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I need to show that the leading order inner solution is given by the below. Thus far, I have rescaled and showed the boundary layer is of order $\epsilon^{\frac{3}{4}}$. Hence at leading order I then try to solve the second order ode $Y_0''(X) + X^{2}Y_0'(X)=0$ but don't get the result. Please help

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Here is what I did, our layer thicknesses don't agree:

We know the boundary layer is at $x = 0,$ so let $\xi = x \epsilon^{-a}.$ Then the equation is, $$\epsilon^{1-2a} y''(\xi) + \xi^2 \epsilon^a - \xi^3 \epsilon^{3a} y = 0.$$

If $a = 1/3,$ then the first two terms balance and the third term is of higher order, so the thickness is ${\cal O}(\epsilon^{1/3}).$ The equation has now become, $$ y'' + \xi^2 y' = 0.$$

Doing the typical $y(\xi) \sim Y_0(\xi) + \cdots,$ the equation you stated is now the problem to solve. Clearly one may integrate this straight up one time to get $$Y'_0 = c_2 e^{-\xi^3 / 3}$$ This is just your basic use of an integrating factor, $e^{\xi^3/3}.$ So then integrate it one more time to get $Y_0(\xi) = c_1 + c_2\int_0^\xi e^{-t^3/3} \; dt.$

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  • $\begingroup$ Matching with the boundary condition will give $c_1 = \alpha$, I forgot to mention. $\endgroup$ – Merkh May 25 '16 at 21:28
  • $\begingroup$ Sorry not sure what I was doing with 3/4! You are definitely right with 1/3! $\endgroup$ – anon May 25 '16 at 22:00
  • $\begingroup$ How do I find c2? $\endgroup$ – anon May 25 '16 at 22:09
  • $\begingroup$ By matching the inner and outer layers. It probably suffices to require that $\lim_{\xi \to \infty} Y_0(\xi) = \lim_{x\to 0} y_{outer}(x).$ I say "probably" based on the difficulty level of the question and because only matching to first order is needed. $\endgroup$ – Merkh May 25 '16 at 23:23

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