are respectively of sizes , O(1), and O(1). The leading-order balance on this timescale, valid in the distinguished limit , is therefore given by the second and fourth terms, i.e., This has solution for some constant . Applying the boundary condition , we would have ; applying the boundary condition , we would have . It is therefore impossible to satisfy both boundary conditions, so is not a valid approximation to make across the whole of the domain (i.e. this is a singular perturbation problem). From this we infer that there must be a boundary layer at one of the endpoints of the domain where needs to be included. This region will be where is no longer negligible compared to the independent variable , i.e. and are of comparable size, i.e. the boundary layer is adjacent to . Therefore, the other boundary condition applies in this outer region, so , i.e. is an accurate approximate solution to the original boundary value problem in this outer region. It is the leading-order solution. Inner solution, valid for t = O(ε) In the inner region, and are both tiny, but of comparable size, so define the new O(1) time variable . Rescale the original boundary value problem by replacing with , and the problem becomes which, after multiplying by and taking , is Alternatively, consider that when has reduced to size , then is still of size O(1) (using the expression for ), and so the four terms on the left hand side of the original equation are respectively of sizes , , O(1) and O(1). The leading-order balance on this timescale, valid in the distinguished limit , is therefore given by the first and second terms, i.e. This has solution for some constants and . Since applies in this inner region, this gives , so an accurate approximate solution to the original boundary value problem in this inner region (it is the leading-order solution) is Matching We use matching to find the value of the constant . The idea of matching is that the inner and outer solutions should agree for values of in an intermediate (or overlap) region, i.e. where . We need the outer limit of the inner solution to match the inner limit of the outer solution, i.e., which gives . The above problem is the simplest of the simple problems dealing with matched asymptotic expansions. One can immediately