1 初期値問題と二点境界値問題 定義 1.1. a < b a<b a < b P , Q , f : [ a , b ] → R P,Q,f:[a,b]\to\R P , Q , f : [ a , b ] → R α a , β a , α b , β b \alpha_a,\beta_a,\alpha_b,\beta_b α a , β a , α b , β b
y ′ ′ + P ( x ) y ′ + Q ( x ) y = f ( x ) y''+P(x)y'+Q(x)y=f(x) y ′′ + P ( x ) y ′ + Q ( x ) y = f ( x ) (1.1.1) \text{(1.1.1)} (1.1.1) に対して
B a [ y ] = α a y ( a ) + β a y ′ ( a ) = 0 , B b [ y ] = α b y ( b ) + β b y ′ ( b ) = 0 \begin{aligned}
B_a[y]&=\alpha_a y(a)+\beta_a y'(a)=0,\\
B_b[y]&=\alpha_b y(b)+\beta_b y'(b)=0
\end{aligned} B a [ y ] B b [ y ] = α a y ( a ) + β a y ′ ( a ) = 0 , = α b y ( b ) + β b y ′ ( b ) = 0 (1.1.2) \text{(1.1.2)} (1.1.2) を課す。ここで
( α a , β a ) ≠ ( 0 , 0 ) , ( α b , β b ) ≠ ( 0 , 0 ) (\alpha_a,\beta_a)\ne(0,0),\qquad
(\alpha_b,\beta_b)\ne(0,0) ( α a , β a ) = ( 0 , 0 ) , ( α b , β b ) = ( 0 , 0 ) とする。式 (1.1.1) と式 (1.1.2) を満たすy ∈ C 2 ( [ a , b ] ; R ) y\in C^2([a,b];\R) y ∈ C 2 ([ a , b ] ; R ) 斉次分離型二点境界値問題 (two-point boundary value problem with homogeneous separated boundary conditions )式 (1.1.2) の右辺が零であることを指し、微分方程式の右辺f f f
斉次方程式y ′ ′ + P y ′ + Q y = 0 y''+Py'+Qy=0 y ′′ + P y ′ + Q y = 0 y ( a ) y(a) y ( a ) y ′ ( a ) y'(a) y ′ ( a ) §E13.9 定理 5.2 により、解がただ一つ存在する。一方、境界値問題では、B a [ y ] = 0 B_a[y]=0 B a [ y ] = 0 ( y ( a ) , y ′ ( a ) ) (y(a),y'(a)) ( y ( a ) , y ′ ( a )) R 2 \R^2 R 2 B b [ y ] = 0 B_b[y]=0 B b [ y ] = 0
命題 1.2. a < b a<b a < b P , Q , f ∈ C ( [ a , b ] ; R ) P,Q,f\in C([a,b];\R) P , Q , f ∈ C ([ a , b ] ; R ) ( α a , β a ) (\alpha_a,\beta_a) ( α a , β a ) ( α b , β b ) (\alpha_b,\beta_b) ( α b , β b ) ( 0 , 0 ) (0,0) ( 0 , 0 ) C 2 ( [ a , b ] ; R ) C^2([a,b];\R) C 2 ([ a , b ] ; R )
L [ y ] = y ′ ′ + P y ′ + Q y L[y]=y''+Py'+Qy L [ y ] = y ′′ + P y ′ + Q y とし、斉次境界値問題
L [ y ] = 0 , B a [ y ] = B b [ y ] = 0 L[y]=0,\qquad B_a[y]=B_b[y]=0 L [ y ] = 0 , B a [ y ] = B b [ y ] = 0 (1.2.1) \text{(1.2.1)} (1.2.1) を満たすy ∈ C 2 ( [ a , b ] ; R ) y\in C^2([a,b];\R) y ∈ C 2 ([ a , b ] ; R ) N N N 式 (1.1.1) 、式 (1.1.2) が解y p y_p y p { y p + y h ∣ y h ∈ N } \{y_p+y_h\mid y_h\in N\} { y p + y h ∣ y h ∈ N } y y y y = y p + y h y=y_p+y_h y = y p + y h y h ∈ N y_h\in N y h ∈ N N = { 0 } N=\{0\} N = { 0 } N N N y h y_h y h y p y_p y p C ∈ R C\in\R C ∈ R y p + C y h y_p+Cy_h y p + C y h
証明. y y y L [ y − y p ] = 0 L[y-y_p]=0 L [ y − y p ] = 0 B a [ y − y p ] = B b [ y − y p ] = 0 B_a[y-y_p]=B_b[y-y_p]=0 B a [ y − y p ] = B b [ y − y p ] = 0 y − y p ∈ N y-y_p\in N y − y p ∈ N y h ∈ N y_h\in N y h ∈ N
L [ y p + y h ] = f , B a [ y p + y h ] = B b [ y p + y h ] = 0 L[y_p+y_h]=f,\qquad
B_a[y_p+y_h]=B_b[y_p+y_h]=0 L [ y p + y h ] = f , B a [ y p + y h ] = B b [ y p + y h ] = 0 である。y = y p + y h y=y_p+y_h y = y p + y h y h y_h y h y − y p y-y_p y − y p N = { 0 } N=\{0\} N = { 0 } y , y ~ y,\tilde y y , y ~ y − y ~ ∈ N y-\tilde y\in N y − y ~ ∈ N y = y ~ y=\tilde y y = y ~ y h ∈ N y_h\in N y h ∈ N C ∈ R C\in\R C ∈ R C y h ∈ N Cy_h\in N C y h ∈ N y p + C y h y_p+Cy_h y p + C y h C C C ▨
2 固有値問題 定義 2.1. a < b a<b a < b Λ ⊆ C \Lambda\subseteq\C Λ ⊆ C λ ∈ Λ \lambda\in\Lambda λ ∈ Λ P λ , Q λ ∈ L 1 ( [ a , b ] ; C ) P_\lambda,Q_\lambda\in L^1([a,b];\C) P λ , Q λ ∈ L 1 ([ a , b ] ; C ) ( 0 , 0 ) (0,0) ( 0 , 0 ) ( α a ( λ ) , β a ( λ ) ) (\alpha_a(\lambda),\beta_a(\lambda)) ( α a ( λ ) , β a ( λ )) ( α b ( λ ) , β b ( λ ) ) (\alpha_b(\lambda),\beta_b(\lambda)) ( α b ( λ ) , β b ( λ ))
L λ [ y ] = y ′ ′ + P λ y ′ + Q λ y , B a , λ [ y ] = α a ( λ ) y ( a ) + β a ( λ ) y ′ ( a ) , B b , λ [ y ] = α b ( λ ) y ( b ) + β b ( λ ) y ′ ( b ) L_\lambda[y]=y''+P_\lambda y'+Q_\lambda y,\qquad
B_{a,\lambda}[y]=\alpha_a(\lambda)y(a)+\beta_a(\lambda)y'(a),\qquad
B_{b,\lambda}[y]=\alpha_b(\lambda)y(b)+\beta_b(\lambda)y'(b) L λ [ y ] = y ′′ + P λ y ′ + Q λ y , B a , λ [ y ] = α a ( λ ) y ( a ) + β a ( λ ) y ′ ( a ) , B b , λ [ y ] = α b ( λ ) y ( b ) + β b ( λ ) y ′ ( b ) とおく。λ ∈ Λ \lambda\in\Lambda λ ∈ Λ y ∈ C 1 ( [ a , b ] ; C ) y\in C^1([a,b];\C) y ∈ C 1 ([ a , b ] ; C ) y ′ ∈ A C ( [ a , b ] ; C ) y'\in AC([a,b];\C) y ′ ∈ A C ([ a , b ] ; C )
L λ [ y ] = 0 ほとんど至る所 , B a , λ [ y ] = B b , λ [ y ] = 0 L_\lambda[y]=0\quad\text{ほとんど至る所},\qquad B_{a,\lambda}[y]=B_{b,\lambda}[y]=0 L λ [ y ] = 0 ほとんど至る所 , B a , λ [ y ] = B b , λ [ y ] = 0 を満たす零でないy y y λ \lambda λ 固有値 (eigenvalue of a boundary value problem )y y y λ \lambda λ 固有関数 (eigenfunction of a boundary value problem )
例 2.2. L > 0 L>0 L > 0 λ \lambda λ
y ′ ′ + λ y = 0 , y ( 0 ) = y ( L ) = 0 y''+\lambda y=0,\qquad
y(0)=y(L)=0 y ′′ + λ y = 0 , y ( 0 ) = y ( L ) = 0 を考える。
μ > 0 \mu>0 μ > 0 λ = μ 2 \lambda=\mu^2 λ = μ 2
y = A cos ( μ x ) + B sin ( μ x ) y=A\cos(\mu x)+B\sin(\mu x) y = A cos ( μx ) + B sin ( μx ) である。y ( 0 ) = 0 y(0)=0 y ( 0 ) = 0 A = 0 A=0 A = 0 sin ( μ L ) = 0 \sin(\mu L)=0 sin ( μL ) = 0 μ > 0 \mu>0 μ > 0 μ L ∈ { n π ∣ n = 1 , 2 , … } \mu L\in\{n\pi\mid n=1,2,\ldots\} μL ∈ { nπ ∣ n = 1 , 2 , … }
λ n = ( n π L ) 2 , n = 1 , 2 , … \lambda_n=\left(\frac{n\pi}{L}\right)^2,\qquad
n=1,2,\ldots λ n = ( L nπ ) 2 , n = 1 , 2 , … であり、λ n \lambda_n λ n
y n ( x ) = sin ( n π x L ) y_n(x)=\sin\left(\frac{n\pi x}{L}\right) y n ( x ) = sin ( L nπ x ) の零でない定数倍である。
λ = 0 \lambda=0 λ = 0 y = A + B x y=A+Bx y = A + B x A = B = 0 A=B=0 A = B = 0 μ > 0 \mu>0 μ > 0 λ = − μ 2 \lambda=-\mu^2 λ = − μ 2
y = A cosh ( μ x ) + B sinh ( μ x ) y=A\cosh(\mu x)+B\sinh(\mu x) y = A cosh ( μx ) + B sinh ( μx ) である。y ( 0 ) = 0 y(0)=0 y ( 0 ) = 0 A = 0 A=0 A = 0 y ( L ) = 0 y(L)=0 y ( L ) = 0 sinh ( μ L ) > 0 \sinh(\mu L)>0 sinh ( μL ) > 0 B = 0 B=0 B = 0
3 Green 関数の構成 定義 3.1. a < b a<b a < b P , Q ∈ L 1 ( [ a , b ] ; R ) P,Q\in L^1([a,b];\R) P , Q ∈ L 1 ([ a , b ] ; R ) f ∈ L 1 ( [ a , b ] ; C ) f\in L^1([a,b];\C) f ∈ L 1 ([ a , b ] ; C ) ( α a , β a ) (\alpha_a,\beta_a) ( α a , β a ) ( α b , β b ) (\alpha_b,\beta_b) ( α b , β b ) ( 0 , 0 ) (0,0) ( 0 , 0 ) a a a y : [ a , b ] → C y:[a,b]\to\C y : [ a , b ] → C
B a [ y ] = α a y ( a ) + β a y ′ ( a ) B_a[y]=\alpha_a y(a)+\beta_a y'(a) B a [ y ] = α a y ( a ) + β a y ′ ( a ) とおき、b b b y : [ a , b ] → C y:[a,b]\to\C y : [ a , b ] → C
B b [ y ] = α b y ( b ) + β b y ′ ( b ) B_b[y]=\alpha_b y(b)+\beta_b y'(b) B b [ y ] = α b y ( b ) + β b y ′ ( b ) とおく。ここでy ′ ( a ) y'(a) y ′ ( a ) y ′ ( b ) y'(b) y ′ ( b ) y ∈ C 1 ( [ a , b ] ; C ) y\in C^1([a,b];\C) y ∈ C 1 ([ a , b ] ; C ) y ′ ∈ A C ( [ a , b ] ; C ) y'\in AC([a,b];\C) y ′ ∈ A C ([ a , b ] ; C )
y ′ ′ + P y ′ + Q y = f ほとんど至る所 , B a [ y ] = B b [ y ] = 0 y''+Py'+Qy=f\quad\text{ほとんど至る所},\qquad
B_a[y]=B_b[y]=0 y ′′ + P y ′ + Q y = f ほとんど至る所 , B a [ y ] = B b [ y ] = 0 を満たすy y y 可積分係数の斉次分離型二点境界値問題 (two-point boundary value problem with integrable coefficients and homogeneous separated boundary conditions )y y y f = 0 f=0 f = 0 境界核 (boundary kernel )
補題 3.2. a < b a<b a < b P , Q ∈ L 1 ( [ a , b ] ; R ) P,Q\in L^1([a,b];\R) P , Q ∈ L 1 ([ a , b ] ; R ) ( α a , β a ) (\alpha_a,\beta_a) ( α a , β a ) ( α b , β b ) (\alpha_b,\beta_b) ( α b , β b ) ( 0 , 0 ) (0,0) ( 0 , 0 ) w ∈ C 1 ( [ a , b ] ; R ) w\in C^1([a,b];\R) w ∈ C 1 ([ a , b ] ; R ) w ′ ∈ A C ( [ a , b ] ; R ) w'\in AC([a,b];\R) w ′ ∈ A C ([ a , b ] ; R ) w ′ ′ + P w ′ + Q w = 0 w''+Pw'+Qw=0 w ′′ + P w ′ + Qw = 0 w w w
u ( a ) = β a , u ′ ( a ) = − α a , v ( b ) = β b , v ′ ( b ) = − α b u(a)=\beta_a,\qquad u'(a)=-\alpha_a,\qquad
v(b)=\beta_b,\qquad v'(b)=-\alpha_b u ( a ) = β a , u ′ ( a ) = − α a , v ( b ) = β b , v ′ ( b ) = − α b を満たすu , v u,v u , v u , v u,v u , v B a [ u ] = B b [ v ] = 0 B_a[u]=B_b[v]=0 B a [ u ] = B b [ v ] = 0 W = u v ′ − u ′ v W=uv'-u'v W = u v ′ − u ′ v
この分離型二点境界値問題の境界核は自明である。
あるx 0 ∈ [ a , b ] x_0\in[a,b] x 0 ∈ [ a , b ] W ( x 0 ) ≠ 0 W(x_0)\ne0 W ( x 0 ) = 0
任意のx ∈ [ a , b ] x\in[a,b] x ∈ [ a , b ] W ( x ) ≠ 0 W(x)\ne0 W ( x ) = 0
証明. §E13.9 命題 4.1 をn = 2 n=2 n = 2 K = R \mathbb K=\R K = R I = [ a , b ] I=[a,b] I = [ a , b ] a a a b b b u , v u,v u , v ( β a , − α a ) (\beta_a,-\alpha_a) ( β a , − α a ) ( β b , − α b ) (\beta_b,-\alpha_b) ( β b , − α b ) u , v u,v u , v
B a [ u ] = α a β a − β a α a = 0 , B b [ v ] = α b β b − β b α b = 0 , W ( a ) = α a v ( a ) + β a v ′ ( a ) = B a [ v ] , W ( b ) = − α b u ( b ) − β b u ′ ( b ) = − B b [ u ] \begin{aligned}
B_a[u]&=\alpha_a\beta_a-\beta_a\alpha_a=0,&
B_b[v]&=\alpha_b\beta_b-\beta_b\alpha_b=0,\\
W(a)&=\alpha_av(a)+\beta_av'(a)=B_a[v],&
W(b)&=-\alpha_bu(b)-\beta_bu'(b)=-B_b[u]
\end{aligned} B a [ u ] W ( a ) = α a β a − β a α a = 0 , = α a v ( a ) + β a v ′ ( a ) = B a [ v ] , B b [ v ] W ( b ) = α b β b − β b α b = 0 , = − α b u ( b ) − β b u ′ ( b ) = − B b [ u ] である。
(1) ⇒ \Rightarrow ⇒ (2) v v v B b [ v ] = 0 B_b[v]=0 B b [ v ] = 0 B a [ v ] ≠ 0 B_a[v]\ne0 B a [ v ] = 0 W ( a ) = B a [ v ] ≠ 0 W(a)=B_a[v]\ne0 W ( a ) = B a [ v ] = 0
(2) ⇒ \Rightarrow ⇒ (3) §E13.9 命題 5.5 をu , v u,v u , v t 0 = x 0 t_0=x_0 t 0 = x 0
(3) ⇒ \Rightarrow ⇒ (1) W W W u , v u,v u , v K = C \mathbb K=\C K = C §E13.9 命題 5.5 により、その解空間の基本解系である。境界核の元y y y c 1 , c 2 ∈ C c_1,c_2\in\C c 1 , c 2 ∈ C y = c 1 u + c 2 v y=c_1u+c_2v y = c 1 u + c 2 v B a [ u ] = B b [ v ] = 0 B_a[u]=B_b[v]=0 B a [ u ] = B b [ v ] = 0
0 = B a [ y ] = c 2 B a [ v ] = c 2 W ( a ) , 0 = B b [ y ] = c 1 B b [ u ] = − c 1 W ( b ) 0=B_a[y]=c_2B_a[v]=c_2W(a),\qquad
0=B_b[y]=c_1B_b[u]=-c_1W(b) 0 = B a [ y ] = c 2 B a [ v ] = c 2 W ( a ) , 0 = B b [ y ] = c 1 B b [ u ] = − c 1 W ( b ) であり、W ( a ) ≠ 0 W(a)\ne0 W ( a ) = 0 W ( b ) ≠ 0 W(b)\ne0 W ( b ) = 0 c 1 = c 2 = 0 c_1=c_2=0 c 1 = c 2 = 0 y = 0 y=0 y = 0 ▨
定義 3.3. a < b a<b a < b P , Q ∈ L 1 ( [ a , b ] ; R ) P,Q\in L^1([a,b];\R) P , Q ∈ L 1 ([ a , b ] ; R ) ( α a , β a ) (\alpha_a,\beta_a) ( α a , β a ) ( α b , β b ) (\alpha_b,\beta_b) ( α b , β b ) ( 0 , 0 ) (0,0) ( 0 , 0 ) u , v u,v u , v 補題 3.2 のu , v u,v u , v W = u v ′ − u ′ v W=uv'-u'v W = u v ′ − u ′ v W W W [ a , b ] [a,b] [ a , b ]
G ( x , s ) = { u ( x ) v ( s ) W ( s ) , a ≤ x ≤ s ≤ b , u ( s ) v ( x ) W ( s ) , a ≤ s ≤ x ≤ b G(x,s)=
\begin{cases}
\dfrac{u(x)v(s)}{W(s)},&a\le x\le s\le b,\\[3mm]
\dfrac{u(s)v(x)}{W(s)},&a\le s\le x\le b
\end{cases} G ( x , s ) = ⎩ ⎨ ⎧ W ( s ) u ( x ) v ( s ) , W ( s ) u ( s ) v ( x ) , a ≤ x ≤ s ≤ b , a ≤ s ≤ x ≤ b (3.3.1) \text{(3.3.1)} (3.3.1) で定める関数G : [ a , b ] 2 → R G:[a,b]^2\to\R G : [ a , b ] 2 → R Green 関数 (Green's function )
定理 3.4. a < b a<b a < b P , Q ∈ L 1 ( [ a , b ] ; R ) P,Q\in L^1([a,b];\R) P , Q ∈ L 1 ([ a , b ] ; R ) ( α a , β a ) (\alpha_a,\beta_a) ( α a , β a ) ( α b , β b ) (\alpha_b,\beta_b) ( α b , β b ) ( 0 , 0 ) (0,0) ( 0 , 0 ) 補題 3.2 のu , v u,v u , v W = u v ′ − u ′ v W=uv'-u'v W = u v ′ − u ′ v G G G G G G [ a , b ] 2 [a,b]^2 [ a , b ] 2 x ≠ s x\ne s x = s G ( ⋅ , s ) G(\,\cdot\,,s) G ( ⋅ , s ) x x x x ∈ { a , b } x\in\{a,b\} x ∈ { a , b }
∂ x G ( x , s ) = { u ′ ( x ) v ( s ) W ( s ) , x < s , u ( s ) v ′ ( x ) W ( s ) , s < x \partial_xG(x,s)=
\begin{cases}
\dfrac{u'(x)v(s)}{W(s)},&x<s,\\[3mm]
\dfrac{u(s)v'(x)}{W(s)},&s<x
\end{cases} ∂ x G ( x , s ) = ⎩ ⎨ ⎧ W ( s ) u ′ ( x ) v ( s ) , W ( s ) u ( s ) v ′ ( x ) , x < s , s < x である。任意のs ∈ ( a , b ) s\in(a,b) s ∈ ( a , b )
∂ x G ( s + 0 , s ) − ∂ x G ( s − 0 , s ) = 1 , B a [ G ( ⋅ , s ) ] = B b [ G ( ⋅ , s ) ] = 0 \partial_xG(s+0,s)-\partial_xG(s-0,s)=1,\qquad
B_a[G(\,\cdot\,,s)]=B_b[G(\,\cdot\,,s)]=0 ∂ x G ( s + 0 , s ) − ∂ x G ( s − 0 , s ) = 1 , B a [ G ( ⋅ , s )] = B b [ G ( ⋅ , s )] = 0 が成り立つ。ここで∂ x G ( s ± 0 , s ) \partial_xG(s\pm0,s) ∂ x G ( s ± 0 , s ) x → s ± 0 x\to s\pm0 x → s ± 0 G G G
さらに、任意のf ∈ L 1 ( [ a , b ] ; C ) f\in L^1([a,b];\C) f ∈ L 1 ([ a , b ] ; C ) y y y x ∈ [ a , b ] x\in[a,b] x ∈ [ a , b ]
y ( x ) = ∫ a b G ( x , s ) f ( s ) d s , y ′ ( x ) = ∫ a b ∂ x G ( x , s ) f ( s ) d s y(x)=\int_a^bG(x,s)f(s)\,ds,\qquad
y'(x)=\int_a^b\partial_xG(x,s)f(s)\,ds y ( x ) = ∫ a b G ( x , s ) f ( s ) d s , y ′ ( x ) = ∫ a b ∂ x G ( x , s ) f ( s ) d s が成り立つ。端点x ∈ { a , b } x\in\{a,b\} x ∈ { a , b } y ′ ( x ) y'(x) y ′ ( x ) f f f y y y f f f y y y P , Q , f P,Q,f P , Q , f y ∈ C 2 ( [ a , b ] ; C ) y\in C^2([a,b];\C) y ∈ C 2 ([ a , b ] ; C )
証明. 補題 3.2 によりW W W [ a , b ] [a,b] [ a , b ] u , v , W u,v,W u , v , W G G G u ( x ) v ( s ) / W ( s ) u(x)v(s)/W(s) u ( x ) v ( s ) / W ( s ) u ( s ) v ( x ) / W ( s ) u(s)v(x)/W(s) u ( s ) v ( x ) / W ( s ) [ a , b ] 2 [a,b]^2 [ a , b ] 2 x = s x=s x = s G G G [ a , b ] 2 [a,b]^2 [ a , b ] 2 s ∈ [ a , b ] s\in[a,b] s ∈ [ a , b ] G ( ⋅ , s ) G(\,\cdot\,,s) G ( ⋅ , s ) [ a , s ] [a,s] [ a , s ] u u u v ( s ) / W ( s ) v(s)/W(s) v ( s ) / W ( s ) [ s , b ] [s,b] [ s , b ] v v v u ( s ) / W ( s ) u(s)/W(s) u ( s ) / W ( s ) ∂ x G \partial_xG ∂ x G s ∈ ( a , b ) s\in(a,b) s ∈ ( a , b )
∂ x G ( s + 0 , s ) − ∂ x G ( s − 0 , s ) = u ( s ) v ′ ( s ) − u ′ ( s ) v ( s ) W ( s ) = 1 \partial_xG(s+0,s)-\partial_xG(s-0,s)=\frac{u(s)v'(s)-u'(s)v(s)}{W(s)}=1 ∂ x G ( s + 0 , s ) − ∂ x G ( s − 0 , s ) = W ( s ) u ( s ) v ′ ( s ) − u ′ ( s ) v ( s ) = 1 であり、G ( ⋅ , s ) G(\,\cdot\,,s) G ( ⋅ , s ) a a a u u u b b b v v v
B a [ G ( ⋅ , s ) ] = v ( s ) W ( s ) B a [ u ] = 0 , B b [ G ( ⋅ , s ) ] = u ( s ) W ( s ) B b [ v ] = 0 B_a[G(\,\cdot\,,s)]=\frac{v(s)}{W(s)}B_a[u]=0,\qquad
B_b[G(\,\cdot\,,s)]=\frac{u(s)}{W(s)}B_b[v]=0 B a [ G ( ⋅ , s )] = W ( s ) v ( s ) B a [ u ] = 0 , B b [ G ( ⋅ , s )] = W ( s ) u ( s ) B b [ v ] = 0 である。
f ∈ L 1 ( [ a , b ] ; C ) f\in L^1([a,b];\C) f ∈ L 1 ([ a , b ] ; C ) 1 / W 1/W 1/ W u f / W uf/W u f / W v f / W vf/W v f / W
I 1 ( x ) = ∫ a x u ( s ) f ( s ) W ( s ) d s , I 2 ( x ) = ∫ x b v ( s ) f ( s ) W ( s ) d s , y = v I 1 + u I 2 I_1(x)=\int_a^x\frac{u(s)f(s)}{W(s)}\,ds,\qquad
I_2(x)=\int_x^b\frac{v(s)f(s)}{W(s)}\,ds,\qquad
y=vI_1+uI_2 I 1 ( x ) = ∫ a x W ( s ) u ( s ) f ( s ) d s , I 2 ( x ) = ∫ x b W ( s ) v ( s ) f ( s ) d s , y = v I 1 + u I 2 とおくと、G G G x ∈ [ a , b ] x\in[a,b] x ∈ [ a , b ] y ( x ) = ∫ a b G ( x , s ) f ( s ) d s y(x)=\int_a^bG(x,s)f(s)\,ds y ( x ) = ∫ a b G ( x , s ) f ( s ) d s u , v u,v u , v K = C \mathbb K=\C K = C §E13.9 命題 5.5 により、その解空間の基本解系である。§E13.11 系 5.1 をK = C \mathbb K=\C K = C I = [ a , b ] I=[a,b] I = [ a , b ] y 1 = u y_1=u y 1 = u y 2 = v y_2=v y 2 = v t 0 = a t_0=a t 0 = a R = f R=f R = f
y p ( x ) = − u ( x ) ∫ a x v ( s ) f ( s ) W ( s ) d s + v ( x ) I 1 ( x ) = y ( x ) − I 2 ( a ) u ( x ) y_p(x)=-u(x)\int_a^x\frac{v(s)f(s)}{W(s)}\,ds+v(x)I_1(x)=y(x)-I_2(a)u(x) y p ( x ) = − u ( x ) ∫ a x W ( s ) v ( s ) f ( s ) d s + v ( x ) I 1 ( x ) = y ( x ) − I 2 ( a ) u ( x ) はC 1 C^1 C 1 y p ′ y_p' y p ′ y p ′ ′ + P y p ′ + Q y p = f y_p''+Py_p'+Qy_p=f y p ′′ + P y p ′ + Q y p = f u u u y = y p + I 2 ( a ) u y=y_p+I_2(a)u y = y p + I 2 ( a ) u C 1 C^1 C 1 y ′ y' y ′ y ′ ′ + P y ′ + Q y = f y''+Py'+Qy=f y ′′ + P y ′ + Q y = f
§E13.2 定理 2.5 によりI 1 , I 2 I_1,I_2 I 1 , I 2 §E13.2 系 2.7 (2) によりu , v u,v u , v §E13.2 定理 3.1 (2) によりy y y
y ′ = v ′ I 1 + v u f W + u ′ I 2 − u v f W = v ′ I 1 + u ′ I 2 y'=v'I_1+v\frac{uf}{W}+u'I_2-u\frac{vf}{W}=v'I_1+u'I_2 y ′ = v ′ I 1 + v W u f + u ′ I 2 − u W v f = v ′ I 1 + u ′ I 2 である。右辺は連続であるから、§E13.2 系 2.7 (1) により、端点の片側導関数を含めて全点でy ′ = v ′ I 1 + u ′ I 2 y'=v'I_1+u'I_2 y ′ = v ′ I 1 + u ′ I 2 ∂ x G \partial_xG ∂ x G x ∈ [ a , b ] x\in[a,b] x ∈ [ a , b ] ∫ a b ∂ x G ( x , s ) f ( s ) d s \int_a^b\partial_xG(x,s)f(s)\,ds ∫ a b ∂ x G ( x , s ) f ( s ) d s
I 1 ( a ) = I 2 ( b ) = 0 I_1(a)=I_2(b)=0 I 1 ( a ) = I 2 ( b ) = 0
y ( a ) = I 2 ( a ) u ( a ) , y ′ ( a ) = I 2 ( a ) u ′ ( a ) , y ( b ) = I 1 ( b ) v ( b ) , y ′ ( b ) = I 1 ( b ) v ′ ( b ) \begin{aligned}
y(a)&=I_2(a)u(a),&y'(a)&=I_2(a)u'(a),\\
y(b)&=I_1(b)v(b),&y'(b)&=I_1(b)v'(b)
\end{aligned} y ( a ) y ( b ) = I 2 ( a ) u ( a ) , = I 1 ( b ) v ( b ) , y ′ ( a ) y ′ ( b ) = I 2 ( a ) u ′ ( a ) , = I 1 ( b ) v ′ ( b ) であり、B a [ y ] = I 2 ( a ) B a [ u ] = 0 B_a[y]=I_2(a)B_a[u]=0 B a [ y ] = I 2 ( a ) B a [ u ] = 0 B b [ y ] = I 1 ( b ) B b [ v ] = 0 B_b[y]=I_1(b)B_b[v]=0 B b [ y ] = I 1 ( b ) B b [ v ] = 0 y y y
f f f I 1 , I 2 I_1,I_2 I 1 , I 2 y y y u , v , W u,v,W u , v , W f f f I 1 , I 2 I_1,I_2 I 1 , I 2 y y y P , Q , f P,Q,f P , Q , f §E13.9 系 4.3 をn = 2 n=2 n = 2 K = C \mathbb K=\C K = C I = [ a , b ] I=[a,b] I = [ a , b ] p 1 = P p_1=P p 1 = P p 0 = Q p_0=Q p 0 = Q r = f r=f r = f y ∈ C 2 ( [ a , b ] ; C ) y\in C^2([a,b];\C) y ∈ C 2 ([ a , b ] ; C ) y ′ ′ + P y ′ + Q y = f y''+Py'+Qy=f y ′′ + P y ′ + Q y = f ▨
例 3.5. [ 0 , 1 ] [0,1] [ 0 , 1 ] f ( x ) = 1 f(x)=1 f ( x ) = 1 1 / 2 < x ≤ 1 1/2<x\le1 1/2 < x ≤ 1 f ( x ) = 0 f(x)=0 f ( x ) = 0 0 ≤ x ≤ 1 / 2 0\le x\le1/2 0 ≤ x ≤ 1/2 y ′ ′ = f y''=f y ′′ = f y ( 0 ) = y ( 1 ) = 0 y(0)=y(1)=0 y ( 0 ) = y ( 1 ) = 0 P = Q = 0 P=Q=0 P = Q = 0 ( α 0 , β 0 ) = ( α 1 , β 1 ) = ( 1 , 0 ) (\alpha_0,\beta_0)=(\alpha_1,\beta_1)=(1,0) ( α 0 , β 0 ) = ( α 1 , β 1 ) = ( 1 , 0 ) 補題 3.2 のu , v u,v u , v u ( x ) = − x u(x)=-x u ( x ) = − x v ( x ) = 1 − x v(x)=1-x v ( x ) = 1 − x W = 1 W=1 W = 1 W W W 定理 3.4 により、一意解は
y ( x ) = ∫ 1 / 2 1 G ( x , s ) d s = { − x / 8 , 0 ≤ x ≤ 1 / 2 , − x / 8 + ( x − 1 / 2 ) 2 / 2 , 1 / 2 ≤ x ≤ 1 y(x)=\int_{1/2}^1G(x,s)\,ds
=\begin{cases}
-x/8,&0\leq x\leq1/2,\\[1mm]
-x/8+(x-1/2)^2/2,&1/2\leq x\leq1
\end{cases} y ( x ) = ∫ 1/2 1 G ( x , s ) d s = { − x /8 , − x /8 + ( x − 1/2 ) 2 /2 , 0 ≤ x ≤ 1/2 , 1/2 ≤ x ≤ 1 である。実際、x ≤ 1 / 2 x\leq1/2 x ≤ 1/2 − x ∫ 1 / 2 1 ( 1 − s ) d s = − x / 8 -x\int_{1/2}^1(1-s)\,ds=-x/8 − x ∫ 1/2 1 ( 1 − s ) d s = − x /8 x ≥ 1 / 2 x\geq1/2 x ≥ 1/2 − ( 1 − x ) ∫ 1 / 2 x s d s − x ∫ x 1 ( 1 − s ) d s = − x / 8 + ( x − 1 / 2 ) 2 / 2 -(1-x)\int_{1/2}^x s\,ds-x\int_x^1(1-s)\,ds=-x/8+(x-1/2)^2/2 − ( 1 − x ) ∫ 1/2 x s d s − x ∫ x 1 ( 1 − s ) d s = − x /8 + ( x − 1/2 ) 2 /2 − 1 / 8 -1/8 − 1/8 − 1 / 8 + x − 1 / 2 -1/8+x-1/2 − 1/8 + x − 1/2 y ∈ C 1 y\in C^1 y ∈ C 1 y ′ ∈ A C y'\in AC y ′ ∈ A C y ′ y' y ′ x = 1 / 2 x=1/2 x = 1/2 0 0 0 1 1 1 y y y
4 連続係数の場合 補題 4.1. a < b a<b a < b P , Q ∈ C ( [ a , b ] ; R ) P,Q\in C([a,b];\R) P , Q ∈ C ([ a , b ] ; R )
f ∈ C ( [ a , b ] ; C ) f\in C([a,b];\C) f ∈ C ([ a , b ] ; C ) y ∈ C 1 ( [ a , b ] ; C ) y\in C^1([a,b];\C) y ∈ C 1 ([ a , b ] ; C ) y ′ ∈ A C ( [ a , b ] ; C ) y'\in AC([a,b];\C) y ′ ∈ A C ([ a , b ] ; C ) y ′ ′ + P y ′ + Q y = f y''+Py'+Qy=f y ′′ + P y ′ + Q y = f y ∈ C 2 ( [ a , b ] ; C ) y\in C^2([a,b];\C) y ∈ C 2 ([ a , b ] ; C ) y ′ ′ + P y ′ + Q y = f y''+Py'+Qy=f y ′′ + P y ′ + Q y = f [ a , b ] [a,b] [ a , b ] 実数の組( α a , β a ) (\alpha_a,\beta_a) ( α a , β a ) ( α b , β b ) (\alpha_b,\beta_b) ( α b , β b ) ( 0 , 0 ) (0,0) ( 0 , 0 ) 式 (1.2.1) を満たすy 1 , y 2 ∈ C 2 ( [ a , b ] ; R ) y_1,y_2\in C^2([a,b];\R) y 1 , y 2 ∈ C 2 ([ a , b ] ; R ) y 1 + i y 2 y_1+iy_2 y 1 + i y 2 式 (1.2.1) を満たすy ∈ C 2 ( [ a , b ] ; R ) y\in C^2([a,b];\R) y ∈ C 2 ([ a , b ] ; R )
証明. 主張 (1) を示す。§E13.9 系 4.3 をn = 2 n=2 n = 2 K = C \mathbb K=\C K = C I = [ a , b ] I=[a,b] I = [ a , b ] p 1 = P p_1=P p 1 = P p 0 = Q p_0=Q p 0 = Q r = f r=f r = f y ∈ C 2 ( [ a , b ] ; C ) y\in C^2([a,b];\C) y ∈ C 2 ([ a , b ] ; C ) y ′ ′ + P y ′ + Q y = f y''+Py'+Qy=f y ′′ + P y ′ + Q y = f
主張 (2) を示す。y y y ℜ y \Re y ℜ y ℑ y \Im y ℑ y C 1 ( [ a , b ] ; R ) C^1([a,b];\R) C 1 ([ a , b ] ; R ) P , Q P,Q P , Q ℜ y \Re y ℜ y ℑ y \Im y ℑ y B a [ ℜ y ] = B b [ ℜ y ] = 0 B_a[\Re y]=B_b[\Re y]=0 B a [ ℜ y ] = B b [ ℜ y ] = 0 B a [ ℑ y ] = B b [ ℑ y ] = 0 B_a[\Im y]=B_b[\Im y]=0 B a [ ℑ y ] = B b [ ℑ y ] = 0 主張 (1) をf = 0 f=0 f = 0 式 (1.2.1) を満たすC 2 ( [ a , b ] ; R ) C^2([a,b];\R) C 2 ([ a , b ] ; R ) 式 (1.2.1) を満たすy 1 , y 2 ∈ C 2 ( [ a , b ] ; R ) y_1,y_2\in C^2([a,b];\R) y 1 , y 2 ∈ C 2 ([ a , b ] ; R ) §E13.9 系 4.3 をn = 2 n=2 n = 2 K = R \mathbb K=\R K = R I = [ a , b ] I=[a,b] I = [ a , b ] p 1 = P p_1=P p 1 = P p 0 = Q p_0=Q p 0 = Q r = 0 r=0 r = 0 y 1 ′ , y 2 ′ y_1',y_2' y 1 ′ , y 2 ′ y 1 , y 2 y_1,y_2 y 1 , y 2 y 1 + i y 2 y_1+iy_2 y 1 + i y 2 ▨
補題 4.2. a < b a<b a < b P , Q ∈ C ( [ a , b ] ; R ) P,Q\in C([a,b];\R) P , Q ∈ C ([ a , b ] ; R ) ( α a , β a ) (\alpha_a,\beta_a) ( α a , β a ) ( α b , β b ) (\alpha_b,\beta_b) ( α b , β b ) ( 0 , 0 ) (0,0) ( 0 , 0 ) L [ w ] = w ′ ′ + P w ′ + Q w L[w]=w''+Pw'+Qw L [ w ] = w ′′ + P w ′ + Qw [ a , b ] [a,b] [ a , b ]
L [ u ] = 0 , u ( a ) = β a , u ′ ( a ) = − α a L[u]=0,\qquad
u(a)=\beta_a,\qquad u'(a)=-\alpha_a L [ u ] = 0 , u ( a ) = β a , u ′ ( a ) = − α a を満たすu ∈ C 2 ( [ a , b ] ; R ) u\in C^2([a,b];\R) u ∈ C 2 ([ a , b ] ; R ) [ a , b ] [a,b] [ a , b ]
L [ v ] = 0 , v ( b ) = β b , v ′ ( b ) = − α b L[v]=0,\qquad
v(b)=\beta_b,\qquad v'(b)=-\alpha_b L [ v ] = 0 , v ( b ) = β b , v ′ ( b ) = − α b を満たすv ∈ C 2 ( [ a , b ] ; R ) v\in C^2([a,b];\R) v ∈ C 2 ([ a , b ] ; R ) u , v u,v u , v 補題 3.2 のu , v u,v u , v B a [ u ] = B b [ v ] = 0 B_a[u]=B_b[v]=0 B a [ u ] = B b [ v ] = 0 式 (1.2.1) が自明解だけをもつならば、
W ( x ) = u ( x ) v ′ ( x ) − u ′ ( x ) v ( x ) W(x)=u(x)v'(x)-u'(x)v(x) W ( x ) = u ( x ) v ′ ( x ) − u ′ ( x ) v ( x ) は[ a , b ] [a,b] [ a , b ]
証明. §E13.9 系 4.3 をn = 2 n=2 n = 2 K = R \mathbb K=\R K = R I = [ a , b ] I=[a,b] I = [ a , b ] p 1 = P p_1=P p 1 = P p 0 = Q p_0=Q p 0 = Q r = 0 r=0 r = 0 L [ w ] = 0 L[w]=0 L [ w ] = 0 w ∈ C 2 ( [ a , b ] ; R ) w\in C^2([a,b];\R) w ∈ C 2 ([ a , b ] ; R ) w ∈ C 1 ( [ a , b ] ; R ) w\in C^1([a,b];\R) w ∈ C 1 ([ a , b ] ; R ) w ′ ∈ A C ( [ a , b ] ; R ) w'\in AC([a,b];\R) w ′ ∈ A C ([ a , b ] ; R ) L [ w ] = 0 L[w]=0 L [ w ] = 0 w w w 補題 3.2 により、この補題のu , v u,v u , v u , v u,v u , v B a [ u ] = B b [ v ] = 0 B_a[u]=B_b[v]=0 B a [ u ] = B b [ v ] = 0 式 (1.2.1) が自明解だけをもつならば、補題 4.1 (2) により境界核は自明であり、補題 3.2 によりW W W ▨
系 4.3. a < b a<b a < b P , Q ∈ C ( [ a , b ] ; R ) P,Q\in C([a,b];\R) P , Q ∈ C ([ a , b ] ; R ) ( α a , β a ) (\alpha_a,\beta_a) ( α a , β a ) ( α b , β b ) (\alpha_b,\beta_b) ( α b , β b ) ( 0 , 0 ) (0,0) ( 0 , 0 ) 式 (1.2.1) が自明解だけをもつならば、この分離型二点境界値問題の境界核は自明であり、任意の連続関数f : [ a , b ] → R f:[a,b]\to\R f : [ a , b ] → R 式 (1.1.1) 、式 (1.1.2) は一意なC 2 ( [ a , b ] ; R ) C^2([a,b];\R) C 2 ([ a , b ] ; R ) 式 (3.3.1) の Green 関数を用いて
y ( x ) = ∫ a b G ( x , s ) f ( s ) d s y(x)=\int_a^bG(x,s)f(s)\,ds y ( x ) = ∫ a b G ( x , s ) f ( s ) d s で与えられる。
証明. 補題 4.1 (2) により境界核は自明である。連続関数f f f 定理 3.4 により、y ( x ) = ∫ a b G ( x , s ) f ( s ) d s y(x)=\int_a^bG(x,s)f(s)\,ds y ( x ) = ∫ a b G ( x , s ) f ( s ) d s C 2 C^2 C 2 式 (1.1.1) 、式 (1.1.2) を満たすz ∈ C 2 ( [ a , b ] ; R ) z\in C^2([a,b];\R) z ∈ C 2 ([ a , b ] ; R ) §E13.9 系 4.3 をn = 2 n=2 n = 2 K = R \mathbb K=\R K = R I = [ a , b ] I=[a,b] I = [ a , b ] p 1 = P p_1=P p 1 = P p 0 = Q p_0=Q p 0 = Q r = f r=f r = f z ′ z' z ′ z z z z z z z = y z=y z = y ▨
5 演習 問題 5.1 (Green 関数の構成と検算). 境界値問題
y ′ ′ = 1 , y ( 0 ) = y ( 1 ) = 0 y''=1,\qquad y(0)=y(1)=0 y ′′ = 1 , y ( 0 ) = y ( 1 ) = 0 について、Green 関数を構成し、積分表示から解を求めよ。また、境界条件と微分方程式を直接に確かめよ。
解答. 左端境界条件を満たす斉次方程式の解としてu ( x ) = − x u(x)=-x u ( x ) = − x v ( x ) = 1 − x v(x)=1-x v ( x ) = 1 − x
W ( u , v ) = u v ′ − u ′ v = 1 W(u,v)=uv'-u'v=1 W ( u , v ) = u v ′ − u ′ v = 1 であるから、Green 関数は
G ( x , s ) = { − x ( 1 − s ) , x ≤ s , − s ( 1 − x ) , s ≤ x G(x,s)=
\begin{cases}
-x(1-s),&x\leq s,\\
-s(1-x),&s\leq x
\end{cases} G ( x , s ) = { − x ( 1 − s ) , − s ( 1 − x ) , x ≤ s , s ≤ x となる。したがって、
y ( x ) = ∫ 0 1 G ( x , s ) d s = − ∫ 0 x s ( 1 − x ) d s − ∫ x 1 x ( 1 − s ) d s = x ( x − 1 ) 2 . \begin{aligned}
y(x)
&=\int_0^1G(x,s)\,ds\\
&=-\int_0^x s(1-x)\,ds-\int_x^1x(1-s)\,ds
=\frac{x(x-1)}2.
\end{aligned} y ( x ) = ∫ 0 1 G ( x , s ) d s = − ∫ 0 x s ( 1 − x ) d s − ∫ x 1 x ( 1 − s ) d s = 2 x ( x − 1 ) . 実際にy ( 0 ) = y ( 1 ) = 0 y(0)=y(1)=0 y ( 0 ) = y ( 1 ) = 0 y ′ ′ = 1 y''=1 y ′′ = 1 A + B x A+Bx A + B x 0 0 0 A = B = 0 A=B=0 A = B = 0 ▨
問題 5.2. Q : [ 0 , 2 ] → R Q:[0,2]\to\R Q : [ 0 , 2 ] → R 0 ≤ x ≤ 1 0\le x\le1 0 ≤ x ≤ 1 Q ( x ) = 0 Q(x)=0 Q ( x ) = 0 1 < x ≤ 2 1<x\le2 1 < x ≤ 2 Q ( x ) = − 1 Q(x)=-1 Q ( x ) = − 1 f ∈ L 1 ( [ 0 , 2 ] ; C ) f\in L^1([0,2];\C) f ∈ L 1 ([ 0 , 2 ] ; C )
y ′ ′ + Q y = f ほとんど至る所 , y ( 0 ) = y ( 2 ) = 0 y''+Qy=f\quad\text{ほとんど至る所},\qquad y(0)=y(2)=0 y ′′ + Q y = f ほとんど至る所 , y ( 0 ) = y ( 2 ) = 0 について、補題 3.2 のu , v , W u,v,W u , v , W G G G [ 0 , 2 ] 2 [0,2]^2 [ 0 , 2 ] 2 G ≤ 0 G\le0 G ≤ 0 u u u v v v x = 1 x=1 x = 1
解答. この問題はP = 0 P=0 P = 0 ( α 0 , β 0 ) = ( α 2 , β 2 ) = ( 1 , 0 ) (\alpha_0,\beta_0)=(\alpha_2,\beta_2)=(1,0) ( α 0 , β 0 ) = ( α 2 , β 2 ) = ( 1 , 0 )
u ( x ) = { − x , 0 ≤ x ≤ 1 , − e x − 1 , 1 ≤ x ≤ 2 , v ( x ) = { sinh 1 + ( 1 − x ) cosh 1 , 0 ≤ x ≤ 1 , sinh ( 2 − x ) , 1 ≤ x ≤ 2 u(x)=
\begin{cases}
-x,&0\le x\le1,\\
-e^{x-1},&1\le x\le2,
\end{cases}
\qquad
v(x)=
\begin{cases}
\sinh1+(1-x)\cosh1,&0\le x\le1,\\
\sinh(2-x),&1\le x\le2
\end{cases} u ( x ) = { − x , − e x − 1 , 0 ≤ x ≤ 1 , 1 ≤ x ≤ 2 , v ( x ) = { sinh 1 + ( 1 − x ) cosh 1 , sinh ( 2 − x ) , 0 ≤ x ≤ 1 , 1 ≤ x ≤ 2 とおく。x = 1 x=1 x = 1 u ( 1 ) = u ′ ( 1 ) = − 1 u(1)=u'(1)=-1 u ( 1 ) = u ′ ( 1 ) = − 1 v ( 1 ) = sinh 1 v(1)=\sinh1 v ( 1 ) = sinh 1 v ′ ( 1 ) = − cosh 1 v'(1)=-\cosh1 v ′ ( 1 ) = − cosh 1 u , v ∈ C 1 ( [ 0 , 2 ] ; R ) u,v\in C^1([0,2];\R) u , v ∈ C 1 ([ 0 , 2 ] ; R ) g , h : [ 0 , 2 ] → R g,h:[0,2]\to\R g , h : [ 0 , 2 ] → R 0 ≤ x ≤ 1 0\le x\le1 0 ≤ x ≤ 1 g ( x ) = h ( x ) = 0 g(x)=h(x)=0 g ( x ) = h ( x ) = 0 1 < x ≤ 2 1<x\le2 1 < x ≤ 2 g ( x ) = − e x − 1 g(x)=-e^{x-1} g ( x ) = − e x − 1 h ( x ) = sinh ( 2 − x ) h(x)=\sinh(2-x) h ( x ) = sinh ( 2 − x ) u ′ , v ′ u',v' u ′ , v ′ [ 0 , 1 ] [0,1] [ 0 , 1 ] [ 1 , 2 ] [1,2] [ 1 , 2 ] §E13.2 系 2.7 (2) を各区間に適用すると、すべてのx ∈ [ 0 , 2 ] x\in[0,2] x ∈ [ 0 , 2 ]
u ′ ( x ) = u ′ ( 0 ) + ∫ 0 x g ( s ) d s , v ′ ( x ) = v ′ ( 0 ) + ∫ 0 x h ( s ) d s u'(x)=u'(0)+\int_0^xg(s)\,ds,\qquad
v'(x)=v'(0)+\int_0^xh(s)\,ds u ′ ( x ) = u ′ ( 0 ) + ∫ 0 x g ( s ) d s , v ′ ( x ) = v ′ ( 0 ) + ∫ 0 x h ( s ) d s が成り立つ。§E13.2 定理 2.5 によりu ′ , v ′ u',v' u ′ , v ′ 0 < x < 1 0<x<1 0 < x < 1 u ′ ′ + Q u = 0 u''+Qu=0 u ′′ + Q u = 0 v ′ ′ + Q v = 0 v''+Qv=0 v ′′ + Q v = 0 1 < x < 2 1<x<2 1 < x < 2 u ′ ′ + Q u = − e x − 1 + e x − 1 = 0 u''+Qu=-e^{x-1}+e^{x-1}=0 u ′′ + Q u = − e x − 1 + e x − 1 = 0 v ′ ′ + Q v = sinh ( 2 − x ) − sinh ( 2 − x ) = 0 v''+Qv=\sinh(2-x)-\sinh(2-x)=0 v ′′ + Q v = sinh ( 2 − x ) − sinh ( 2 − x ) = 0 u ( 0 ) = 0 u(0)=0 u ( 0 ) = 0 u ′ ( 0 ) = − 1 u'(0)=-1 u ′ ( 0 ) = − 1 v ( 2 ) = 0 v(2)=0 v ( 2 ) = 0 v ′ ( 2 ) = − 1 v'(2)=-1 v ′ ( 2 ) = − 1 補題 3.2 の一意性により、u , v u,v u , v u , v u,v u , v
0 ≤ x ≤ 1 0\le x\le1 0 ≤ x ≤ 1
W = u v ′ − u ′ v = x cosh 1 + sinh 1 + ( 1 − x ) cosh 1 = sinh 1 + cosh 1 = e W=uv'-u'v=x\cosh1+\sinh1+(1-x)\cosh1=\sinh1+\cosh1=e W = u v ′ − u ′ v = x cosh 1 + sinh 1 + ( 1 − x ) cosh 1 = sinh 1 + cosh 1 = e であり、1 ≤ x ≤ 2 1\le x\le2 1 ≤ x ≤ 2
W = e x − 1 cosh ( 2 − x ) + e x − 1 sinh ( 2 − x ) = e x − 1 e 2 − x = e W=e^{x-1}\cosh(2-x)+e^{x-1}\sinh(2-x)=e^{x-1}e^{2-x}=e W = e x − 1 cosh ( 2 − x ) + e x − 1 sinh ( 2 − x ) = e x − 1 e 2 − x = e である。W W W 補題 3.2 により境界核は自明であり、Green 関数は
G ( x , s ) = { u ( x ) v ( s ) / e , 0 ≤ x ≤ s ≤ 2 , u ( s ) v ( x ) / e , 0 ≤ s ≤ x ≤ 2 G(x,s)=
\begin{cases}
u(x)v(s)/e,&0\le x\le s\le2,\\
u(s)v(x)/e,&0\le s\le x\le2
\end{cases} G ( x , s ) = { u ( x ) v ( s ) / e , u ( s ) v ( x ) / e , 0 ≤ x ≤ s ≤ 2 , 0 ≤ s ≤ x ≤ 2 である。[ 0 , 2 ] [0,2] [ 0 , 2 ] u ≤ 0 u\le0 u ≤ 0 0 ≤ x ≤ 1 0\le x\le1 0 ≤ x ≤ 1 v ( x ) ≥ sinh 1 > 0 v(x)\ge\sinh1>0 v ( x ) ≥ sinh 1 > 0 1 ≤ x ≤ 2 1\le x\le2 1 ≤ x ≤ 2 v ( x ) = sinh ( 2 − x ) ≥ 0 v(x)=\sinh(2-x)\ge0 v ( x ) = sinh ( 2 − x ) ≥ 0 v ≥ 0 v\ge0 v ≥ 0 [ 0 , 2 ] 2 [0,2]^2 [ 0 , 2 ] 2 G ≤ 0 G\le0 G ≤ 0
x = 1 x=1 x = 1 u ′ u' u ′ 0 0 0 − e 0 = − 1 -e^0=-1 − e 0 = − 1 v ′ v' v ′ 0 0 0 sinh 1 ≠ 0 \sinh1\ne0 sinh 1 = 0 u ′ u' u ′ v ′ v' v ′ x = 1 x=1 x = 1 u u u v v v x = 1 x=1 x = 1 ▨
問題 5.3 (固有値における可解性). 区間[ 0 , π ] [0,\pi] [ 0 , π ]
y ′ ′ + y = sin x , y ( 0 ) = y ( π ) = 0 y''+y=\sin x,\qquad y(0)=y(\pi)=0 y ′′ + y = sin x , y ( 0 ) = y ( π ) = 0 が解をもたないことを示せ。また、右辺を零に置き換えた斉次境界値問題のすべての解を求めよ。
解答. 非同次方程式の一般解は
y ( x ) = A cos x + B sin x − x cos x 2 y(x)=A\cos x+B\sin x-\frac{x\cos x}{2} y ( x ) = A cos x + B sin x − 2 x cos x である。実際、− x cos x / 2 -x\cos x/2 − x cos x /2 sin x \sin x sin x A = 0 A=0 A = 0
y ( π ) = π 2 ≠ 0 y(\pi)=\frac{\pi}{2}\ne0 y ( π ) = 2 π = 0 である。したがって、二つの境界条件を同時に満たす解は存在しない。
斉次方程式y ′ ′ + y = 0 y''+y=0 y ′′ + y = 0 y = A cos x + B sin x y=A\cos x+B\sin x y = A cos x + B sin x A = 0 A=0 A = 0 B sin π = 0 B\sin\pi=0 B sin π = 0 B ∈ R B\in\R B ∈ R
y ( x ) = B sin x , B ∈ R y(x)=B\sin x,\qquad B\in\R y ( x ) = B sin x , B ∈ R である。特に、境界核は自明でない。▨
問題 5.4 (Green 関数の符号による比較). [ 0 , 1 ] [0,1] [ 0 , 1 ] y ′ ′ − y = f y''-y=f y ′′ − y = f y ( 0 ) = y ( 1 ) = 0 y(0)=y(1)=0 y ( 0 ) = y ( 1 ) = 0 f ∈ C ( [ 0 , 1 ] ; R ) f\in C([0,1];\R) f ∈ C ([ 0 , 1 ] ; R ) 補題 4.2 のu , v , W u,v,W u , v , W 式 (3.3.1) の Green 関数G G G [ 0 , 1 ] 2 [0,1]^{2} [ 0 , 1 ] 2 G ≤ 0 G\le0 G ≤ 0 [ 0 , 1 ] [0,1] [ 0 , 1 ] f ≥ 0 f\ge0 f ≥ 0 y ≤ 0 y\le0 y ≤ 0 f ≡ 1 f\equiv1 f ≡ 1
解答. Dirichlet 条件は境界係数( α 0 , β 0 ) = ( α 1 , β 1 ) = ( 1 , 0 ) (\alpha_{0},\beta_{0})=(\alpha_{1},\beta_{1})=(1,0) ( α 0 , β 0 ) = ( α 1 , β 1 ) = ( 1 , 0 ) u u u u ′ ′ − u = 0 u''-u=0 u ′′ − u = 0 u ( 0 ) = 0 u(0)=0 u ( 0 ) = 0 u ′ ( 0 ) = − 1 u'(0)=-1 u ′ ( 0 ) = − 1 u ( x ) = − sinh x u(x)=-\sinh x u ( x ) = − sinh x v v v v ′ ′ − v = 0 v''-v=0 v ′′ − v = 0 v ( 1 ) = 0 v(1)=0 v ( 1 ) = 0 v ′ ( 1 ) = − 1 v'(1)=-1 v ′ ( 1 ) = − 1 v ( x ) = sinh ( 1 − x ) v(x)=\sinh(1-x) v ( x ) = sinh ( 1 − x )
W = u v ′ − u ′ v = sinh x cosh ( 1 − x ) + cosh x sinh ( 1 − x ) = sinh 1 W=uv'-u'v=\sinh x\cosh(1-x)+\cosh x\sinh(1-x)=\sinh1 W = u v ′ − u ′ v = sinh x cosh ( 1 − x ) + cosh x sinh ( 1 − x ) = sinh 1 である。斉次問題の解A cosh x + B sinh x A\cosh x+B\sinh x A cosh x + B sinh x y ( 0 ) = 0 y(0)=0 y ( 0 ) = 0 A = 0 A=0 A = 0 y ( 1 ) = 0 y(1)=0 y ( 1 ) = 0 B sinh 1 = 0 B\sinh1=0 B sinh 1 = 0 B = 0 B=0 B = 0
G ( x , s ) = { − sinh x sinh ( 1 − s ) sinh 1 , 0 ≤ x ≤ s ≤ 1 , − sinh s sinh ( 1 − x ) sinh 1 , 0 ≤ s ≤ x ≤ 1 G(x,s)=
\begin{cases}
-\dfrac{\sinh x\,\sinh(1-s)}{\sinh1},&0\le x\le s\le1,\\[3mm]
-\dfrac{\sinh s\,\sinh(1-x)}{\sinh1},&0\le s\le x\le1
\end{cases} G ( x , s ) = ⎩ ⎨ ⎧ − sinh 1 sinh x sinh ( 1 − s ) , − sinh 1 sinh s sinh ( 1 − x ) , 0 ≤ x ≤ s ≤ 1 , 0 ≤ s ≤ x ≤ 1 である。sinh \sinh sinh [ 0 , 1 ] [0,1] [ 0 , 1 ] 0 0 0 G ≤ 0 G\le0 G ≤ 0
系 4.3 により解はただ一つ存在してy ( x ) = ∫ 0 1 G ( x , s ) f ( s ) d s y(x)=\int_{0}^{1}G(x,s)f(s)\,ds y ( x ) = ∫ 0 1 G ( x , s ) f ( s ) d s f ≥ 0 f\ge0 f ≥ 0 0 0 0 y ( x ) ≤ 0 y(x)\le0 y ( x ) ≤ 0
f ≡ 1 f\equiv1 f ≡ 1 ∫ 0 x sinh s d s = cosh x − 1 \int_{0}^{x}\sinh s\,ds=\cosh x-1 ∫ 0 x sinh s d s = cosh x − 1 ∫ x 1 sinh ( 1 − s ) d s = cosh ( 1 − x ) − 1 \int_{x}^{1}\sinh(1-s)\,ds=\cosh(1-x)-1 ∫ x 1 sinh ( 1 − s ) d s = cosh ( 1 − x ) − 1
y ( x ) = − sinh ( 1 − x ) ( cosh x − 1 ) + sinh x ( cosh ( 1 − x ) − 1 ) sinh 1 = − 1 + sinh x + sinh ( 1 − x ) sinh 1 y(x)=-\frac{\sinh(1-x)(\cosh x-1)+\sinh x(\cosh(1-x)-1)}{\sinh1}=-1+\frac{\sinh x+\sinh(1-x)}{\sinh1} y ( x ) = − sinh 1 sinh ( 1 − x ) ( cosh x − 1 ) + sinh x ( cosh ( 1 − x ) − 1 ) = − 1 + sinh 1 sinh x + sinh ( 1 − x ) である。ここで加法定理sinh ( 1 − x ) cosh x + sinh x cosh ( 1 − x ) = sinh 1 \sinh(1-x)\cosh x+\sinh x\cosh(1-x)=\sinh1 sinh ( 1 − x ) cosh x + sinh x cosh ( 1 − x ) = sinh 1 y ′ ′ − y = 1 y''-y=1 y ′′ − y = 1 y ( 0 ) = y ( 1 ) = 0 y(0)=y(1)=0 y ( 0 ) = y ( 1 ) = 0 ▨
問題 5.5. f ∈ L 1 ( [ 0 , 1 ] ; C ) f\in L^1([0,1];\C) f ∈ L 1 ([ 0 , 1 ] ; C ) ∥ f ∥ 1 = ∫ 0 1 ∣ f ( s ) ∣ d s \|f\|_1=\int_0^1|f(s)|\,ds ∥ f ∥ 1 = ∫ 0 1 ∣ f ( s ) ∣ d s
y ′ ′ = f ほとんど至る所 , y ( 0 ) = 0 , y ′ ( 1 ) + y ( 1 ) = 0 y''=f\quad\text{ほとんど至る所},\qquad y(0)=0,\qquad y'(1)+y(1)=0 y ′′ = f ほとんど至る所 , y ( 0 ) = 0 , y ′ ( 1 ) + y ( 1 ) = 0 について、補題 3.2 のu , v , W u,v,W u , v , W G G G y y y x ∈ [ 0 , 1 ] x\in[0,1] x ∈ [ 0 , 1 ]
∣ y ( x ) ∣ ≤ 1 2 ∥ f ∥ 1 , ∣ y ′ ( x ) ∣ ≤ ∥ f ∥ 1 |y(x)|\le\frac12\|f\|_1,\qquad |y'(x)|\le\|f\|_1 ∣ y ( x ) ∣ ≤ 2 1 ∥ f ∥ 1 , ∣ y ′ ( x ) ∣ ≤ ∥ f ∥ 1 を満たすことを示せ。
解答. この問題はP = Q = 0 P=Q=0 P = Q = 0 ( α 0 , β 0 ) = ( 1 , 0 ) (\alpha_0,\beta_0)=(1,0) ( α 0 , β 0 ) = ( 1 , 0 ) ( α 1 , β 1 ) = ( 1 , 1 ) (\alpha_1,\beta_1)=(1,1) ( α 1 , β 1 ) = ( 1 , 1 ) u u u u ′ ′ = 0 u''=0 u ′′ = 0 u ( 0 ) = 0 u(0)=0 u ( 0 ) = 0 u ′ ( 0 ) = − 1 u'(0)=-1 u ′ ( 0 ) = − 1 u ( x ) = − x u(x)=-x u ( x ) = − x v v v v ′ ′ = 0 v''=0 v ′′ = 0 v ( 1 ) = 1 v(1)=1 v ( 1 ) = 1 v ′ ( 1 ) = − 1 v'(1)=-1 v ′ ( 1 ) = − 1 v ( x ) = 2 − x v(x)=2-x v ( x ) = 2 − x
W = u v ′ − u ′ v = x + ( 2 − x ) = 2 W=uv'-u'v=x+(2-x)=2 W = u v ′ − u ′ v = x + ( 2 − x ) = 2 は零でないので、補題 3.2 により境界核は自明であり、Green 関数は
G ( x , s ) = { − x ( 2 − s ) 2 , 0 ≤ x ≤ s ≤ 1 , − s ( 2 − x ) 2 , 0 ≤ s ≤ x ≤ 1 G(x,s)=
\begin{cases}
-\dfrac{x(2-s)}2,&0\le x\le s\le1,\\[3mm]
-\dfrac{s(2-x)}2,&0\le s\le x\le1
\end{cases} G ( x , s ) = ⎩ ⎨ ⎧ − 2 x ( 2 − s ) , − 2 s ( 2 − x ) , 0 ≤ x ≤ s ≤ 1 , 0 ≤ s ≤ x ≤ 1 である。定理 3.4 により解y y y x ∈ [ 0 , 1 ] x\in[0,1] x ∈ [ 0 , 1 ]
y ( x ) = ∫ 0 1 G ( x , s ) f ( s ) d s , y ′ ( x ) = ∫ 0 1 ∂ x G ( x , s ) f ( s ) d s , ∂ x G ( x , s ) = { − 2 − s 2 , x < s , s 2 , s < x y(x)=\int_0^1G(x,s)f(s)\,ds,\qquad
y'(x)=\int_0^1\partial_xG(x,s)f(s)\,ds,\qquad
\partial_xG(x,s)=
\begin{cases}
-\dfrac{2-s}2,&x<s,\\[3mm]
\dfrac s2,&s<x
\end{cases} y ( x ) = ∫ 0 1 G ( x , s ) f ( s ) d s , y ′ ( x ) = ∫ 0 1 ∂ x G ( x , s ) f ( s ) d s , ∂ x G ( x , s ) = ⎩ ⎨ ⎧ − 2 2 − s , 2 s , x < s , s < x が成り立つ。0 ≤ x ≤ s ≤ 1 0\le x\le s\le1 0 ≤ x ≤ s ≤ 1 0 ≤ x ( 2 − s ) ≤ s ( 2 − s ) ≤ 1 0\le x(2-s)\le s(2-s)\le1 0 ≤ x ( 2 − s ) ≤ s ( 2 − s ) ≤ 1 0 ≤ s ≤ x ≤ 1 0\le s\le x\le1 0 ≤ s ≤ x ≤ 1 0 ≤ s ( 2 − x ) ≤ x ( 2 − x ) ≤ 1 0\le s(2-x)\le x(2-x)\le1 0 ≤ s ( 2 − x ) ≤ x ( 2 − x ) ≤ 1 ∣ G ∣ ≤ 1 / 2 |G|\le1/2 ∣ G ∣ ≤ 1/2 s ∈ [ 0 , 1 ] s\in[0,1] s ∈ [ 0 , 1 ] 1 / 2 ≤ ( 2 − s ) / 2 ≤ 1 1/2\le(2-s)/2\le1 1/2 ≤ ( 2 − s ) /2 ≤ 1 0 ≤ s / 2 ≤ 1 / 2 0\le s/2\le1/2 0 ≤ s /2 ≤ 1/2 s ≠ x s\ne x s = x ∣ ∂ x G ( x , s ) ∣ ≤ 1 |\partial_xG(x,s)|\le1 ∣ ∂ x G ( x , s ) ∣ ≤ 1
∣ y ( x ) ∣ ≤ ∫ 0 1 ∣ G ( x , s ) ∣ ∣ f ( s ) ∣ d s ≤ 1 2 ∥ f ∥ 1 , ∣ y ′ ( x ) ∣ ≤ ∫ 0 1 ∣ ∂ x G ( x , s ) ∣ ∣ f ( s ) ∣ d s ≤ ∥ f ∥ 1 |y(x)|\le\int_0^1|G(x,s)||f(s)|\,ds\le\frac12\|f\|_1,\qquad
|y'(x)|\le\int_0^1|\partial_xG(x,s)||f(s)|\,ds\le\|f\|_1 ∣ y ( x ) ∣ ≤ ∫ 0 1 ∣ G ( x , s ) ∣∣ f ( s ) ∣ d s ≤ 2 1 ∥ f ∥ 1 , ∣ y ′ ( x ) ∣ ≤ ∫ 0 1 ∣ ∂ x G ( x , s ) ∣∣ f ( s ) ∣ d s ≤ ∥ f ∥ 1 である。▨
問題 5.6. f ∈ L 1 ( [ 0 , 1 ] ; C ) f\in L^1([0,1];\C) f ∈ L 1 ([ 0 , 1 ] ; C )
y ′ ′ = f ほとんど至る所 , y ′ ( 0 ) = y ′ ( 1 ) = 0 y''=f\quad\text{ほとんど至る所},\qquad y'(0)=y'(1)=0 y ′′ = f ほとんど至る所 , y ′ ( 0 ) = y ′ ( 1 ) = 0 について、補題 3.2 のu , v u,v u , v W W W ∫ 0 1 f ( s ) d s = 0 \int_0^1f(s)\,ds=0 ∫ 0 1 f ( s ) d s = 0
解答. この問題はP = Q = 0 P=Q=0 P = Q = 0 ( α 0 , β 0 ) = ( α 1 , β 1 ) = ( 0 , 1 ) (\alpha_0,\beta_0)=(\alpha_1,\beta_1)=(0,1) ( α 0 , β 0 ) = ( α 1 , β 1 ) = ( 0 , 1 ) u u u u ′ ′ = 0 u''=0 u ′′ = 0 u ( 0 ) = 1 u(0)=1 u ( 0 ) = 1 u ′ ( 0 ) = 0 u'(0)=0 u ′ ( 0 ) = 0 v v v v ′ ′ = 0 v''=0 v ′′ = 0 v ( 1 ) = 1 v(1)=1 v ( 1 ) = 1 v ′ ( 1 ) = 0 v'(1)=0 v ′ ( 1 ) = 0 u = v = 1 u=v=1 u = v = 1 W = u v ′ − u ′ v = 0 W=uv'-u'v=0 W = u v ′ − u ′ v = 0 補題 3.2 により境界核は自明でなく、実際に定数関数1 1 1
y y y y ′ y' y ′ y ′ ′ = f y''=f y ′′ = f §E13.2 定理 2.5 により
0 = y ′ ( 1 ) − y ′ ( 0 ) = ∫ 0 1 f ( s ) d s 0=y'(1)-y'(0)=\int_0^1f(s)\,ds 0 = y ′ ( 1 ) − y ′ ( 0 ) = ∫ 0 1 f ( s ) d s である。逆に∫ 0 1 f ( s ) d s = 0 \int_0^1f(s)\,ds=0 ∫ 0 1 f ( s ) d s = 0 §E13.9 命題 4.1 により、y 0 ′ ′ = f y_0''=f y 0 ′′ = f y 0 ( 0 ) = y 0 ′ ( 0 ) = 0 y_0(0)=y_0'(0)=0 y 0 ( 0 ) = y 0 ′ ( 0 ) = 0 y 0 ∈ C 1 ( [ 0 , 1 ] ; C ) y_0\in C^1([0,1];\C) y 0 ∈ C 1 ([ 0 , 1 ] ; C ) y 0 ′ y_0' y 0 ′ §E13.2 定理 2.5 によりy 0 ′ ( 1 ) = y 0 ′ ( 0 ) + ∫ 0 1 f ( s ) d s = 0 y_0'(1)=y_0'(0)+\int_0^1f(s)\,ds=0 y 0 ′ ( 1 ) = y 0 ′ ( 0 ) + ∫ 0 1 f ( s ) d s = 0 y 0 y_0 y 0
解y y y w = y − y 0 w=y-y_0 w = y − y 0 w ′ w' w ′ w ′ ′ = 0 w''=0 w ′′ = 0 §E13.2 定理 2.5 によりw ′ w' w ′ w ′ ( 0 ) = 0 w'(0)=0 w ′ ( 0 ) = 0 §E13.2 系 2.7 (2) によりw ( x ) = w ( 0 ) + ∫ 0 x w ′ ( s ) d s = w ( 0 ) w(x)=w(0)+\int_0^xw'(s)\,ds=w(0) w ( x ) = w ( 0 ) + ∫ 0 x w ′ ( s ) d s = w ( 0 ) { y 0 + c ∣ c ∈ C } \{y_0+c\mid c\in\C\} { y 0 + c ∣ c ∈ C } ▨