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I am working on a physical problem using matched asymptotic expansions. In the first order terms I encountered some issues, which made me realise that I may be doing conceptually something wrong on lower order terms. The reason for this is that on physical ground I can guess the solution. This solution does not have true boundary layers at this order, but I don't think that this should be an issue here, as my question is related to the more general approach.

My domain runs from $-\frac{1}{2}$ to $\frac{1}{2}$ and the boundary layer coordinates are defined as below

$ \xi = A^{-1} ( \frac{1}{2} + x ) \\ \zeta = A^{-1} ( \frac{1}{2} - x ) $

Where $A << 1$ (often in these type of problems $\epsilon$)

Coordinates definition

My problem is actually two-dimensional and also has two variables to solve for (i. e. not only $f$ but also $g$). In this problem the $y$-range is smaller than the $x$-range, therefor, this approach is used.

The equation I am trying to solve is $ A^2 f_{xx}+f_{yy}=-A (c(x)g_y)_y+A^3(c(x)g_x)_x $

Where the function $c(x)$ is a predefined linear function ($1+\alpha x$). For $g$ I have different kinds of equations, but in the most simple cases it will be $g_0=0$ and $g_1=0$.

The equation for $g$ is: $ g_{yyyy} + 2 A^2 g_{xxyy} + A^4 g_{xxxx} - \beta A^{n} ( c(x) f_{yy} + A c^2 g_{yy} + A^2 (c f_{xx} + f_xc_x) + A^3 (c^2 g_{xx} + 2 c c_x g_x) )=0$ And $n=0,-1,-2$ with zero value and zero gradient boundary conditions everywhere, except at the top, where $g_yy=0$. For $n=0$ $g_0=0$

Which gives equations $ f_{yy} = 0 \\ \nabla^2 \widetilde{f} = 0\\ \nabla^2 \widehat{f} = 0$ Where the Laplacian is defined in the local coordinate system.

I have BCs $f(-\frac{1}{2},y)=-\frac{1}{2}\\f(\frac{1}{2},y)=\frac{1}{2}\\f_y(x,0)=0\\f_y(x,1)=0$

The solution is straightforward, because $f(x,y)=x$ statisfies all boundary conditions. However, my question is: How can this be done formally in a matched asymptotic solutions approach? (or did I make a mistake up until this point?)

My approach The general solution satisfying the above equations, are as follows $f=c_1(x)\\ \widetilde{f}=-\frac{1}{2}+B_1\xi\\ \widehat{f}=\frac{1}{2}+B_2\zeta$ Where $B_1$ and $B_2$ are constants, and $c_1(x)$ is an arbitrary function. This arbitrary function is giving me a headache. I tried intermediate coordinates ($0<\delta<1$)

$q=A^{-\delta}(\frac{1}{2}+x)\\ r=A^{-\delta}(\frac{1}{2}-x)$

Such that $x=-\frac{1}{2}+A^\delta q$ and $x=\frac{1}{2}-A^\delta r$ and $ \xi=A^{-1+\delta}q$ and $\zeta=A^{-1+\delta}r$

I tried some matching now $ f=c_1(-\frac{1}{2}+A^\delta)=c_1(-\frac{1}{2})+c'_1(-\frac{1}{2})A^\delta q + h.o.t.$

$\widetilde{f}=-\frac{1}{2}+B_1A^{-1+\delta}q $

I thought the formal approach is now to match terms in equal order $A$ to solve the coefficients. Does it make sense to equate terms in equal order of $q$? This will give that second and higher order derivatives of the function $c_1$ are zero, giving me a linear equation. However, it looks like $B_1$ will depend on $A$. This does not really make sense, because then these terms should appear in higher order.

Any comments, hints, suggestions, remarks are very welcome here, as I have to be missing some important point here.

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    Thanks for your answers until now. I will have a look into your hint of the last comment and if I find anything of interest I will edit my question or post it as an answer.2012-05-15

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