14.4 Singular perturbation and boundary layers

When the small parameter multiplies the highest derivative, setting \(\varepsilon = 0\) lowers the order of the equation, and the reduced problem cannot satisfy all the boundary conditions. The solution then changes rapidly in a thin boundary layer.

Example 14.4.1. For \(\varepsilon y'' + y' + y = 0\), \(y(0) = 0\), \(y(1) = 1\), with \(0 < \varepsilon \ll 1\), the reduced equation \(y' + y = 0\) is first order, and can satisfy only one condition. The outer solution satisfying \(y(1) = 1\) is \(y_{\text {out}} = e^{1-t}\), which has \(y(0) = e \neq 0\). Near \(t = 0\), in the stretched variable \(\xi = t/\varepsilon \), the equation becomes \(y_{\xi \xi } + y_\xi + \varepsilon y = 0\), with leading-order solution \(y_{\text {in}} = A(1 - e^{-\xi })\); matching \(y_{\text {in}}(\xi \to \infty ) = y_{\text {out}}(0) = e\) gives \(A = e\). The composite approximation, inner plus outer minus their common part, \[ y \approx e^{1-t} - e^{1 - t/\varepsilon } , \] is uniformly accurate to \(O(\varepsilon )\) on \([0, 1]\). The layer has width \(O(\varepsilon )\), and its position is fixed by the sign of the coefficient of \(y'\): with \(-y'\) instead of \(+y'\) the layer would be at \(t = 1\) (Figure 19).

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Figure 19: The boundary value problem \(\varepsilon y'' + y' + y = 0\), \(y(0) = 0\), \(y(1) = 1\), with \(\varepsilon = 0.05\). Away from \(t = 0\) the exact solution follows the outer solution \(e^{1-t}\) of the reduced equation; in a layer of width of order \(\varepsilon \) at \(t = 0\) it rises steeply from \(0\) to meet it. The composite approximation \(e^{1-t} - e^{1-t/\varepsilon }\) differs from the exact solution by at most \(0.086\) here.

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