Q5.3-20E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

In Project C of Chapter 4, it was shown that the simple pendulum equation $\mathbf{\theta }\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{+}\mathbf{sin\theta }\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{0}$ has periodic solutions when the initial displacement and velocity show that the period of the solution may depend on the initial conditions by using the vectorized Runge–Kutta algorithm with h = 0.02 to approximate the solutions to the simple pendulum problem on

[0, 4] for the initial conditions:

$$\begin{gathered} \mathbf{(}\mathbf{a}\mathbf{)}\mathbf{ }\mathbf{ }\mathbf{\theta }\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{1}\mathbf{,}\mathbf{\theta }\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0} \\ \mathbf{(}\mathbf{b}\mathbf{)}\mathbf{ }\mathbf{ }\mathbf{\theta }\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{5}\mathbf{,}\mathbf{\theta }\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0} \\ \mathbf{(}\mathbf{c}\mathbf{)}\mathbf{ }\mathbf{ }\mathbf{\theta }\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{,}\mathbf{\theta }\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0} \end{gathered}$$

[Hint: Approximate the length of time it takes to reach].

Step-by-Step Solution

Answer

(a) The period is 6.28.

 

(b) The period is 6.4.

 

(c) The period is about 6.68.

Transform the equation

The equation is $\mathrm{\theta }\text{'}\text{'}\left(\mathrm{t}\right)+\mathrm{sin\theta }\left(\mathrm{t}\right)=0$

 

The system can be written as:

$$\begin{gathered} {\mathrm{x}}_1=\mathrm{\theta } \\ {\mathrm{x}}_2=\mathrm{\theta }\text{'}=\mathrm{x}{\text{'}}_1 \end{gathered}$$

The transform equation is:

  $$\begin{gathered} \mathbf{x}{\mathbf{\text{'}}}_{\mathbf{1}}\mathbf{=}{\mathbf{x}}_{\mathbf{2}} \\ \mathbf{x}{\mathbf{\text{'}}}_{\mathbf{2}}\mathbf{=}\mathbf{-}{\mathbf{sinx}}_{\mathbf{1}} \end{gathered}$$

The initial conditions are:

$$\begin{gathered} {\mathrm{x}}_1\left(0\right)=\mathrm{\theta }\left(0\right)=0.1,0.5,1 \\ {\mathrm{x}}_2\left(0\right)=\mathrm{\theta }\text{'}\left(0\right)=0,0,0 \end{gathered}$$

Apply Runge –Kutta method.

For the solution, apply the Runge-Kutta method in Matlab. For h=0.02

h	$\theta$ for part a	$\theta$ for part b	$\theta$ for part c
0	0.1	0.5	1
0.02	0.0999	0.4999	0.99983
0.04	0.09992	0.4996	0.99932
0.06	0.0998	0.49913	0.9984
0.2	0.980	0.492251	0.9863
1	0.054	0.2777	0.6000
2	-0.0415	-0.194	-0.306

Continuing this procedure for value h=4 and getting the result -0.0655,-0.3507,-0.82576, respectively.

 

From the table and all the results, one gets the first time $-\mathrm{\theta }\left(0\right)$ for part (a) is 3.14.this will be half a period. So, the period of $\theta$ is about 6.28.

 

The first time $-\mathrm{\theta }\left(0\right)$ for part (b) is 3.2.and the period of $\theta$ is about 6.4.

 

The first time $-\mathrm{\theta }\left(0\right)$ for part (c) is 3.34, and the period of $\theta$ are about 6.68.

 

Thus, this is the required result.