# Matched asymptotic expansions with an unknown outer function

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 $\epsilon$)

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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Your boundary layer solutions are not boundary layers. I would expect the correct answers to look more like $\tilde{f}\sim\cos(ky)\exp(-k(x+1/2))$. Sorry, I do not have much time at the moment, but can have a closer look in a few weeks time, if it is at all helpful. Can you specify your equation fully, so that there won't be any unpleasant surprise from the rhs terms. – Aleksey Pichugin May 15 '12 at 8:51
@AlekseyPichugin Thanks for your comment. I think I do not have boundary layers at the lowest order. Is it possible to have them at second order but not at first order? The point is that it is a coupled system with two equations. I will extend my question with those to be complete. – Bernhard May 15 '12 at 9:02
Clearly, your leading order "regular" solution $f=x$ satisfies the leading order equation and all boundary conditions. This means that boundary layers can only enter the problem at a higher order (if at all). This is not uncommon, think about all engineering theories of rods, plates and shells - these are only possible because the leading order solution does not depend on boundary layers. In fact, in plate bending theory the boundary layers do contribute at the leading order, which led to a huge controversy regarding the correct choice of boundary conditions. – Aleksey Pichugin May 15 '12 at 9:55
One more thing, since $g$ enters the next order, I do need some kind of equation for $g$, just to be able to write down a sketch of the solution. – Aleksey Pichugin May 15 '12 at 9:58
@AlekseyPichugin Thanks again for your comments. Basically, for the easiest type of equation it looks for the boundary solution $\nabla^4 g = 0$ (see edit) – Bernhard May 15 '12 at 10:59