system: [SYSTEM]

user: 4. 定理的证明 4.1. 定理1的证明 方程两边点乘 Δ u 再积分： ⇒ Δ u ⋅ ∂ t u = Δ u ⋅ ( u ∧ Δ u ) = 0 ⇒ ∫ R 2 Δ u ⋅ ∂ t u d x = − 1 2 d d t ∫ R 2 | ∇ u | 2 d x 0 = 0 ⇒ ∫ R 2 | ∇ u | 2 d x = C ⇒ E ( u ( t ) ) = E ( u o ) 4.2. 定理2的证明 先用数学归纳法可求得 R k = ( cos k π 2 − sin k π 2 0 sin k π 2 cos k π 2 0 0 0 0 ) K = 1 时显然成立，若 K = n ( n > 1 ) 时成立，则 R n + 1 = ( cos n π 2 − sin n π 2 0 sin n π 2 cos n π 2 0 0 0 0 ) ( 0 − 1 0 1 0 0 0 0 0 ) = ( − sin n π 2 − cos n π 2 0 cos n π 2 − sin n π 2 0 0 0 0 ) = ( cos ( n + 1 ) π 2 − sin ( n + 1 ) π 2 0 sin ( n + 1 ) π 2 cos ( n + 1 ) π 2 0 0 0 0 ) 再计算 e k θ R ： e k θ R = ∑ h = 0 ∞ 1 n ! ( k θ R ) n = ∑ h = 0 ∞ 1 n ! ( k θ ) n ( cos n π 2 − sin n π 2 0 sin n π 2 cos n π 2 0 0 0 0 ) 分别计算得： ∑ h = 0 ∞ 1 n ! ( k θ ) n cos n π 2 = ∑ h = 0 ∞ 1 ( 2 n ) ! ( k θ ) 2 n ( − 1 ) n = cos k θ 同理可得： e k θ R = ( cos k θ − sin k θ 0 sin k θ cos k θ 0 0 0 1 ) 记： A = e k θ R = ( cos k θ − sin k θ 0 sin k θ cos k θ 0 0 0 1 ) ， ， V = A Q deg V = 1 4 π ∫ R 2 | V x 1 V V x 2 | d x = 1 4 π ∫ R 2 | ( A Q ) x 1 A Q ( A Q ) x 2 | d x = 1 4 π ∫ R 2 | A θ θ x 1 Q + A Q r R x 1 A Q A θ θ x 2 Q + A Q r R x 2 | d x | A θ θ x 1 Q + A Q r R x 1 A Q A θ θ x 2 Q + A Q r R x 2 | = | − sin θ r A θ Q + A Q r cos θ A Q cos θ r A θ Q + A Q r sin θ | = | − sin θ r A θ Q A Q cos θ r A θ Q + A Q r sin θ | + | A Q r cos θ A Q cos θ r A θ Q + A Q r sin θ | = | − sin θ r A θ Q A Q A Q r sin θ | + | A Q r cos θ A Q cos θ r A θ Q | = ( sin θ ) 2 r | − A θ Q A Q A Q r | + ( sin θ ) 2 r | A Q r A Q A θ Q | = 1 r | A Q r A Q A θ Q | deg V = 1 4π ∫ R 2 1 r | A Q r A Q A θ Q | d x 记： Q r = ( 2 r k 1 + r 2 k 0 r 2 k − 1 r 2 k + 1 ) r = ( M r 0 N r ) ， A θ = ( − k sin k θ − k cos k θ 0 k cos k θ − k sin k θ 0 0 0 0 ) deg V = 1 4 π ∫ 0 + ∞ ∫ 0 2π | cos k θ M r cos k θ M − k sin k θ M sin k θ M r sin k θ M k cos k θ M N r N 0 | d r d θ = 1 4 π ∫ 0 + ∞ ∫ 0 2π | 0 cos k θ M − k sin k θ M 0 sin k θ M k cos k θ M N r − M r M N N 0 | d r d θ = 1 4 π ∫ 0 + ∞ ∫ 0 2π | 0 0 − k sin k θ M 0 1 sin k θ M k cos k θ M N r − M r M N N 0 | d r d θ = k 2 ∫ 0 + ∞ M 2 N r − M N M r d r = k 2 ( M 2 N | 0 + ∞ − 3 ∫ 0 + ∞ M N M r d r ) = − 3 k 2 ∫ 0 + ∞ M N M r d r = − 3 k 4 ∫ 0 + ∞ ( M 2 ) r N d r = 3 k 4 ∫ 0 + ∞ M 2 N r d r = 3 k 4 ∫ 0 + ∞ 4 r 2 k ( 1 + r 2 k ) 2 ⋅ 4 k r 2 k − 1 ( 1 + r 2 k ) 2 d r = 12 k 2 ∫ 0 + ∞ r 4 k − 1 ( 1 + r 2 k ) 4 d r = 12 k 2 ∫ 0 + ∞ s 4 k − 1 2 k ( 1 + s ) 4 ⋅ 1 2 k ⋅ s 1 2 k − 1 d s = 6 k ∫ 0 + ∞ s ( 1 + s ) 4 d s = 6 k ∫ 0 + ∞ 1 ( 1 + x ) 3 − 1 ( 1 + x ) 4 d x = 6 k × ( 1 3 ( 1 + x ) 3 − 1 2 ( 1 + x ) 2 | 0 + ∞ ) = k 4.3. 定理3的证明 由定理2知： V = ( 2 r k 1 + r 2 k cos ( k θ ) 2 r k 1 + r 2 k sin ( k θ ) r 2 k − 1 r 2 k + 1 ) = ( A cos ( k θ ) A sin ( k θ ) B ) ∂ u ∂ x 1 = ∂ u ∂ r ⋅ ∂ r ∂ x 1 + ∂ u ∂ θ ⋅ ∂ θ ∂ x 1 = cos θ ⋅ ∂ u ∂ r − sin θ r ∂ u ∂ θ ∂ u ∂ x 2 = ∂ u ∂ r ⋅ ∂ r ∂ x 2 + ∂ u ∂ θ ⋅ ∂ θ ∂ x 2 = sin θ ⋅ ∂ u ∂ r + cos θ r ∂ u ∂ θ | ∇ u | 2 = | ( ∂ u ∂ x 1 ， ∂ u ∂ x 2 ， ∂ u ∂ t ) | 2 = | ∂ u ∂ x 1 | 2 + | ∂ u ∂ x 2 | 2 + | ∂ u ∂ t | 2 = | cos θ ⋅ ∂ u ∂ r − sin θ r ∂ u ∂ θ | 2 + | sin θ ⋅ ∂ u ∂ r + cos θ r ∂ u ∂ θ | 2 = | ∂ u ∂ r | 2 + | ∂ u ∂ θ ⋅ 1 r | 2 E ( V ) = ∫ R 2 | ∇ V | 2 d x = ∫ R 2 | ∂ V ∂ r | 2 + | ∂ V ∂ θ ⋅ 1 r | 2 d x | ∂ V ∂ r | 2 = A r 2 cos 2 ( k θ ) + A r 2 sin 2 ( k θ ) + B r 2 = A r 2 + B r 2 | ∂ V ∂ θ ⋅ 1 r | 2 = 1 r 2 ( A 2 k 2 sin 2 ( k θ ) + A 2 k 2 cos 2 ( k θ ) + 0 ) = k 2 r 2 A 2 E ( V ) = ∫ R 2 A r 2 + B r 2 + k 2 r 2 A 2 d x = ∫ R 2 8 k 2 r 2 k − 2 ( 1 + r 2 k ) 2 d x = 16 π k 2 ∫ 0 + ∞ r 2 k − 2 ( 1 + r 2 k ) 2 ⋅ r d r 时，显然 E ( V ) = 0 k > 0 时，令 S = r 2 k ，则 d r = 1 2 k S 1 2 k − 1 d s ，带入得： 16 π k 2 ∫ 0 + ∞ 1 2 k ( 1 + s ) 2 d s = 8 k π k < 0 时，令 S = r 2 k ，则 d r = 1 2 k S 1 2 k − 1 d s ，带入得： 16 π k 2 ∫ + ∞ 0 1 2 k ( 1 + s ) 2 d s = − 8 k π 综上： E ( V ) = 8 π | k | 。