個体群動態の数理

(1)

個体群動態の数理

• 科目ナンバリングコード：2223011A3

• 開設科目名：個体群動態の数理

• 講義コード：4802000

• 開講期・曜日・時限・教室：前期水曜日１・２時限情報科学講義室（Ｇ３０２）

• 対象学生：3回生

奈良女子大学理学部・化学生物環境学科環境科学コース高須夫悟

(2)

平衡点の局所安定性解析

2 変数常微分方程式の平衡点の局所安定性解析

Lotka Volterra 競争モデルでは

平衡点 (n₁*, n₂*) は次式を満たす

時間的に変化しない状態を平衡状態（平衡点）と呼ぶ。

dn₁

dt = f₁(n₁, n₂)

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dn₂

dt = f₂(n₁, n₂)

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f₁(n₁, n₂) = r₁

✓

1 n₁ + ↵₁₂n₂ K₁

◆ n₁

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f₂(n₁, n₂) = r₂

✓

1 ↵₂₁n₁ + n₂ K₂

◆ n₂

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dn₁

dt = f₁(n^⇤₁, n^⇤₂) = 0

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dn₂

dt = f₂(n^⇤₁, n^⇤₂) = 0

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(3)

安定性解析

1

平衡点からの微小なずれを h₁(t), h₂(t) とする

n₁(t) = n₁* + h₁(t), n₂(t) = n₂* + h₂(t)

これを元の式に代入して、ずれ h₁, h₂ に関する式に書き直す

dn₁

dt = f₁(n₁, n₂)

dn₂

dt = f₂(n₁, n₂)

dn₁

dt = dh₁

dt = f₁(n^⇤₁ + h₁, n^⇤₂ + h₂)

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dn₂

dt = dh₂

dt = f₂(n^⇤₁ + h₁, n^⇤₂ + h₂)

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(4)

安定性解析

2

2 変数関数のテイラー展開をして、ずれが微少量であることから h₁, h₂ の

2 次以上の項を無視すると、次式を得る。

ベクトルと行列表示では右式となる

をヤコビ行列というヤコビ行列に平衡点の値を代入した行列をコミュニティ行列と呼ぶ

dh₁

dt = @f₁(n^⇤₁, n^⇤₂)

@n₁ h₁ + @f₁(n^⇤₁, n^⇤₂)

@n₂ h₂

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dh₂

dt = @f₂(n^⇤₁, n^⇤₂)

@n₁ h₁ + @f₂(n^⇤₁, n^⇤₂)

@n₂ h₂

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@f₁

@n₁

@f₁

@n₂

@f₂

@n1

@f₂

@n2

!

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d dt

✓ h₁ h₂

◆

=

@f₁(n^⇤₁,n^⇤₂)

@n1

@f₁(n^⇤₁,n^⇤₂)

@n2

@f2(n^⇤₁,n^⇤₂)

@n₁

@f2(n^⇤₁,n^⇤₂)

@n₂

! ✓ h₁ h₂

◆

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(5)

線型常微分方程式の解

線型常微分方程式は解ける！

行列 A は定数行列

解 h₁(t), h₂(t) は行列 A の固有値 λ₁, λ₂ を用いて次のように書ける

λ₁ ≠ λ₂ の時

λ1 = λ2 = λ の時

局所安定性解析では、h₁, h₂ が時間とともにゼロに収束するかしないかに注目

c_ij はいずれも定数（初期条件で決まる）

d dt

✓ h₁ h₂

◆

= A

✓ h₁ h₂

◆

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✓ h₁ h₂

◆

=

✓ c₁₁ c₂₁

◆

exp[ ₁t] +

✓ c₂₁ c₂₂

◆

exp[ ₂t]

<latexit sha1_base64="DqNjoOfFNtbR472a4AHSk6sIfec=">AAADM3ichVFNa9RAGH4Tv+r60dVehF6CS6UiLJNYUASl0IvHfrhtcWcJk9nZ3aGzSZjMLq4hf6B/oIeeKoiUnvsLeunJm4fiTbyIeKrgxYNvsoGibe2EzDzzzPs888xMECuZGEKOLPvS5StXr01cr9y4eev2ZPXO3dUkGmguGjxSkV4PWCKUDEXDSKPEeqwF6wdKrAUbC/n62lDoREbhKzOKRavPuqHsSM4MUn71I1WiY2ZpILoyTJnWbJSlPOv5rkOp0/M9KsJ2yVMtuz3z0HnuVM5WcT913SwXIvLc7AwtFW/iJlUYsM1wD9N6dL6Xd+LlXejloZfj+NUaqZOiOaeBW4IalG0xqn4ACm2IgMMA+iAgBINYAYMEvya4QCBGrgUpchqRLNYFZFBB7QCrBFYwZDew7+KsWbIhznPPpFBz3EXhr1HpwAz5RHbJMTkke+Qb+X2uV1p45FlGOAZjrYj9yc17K78uVPVxNNA7Uf03s4EOPC2ySsweF0x+Cj7WD99uHa88W55JH5B35Dvm3yFH5ABPEA5/8vdLYnkbKvgA7r/XfRqsenX3cd1bmqvNvyifYgKm4T7M4n0/gXl4CYvQAG7NWa8tbrXtffuz/cX+Oi61rVIzBX81+8cfHSfYsw==</latexit>

✓ h₁ h₂

◆

=

✓ c₁₁ c₂₁

◆

exp[ t] +

✓ c₂₁ c₂₂

◆

texp[ t]

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(6)

固有値と局所安定性

固有値が実数の時（2 つの固有値 λ₁, λ₂ は共に実数）：

λ₁ < 0 かつ λ₂ < 0 の時、ずれ h₁, h₂ はゼロに収束

λ₁, λ₂ のいずれかが正の時、ずれ h₁, h₂ は発散（線形近似が成立しなくなる）

λ1 ≠ λ2 の時

（λ₁ = λ₂でも同じ議論が成り立つ）

平衡点からの微小なずれ h₁, h₂ の時間変化は線形近似により次で与えられた

✓ h₁ h₂

◆

=

✓ c₁₁ c₂₁

◆

exp[ ₁t] +

✓ c₂₁ c₂₂

◆

exp[ ₂t]

(7)

固有値が複素数の時（2 つの固有値は複素共役 λ₁, λ₂ = a ± b i ）：

固有値の実部 a = Re λ₁ = Re λ₂ が負の時、ずれ h₁, h₂ はゼロに収束

固有値の実部 a が正の時、ずれ h₁, h₂ は発散（線形近似が成立しなくなる）

λ₁ ≠ λ₂ の時

（λ₁ = λ₂でも同じ議論が成り立つ）

✓ h₁ h₂

◆

=

✓ c₁₁ c₂₁

◆

exp[ ₁t] +

✓ c₂₁ c₂₂

◆

exp[ ₂t]

exp[ix] = cosx + i sinx

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(8)

まとめ

一般に多変数の常微分方程式の平衡点の安定性は、コミュニティ行列の固有値で決まる。

平衡点からのずれに関する線型微分方程式

コミュニティ行列 A のすべての固有値 λ に対して

Re λ < 0 であれば、平衡点は局所安定

Re λ > 0 となる固有値が 1 つでも存在すれば、平衡点は不安定平衡点の近傍で

線形近似

固有値が複素数であれば平衡点の近傍で振動が起こる

dn₁

dt = f₁(n₁, n₂)

dn₂

dt = f₂(n₁, n₂)

d dt

✓ h₁ h₂

◆

= A

✓ h₁ h₂

◆

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(9)

Lotka Volterra

の競争モデル

ヤコビ行列 J は

平衡点は 4 つ存在

(0, 0), (K₁, 0), (0, K₂),

✓K₁ ↵₁₂K₂

1 ↵₁₂↵₂₁ , K₂ ↵₂₁K₁ 1 ↵₁₂↵₂₁

◆

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J =

@f₁

@n₁

@f₁

@n₂

@f₂

@n₁

@f₂

@n₂

!

= r₁ ^K¹ ²ⁿ_K¹ ^↵¹²ⁿ²

1 r₁^↵¹²_Kⁿ¹

1

r₂ ^↵²¹_Kⁿ²

2 r₂^K² ^↵²¹_Kⁿ¹ ²ⁿ²

2

!

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dn₁

dt = r₁

✓

1 n₁ + ↵12n₂ K₁

◆

n₁ = f₁

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dn₂

dt = r₂

✓

1 ↵₂₁n₁ + n₂ K₂

◆

n₂ = f₂

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(10)

平衡点

(0, 0)

平衡点 (n₁*, n₂*) = (0, 0) をヤコビ行列に代入すると、コミュニティ行列は

固有値は λ = r₁, r₂ > 0

より

2 つの固有値は実数であり、共に正なので平衡点 (0, 0) は不安定。

A =

✓ r₁ 0 0 r₂

◆

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| I A| = r₁ 0

0 r₂ = ( r₁)( r₂) = 0

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(11)

平衡点

( K

1

, 0)

平衡点 (n₁*, n₂*) = (K₁, 0) をヤコビ行列に代入すると、コミュニティ行列は

固有値は λ = –r₁< 0, (1–α₂₁ K₁/K₂)r₂

平衡点 (K₁, 0) は不安定 K₂ > α₂₁ K₁ の時

平衡点 (K₁, 0) は局所的に安定 K₂ < α₂₁ K₁ の時

A = r₁ ↵₁₂r₁ 0 r₂ ⇣

1 ^↵²¹_K^K¹

2

⌘

!

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(12)

平衡点

(0, K

2

)

平衡点 (n₁*, n₂*) = (0, K₂) をヤコビ行列に代入すると、コミュニティ行列は

固有値は λ = –r₂< 0, (1– α₁₂ K₂/K₁)r₁

平衡点 (0, K₂) は不安定 K₁ > α₁₂ K₂ の時

平衡点 (0, K₂) は局所的に安定 K₁ < α₁₂ K₂ の時

A = r₁ ⇣

1 ^↵¹²_K^K²

1

⌘ 0 r₂↵₂₁ r₂

!

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(13)

内部平衡点

平衡点 (n₁*, n₂*) = をヤコビ行列に代入すると、

コミュニティ行列 A を得る

行列 A の固有値 λ の実部の符号で安定性が決まる

✓K₁ ↵₁₂K₂

1 ↵₁₂↵₂₁ , K₂ ↵₂₁K₁ 1 ↵₁₂↵₂₁

◆

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A = r₁ ₍₁^K¹_↵ ^↵¹²^K²

12↵₂₁)K₁ r₁₍₁ ^K_↵¹^+↵¹²^K²

12↵₂₁)K₁

r₂ ₍₁^↵²¹_↵^K¹ ^K²

12↵₂₁)K₂ r₂₍₁^↵²¹_↵^K¹ ^K²

12↵₂₁)K₂

!

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(14)

T = trace A = a + d D = det A = ad – bc

行列の固有値（実数を含む）の実部が負であるための

必要十分条件は

T < 0 かつ D > 0

内部平衡点が正、かつ K₁ > α₁₂ K₂, K₂ > α₂₁ K₁ の時、上記の条件を満たす

内部平衡点は局所的に安定 Lotka Volterra 競争モデルの

A =

✓ a b c d

◆

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(15)

具体例

Saccharomyces cerevisiae : r₁ = 0.218 h^–1, K₁ = 13.0, α₁₂ = 3.15 Schizosaccharomyces kephir : r₂ = 0.061 h^–1, K₂ = 5.8, α₂₁ = 0.439

K₁ / α₁₂ = 4.127 < K₂ = 5.8 K₂ / α₂₁= 13.212 > K₁ =13.0

Saccharomyces は絶滅する。もしくは、

単独培養 pure culture のデータからパラメータ値を推定

初期値に依存してどちらかが絶滅する可能性

(16)

変数のスケール変換

ダイナミクス（微分方程式・差分式等）の解析においては、変数を適当にスケール変換してから、問題に取り組む方が見通しが良い。

例えば、Lotka Volterra の競争モデル

元の変数とパラメータを定数倍しただけなので、

定性的な性質は保持されている。

パラメータ数が少ない分、計算間違いが減る。

解析の見通しが良い。

dn₂

dt = r₂

✓

1 ↵₂₁n₁ + n₂ K₂

◆ n₂

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dn₁

dt = r₁

✓

1 n₁ + ↵₁₂n₂ K₁

◆ n₁

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n₁

K₁ ! u₁

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n₂

K₂ ! u₂

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r<latexit sha1_base64="Khulaxxmhs11htnWbX21JJOmrlg=">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</latexit> ₁t ! ⌧

↵₁₂ K₂

K₁ ! ¹²

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↵₂₁ K₁

K₂ ! ²¹

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du₁

d⌧ = (1 u₁ ₁₂u₂)u₁

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du₂

d⌧ = ⇢(1 ₂₁u₁ u₂)u₂

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(17)

問題

1

次の微分方程式の解を求めよ。初期値は x(0) = x₀ とする。

1)

2)

3)

(18)

問題

2

次の Lotka Volterra 競争モデルについて、設問に答えよ

1）平衡点をすべて求めよ

2）すべての平衡点の局所安定性を判定せよ

3）相平面解析により、系の振る舞いを視覚的に調べよ 4）数値計算により、系の振る舞いを概観せよ

dn₁

dt = 0.5

✓

1 n₁ + 0.5n₂ 90

◆ n₁

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dn₂ dt =

✓

1 n₁ + n₂ 200

◆ n₂

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(19)

問題

3

T = trace A = a + d, D = det A = ad – bc

行列の固有値（実数を含む）の実部が負であるための

必要十分条件は T < 0 かつ D > 0 であることを示せ（a, b, c, d は実数とする）。

固有値が実数・複素数の場合に分けて考える。

A =

✓ a b c d

◆

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