how to find the thickness of the boundary layer near $0$?

Question

Given the BVP

$$ \begin{cases} \varepsilon y'' + \sqrt{x} y' - y =0 \\ y(0)=0 \\ y(1) = \mathrm{e}^2 \end{cases} $$

How do we find the thickness of the boundary layer near $x = 0$ ?

Answer 1

You just have to transform to a stretched variable near the boundary layer. Let's assume that the boundary layer thickness is of the form $\epsilon^{\gamma}$. Then we introduce our stretched variable \begin{equation} \xi = \frac{x}{\epsilon^{\gamma}} \end{equation} If we change coordinates in the ODE from $x$ to $\xi$, we have the new, scaled equation \begin{equation} \epsilon^{1 - 2 \gamma} \frac{d^2 y}{d \xi^2} + \epsilon^{\gamma /2 - \gamma} \sqrt{\xi} \frac{d y}{d \xi} - y = 0 \end{equation} From here, we simply match the powers of $\epsilon$ in the highest derivative term to one of the other terms to ensure a detailed balance. Hence, we must have that \begin{equation} \epsilon^{1 - 2 \gamma}= \epsilon^{-\gamma /2 } \end{equation} or $\gamma = 2/3$. So, the boundary layer thickness is \begin{equation} \mathcal{O}(\epsilon^{2/3}) \end{equation}

Answer 2

Set $Y(X)=y(\delta X)$ for the inner solution, then $$ Y'(X)=δy'(\delta X), Y''(X)=δ^2y''(δX),\\~\\\qquad εY''(X)=δ^2εy''(δX)=-δ^2\sqrt{δX}y'(δX)+δ^2y(δX)\\ =-δ^{3/2}\sqrt{X}Y'(X)+δ^2Y(X) $$ Non-trivial balancing occurs when $ε=δ^{3/2}$ or $δ=ε^{2/3}$.