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If there is a boundary layer at $x=0$ and I have found the outer solutions $y^{left}_{out}$ and $y^{right}_{out}$, and the inner solution $y_{in}$. Than how can I put them together to get a uniformly valid solution $y_{unif}$? Is that just

$$y_{unif}=y_{in}+y^{left}_{out}+y^{right}_{out}-y^{left}_{mathc}-y^{right}_{match}$$

I had try this $y_{unif}$ and plot it together with exact solution, they don't match at all and I play around with those terms and get the following

$$y_{unif}=y_{in}+{y^{left}_{out}+y^{right}_{out} \over 2}-{y^{left}_{mathc}+y^{right}_{match} \over 2}$$

This matched with my exact solution pretty well. So my question is which one (or none of them) is collect and why?

Lutz Lehmann
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SamC
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1 Answers1

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What the inner solution provides is a sharp sigmoid function that is almost a step function, which serves to switch the integration constant in the outer solution. To be able to combine the solutions in this additive fashion assumes that adding a constant to an outer solution gives another outer solution. Then $$ y_{\rm out}^{\rm right}-y_{\rm out}^{\rm left}=K=const. $$ and $K$ is the width of the jump provided by the inner solution. In the usual examples, the jump is symmetric from $L-K/2$ to $L$ at the central point and then to $L+K/2$. The constants are fixed so that $L-K/2=y_{\rm match}^{\rm left}$ and $L+K/2=y_{\rm match}^{\rm right}$

Thus you could formulate the full solution as $$ y_{\rm unif}=y_{\rm out}^{\rm left}+(y_{\rm in}-y_{\rm match}^{\rm left}) $$ or equivalently as $$ y_{\rm unif}=y_{\rm out}^{\rm right}+(y_{\rm in}-y_{\rm match}^{\rm right}) $$ or you can take the mean of both formulas $$ y_{\rm unif}=\frac{y_{\rm out}^{\rm left}+y_{\rm out}^{\rm right}}2+\left(y_{\rm in}-\frac{y_{\rm match}^{\rm left}+y_{\rm match}^{\rm right}}2\right) $$

Lutz Lehmann
  • 131,652