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# Igor_pro_ODE_japanese_ver2.0

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### Igor_pro_ODE_japanese_ver2.0

1. 1. 131208 ver 2.0 IGOR PROの使い方（解析編） 常微分方程式（ODE; Ordinary Differential Equation） Presented by Satoshi Kume, Ph.D. 左の反応モデルの連立微分方程式 反応モデル① A k1 ⎧ d[A] ⎪ dt = −k1[A] ⎨ ⎪ d[B] = k [A] 1 ⎩ dt B 1.0 反応モデル①のスクリプト 0.8 0.6 € [A] [B] € 0.4 0.2 0 0 20 40 60 80 [A]=1, [B]=0, k1=0.05 €
2. 2. 反応モデル② A k1 上記反応モデルの連立微分方程式 k2 B ⎧ d[A] ⎪ dt = −k1[A] ⎪ ⎪ d[B] ⎨ = k1[A] − k 2 [B] dt ⎪ ⎪ d[C] ⎪ dt = k2 [B] ⎩ C 1.0 反応モデル②のスクリプト 0.8 € € 0.6 [A] [B] [C] 0.4 0.2 0 0 20 40 60 80 [A]=1, [B]=0, [C]=0, k1=0.06, k2=0.04 €
3. 3. 反応モデル③ A k1 B k2 上記反応モデルの連立微分方程式 k3 C D 1.0 € 0.8 € € [A] [B] [C] [D] 0.6 0.4 0.2 0 0 50 100 150 [A]=1, [B]=0, [C]=0, [D]=0, k1=0.06, k2=0.04 k3=0.04 ⎧ d[A] ⎪ dt = −k1[A] ⎪ ⎪ d[B] = k [A] − k [B] 1 2 ⎪ dt ⎨ ⎪ d[C] = k [B] − k [C] 2 3 ⎪ dt ⎪ d[C] ⎪ = k3 [C] ⎩ dt 反応モデル③のスクリプト €
4. 4. 反応モデル④ A + B k1 k-1 左の反応モデルの連立微分方程式 k2 C D 1.0 ⎧ d[A] d[B] ⎪ dt = dt = −k1[A][B] + k−1[C] ⎪ ⎪ d[C] ⎨ = k1[A][B] − k−1[C] − k 2 [C] ⎪ dt ⎪ d[D] ⎪ dt = k2 [C] ⎩ 0.8 反応モデル④のスクリプト € 0.6 € € [A] or [B] [C] [D] 0.4 0.2 0 0 50 100 150 [A]=1, [B]=1, [C]=0, [D]=0, k1=0.1, k-1=0.05, k2=0.03 €
5. 5. 反応モデル⑤ A + B k1 k-1 上記反応モデルの連立微分方程式 k2 C k-2 D 1.0 [A] or [B] [C] [D] 0.8 € € € 0.6 反応モデル⑤のスクリプト € 0.4 0.2 0 0 50 100 ⎧ d[A] d[B] ⎪ dt = dt = −k1[A][B] + k−1[C] ⎪ ⎪ d[C] ⎨ = k1[A][B] − k−1[C] − k 2 [C] + k−2 [D] ⎪ dt ⎪ d[D] ⎪ dt = k2 [C] − k −2 [D] ⎩ 150 [A]=1, [B]=1, [C]=0, [D]=0, k1=0.12, k-1=0.06, k2=0.05, k-2=0.02
6. 6. € € 反応モデル⑥ 上記反応モデルの連立微分方程式 k1 ⎧ d[A] ⎪ dt = −k1[A] − k 2 [A] + k−1[B] + k −2 [DA ] ⎪ ⎪ d[B] = k [A] − k [B] − k [B] 1 −1 2 ⎪ dt ⎨ ⎪ d[DA ] = k [A] − k [D ] − k [D ] 2 −2 A 3 A ⎪ dt ⎪ d[D ] B ⎪ = k2 [B] + k 3 [DA ] ⎩ dt A k-2 k-1 k2 DA € 1.0 B k2 k3 DB 0.8 [A] [B] [DA] [DB] 0.6 € 0.4 反応モデル⑥のスクリプト € 0.2 0 0 20 40 60 80 [A]=1, [B]=0, [DA]=0, [DB]=0, k1=0.08, k-1=0.04, k2=0.08, k-2=0.5, k3=0.3
7. 7. € 酵素化学への常微分方程式の応用 酵素反応 不可逆モデル E+S € k1 k-1 ES k2 左の反応モデルの連立微分方程式 E+ P 酵素反応 可逆モデル E+S € k1 k-1 € ES 左の反応モデルの連立微分方程式 € k2 k-2 ⎧ d[S] ⎪ dt = −k1[E][S] + k −1[ES] ⎪ ⎪ d[E] = −k [E][S] + ( k + k )[ES] 1 −1 2 ⎪ dt ⎨ ⎪ d[ES] = k [E][S] − ( k + k )[ES] 1 −1 2 ⎪ dt ⎪ d[P] ⎪ = k2 [ES] ⎩ dt E+ P ⎧ d[S] ⎪ dt = −k1[E][S] + k −1[ES] ⎪ ⎪ d[E] = −k [E][S] + ( k + k )[ES] − k [E][P] 1 −1 2 −2 ⎪ dt ⎨ ⎪ d[ES] = k [E][S] − ( k + k )[ES] + k [E][P] 1 −1 2 −2 ⎪ dt ⎪ d[P] ⎪ = k2 [ES] − k −2 [E][P] ⎩ dt
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