問題冊子は英語と数学の合冊で、英語2問と数学2問の計4問すべてが必答です（試験時間は2時間）。ここでは数学の2問を扱います。第1問は前半が実2次形式の対角化という定型計算で、後半は同じ実対称行列を線形力学系 d x / d t = A x d\boldsymbol{x}/dt = A\boldsymbol{x} d x / d t = A x δ ( x − 1 ) \delta(x-1) δ ( x − 1 ) σ → 0 \sigma \to 0 σ → 0
問題 分野 主題 第1問 線形代数・微分方程式 ユニタリ行列と正規直交系、実2次形式の標準形、勾配系の漸近安定性 第2問 微分方程式・フーリエ解析・確率統計 フォッカー–プランク方程式、ガウス解の族、デルタ関数初期条件
4つの設問は独立ではなく、実対称行列という共通の題材でつながっています。設問(1)は n n n n n n u j \boldsymbol{u}_j u j U = ( u 1 , u 2 , … , u n ) U = (\boldsymbol{u}_1, \boldsymbol{u}_2, \dots, \boldsymbol{u}_n) U = ( u 1 , u 2 , … , u n ) U U U { u j } \{\boldsymbol{u}_j\} { u j } A A A n n n x = ( x 1 , … , x n ) T \boldsymbol{x} = (x_1,\dots,x_n)^T x = ( x 1 , … , x n ) T T T T
ϕ ( x ) = x T A x \phi(\boldsymbol{x}) = \boldsymbol{x}^T A \boldsymbol{x} ϕ ( x ) = x T A x を、直交変換 x = P y \boldsymbol{x} = P\boldsymbol{y} x = P y
ϕ = y T B y = ∑ i = 1 n λ i y i 2 , B = P T A P \phi = \boldsymbol{y}^T B \boldsymbol{y} = \sum_{i=1}^{n} \lambda_i y_i^2,
\qquad B = P^T A P ϕ = y T B y = i = 1 ∑ n λ i y i 2 , B = P T A P に直す操作を、具体的な 3 変数の2次形式に適用します。ここで B B B A A A
d x d t = A x , d x d t = ( d x 1 d t , … , d x n d t ) T , ∇ x Φ = ( ∂ Φ ∂ x 1 , … , ∂ Φ ∂ x n ) T \frac{d\boldsymbol{x}}{dt} = A\boldsymbol{x},
\qquad
\frac{d\boldsymbol{x}}{dt} = \left(\frac{dx_1}{dt},\dots,\frac{dx_n}{dt}\right)^T,
\qquad
\nabla_{\boldsymbol{x}}\Phi = \left(\frac{\partial \Phi}{\partial x_1},\dots,\frac{\partial \Phi}{\partial x_n}\right)^T d t d x = A x , d t d x = ( d t d x 1 , … , d t d x n ) T , ∇ x Φ = ( ∂ x 1 ∂ Φ , … , ∂ x n ∂ Φ ) T を扱い、これがスカラー関数 Φ \Phi Φ d x / d t = − ∇ x Φ ( x ) d\boldsymbol{x}/dt = -\nabla_{\boldsymbol{x}}\Phi(\boldsymbol{x}) d x / d t = − ∇ x Φ ( x ) Φ \Phi Φ
成分は複素数とし、C n \mathbb{C}^n C n ⟨ a , b ⟩ = ∑ i = 1 n a i ‾ b i \langle \boldsymbol{a}, \boldsymbol{b}\rangle = \sum_{i=1}^n \overline{a_i}\, b_i ⟨ a , b ⟩ = ∑ i = 1 n a i b i U † U = I n U^{\dagger} U = I_n U † U = I n U † = U ‾ T U^{\dagger} = \overline{U}^{\,T} U † = U T ( U † ) j i = U i j ‾ (U^{\dagger})_{ji} = \overline{U_{ij}} ( U † ) j i = U ij
U = ( u 1 , … , u n ) U = (\boldsymbol{u}_1,\dots,\boldsymbol{u}_n) U = ( u 1 , … , u n ) U U U ( i , j ) (i,j) ( i , j ) u j \boldsymbol{u}_j u j i i i U i j = ( u j ) i U_{ij} = (\boldsymbol{u}_j)_i U ij = ( u j ) i U † U U^{\dagger}U U † U
( U † U ) j k = ∑ i = 1 n ( U † ) j i U i k = ∑ i = 1 n U i j ‾ U i k = ∑ i = 1 n ( u j ) i ‾ ( u k ) i = ⟨ u j , u k ⟩ (U^{\dagger}U)_{jk} = \sum_{i=1}^{n} (U^{\dagger})_{ji}\, U_{ik}
= \sum_{i=1}^{n} \overline{U_{ij}}\, U_{ik}
= \sum_{i=1}^{n} \overline{(\boldsymbol{u}_j)_i}\, (\boldsymbol{u}_k)_i
= \langle \boldsymbol{u}_j, \boldsymbol{u}_k \rangle ( U † U ) j k = i = 1 ∑ n ( U † ) j i U ik = i = 1 ∑ n U ij U ik = i = 1 ∑ n ( u j ) i ( u k ) i = ⟨ u j , u k ⟩ となります。行列の積の ( j , k ) (j,k) ( j , k ) j j j k k k U U U ( I n ) j k = δ j k (I_n)_{jk} = \delta_{jk} ( I n ) j k = δ j k
⟨ u j , u k ⟩ = δ j k ( j , k = 1 , … , n ) \langle \boldsymbol{u}_j, \boldsymbol{u}_k \rangle = \delta_{jk}
\qquad (j,k = 1,\dots,n) ⟨ u j , u k ⟩ = δ j k ( j , k = 1 , … , n ) が成り立ちます。j = k j = k j = k ∥ u j ∥ = 1 \lVert \boldsymbol{u}_j \rVert = 1 ∥ u j ∥ = 1 j ≠ k j \neq k j = k u j ⊥ u k \boldsymbol{u}_j \perp \boldsymbol{u}_k u j ⊥ u k { u j } \{\boldsymbol{u}_j\} { u j }
さらに、この n n n ∑ j c j u j = 0 \sum_j c_j \boldsymbol{u}_j = \boldsymbol{0} ∑ j c j u j = 0 u k \boldsymbol{u}_k u k ∑ j c j δ k j = c k = 0 \sum_j c_j \delta_{kj} = c_k = 0 ∑ j c j δ k j = c k = 0 k k k n n n n n n { u j } \{\boldsymbol{u}_j\} { u j } C n \mathbb{C}^n C n
なお上の計算は同値変形ですから逆も成り立ち、列ベクトルが正規直交系をなすことと U U U U † = U T U^{\dagger} = U^T U † = U T
ϕ ( x ) = 4 x 1 2 + 2 x 2 2 + 2 x 3 2 − 2 x 1 x 2 + 2 x 2 x 3 − 2 x 3 x 1 \phi(\boldsymbol{x}) = 4x_1^2 + 2x_2^2 + 2x_3^2 - 2x_1x_2 + 2x_2x_3 - 2x_3x_1 ϕ ( x ) = 4 x 1 2 + 2 x 2 2 + 2 x 3 2 − 2 x 1 x 2 + 2 x 2 x 3 − 2 x 3 x 1 x T A x = ∑ i , j A i j x i x j \boldsymbol{x}^T A \boldsymbol{x} = \sum_{i,j} A_{ij}x_ix_j x T A x = ∑ i , j A ij x i x j A A A x i 2 x_i^2 x i 2 A i i A_{ii} A ii x i x j x_ix_j x i x j i ≠ j i \neq j i = j A i j + A j i = 2 A i j A_{ij} + A_{ji} = 2A_{ij} A ij + A j i = 2 A ij A 11 = 4 A_{11}=4 A 11 = 4 A 22 = A 33 = 2 A_{22}=A_{33}=2 A 22 = A 33 = 2 A 12 = A 21 = − 1 A_{12}=A_{21}=-1 A 12 = A 21 = − 1 A 23 = A 32 = 1 A_{23}=A_{32}=1 A 23 = A 32 = 1 A 13 = A 31 = − 1 A_{13}=A_{31}=-1 A 13 = A 31 = − 1
A = ( 4 − 1 − 1 − 1 2 1 − 1 1 2 ) A = \begin{pmatrix} 4 & -1 & -1 \\ -1 & 2 & 1 \\ -1 & 1 & 2 \end{pmatrix} A = 4 − 1 − 1 − 1 2 1 − 1 1 2 固有値を求めます。
det ( A − λ I ) = ( 4 − λ ) [ ( 2 − λ ) 2 − 1 ] + ( λ − 1 ) + ( λ − 1 ) = ( 4 − λ ) ( 1 − λ ) ( 3 − λ ) − 2 ( 1 − λ ) = ( 1 − λ ) [ ( 4 − λ ) ( 3 − λ ) − 2 ] = ( 1 − λ ) ( λ 2 − 7 λ + 10 ) = ( 1 − λ ) ( λ − 2 ) ( λ − 5 ) \begin{aligned}
\det(A - \lambda I)
&= (4-\lambda)\bigl[(2-\lambda)^2 - 1\bigr] + (\lambda - 1) + (\lambda - 1) \\
&= (4-\lambda)(1-\lambda)(3-\lambda) - 2(1-\lambda) \\
&= (1-\lambda)\bigl[(4-\lambda)(3-\lambda) - 2\bigr] \\
&= (1-\lambda)(\lambda^2 - 7\lambda + 10) = (1-\lambda)(\lambda-2)(\lambda-5)
\end{aligned} det ( A − λ I ) = ( 4 − λ ) [ ( 2 − λ ) 2 − 1 ] + ( λ − 1 ) + ( λ − 1 ) = ( 4 − λ ) ( 1 − λ ) ( 3 − λ ) − 2 ( 1 − λ ) = ( 1 − λ ) [ ( 4 − λ ) ( 3 − λ ) − 2 ] = ( 1 − λ ) ( λ 2 − 7 λ + 10 ) = ( 1 − λ ) ( λ − 2 ) ( λ − 5 ) よって固有値は λ 1 = 1 \lambda_1 = 1 λ 1 = 1 λ 2 = 2 \lambda_2 = 2 λ 2 = 2 λ 3 = 5 \lambda_3 = 5 λ 3 = 5 4 + 2 + 2 = 8 = 1 + 2 + 5 4+2+2 = 8 = 1+2+5 4 + 2 + 2 = 8 = 1 + 2 + 5 10 = 1 ⋅ 2 ⋅ 5 10 = 1\cdot 2\cdot 5 10 = 1 ⋅ 2 ⋅ 5
各固有ベクトルを求めます。λ = 1 \lambda = 1 λ = 1 ( A − I ) v = 0 (A-I)\boldsymbol{v} = \boldsymbol{0} ( A − I ) v = 0 − v 1 + v 2 + v 3 = 0 -v_1+v_2+v_3=0 − v 1 + v 2 + v 3 = 0 3 v 1 − v 2 − v 3 = 0 3v_1 - v_2 - v_3 = 0 3 v 1 − v 2 − v 3 = 0 3 v 1 = v 1 3v_1 = v_1 3 v 1 = v 1 v 1 = 0 v_1 = 0 v 1 = 0 v 2 = − v 3 v_2 = -v_3 v 2 = − v 3 λ = 2 \lambda = 2 λ = 2 − v 1 + v 3 = 0 -v_1+v_3 = 0 − v 1 + v 3 = 0 − v 1 + v 2 = 0 -v_1+v_2=0 − v 1 + v 2 = 0 v 1 = v 2 = v 3 v_1=v_2=v_3 v 1 = v 2 = v 3 λ = 5 \lambda = 5 λ = 5 v 1 + v 2 + v 3 = 0 v_1+v_2+v_3=0 v 1 + v 2 + v 3 = 0 − v 1 − 3 v 2 + v 3 = 0 -v_1-3v_2+v_3=0 − v 1 − 3 v 2 + v 3 = 0 − 2 v 2 + 2 v 3 = 0 -2v_2+2v_3=0 − 2 v 2 + 2 v 3 = 0 v 2 = v 3 v_2=v_3 v 2 = v 3 v 1 = − 2 v 2 v_1 = -2v_2 v 1 = − 2 v 2
u 1 = 1 2 ( 0 1 − 1 ) , u 2 = 1 3 ( 1 1 1 ) , u 3 = 1 6 ( 2 − 1 − 1 ) \boldsymbol{u}_1 = \frac{1}{\sqrt 2}\begin{pmatrix} 0 \\ 1 \\ -1 \end{pmatrix},
\qquad
\boldsymbol{u}_2 = \frac{1}{\sqrt 3}\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix},
\qquad
\boldsymbol{u}_3 = \frac{1}{\sqrt 6}\begin{pmatrix} 2 \\ -1 \\ -1 \end{pmatrix} u 1 = 2 1 0 1 − 1 , u 2 = 3 1 1 1 1 , u 3 = 6 1 2 − 1 − 1 とします（u 3 \boldsymbol{u}_3 u 3 u 1 ⋅ u 2 = u 2 ⋅ u 3 = u 3 ⋅ u 1 = 0 \boldsymbol{u}_1\cdot\boldsymbol{u}_2 = \boldsymbol{u}_2\cdot\boldsymbol{u}_3 = \boldsymbol{u}_3\cdot\boldsymbol{u}_1 = 0 u 1 ⋅ u 2 = u 2 ⋅ u 3 = u 3 ⋅ u 1 = 0
P = ( u 1 , u 2 , u 3 ) = ( 0 1 / 3 2 / 6 1 / 2 1 / 3 − 1 / 6 − 1 / 2 1 / 3 − 1 / 6 ) P = (\boldsymbol{u}_1, \boldsymbol{u}_2, \boldsymbol{u}_3)
= \begin{pmatrix}
0 & 1/\sqrt3 & 2/\sqrt6 \\
1/\sqrt2 & 1/\sqrt3 & -1/\sqrt6 \\
-1/\sqrt2 & 1/\sqrt3 & -1/\sqrt6
\end{pmatrix} P = ( u 1 , u 2 , u 3 ) = 0 1/ 2 − 1/ 2 1/ 3 1/ 3 1/ 3 2/ 6 − 1/ 6 − 1/ 6 は設問(1)（の実版）より P T P = I P^TP = I P T P = I det P = + 1 \det P = +1 det P = + 1 u 3 \boldsymbol{u}_3 u 3 P P P P P P A P = ( λ 1 u 1 , λ 2 u 2 , λ 3 u 3 ) = P d i a g ( 1 , 2 , 5 ) AP = (\lambda_1\boldsymbol{u}_1, \lambda_2\boldsymbol{u}_2, \lambda_3\boldsymbol{u}_3) = P\,\mathrm{diag}(1,2,5) A P = ( λ 1 u 1 , λ 2 u 2 , λ 3 u 3 ) = P diag ( 1 , 2 , 5 )
B = P − 1 A P = P T A P = ( 1 0 0 0 2 0 0 0 5 ) B = P^{-1}AP = P^TAP = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 5 \end{pmatrix} B = P − 1 A P = P T A P = 1 0 0 0 2 0 0 0 5 となり、x = P y \boldsymbol{x} = P\boldsymbol{y} x = P y
ϕ = ( P y ) T A ( P y ) = y T ( P T A P ) y = y 1 2 + 2 y 2 2 + 5 y 3 2 \phi = (P\boldsymbol{y})^T A (P\boldsymbol{y}) = \boldsymbol{y}^T(P^TAP)\boldsymbol{y}
= y_1^2 + 2y_2^2 + 5y_3^2 ϕ = ( P y ) T A ( P y ) = y T ( P T A P ) y = y 1 2 + 2 y 2 2 + 5 y 3 2 が標準形です。答えは A A A P P P λ 1 = 1 , λ 2 = 2 , λ 3 = 5 \lambda_1 = 1,\ \lambda_2 = 2,\ \lambda_3 = 5 λ 1 = 1 , λ 2 = 2 , λ 3 = 5 ϕ = y 1 2 + 2 y 2 2 + 5 y 3 2 \phi = y_1^2 + 2y_2^2 + 5y_3^2 ϕ = y 1 2 + 2 y 2 2 + 5 y 3 2
検算します。x = ( 1 , 0 , 0 ) T \boldsymbol{x} = (1,0,0)^T x = ( 1 , 0 , 0 ) T ϕ = 4 \phi = 4 ϕ = 4 y = P − 1 x = P T x = ( 0 , 1 / 3 , 2 / 6 ) T \boldsymbol{y} = P^{-1}\boldsymbol{x} = P^T\boldsymbol{x} = (0,\ 1/\sqrt3,\ 2/\sqrt6)^T y = P − 1 x = P T x = ( 0 , 1/ 3 , 2/ 6 ) T 0 + 2 ⋅ 1 3 + 5 ⋅ 4 6 = 2 3 + 10 3 = 4 0 + 2\cdot\frac13 + 5\cdot\frac46 = \frac23 + \frac{10}{3} = 4 0 + 2 ⋅ 3 1 + 5 ⋅ 6 4 = 3 2 + 3 10 = 4 ϕ \phi ϕ ϕ ( x ) = 0 \phi(\boldsymbol{x}) = 0 ϕ ( x ) = 0 x = 0 \boldsymbol{x} = \boldsymbol{0} x = 0
答えは
Φ ( x ) = − 1 2 x T A x = − 1 2 ∑ i , j = 1 n A i j x i x j \Phi(\boldsymbol{x}) = -\frac12 \boldsymbol{x}^T A \boldsymbol{x} = -\frac12 \sum_{i,j=1}^{n} A_{ij}x_ix_j Φ ( x ) = − 2 1 x T A x = − 2 1 i , j = 1 ∑ n A ij x i x j （加法定数を除く）です。以下で確認します。任意の（対称とは限らない）n n n A A A
∂ ∂ x i ( ∑ j , k A j k x j x k ) = ∑ k A i k x k + ∑ j A j i x j = ( A x ) i + ( A T x ) i \frac{\partial}{\partial x_i}\Bigl(\sum_{j,k} A_{jk}x_jx_k\Bigr)
= \sum_{k} A_{ik}x_k + \sum_{j} A_{ji}x_j
= (A\boldsymbol{x})_i + (A^T\boldsymbol{x})_i ∂ x i ∂ ( j , k ∑ A j k x j x k ) = k ∑ A ik x k + j ∑ A j i x j = ( A x ) i + ( A T x ) i です（x j x_j x j x k x_k x k A T = A A^T = A A T = A 2 ( A x ) i 2(A\boldsymbol{x})_i 2 ( A x ) i ∇ x ( x T A x ) = 2 A x \nabla_{\boldsymbol{x}}(\boldsymbol{x}^TA\boldsymbol{x}) = 2A\boldsymbol{x} ∇ x ( x T A x ) = 2 A x
− ∇ x Φ = 1 2 ∇ x ( x T A x ) = A x = d x d t -\nabla_{\boldsymbol{x}}\Phi = \frac12\nabla_{\boldsymbol{x}}\bigl(\boldsymbol{x}^TA\boldsymbol{x}\bigr) = A\boldsymbol{x} = \frac{d\boldsymbol{x}}{dt} − ∇ x Φ = 2 1 ∇ x ( x T A x ) = A x = d t d x が成り立ちます。これで Φ \Phi Φ
対称性がここで本質的に効いていることを補足します。R n \mathbb{R}^n R n F ( x ) = A x \boldsymbol{F}(\boldsymbol{x}) = A\boldsymbol{x} F ( x ) = A x F = − ∇ Φ \boldsymbol{F} = -\nabla\Phi F = − ∇Φ ∂ F i / ∂ x j = ∂ F j / ∂ x i \partial F_i/\partial x_j = \partial F_j/\partial x_i ∂ F i / ∂ x j = ∂ F j / ∂ x i ∂ 2 Φ / ∂ x i ∂ x j \partial^2\Phi/\partial x_i\partial x_j ∂ 2 Φ/ ∂ x i ∂ x j R n \mathbb{R}^n R n Φ \Phi Φ ∂ F i / ∂ x j = A i j \partial F_i/\partial x_j = A_{ij} ∂ F i / ∂ x j = A ij A i j = A j i A_{ij} = A_{ji} A ij = A j i A A A A A A
また Φ \Phi Φ − ∇ Φ 1 = − ∇ Φ 2 -\nabla\Phi_1 = -\nabla\Phi_2 − ∇ Φ 1 = − ∇ Φ 2 ∇ ( Φ 1 − Φ 2 ) = 0 \nabla(\Phi_1-\Phi_2) = \boldsymbol{0} ∇ ( Φ 1 − Φ 2 ) = 0 R n \mathbb{R}^n R n Φ 1 − Φ 2 \Phi_1 - \Phi_2 Φ 1 − Φ 2 Φ ( 0 ) = 0 \Phi(\boldsymbol{0}) = 0 Φ ( 0 ) = 0 Φ = − ϕ / 2 \Phi = -\phi/2 Φ = − ϕ /2 A A A { u i } \{\boldsymbol{u}_i\} { u i } λ i \lambda_i λ i x = ∑ i c i u i \boldsymbol{x} = \sum_i c_i\boldsymbol{u}_i x = ∑ i c i u i
Φ ( x ) = − 1 2 ∑ i = 1 n λ i c i 2 \Phi(\boldsymbol{x}) = -\frac12\sum_{i=1}^{n}\lambda_i c_i^2 Φ ( x ) = − 2 1 i = 1 ∑ n λ i c i 2 と書けます。これを設問(4)で使います。
前半を示します。Φ \Phi Φ C ∞ C^\infty C ∞ x ( t ) \boldsymbol{x}(t) x ( t ) x ( t ) = e A t x ( 0 ) \boldsymbol{x}(t) = e^{At}\boldsymbol{x}(0) x ( t ) = e A t x ( 0 ) t t t C 1 C^1 C 1
d Φ ( x ( t ) ) d t = ∑ i = 1 n ∂ Φ ∂ x i d x i d t = ∇ x Φ ⋅ ( − ∇ x Φ ) = − ∥ ∇ x Φ ∥ 2 = − ∥ A x ( t ) ∥ 2 ≤ 0 \frac{d\Phi(\boldsymbol{x}(t))}{dt}
= \sum_{i=1}^{n}\frac{\partial \Phi}{\partial x_i}\frac{dx_i}{dt}
= \nabla_{\boldsymbol{x}}\Phi \cdot \left(-\nabla_{\boldsymbol{x}}\Phi\right)
= -\bigl\lVert \nabla_{\boldsymbol{x}}\Phi \bigr\rVert^2
= -\lVert A\boldsymbol{x}(t)\rVert^2 \le 0 d t d Φ ( x ( t )) = i = 1 ∑ n ∂ x i ∂ Φ d t d x i = ∇ x Φ ⋅ ( − ∇ x Φ ) = − ∇ x Φ 2 = − ∥ A x ( t ) ∥ 2 ≤ 0 となります。等号が成り立つのは A x ( t ) = 0 A\boldsymbol{x}(t) = \boldsymbol{0} A x ( t ) = 0 Φ \Phi Φ
後半を示します。仮定は「任意の x ≠ 0 \boldsymbol{x} \neq \boldsymbol{0} x = 0 Φ ( x ) > 0 \Phi(\boldsymbol{x}) > 0 Φ ( x ) > 0 A A A λ 1 , … , λ n \lambda_1,\dots,\lambda_n λ 1 , … , λ n { u i } \{\boldsymbol{u}_i\} { u i } x = u i \boldsymbol{x} = \boldsymbol{u}_i x = u i ≠ 0 \neq \boldsymbol{0} = 0
0 < Φ ( u i ) = − 1 2 u i T A u i = − λ i 2 ∥ u i ∥ 2 = − λ i 2 0 < \Phi(\boldsymbol{u}_i) = -\frac12 \boldsymbol{u}_i^T A \boldsymbol{u}_i = -\frac{\lambda_i}{2}\lVert\boldsymbol{u}_i\rVert^2 = -\frac{\lambda_i}{2} 0 < Φ ( u i ) = − 2 1 u i T A u i = − 2 λ i ∥ u i ∥ 2 = − 2 λ i より λ i < 0 \lambda_i < 0 λ i < 0 i i i Φ \Phi Φ − A -A − A
そこで c i ( t ) = u i ⋅ x ( t ) c_i(t) = \boldsymbol{u}_i\cdot\boldsymbol{x}(t) c i ( t ) = u i ⋅ x ( t ) A A A
d c i d t = u i ⋅ A x = ( A T u i ) ⋅ x = ( A u i ) ⋅ x = λ i c i \frac{dc_i}{dt} = \boldsymbol{u}_i\cdot A\boldsymbol{x} = (A^T\boldsymbol{u}_i)\cdot\boldsymbol{x} = (A\boldsymbol{u}_i)\cdot\boldsymbol{x} = \lambda_i c_i d t d c i = u i ⋅ A x = ( A T u i ) ⋅ x = ( A u i ) ⋅ x = λ i c i なので c i ( t ) = c i ( 0 ) e λ i t c_i(t) = c_i(0)e^{\lambda_i t} c i ( t ) = c i ( 0 ) e λ i t { u i } \{\boldsymbol{u}_i\} { u i }
∥ x ( t ) ∥ 2 = ∑ i = 1 n c i ( 0 ) 2 e 2 λ i t ≤ e 2 λ max t ∑ i = 1 n c i ( 0 ) 2 = e 2 λ max t ∥ x ( 0 ) ∥ 2 , λ max ≡ max i λ i < 0 \lVert \boldsymbol{x}(t)\rVert^2 = \sum_{i=1}^{n} c_i(0)^2 e^{2\lambda_i t}
\le e^{2\lambda_{\max}t}\sum_{i=1}^{n}c_i(0)^2 = e^{2\lambda_{\max}t}\lVert\boldsymbol{x}(0)\rVert^2,
\qquad \lambda_{\max} \equiv \max_i \lambda_i < 0 ∥ x ( t ) ∥ 2 = i = 1 ∑ n c i ( 0 ) 2 e 2 λ i t ≤ e 2 λ m a x t i = 1 ∑ n c i ( 0 ) 2 = e 2 λ m a x t ∥ x ( 0 ) ∥ 2 , λ m a x ≡ i max λ i < 0 が成り立ち、
∥ x ( t ) ∥ ≤ e λ max t ∥ x ( 0 ) ∥ → t → ∞ 0 \lVert \boldsymbol{x}(t)\rVert \le e^{\lambda_{\max}t}\,\lVert\boldsymbol{x}(0)\rVert \xrightarrow{\ t\to\infty\ } 0 ∥ x ( t )∥ ≤ e λ m a x t ∥ x ( 0 )∥ t → ∞ 0 です。したがって任意の初期条件からの解軌道は x = 0 \boldsymbol{x} = \boldsymbol{0} x = 0 ∣ λ max ∣ \lvert\lambda_{\max}\rvert ∣ λ m a x ∣ Φ ( x ( t ) ) = − 1 2 ∑ i λ i c i ( 0 ) 2 e 2 λ i t → 0 \Phi(\boldsymbol{x}(t)) = -\frac12\sum_i\lambda_ic_i(0)^2e^{2\lambda_it} \to 0 Φ ( x ( t )) = − 2 1 ∑ i λ i c i ( 0 ) 2 e 2 λ i t → 0
この結論はリアプノフの定理の内容そのものです。Φ \Phi Φ Φ ( 0 ) = 0 \Phi(\boldsymbol{0})=0 Φ ( 0 ) = 0 x ≠ 0 \boldsymbol{x}\neq\boldsymbol{0} x = 0 Φ > 0 \Phi>0 Φ > 0 Φ \Phi Φ − A -A − A x ≠ 0 \boldsymbol{x}\neq\boldsymbol{0} x = 0 A x ≠ 0 A\boldsymbol{x}\neq\boldsymbol{0} A x = 0 d Φ / d t < 0 d\Phi/dt < 0 d Φ/ d t < 0 d x / d t = − ∇ Φ d\boldsymbol{x}/dt = -\nabla\Phi d x / d t = − ∇Φ λ i > 0 \lambda_i > 0 λ i > 0 Φ ( u i ) < 0 \Phi(\boldsymbol{u}_i) < 0 Φ ( u i ) < 0 c i ( 0 ) ≠ 0 c_i(0)\neq 0 c i ( 0 ) = 0 ∣ c i ( t ) ∣ = ∣ c i ( 0 ) ∣ e λ i t \lvert c_i(t)\rvert = \lvert c_i(0)\rvert e^{\lambda_i t} ∣ c i ( t )∣ = ∣ c i ( 0 )∣ e λ i t
1変数の分布関数 f ( x , t ) f(x,t) f ( x , t )
∂ f ∂ t = ∂ ∂ x ( x f ) + D ∂ 2 f ∂ x 2 \frac{\partial f}{\partial t} = \frac{\partial}{\partial x}\bigl(xf\bigr) + D\frac{\partial^2 f}{\partial x^2} ∂ t ∂ f = ∂ x ∂ ( x f ) + D ∂ x 2 ∂ 2 f を、D D D x → ± ∞ x \to \pm\infty x → ± ∞ f f f ∂ f / ∂ x \partial f/\partial x ∂ f / ∂ x 0 0 0 − ∂ ( v f ) / ∂ x -\partial(vf)/\partial x − ∂ ( v f ) / ∂ x v ( x ) = − x v(x) = -x v ( x ) = − x D D D 1 1 1
「十分速やかに」の意味は、以下の部分積分で現れる境界項がすべて消えること、具体的には x → ± ∞ x \to \pm\infty x → ± ∞ f → 0 f \to 0 f → 0 ∂ f / ∂ x → 0 \partial f/\partial x \to 0 ∂ f / ∂ x → 0 x f → 0 xf \to 0 x f → 0 x ∂ f / ∂ x → 0 x\,\partial f/\partial x \to 0 x ∂ f / ∂ x → 0 x 2 f → 0 x^2 f \to 0 x 2 f → 0 g g g g ( x , σ ) = 1 2 π σ e − x 2 / ( 2 σ 2 ) g(x,\sigma) = \frac{1}{\sqrt{2\pi}\,\sigma}e^{-x^2/(2\sigma^2)} g ( x , σ ) = 2 π σ 1 e − x 2 / ( 2 σ 2 )
f f f t t t C 1 C^1 C 1 ∂ f / ∂ t \partial f/\partial t ∂ f / ∂ t t t t
d I d t = ∫ − ∞ ∞ ∂ f ∂ t d x = ∫ − ∞ ∞ ∂ ∂ x ( x f + D ∂ f ∂ x ) d x = [ x f + D ∂ f ∂ x ] x = − ∞ x = + ∞ \frac{dI}{dt} = \int_{-\infty}^{\infty}\frac{\partial f}{\partial t}\,dx
= \int_{-\infty}^{\infty}\frac{\partial}{\partial x}\left(xf + D\frac{\partial f}{\partial x}\right)dx
= \left[\,xf + D\frac{\partial f}{\partial x}\,\right]_{x=-\infty}^{x=+\infty} d t d I = ∫ − ∞ ∞ ∂ t ∂ f d x = ∫ − ∞ ∞ ∂ x ∂ ( x f + D ∂ x ∂ f ) d x = [ x f + D ∂ x ∂ f ] x = − ∞ x = + ∞ です。ここで、方程式の右辺全体が x x x x f → 0 xf \to 0 x f → 0 ∂ f / ∂ x → 0 \partial f/\partial x \to 0 ∂ f / ∂ x → 0
d I d t = 0 \frac{dI}{dt} = 0 d t d I = 0 が示されました。言い換えると、この方程式は流れ j = − ( x f + D ∂ f / ∂ x ) j = -\left(xf + D\,\partial f/\partial x\right) j = − ( x f + D ∂ f / ∂ x ) ∂ f / ∂ t + ∂ j / ∂ x = 0 \partial f/\partial t + \partial j/\partial x = 0 ∂ f / ∂ t + ∂ j / ∂ x = 0 I I I I = 1 I = 1 I = 1
g ( x , σ ) g(x,\sigma) g ( x , σ ) ∂ g / ∂ x = − ( x / σ 2 ) g \partial g/\partial x = -(x/\sigma^2)g ∂ g / ∂ x = − ( x / σ 2 ) g ∂ 2 g / ∂ x 2 = ( x 2 / σ 4 − 1 / σ 2 ) g \partial^2 g/\partial x^2 = \left(x^2/\sigma^4 - 1/\sigma^2\right)g ∂ 2 g / ∂ x 2 = ( x 2 / σ 4 − 1/ σ 2 ) g
∂ ∂ x ( x g ) + D ∂ 2 g ∂ x 2 = g + x ∂ g ∂ x + D ( x 2 σ 4 − 1 σ 2 ) g = [ ( 1 − D σ 2 ) − x 2 σ 2 ( 1 − D σ 2 ) ] g = ( 1 − D σ 2 ) ( 1 − x 2 σ 2 ) g \begin{aligned}
\frac{\partial}{\partial x}(xg) + D\frac{\partial^2 g}{\partial x^2}
&= g + x\frac{\partial g}{\partial x} + D\left(\frac{x^2}{\sigma^4} - \frac{1}{\sigma^2}\right)g \\
&= \left[\left(1 - \frac{D}{\sigma^2}\right) - \frac{x^2}{\sigma^2}\left(1 - \frac{D}{\sigma^2}\right)\right]g
= \left(1 - \frac{D}{\sigma^2}\right)\left(1 - \frac{x^2}{\sigma^2}\right)g
\end{aligned} ∂ x ∂ ( xg ) + D ∂ x 2 ∂ 2 g = g + x ∂ x ∂ g + D ( σ 4 x 2 − σ 2 1 ) g = [ ( 1 − σ 2 D ) − σ 2 x 2 ( 1 − σ 2 D ) ] g = ( 1 − σ 2 D ) ( 1 − σ 2 x 2 ) g となります。g > 0 g > 0 g > 0 1 − x 2 / σ 2 1 - x^2/\sigma^2 1 − x 2 / σ 2 0 0 0 x x x 0 0 0 1 − D / σ 2 = 0 1 - D/\sigma^2 = 0 1 − D / σ 2 = 0 σ > 0 \sigma > 0 σ > 0
σ = D \sigma = \sqrt{D} σ = D です。
別の見方をすると、定常条件 ∂ x ( x f + D f ′ ) = 0 \partial_x\left(xf + Df'\right) = 0 ∂ x ( x f + D f ′ ) = 0 x f + D f ′ = const xf + Df' = \text{const} x f + D f ′ = const x → ∞ x \to \infty x → ∞ 0 0 0 0 0 0 0 0 0 f ′ / f = − x / D f'/f = -x/D f ′ / f = − x / D f ∝ e − x 2 / ( 2 D ) f \propto e^{-x^2/(2D)} f ∝ e − x 2 / ( 2 D ) σ 2 = D \sigma^2 = D σ 2 = D
次元の確認をしておきます。方程式の第1項 ∂ x ( x f ) \partial_x(xf) ∂ x ( x f ) ∂ f / ∂ t \partial f/\partial t ∂ f / ∂ t D D D [ x ] 2 [x]^2 [ x ] 2 σ 2 = D \sigma^2 = D σ 2 = D [ x ] 2 [x]^2 [ x ] 2 σ 2 = D \sigma^2 = D σ 2 = D
u = x − X ( t ) u = x - X(t) u = x − X ( t ) f ( x , t ) = g ( u , σ ( t ) ) f(x,t) = g(u,\sigma(t)) f ( x , t ) = g ( u , σ ( t ))
∂ g ∂ u = − u σ 2 g , ∂ 2 g ∂ u 2 = ( u 2 σ 4 − 1 σ 2 ) g , ∂ g ∂ σ = ( u 2 σ 3 − 1 σ ) g \frac{\partial g}{\partial u} = -\frac{u}{\sigma^2}g,
\qquad
\frac{\partial^2 g}{\partial u^2} = \left(\frac{u^2}{\sigma^4} - \frac{1}{\sigma^2}\right)g,
\qquad
\frac{\partial g}{\partial \sigma} = \left(\frac{u^2}{\sigma^3} - \frac{1}{\sigma}\right)g ∂ u ∂ g = − σ 2 u g , ∂ u 2 ∂ 2 g = ( σ 4 u 2 − σ 2 1 ) g , ∂ σ ∂ g = ( σ 3 u 2 − σ 1 ) g です（3つ目は ln g = − ln ( 2 π σ ) − u 2 / ( 2 σ 2 ) \ln g = -\ln(\sqrt{2\pi}\sigma) - u^2/(2\sigma^2) ln g = − ln ( 2 π σ ) − u 2 / ( 2 σ 2 ) σ \sigma σ ∂ u / ∂ t = − X ˙ \partial u/\partial t = -\dot X ∂ u / ∂ t = − X ˙ ∂ u / ∂ x = 1 \partial u/\partial x = 1 ∂ u / ∂ x = 1
∂ f ∂ t = − X ˙ ∂ g ∂ u + σ ˙ ∂ g ∂ σ = [ X ˙ u σ 2 + σ ˙ ( u 2 σ 3 − 1 σ ) ] g , ∂ ∂ x ( x f ) + D ∂ 2 f ∂ x 2 = f + x ∂ g ∂ u + D ∂ 2 g ∂ u 2 = [ 1 − ( u + X ) u σ 2 + D ( u 2 σ 4 − 1 σ 2 ) ] g \begin{aligned}
\frac{\partial f}{\partial t}
&= -\dot X\frac{\partial g}{\partial u} + \dot\sigma\frac{\partial g}{\partial \sigma}
= \left[\frac{\dot X u}{\sigma^2} + \dot\sigma\left(\frac{u^2}{\sigma^3} - \frac{1}{\sigma}\right)\right]g, \\[2pt]
\frac{\partial}{\partial x}(xf) + D\frac{\partial^2 f}{\partial x^2}
&= f + x\frac{\partial g}{\partial u} + D\frac{\partial^2 g}{\partial u^2}
= \left[1 - \frac{(u+X)u}{\sigma^2} + D\left(\frac{u^2}{\sigma^4} - \frac{1}{\sigma^2}\right)\right]g
\end{aligned} ∂ t ∂ f ∂ x ∂ ( x f ) + D ∂ x 2 ∂ 2 f = − X ˙ ∂ u ∂ g + σ ˙ ∂ σ ∂ g = [ σ 2 X ˙ u + σ ˙ ( σ 3 u 2 − σ 1 ) ] g , = f + x ∂ u ∂ g + D ∂ u 2 ∂ 2 g = [ 1 − σ 2 ( u + X ) u + D ( σ 4 u 2 − σ 2 1 ) ] g となります（x = u + X x = u + X x = u + X g g g u u u g ≠ 0 g \neq 0 g = 0 1 , u , u 2 1, u, u^2 1 , u , u 2 u u u x x x u u u u 0 , u 1 , u 2 u^0, u^1, u^2 u 0 , u 1 , u 2
u 2 : σ ˙ σ 3 = − 1 σ 2 + D σ 4 ⟺ σ σ ˙ = D − σ 2 , u 1 : X ˙ σ 2 = − X σ 2 ⟺ X ˙ = − X , u 0 : − σ ˙ σ = 1 − D σ 2 ⟺ σ σ ˙ = D − σ 2 \begin{aligned}
u^2 &: \quad \frac{\dot\sigma}{\sigma^3} = -\frac{1}{\sigma^2} + \frac{D}{\sigma^4}
&&\Longleftrightarrow\quad \sigma\dot\sigma = D - \sigma^2, \\
u^1 &: \quad \frac{\dot X}{\sigma^2} = -\frac{X}{\sigma^2}
&&\Longleftrightarrow\quad \dot X = -X, \\
u^0 &: \quad -\frac{\dot\sigma}{\sigma} = 1 - \frac{D}{\sigma^2}
&&\Longleftrightarrow\quad \sigma\dot\sigma = D - \sigma^2
\end{aligned} u 2 u 1 u 0 : σ 3 σ ˙ = − σ 2 1 + σ 4 D : σ 2 X ˙ = − σ 2 X : − σ σ ˙ = 1 − σ 2 D ⟺ σ σ ˙ = D − σ 2 , ⟺ X ˙ = − X , ⟺ σ σ ˙ = D − σ 2 を得ます。u 0 u^0 u 0 u 2 u^2 u 2
d X d t = − X , d σ d t = D σ − σ \frac{dX}{dt} = -X,
\qquad
\frac{d\sigma}{dt} = \frac{D}{\sigma} - \sigma d t d X = − X , d t d σ = σ D − σ です。第2式は d ( σ 2 ) / d t = 2 ( D − σ 2 ) d(\sigma^2)/dt = 2(D - \sigma^2) d ( σ 2 ) / d t = 2 ( D − σ 2 ) X = 0 X = 0 X = 0 σ = D \sigma = \sqrt D σ = D 0 0 0 I = 1 I = 1 I = 1 ⟨ x ⟩ = ∫ x f d x \langle x\rangle = \int xf\,dx ⟨ x ⟩ = ∫ x f d x ⟨ x 2 ⟩ = ∫ x 2 f d x \langle x^2\rangle = \int x^2f\,dx ⟨ x 2 ⟩ = ∫ x 2 f d x
d ⟨ x ⟩ d t = − ⟨ x ⟩ , d ⟨ x 2 ⟩ d t = − 2 ⟨ x 2 ⟩ + 2 D \frac{d\langle x\rangle}{dt} = -\langle x\rangle,
\qquad
\frac{d\langle x^2\rangle}{dt} = -2\langle x^2\rangle + 2D d t d ⟨ x ⟩ = − ⟨ x ⟩ , d t d ⟨ x 2 ⟩ = − 2 ⟨ x 2 ⟩ + 2 D です。実際 ∫ x ∂ x ( x f ) d x = − ∫ x f d x \int x\,\partial_x(xf)\,dx = -\int xf\,dx ∫ x ∂ x ( x f ) d x = − ∫ x f d x ∫ x ∂ x 2 f d x = 0 \int x\,\partial_x^2 f\,dx = 0 ∫ x ∂ x 2 f d x = 0 ∫ x 2 ∂ x ( x f ) d x = − 2 ⟨ x 2 ⟩ \int x^2\partial_x(xf)\,dx = -2\langle x^2\rangle ∫ x 2 ∂ x ( x f ) d x = − 2 ⟨ x 2 ⟩ ∫ x 2 ∂ x 2 f d x = 2 ∫ f d x = 2 \int x^2\partial_x^2f\,dx = 2\int f\,dx = 2 ∫ x 2 ∂ x 2 f d x = 2 ∫ f d x = 2 V = ⟨ x 2 ⟩ − ⟨ x ⟩ 2 V = \langle x^2\rangle - \langle x\rangle^2 V = ⟨ x 2 ⟩ − ⟨ x ⟩ 2 d V / d t = − 2 V + 2 D dV/dt = -2V + 2D d V / d t = − 2 V + 2 D ⟨ x ⟩ = X \langle x\rangle = X ⟨ x ⟩ = X V = σ 2 V = \sigma^2 V = σ 2
初期条件 f ( x , 0 ) = δ ( x − 1 ) f(x,0) = \delta(x-1) f ( x , 0 ) = δ ( x − 1 ) δ ( x − 1 ) = lim σ → 0 g ( x − 1 , σ ) \delta(x-1) = \lim_{\sigma\to 0}g(x-1,\sigma) δ ( x − 1 ) = lim σ → 0 g ( x − 1 , σ ) X ( 0 ) = 1 X(0) = 1 X ( 0 ) = 1 σ ( 0 ) → + 0 \sigma(0) \to +0 σ ( 0 ) → + 0 σ = 0 \sigma = 0 σ = 0 σ \sigma σ σ ( 0 ) = ε > 0 \sigma(0) = \varepsilon > 0 σ ( 0 ) = ε > 0 ε → + 0 \varepsilon \to +0 ε → + 0
s = σ 2 s = \sigma^2 s = σ 2 d s / d t = 2 ( D − s ) ds/dt = 2(D - s) d s / d t = 2 ( D − s ) s ( 0 ) = ε 2 s(0) = \varepsilon^2 s ( 0 ) = ε 2
s ( t ) = D + ( ε 2 − D ) e − 2 t → ε → + 0 σ 2 ( t ) = D ( 1 − e − 2 t ) s(t) = D + \left(\varepsilon^2 - D\right)e^{-2t}
\quad\xrightarrow{\ \varepsilon\to+0\ }\quad
\sigma^2(t) = D\left(1 - e^{-2t}\right) s ( t ) = D + ( ε 2 − D ) e − 2 t ε → + 0 σ 2 ( t ) = D ( 1 − e − 2 t ) です。X X X X ( t ) = X ( 0 ) e − t = e − t X(t) = X(0)e^{-t} = e^{-t} X ( t ) = X ( 0 ) e − t = e − t t > 0 t > 0 t > 0
f ( x , t ) = g ( x − e − t , D ( 1 − e − 2 t ) ) = 1 2 π D ( 1 − e − 2 t ) exp [ − ( x − e − t ) 2 2 D ( 1 − e − 2 t ) ] f(x,t) = g\Bigl(x - e^{-t},\ \sqrt{D\left(1-e^{-2t}\right)}\Bigr)
= \frac{1}{\sqrt{2\pi D\left(1-e^{-2t}\right)}}
\exp\left[-\frac{\left(x - e^{-t}\right)^2}{2D\left(1-e^{-2t}\right)}\right] f ( x , t ) = g ( x − e − t , D ( 1 − e − 2 t ) ) = 2 π D ( 1 − e − 2 t ) 1 exp [ − 2 D ( 1 − e − 2 t ) ( x − e − t ) 2 ] です。t > 0 t>0 t > 0 σ 2 ( t ) > 0 \sigma^2(t) > 0 σ 2 ( t ) > 0 x → ± ∞ x \to \pm\infty x → ± ∞ ∫ f d x = 1 \int f\,dx = 1 ∫ f d x = 1 t t t t → + 0 t \to +0 t → + 0 X → 1 X \to 1 X → 1 σ → 0 \sigma \to 0 σ → 0 φ \varphi φ ∫ f ( x , t ) φ ( x ) d x → φ ( 1 ) \int f(x,t)\varphi(x)dx \to \varphi(1) ∫ f ( x , t ) φ ( x ) d x → φ ( 1 ) f ( ⋅ , t ) → δ ( x − 1 ) f(\cdot,t) \to \delta(x-1) f ( ⋅ , t ) → δ ( x − 1 )
t → ∞ t \to \infty t → ∞ e − t → 0 e^{-t} \to 0 e − t → 0 e − 2 t → 0 e^{-2t}\to 0 e − 2 t → 0
f ( x , t ) ⟶ 1 2 π D e − x 2 / ( 2 D ) = g ( x , D ) f(x,t) \longrightarrow \frac{1}{\sqrt{2\pi D}}\,e^{-x^2/(2D)} = g\left(x,\sqrt D\right) f ( x , t ) ⟶ 2 π D 1 e − x 2 / ( 2 D ) = g ( x , D ) となり、設問(2)で求めた定常解（σ = D \sigma = \sqrt D σ = D x x x e − t e^{-t} e − t D e − 2 t De^{-2t} D e − 2 t O ( e − t ) O(e^{-t}) O ( e − t ) x = 1 x=1 x = 1 e − t e^{-t} e − t 0 0 0 D \sqrt D D
以上はガウス分布という形を仮定した議論なので、解がこれ以外にないことをフーリエ変換で確かめておきます。f ^ ( k , t ) = ∫ − ∞ ∞ f ( x , t ) e − i k x d x \hat f(k,t) = \int_{-\infty}^{\infty}f(x,t)e^{-ikx}dx f ^ ( k , t ) = ∫ − ∞ ∞ f ( x , t ) e − ik x d x ∂ f ^ / ∂ k = − i ∫ x f e − i k x d x \partial\hat f/\partial k = -i\int xfe^{-ikx}dx ∂ f ^ / ∂ k = − i ∫ x f e − ik x d x
∫ ∂ ∂ x ( x f ) e − i k x d x = i k ∫ x f e − i k x d x = − k ∂ f ^ ∂ k , ∫ D ∂ 2 f ∂ x 2 e − i k x d x = − D k 2 f ^ \int \frac{\partial}{\partial x}(xf)e^{-ikx}dx = ik\int xfe^{-ikx}dx = -k\frac{\partial \hat f}{\partial k},
\qquad
\int D\frac{\partial^2f}{\partial x^2}e^{-ikx}dx = -Dk^2\hat f ∫ ∂ x ∂ ( x f ) e − ik x d x = ik ∫ x f e − ik x d x = − k ∂ k ∂ f ^ , ∫ D ∂ x 2 ∂ 2 f e − ik x d x = − D k 2 f ^ なので、方程式は1階の偏微分方程式
∂ f ^ ∂ t + k ∂ f ^ ∂ k = − D k 2 f ^ \frac{\partial \hat f}{\partial t} + k\frac{\partial \hat f}{\partial k} = -Dk^2\hat f ∂ t ∂ f ^ + k ∂ k ∂ f ^ = − D k 2 f ^ になります。特性曲線 k ( t ) = k 0 e t k(t) = k_0e^{t} k ( t ) = k 0 e t d d t f ^ ( k 0 e t , t ) = − D k 0 2 e 2 t f ^ \frac{d}{dt}\hat f\bigl(k_0e^t,t\bigr) = -Dk_0^2e^{2t}\hat f d t d f ^ ( k 0 e t , t ) = − D k 0 2 e 2 t f ^ k 0 = k e − t k_0 = ke^{-t} k 0 = k e − t
f ^ ( k , t ) = f ^ ( k e − t , 0 ) exp [ − D k 2 2 ( 1 − e − 2 t ) ] \hat f(k,t) = \hat f\left(ke^{-t},0\right)\exp\left[-\frac{Dk^2}{2}\left(1-e^{-2t}\right)\right] f ^ ( k , t ) = f ^ ( k e − t , 0 ) exp [ − 2 D k 2 ( 1 − e − 2 t ) ] という一般解を得ます。初期条件 f ^ ( k , 0 ) = ∫ δ ( x − 1 ) e − i k x d x = e − i k \hat f(k,0) = \int\delta(x-1)e^{-ikx}dx = e^{-ik} f ^ ( k , 0 ) = ∫ δ ( x − 1 ) e − ik x d x = e − ik
f ^ ( k , t ) = exp [ − i k e − t − D k 2 2 ( 1 − e − 2 t ) ] \hat f(k,t) = \exp\left[-ike^{-t} - \frac{Dk^2}{2}\left(1-e^{-2t}\right)\right] f ^ ( k , t ) = exp [ − ik e − t − 2 D k 2 ( 1 − e − 2 t ) ] で、これは平均 e − t e^{-t} e − t D ( 1 − e − 2 t ) D(1-e^{-2t}) D ( 1 − e − 2 t ) f ( x , t ) f(x,t) f ( x , t ) t → ∞ t\to\infty t → ∞ f ^ → e − D k 2 / 2 \hat f \to e^{-Dk^2/2} f ^ → e − D k 2 /2 g ( x , D ) g(x,\sqrt D) g ( x , D )
出典: 東京大学大学院理学系研究科 物理学専攻 平成17年度 修士課程 入学試験問題 数学。問題文は要約して引用しています。