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The equation is $\frac{dA}{d\tau} = \frac{3}{2}iA^2 A^*$, where $A: \mathbb{R}\rightarrow\mathbb{C}$.

Context:

The function $A$ arises in a perturbation methods problem, using the method of multiple scales. The original equation is the undamped Duffing equation, $y^{\prime\prime}+y+\varepsilon y^3 = 0, \ y(0)=1, \ y^\prime(0)=0.$ It is determined that $y$ can be approximated as $y(t) = A(\tau)e^{it}+A^*(\tau)e^{-it}$. Under this framework, the original equation arises as a solvability condition for preventing secular terms, and the solution is $A = \frac{1}{2}e^{\frac{3}{8}i\tau}$ after imposing the intial conditions.

My question is, how does one reach the given solution?

Attempt:

My attempt so far has been to deal with this the way I would if A was a real-valued function, which is to separate the equation into $\frac{dA}{A^2A^*} = \frac{3}{2}id\tau$, then integrate. The problem I run into is that the left hand side cannot be integrated like a real-valued function since it involves a conjugate. In $\mathbb{R}$, the $^*$ operator is an identity, so the LHS becomes $\frac{dA}{A^3}$, and solving the equation gives $A$ as a square root function of $\tau$. I thought of using contour integration, but there's no domain to integrate over. When doing the method of separation with a real-valued function, I would use an indefinite integral. However, I don't know how to use an indefinite integral with a complex-valued integrand, if it's even possible.

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Note that if $s(\tau) = A(\tau)\overline{A}(\tau)$, then $s'(\tau) = A(\tau) \overline{A'(\tau)}+A'(\tau) \overline{A(\tau)} = 0$, hence we can treat $A \overline{A} = |A|^2$ as a constant and the equation becomes $A' = i{3 \over 2}|A(0)|^2 A$ which has solution $A(\tau) = A(0) e^{i{3 \over 2}|A(0)|^2 \tau}$.

Presumably in the case above, you have $A(0) = {1 \over 2}$.

Addendum:

Note that $A'(\tau) = i{3 \over 2} A(\tau)^2 \overline{A(\tau)}$, so $A(\tau)\overline{A'(\tau)} = -i{3 \over 2}|A(\tau)|^4$, and $A'(\tau) \overline{A(\tau)} = i{3 \over 2}|A(\tau)|^4$. Summing gives zero.

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  • $\begingroup$ Why is $s^\prime(\tau)=0$? I'm getting $s^\prime(\tau) = 2uu^\prime + 2vv^\prime$, where $A=u+iv$. $\endgroup$ – Alex Jones Mar 25 '17 at 3:58
  • $\begingroup$ @AlexanderJ93: Let me check, wine & maths. don't mix well... $\endgroup$ – copper.hat Mar 25 '17 at 4:03
  • $\begingroup$ @AlexanderJ93: I think my computation is correct, I presume by $z^*$ you mean the conjugate? Just substitute the formula for $A'(\tau)$ into the expression above for $s'(\tau)$. $\endgroup$ – copper.hat Mar 25 '17 at 4:04
  • $\begingroup$ Yes, we're talking about the same thing. But $|A|^2 = u^2+v^2$ and $[|A|^2]^\prime = 2uu^\prime+2vv^\prime$, as far as I can tell. I get the same expression for $s^\prime$ as you are, but I'm not getting it to be 0. Can you show me what steps you're using to get there? $\endgroup$ – Alex Jones Mar 25 '17 at 4:07
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    $\begingroup$ @AlexanderJ93: Sometimes a cigar is just a cigar :-). $\endgroup$ – copper.hat Mar 25 '17 at 4:19

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