This is a mathematica exercise that I have to do, where $y(x) = x - \epsilon \sin(2y)$ and it wants me to express the solution $y$ of the equation as a power series in $ \epsilon$.
Perturbation problem
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1And what is the question? – 2012-06-15
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3Note that this is the Kepler equation; much has been written about this. – 2012-06-15
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0Thank you. I'll look this up. – 2012-06-15
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0I don't see any derivatives, so I removed the "differential equations" tag. – 2012-06-15
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0I am eager to know a solution... – 2012-10-10
1 Answers
We're looking for a perturbative expansion of the solution $y(x;\epsilon)$ to $$ y(x) = x-\epsilon \sin(2y)$$ I don't know if you're asking for an expansion to all orders. If so, I have no closed form to offer. But, for illustration, the first four terms in the series may be found as follows by use of the addition theorem (this procedure may be continued to any desired order). Expand $$y(x;\epsilon)=y_o(x)+\epsilon y_1(x)+\epsilon^2 y_2(x)+\epsilon^3 y_3(x) + o(\epsilon^3).$$ Then clearly, $y_o(x)=x$ and $$\epsilon \,y_1(x) +o(\epsilon)=-\epsilon\sin(2x+\epsilon 2y_1+o(\epsilon))=-\epsilon\sin(2x)\underbrace{\cos(\epsilon\, 2y_1+o(\epsilon))}_{\sim 1}\\ \phantom{tttttttttttttttttttttttttttttttttttttttttttttttttt}+\epsilon\cos(2x)\underbrace{\sin(\epsilon\, 2y_1+o(\epsilon))}_{\sim 2\epsilon y_1 = O(\epsilon)}. \\$$ This implies $y_1(x)= -\sin(2x)$. Similarly, at the next two orders we find $y_2(x)=2\sin(2x)\cos(2x)=\sin(4x)$ and $y_3(x)=2\sin^3(x)-2\cos(2x)\sin(4x)$, if I haven't made a mistake in the algebra. Hence $$y(x;\epsilon) = x - \epsilon \sin(x) + \epsilon^2 \sin(2x)\cos(x)+\epsilon^3 (2\sin^3(x)-2\cos(2x)\sin(4x)) +o(\epsilon^3). $$