asymptotic expansion/approximation

Question

Find the small solution of $$y''-y\left ( 1-y^{2} \right )=0 \text{ with } y\left ( 0 \right )=\epsilon \ll 1$$

Making a pun, I decided that $$y^{3}\left ( 0 \right )\ll y\left ( 0 \right )$$ so neglect $$y^{3}\left ( 0 \right )$$ This gives $$y''_{0}-y_{0}=0$$

The general solution is $$c_{1}e^{-x}+c_{2}e^{x}$$ but I shall drop the non-decaying term to obtain $$c_{2}e^{-x}$$

$$y_{0}=\epsilon e^{-x}$$

obviously, the solution y has the form $$y=y_{0}+y_{1} $$where $$y_{0}\gg y_{1}$$

Substituting $$y=y_{0}+y_{1}$$ into the original equation I do not arrive at what my text states. I desparately need a leg up to bridge this gap. The whole numerical and asymptotic approach to me is extremely'fuzzy' Thanks in advance

Answer 1

We're going to substitute $y=y_0 + y_1$ into the differential equation then throw away most (but not too many) of the smallest terms. This last part is a bit tricky but it will start to make more sense as you become more familiar with it.

So, suppose that $y = y_0 + y_1$ where $y_1 \ll y_0$ (and $y_0$ is small as well). Substituting this into the equation

$$ 0 = y'' - y(1-y^2) $$

yields

$$ \begin{align} 0 &= (y_0 + y_1)'' - (y_0 + y_1) \Bigl[ 1 - (y_0 + y_1)^2 \Bigr] \\ &= (y_0 + y_1)'' - (y_0 + y_1) \Bigl[ 1 - y_0^2 - 2y_0y_1 - y_1^2 \Bigr]. \tag{1} \end{align} $$

Let's look inside of the brackets first, since there are a lot of very small terms there. The term $y_0^2$ is small since $y_0$ is small, but then $2y_0y_1$ is even smaller since $y_1 \ll y_0$, and thus $y_1^2$ is smaller still. We could throw all of these away, but then we would be left with the equation

$$ 0 = (y_0 + y_1)'' - (y_0 + y_1) $$

which is exactly the one we solved to find $y_0$. Since throwing all three of the small terms away doesn't give us any more information than we already have, let's try keeping the largest of them. Again, we know that

$$ y_0^2 \gg 2y_0y_1 \gg y_1^2, $$

so let's only keep $y_0^2$. Equation $(1)$ is then approximated by

$$ 0 \approx (y_0 + y_1)'' - (y_0 + y_1) \Bigl[ 1 - y_0^2 \Bigr], \tag{2} $$

and now we simplify a little:

$$ \begin{align} 0 &\approx (y_0 + y_1)'' - (y_0 + y_1) \Bigl[ 1 - y_0^2 \Bigr] \\ &= y_0'' + y_1'' - y_0 - y_1 + y_0^3 + y_0^2y_1 \\ &= y_1'' - y_1 + y_0^3 + y_0^2y_1, \end{align} $$

where in the last line we used the fact that $y_0'' - y_0 = 0$. Now $y_0^3$ is small but $y_0^2y_1$ is even smaller, so let's throw the smaller away. We then arrive at the final approximation to $(1)$,

$$ \begin{align} 0 &\approx y_1'' - y_1 + y_0^3 \\ &= y_1'' - y_1 + \epsilon^3 e^{-3x}. \tag{3} \end{align} $$

Solving this yields

$$ y_1(x) = -\frac{\epsilon^3}{8} e^{-3x} + Ae^{-x} + Be^x. $$

We need to have $y_1 \ll y_0$, and this can't be true if $B \neq 0$, so we take $B = 0$. Further, if $A \neq 0$ then $y_1(x) \sim Ae^{-x}$ for large $x$, so we won't have $y_1 \ll y_0$ in that case either. Thus we take $A = B = 0$, and we are left with

$$ y_1(x) = -\frac{\epsilon^3}{8} e^{-3x}, $$

which is $\ll y_0$ both as $\epsilon \to 0$ and as $x \to \infty$.

Note that since $y = y_0 + y_1$ and $y(0) = y_0(0) = \epsilon$ we need to have $y_1(0) \ll \epsilon$, and this is indeed true since $y_1(0) = -\epsilon^3/8$.

Thus

$$ y(x) \approx \epsilon \left( e^{-x} - \frac{\epsilon^2}{8} e^{-3x} + \cdots \right). $$

Note also that there is a typo in your exercise, the $+$ should be a $-$.

Answer 2

With this initial value and the "normal" look of the differential equation, one can expect that the solution staying small means that it is $O(ϵ)$ for all times $x\in[0,\infty)$.

Setting $y=ϵz$ one expects $z$ to stay bounded. $z$ now satisfies the scaled differential equation $$ z''-z+ϵ^2z^3=0\text{ with } z(0)=1, $$ which now has a clear perturbation term. For the limit case $ϵ=0$, you get $z''-z=0$ with solution $$ z(x)=ce^x+(1-c)e^{-x} $$ For this to stay bounded we need $c=0$ so that the growing term vanishes.

For the next best approximation we look for a perturbation in the size of the perturbation parameter, $z=e^{-x}+ϵ^2z_1$. Insertion into the differential equation reduces to $$ z_1''-z_1+(e^{-x}+ϵ^2z_1)^3=0. $$ Again removing the small parts containing $ϵ$ gives $$ z_1''-z_1=-e^{-3x}\text{ with }z_1(0)=0. $$ The bounded solutions of that have the form $z_1(x)=Ae^{-x}+Be^{-3x}$ which results in the equations for the coefficients $$ (9B-B)e^{-3x}=-e^{-3x}\text{ and }A+B=0 $$ so that the new approximation is $$ z(x)=(1+\tfrac18ϵ^2)e^{-x}-\tfrac18ϵ^2e^{-3x}+O(ϵ^4) $$ or $$ y(x)=ϵ(1+\tfrac18ϵ^2)e^{-x}-\tfrac18ϵ^3e^{-3x}+O(ϵ^5) $$