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I am required to find the leading order outer and inner solutions and then the constants by asymptotic matching.

I have shown there exists a boundary layer at x=0 and hence have use the condition$ y(1)=1$ to find the leader order outer solution to be $y_0(x)$ ~ $e^{\frac{1-x^{\frac{2}{3}}}{8}}$. I then rescaled writing $x= \epsilon^{\frac{3}{4}}X$ and found the leading order inner solution to be $Y_0(X)$ ~ $A + c \int_{0}^{X}{e^{-9s^{\frac{4}{3}}}} ds$. I then used the boundary condition $y(0)=1$ to find A=1 by taking the limit as X tends to infinity. Is this right?

My main issue is how to find $c$ using asymptotic matching. In previous examples I have seen, it has been easy to approximate the integral in the inner solution and hence match with the outer solution. Any help would be greatly appreciated.

Thanks

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  • $\begingroup$ You probably have to treat the integral $\int_0^\infty\exp(-9s^{\frac43})ds$ as a constant that is known to exist but with little information on its value. $\endgroup$ – LutzL May 18 '16 at 18:30
  • $\begingroup$ Oh I see.... Will try that $\endgroup$ – anon May 18 '16 at 22:04
  • $\begingroup$ You can't use $Y(1)=1$. The outer solution has already used that boundary condition. For the inner solution, you need a boundary condition in the boundary layer, i.e. you need to know $Y(0)$. Also note that $y(1)=Y(\epsilon^{4/3})$, not $Y(1)$. $\endgroup$ – David May 18 '16 at 22:44
  • $\begingroup$ Sorry yes, that's what I meant! Any tips for how to evaluate / approximate th integral? $\endgroup$ – anon May 18 '16 at 23:10
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    $\begingroup$ It's a Gaussian integral, so $\int^\infty_0\exp(-(3s^{2/3})^2)=\Gamma(3/4)/(4\sqrt{3})$. $\endgroup$ – David May 19 '16 at 0:15

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