Q16 E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

Use the energy integral lemma to show that every solution to the Duffing equation (18) is bounded; that is $\left|\mathrm{y}\left(\mathrm{t}\right)\right|⩽\mathrm{M}$, for some M. [Hint: First argue that for $\frac{{\mathrm{y}}^2}{2}+\frac{{\mathrm{y}}^4}{4}⩽\mathrm{K}$ some K.]

Step-by-Step Solution

Answer

Therefore, the given statement is true. The solution to the Duffing equation is bounded and that is,  $\left|\mathrm{y}\right|⩽\mathrm{M}$for some M is true.

Step 1: General form

The Energy Integral Lemma:

Let y(t) 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,  $\mathrm{f}\left(\mathrm{y}\right)$that is, $\mathrm{f}\left(\mathrm{y}\right)=\frac{\mathrm{d}}{\mathrm{dy}}\mathrm{F}\left(\mathrm{y}\right)$. 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. $\frac{\mathrm{d}}{\mathrm{dt}}\mathrm{E}\left(\mathrm{t}\right)=0$ 

 Duffing equation: (18)

 $\mathrm{y}+\mathrm{y}+{\mathrm{y}}^3=\mathrm{y}+\left(1+{\mathrm{y}}^2\right)\mathrm{y}=0$ … (1)

Step 2: Verify the equations

Given that, the Duffing equation is $\mathrm{y}+\left(1+{\mathrm{y}}^2\right)\mathrm{y}=0$

Then, $\mathrm{y}=-\left(\mathrm{y}+{\mathrm{y}}^3\right)$ . So, $\mathrm{f}\left(\mathrm{y}\right)=-\left(\mathrm{y}+{\mathrm{y}}^3\right)$

 

Now find the F(y).

 

$$\begin{gathered} \mathrm{f}\left(\mathrm{y}\right)=\frac{\mathrm{d}}{\mathrm{dy}}\mathrm{F}\left(\mathrm{y}\right) \\ -\left(\mathrm{y}+{\mathrm{y}}^3\right)=\frac{\mathrm{d}}{\mathrm{dy}}\left(-\frac{{\mathrm{y}}^2}{2}-\frac{{\mathrm{y}}^4}{4}\right) \\ \mathrm{F}\left(\mathrm{y}\right)=-\frac{{\mathrm{y}}^2}{2}-\frac{{\mathrm{y}}^4}{4} \end{gathered}$$

 

Then, find the quantity.

$$\begin{gathered} \mathrm{E}=\frac{1}{2}{\left(\mathrm{y}\text{'}\left(\mathrm{t}\right)\right)}^2-\mathrm{F}\left(\mathrm{y}\left(\mathrm{t}\right)\right) \\ =\frac{1}{2}{\left(\mathrm{y}\text{'}\left(\mathrm{t}\right)\right)}^2+\frac{{\mathrm{y}}^2}{2}+\frac{{\mathrm{y}}^4}{4} \end{gathered}$$

 

So, $\frac{1}{2}{\left(\mathrm{y}\text{'}\left(\mathrm{t}\right)\right)}^2+\frac{{\mathrm{y}}^2}{2}+\frac{{\mathrm{y}}^4}{4}=\mathrm{constant}$ 

Step 3: Verify the equation

Since the above equations are square terms. Then,

$$\begin{gathered} \frac{1}{2}{\left(\mathrm{y}\text{'}\left(\mathrm{t}\right)\right)}^2+\frac{{\mathrm{y}}^2}{2}+\frac{{\mathrm{y}}^4}{4}⩾0 \\ \frac{1}{2}{\left(\mathrm{y}\text{'}\left(\mathrm{t}\right)\right)}^2+\frac{{\mathrm{y}}^4}{4}⩾0 \end{gathered}$$

 And,

 $$\begin{gathered} \frac{{\mathrm{y}}^2}{2}⩽\mathrm{K} \\ {\mathrm{y}}^2⩽2\mathrm{K} \\ \left|\mathrm{y}\right|⩽\sqrt{2\mathrm{K}}=\mathrm{M} \end{gathered}$$

So, $\left|\mathrm{y}\right|⩽\mathrm{M}$