Double Pendulum

Equations:

$$\begin{align}\dot{x}_{0}&=x_{2}\\\dot{x}_{1}&=x_{3}\\\dot{x}_{2}&=\frac{9.81 \cdot par_{0} \cdot \sin {x_{0}} - 1.0 \cdot par_{1} \cdot x_{2}^{2} \cdot \sin {x_{1}} \cdot \cos {x_{1}} - 1.0 \cdot par_{1} \cdot x_{2}^{2} \cdot \sin {x_{1}} - 2.0 \cdot par_{1} \cdot x_{2} \cdot x_{3} \cdot \sin {x_{1}} - 1.0 \cdot par_{1} \cdot x_{3}^{2} \cdot \sin {x_{1}} - 9.81 \cdot par_{1} \cdot \sin {x_{0}} \cdot \cos^{2} {x_{1}} + 9.81 \cdot par_{1} \cdot \sin {x_{0}} - 9.81 \cdot par_{1} \cdot \sin {x_{1}} \cdot \cos {x_{0}} \cdot \cos {x_{1}}}{- 1.0 \cdot par_{0} + par_{1} \cdot \cos^{2} {x_{1}} - 1.0 \cdot par_{1}}\\\dot{x}_{3}&=\frac{1.0 \cdot par_{0} \cdot x_{2}^{2} \cdot \sin {x_{1}} - 9.81 \cdot par_{0} \cdot \sin {x_{0}} + 9.81 \cdot par_{0} \cdot \sin {x_{1}} \cdot \cos {x_{0}} + 2.0 \cdot par_{1} \cdot x_{2}^{2} \cdot \sin {x_{1}} \cdot \cos {x_{1}} + 2.0 \cdot par_{1} \cdot x_{2}^{2} \cdot \sin {x_{1}} + 2.0 \cdot par_{1} \cdot x_{2} \cdot x_{3} \cdot \sin {x_{1}} \cdot \cos {x_{1}} + 2.0 \cdot par_{1} \cdot x_{2} \cdot x_{3} \cdot \sin {x_{1}} + 1.0 \cdot par_{1} \cdot x_{3}^{2} \cdot \sin {x_{1}} \cdot \cos {x_{1}} + 1.0 \cdot par_{1} \cdot x_{3}^{2} \cdot \sin {x_{1}} + 9.81 \cdot par_{1} \cdot \sin {x_{0}} \cdot \cos^{2} {x_{1}} - 9.81 \cdot par_{1} \cdot \sin {x_{0}} + 9.81 \cdot par_{1} \cdot \sin {x_{1}} \cdot \cos {x_{0}} \cdot \cos {x_{1}} + 9.81 \cdot par_{1} \cdot \sin {x_{1}} \cdot \cos {x_{0}}}{- 1.0 \cdot par_{0} + par_{1} \cdot \cos^{2} {x_{1}} - 1.0 \cdot par_{1}}\end{align}$$

Description:

Equations of motion for the double pendulum model derived automatically with the program Maple.

Keywords:

chaotic

Initial value problems:

IVP 126

Class: PINLB

Initial conditions:

$$\begin{align}x_{0}(0)&=2.355\\x_{1}(0)&=-1.727\\x_{2}(0)&=0.43\\x_{3}(0)&=0.67\end{align}$$

Parameters:

$$\begin{align}par_{0}&=1.0\\par_{1}&=1.0\end{align}$$

Exact Solution:

No exact solution provided.

Problem has not been tested yet.

IVP 127

Class: PINLB

Initial conditions:

$$\begin{align}x_{0}(0)&=\left[ 1.855, 2.855\right]\\x_{1}(0)&=-1.727\\x_{2}(0)&=0.43\\x_{3}(0)&=0.67\end{align}$$

Parameters:

$$\begin{align}par_{0}&=1.0\\par_{1}&=1.0\end{align}$$

Exact Solution:

No exact solution provided.

Problem has not been tested yet.

IVP 128

Class: PINLB

Initial conditions:

$$\begin{align}x_{0}(0)&=2.355\\x_{1}(0)&=-1.727\\x_{2}(0)&=0.43\\x_{3}(0)&=0.67\end{align}$$

Parameters:

$$\begin{align}par_{0}&=\left[ 0.5, 1.5\right]\\par_{1}&=1.0\end{align}$$

Exact Solution:

No exact solution provided.

Problem has not been tested yet.

IVP 129

Class: PINLB

Initial conditions:

$$\begin{align}x_{0}(0)&=\left[ 2.255, 2.455\right]\\x_{1}(0)&=-1.727\\x_{2}(0)&=0.43\\x_{3}(0)&=0.67\end{align}$$

Parameters:

$$\begin{align}par_{0}&=\left[ 0.9, 1.1\right]\\par_{1}&=1.0\end{align}$$

Exact Solution:

No exact solution provided.

Problem has not been tested yet.

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