Logistic Models with Differential Equations · 微分方程中的逻辑斯模型
| English | 中文 | Pinyin · 拼音 |
|---|---|---|
| logistic/ləˈdʒɪstɪk/ | 逻辑斯蒂 | luó jí sī dì |
| carrying capacity/ˈkæriɪŋ kəˈpæsɪti/ | 环境容纳量 | huán jìng róng nà liàng |
Growth that hits a ceiling
- Pure exponential growth rises forever — unrealistic for real populations with limited resources.
- The logistic 逻辑斯蒂 model fixes this: growth that slows as it nears a maximum.
- It starts nearly exponential, then levels off toward a carrying capacity 环境容纳量 $L$.
- It's the standard model for constrained growth.
会撞上"天花板"的增长
- 纯指数增长永远上升——对资源有限的真实种群不现实。
- 逻辑斯蒂模型修正了这点:增长在接近最大值时放缓。
- 它开始时几乎指数,然后向一个承载量 $L$ 趋平。
- 它是受限增长的标准模型。
The logistic differential equation
-
$$\dfrac{dP}{dt}=kP\Big(1-\dfrac{P}{L}\Big)$$
- The $kP$ part is exponential-like growth; the $\big(1-\tfrac{P}{L}\big)$ factor brakes it as $P$ nears $L$.
- When $P$ is small, $1-\tfrac{P}{L}\approx1$: nearly exponential. When $P\approx L$, the factor $\to0$: growth stops.
- $L$ is the carrying capacity — the population's ceiling.
逻辑斯蒂微分方程
-
$$\dfrac{dP}{dt}=kP\Big(1-\dfrac{P}{L}\Big)$$
- $kP$ 部分是类指数增长;$\big(1-\tfrac{P}{L}\big)$ 因子在 $P$ 接近 $L$ 时刹车。
- 当 $P$ 小时,$1-\tfrac{P}{L}\approx1$:几乎指数。当 $P\approx L$,因子 $\to0$:增长停止。
- $L$ 是承载量——种群的上限。
The logistic differential equation is $\dfrac{dP}{dt}=$ · 逻辑斯微分方程为 $\dfrac{dP}{dt}=$
The braking factor $(1-P/L)$ slows growth near $L$. · 制动因子 $(1-P/L)$ 减缓了靠近 $L$ 时的增长。
The maximum sustainable value $L$ that $P$ approaches is the ____ capacity. · 最大可持续值 $L$ 即 $P$ 接近于 ____ 容量。
The carrying capacity is the ceiling. · 承载能力是上限。
Long-term behavior
- As $t\to\infty$, $P$ approaches the carrying capacity $L$ (for $0
- $P=L$ is a stable equilibrium; $P=0$ is an unstable one.
- Below $L$ the population rises toward $L$; above $L$ it falls back to $L$.
- So $L$ is the value the model settles at, regardless of the (positive) start.
长期行为
- 当 $t\to\infty$,$P$ 趋近承载量 $L$(对 $0
- $P=L$ 是稳定平衡;$P=0$ 是不稳定平衡。
- 在 $L$ 以下,种群升向 $L$;在 $L$ 以上,回落到 $L$。
- 所以 $L$ 是模型安定到的值,与(正的)起点无关。
The S-shaped logistic curve · S形逻辑曲线
y = a·tanh(bx) + d (S-shape) · y = a·tanh(bx) + d (S形)
A logistic curve rises fast near the middle and flattens toward the carrying capacity $L$ — steepest at $P=L/2$. · 逻辑斯曲线在中部附近快速上升，并趋近于承载能力 $L$ —— 最陡处位于 $P=L/2$。
For · 支持 $\dfrac{dP}{dt}=0.1P(1-\tfrac{P}{2000})$, what is the carrying capacity $L$? · 对于 $\dfrac{dP}{dt}=0.1P(1-\tfrac{P}{2000})$，承载能力 $L$ 是多少？
$L=2000$, the long-term limit. · $L=2000$，长期极限。
As $t\to\infty$ (with $0
$P\to L$ from below. · 从下方趋近 $P\to L$。
Fastest growth at half capacity
- The population grows fastest when $P=\dfrac{L}{2}$ — the midpoint of the S-curve.
- There the logistic curve has its inflection point: growth accelerates below $L/2$, decelerates above.
- Before $L/2$ the rate is climbing; after, it's easing off toward zero.
- This half-capacity peak is a favorite exam fact.
半承载量处增长最快
- 当 $P=\dfrac{L}{2}$ 时种群增长最快——S 形曲线的中点。
- 那里逻辑斯蒂曲线有它的拐点:低于 $L/2$ 增长加速,高于则减速。
- $L/2$ 之前速率在攀升;之后向零趋缓。
- 这个半承载量峰值是考试爱考的事实。
For the same model ($L=2000$), at what population is growth fastest? · 对于同一模型（$L=2000$），种群数量在何处增长最快？
Fastest at $P=L/2=1000$. · 在 $P=L/2=1000$ 时最快。
Growth is fastest exactly at the carrying capacity $P=L$. · 增长恰好发生在承载能力 $P=L$ 处最快。
At $P=L$ growth stops; the fastest rate is at $P=L/2$. · 在 $P=L$ 处增长停止；最快速度出现在 $P=L/2$。
The carrying capacity $L$ is where growth stops ($\frac{dP}{dt}=0$ because the $\big(1-\tfrac{P}{L}\big)$ factor is $0$), not where it's fastest. The maximum rate is at $P=\tfrac{L}{2}$. Don't confuse the two: $L$ is the ceiling the population approaches; $\tfrac L2$ is where it's climbing quickest.
承载量 $L$ 是增长停止之处($\frac{dP}{dt}=0$,因为 $\big(1-\tfrac{P}{L}\big)$ 因子为 $0$),而非最快之处。最大速率在 $P=\tfrac{L}{2}$。别把两者搞混:$L$ 是种群趋近的上限;$\tfrac L2$ 是它爬升最快之处。
A fish population obeys $\dfrac{dP}{dt}=0.1P\big(1-\tfrac{P}{2000}\big)$.
- Carrying capacity: $L=2000$ fish — the long-term limit.
- Fastest growth at $P=\tfrac{L}{2}=1000$ fish.
- As $t\to\infty$, $P\to 2000$ from below (if it started under $2000$).
一个鱼群服从 $\dfrac{dP}{dt}=0.1P\big(1-\tfrac{P}{2000}\big)$。
- 承载量:$L=2000$ 条鱼——长期极限。
- 增长最快在 $P=\tfrac{L}{2}=1000$ 条鱼。
- 当 $t\to\infty$,$P$ 从下方 $\to 2000$(若起始低于 $2000$)。
The logistic model $\frac{dP}{dt}=kP\big(1-\frac{P}{L}\big)$ describes growth limited by a carrying capacity $L$: near-exponential when small, leveling off toward $L$ as $t\to\infty$. Growth is fastest at $P=\tfrac L2$ (the inflection point), and stops at $P=L$.
逻辑斯蒂模型 $\frac{dP}{dt}=kP\big(1-\frac{P}{L}\big)$ 描述受承载量 $L$ 限制的增长:小时几乎指数,随 $t\to\infty$ 向 $L$ 趋平。增长在 $P=\tfrac L2$(拐点)处最快,在 $P=L$ 处停止。