비선형 1계 미분방정식의 경계의 직선화
📂편미분방정식 비선형 1계 미분방정식의 경계의 직선화 빌드업 비선형 1계 미분 방정식 의 특성 방정식 을 쉽게 풀기 위한 방법 중 하나는 정의역 Ω \Omega Ω ∂ Ω \partial \Omega ∂ Ω Γ \Gamma Γ x 0 x^{0} x 0 경계의 직선화 straightening the boundary 라고 한다.
Ω ⊂ R n \Omega \subset \mathbb{R}^{n} Ω ⊂ R n 열린 집합 이고 ∂ Ω \partial \Omega ∂ Ω C 2 C^{2} C 2 편미분방정식 F ∈ C 1 ( R n × R × Ω ˉ ) F \in C^{1}(\mathbb{R}^{n} \times \mathbb{R} \times \bar \Omega) F ∈ C 1 ( R n × R × Ω ˉ ) 경계조건 이 주어졌다고 하자.
{ F ( D u , u , x ) = 0 in Ω u = g on Γ
\begin{equation}
\left\{
\begin{aligned}
F(Du,\ u,\ x)&=0 && \text{in } \Omega
\\ u&=g && \text{on } \Gamma
\end{aligned}
\right.
\end{equation}
{ F ( D u , u , x ) u = 0 = g in Ω on Γ
이때 Γ ⊂ ∂ Ω \Gamma \subset \partial \Omega Γ ⊂ ∂ Ω g : Γ → R g : \Gamma \to \mathbb{R} g : Γ → R
설명 경계위의 고정된 점 x 0 ∈ ∂ Ω x^{0} \in \partial \Omega x 0 ∈ ∂ Ω Φ : R n → R n \Phi\ :\ \mathbb{R}^{n} \to \mathbb{R}^{n} Φ : R n → R n
{ Φ ( Ω ) : = V Φ ( x ) : = ( Φ 1 ( x ) , ⋯ , Φ n ( x ) ) = ( y 1 , ⋯ , y n ) = y , y ∈ V y i = Φ i ( x ) : = x i , x ∈ R n ( i = 1 , ⋯ , n − 1 ) y n = Φ n ( x ) : = x n − γ ( x 1 , ⋯ , x n − 1 ) , x ∈ R n
\left\{
\begin{align*}
\Phi (\Omega) &:= V \\
\Phi (x) &:= \left( \Phi^{1}(x),\ \cdots, \Phi^{n}(x) \right) = (y_{1},\ \cdots,\ y_{n})=y, \quad y \in V \\
y_{i}=\Phi^{i}(x) &:= x_{i}, \quad x \in \mathbb{R}^{n}\ (i=1,\cdots, n-1) \\
y_{n}=\Phi^{n}(x) &:= x_{n}-\gamma (x_{1},\cdots,x_{n-1}), \quad x \in \mathbb{R}^{n}
\end{align*}
\right.
⎩ ⎨ ⎧ Φ ( Ω ) Φ ( x ) y i = Φ i ( x ) y n = Φ n ( x ) := V := ( Φ 1 ( x ) , ⋯ , Φ n ( x ) ) = ( y 1 , ⋯ , y n ) = y , y ∈ V := x i , x ∈ R n ( i = 1 , ⋯ , n − 1 ) := x n − γ ( x 1 , ⋯ , x n − 1 ) , x ∈ R n
즉 Φ \Phi Φ n n n 0 0 0 Ψ \Psi Ψ
{ l Ψ ( y ) : = Φ − 1 ( y ) Ψ i ( y ) = x i = y i , y ∈ R n ( i = 1 , ⋯ , n − 1 ) Ψ n ( y ) = x n = y n + γ ( x 1 , ⋯ , x n − 1 ) , x ∈ R n
\left\{
\begin{align*}
{l}\Psi (y) &:= \Phi^{{}-1}(y) \\
\Psi^{i}(y) &= x_{i}=y_{i}, \quad y \in \mathbb{R}^{n}\ (i=1,\cdots, n-1) \\
\Psi^{n}(y) &= x_{n}=y_{n}+\gamma (x_{1},\cdots,x_{n-1}), \quad x \in \mathbb{R}^{n}
\end{align*}
\right.
⎩ ⎨ ⎧ l Ψ ( y ) Ψ i ( y ) Ψ n ( y ) := Φ − 1 ( y ) = x i = y i , y ∈ R n ( i = 1 , ⋯ , n − 1 ) = x n = y n + γ ( x 1 , ⋯ , x n − 1 ) , x ∈ R n
이를 그림으로 나타내면 아래와 같다.
이제 Γ ⊂ ∂ Ω \Gamma \subset \partial \Omega Γ ⊂ ∂ Ω g ∈ C ( Γ ) g\in C(\Gamma) g ∈ C ( Γ ) x 0 ∈ Γ x^{0} \in \Gamma x 0 ∈ Γ u ∈ C 1 ( Ω ) ∩ C ( Ω ˉ ) u \in C^{1} (\Omega)\cap C(\bar \Omega) u ∈ C 1 ( Ω ) ∩ C ( Ω ˉ ) ( 1 ) (1) ( 1 ) v v v
v ( y ) : = u ( Ψ ( y ) ) ∀ y ∈ V
v(y) := u(\Psi (y)) \quad \forall\ y\in V
v ( y ) := u ( Ψ ( y )) ∀ y ∈ V
다시 말해 V V V u u u v v v
u ( x ) = v ( Φ ( x ) ) ∀ x ∈ Ω
u(x)=v(\Phi (x)) \quad \forall\ x\in \Omega
u ( x ) = v ( Φ ( x )) ∀ x ∈ Ω
이제 D u , u , x Du, u, x D u , u , x V V V u x i u_{x_{i}} u x i
u x i ( x ) = ∑ k = 1 n v y k ( Φ ( x ) ) Φ x i k ( x )
u_{x_{i}}(x)=\sum \limits _{k=1}^{n} v_{y_{k}} \left( \Phi (x) \right) \Phi^{k}_{x_{i}}(x)
u x i ( x ) = k = 1 ∑ n v y k ( Φ ( x ) ) Φ x i k ( x )
따라서
D u ( x ) = ( ∑ k = 1 n v y k ( Φ ( x ) ) Φ x 1 k ( x ) , ⋯ , ∑ k = 1 n v y k ( Φ ( x ) ) Φ x n k ( x ) ) = ( v y 1 Φ x 1 1 + ⋯ + v y n Φ x 1 n , ⋯ , v y 1 Φ x n 1 + ⋯ + v y n Φ x n n ) = ( v y 1 v y 2 ⋯ v y n ) ( Φ x 1 1 Φ x 2 1 ⋯ Φ x n 1 Φ x 1 2 Φ x 2 2 ⋯ Φ x n 2 ⋮ ⋮ ⋱ ⋮ Φ x 1 n Φ x 2 n ⋯ Φ x n n ) = D v ( Φ ( x ) ) D Φ ( x )
\begin{align*}
Du(x) &= \left( \sum \limits _{k=1}^{n} v_{y_{k}} \left( \Phi (x) \right) \Phi^{k}_{x_{1}}(x),\ \cdots,\ \sum \limits _{k=1}^{n} v_{y_{k}} \left( \Phi (x) \right) \Phi^{k}_{x_{n}}(x) \right)
\\ &= (v_{y_{1}}\Phi^{1}_{x_{1}}+\cdots+v_{y_{n}}\Phi^{n}_{x_{1}},\ \cdots ,\ v_{y_{1}}\Phi^{1}_{x_{n}}+\cdots+v_{y_{n}}\Phi^{n}_{x_{n}} )
\\ &= \begin{pmatrix}
v_{y_{1}} & v_{y_{2}} & \cdots & v_{y_{n}}
\end{pmatrix} \begin{pmatrix}
\Phi^{1}_{x_{1}} & \Phi^{1}_{x_{2}} & \cdots &\Phi^{1}_{x_{n}}
\\ \Phi^{2}_{x_{1}} & \Phi^{2}_{x_{2}} & \cdots & \Phi^{2}_{x_{n}}
\\ \vdots & \vdots & \ddots & \vdots
\\ \Phi^{n}_{x_{1}} & \Phi^{n}_{x_{2}} & \cdots & \Phi^{n}_{x_{n}}
\end{pmatrix}
\\ &= Dv\left( \Phi (x) \right) D\Phi (x)
\end{align*}
D u ( x ) = ( k = 1 ∑ n v y k ( Φ ( x ) ) Φ x 1 k ( x ) , ⋯ , k = 1 ∑ n v y k ( Φ ( x ) ) Φ x n k ( x ) ) = ( v y 1 Φ x 1 1 + ⋯ + v y n Φ x 1 n , ⋯ , v y 1 Φ x n 1 + ⋯ + v y n Φ x n n ) = ( v y 1 v y 2 ⋯ v y n ) Φ x 1 1 Φ x 1 2 ⋮ Φ x 1 n Φ x 2 1 Φ x 2 2 ⋮ Φ x 2 n ⋯ ⋯ ⋱ ⋯ Φ x n 1 Φ x n 2 ⋮ Φ x n n = D v ( Φ ( x ) ) D Φ ( x )
혹은
D u ( Ψ ( y ) ) = D v ( y ) D Φ ( Ψ ( y ) )
Du(\Psi (y)) = Dv(y) D\Phi (\Psi (y))
D u ( Ψ ( y )) = D v ( y ) D Φ ( Ψ ( y ))
그러면 ( 1 ) (1) ( 1 )
F ( D u ( Ψ ( y ) ) , u ( Ψ ( y ) ) , Ψ ( y ) ) = F ( D v ( y ) D Φ ( Ψ ( y ) ) , v ( y ) , Ψ ( y ) ) = 0
F\Big( Du(\Psi (y) ), u( \Psi (y) ), \Psi (y) \Big) = F\Big( Dv(y)D\Phi (\Psi (y)), v(y), \Psi (y) \Big)=0
F ( D u ( Ψ ( y )) , u ( Ψ ( y )) , Ψ ( y ) ) = F ( D v ( y ) D Φ ( Ψ ( y )) , v ( y ) , Ψ ( y ) ) = 0
이제 다음과 같이 비선형 1계 편미분 방정식을 정의하자.
G ( q , w , y ) : = F ( q D Φ ( Ψ ( y ) ) , w , Ψ ( y ) ) , ∀ ( q , w , y ) ∈ R n × R × V ˉ
G(q, w, y):=F \Big (q D\Phi (\Psi (y) ), w, \Psi (y) \Big), \quad \forall (q, w, y)\in\mathbb{R}^{n}\times\mathbb{R}\times\bar V
G ( q , w , y ) := F ( q D Φ ( Ψ ( y )) , w , Ψ ( y ) ) , ∀ ( q , w , y ) ∈ R n × R × V ˉ
그러면 G ∈ C 1 ( R n × R × V ˉ ) G\in C^{1}(\mathbb{R}^{n} \times \mathbb{R} \times \bar V) G ∈ C 1 ( R n × R × V ˉ )
G ( D v ( y ) , v ( y ) , y ) = 0 , ∀ y ∈ V
G \Big( Dv(y),\ v(y),\ y \Big)=0, \quad \forall y\in V
G ( D v ( y ) , v ( y ) , y ) = 0 , ∀ y ∈ V
그리고 Δ : = Φ ( Γ ) \Delta:=\Phi (\Gamma) Δ := Φ ( Γ ) h ( y ) : = g ( Φ ( y ) ) y ∈ Δ h(y):=g(\Phi (y)) \quad y \in \Delta h ( y ) := g ( Φ ( y )) y ∈ Δ Δ \Delta Δ Δ ⊂ ∂ V \Delta \subset \partial V Δ ⊂ ∂ V Δ \Delta Δ y 0 y^{0} y 0 v ∈ C 1 ( V ) ∩ C ( V ˉ ) v \in C^{1}(V) \cap C(\bar V) v ∈ C 1 ( V ) ∩ C ( V ˉ )
{ G ( D v , v , y ) = 0 in V v = h on Δ ⊂ ∂ V
\left\{
\begin{align*}
G(Dv,\ v,\ y) &= 0 && \text{in } V \\
v &= h && \text{on } \Delta \subset \partial V
\end{align*}
\right.
{ G ( D v , v , y ) v = 0 = h in V on Δ ⊂ ∂ V
이는 ( 1 ) (1) ( 1 )
