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Consider the nonlinear differential equation of the pendulum $$\frac{d^2\theta}{dt^2}+\sin \theta=0$$ with $\theta(0)=\frac{\pi}3$ and $\theta'(0)=0$. Using the series method, find the first four nonzero terms of the solution.

Here is what I found from Maple so far: enter image description here

Text-only (for the series solution): \begin{align} \theta(t)&=\frac{\pi}3-\frac 12\sin\left(\frac{\pi}3\right)t^2+\frac 1{24}\sin\left(\frac{\pi}3\right)\cos\left(\frac{\pi}3\right)t^4+O(t^6)\\ &=\frac{\pi}3-\frac{\sqrt{3}}4t^2+\frac{\sqrt{3}}{96}t^4+O(t^6) \end{align}

But how can one find this solution by hand, using the series method?

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    $\begingroup$ 1. Note that the condition $\theta'(0)=0$ implies that the solution $t\mapsto\theta(t)$ is even. 2. Let $\theta(t)=\frac\pi3+u(t)$ with $u(t)=at^2+bt^4+ct^6+O(t^8)$, then $\theta''(t)=u''(t)=2a+12bt^2+30ct^4+O(t^6)$. 3. Expand $\sin(\theta(t))$ into powers of $t$ up to $t^4$, using $\sin(\theta(t))=\frac{\sqrt3}2\cos(u(t))+\frac12\sin(u(t))$, $\cos(u(t))=1-\frac12a^2t^4+O(t^6)$, $\sin(u(t))=at^2+O(t^6)$. 4. Equate the coefficients of $1$, $t^2$, $t^4$, and deduce the values of $(a,b,c)$. (5. If one of these is zero, weep and restart with the expansion up to order $t^6$. Otherwise, rejoice.) $\endgroup$ – Did May 20 '15 at 6:50
  • $\begingroup$ What are these $c_k$s? It seems that $c_2=a$, $c_4=b$, $c_6=c$, from which the values of $a$, $b$, $c$ follow. $\endgroup$ – Did May 20 '15 at 16:35
  • $\begingroup$ Suggestion: read carefully and slowly my first comment, and then come back if ever something is still unclear. $\endgroup$ – Did May 20 '15 at 16:47
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Expanding on @Did's comment:

The initial condition $θ′(0)=0$ implies that the solution $t↦θ(t)$ is even. Express the series solution as $\theta(t)=c_0+u(t)$, with $u(t):=c_2t^2+c_4t^4+c_6t^6+O(t^8)$. The initial condition $\theta(0)=\frac{\pi}3$ shows that $$c_0=\frac{\pi}3.$$ Thus, $\theta(t)=\frac{\pi}3+u(t)$. This also means \begin{align} \sin(\theta(t))&=\sin\left(\frac{\pi}3+u(t)\right) \\ &=\sin \left(\frac{\pi}3\right) \cos (u(t))+\cos \left(\frac{\pi}3\right) \sin( u(t)) \\ &= \frac{\sqrt{3}}2 \cos (u(t))+\frac 12 \sin (u(t)) \end{align} Now, $$\frac{\sqrt{3}}2\cos(u(t))=\frac{\sqrt{3}}2-\frac{\sqrt{3}}4 c_2 t^4 + O(t^6)$$ and $$\frac 12\sin(u(t))=\frac 12c_2t^2+\frac 12c_4t^4+O(t^6).$$ Therefore, $$\sin(\theta(t))=\frac{\sqrt{3}}2+\frac 12c_2t^2+\left(-\frac{\sqrt{3}}4c_2+\frac 12c_4\right)t^4+O(t^6)$$

Now, the ODE $\theta''(t)=-\sin(\theta(t))$, written in series expansion form, becomes $$2c_2+12c_4t^2+30c_6t^4+O(t^6)=-\left[\frac{\sqrt{3}}2+\frac 12c_2t^2-\left(\frac 12c_4-\frac{\sqrt{3}}4 c_2\right)t^4+O(t^6) \right]$$

Equating the coefficients gives $$c_2=-\frac{\sqrt{3}}4,\quad c_4=\frac{\sqrt{3}}{96},\quad c_6=\frac {\sqrt{3}}{5760}+\frac 1{160}.$$

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