1
$\begingroup$

Let us look at an example, the boundary layer problem $$\epsilon y'' + (1+x) y'+ y = 0$, \quad y(0) = y(1) = 1.$$ We will substitute $y(x) = \exp\left[\frac{1}{\delta} \sum_{n=0}^\infty \delta^n S_n(x) \right]$ into the equation and divide by $y$, and we get $$\left[\frac{\epsilon}{\delta} S_0'' + \epsilon S_1 '' + \cdots + \frac{\epsilon}{\delta^2} S_0'^2 + 2\frac{\epsilon}{\delta}S_0' S_1' \cdots\right] + \left[\frac{1}{\delta}(1+x) S_0' + \cdots\right] + 1 = 0 \quad (\star)$$

Two largest terms $\frac{\epsilon}{\delta^2} S_0'^2$ and $\frac{1}{\delta}(1+x) S_0'$ must be balanced, so we take $\delta = \epsilon$ and at $O(1/\epsilon)$, we have $$S_0'^2 + (1+x)S_0' \sim 0.$$

There are two cases, $S_0' \sim 0$ or $S_0' \sim -(1+x)$.

For the first case, we get $S_0 = A$ a constant, and look at $(\star)$ at $O(1)$ level, we get $$\frac{1}{\epsilon} S_0'^2 + \frac{1}{\epsilon} (1+x) S_0' + S_0'' + 2S_0' S_1' + (1+x) S_1' +1 \sim 0$$ which reduces to $$(1+x)S_1' \sim -1$$ so $S_1 \sim -\log(1+x) + B$. Now this will give a solution $$y_1 = \exp\left[\frac{1}{\epsilon}S_0 + S_1 + O (\epsilon)\right] \sim \exp\left[\frac{1}{\epsilon}A + \Big(-\log(1+x) + B\Big)\right] \text{ as }\epsilon \rightarrow 0.$$

The next step, the book says the above is $\sim \frac{C}{1+x}$ where $e^{\frac{1}{\epsilon}A}$ is obsorbed into the constant $C$, but how is this possible since this term is not small as $\epsilon$ goes to zero.

For completeness, I will add the calculation for the second case where $S_0' = -(1+x)$. Then we have $S_0 = -x^2/2 - x + B$. Again looking at $O(1)$ terms, $$\frac{1}{\epsilon} S_0'^2 + \frac{1}{\epsilon} (1+x) S_0' + S_0'' + 2S_0' S_1' + (1+x) S_1' +1 \sim 0$$ this reduces to $$-1 -2(1+x)S_1' + 1 \sim 0$$ so $S_1' \sim 0$ and $S_1 = C_1$. So we get a second solution $$y_2 \sim \exp\left[\frac{1}{\epsilon}\left( -\frac{x^2}{2} - x + B \right) + C_1 \right]$$

Again, the book claims the above is $\sim > D\exp\left[\frac{1}{\epsilon}\left( -\frac{x^2}{2} - x \right) > \right]$, so the constant $e^{B/\epsilon}$ is also absorbed into the constant $D$.

$\endgroup$
2
  • $\begingroup$ You started with: "For the first case" Are you going for the second case $S_0'=-(1+x)$ also? $\endgroup$ – Diger Jul 17 '18 at 19:34
  • $\begingroup$ The second case is just like this, they included another $e^{C/\epsilon}$ type of term into a constant just like here... $\endgroup$ – Xiao Jul 17 '18 at 19:36
1
$\begingroup$

You do the integration for fixed $ϵ$, or in other words, the integration constants $A,B,C$ etc. can depend on $ϵ$, they are only constant relative to $x$. The constants are fixed by the boundary conditions.

Relative to https://math.stackexchange.com/a/2807460/115115 where $f(x)=x+1$ and the anti-derivatives with $F(0)=0=G(0)$ are $F(x)=\int f dx=\frac12x^2+x$, $G(x)=\int \frac1fdx=\ln(x+1)$, the solution form is $$ y(x)\approx Ae^{-G(x)}+\frac{B}{f(x)}e^{-\frac1εF(x)+G(x)}=\frac{A}{x+1}+Be^{-\frac1{2ε}(x^2+2x)}. $$ To satisfy the initial conditions \begin{align} 1=y(0)&=A+B\\ 1=y(1)&=\frac A2+Be^{-\frac3{2ε}} \end{align} we get approximatively $A=2$ and $B=-1$, as $ e^{-\frac3{2ε}}\in O(ε^k)$ for any $k>0$. The solution that closely approximates the solution and the boundary conditions is thus $$ y(x)\approx \frac{2}{x+1}-e^{-\frac1{2ε}(x^2+2x)} $$

$\endgroup$
6
  • $\begingroup$ Thank you for the answer, but then how would I make sense of $ \sim \frac{A_\epsilon}{1+x}$ as $\epsilon \rightarrow 0^+$ if $A_\epsilon$ is going to infinity, since essentially, I would like to solve for the constant $A_\epsilon$ with the boundary conditions. And Also I think in the solution there should not be a $(x+1)^2$ term, maybe $G(x) = -\log(1+x)$? $\endgroup$ – Xiao Jul 17 '18 at 20:32
  • $\begingroup$ In my parametrization $A=2$, $B=-1$ are independent of $ε$. The $(1+x)^2$ factor is a result of the next term in the expansion of $S$, relative to the rapidly falling exponential term the difference to the constant $1$ is not that important for very small $ε$, for "largish" small $ε$ there may be a visible difference from including this term. $\endgroup$ – Lutz Lehmann Jul 17 '18 at 20:39
  • $\begingroup$ Thanks! In your linked answer, how did you obtain the general solution formula? It seems that you also dropped the ${C/\epsilon}$ type of constant inside the exponential? $\endgroup$ – Xiao Jul 18 '18 at 0:11
  • $\begingroup$ You are free to fix the integration constants in the exponential any way you wish as any variability they contain is also contained in the constant factor in front of the exponential. $\endgroup$ – Lutz Lehmann Jul 18 '18 at 4:48
  • $\begingroup$ What is amazing is that an exact solution of the DE can be found (by a CAS !!) $\endgroup$ – Claude Leibovici Jul 18 '18 at 5:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.