Thickness of the Boundary Layer

Question

Given an ODE

$$\epsilon y''+2xy'=x \cos(x)$$

with boundary condition $y(\pm {\pi \over 2})=2$

Where is the boundary layer and what is the thickness of it?

Answer 1

Outer solution

You have the outer equation $$ y'=\tfrac12\cos x\implies y(x)=C+\tfrac12\sin x $$ Provided there are no jumps (rapid changes) at the outer boundaries, the left boundary condition leads to $y(x)=\tfrac52+\tfrac12\sin x$, the right boundary condition to $y(x)=\tfrac32+\tfrac12\sin x$

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Where the inner or boundary layer is not

For the inner parametrization $x=x_0+δ X$ and $Y(X)=y(x_0+δ X)$ the equation reads $$ ϵY''+2δ(x_0+δ X)Y'=δ^2(x_0+δ X)\cos(x_0+δ X). $$ For $x_0\notin\{-\frac\pi2,0,\frac\pi2\}$ a first reduction of infinitesimal versus appreciable quantities leads to $$ ϵY''+2δx_0Y'=δ^2x_0\cos(x_0) $$ where the only interesting case is for $δ\sim ϵ$. Taking $ϵ=2δ·|x_0|$ leads to the observation that the solution to the inner equation $Y''\pm Y'=0$ must be locally constant, since the exponential term of it is unbounded. However, to be locally (nearly) constant is equivalent to being continuous, which is satisfied by the outer solutions.

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For $x_0=\pm\frac\pi2$ the right side reduces to $\mp δ^3x_0^2$, the reduced equation thus corresponds to the former case. The exponential term of the solution is still not bounded inside the interval, thus providing no jump.

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Inner layer

For $x_0=0$, $x=δX$, a first reduction of the singular equation removing non-essential infinitesimals reads as $$ ϵY''+2δ^2 XY'=δ^3 X. $$ Since $δ^3\ll δ^2$ the only non-trival case appears for $ϵ=δ^2$. The reduced inner equation is $$ Y''+2XY'=0\implies Y'=C·e^{-X^2}\implies Y(X)=C·\int_0^Xe^{-s^2}ds+D $$ with limits $Y(-\infty)=D-C·\frac{\sqrt{\pi}}2$ and $Y(\infty)=D+C·\frac{\sqrt{\pi}}2$.

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Approximation formula

For the jump from $\frac52$ to $\frac32$ one needs $D=2$ and $C=-\frac{1}{\sqrt{\pi}}$ giving the total approximation as $$ y_{approx}(x)=2+\tfrac12\bigl(\sin x-\text{erf}(x/\sqrtϵ)\bigr) $$

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The exponent for the boundary layer thickness is thus $\frac12$.