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some typos.
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Fang Liu
2023-06-04 16:24:38 +08:00
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quarto/ODEs/differential_equations.qmd
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@@ -16,7 +16,7 @@ using ModelingToolkit
---
The [`DifferentialEquations`](https://github.com/SciML/DifferentialEquations.jl) suite of packages contains solvers for a wide range of various differential equations. This section just briefly touches touch on ordinary differential equations (ODEs), and so relies only on `OrdinaryDiffEq` part of the suite. For more detail on this type and many others covered by the suite of packages, there are many other resources, including the [documentation](https://diffeq.sciml.ai/stable/) and accompanying [tutorials](https://github.com/SciML/SciMLTutorials.jl).
The [`DifferentialEquations`](https://github.com/SciML/DifferentialEquations.jl) suite of packages contains solvers for a wide range of various differential equations. This section just briefly touches on ordinary differential equations (ODEs), and so relies only on `OrdinaryDiffEq` part of the suite. For more detail on this type and many others covered by the suite of packages, there are many other resources, including the [documentation](https://diffeq.sciml.ai/stable/) and accompanying [tutorials](https://github.com/SciML/SciMLTutorials.jl).
## SIR Model
@@ -266,7 +266,7 @@ $$
\frac{di}{ds} = \frac{di/dt}{ds/dt} = \frac{b \cdot s(t) \cdot i(t) - k \cdot i(t)}{-b \cdot s(t) \cdot i(t)} = -1 + \frac{1}{c \cdot s}
$$
This equation does not depend on $t$; $s$ is the dependent variable. It could be solved numerically, but in this case affords an algebraic solution: $i = -s + (1/c) \log(s) + q$, where $q$ is some constant. The quantity $q = i + s - (1/c) \log(s)$ does not depend on time, so is the same at time $t=0$ as it is as $t \rightarrow \infty$. At $t=0$ we have $s(0) \approx 1$ and $i(0) \approx 0$, whereas $t \rightarrow \infty$, $i(t) \rightarrow 0$ and $s(t)$ goes to the steady state value, which can be estimated. Solving with $t=0$, we see $q=0 + 1 - (1/c)\log(1) = 1$. In the limit them $1 = 0 + s_{\infty} - (1/c)\log(s_\infty)$ or $c = \log(s_\infty)/(1-s_\infty)$.
This equation does not depend on $t$; $s$ is the dependent variable. It could be solved numerically, but in this case affords an algebraic solution: $i = -s + (1/c) \log(s) + q$, where $q$ is some constant. The quantity $q = i + s - (1/c) \log(s)$ does not depend on time, so is the same at time $t=0$ as it is as $t \rightarrow \infty$. At $t=0$ we have $s(0) \approx 1$ and $i(0) \approx 0$, whereas $t \rightarrow \infty$, $i(t) \rightarrow 0$ and $s(t)$ goes to the steady state value, which can be estimated. Solving with $t=0$, we see $q=0 + 1 - (1/c)\log(1) = 1$. In the limit then $1 = 0 + s_{\infty} - (1/c)\log(s_\infty)$ or $c = \log(s_\infty)/(s_\infty - 1)$.
## Trajectory with drag