# Asymptotic Expansion, Regular Perturbation

Regular perturbation. Find the first two terms in an asymptotic expansion of the small parameter $ϵ$ of the solution of $$xy'+y=ϵy^{1/2},\quad x>0,\quad y(1)=1.$$ Explain why the expansion is not valid as $x\to\infty$. What form of rescaling would be necessary to examine behaviour for large $x$?

I've been learning about how to construct asymptotic solutions to regularly perturbed DEs, but I'm unsure of how to treat the $$y^{1/2}$$ on the right hand side.

So I divide through by $x$, and set an expansion for $y$ in terms of $y_i$, but how do I do the square root of a series?

Thank you for any help given!

• plug in the series and expand the square root up to the order u want.example: $\sqrt{1+\epsilon y_1+\epsilon^2 y_2+\mathcal{O}(\epsilon^3)}=1+\frac{1}{2}(\epsilon y_1+\epsilon^2 y_2)-\frac{1}{8}\epsilon^2y_1+\mathcal{O}(\epsilon^3$) Mar 3, 2016 at 21:40

The lowest order solution is obviously $y=x^{-1}$ leading to the first order equation $$(xy_1)'=x^{-1/2}, \; y_1(1)=0 \implies xy_1=2(x^{1/2}-1) \implies y_1=2(x^{-1/2}-x^{-1})$$ In the next approximation one gets $$(x(y_0+ϵy_1+ϵ^2y_2))'=ϵy_0^{1/2}(1+\tfrac12ϵy_0^{-1}y_1) \\ (xy_2)'=1-x^{-1/2}, y_2(1)=0 \\ y_2=1-2x^{-1/2}+x^{-1}$$ Thus the first three terms of the asymptotic expansion are $$y(x)=x^{-1}+2ϵ(x^{-1/2}-x^{-1})+ϵ^2(1-2x^{-1/2}+x^{-1})$$

For the exact solution you get $$(\sqrt{xy})'=\frac12\frac{(xy)'}{\sqrt{xy}}=\frac12ϵx^{-1/2} \implies \sqrt{xy}=ϵx^{1/2}+1-ϵ \\ \implies y=x^{-1}+2ϵ(x^{-1/2}-x^{-1})+ϵ^2(x^{-1}-2x^{-1/2}+1)$$

So the above three term expansion is already the exact solution.

Update: The equation I solved was $xy'+y=ϵxy^{1/2},\quad x>0,\quad y(1)=1.$ @LutzL's solution is correct.

Please follow the suggestion of @tired. I did this as an exercise, the solution should be (sorry I used $t$ instead of $x$). $$\frac{1}{t} - \epsilon \frac{2}{3} \left(\left(\sqrt{\frac{1}{t}}-t\right) \sqrt{\frac{1}{t}}\right)$$

The figure below shows good agreement for $\epsilon = 0.1$ (red dashed curve is the numerical solution, blue curve is the perturbation solution). With regard to the later part of the question, the closed form of the solution by @LutzL doesn't seem to have issue when ${x \to ∞}$. Nevertheless, I think that what it may be getting at is that this expansion is asymptotic, so $$\epsilon (\frac{1}{√x}-\frac{1}{x}) = o(\frac{1}{x})$$ as ${\epsilon \to 0}$. However, for $x=O(\frac{1}{\epsilon^2})$ this is not the case, so the expansion is invalid as ${x \to ∞}$. Thus mutliscale perturbation theory is required, which is enough to say for the question.

I'm not sure if this is right, but this is my answer for when I hand this in (DEs II has got lousy).