Question

For Duffing's equation given in Problem 13, the behaviour of the solutions changes as rchanges sign. When$\mathit{r}\mathbf{>}\mathbf{0}$, the restoring force$\mathit{k}\mathit{y}\mathbf{+}\mathit{r}{\mathit{y}}^{\mathbf{3}}$becomes stronger than for the linear spring$\mathbf{(}\mathit{r}\mathbf{=}\mathbf{0}\mathbf{)}$. Such a spring is called hard. When$\mathit{r}\mathbf{<}\mathbf{0}$, the restoring force becomes weaker than the linear spring and the spring is called soft. Pendulums act like soft springs.

(a) Redo Problem 13 with$\mathit{r}\mathbf{=}\mathbf{-}\mathbf{1}$. Notice that for the initial conditions,$\mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}$ the soft and hard springs appear to respond in the same way forsmall.

(b) Keeping$\mathit{k}\mathbf{=}\mathit{A}\mathbf{=}\mathbf{1}$and,$\mathit{\omega }\mathbf{=}\mathbf{10}$ change the initial conditions to$\mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}$and${\mathit{y}}^{\mathbf{\text{'}}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$. Now redo Problem 13 with$\mathit{r}\mathbf{=}\mathbf{\pm }\mathbf{1}$.

(c) Based on the results of part (b), is there a difference between the behavior of soft and hard springs forsmall? Describe.

Short answer

(a) The required polynomial is,$p_3\left(t\right)=t+\frac{t^2}{2}-\frac{t^3}{6}$

(b) For$r=1,p_3\left(t\right)=1-\frac{t^2}{2}-\frac{33t^4}{8}$ and for$r=-1,p_3\left(t\right)=1+\frac{t^2}{2}-\frac{101t^3}{24}$

(c) For small values of t , both the soft and hard springs show the same behaviour.

Step-by-step solution

Step 1:To Find the Taylor polynomial of degree

The formula for the Taylor polynomial of degree n centered at$x_0$, approximating a function$f\left(x\right)$possessing n derivatives at,$x_0$is given by

$p_n\left(x\right)=f\left(x_0\right)+f^{\text{'}}\left(x_0\right)\times (x-x_0)+f^{\text{'}\text{'}}\left(x_0\right)\times \frac{{(x-x_0)}^2}{2!}+\cdots +f^n\left(x_0\right)\times \frac{{(x-x_0)}^n}{n!}$

The differential equation is given as

$y^{\text{'}\text{'}}+ky+r{y}^3=A\cos \omega t$

By substituting given values $k=A=1,r=-1$and $\omega =10$, differential equation becomes

$y^{\text{'}\text{'}}+y-y^3=\cos 10t$

It is given that for the function $y\left(x\right)$,

$y\left(0\right)=0\text{and}{y}^{\text{'}}\left(0\right)=1$

The Taylor's polynomial cantered around$x_0=0$ is given by

$p_n\left(x\right)=y\left(0\right)+y^{\text{'}}\left(0\right)\times (x-0)+y^{\text{'}\text{'}}\left(0\right)\times \frac{{(x-0)}^2}{2!}+\cdots +y^n\left(0\right)\times \frac{{(x-0)}^n}{n!}$

We need the value of$y\left(0\right),y^{\text{'}}\left(0\right),y^{\text{'}\text{'}}\left(0\right)$ and $y^{\text{'}\text{'}\text{'}}\left(0\right)$etc for finding the behaviour of spring. The first two are provided by the initial conditions.

The value of $y^{\text{'}\text{'}}\left(0\right)$can be deduced from the differential equation itself and the values of the lower derivatives.

$\begin{array}{l}{y}^{\text{'}\text{'}}\left(0\right)=\cos 10t-y\left(0\right)+y^3\left(0\right)\\ y^{\text{'}\text{'}}\left(0\right)=1\end{array}$

Now since$y^{\text{'}\text{'}}=\cos 10t-y+y^3$ holds for some interval around $x_0=0$, we can differentiate both sides to derive,

$\begin{array}{c}{y}^{\text{'}\text{'}\text{'}}=-10\sin 10t-y^{\text{'}}+3y^2{y}^{\text{'}}\\ y^{\text{'}\text{'}\text{'}}\left(0\right)=-10\sin 10t-y^{\text{'}}\left(0\right)\left(1-3y^2\left(0\right)\right)\\ =-1\end{array}$

The Taylor polynomial for the soft spring when$r<0$ in the form of equation is given by.$p_3\left(t\right)=t+\frac{t^2}{2}-\frac{t^3}{6}\cdots \cdots$

Step 2:To Find the oscillatory equation

$x_0=0$First, we will find the oscillatory equation in case where $r>0$,the restoring force is stronger and the spring is hard. The formula for the Taylor polynomial of degree centered at $x_0$, approximating a function $f\left(x\right)$possessing n derivatives at ,$x_0$is given by,

$p_n\left(x\right)=f\left(x_0\right)+f^{\text{'}}\left(x_0\right)\times (x-x_0)+f^{\text{'}\text{'}}\left(x_0\right)\times \frac{{(x-x_0)}^2}{2!}+\cdots +f^n\left(x_0\right)\times \frac{{(x-x_0)}^n}{n!}$

The differential equation is given as

$y^{\text{'}\text{'}}+ky+r{y}^3=A\cos \omega t$

By substituting given values $k=A=1,r=1$and $\omega =10$, differential equation becomes

$y^{\text{'}\text{'}}+y+y^3=\cos 10t$

It is given that for the function $y\left(x\right)$,

$y\left(0\right)=1\text{and}{y}^{\text{'}}\left(0\right)=0$

The Taylor's polynomial cantered around $x_0=0$is given by

$p_n\left(x\right)=y\left(0\right)+y^{\text{'}}\left(0\right)\times (x-0)+y^{\text{'}\text{'}}\left(0\right)\times \frac{{(x-0)}^2}{2!}+\cdots +y^n\left(0\right)\times \frac{{(x-0)}^n}{n!}$

We need the value of $y\left(0\right),y^{\text{'}}\left(0\right),y^{\text{'}\text{'}}\left(0\right)$and $y^{\text{'}\text{'}\text{'}}\left(0\right)$etc for finding the behaviour of spring. The first two are provided by the initial conditions.

The value of $y^{\text{'}\text{'}}\left(0\right)$can be deduced from the differential equation itself and the values of the lower derivatives.

$\begin{array}{l}{y}^{\text{'}\text{'}}\left(0\right)=\cos 10t-y\left(0\right)-y^3\left(0\right)\\ y^{\text{'}\text{'}}\left(0\right)=-1\end{array}$

Now since $y^{\text{'}\text{'}}=\cos 10t-y-y^3$holds for some interval around $x_0=0$, we can differentiate both sides to derive,

$\begin{array}{c}{y}^{\text{'}\text{'}\text{'}}=-10\sin 10t-y^{\text{'}}-3y^2{y}^{\text{'}}\\ y^{\text{'}\text{'}\text{'}}\left(0\right)=-10\sin 10t-y^{\text{'}}\left(0\right)\left(1+3y^2\left(0\right)\right)\\ =0\end{array}$

Similarly,

$\begin{array}{c}{y}^4=-100\cos 10t-y^{\text{'}\text{'}}-6y{\left(y^{\text{'}}\right)}^2\\ y^4\left(0\right)=-100\cos 10t-y^{\text{'}\text{'}}\left(0\right)-6y\left(0\right){\left(y^{\text{'}}\left(0\right)\right)}^2\\ y^4\left(0\right)=-99\end{array}$

The Taylor polynomial for the hard spring when $r>0$in the form of equation is given by,

$p_3\left(t\right)=1-\frac{t^2}{2}-\frac{33t^4}{8}\cdots \cdots$

Now, we will find the oscillatory equation in case where ,the restoring force is weaker and the spring is soft. The formula for the Taylor polynomial of degree n

centered at$x_0$, approximating a function $f\left(x\right)$possessing n derivatives at , $x_0$is given by

$p_n\left(x\right)=f\left(x_0\right)+f^{\text{'}}\left(x_0\right)\times (x-x_0)+f^{\text{'}\text{'}}\left(x_0\right)\times \frac{{(x-x_0)}^2}{2!}+\cdots +f^n\left(x_0\right)\times \frac{{(x-x_0)}^n}{n!}$

The differential equation is given as

$y^{\text{'}\text{'}}+ky+r{y}^3=A\cos \omega t$

By substituting given values$k=A=1,r=-1$and $\omega =10$, differential equation becomes

$y^{\text{'}\text{'}}+y-y^3=\cos 10t$

It is given that for the function , $y\left(x\right)$.$y\left(0\right)=1\text{and}{y}^{\text{'}}\left(0\right)=0$

The Taylor's polynomial centered around $x_0=0$is given by

$p_n\left(x\right)=y\left(0\right)+y^{\text{'}}\left(0\right)\times (x-0)+y^{\text{'}\text{'}}\left(0\right)\times \frac{{(x-0)}^2}{2!}+\cdots +y^n\left(0\right)\times \frac{{(x-0)}^n}{n!}$

We need the value of $y\left(0\right),y^{\text{'}}\left(0\right),y^{\text{'}\text{'}}\left(0\right)$and$y^{\text{'}\text{'}\text{'}}\left(0\right)$etc. for finding the behaviour of spring. The first two are provided by the initial conditions.

The value of $y^{\text{'}\text{'}}\left(0\right)$can be deduced from the differential equation itself and the values of the lower derivatives.

$\begin{array}{l}{y}^{\text{'}\text{'}}\left(0\right)=\cos 10t-y\left(0\right)+y^3\left(0\right)\\ y^{\text{'}\text{'}}\left(0\right)=1\end{array}$

Now since$y^{\text{'}\text{'}}=\cos 10t-y+y^3$holds for some interval around $x_0=0$, we can differentiate both sides to derive,

$\begin{array}{c}{y}^{\text{'}\text{'}\text{'}}=-10\sin 10t-y^{\text{'}}+3y^2\left(y^{\text{'}}\right)\\ y^{\text{'}\text{'}\text{'}}\left(0\right)=-10\sin 10t-y^{\text{'}}\left(0\right)\left(1-3y^2\left(0\right)\right)\\ =0\end{array}$

Similarly,

$\begin{array}{c}{y}^4=-100\cos 10t-y^{\text{'}\text{'}}+6y{\left(y^{\text{'}}\right)}^2\\ y^4\left(0\right)=-100\cos 10t-y^{\text{'}\text{'}}\left(0\right)+6y\left(0\right)\left(y^{\text{'}}{\left(0\right)}^2\right)\\ =-101\end{array}$

The Taylor polynomial for the hard spring when$r>0$in the form of equation is given by

$p_3\left(t\right)=1+\frac{t^2}{2}-\frac{101t^4}{24}\cdots \cdots$

Step 3:The initial condition

From problem 13 and $14\left(a\right)$, the solution describing the behaviour of both soft and hard spring, for the initial conditions $y\left(0\right)=0$and $y^{\text{'}}\left(0\right)=1$, is approximated by the polynomial

$p_3\left(t\right)=t+\frac{t^2}{2}-\frac{t^3}{6}+\cdots$

For small , the higher order terms will be negligible and the polynomial can be written as

$p_3\left(t\right)\approx t$

Similarly, from problem $14(\text{b})$, the solution describing the behaviour of soft and hard spring, for the initial conditions$y\left(0\right)=1$ and$y^{\text{'}}\left(0\right)=0$ , are approximated by the polynomial,

$\begin{array}{ll}{p}_3\left(t\right)=1-\frac{t^2}{2}-\frac{33t^4}{8}+\cdots & \left[\text{ForHardSpring}\right]\\ p_3\left(t\right)=1+\frac{t^2}{2}-\frac{101t^4}{24}+\cdots & \left[\text{ForSoftSpring}\right]\end{array}$

For small t , the higher order terms will be negligible and both the polynomial can be written as

$p_3\left(t\right)\approx 1$

We can see from the above analysis that, in both the cases the soft and hard spring show same behaviour for small , the difference however is that for the initial conditions $y\left(0\right)=0$and $y^{\text{'}}\left(0\right)=1$, the springs behave in direct proportion to the small , whereas for the initial conditions$y\left(0\right)=$ 1 and , $y^{\text{'}}\left(0\right)=0$the springs behave as constant with value 1 , to the small t

Step 4:Final proof

(a) the required Taylor’s polynomial is,$p_3\left(t\right)=t+\frac{t^2}{2}-\frac{t^3}{6}$

(b) For $r=1,p_3\left(t\right)=1-\frac{t^2}{2}-\frac{33t^4}{8}$and for$r=-1,p_3\left(t\right)=1+\frac{t^2}{2}-\frac{101t^3}{24}$

(c) For small values of t , both the soft and hard springs show the same behaviour.