Q10E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Use the result of Problem 8 to prove that if the pendulum in Figure 4.18 on page 208 is released from rest at the angle $\mathbf{0}\mathbf{<}\mathbf{\alpha }\mathbf{<}\mathbf{\pi }$, then $\left|\mathbf{\theta }\left(\mathbf{t}\right)\right|\le \mathbf{\alpha }$for all t.

Step-by-Step Solution

Answer

Therefore, the given statement is true. The pendulum in Figure 4.18 is released from rest at the angle $0<\mathrm{\alpha }<\mathrm{\pi }$, then $\left|\mathrm{\theta }\left(\mathrm{t}\right)\right|\le \mathrm{\alpha }$ for all t is true.

 

Step 1: The Energy Integral Lemma

Let $\mathrm{y}\left(\mathrm{t}\right)$be a solution to the differential equation $\mathrm{y}"=\mathrm{f}\left(\mathrm{y}\right)$, where f(y) is a continuous function that does not depend on y’ or the independent variable t.

Let F(y) is an indefinite integral of $\mathrm{f}\left(\mathrm{y}\right)$i.e. $\mathrm{f}\left(\mathrm{y}\right)=\frac{\mathrm{d}}{\mathrm{dy}}\mathrm{F}\left(\mathrm{y}\right)$. Then the quantity $\mathrm{E}\left(\mathrm{t}\right)=\frac{1}{2}\mathrm{y}\text{'}{\left(\mathrm{t}\right)}^2-\mathrm{F}\left(\mathrm{y}\left(\mathrm{t}\right)\right)$ is constant; i.e. $\frac{\mathrm{d}}{\mathrm{dt}}\mathrm{E}\left(\mathrm{t}\right)=0$ .

 

Step 2: Prove the given equation

Referring from Problem 8:   $\frac{1}{2}\mathrm{\theta }\text{'}{\left(\mathrm{t}\right)}^2-\frac{\mathrm{g}}{\ell }\mathrm{cos\theta }=\mathrm{constant}\dots \left(1\right)$

To prove: the pendulum is released from rest at the angle $0<\mathrm{\alpha }<\mathrm{\pi }$, then $\left|\mathrm{\theta }\left(\mathrm{t}\right)\right|\le \mathrm{\alpha }$ for all t.

 

The initial conditions are $\mathrm{\theta }\left(0\right)=\mathrm{\alpha },\mathrm{\theta }\text{'}\left(0\right)=0$ .

Let us take $\mathrm{constant} =\mathrm{k}$. Then, 

 

$\frac{1}{2}\mathrm{\theta }\text{'}{\left(\mathrm{t}\right)}^2-\frac{\mathrm{g}}{\ell }\mathrm{cos\theta }=\mathrm{k}\dots \left(2\right)$  

 

Now, implement the initial conditions here.

$\begin{array}{c}\frac{1}{2}\mathrm{\theta }\text{'}{\left(\mathrm{t}\right)}^2-\frac{\mathrm{g}}{\mathrm{\ell }}\mathrm{cos\theta }=\mathrm{k}\\ \frac{1}{2}\mathrm{\theta }\text{'}{\left(0\right)}^2-\frac{\mathrm{g}}{\mathrm{\ell }}\mathrm{cos\theta }\left(0\right)=\mathrm{k}\\ 0-\frac{\mathrm{g}}{\mathrm{\ell }}\mathrm{cos\alpha }=\mathrm{k}\\ \mathrm{k}=-\frac{\mathrm{g}}{\mathrm{\ell }}\mathrm{cos\alpha }\end{array}$

Now, substitute the value of k in equation (2).

$\begin{array}{c}\frac{1}{2}\theta \text{'}{\left(t\right)}^2-\frac{g}{\ell }\cos \theta =k\\ \frac{1}{2}\theta \text{'}{\left(t\right)}^2-\frac{g}{\ell }\cos \theta =-\frac{g}{\ell }\cos \alpha \end{array}$

Here ${(\mathrm{\theta }\text{'})}^2$ can’t be a negative number. So, we can write this condition as:

$\begin{array}{c}\frac{\mathrm{g}}{\mathrm{\ell }}\mathrm{cos\theta }\ge \frac{\mathrm{g}}{\mathrm{\ell }}\mathrm{cos\alpha }\\ \mathrm{cos\theta }\ge \mathrm{cos\alpha }\end{array}$

Multiply ${\cos }^{-1}$ on both sides.

 

$\begin{array}{c}{\cos }^{-1}\left(\mathrm{cos\theta }\right)\le {\cos }^{-1}\left(\mathrm{cos\alpha }\right)\\ \left|\mathrm{\theta }\left(\mathrm{t}\right)\right|\le \mathrm{\alpha }\end{array}$

 

Hence it is proved that $\left|\mathrm{\theta }\left(\mathrm{t}\right)\right|\le \mathrm{\alpha }$.