Question

Use a Maclaurin series to show that a power series solution of the initial-value problem

$\frac{{\mathit{d}}^{\mathit{2}}\mathit{\theta }}{\mathit{d}{\mathit{t}}^{\mathit{2}}}\mathit{+}\frac{\mathit{g}}{\mathit{l}}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }\mathit{=}\mathit{0}\mathit{,}\mathit{\theta }\mathit{\left(}\mathit{0}\mathit{\right)}\mathit{=}\frac{\mathit{\pi }}{\mathit{6}}\mathit{,}{\mathit{\theta }}^{\mathit{\text{'}}}\mathit{\left(}\mathit{0}\mathit{\right)}\mathit{=}\mathit{0}$

is given by

$\mathit{\theta }\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\frac{\mathbf{\pi }}{\mathbf{6}}\mathbf{-}\frac{\mathbf{g}}{\mathbf{4}\mathbf{l}}{\mathit{t}}^{\mathbf{2}}\mathbf{+}\frac{\sqrt{\mathbf{3}}{\mathbf{g}}^{\mathbf{2}}}{\mathbf{96}{\mathbf{l}}^{\mathbf{2}}}{\mathit{t}}^{\mathbf{4}}\mathbf{+}\mathbf{\cdots }$

[Hint: See Example 3 in Section 4.10.]

Short answer

Find power derivatives of $\theta$at$t=0$ and then substitute them in the Maclaurin series to find the power series solution of the initial-value problem

Step-by-step solution

Step 1:Definition of Non-linear and Linear Spring

NONLINEAR SPRINGS The mathematical model has the form

$m\frac{d^{2}x}{d{t}^2}+F\left(x\right)=0$

where$F\left(x\right)=kx$. Becausedenotes the displacement of the mass from its equilibrium position,$F\left(x\right)=kx$is Hooke's law-that is, the force exerted by the spring that tends to restore the mass to the equilibrium position. A spring acting under a linear restoring force$F\left(x\right)=kx$is naturally referred to as a linear spring.

A spring whose mathematical model incorporates a nonlinear restorative force, such as

$m\frac{d^{2}x}{d{t}^2}+k{x}^3=0\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}or\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}m\frac{d^{2}x}{d{t}^2}+kx+k_1{x}^3=0$

is called a nonlinear spring

Taylor’s series expansion

Let's rewrite the given initial-value problem as

${\theta }^{\text{'}\text{'}}+\frac{g}{l}\sin \theta =0,\theta \left(0\right)=\frac{\pi }{6},{\theta }^{\text{'}}\left(0\right)=0$

If we assume that the solution $\theta \left(t\right)$of the problem is analytic at 0, then $\theta \left(t\right)$possesses a Taylor series expansion centered at0 .

$\theta \left(t\right)=\theta \left(0\right)+\frac{{\theta }^{\text{'}}\left(0\right)}{1!}t+\frac{{\theta }^{\text{'}\text{'}}\left(0\right)}{2!}{t}^2+\frac{{\theta }^{\text{'}\text{'}\text{'}}\left(0\right)}{3!}{t}^3+\frac{{\theta }^{\text{'}\text{'}\text{'}\text{'}}\left(0\right)}{4!}{t}^4+\cdots$

The values of the first and second terms in the above series are already known since those are the specified initial conditions. Moreover, the differential equation itself defines the value of the second derivative at 0 , i.e.,

${\theta }^{\text{'}\text{'}}\left(0\right)=-\frac{g}{l}\sin \theta \left(0\right)$

$=-\frac{g}{l}\sin \left(\frac{\pi }{6}\right)$

$=-\frac{g}{l}\left(\frac{1}{2}\right)$

$=-\frac{g}{2l}$

Find the derivatives

Now, we can find the expressions for the higher derivatives by calculating the successive derivatives of the differential equation as

$$\begin{gathered} {\theta }^{\text{'}\text{'}\text{'}}\left(t\right)=\frac{d}{dt}{\theta }^{\text{'}\text{'}}\left(t\right) \\ =\frac{d}{dt}(-\frac{g}{l}\sin \theta ) \\ =-\frac{g}{l}\cos \theta \cdot {\theta }^{\text{'}}\left(t\right) \end{gathered}$$

${\theta }^{\text{'}\text{'}\text{'}}\left(0\right)=-\frac{g}{l}\cos \theta \left(0\right)\cdot {\theta }^{\text{'}}\left(0\right)$

$=0\text{since}{\theta }^{\text{'}}\left(0\right)=0$

Also, the fourth derivative is

$$\begin{gathered} {\theta }^{\text{'}\text{'}\text{'}\text{'}}\left(t\right)=\frac{d}{dt}{\theta }^{\text{'}\text{'}\text{'}}\left(t\right) \\ =\frac{d}{dt}\left(-\frac{g}{l}\cos \theta \cdot {\theta }^{\text{'}}\left(t\right)\right) \end{gathered}$$

$=-\frac{g}{l}\left(-\sin \theta \cdot {\left({\theta }^{\text{'}}\left(t\right)\right)}^2+\cos \theta \cdot {\theta }^{\text{'}\text{'}}\left(t\right)\right)$

${\theta }^{\text{'}\text{'}\text{'}\text{'}}\left(0\right)=-\frac{g}{l}\left(-\sin \theta \left(0\right){\left({\theta }^{\text{'}}\left(0\right)\right)}^2+\cos \theta \left(0\right){\theta }^{\text{'}\text{'}}\left(0\right)\right)$

$=-\frac{g}{l}[0+\cos \left(\frac{\pi }{6}\right)\cdot (-\frac{g}{2l})]$

$=-\frac{g}{l}[\frac{\sqrt{3}}{2}\cdot (-\frac{g}{2l})]$

$=\frac{\sqrt{3}{g}^2}{4l^2}$

Since,

${\theta }^{\text{'}\text{'}}\left(0\right)=-\frac{g}{2l}$

Substituting the values of derivatives in the Taylor series expansion centered at 0 then gives power series solution of the initial-value problem as

$\begin{array}{rcl}\theta \left(t\right)& =& \theta \left(0\right)+\frac{{\theta }^{\text{'}}\left(0\right)}{1!}t+\frac{{\theta }^{\text{'}\text{'}}\left(0\right)}{2!}{t}^2+\frac{{\theta }^{\text{'}\text{'}\text{'}}\left(0\right)}{3!}{t}^3+\frac{{\theta }^{\text{'}\text{'}\text{'}}\left(0\right)}{4!}{t}^4+\cdots \\ & =& \frac{\pi }{6}+0+\frac{-\frac{g}{2l}}{2}{t}^2+0+\frac{\frac{\sqrt{3}{g}^2}{4l^2}}{24}{t}^4+\cdots \end{array}$

$\theta \left(t\right)=\frac{\pi }{6}-\frac{g}{4l}{t}^2+\frac{\sqrt{3}{g}^2}{96l^2}{t}^4+\cdots$

Result

Find power derivatives of $\theta$at $t=0$and then substitute them in the Maclaurin series to find the power series solution of the initial-value problem.