ( 0 ) = ξ \boldsymbol x(0)=\boldsymbol\xi x ( 0 ) = ξ ξ ↦ x ξ \boldsymbol\xi\mapsto\boldsymbol x_{\boldsymbol\xi} ξ ↦ x ξ x ↦ x ( 0 ) \boldsymbol x\mapsto\boldsymbol x(0) x ↦ x ( 0 ) x ξ ( T ) = M ξ + M c \boldsymbol x_{\boldsymbol\xi}(T)=M\boldsymbol\xi+M\boldsymbol c x ξ ( T ) = M ξ + M c x ξ ( 0 ) = ξ \boldsymbol x_{\boldsymbol\xi}(0)=\boldsymbol\xi x ξ ( 0 ) = ξ x ξ ( T ) = ρ x ξ ( 0 ) \boldsymbol x_{\boldsymbol\xi}(T)=\rho\,\boldsymbol x_{\boldsymbol\xi}(0) x ξ ( T ) = ρ x ξ ( 0 ) ( ρ I − M ) ξ = M c (\rho I-M)\boldsymbol\xi=M\boldsymbol c ( ρ I − M ) ξ = M c ▨
定理 4.4 (周期系の Fredholm の交代定理). K ∈ { R , C } \mathbb K\in\{\R,\C\} K ∈ { R , C } d ∈ N ≥ 1 d\in\NN d ∈ N ≥ 1 T > 0 T>0 T > 0 A ∈ L 1 ( [ 0 , T ] ; K d × d ) A\in L^1([0,T];\mathbb K^{d\times d}) A ∈ L 1 ([ 0 , T ] ; K d × d ) b ∈ L 1 ( [ 0 , T ] ; K d ) \boldsymbol b\in L^1([0,T];\mathbb K^d) b ∈ L 1 ([ 0 , T ] ; K d ) ρ ∈ K ∖ { 0 } \rho\in\mathbb K\setminus\{0\} ρ ∈ K ∖ { 0 } Φ 0 \Phi_0 Φ 0 x ′ = A ( t ) x \boldsymbol x'=A(t)\boldsymbol x x ′ = A ( t ) x [ 0 , T ] [0,T] [ 0 , T ] Φ 0 ( 0 ) = I \Phi_0(0)=I Φ 0 ( 0 ) = I M = Φ 0 ( T ) M=\Phi_0(T) M = Φ 0 ( T )
x ′ = A ( t ) x + b ( t ) ほとんど至る所 , x ( T ) = ρ x ( 0 ) \boldsymbol x'=A(t)\boldsymbol x+\boldsymbol b(t)\quad\text{ほとんど至る所},\qquad\boldsymbol x(T)=\rho\,\boldsymbol x(0) x ′ = A ( t ) x + b ( t ) ほとんど至る所 , x ( T ) = ρ x ( 0 ) の解とは、これを満たすx ∈ A C ( [ 0 , T ] ; K d ) \boldsymbol x\in AC([0,T];\mathbb K^d) x ∈ A C ([ 0 , T ] ; K d ) b = 0 \boldsymbol b=\boldsymbol 0 b = 0 N ρ \mathcal N_\rho N ρ
ψ ′ = − A ( t ) ∗ ψ ほとんど至る所 , ψ ( T ) = ρ ‾ − 1 ψ ( 0 ) \boldsymbol\psi'=-A(t)^*\boldsymbol\psi\quad\text{ほとんど至る所},\qquad\boldsymbol\psi(T)=\overline\rho^{\,-1}\boldsymbol\psi(0) ψ ′ = − A ( t ) ∗ ψ ほとんど至る所 , ψ ( T ) = ρ − 1 ψ ( 0 ) を満たすψ ∈ A C ( [ 0 , T ] ; K d ) \boldsymbol\psi\in AC([0,T];\mathbb K^d) ψ ∈ A C ([ 0 , T ] ; K d ) N ρ ∗ \mathcal N^*_\rho N ρ ∗
境界値問題が解をもつことは、すべてのψ ∈ N ρ ∗ \boldsymbol\psi\in\mathcal N^*_\rho ψ ∈ N ρ ∗ ∫ 0 T ψ ( t ) ∗ b ( t ) d t = 0 \int_0^T\boldsymbol\psi(t)^*\boldsymbol b(t)\,dt=0 ∫ 0 T ψ ( t ) ∗ b ( t ) d t = 0
dim K N ρ = dim K N ρ ∗ = dim K ker ( M − ρ I ) \dim_{\mathbb K}\mathcal N_\rho=\dim_{\mathbb K}\mathcal N^*_\rho=\dim_{\mathbb K}\ker(M-\rho I) dim K N ρ = dim K N ρ ∗ = dim K ker ( M − ρ I ) 境界値問題が解x p \boldsymbol x_p x p x p + N ρ \boldsymbol x_p+\mathcal N_\rho x p + N ρ
特に、A ∈ L l o c 1 ( R ; K d × d ) A\in L^1_{\mathrm{loc}}(\R;\mathbb K^{d\times d}) A ∈ L loc 1 ( R ; K d × d ) b ∈ L l o c 1 ( R ; K d ) \boldsymbol b\in L^1_{\mathrm{loc}}(\R;\mathbb K^d) b ∈ L loc 1 ( R ; K d ) t t t A ( t + T ) = A ( t ) A(t+T)=A(t) A ( t + T ) = A ( t ) b ( t + T ) = b ( t ) \boldsymbol b(t+T)=\boldsymbol b(t) b ( t + T ) = b ( t ) x ′ = A ( t ) x + b ( t ) \boldsymbol x'=A(t)\boldsymbol x+\boldsymbol b(t) x ′ = A ( t ) x + b ( t ) R \R R T T T [ 0 , T ] [0,T] [ 0 , T ] ρ = 1 \rho=1 ρ = 1 T T T ψ ′ = − A ( t ) ∗ ψ \boldsymbol\psi'=-A(t)^*\boldsymbol\psi ψ ′ = − A ( t ) ∗ ψ R \R R T T T ψ \boldsymbol\psi ψ ∫ 0 T ψ ∗ b d t = 0 \int_0^T\boldsymbol\psi^*\boldsymbol b\,dt=0 ∫ 0 T ψ ∗ b d t = 0
証明. c = ∫ 0 T Φ 0 ( s ) − 1 b ( s ) d s \boldsymbol c=\int_0^T\Phi_0(s)^{-1}\boldsymbol b(s)\,ds c = ∫ 0 T Φ 0 ( s ) − 1 b ( s ) d s 補題 4.3 により、境界値問題の解は( ρ I − M ) ξ = M c (\rho I-M)\boldsymbol\xi=M\boldsymbol c ( ρ I − M ) ξ = M c ξ \boldsymbol\xi ξ x ξ \boldsymbol x_{\boldsymbol\xi} x ξ M c ∈ im ( ρ I − M ) M\boldsymbol c\in\operatorname{im}(\rho I-M) M c ∈ im ( ρ I − M ) b = 0 \boldsymbol b=\boldsymbol 0 b = 0 c = 0 \boldsymbol c=\boldsymbol 0 c = 0 x ξ = Φ 0 ξ \boldsymbol x_{\boldsymbol\xi}=\Phi_0\boldsymbol\xi x ξ = Φ 0 ξ ξ ↦ Φ 0 ξ \boldsymbol\xi\mapsto\Phi_0\boldsymbol\xi ξ ↦ Φ 0 ξ ker ( ρ I − M ) \ker(\rho I-M) ker ( ρ I − M ) N ρ \mathcal N_\rho N ρ
補題 4.2 (2) によりΨ 0 = ( Φ 0 ∗ ) − 1 \Psi_0=(\Phi_0^*)^{-1} Ψ 0 = ( Φ 0 ∗ ) − 1 Ψ 0 ( 0 ) = I \Psi_0(0)=I Ψ 0 ( 0 ) = I §E13.9 定理 2.2 により随伴系の各解はψ = Ψ 0 ζ \boldsymbol\psi=\Psi_0\boldsymbol\zeta ψ = Ψ 0 ζ ζ = ψ ( 0 ) \boldsymbol\zeta=\boldsymbol\psi(0) ζ = ψ ( 0 ) ψ ( T ) = ( M ∗ ) − 1 ζ \boldsymbol\psi(T)=(M^*)^{-1}\boldsymbol\zeta ψ ( T ) = ( M ∗ ) − 1 ζ ψ ∈ N ρ ∗ \boldsymbol\psi\in\mathcal N^*_\rho ψ ∈ N ρ ∗ M ∗ ζ = ρ ‾ ζ M^*\boldsymbol\zeta=\overline\rho\,\boldsymbol\zeta M ∗ ζ = ρ ζ ζ ∈ ker ( ( ρ I − M ) ∗ ) \boldsymbol\zeta\in\ker\bigl((\rho I-M)^*\bigr) ζ ∈ ker ( ( ρ I − M ) ∗ ) ζ ↦ Ψ 0 ζ \boldsymbol\zeta\mapsto\Psi_0\boldsymbol\zeta ζ ↦ Ψ 0 ζ ker ( ( ρ I − M ) ∗ ) \ker\bigl((\rho I-M)^*\bigr) ker ( ( ρ I − M ) ∗ ) N ρ ∗ \mathcal N^*_\rho N ρ ∗ ψ = Ψ 0 ζ ∈ N ρ ∗ \boldsymbol\psi=\Psi_0\boldsymbol\zeta\in\mathcal N^*_\rho ψ = Ψ 0 ζ ∈ N ρ ∗ ψ ( t ) ∗ = ζ ∗ Φ 0 ( t ) − 1 \boldsymbol\psi(t)^*=\boldsymbol\zeta^*\Phi_0(t)^{-1} ψ ( t ) ∗ = ζ ∗ Φ 0 ( t ) − 1
∫ 0 T ψ ∗ b d t = ζ ∗ c , ζ ∗ M c = ( M ∗ ζ ) ∗ c = ρ ζ ∗ c \int_0^T\boldsymbol\psi^*\boldsymbol b\,dt=\boldsymbol\zeta^*\boldsymbol c,\qquad
\boldsymbol\zeta^*M\boldsymbol c=(M^*\boldsymbol\zeta)^*\boldsymbol c=\rho\,\boldsymbol\zeta^*\boldsymbol c ∫ 0 T ψ ∗ b d t = ζ ∗ c , ζ ∗ M c = ( M ∗ ζ ) ∗ c = ρ ζ ∗ c である。
正方行列N ∈ K d × d N\in\mathbb K^{d\times d} N ∈ K d × d im N \operatorname{im}N im N ker N ∗ \ker N^* ker N ∗ ζ \boldsymbol\zeta ζ ζ ∗ v = 0 \boldsymbol\zeta^*\boldsymbol v=0 ζ ∗ v = 0 v \boldsymbol v v V V V ζ ∈ ker N ∗ \boldsymbol\zeta\in\ker N^* ζ ∈ ker N ∗ ζ ∗ N v = ( N ∗ ζ ) ∗ v = 0 \boldsymbol\zeta^*N\boldsymbol v=(N^*\boldsymbol\zeta)^*\boldsymbol v=0 ζ ∗ N v = ( N ∗ ζ ) ∗ v = 0 im N ⊆ V \operatorname{im}N\subseteq V im N ⊆ V ker N ∗ \ker N^* ker N ∗ ζ 1 , … , ζ k \boldsymbol\zeta_1,\ldots,\boldsymbol\zeta_k ζ 1 , … , ζ k ζ 1 ∗ , … , ζ k ∗ \boldsymbol\zeta_1^*,\ldots,\boldsymbol\zeta_k^* ζ 1 ∗ , … , ζ k ∗ k × d k\times d k × d k k k V V V dim V = d − k = rank N ∗ = rank N = dim im N \dim V=d-k=\operatorname{rank}N^*=\operatorname{rank}N=\dim\operatorname{im}N dim V = d − k = rank N ∗ = rank N = dim im N
主張 (1) を示す。N = ρ I − M N=\rho I-M N = ρ I − M ζ ∈ ker ( ( ρ I − M ) ∗ ) \boldsymbol\zeta\in\ker\bigl((\rho I-M)^*\bigr) ζ ∈ ker ( ( ρ I − M ) ∗ ) ζ ∗ M c = 0 \boldsymbol\zeta^*M\boldsymbol c=0 ζ ∗ M c = 0 ρ ≠ 0 \rho\ne0 ρ = 0 ζ ∈ ker ( ( ρ I − M ) ∗ ) \boldsymbol\zeta\in\ker\bigl((\rho I-M)^*\bigr) ζ ∈ ker ( ( ρ I − M ) ∗ ) ζ ∗ c = 0 \boldsymbol\zeta^*\boldsymbol c=0 ζ ∗ c = 0 ψ ∈ N ρ ∗ \boldsymbol\psi\in\mathcal N^*_\rho ψ ∈ N ρ ∗ ∫ 0 T ψ ∗ b d t = 0 \int_0^T\boldsymbol\psi^*\boldsymbol b\,dt=0 ∫ 0 T ψ ∗ b d t = 0
主張 (2) を示す。第一段によりdim N ρ = dim ker ( M − ρ I ) \dim\mathcal N_\rho=\dim\ker(M-\rho I) dim N ρ = dim ker ( M − ρ I ) dim N ρ ∗ = dim ker ( ( ρ I − M ) ∗ ) = d − rank ( ρ I − M ) = dim ker ( M − ρ I ) \dim\mathcal N^*_\rho=\dim\ker\bigl((\rho I-M)^*\bigr)=d-\operatorname{rank}(\rho I-M)=\dim\ker(M-\rho I) dim N ρ ∗ = dim ker ( ( ρ I − M ) ∗ ) = d − rank ( ρ I − M ) = dim ker ( M − ρ I )
主張 (3) を示す。境界値問題の二つの解の差はb = 0 \boldsymbol b=\boldsymbol 0 b = 0 x p \boldsymbol x_p x p N ρ \mathcal N_\rho N ρ
主張 (4) を示す。R \R R T T T x ( T ) = x ( 0 ) \boldsymbol x(T)=\boldsymbol x(0) x ( T ) = x ( 0 ) y \boldsymbol y y ρ = 1 \rho=1 ρ = 1 §E13.9 定理 1.1 によりx ( 0 ) = y ( 0 ) \boldsymbol x(0)=\boldsymbol y(0) x ( 0 ) = y ( 0 ) R \R R x \boldsymbol x x [ 0 , T ] [0,T] [ 0 , T ] x ∣ [ 0 , T ] = y \boldsymbol x|_{[0,T]}=\boldsymbol y x ∣ [ 0 , T ] = y z ( t ) = x ( t + T ) \boldsymbol z(t)=\boldsymbol x(t+T) z ( t ) = x ( t + T ) A A A b \boldsymbol b b z ( 0 ) = y ( T ) = y ( 0 ) = x ( 0 ) \boldsymbol z(0)=\boldsymbol y(T)=\boldsymbol y(0)=\boldsymbol x(0) z ( 0 ) = y ( T ) = y ( 0 ) = x ( 0 ) z = x \boldsymbol z=\boldsymbol x z = x x \boldsymbol x x T T T T T T [ 0 , T ] [0,T] [ 0 , T ] R \R R − A ∗ -A^* − A ∗ T T T b = 0 \boldsymbol b=\boldsymbol 0 b = 0 R \R R T T T N 1 ∗ \mathcal N^*_1 N 1 ∗ 主張 (1) をρ = 1 \rho=1 ρ = 1 ▨
系 4.5. K ∈ { R , C } \mathbb K\in\{\R,\C\} K ∈ { R , C } d ∈ N ≥ 1 d\in\NN d ∈ N ≥ 1 T > 0 T>0 T > 0 K d \mathbb K^d K d ∥ ⋅ ∥ \lVert\cdot\rVert ∥ ⋅ ∥ A ∈ L 1 ( [ 0 , T ] ; K d × d ) A\in L^1([0,T];\mathbb K^{d\times d}) A ∈ L 1 ([ 0 , T ] ; K d × d ) ρ ∈ K ∖ { 0 } \rho\in\mathbb K\setminus\{0\} ρ ∈ K ∖ { 0 } Φ 0 \Phi_0 Φ 0 x ′ = A ( t ) x \boldsymbol x'=A(t)\boldsymbol x x ′ = A ( t ) x [ 0 , T ] [0,T] [ 0 , T ] Φ 0 ( 0 ) = I \Phi_0(0)=I Φ 0 ( 0 ) = I M = Φ 0 ( T ) M=\Phi_0(T) M = Φ 0 ( T ) ρ \rho ρ M M M b ∈ L 1 ( [ 0 , T ] ; K d ) \boldsymbol b\in L^1([0,T];\mathbb K^d) b ∈ L 1 ([ 0 , T ] ; K d ) x ′ = A ( t ) x + b ( t ) \boldsymbol x'=A(t)\boldsymbol x+\boldsymbol b(t) x ′ = A ( t ) x + b ( t ) x ( T ) = ρ x ( 0 ) \boldsymbol x(T)=\rho\boldsymbol x(0) x ( T ) = ρ x ( 0 ) A C ( [ 0 , T ] ; K d ) AC([0,T];\mathbb K^d) A C ([ 0 , T ] ; K d ) t ∈ [ 0 , T ] t\in[0,T] t ∈ [ 0 , T ]
x ( t ) = ∫ 0 T G ( t , s ) b ( s ) d s , G ( t , s ) = { Φ 0 ( t ) ( ρ I − M ) − 1 ρ Φ 0 ( s ) − 1 ( 0 ≤ s ≤ t ≤ T ) , Φ 0 ( t ) ( ρ I − M ) − 1 M Φ 0 ( s ) − 1 ( 0 ≤ t < s ≤ T ) \boldsymbol x(t)=\int_0^TG(t,s)\boldsymbol b(s)\,ds,\qquad
G(t,s)=\begin{cases}\Phi_0(t)(\rho I-M)^{-1}\rho\,\Phi_0(s)^{-1}&(0\le s\le t\le T),\\[2pt]\Phi_0(t)(\rho I-M)^{-1}M\,\Phi_0(s)^{-1}&(0\le t<s\le T)\end{cases} x ( t ) = ∫ 0 T G ( t , s ) b ( s ) d s , G ( t , s ) = { Φ 0 ( t ) ( ρ I − M ) − 1 ρ Φ 0 ( s ) − 1 Φ 0 ( t ) ( ρ I − M ) − 1 M Φ 0 ( s ) − 1 ( 0 ≤ s ≤ t ≤ T ) , ( 0 ≤ t < s ≤ T ) と表される。さらにK G = sup t , s ∈ [ 0 , T ] ∥ G ( t , s ) ∥ < ∞ K_G=\sup_{t,s\in[0,T]}\lVert G(t,s)\rVert<\infty K G = sup t , s ∈ [ 0 , T ] ∥ G ( t , s )∥ < ∞ sup t ∈ [ 0 , T ] ∥ x ( t ) ∥ ≤ K G ∫ 0 T ∥ b ( s ) ∥ d s \sup_{t\in[0,T]}\lVert\boldsymbol x(t)\rVert\le K_G\int_0^T\lVert\boldsymbol b(s)\rVert\,ds sup t ∈ [ 0 , T ] ∥ x ( t )∥ ≤ K G ∫ 0 T ∥ b ( s )∥ d s
証明. ρ I − M \rho I-M ρ I − M c = ∫ 0 T Φ 0 ( s ) − 1 b ( s ) d s \boldsymbol c=\int_0^T\Phi_0(s)^{-1}\boldsymbol b(s)\,ds c = ∫ 0 T Φ 0 ( s ) − 1 b ( s ) d s 補題 4.3 により、境界値問題の解は( ρ I − M ) ξ = M c (\rho I-M)\boldsymbol\xi=M\boldsymbol c ( ρ I − M ) ξ = M c ξ = ( ρ I − M ) − 1 M c \boldsymbol\xi=(\rho I-M)^{-1}M\boldsymbol c ξ = ( ρ I − M ) − 1 M c ξ \boldsymbol\xi ξ x ξ \boldsymbol x_{\boldsymbol\xi} x ξ c \boldsymbol c c [ 0 , t ] [0,t] [ 0 , t ] ( t , T ] (t,T] ( t , T ]
x ( t ) = ∫ 0 t Φ 0 ( t ) [ ( ρ I − M ) − 1 M + I ] Φ 0 ( s ) − 1 b ( s ) d s + ∫ t T Φ 0 ( t ) ( ρ I − M ) − 1 M Φ 0 ( s ) − 1 b ( s ) d s \boldsymbol x(t)=\int_0^t\Phi_0(t)\bigl[(\rho I-M)^{-1}M+I\bigr]\Phi_0(s)^{-1}\boldsymbol b(s)\,ds+\int_t^T\Phi_0(t)(\rho I-M)^{-1}M\Phi_0(s)^{-1}\boldsymbol b(s)\,ds x ( t ) = ∫ 0 t Φ 0 ( t ) [ ( ρ I − M ) − 1 M + I ] Φ 0 ( s ) − 1 b ( s ) d s + ∫ t T Φ 0 ( t ) ( ρ I − M ) − 1 M Φ 0 ( s ) − 1 b ( s ) d s である。( ρ I − M ) − 1 M + I = ( ρ I − M ) − 1 ( M + ρ I − M ) = ( ρ I − M ) − 1 ρ (\rho I-M)^{-1}M+I=(\rho I-M)^{-1}(M+\rho I-M)=(\rho I-M)^{-1}\rho ( ρ I − M ) − 1 M + I = ( ρ I − M ) − 1 ( M + ρ I − M ) = ( ρ I − M ) − 1 ρ Φ 0 \Phi_0 Φ 0 Φ 0 − 1 \Phi_0^{-1} Φ 0 − 1 [ 0 , T ] [0,T] [ 0 , T ] K G < ∞ K_G<\infty K G < ∞ t ∈ [ 0 , T ] t\in[0,T] t ∈ [ 0 , T ] §E13.2 補題 1.3 (2) を可積分関数s ↦ G ( t , s ) b ( s ) s\mapsto G(t,s)\boldsymbol b(s) s ↦ G ( t , s ) b ( s ) ∥ ⋅ ∥ \lVert\cdot\rVert ∥ ⋅ ∥ ∥ x ( t ) ∥ ≤ ∫ 0 T ∥ G ( t , s ) ∥ ∥ b ( s ) ∥ d s ≤ K G ∫ 0 T ∥ b ( s ) ∥ d s \lVert\boldsymbol x(t)\rVert\le\int_0^T\lVert G(t,s)\rVert\lVert\boldsymbol b(s)\rVert\,ds\le K_G\int_0^T\lVert\boldsymbol b(s)\rVert\,ds ∥ x ( t )∥ ≤ ∫ 0 T ∥ G ( t , s )∥ ∥ b ( s )∥ d s ≤ K G ∫ 0 T ∥ b ( s )∥ d s ▨
証明. 主張 (1) を示す。x ∈ C T \boldsymbol x\in C_T x ∈ C T sup t ∈ R ∣ x ( t ) ∣ = max t ∈ [ 0 , T ] ∣ x ( t ) ∣ < ∞ \sup_{t\in\R}\lvert\boldsymbol x(t)\rvert=\max_{t\in[0,T]}\lvert\boldsymbol x(t)\rvert<\infty sup t ∈ R ∣ x ( t )∣ = max t ∈ [ 0 , T ] ∣ x ( t )∣ < ∞ C T C_T C T R \R R C b ( R ; K d ) C_b(\R;\mathbb K^d) C b ( R ; K d ) K d \mathbb K^d K d ∣ ⋅ ∣ \lvert\cdot\rvert ∣ ⋅ ∣ ∥ ⋅ ∥ ∞ \lVert\cdot\rVert_\infty ∥ ⋅ ∥ ∞ C T C_T C T ( x n ) (\boldsymbol x_n) ( x n ) C b ( R ; K d ) C_b(\R;\mathbb K^d) C b ( R ; K d ) x \boldsymbol x x t ∈ R t\in\R t ∈ R x ( t + T ) = lim n x n ( t + T ) = lim n x n ( t ) = x ( t ) \boldsymbol x(t+T)=\lim_n\boldsymbol x_n(t+T)=\lim_n\boldsymbol x_n(t)=\boldsymbol x(t) x ( t + T ) = lim n x n ( t + T ) = lim n x n ( t ) = x ( t ) x ∈ C T \boldsymbol x\in C_T x ∈ C T C T C_T C T §E13.4 補題 3.2 により( C T , ∥ ⋅ ∥ ∞ ) (C_T,\lVert\cdot\rVert_\infty) ( C T , ∥ ⋅ ∥ ∞ ) ( x n ) (\boldsymbol x_n) ( x n ) C T 1 C^1_T C T 1 C T C_T C T x n → x \boldsymbol x_n\to\boldsymbol x x n → x x n ′ → w \boldsymbol x_n'\to\boldsymbol w x n ′ → w §E13.2 補題 1.3 (2) によりx n ( t ) = x n ( 0 ) + ∫ 0 t x n ′ \boldsymbol x_n(t)=\boldsymbol x_n(0)+\int_0^t\boldsymbol x_n' x n ( t ) = x n ( 0 ) + ∫ 0 t x n ′ x ( t ) = x ( 0 ) + ∫ 0 t w \boldsymbol x(t)=\boldsymbol x(0)+\int_0^t\boldsymbol w x ( t ) = x ( 0 ) + ∫ 0 t w w \boldsymbol w w x ∈ C 1 \boldsymbol x\in C^1 x ∈ C 1 x ′ = w \boldsymbol x'=\boldsymbol w x ′ = w ∥ x n − x ∥ C 1 → 0 \lVert\boldsymbol x_n-\boldsymbol x\rVert_{C^1}\to0 ∥ x n − x ∥ C 1 → 0 ∥ A ∥ ∞ = max t ∈ [ 0 , T ] ∥ A ( t ) ∥ \lVert A\rVert_\infty=\max_{t\in[0,T]}\lVert A(t)\rVert ∥ A ∥ ∞ = max t ∈ [ 0 , T ] ∥ A ( t )∥ L x L\boldsymbol x L x T T T ∥ L x ∥ ∞ ≤ max ( 1 , ∥ A ∥ ∞ ) ∥ x ∥ C 1 \lVert L\boldsymbol x\rVert_\infty\le\max(1,\lVert A\rVert_\infty)\lVert\boldsymbol x\rVert_{C^1} ∥ L x ∥ ∞ ≤ max ( 1 , ∥ A ∥ ∞ ) ∥ x ∥ C 1
b ∈ C ( R ; K d ) \boldsymbol b\in C(\R;\mathbb K^d) b ∈ C ( R ; K d ) x \boldsymbol x x x ′ = A x + b \boldsymbol x'=A\boldsymbol x+\boldsymbol b x ′ = A x + b x \boldsymbol x x A x + b A\boldsymbol x+\boldsymbol b A x + b §E13.2 系 2.7 (1) によりx \boldsymbol x x C 1 C^1 C 1 t ∈ R t\in\R t ∈ R x ′ ( t ) = A ( t ) x ( t ) + b ( t ) \boldsymbol x'(t)=A(t)\boldsymbol x(t)+\boldsymbol b(t) x ′ ( t ) = A ( t ) x ( t ) + b ( t ) − A ∗ -A^* − A ∗ C 1 C^1 C 1 C 1 C^1 C 1 §E13.2 系 2.7 (2) により局所絶対連続である。従ってN \mathcal N N x ′ = A ( t ) x \boldsymbol x'=A(t)\boldsymbol x x ′ = A ( t ) x T T T N ∗ \mathcal N^* N ∗ T T T
主張 (2) を示す。R \R R [ 0 , T ] [0,T] [ 0 , T ] [ 0 , T ] [0,T] [ 0 , T ] 定理 4.4 のM M M 定理 4.4 (4) の制限写像によりN \mathcal N N N ∗ \mathcal N^* N ∗ N 1 \mathcal N_1 N 1 N 1 ∗ \mathcal N^*_1 N 1 ∗ 定理 4.4 (2) により次元はともにdim ker ( M − I ) ≤ d \dim\ker(M-I)\le d dim ker ( M − I ) ≤ d
主張 (3) を示す。b ∈ C T \boldsymbol b\in C_T b ∈ C T im L \im L im L x ′ = A x + b \boldsymbol x'=A\boldsymbol x+\boldsymbol b x ′ = A x + b T T T 定理 4.4 (4) により、b ∈ im L \boldsymbol b\in\im L b ∈ im L ψ ∈ N ∗ \boldsymbol\psi\in\mathcal N^* ψ ∈ N ∗ ℓ ψ ( b ) = ∫ 0 T ψ ∗ b d t \ell_{\boldsymbol\psi}(\boldsymbol b)=\int_0^T\boldsymbol\psi^*\boldsymbol b\,dt ℓ ψ ( b ) = ∫ 0 T ψ ∗ b d t ∣ ℓ ψ ( b ) ∣ ≤ T ∥ ψ ∥ ∞ ∥ b ∥ ∞ \lvert\ell_{\boldsymbol\psi}(\boldsymbol b)\rvert\le T\lVert\boldsymbol\psi\rVert_\infty\lVert\boldsymbol b\rVert_\infty ∣ ℓ ψ ( b )∣ ≤ T ∥ ψ ∥ ∞ ∥ b ∥ ∞ ℓ ψ \ell_{\boldsymbol\psi} ℓ ψ im L = ⋂ ψ ker ℓ ψ \im L=\bigcap_{\boldsymbol\psi}\ker\ell_{\boldsymbol\psi} im L = ⋂ ψ ker ℓ ψ k = dim N ∗ k=\dim\mathcal N^* k = dim N ∗ k ≥ 1 k\ge1 k ≥ 1 N ∗ \mathcal N^* N ∗ ψ 1 , … , ψ k \boldsymbol\psi_1,\ldots,\boldsymbol\psi_k ψ 1 , … , ψ k Γ i j = ∫ 0 T ψ i ∗ ψ j d t \Gamma_{ij}=\int_0^T\boldsymbol\psi_i^*\boldsymbol\psi_j\,dt Γ ij = ∫ 0 T ψ i ∗ ψ j d t a ∈ K k \boldsymbol a\in\mathbb K^k a ∈ K k a ∗ Γ a = ∫ 0 T ∣ ∑ j a j ψ j ∣ 2 d t \boldsymbol a^*\Gamma\boldsymbol a=\int_0^T\bigl\lvert\sum_ja_j\boldsymbol\psi_j\bigr\rvert^2dt a ∗ Γ a = ∫ 0 T ∑ j a j ψ j 2 d t 0 0 0 ∑ j a j ψ j \sum_ja_j\boldsymbol\psi_j ∑ j a j ψ j 0 \boldsymbol 0 0 a = 0 \boldsymbol a=\boldsymbol 0 a = 0 Γ \Gamma Γ ℓ ( b ) = ( ℓ ψ 1 ( b ) , … , ℓ ψ k ( b ) ) T \boldsymbol\ell(\boldsymbol b)=(\ell_{\boldsymbol\psi_1}(\boldsymbol b),\ldots,\ell_{\boldsymbol\psi_k}(\boldsymbol b))^{\mathsf T} ℓ ( b ) = ( ℓ ψ 1 ( b ) , … , ℓ ψ k ( b ) ) T Q b = ∑ j ( Γ − 1 ℓ ( b ) ) j ψ j Q\boldsymbol b=\sum_j\bigl(\Gamma^{-1}\boldsymbol\ell(\boldsymbol b)\bigr)_j\boldsymbol\psi_j Q b = ∑ j ( Γ − 1 ℓ ( b ) ) j ψ j Q Q Q N ∗ \mathcal N^* N ∗ ψ = ∑ j a j ψ j \boldsymbol\psi=\sum_ja_j\boldsymbol\psi_j ψ = ∑ j a j ψ j ℓ ( ψ ) = Γ a \boldsymbol\ell(\boldsymbol\psi)=\Gamma\boldsymbol a ℓ ( ψ ) = Γ a Q ψ = ψ Q\boldsymbol\psi=\boldsymbol\psi Q ψ = ψ ℓ ( b − Q b ) = ℓ ( b ) − Γ Γ − 1 ℓ ( b ) = 0 \boldsymbol\ell(\boldsymbol b-Q\boldsymbol b)=\boldsymbol\ell(\boldsymbol b)-\Gamma\Gamma^{-1}\boldsymbol\ell(\boldsymbol b)=\boldsymbol 0 ℓ ( b − Q b ) = ℓ ( b ) − Γ Γ − 1 ℓ ( b ) = 0 b − Q b ∈ im L \boldsymbol b-Q\boldsymbol b\in\im L b − Q b ∈ im L ψ = ∑ j a j ψ j ∈ im L \boldsymbol\psi=\sum_ja_j\boldsymbol\psi_j\in\im L ψ = ∑ j a j ψ j ∈ im L Γ a = ℓ ( ψ ) = 0 \Gamma\boldsymbol a=\boldsymbol\ell(\boldsymbol\psi)=\boldsymbol 0 Γ a = ℓ ( ψ ) = 0 ψ = 0 \boldsymbol\psi=\boldsymbol 0 ψ = 0 C T = im L ⊕ N ∗ C_T=\im L\oplus\mathcal N^* C T = im L ⊕ N ∗ Q Q Q N ∗ \mathcal N^* N ∗ k = 0 k=0 k = 0 im L = C T \im L=C_T im L = C T Q = 0 Q=0 Q = 0
主張 (4) を示す。N \mathcal N N ϕ 1 , … , ϕ k \boldsymbol\phi_1,\ldots,\boldsymbol\phi_k ϕ 1 , … , ϕ k H i j = ∫ 0 T ϕ i ∗ ϕ j d t H_{ij}=\int_0^T\boldsymbol\phi_i^*\boldsymbol\phi_j\,dt H ij = ∫ 0 T ϕ i ∗ ϕ j d t Π x = ∑ j ( H − 1 m ( x ) ) j ϕ j \Pi\boldsymbol x=\sum_j\bigl(H^{-1}\boldsymbol m(\boldsymbol x)\bigr)_j\boldsymbol\phi_j Π x = ∑ j ( H − 1 m ( x ) ) j ϕ j m ( x ) i = ∫ 0 T ϕ i ∗ x d t \boldsymbol m(\boldsymbol x)_i=\int_0^T\boldsymbol\phi_i^*\boldsymbol x\,dt m ( x ) i = ∫ 0 T ϕ i ∗ x d t C T 1 = N ⊕ X 1 C^1_T=\mathcal N\oplus X_1 C T 1 = N ⊕ X 1 N \mathcal N N k = 0 k=0 k = 0 Π = 0 \Pi=0 Π = 0 X 1 = C T 1 X_1=C^1_T X 1 = C T 1 ∣ m ( x ) i ∣ ≤ T ∥ ϕ i ∥ ∞ ∥ x ∥ ∞ \lvert\boldsymbol m(\boldsymbol x)_i\rvert\le T\lVert\boldsymbol\phi_i\rVert_\infty\lVert\boldsymbol x\rVert_\infty ∣ m ( x ) i ∣ ≤ T ∥ ϕ i ∥ ∞ ∥ x ∥ ∞ C Π ≥ 0 C_\Pi\ge0 C Π ≥ 0 ∥ Π x ∥ ∞ ≤ C Π ∥ x ∥ ∞ \lVert\Pi\boldsymbol x\rVert_\infty\le C_\Pi\lVert\boldsymbol x\rVert_\infty ∥ Π x ∥ ∞ ≤ C Π ∥ x ∥ ∞ ker L ∩ X 1 = N ∩ X 1 = { 0 } \ker L\cap X_1=\mathcal N\cap X_1=\{\boldsymbol 0\} ker L ∩ X 1 = N ∩ X 1 = { 0 } L ∣ X 1 L|_{X_1} L ∣ X 1
Φ 0 \Phi_0 Φ 0 Φ 0 ( 0 ) = I \Phi_0(0)=I Φ 0 ( 0 ) = I K Φ = max [ 0 , T ] ∥ Φ 0 ∥ K_\Phi=\max_{[0,T]}\lVert\Phi_0\rVert K Φ = max [ 0 , T ] ∥ Φ 0 ∥ K Φ ′ = max [ 0 , T ] ∥ Φ 0 − 1 ∥ K'_\Phi=\max_{[0,T]}\lVert\Phi_0^{-1}\rVert K Φ ′ = max [ 0 , T ] ∥ Φ 0 − 1 ∥ ( I − M ) R v = v (I-M)R\boldsymbol v=\boldsymbol v ( I − M ) R v = v v ∈ im ( I − M ) \boldsymbol v\in\operatorname{im}(I-M) v ∈ im ( I − M ) R ∈ K d × d R\in\mathbb K^{d\times d} R ∈ K d × d im ( I − M ) \operatorname{im}(I-M) im ( I − M ) 0 \boldsymbol 0 0 b ∈ im L \boldsymbol b\in\im L b ∈ im L c ( b ) = ∫ 0 T Φ 0 ( s ) − 1 b ( s ) d s \boldsymbol c(\boldsymbol b)=\int_0^T\Phi_0(s)^{-1}\boldsymbol b(s)\,ds c ( b ) = ∫ 0 T Φ 0 ( s ) − 1 b ( s ) d s L x = b L\boldsymbol x=\boldsymbol b L x = b x ∈ C T 1 \boldsymbol x\in C^1_T x ∈ C T 1 [ 0 , T ] [0,T] [ 0 , T ] ρ = 1 \rho=1 ρ = 1 補題 4.3 により( I − M ) x ( 0 ) = M c ( b ) (I-M)\boldsymbol x(0)=M\boldsymbol c(\boldsymbol b) ( I − M ) x ( 0 ) = M c ( b ) M c ( b ) ∈ im ( I − M ) M\boldsymbol c(\boldsymbol b)\in\operatorname{im}(I-M) M c ( b ) ∈ im ( I − M )
x b ( t ) = Φ 0 ( t ) [ R M c ( b ) + ∫ 0 t Φ 0 ( s ) − 1 b ( s ) d s ] ( t ∈ R ) \boldsymbol x_{\boldsymbol b}(t)=\Phi_0(t)\left[RM\boldsymbol c(\boldsymbol b)+\int_0^t\Phi_0(s)^{-1}\boldsymbol b(s)\,ds\right]\qquad(t\in\R) x b ( t ) = Φ 0 ( t ) [ R M c ( b ) + ∫ 0 t Φ 0 ( s ) − 1 b ( s ) d s ] ( t ∈ R ) とおくと、§E13.11 定理 2.1 によりx b \boldsymbol x_{\boldsymbol b} x b x ′ = A x + b \boldsymbol x'=A\boldsymbol x+\boldsymbol b x ′ = A x + b R \R R ( I − M ) R M c ( b ) = M c ( b ) (I-M)RM\boldsymbol c(\boldsymbol b)=M\boldsymbol c(\boldsymbol b) ( I − M ) R M c ( b ) = M c ( b ) 補題 4.3 をξ = R M c ( b ) \boldsymbol\xi=RM\boldsymbol c(\boldsymbol b) ξ = R M c ( b ) ρ = 1 \rho=1 ρ = 1 x b ( T ) = x b ( 0 ) \boldsymbol x_{\boldsymbol b}(T)=\boldsymbol x_{\boldsymbol b}(0) x b ( T ) = x b ( 0 ) 定理 4.4 (4) の制限写像の全単射性と初期値問題の一意性によりx b \boldsymbol x_{\boldsymbol b} x b T T T x b ∈ C T 1 \boldsymbol x_{\boldsymbol b}\in C^1_T x b ∈ C T 1 L x b = b L\boldsymbol x_{\boldsymbol b}=\boldsymbol b L x b = b t ∈ [ 0 , T ] t\in[0,T] t ∈ [ 0 , T ]
∣ x b ( t ) ∣ ≤ K Φ ( ∥ R ∥ ∥ M ∥ + 1 ) T K Φ ′ ∥ b ∥ ∞ \lvert\boldsymbol x_{\boldsymbol b}(t)\rvert\le K_\Phi\bigl(\lVert R\rVert\lVert M\rVert+1\bigr)TK'_\Phi\lVert\boldsymbol b\rVert_\infty ∣ x b ( t )∣ ≤ K Φ ( ∥ R ∥ ∥ M ∥ + 1 ) T K Φ ′ ∥ b ∥ ∞ であるから、C 1 = K Φ ( ∥ R ∥ ∥ M ∥ + 1 ) T K Φ ′ C_1=K_\Phi(\lVert R\rVert\lVert M\rVert+1)TK'_\Phi C 1 = K Φ (∥ R ∥ ∥ M ∥ + 1 ) T K Φ ′ ∥ x b ∥ ∞ ≤ C 1 ∥ b ∥ ∞ \lVert\boldsymbol x_{\boldsymbol b}\rVert_\infty\le C_1\lVert\boldsymbol b\rVert_\infty ∥ x b ∥ ∞ ≤ C 1 ∥ b ∥ ∞ x = x b − Π x b \boldsymbol x=\boldsymbol x_{\boldsymbol b}-\Pi\boldsymbol x_{\boldsymbol b} x = x b − Π x b x ∈ X 1 \boldsymbol x\in X_1 x ∈ X 1 L x = b L\boldsymbol x=\boldsymbol b L x = b L ∣ X 1 L|_{X_1} L ∣ X 1 ∥ x ∥ ∞ ≤ ( 1 + C Π ) C 1 ∥ b ∥ ∞ \lVert\boldsymbol x\rVert_\infty\le(1+C_\Pi)C_1\lVert\boldsymbol b\rVert_\infty ∥ x ∥ ∞ ≤ ( 1 + C Π ) C 1 ∥ b ∥ ∞
x ∈ X 1 \boldsymbol x\in X_1 x ∈ X 1 x = x L x − Π x L x \boldsymbol x=\boldsymbol x_{L\boldsymbol x}-\Pi\boldsymbol x_{L\boldsymbol x} x = x L x − Π x L x ∥ x ∥ ∞ ≤ ( 1 + C Π ) C 1 ∥ L x ∥ ∞ \lVert\boldsymbol x\rVert_\infty\le(1+C_\Pi)C_1\lVert L\boldsymbol x\rVert_\infty ∥ x ∥ ∞ ≤ ( 1 + C Π ) C 1 ∥ L x ∥ ∞ x ′ = A x + L x \boldsymbol x'=A\boldsymbol x+L\boldsymbol x x ′ = A x + L x ∥ x ′ ∥ ∞ ≤ ∥ A ∥ ∞ ∥ x ∥ ∞ + ∥ L x ∥ ∞ \lVert\boldsymbol x'\rVert_\infty\le\lVert A\rVert_\infty\lVert\boldsymbol x\rVert_\infty+\lVert L\boldsymbol x\rVert_\infty ∥ x ′ ∥ ∞ ≤ ∥ A ∥ ∞ ∥ x ∥ ∞ + ∥ L x ∥ ∞ C = ( 1 + ∥ A ∥ ∞ ) ( 1 + C Π ) C 1 + 1 C=(1+\lVert A\rVert_\infty)(1+C_\Pi)C_1+1 C = ( 1 + ∥ A ∥ ∞ ) ( 1 + C Π ) C 1 + 1 ∥ x ∥ C 1 ≤ C ∥ L x ∥ ∞ \lVert\boldsymbol x\rVert_{C^1}\le C\lVert L\boldsymbol x\rVert_\infty ∥ x ∥ C 1 ≤ C ∥ L x ∥ ∞ ▨
命題 4.7. K ∈ { R , C } \mathbb K\in\{\R,\C\} K ∈ { R , C } d ∈ N ≥ 1 d\in\NN d ∈ N ≥ 1 A ∈ K d × d A\in\mathbb K^{d\times d} A ∈ K d × d T > 0 T>0 T > 0 ω = 2 π / T \omega=2\pi/T ω = 2 π / T x ′ = A x \boldsymbol x'=A\boldsymbol x x ′ = A x T T T x : R → C d \boldsymbol x\colon\R\to\C^d x : R → C d Z \mathcal Z Z ψ ′ = − A ∗ ψ \boldsymbol\psi'=-A^*\boldsymbol\psi ψ ′ = − A ∗ ψ T T T ψ : R → C d \boldsymbol\psi\colon\R\to\C^d ψ : R → C d Z ∗ \mathcal Z^* Z ∗
ker ( A − i n ω I ) ≠ { 0 } \ker(A-in\omega I)\ne\{\boldsymbol 0\} ker ( A − inω I ) = { 0 } n ∈ Z n\in\Z n ∈ Z Z \mathcal Z Z n n n v n ∈ ker ( A − i n ω I ) \boldsymbol v_n\in\ker(A-in\omega I) v n ∈ ker ( A − inω I ) x ( t ) = ∑ n e i n ω t v n \boldsymbol x(t)=\sum_ne^{in\omega t}\boldsymbol v_n x ( t ) = ∑ n e inω t v n Z \mathcal Z Z dim C Z = ∑ n ∈ Z dim C ker ( A − i n ω I ) \dim_{\C}\mathcal Z=\sum_{n\in\Z}\dim_{\C}\ker(A-in\omega I) dim C Z = ∑ n ∈ Z dim C ker ( A − inω I ) Z ∗ \mathcal Z^* Z ∗ n n n w n ∈ ker ( A ∗ + i n ω I ) \boldsymbol w_n\in\ker(A^*+in\omega I) w n ∈ ker ( A ∗ + inω I ) ψ ( t ) = ∑ n e i n ω t w n \boldsymbol\psi(t)=\sum_ne^{in\omega t}\boldsymbol w_n ψ ( t ) = ∑ n e inω t w n Z ∗ \mathcal Z^* Z ∗ dim C Z ∗ = dim C Z \dim_{\C}\mathcal Z^*=\dim_{\C}\mathcal Z dim C Z ∗ = dim C Z K = R \mathbb K=\R K = R Z \mathcal Z Z Z R \mathcal Z_{\R} Z R R \R R dim R Z R = dim C Z \dim_{\R}\mathcal Z_{\R}=\dim_{\C}\mathcal Z dim R Z R = dim C Z v ∈ ker ( A − i n ω I ) \boldsymbol v\in\ker(A-in\omega I) v ∈ ker ( A − inω I ) ℜ ( e i n ω t v ) \Re\bigl(e^{in\omega t}\boldsymbol v\bigr) ℜ ( e inω t v ) ℑ ( e i n ω t v ) \Im\bigl(e^{in\omega t}\boldsymbol v\bigr) ℑ ( e inω t v ) Z ∗ \mathcal Z^* Z ∗ Z R ∗ \mathcal Z^*_{\R} Z R ∗ 特に、i ω i\omega iω − i ω -i\omega − iω A A A A A A i ω Z i\omega\Z iω Z ± i ω \pm i\omega ± iω dim C Z = dim C Z ∗ = 2 \dim_{\C}\mathcal Z=\dim_{\C}\mathcal Z^*=2 dim C Z = dim C Z ∗ = 2 K = R \mathbb K=\R K = R dim R Z R = dim R Z R ∗ = 2 \dim_{\R}\mathcal Z_{\R}=\dim_{\R}\mathcal Z^*_{\R}=2 dim R Z R = dim R Z R ∗ = 2
証明. §E13.12 定理 2.2 (3) をI = R I=\R I = R t ↦ e t A t\mapsto e^{tA} t ↦ e t A x ′ = A x \boldsymbol x'=A\boldsymbol x x ′ = A x e 0 A = I e^{0A}=I e 0 A = I e T A e^{TA} e T A 命題 1.6 (1) をρ = 1 \rho=1 ρ = 1 x ↦ x ( 0 ) \boldsymbol x\mapsto\boldsymbol x(0) x ↦ x ( 0 ) Z \mathcal Z Z ker ( e T A − I ) \ker(e^{TA}-I) ker ( e T A − I ) e T β = 1 e^{T\beta}=1 e T β = 1 β ∈ i ω Z \beta\in i\omega\Z β ∈ iω Z 系 2.2 (2) をT A TA T A ρ = 1 \rho=1 ρ = 1
ker ( e T A − I ) = ⨁ n ∈ Z ker ( T A − T i n ω I ) = ⨁ n ∈ Z ker ( A − i n ω I ) \ker(e^{TA}-I)=\bigoplus_{n\in\Z}\ker(TA-Tin\omega I)=\bigoplus_{n\in\Z}\ker(A-in\omega I) ker ( e T A − I ) = n ∈ Z ⨁ ker ( T A − T inω I ) = n ∈ Z ⨁ ker ( A − inω I ) であり、右辺で{ 0 } \{\boldsymbol 0\} { 0 } A A A v ∈ ker ( A − i n ω I ) \boldsymbol v\in\ker(A-in\omega I) v ∈ ker ( A − inω I ) 系 3.1 (3) によりe t A v = e i n ω t v e^{tA}\boldsymbol v=e^{in\omega t}\boldsymbol v e t A v = e inω t v x ( 0 ) = ∑ n v n \boldsymbol x(0)=\sum_n\boldsymbol v_n x ( 0 ) = ∑ n v n x ( t ) = e t A x ( 0 ) = ∑ n e i n ω t v n \boldsymbol x(t)=e^{tA}\boldsymbol x(0)=\sum_ne^{in\omega t}\boldsymbol v_n x ( t ) = e t A x ( 0 ) = ∑ n e inω t v n 主張 (1) は示された。
主張 (2) を示す。前段を定数行列− A ∗ -A^* − A ∗ ker ( − A ∗ − i n ω I ) = ker ( A ∗ + i n ω I ) \ker(-A^*-in\omega I)=\ker(A^*+in\omega I) ker ( − A ∗ − inω I ) = ker ( A ∗ + inω I ) A ∗ + i n ω I = ( A − i n ω I ) ∗ A^*+in\omega I=(A-in\omega I)^* A ∗ + inω I = ( A − inω I ) ∗ dim ker ( A ∗ + i n ω I ) = d − rank ( A − i n ω I ) = dim ker ( A − i n ω I ) \dim\ker(A^*+in\omega I)=d-\operatorname{rank}(A-in\omega I)=\dim\ker(A-in\omega I) dim ker ( A ∗ + inω I ) = d − rank ( A − inω I ) = dim ker ( A − inω I ) n n n
主張 (3) を示す。命題 1.6 (3) をρ = 1 \rho=1 ρ = 1 dim R Z R = dim C Z \dim_{\R}\mathcal Z_{\R}=\dim_{\C}\mathcal Z dim R Z R = dim C Z A A A ℜ ( e i n ω t v ) \Re\bigl(e^{in\omega t}\boldsymbol v\bigr) ℜ ( e inω t v ) ℑ ( e i n ω t v ) \Im\bigl(e^{in\omega t}\boldsymbol v\bigr) ℑ ( e inω t v ) Z R \mathcal Z_{\R} Z R x ∈ Z R \boldsymbol x\in\mathcal Z_{\R} x ∈ Z R x = ℜ x = ∑ n ℜ ( e i n ω t v n ) \boldsymbol x=\Re\boldsymbol x=\sum_n\Re\bigl(e^{in\omega t}\boldsymbol v_n\bigr) x = ℜ x = ∑ n ℜ ( e inω t v n ) − A ∗ = − A T -A^*=-A^{\mathsf T} − A ∗ = − A T Z R ∗ \mathcal Z^*_{\R} Z R ∗
主張 (4) を示す。± i ω \pm i\omega ± iω dim ker ( A ∓ i ω I ) ≥ 1 \dim\ker(A\mp i\omega I)\ge1 dim ker ( A ∓ iω I ) ≥ 1 §E3.26 定理 2.1 によりdim ker ( A ∓ i ω I ) ≤ dim G ± i ω ( A ) = 1 \dim\ker(A\mp i\omega I)\le\dim G_{\pm i\omega}(A)=1 dim ker ( A ∓ iω I ) ≤ dim G ± iω ( A ) = 1 n n n ker ( A − i n ω I ) = { 0 } \ker(A-in\omega I)=\{\boldsymbol 0\} ker ( A − inω I ) = { 0 } 主張 (1) 、主張 (2) 、主張 (3) により結論を得る。▨
例 4.8. m ∈ N ≥ 1 m\in\NN m ∈ N ≥ 1 A m = ( 0 1 − m 2 0 ) A_m=\begin{pmatrix}0&1\\-m^2&0\end{pmatrix} A m = ( 0 − m 2 1 0 ) Y ′ = A m Y \boldsymbol Y'=A_m\boldsymbol Y Y ′ = A m Y ( cos m t , − m sin m t ) T (\cos mt,-m\sin mt)^{\mathsf T} ( cos m t , − m sin m t ) T ( m − 1 sin m t , cos m t ) T (m^{-1}\sin mt,\cos mt)^{\mathsf T} ( m − 1 sin m t , cos m t ) T Φ 0 ( t ) \Phi_0(t) Φ 0 ( t ) det Φ 0 ( t ) = 1 \det\Phi_0(t)=1 det Φ 0 ( t ) = 1 Φ 0 ( 0 ) = I \Phi_0(0)=I Φ 0 ( 0 ) = I 2 π 2\pi 2 π 2 π 2\pi 2 π Y ′ = A m Y \boldsymbol Y'=A_m\boldsymbol Y Y ′ = A m Y I I I ψ ′ = − A m T ψ \boldsymbol\psi'=-A_m^{\mathsf T}\boldsymbol\psi ψ ′ = − A m T ψ ψ 1 ′ = m 2 ψ 2 \psi_1'=m^2\psi_2 ψ 1 ′ = m 2 ψ 2 ψ 2 ′ = − ψ 1 \psi_2'=-\psi_1 ψ 2 ′ = − ψ 1 ( m sin m t , cos m t ) T (m\sin mt,\cos mt)^{\mathsf T} ( m sin m t , cos m t ) T ( − m cos m t , sin m t ) T (-m\cos mt,\sin mt)^{\mathsf T} ( − m cos m t , sin m t ) T t = 0 t=0 t = 0 ( 0 , 1 ) T (0,1)^{\mathsf T} ( 0 , 1 ) T ( − m , 0 ) T (-m,0)^{\mathsf T} ( − m , 0 ) T §E13.9 定理 2.2 により、随伴系の実数値の解の全体はこの二つで張られ、そのすべてが2 π 2\pi 2 π ψ 2 \psi_2 ψ 2 cos m t \cos mt cos m t sin m t \sin mt sin m t
m = 1 m=1 m = 1 y ′ ′ + y = cos t y''+y=\cos t y ′′ + y = cos t 2 π 2\pi 2 π C 2 C^2 C 2 y y y C 2 C^2 C 2 §E13.2 系 2.7 (2) によりy ′ y' y ′ §E13.9 命題 4.1 によりY = ( y , y ′ ) T \boldsymbol Y=(y,y')^{\mathsf T} Y = ( y , y ′ ) T Y ′ = A 1 Y + b \boldsymbol Y'=A_1\boldsymbol Y+\boldsymbol b Y ′ = A 1 Y + b b ( t ) = ( 0 , cos t ) T \boldsymbol b(t)=(0,\cos t)^{\mathsf T} b ( t ) = ( 0 , cos t ) T y y y 2 π 2\pi 2 π Y \boldsymbol Y Y 2 π 2\pi 2 π 2 π 2\pi 2 π ψ ( t ) = ( sin t , cos t ) T \boldsymbol\psi(t)=(\sin t,\cos t)^{\mathsf T} ψ ( t ) = ( sin t , cos t ) T
∫ 0 2 π ψ ( t ) T b ( t ) d t = ∫ 0 2 π cos 2 t d t = π ≠ 0 \int_0^{2\pi}\boldsymbol\psi(t)^{\mathsf T}\boldsymbol b(t)\,dt=\int_0^{2\pi}\cos^2t\,dt=\pi\ne0 ∫ 0 2 π ψ ( t ) T b ( t ) d t = ∫ 0 2 π cos 2 t d t = π = 0 であるから、定理 4.4 (4) により一階の系の2 π 2\pi 2 π y ′ ′ + y = cos t y''+y=\cos t y ′′ + y = cos t 2 π 2\pi 2 π C 2 C^2 C 2
m = 2 m=2 m = 2 y ′ ′ + 4 y = cos t y''+4y=\cos t y ′′ + 4 y = cos t Y ′ = A 2 Y + b \boldsymbol Y'=A_2\boldsymbol Y+\boldsymbol b Y ′ = A 2 Y + b b ( t ) = ( 0 , cos t ) T \boldsymbol b(t)=(0,\cos t)^{\mathsf T} b ( t ) = ( 0 , cos t ) T I I I 1 1 1 2 π 2\pi 2 π ψ \boldsymbol\psi ψ ψ T b = ψ 2 cos t \boldsymbol\psi^{\mathsf T}\boldsymbol b=\psi_2\cos t ψ T b = ψ 2 cos t ψ 2 \psi_2 ψ 2 cos 2 t \cos2t cos 2 t sin 2 t \sin2t sin 2 t ∫ 0 2 π cos 2 t cos t d t = ∫ 0 2 π sin 2 t cos t d t = 0 \int_0^{2\pi}\cos2t\cos t\,dt=\int_0^{2\pi}\sin2t\cos t\,dt=0 ∫ 0 2 π cos 2 t cos t d t = ∫ 0 2 π sin 2 t cos t d t = 0 Y p ( t ) = ( 1 3 cos t , − 1 3 sin t ) T \boldsymbol Y_p(t)=(\tfrac13\cos t,-\tfrac13\sin t)^{\mathsf T} Y p ( t ) = ( 3 1 cos t , − 3 1 sin t ) T 2 π 2\pi 2 π 定理 4.4 (3) と定理 4.4 (4) をT = 2 π T=2\pi T = 2 π ρ = 1 \rho=1 ρ = 1 x p = Y p ∣ [ 0 , 2 π ] \boldsymbol x_p=\boldsymbol Y_p|_{[0,2\pi]} x p = Y p ∣ [ 0 , 2 π ] 2 π 2\pi 2 π Y p \boldsymbol Y_p Y p 2 π 2\pi 2 π §E13.9 定理 2.2 により同次系の実数値の解はΦ 0 \Phi_0 Φ 0 Φ 0 \Phi_0 Φ 0 2 π 2\pi 2 π ( cos 2 t , 1 2 sin 2 t ) (\cos2t,\tfrac12\sin2t) ( cos 2 t , 2 1 sin 2 t ) 2 π 2\pi 2 π cos 2 t \cos2t cos 2 t sin 2 t \sin2t sin 2 t §E13.9 命題 4.1 により、y ′ ′ + 4 y = cos t y''+4y=\cos t y ′′ + 4 y = cos t 2 π 2\pi 2 π 1 3 cos t + c 1 cos 2 t + c 2 sin 2 t \tfrac13\cos t+c_1\cos2t+c_2\sin2t 3 1 cos t + c 1 cos 2 t + c 2 sin 2 t c 1 , c 2 ∈ R c_1,c_2\in\R c 1 , c 2 ∈ R 1 1 1 定理 4.4 (1) の条件を「1 1 1
5 Hill 方程式と判別式 定義 5.1 (Hill 方程式). T > 0 T>0 T > 0 q ∈ L l o c 1 ( R ; R ) q\in L^1_{\mathrm{loc}}(\R;\R) q ∈ L loc 1 ( R ; R ) t ∈ R t\in\R t ∈ R q ( t + T ) = q ( t ) q(t+T)=q(t) q ( t + T ) = q ( t ) λ ∈ C \lambda\in\C λ ∈ C
− y ′ ′ + q ( t ) y = λ y -y''+q(t)y=\lambda y − y ′′ + q ( t ) y = λ y をHill 方程式 (Hill's equation )y ∈ C 1 ( R ; C ) y\in C^1(\R;\C) y ∈ C 1 ( R ; C ) y ′ ∈ A C l o c ( R ; C ) y'\in AC_{\mathrm{loc}}(\R;\C) y ′ ∈ A C loc ( R ; C ) §E13.9 命題 4.1 により、y y y Y = ( y , y ′ ) T \boldsymbol Y=(y,y')^{\mathsf T} Y = ( y , y ′ ) T T T T
Y ′ = A λ ( t ) Y , A λ ( t ) = ( 0 1 q ( t ) − λ 0 ) \boldsymbol Y'=A_\lambda(t)\boldsymbol Y,\qquad A_\lambda(t)=\begin{pmatrix}0&1\\q(t)-\lambda&0\end{pmatrix} Y ′ = A λ ( t ) Y , A λ ( t ) = ( 0 q ( t ) − λ 1 0 ) の局所絶対連続解であるための必要十分条件である。( θ , θ ′ ) ( 0 ) = ( 1 , 0 ) (\theta,\theta')(0)=(1,0) ( θ , θ ′ ) ( 0 ) = ( 1 , 0 ) θ ( ⋅ , λ ) \theta(\cdot,\lambda) θ ( ⋅ , λ ) ( φ , φ ′ ) ( 0 ) = ( 0 , 1 ) (\varphi,\varphi')(0)=(0,1) ( φ , φ ′ ) ( 0 ) = ( 0 , 1 ) φ ( ⋅ , λ ) \varphi(\cdot,\lambda) φ ( ⋅ , λ )
Φ 0 ( t , λ ) = ( θ ( t , λ ) φ ( t , λ ) θ ′ ( t , λ ) φ ′ ( t , λ ) ) , M ( λ ) = Φ 0 ( T , λ ) , Δ ( λ ) = θ ( T , λ ) + φ ′ ( T , λ ) \Phi_0(t,\lambda)=\begin{pmatrix}\theta(t,\lambda)&\varphi(t,\lambda)\\\theta'(t,\lambda)&\varphi'(t,\lambda)\end{pmatrix},\qquad M(\lambda)=\Phi_0(T,\lambda),\qquad\Delta(\lambda)=\theta(T,\lambda)+\varphi'(T,\lambda) Φ 0 ( t , λ ) = ( θ ( t , λ ) θ ′ ( t , λ ) φ ( t , λ ) φ ′ ( t , λ ) ) , M ( λ ) = Φ 0 ( T , λ ) , Δ ( λ ) = θ ( T , λ ) + φ ′ ( T , λ ) とおく。M ( λ ) M(\lambda) M ( λ ) Y ′ = A λ ( t ) Y \boldsymbol Y'=A_\lambda(t)\boldsymbol Y Y ′ = A λ ( t ) Y Δ = tr M \Delta=\operatorname{tr}M Δ = tr M 判別式 (discriminant )
命題 5.2. T > 0 T>0 T > 0 q ∈ L l o c 1 ( R ; R ) q\in L^1_{\mathrm{loc}}(\R;\R) q ∈ L loc 1 ( R ; R ) t t t q ( t + T ) = q ( t ) q(t+T)=q(t) q ( t + T ) = q ( t ) λ ∈ C \lambda\in\C λ ∈ C θ , φ , M , Δ \theta,\varphi,M,\Delta θ , φ , M , Δ − y ′ ′ + q ( t ) y = λ y -y''+q(t)y=\lambda y − y ′′ + q ( t ) y = λ y
すべてのt ∈ R t\in\R t ∈ R θ ( t , λ ) φ ′ ( t , λ ) − θ ′ ( t , λ ) φ ( t , λ ) = 1 \theta(t,\lambda)\varphi'(t,\lambda)-\theta'(t,\lambda)\varphi(t,\lambda)=1 θ ( t , λ ) φ ′ ( t , λ ) − θ ′ ( t , λ ) φ ( t , λ ) = 1 det M ( λ ) = 1 \det M(\lambda)=1 det M ( λ ) = 1
Y ′ = A λ ( t ) Y \boldsymbol Y'=A_\lambda(t)\boldsymbol Y Y ′ = A λ ( t ) Y ρ 2 − Δ ( λ ) ρ + 1 = 0 \rho^2-\Delta(\lambda)\rho+1=0 ρ 2 − Δ ( λ ) ρ + 1 = 0 1 1 1 Δ ( λ ) ≠ ± 2 \Delta(\lambda)\ne\pm2 Δ ( λ ) = ± 2 ρ \rho ρ ρ − 1 \rho^{-1} ρ − 1 ρ ≠ ρ − 1 \rho\ne\rho^{-1} ρ = ρ − 1 t ∈ R t\in\R t ∈ R u ± ( t + T ) = ρ ± 1 u ± ( t ) u_\pm(t+T)=\rho^{\pm1}u_\pm(t) u ± ( t + T ) = ρ ± 1 u ± ( t ) u + , u − u_+,u_- u + , u − ∣ ρ ∣ < 1 \lvert\rho\rvert<1 ∣ ρ ∣ < 1 C > 0 C>0 C > 0 t ∈ R t\in\R t ∈ R ∣ u + ( t ) ∣ + ∣ u + ′ ( t ) ∣ ≤ C ∣ ρ ∣ t / T \lvert u_+(t)\rvert+\lvert u_+'(t)\rvert\le C\lvert\rho\rvert^{t/T} ∣ u + ( t )∣ + ∣ u + ′ ( t )∣ ≤ C ∣ ρ ∣ t / T 証明. tr A λ = 0 \operatorname{tr}A_\lambda=0 tr A λ = 0 §E13.9 定理 3.1 によりdet Φ 0 ( t , λ ) = det Φ 0 ( 0 , λ ) = 1 \det\Phi_0(t,\lambda)=\det\Phi_0(0,\lambda)=1 det Φ 0 ( t , λ ) = det Φ 0 ( 0 , λ ) = 1 主張 (1) を得る。M ( λ ) M(\lambda) M ( λ ) ρ 2 − ( tr M ) ρ + det M = ρ 2 − Δ ρ + 1 \rho^2-(\operatorname{tr}M)\rho+\det M=\rho^2-\Delta\rho+1 ρ 2 − ( tr M ) ρ + det M = ρ 2 − Δ ρ + 1 主張 (2) を得る。
主張 (3) を示す。二根の積は1 1 1 ρ , ρ − 1 \rho,\rho^{-1} ρ , ρ − 1 ρ = ρ − 1 \rho=\rho^{-1} ρ = ρ − 1 ρ = ± 1 \rho=\pm1 ρ = ± 1 Δ = ρ + ρ − 1 = ± 2 \Delta=\rho+\rho^{-1}=\pm2 Δ = ρ + ρ − 1 = ± 2 Δ ≠ ± 2 \Delta\ne\pm2 Δ = ± 2 M M M ρ , ρ − 1 \rho,\rho^{-1} ρ , ρ − 1 ξ + , ξ − \boldsymbol\xi_+,\boldsymbol\xi_- ξ + , ξ − 命題 1.6 (1) によりY ± ( 0 ) = ξ ± \boldsymbol Y_\pm(0)=\boldsymbol\xi_\pm Y ± ( 0 ) = ξ ± ρ ± 1 \rho^{\pm1} ρ ± 1 Y ± \boldsymbol Y_\pm Y ± u ± u_\pm u ± Y ± = ( u ± , u ± ′ ) T \boldsymbol Y_\pm=(u_\pm,u_\pm')^{\mathsf T} Y ± = ( u ± , u ± ′ ) T u ± ( t + T ) = ρ ± 1 u ± ( t ) u_\pm(t+T)=\rho^{\pm1}u_\pm(t) u ± ( t + T ) = ρ ± 1 u ± ( t ) a u + + b u − = 0 au_++bu_-=0 a u + + b u − = 0 a ξ + + b ξ − = 0 a\boldsymbol\xi_++b\boldsymbol\xi_-=\boldsymbol 0 a ξ + + b ξ − = 0 a = b = 0 a=b=0 a = b = 0 ∣ ρ ∣ < 1 \lvert\rho\rvert<1 ∣ ρ ∣ < 1 K = max s ∈ [ 0 , T ] ( ∣ u + ( s ) ∣ + ∣ u + ′ ( s ) ∣ ) K=\max_{s\in[0,T]}(\lvert u_+(s)\rvert+\lvert u_+'(s)\rvert) K = max s ∈ [ 0 , T ] (∣ u + ( s )∣ + ∣ u + ′ ( s )∣) t ∈ R t\in\R t ∈ R t = n T + s t=nT+s t = n T + s n ∈ Z n\in\Z n ∈ Z s ∈ [ 0 , T ) s\in[0,T) s ∈ [ 0 , T ) Y + ( t ) = ρ n Y + ( s ) \boldsymbol Y_+(t)=\rho^n\boldsymbol Y_+(s) Y + ( t ) = ρ n Y + ( s )
∣ u + ( t ) ∣ + ∣ u + ′ ( t ) ∣ ≤ K ∣ ρ ∣ n = K ∣ ρ ∣ ( t − s ) / T ≤ K ∣ ρ ∣ − 1 ∣ ρ ∣ t / T \lvert u_+(t)\rvert+\lvert u_+'(t)\rvert\le K\lvert\rho\rvert^n=K\lvert\rho\rvert^{(t-s)/T}\le K\lvert\rho\rvert^{-1}\lvert\rho\rvert^{t/T} ∣ u + ( t )∣ + ∣ u + ′ ( t )∣ ≤ K ∣ ρ ∣ n = K ∣ ρ ∣ ( t − s ) / T ≤ K ∣ ρ ∣ − 1 ∣ ρ ∣ t / T であり、C = K ∣ ρ ∣ − 1 C=K\lvert\rho\rvert^{-1} C = K ∣ ρ ∣ − 1 ▨
定理 5.3. T > 0 T>0 T > 0 q ∈ L l o c 1 ( R ; R ) q\in L^1_{\mathrm{loc}}(\R;\R) q ∈ L loc 1 ( R ; R ) t t t q ( t + T ) = q ( t ) q(t+T)=q(t) q ( t + T ) = q ( t ) λ ∈ R \lambda\in\R λ ∈ R M , Δ M,\Delta M , Δ − y ′ ′ + q ( t ) y = λ y -y''+q(t)y=\lambda y − y ′′ + q ( t ) y = λ y M ( λ ) M(\lambda) M ( λ ) Δ ( λ ) ∈ R \Delta(\lambda)\in\R Δ ( λ ) ∈ R
∣ Δ ( λ ) ∣ < 2 \lvert\Delta(\lambda)\rvert<2 ∣ Δ ( λ )∣ < 2 R \R R ∣ Δ ( λ ) ∣ > 2 \lvert\Delta(\lambda)\rvert>2 ∣ Δ ( λ )∣ > 2 R \R R ∣ ρ ∣ > 1 \lvert\rho\rvert>1 ∣ ρ ∣ > 1 ρ ∈ R \rho\in\R ρ ∈ R t t t y ( t + T ) = ρ y ( t ) y(t+T)=\rho y(t) y ( t + T ) = ρ y ( t ) y y y ε ∈ { 1 , − 1 } \varepsilon\in\{1,-1\} ε ∈ { 1 , − 1 } Δ ( λ ) = 2 ε \Delta(\lambda)=2\varepsilon Δ ( λ ) = 2 ε t t t y ( t + T ) = ε y ( t ) y(t+T)=\varepsilon y(t) y ( t + T ) = ε y ( t ) y y y R \R R M ( λ ) = ε I M(\lambda)=\varepsilon I M ( λ ) = ε I 証明. λ ∈ R \lambda\in\R λ ∈ R A λ A_\lambda A λ §E13.9 命題 4.1 の初期値問題の一意性によりθ , φ \theta,\varphi θ , φ M ( λ ) M(\lambda) M ( λ ) Δ ( λ ) ∈ R \Delta(\lambda)\in\R Δ ( λ ) ∈ R §E13.9 定理 5.2 により解の全体は二次元である。
主張 (1) を示す。∣ Δ ∣ < 2 \lvert\Delta\rvert<2 ∣ Δ ∣ < 2 ρ 2 − Δ ρ + 1 = 0 \rho^2-\Delta\rho+1=0 ρ 2 − Δ ρ + 1 = 0 Δ 2 − 4 \Delta^2-4 Δ 2 − 4 ρ , ρ ‾ \rho,\overline\rho ρ , ρ 命題 5.2 (2) により∣ ρ ∣ 2 = ρ ρ ‾ = 1 \lvert\rho\rvert^2=\rho\overline\rho=1 ∣ ρ ∣ 2 = ρ ρ = 1 命題 5.2 (3) の解u ± u_\pm u ± ∣ u ± ( t + n T ) ∣ = ∣ u ± ( t ) ∣ \lvert u_\pm(t+nT)\rvert=\lvert u_\pm(t)\rvert ∣ u ± ( t + n T )∣ = ∣ u ± ( t )∣ sup R ∣ u ± ∣ = max [ 0 , T ] ∣ u ± ∣ < ∞ \sup_{\R}\lvert u_\pm\rvert=\max_{[0,T]}\lvert u_\pm\rvert<\infty sup R ∣ u ± ∣ = max [ 0 , T ] ∣ u ± ∣ < ∞ u + , u − u_+,u_- u + , u −
主張 (2) を示す。∣ Δ ∣ > 2 \lvert\Delta\rvert>2 ∣ Δ ∣ > 2 ∣ ρ ∣ > 1 \lvert\rho\rvert>1 ∣ ρ ∣ > 1 ρ \rho ρ ρ − 1 \rho^{-1} ρ − 1 M M M 命題 1.6 (3) により乗数ρ ± 1 \rho^{\pm1} ρ ± 1 u ± u_\pm u ± 命題 5.2 (3) の証明と同じく一次独立である。u + u_+ u + y = a u + + b u − y=au_++bu_- y = a u + + b u − a ≠ 0 a\ne0 a = 0 u + u_+ u + u + ( t + n T ) = ρ n u + ( t ) u_+(t+nT)=\rho^nu_+(t) u + ( t + n T ) = ρ n u + ( t ) u + ( t 0 ) ≠ 0 u_+(t_0)\ne0 u + ( t 0 ) = 0 t 0 ∈ [ 0 , T ] t_0\in[0,T] t 0 ∈ [ 0 , T ]
∣ y ( t 0 + n T ) ∣ ≥ ∣ a ∣ ∣ ρ ∣ n ∣ u + ( t 0 ) ∣ − ∣ b ∣ ∣ ρ ∣ − n ∣ u − ( t 0 ) ∣ ⟶ ∞ ( n → ∞ ) \lvert y(t_0+nT)\rvert\ge\lvert a\rvert\lvert\rho\rvert^n\lvert u_+(t_0)\rvert-\lvert b\rvert\lvert\rho\rvert^{-n}\lvert u_-(t_0)\rvert\longrightarrow\infty\qquad(n\to\infty) ∣ y ( t 0 + n T )∣ ≥ ∣ a ∣ ∣ ρ ∣ n ∣ u + ( t 0 )∣ − ∣ b ∣ ∣ ρ ∣ − n ∣ u − ( t 0 )∣ ⟶ ∞ ( n → ∞ ) である。a = 0 a=0 a = 0 b ≠ 0 b\ne0 b = 0 u − ( t 0 ) ≠ 0 u_-(t_0)\ne0 u − ( t 0 ) = 0 t 0 t_0 t 0 n → − ∞ n\to-\infty n → − ∞
主張 (3) を示す。Δ = 2 ε \Delta=2\varepsilon Δ = 2 ε ( ρ − ε ) 2 (\rho-\varepsilon)^2 ( ρ − ε ) 2 ε \varepsilon ε M M M 命題 1.6 (3) により零でない実数値の解y y y y ( t + T ) = ε y ( t ) y(t+T)=\varepsilon y(t) y ( t + T ) = ε y ( t ) M = ε I M=\varepsilon I M = ε I 補題 1.2 (3) によりΦ 0 ( t + n T ) = ε n Φ 0 ( t ) \Phi_0(t+nT)=\varepsilon^n\Phi_0(t) Φ 0 ( t + n T ) = ε n Φ 0 ( t ) Φ 0 \Phi_0 Φ 0 R \R R
逆に、すべての解がR \R R u ∈ { θ , φ } u\in\{\theta,\varphi\} u ∈ { θ , φ } K u = sup R ∣ u ∣ K_u=\sup_{\R}\lvert u\rvert K u = sup R ∣ u ∣ c = ∫ 0 T ∣ q ( s ) − λ ∣ d s c=\int_0^T\lvert q(s)-\lambda\rvert\,ds c = ∫ 0 T ∣ q ( s ) − λ ∣ d s q q q T T T [ a , a + T ] [a,a+T] [ a , a + T ] ∫ a a + T ∣ q ( s ) − λ ∣ d s = c \int_a^{a+T}\lvert q(s)-\lambda\rvert\,ds=c ∫ a a + T ∣ q ( s ) − λ ∣ d s = c u u u C 1 C^1 C 1 u ′ ( s a ) = ( u ( a + T ) − u ( a ) ) / T u'(s_a)=\bigl(u(a+T)-u(a)\bigr)/T u ′ ( s a ) = ( u ( a + T ) − u ( a ) ) / T s a ∈ ( a , a + T ) s_a\in(a,a+T) s a ∈ ( a , a + T ) ∣ u ′ ( s a ) ∣ ≤ 2 K u / T \lvert u'(s_a)\rvert\le2K_u/T ∣ u ′ ( s a )∣ ≤ 2 K u / T u ′ u' u ′ u ′ ′ = ( q − λ ) u u''=(q-\lambda)u u ′′ = ( q − λ ) u §E13.2 系 2.6 により、t ∈ [ a , a + T ] t\in[a,a+T] t ∈ [ a , a + T ]
∣ u ′ ( t ) ∣ ≤ ∣ u ′ ( s a ) ∣ + ∣ ∫ s a t ( q ( s ) − λ ) u ( s ) d s ∣ ≤ 2 K u T + c K u \lvert u'(t)\rvert\le\lvert u'(s_a)\rvert+\Bigl\lvert\int_{s_a}^t\bigl(q(s)-\lambda\bigr)u(s)\,ds\Bigr\rvert\le\frac{2K_u}{T}+cK_u ∣ u ′ ( t )∣ ≤ ∣ u ′ ( s a )∣ + ∫ s a t ( q ( s ) − λ ) u ( s ) d s ≤ T 2 K u + c K u である。a ∈ R a\in\R a ∈ R u ′ u' u ′ R \R R Φ 0 \Phi_0 Φ 0 R \R R §E13.9 定理 2.2 によりY ′ = A λ ( t ) Y \boldsymbol Y'=A_\lambda(t)\boldsymbol Y Y ′ = A λ ( t ) Y Y = Φ 0 ξ \boldsymbol Y=\Phi_0\boldsymbol\xi Y = Φ 0 ξ ξ ∈ C 2 \boldsymbol\xi\in\C^2 ξ ∈ C 2 [ 0 , ∞ ) [0,\infty) [ 0 , ∞ ) 定理 3.2 (2) により、絶対値が1 1 1 ε \varepsilon ε M M M ( ρ − ε ) 2 (\rho-\varepsilon)^2 ( ρ − ε ) 2 §E3.26 定理 2.1 によりG ε ( M ) = C 2 G_\varepsilon(M)=\C^2 G ε ( M ) = C 2 定義 2.1 によりker ( M − ε I ) = G ε ( M ) = C 2 \ker(M-\varepsilon I)=G_\varepsilon(M)=\C^2 ker ( M − ε I ) = G ε ( M ) = C 2 M = ε I M=\varepsilon I M = ε I ▨
系 5.4. T > 0 T>0 T > 0 q ∈ L l o c 1 ( R ; R ) q\in L^1_{\mathrm{loc}}(\R;\R) q ∈ L loc 1 ( R ; R ) t t t q ( t + T ) = q ( t ) q(t+T)=q(t) q ( t + T ) = q ( t ) θ , φ , Δ \theta,\varphi,\Delta θ , φ , Δ − y ′ ′ + q ( t ) y = λ y -y''+q(t)y=\lambda y − y ′′ + q ( t ) y = λ y t ∈ R t\in\R t ∈ R λ ↦ θ ( t , λ ) \lambda\mapsto\theta(t,\lambda) λ ↦ θ ( t , λ ) θ ′ ( t , λ ) \theta'(t,\lambda) θ ′ ( t , λ ) φ ( t , λ ) \varphi(t,\lambda) φ ( t , λ ) φ ′ ( t , λ ) \varphi'(t,\lambda) φ ′ ( t , λ ) λ ∈ R \lambda\in\R λ ∈ R Δ \Delta Δ
証明. A 0 = ( 0 1 q 0 ) A_0=\begin{pmatrix}0&1\\q&0\end{pmatrix} A 0 = ( 0 q 1 0 ) A 1 = ( 0 0 − 1 0 ) A_1=\begin{pmatrix}0&0\\-1&0\end{pmatrix} A 1 = ( 0 − 1 0 0 ) A λ = A 0 + λ A 1 A_\lambda=A_0+\lambda A_1 A λ = A 0 + λ A 1 A 0 , A 1 ∈ L l o c 1 ( R ; C 2 × 2 ) A_0,A_1\in L^1_{\mathrm{loc}}(\R;\C^{2\times2}) A 0 , A 1 ∈ L loc 1 ( R ; C 2 × 2 ) ( θ , θ ′ ) T (\theta,\theta')^{\mathsf T} ( θ , θ ′ ) T ( φ , φ ′ ) T (\varphi,\varphi')^{\mathsf T} ( φ , φ ′ ) T b 0 = b 1 = 0 \boldsymbol b_0=\boldsymbol b_1=\boldsymbol 0 b 0 = b 1 = 0 t 0 = 0 t_0=0 t 0 = 0 η = ( 1 , 0 ) T \boldsymbol\eta=(1,0)^{\mathsf T} η = ( 1 , 0 ) T η = ( 0 , 1 ) T \boldsymbol\eta=(0,1)^{\mathsf T} η = ( 0 , 1 ) T §E13.8 命題 5.1 の解であるから、§E13.8 命題 5.1 (3) と§E13.8 命題 5.1 (5) により結論を得る。Δ ( λ ) = θ ( T , λ ) + φ ′ ( T , λ ) \Delta(\lambda)=\theta(T,\lambda)+\varphi'(T,\lambda) Δ ( λ ) = θ ( T , λ ) + φ ′ ( T , λ ) ▨
系 5.5. T > 0 T>0 T > 0 q ∈ L l o c 1 ( R ; R ) q\in L^1_{\mathrm{loc}}(\R;\R) q ∈ L loc 1 ( R ; R ) t t t q ( t + T ) = q ( t ) q(t+T)=q(t) q ( t + T ) = q ( t ) Δ \Delta Δ − y ′ ′ + q ( t ) y = λ y -y''+q(t)y=\lambda y − y ′′ + q ( t ) y = λ y Δ \Delta Δ R \R R { λ ∈ R : ∣ Δ ( λ ) ∣ < 2 } \{\lambda\in\R:\lvert\Delta(\lambda)\rvert<2\} { λ ∈ R : ∣ Δ ( λ )∣ < 2 } { λ ∈ R : ∣ Δ ( λ ) ∣ > 2 } \{\lambda\in\R:\lvert\Delta(\lambda)\rvert>2\} { λ ∈ R : ∣ Δ ( λ )∣ > 2 } R \R R
証明. 系 5.4 によりΔ \Delta Δ C \C C R \R R Δ ∣ R : R → R \Delta|_{\R}\colon\R\to\R Δ ∣ R : R → R ( − 2 , 2 ) (-2,2) ( − 2 , 2 ) R ∖ [ − 2 , 2 ] \R\setminus[-2,2] R ∖ [ − 2 , 2 ] ▨
例 5.6. h ∈ R h\in\R h ∈ R T = π T=\pi T = π q ( t ) = 2 h cos 2 t q(t)=2h\cos2t q ( t ) = 2 h cos 2 t
− y ′ ′ + 2 h cos ( 2 t ) y = a y ( a ∈ R ) -y''+2h\cos(2t)\,y=ay\qquad(a\in\R) − y ′′ + 2 h cos ( 2 t ) y = a y ( a ∈ R ) をMathieu 方程式 (Mathieu equation )h = 0 h=0 h = 0 a > 0 a>0 a > 0 k = a k=\sqrt a k = a θ ( t , a ) = cos k t \theta(t,a)=\cos kt θ ( t , a ) = cos k t φ ( t , a ) = k − 1 sin k t \varphi(t,a)=k^{-1}\sin kt φ ( t , a ) = k − 1 sin k t Δ ( a ) = 2 cos ( π a ) \Delta(a)=2\cos(\pi\sqrt a) Δ ( a ) = 2 cos ( π a ) a = 0 a=0 a = 0 θ = 1 \theta=1 θ = 1 φ = t \varphi=t φ = t Δ ( 0 ) = 2 \Delta(0)=2 Δ ( 0 ) = 2 a < 0 a<0 a < 0 κ = − a \kappa=\sqrt{-a} κ = − a θ = cosh κ t \theta=\cosh\kappa t θ = cosh κ t φ = κ − 1 sinh κ t \varphi=\kappa^{-1}\sinh\kappa t φ = κ − 1 sinh κ t Δ ( a ) = 2 cosh ( π − a ) > 2 \Delta(a)=2\cosh(\pi\sqrt{-a})>2 Δ ( a ) = 2 cosh ( π − a ) > 2
{ a ∈ R : ∣ Δ ( a ) ∣ < 2 } = ( 0 , ∞ ) ∖ { n 2 : n ∈ N ≥ 1 } \{a\in\R:\lvert\Delta(a)\rvert<2\}=(0,\infty)\setminus\{n^2:n\in\NN\} { a ∈ R : ∣ Δ ( a )∣ < 2 } = ( 0 , ∞ ) ∖ { n 2 : n ∈ N ≥ 1 } であり、定理 5.3 (1) によりこの集合のa a a 定理 5.3 (2) によりa < 0 a<0 a < 0 a = n 2 a=n^2 a = n 2 n ∈ N ≥ 1 n\in\NN n ∈ N ≥ 1 θ ( π ) = φ ′ ( π ) = ( − 1 ) n \theta(\pi)=\varphi'(\pi)=(-1)^n θ ( π ) = φ ′ ( π ) = ( − 1 ) n θ ′ ( π ) = φ ( π ) = 0 \theta'(\pi)=\varphi(\pi)=0 θ ′ ( π ) = φ ( π ) = 0 M ( n 2 ) = ( − 1 ) n I M(n^2)=(-1)^nI M ( n 2 ) = ( − 1 ) n I 定理 5.3 (3) によりすべての解が有界である。
a = 0 a=0 a = 0
M ( 0 ) = ( 1 π 0 1 ) ≠ I M(0)=\begin{pmatrix}1&\pi\\0&1\end{pmatrix}\ne I M ( 0 ) = ( 1 0 π 1 ) = I であり、解y ( t ) = t y(t)=t y ( t ) = t 1 1 1 1 1 1 ker ( M ( 0 ) − I ) \ker(M(0)-I) ker ( M ( 0 ) − I ) G 1 ( M ( 0 ) ) = C 2 G_1(M(0))=\C^2 G 1 ( M ( 0 )) = C 2 1 1 1 定理 3.2 (2) の半単純性の条件を外すと、すべての解の有界性は成り立たない。
6 演習 解答. (1) を示す。Φ 0 ( t ) = exp ( a t + sin t ) \Phi_0(t)=\exp(at+\sin t) Φ 0 ( t ) = exp ( a t + sin t ) Φ 0 ′ = ( a + cos t ) Φ 0 \Phi_0'=(a+\cos t)\Phi_0 Φ 0 ′ = ( a + cos t ) Φ 0 Φ 0 ( 0 ) = 1 \Phi_0(0)=1 Φ 0 ( 0 ) = 1 Φ 0 ( 2 π ) = e 2 π a \Phi_0(2\pi)=e^{2\pi a} Φ 0 ( 2 π ) = e 2 π a e 2 π a e^{2\pi a} e 2 π a a + i k a+ik a + ik k ∈ Z k\in\Z k ∈ Z B = a B=a B = a P ( t ) = e sin t P(t)=e^{\sin t} P ( t ) = e s i n t e 2 π B = e 2 π a e^{2\pi B}=e^{2\pi a} e 2 π B = e 2 π a P P P 2 π 2\pi 2 π P ( 0 ) = 1 P(0)=1 P ( 0 ) = 1 Φ 0 ( t ) = P ( t ) e t B \Phi_0(t)=P(t)e^{tB} Φ 0 ( t ) = P ( t ) e tB 定理 3.2 (1) により、すべての解が0 0 0 e 2 π a < 1 e^{2\pi a}<1 e 2 π a < 1 a < 0 a<0 a < 0 1 × 1 1\times1 1 × 1 定理 3.2 (2) により、すべての解が有界であることは、a ≤ 0 a\le0 a ≤ 0 x ( t ) = x ( 0 ) exp ( a t + sin t ) x(t)=x(0)\exp(at+\sin t) x ( t ) = x ( 0 ) exp ( a t + sin t )
(2) を示す。対角行列の指数関数は対角成分の指数関数を並べたものであるから、e T B = diag ( e i π , e c T e i π ) = diag ( − 1 , − e c T ) = M e^{TB}=\operatorname{diag}(e^{i\pi},e^{cT}e^{i\pi})=\operatorname{diag}(-1,-e^{cT})=M e T B = diag ( e iπ , e c T e iπ ) = diag ( − 1 , − e c T ) = M e − t B = diag ( e − i π t / T , e − c t e − i π t / T ) e^{-tB}=\operatorname{diag}\bigl(e^{-i\pi t/T},e^{-ct}e^{-i\pi t/T}\bigr) e − tB = diag ( e − iπ t / T , e − c t e − iπ t / T )
P ( t ) = R ( π t / T ) diag ( 1 , e c t ) diag ( e − i π t / T , e − c t e − i π t / T ) = e − i π t / T R ( π t / T ) P(t)=R(\pi t/T)\operatorname{diag}(1,e^{ct})\operatorname{diag}\bigl(e^{-i\pi t/T},e^{-ct}e^{-i\pi t/T}\bigr)=e^{-i\pi t/T}R(\pi t/T) P ( t ) = R ( π t / T ) diag ( 1 , e c t ) diag ( e − iπ t / T , e − c t e − iπ t / T ) = e − iπ t / T R ( π t / T ) である。R ( ϑ + π ) = − R ( ϑ ) R(\vartheta+\pi)=-R(\vartheta) R ( ϑ + π ) = − R ( ϑ ) e − i π = − 1 e^{-i\pi}=-1 e − iπ = − 1 P ( t + T ) = ( − 1 ) ( − 1 ) e − i π t / T R ( π t / T ) = P ( t ) P(t+T)=(-1)(-1)e^{-i\pi t/T}R(\pi t/T)=P(t) P ( t + T ) = ( − 1 ) ( − 1 ) e − iπ t / T R ( π t / T ) = P ( t ) B B B i π / T i\pi/T iπ / T c + i π / T c+i\pi/T c + iπ / T B B B ▨
問題 6.2. ω > 0 \omega>0 ω > 0 f ∈ C ( R ; C ) f\in C(\R;\C) f ∈ C ( R ; C ) 2 π 2\pi 2 π y ′ ′ + ω 2 y = f ( t ) y''+\omega^2y=f(t) y ′′ + ω 2 y = f ( t ) 2 π 2\pi 2 π C 2 C^2 C 2
ω ∉ N ≥ 1 \omega\notin\NN ω ∈ / N ≥ 1 2 π 2\pi 2 π ω = m ∈ N ≥ 1 \omega=m\in\NN ω = m ∈ N ≥ 1 2 π 2\pi 2 π ∫ 0 2 π f ( t ) cos m t d t = ∫ 0 2 π f ( t ) sin m t d t = 0 \int_0^{2\pi}f(t)\cos mt\,dt=\int_0^{2\pi}f(t)\sin mt\,dt=0 ∫ 0 2 π f ( t ) cos m t d t = ∫ 0 2 π f ( t ) sin m t d t = 0 y p y_p y p y p + a cos m t + b sin m t y_p+a\cos mt+b\sin mt y p + a cos m t + b sin m t a , b ∈ C a,b\in\C a , b ∈ C f f f y p y_p y p a , b ∈ R a,b\in\R a , b ∈ R 解答. §E13.9 命題 4.1 の意味の解y y y y ′ y' y ′ f − ω 2 y f-\omega^2y f − ω 2 y §E13.2 系 2.7 (1) によりy ′ y' y ′ C 1 C^1 C 1 y y y C 2 C^2 C 2 C 2 C^2 C 2 y y y §E13.2 系 2.7 (2) によりy ′ y' y ′ y y y §E13.9 命題 4.1 の意味の解である。従って、y y y 2 π 2\pi 2 π C 2 C^2 C 2 Y = ( y , y ′ ) T \boldsymbol Y=(y,y')^{\mathsf T} Y = ( y , y ′ ) T Y ′ = A Y + b \boldsymbol Y'=A\boldsymbol Y+\boldsymbol b Y ′ = A Y + b A = ( 0 1 − ω 2 0 ) A=\begin{pmatrix}0&1\\-\omega^2&0\end{pmatrix} A = ( 0 − ω 2 1 0 ) b = ( 0 , f ) T \boldsymbol b=(0,f)^{\mathsf T} b = ( 0 , f ) T 2 π 2\pi 2 π Φ 0 ( t ) = ( cos ω t ω − 1 sin ω t − ω sin ω t cos ω t ) \Phi_0(t)=\begin{pmatrix}\cos\omega t&\omega^{-1}\sin\omega t\\-\omega\sin\omega t&\cos\omega t\end{pmatrix} Φ 0 ( t ) = ( cos ω t − ω sin ω t ω − 1 sin ω t cos ω t ) Y ′ = A Y \boldsymbol Y'=A\boldsymbol Y Y ′ = A Y Φ 0 ( 0 ) = I \Phi_0(0)=I Φ 0 ( 0 ) = I M = Φ 0 ( 2 π ) M=\Phi_0(2\pi) M = Φ 0 ( 2 π ) tr M = 2 cos 2 π ω \operatorname{tr}M=2\cos2\pi\omega tr M = 2 cos 2 π ω det M = 1 \det M=1 det M = 1 e ± 2 π i ω e^{\pm2\pi i\omega} e ± 2 π iω 1 1 1 ω ∈ N ≥ 1 \omega\in\NN ω ∈ N ≥ 1
(1) を示す。ω ∉ N ≥ 1 \omega\notin\NN ω ∈ / N ≥ 1 1 1 1 M M M 系 4.5 によりρ = 1 \rho=1 ρ = 1 定理 4.4 (4) により2 π 2\pi 2 π
(2) を示す。ω = m \omega=m ω = m A A A 例 4.8 のA m A_m A m ψ ′ = − A T ψ \boldsymbol\psi'=-A^{\mathsf T}\boldsymbol\psi ψ ′ = − A T ψ 2 π 2\pi 2 π ψ 2 \psi_2 ψ 2 cos m t \cos mt cos m t sin m t \sin mt sin m t A A A ψ ∗ b = ψ 2 ‾ f \boldsymbol\psi^*\boldsymbol b=\overline{\psi_2}f ψ ∗ b = ψ 2 f 定理 4.4 (4) により2 π 2\pi 2 π ∫ 0 2 π f cos m t d t = ∫ 0 2 π f sin m t d t = 0 \int_0^{2\pi}f\cos mt\,dt=\int_0^{2\pi}f\sin mt\,dt=0 ∫ 0 2 π f cos m t d t = ∫ 0 2 π f sin m t d t = 0 y ′ ′ + m 2 y = 0 y''+m^2y=0 y ′′ + m 2 y = 0 a cos m t + b sin m t a\cos mt+b\sin mt a cos m t + b sin m t a , b ∈ C a,b\in\C a , b ∈ C 2 π 2\pi 2 π 定理 4.4 (3) と定理 4.4 (4) により解の全体はy p + a cos m t + b sin m t y_p+a\cos mt+b\sin mt y p + a cos m t + b sin m t f f f y p y_p y p y p y_p y p cos m t \cos mt cos m t sin m t \sin mt sin m t y p + a cos m t + b sin m t y_p+a\cos mt+b\sin mt y p + a cos m t + b sin m t a , b ∈ R a,b\in\R a , b ∈ R ▨
問題 6.3. c > 0 c>0 c > 0 T = 2 T=2 T = 2 q q q j ∈ Z j\in\Z j ∈ Z t ∈ [ 2 j , 2 j + 1 ) t\in[2j,2j+1) t ∈ [ 2 j , 2 j + 1 ) q ( t ) = 0 q(t)=0 q ( t ) = 0 t ∈ [ 2 j + 1 , 2 j + 2 ) t\in[2j+1,2j+2) t ∈ [ 2 j + 1 , 2 j + 2 ) q ( t ) = c q(t)=c q ( t ) = c λ > c \lambda>c λ > c k = λ k=\sqrt\lambda k = λ l = λ − c l=\sqrt{\lambda-c} l = λ − c
Hill 方程式− y ′ ′ + q ( t ) y = λ y -y''+q(t)y=\lambda y − y ′′ + q ( t ) y = λ y Δ ( λ ) = 2 cos k cos l − ( k l + l k ) sin k sin l \Delta(\lambda)=2\cos k\cos l-\bigl(\frac kl+\frac lk\bigr)\sin k\sin l Δ ( λ ) = 2 cos k cos l − ( l k + k l ) sin k sin l
k + l ∈ 2 π N ≥ 1 k+l\in2\pi\NN k + l ∈ 2 π N ≥ 1 sin k ≠ 0 \sin k\ne0 sin k = 0 Δ ( λ ) > 2 \Delta(\lambda)>2 Δ ( λ ) > 2 R \R R 0 < c < 4 π 2 0<c<4\pi^2 0 < c < 4 π 2 n ∈ N ≥ 1 n\in\NN n ∈ N ≥ 1 λ n + λ n − c = 2 π n \sqrt{\lambda_n}+\sqrt{\lambda_n-c}=2\pi n λ n + λ n − c = 2 π n λ n > c \lambda_n>c λ n > c Δ ( λ n ) > 2 \Delta(\lambda_n)>2 Δ ( λ n ) > 2 解答. (1) を示す。κ > 0 \kappa>0 κ > 0 E ( κ ) = ( cos κ κ − 1 sin κ − κ sin κ cos κ ) E(\kappa)=\begin{pmatrix}\cos\kappa&\kappa^{-1}\sin\kappa\\-\kappa\sin\kappa&\cos\kappa\end{pmatrix} E ( κ ) = ( cos κ − κ sin κ κ − 1 sin κ cos κ ) [ 0 , 1 ] [0,1] [ 0 , 1 ] q = 0 q=0 q = 0 Y = ( y , y ′ ) T \boldsymbol Y=(y,y')^{\mathsf T} Y = ( y , y ′ ) T Y ′ = ( 0 1 − k 2 0 ) Y \boldsymbol Y'=\begin{pmatrix}0&1\\-k^2&0\end{pmatrix}\boldsymbol Y Y ′ = ( 0 − k 2 1 0 ) Y ( cos k t , − k sin k t ) T (\cos kt,-k\sin kt)^{\mathsf T} ( cos k t , − k sin k t ) T ( k − 1 sin k t , cos k t ) T (k^{-1}\sin kt,\cos kt)^{\mathsf T} ( k − 1 sin k t , cos k t ) T §E13.9 定理 1.1 の一意性によりY ( 1 ) = E ( k ) Y ( 0 ) \boldsymbol Y(1)=E(k)\boldsymbol Y(0) Y ( 1 ) = E ( k ) Y ( 0 ) [ 1 , 2 ] [1,2] [ 1 , 2 ] q = c q=c q = c Y ′ = ( 0 1 − l 2 0 ) Y \boldsymbol Y'=\begin{pmatrix}0&1\\-l^2&0\end{pmatrix}\boldsymbol Y Y ′ = ( 0 − l 2 1 0 ) Y Y ( 2 ) = E ( l ) Y ( 1 ) \boldsymbol Y(2)=E(l)\boldsymbol Y(1) Y ( 2 ) = E ( l ) Y ( 1 ) M ( λ ) = E ( l ) E ( k ) M(\lambda)=E(l)E(k) M ( λ ) = E ( l ) E ( k )
Δ ( λ ) = tr ( E ( l ) E ( k ) ) = ( cos l cos k − k l sin l sin k ) + ( − l k sin l sin k + cos l cos k ) \Delta(\lambda)=\operatorname{tr}\bigl(E(l)E(k)\bigr)=\Bigl(\cos l\cos k-\frac{k}{l}\sin l\sin k\Bigr)+\Bigl(-\frac lk\sin l\sin k+\cos l\cos k\Bigr) Δ ( λ ) = tr ( E ( l ) E ( k ) ) = ( cos l cos k − l k sin l sin k ) + ( − k l sin l sin k + cos l cos k ) である。
(2) を示す。2 cos ( k + l ) = 2 cos k cos l − 2 sin k sin l 2\cos(k+l)=2\cos k\cos l-2\sin k\sin l 2 cos ( k + l ) = 2 cos k cos l − 2 sin k sin l k l + l k − 2 = ( k − l ) 2 k l \frac kl+\frac lk-2=\frac{(k-l)^2}{kl} l k + k l − 2 = k l ( k − l ) 2
Δ ( λ ) = 2 cos ( k + l ) − ( k − l ) 2 k l sin k sin l \Delta(\lambda)=2\cos(k+l)-\frac{(k-l)^2}{kl}\sin k\sin l Δ ( λ ) = 2 cos ( k + l ) − k l ( k − l ) 2 sin k sin l である。k + l = 2 π n k+l=2\pi n k + l = 2 π n cos ( k + l ) = 1 \cos(k+l)=1 cos ( k + l ) = 1 sin l = sin ( 2 π n − k ) = − sin k \sin l=\sin(2\pi n-k)=-\sin k sin l = sin ( 2 π n − k ) = − sin k Δ ( λ ) = 2 + ( k − l ) 2 k l sin 2 k \Delta(\lambda)=2+\frac{(k-l)^2}{kl}\sin^2k Δ ( λ ) = 2 + k l ( k − l ) 2 sin 2 k k 2 − l 2 = c > 0 k^2-l^2=c>0 k 2 − l 2 = c > 0 k ≠ l k\ne l k = l sin k ≠ 0 \sin k\ne0 sin k = 0 Δ ( λ ) > 2 \Delta(\lambda)>2 Δ ( λ ) > 2 定理 5.3 (2) により零でないすべての解は有界でない。
(3) を示す。g ( λ ) = λ + λ − c g(\lambda)=\sqrt\lambda+\sqrt{\lambda-c} g ( λ ) = λ + λ − c [ c , ∞ ) [c,\infty) [ c , ∞ ) g ( c ) = c < 2 π ≤ 2 π n g(c)=\sqrt c<2\pi\le2\pi n g ( c ) = c < 2 π ≤ 2 π n g ( λ ) → ∞ g(\lambda)\to\infty g ( λ ) → ∞ λ → ∞ \lambda\to\infty λ → ∞ g ( λ n ) = 2 π n g(\lambda_n)=2\pi n g ( λ n ) = 2 π n λ n > c \lambda_n>c λ n > c k = λ n k=\sqrt{\lambda_n} k = λ n l = λ n − c l=\sqrt{\lambda_n-c} l = λ n − c sin k = 0 \sin k=0 sin k = 0 k > 0 k>0 k > 0 k = j π k=j\pi k = j π j ∈ N ≥ 1 j\in\NN j ∈ N ≥ 1 l = 2 π n − k = ( 2 n − j ) π l=2\pi n-k=(2n-j)\pi l = 2 π n − k = ( 2 n − j ) π 0 < l < k 0<l<k 0 < l < k n < j < 2 n n<j<2n n < j < 2 n
c = k 2 − l 2 = π 2 ( j 2 − ( 2 n − j ) 2 ) = 4 π 2 n ( j − n ) ≥ 4 π 2 c=k^2-l^2=\pi^2\bigl(j^2-(2n-j)^2\bigr)=4\pi^2n(j-n)\ge4\pi^2 c = k 2 − l 2 = π 2 ( j 2 − ( 2 n − j ) 2 ) = 4 π 2 n ( j − n ) ≥ 4 π 2 となる。これはc < 4 π 2 c<4\pi^2 c < 4 π 2 sin k ≠ 0 \sin k\ne0 sin k = 0 (2) によりΔ ( λ n ) > 2 \Delta(\lambda_n)>2 Δ ( λ n ) > 2 ▨